Published by Rankguru Technology Solutions Private Limited
IL Achiever Series Physics for JEE Module 2
ISBN 978-81-985634-1-5 [SECOND
EDITION]
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Key Features of the Book
Chapter Outline
1.1 Electric Charges
1.2 Coulomb’s Law
1.3 Forces between Multiple Charges
This outlines topics or learning outcomes students can gain from studying the chapter. It sets a framework for study and a roadmap for learning.
This section highlights the essential takeaways from the discussion while also serving as a mnemonic device to help students recall key concepts.
Key Insights:
■ The accuracy of a measurement is a measure of how close the measured value is to the true value of the quantity. Precision tells us to what resolution or limit the quantity is measured.
Application
A body dropped freely from a multistory building can reach the ground in t1 seconds. It is stopped in its path after t2 seconds and again dropped freely from the point. Find the further time taken by it to reach the ground.
This illustrates the application of the discussed concepts to a physical model.
Solved Examples
Specific problems are presented along with their solutions, explaining the application of principles covered in the textbook.
Solved example
1. Two large non-conducting sheets A and B, having charge densities + 2σ C/m2 and – σ C/m 2 are kept parallel to each other. Find electric field at point P
Sol. EQ A Q A Q pA 2 22 3 0020
Try yourself:
1. How much positive and negative charge is there in a cup of water? Assume that mass of one cup of water is 250 g.
Ans: 1.34 7×10 C
Try Yourself enables the student to practice the concept learned immediately.
This comprehensive set of questions enables students to assess their learning. It helps them to identify areas for improvement and consolidate their mastery of the topic through active recall and practical application.
Organised as per the topics covered in the chapter and divided into three levels, this series of questions enables rigorous practice and application of learning. These questions deepen the understanding of concepts and strengthen the interpretation of theoretical learning.
JEE MAIN LEVEL
LEVEL 1, 2, and 3
TEST YOURSELF
1. If a charge on the body is –1 nC, then how many excess electrons are present on the body?
This comprehensive test is modelled after the JEE exam format to evaluate students’ proficiency across all topics covered, replicating the structure and rigour of the JEE examination. By taking this chapter test, students undergo a final evaluation, identifying their strengths and areas needing improvement.
Level 1 questions test the fundamentals and help fortify the basics of concepts. Level 2 questions are higher in complexity and require deeper understanding of concepts. Level 3 questions perk up the rigour further with more complex and multi-concept questions.
This section contains special question types that focus on in-depth knowledge of concepts, analytical reasoning, and problem-solving skills needed to succeed in JEE Advanced.
Handpicked previous JEE questions familiarise students with the various question types, styles, and recent trends in JEE examinations, enhancing students’ overall preparedness for JEE.
Chapter Outline
5.1 The Law of Inertia
5.2 Newton’s First Law of Motion
5.3 Newton’s Second Law of Motion
5.4 Newton’s Third Law of Motion
5.5 Conservation of Linear Momentum
5.6 Equilibrium of a Particle
5.7 Common Forces in Mechanics
5.8 Solving Problems in Mechanics
5.9 Friction
5.1 THE LAW OF INERTIA
■ Inertia: The inability of a body to change its state of rest, uniform motion, or direction by itself.
■ Inertia means resistance to change.
■ Mass is the measure of inertia.
■ The heavier the body, the greater the force required to change its state, hence greater its inertia.
5.1.1 Types of Inertia
■ Inertia of a body is of three types:
Inertia of rest
Inertia of motion
Inertia of direction
LAWS OF MOTION CHAPTER 5
Inertia of Rest
■ It is the inability of a body to change its state of rest by itself.
■ Inertia of Motion: A body in uniform motion cannot accelerate or retard on its own.
■ Inertia of Direction: A body continues moving in the same direction unless acted upon by an external force.
5.2 NEWTON’S FIRST LAW OF MOTION
■ Newton’s First Law of Motion: "Every body continues in its state of rest or uniform motion in a straight line unless compelled by an external force to act otherwise."
■ Importance of the First Law: Defines inertia and force. Concludes that the acceleration of an object is zero if no external force acts on it.
■ Definition of Force: Force is the physical quantity that changes or tries to change the state of rest or uniform motion of a body.
Force can push or pull a body, stop or change its motion, or change its shape or direction.
5.3 NEWTON’S SECOND LAW OF MOTION
■ Newton’s Second Law of Motion: The time rate of change of momentum of a body is directly proportional to the net external force acting on the body and occurs in the direction of the applied force.
5.3.1 Linear Momentum
■ It is the product of the mass of the body and its velocity.
Linear momentum = mass × velocity
pmv
■ Linear momentum is a vector quantity. Its direction is same as the direction of velocity of the body. SI unit:kgms –1 and the c.g.s unit of linear momentum is gcms–1. The dimensional formula of linear momentum is [M1L1T–1]
■ If a body of mass m changes its velocity from u to v , then the change in the momentum of body is () fi pppmvmu ∆=−=−
Solved example
1. x-component of velocity of a particle of mass 2 kg is 3 m/s and y-component of its velocity is 4 m/s. Find the magnitude of its linear momentum.
Sol. () ˆˆ 34m/svij
Try yourself:
1. A particle of mass 1 kg is moving along positive x -axis with velocity 2 m/s. After some time it starts moving along positive y -axis, with velocity 23 m/s. Find the magnitude of change of linear momentum.
Ans: 4 kgm/s
■ Consider a body of constant mass ‘m’ moving with a velocity ‘v’. Let an external force ‘F’ is acting on it in the direction of the velocity, its velocity changes from v to v + ∆v during an interval “∆t”.
The linear momentum ‘mv’ changes to ‘m(v + ∆v)’.
The change in momentum ∆p = m∆v
The rate of change of momentum
According to the second law, F ∝ rate of change of momentum ∆
∆ FKp t , where K is the proportionality constant.
As the time interval ∆t → 0, the term ∆
becomes the derivativ e of ‘p’ w.r.t ‘t’ i.e.
dp dt . Thus = FKdp dt but
[For constant mass]
The second law can be written as F = Kma
If m = 1 unit, a = 1 unit, F = 1 unit, then K = 1
∴ F = ma
[Here F = Resultant external force]
In vector notation, = Fma
■ Therefore, we can write, F x = ma x, F y = ma y, F z = ma z
Unit of force: SI unit of force is newton.
■ Definition of newton: 1 newton is the force that causes an acceleration of 1 ms –2 on a body of mass 1 kg in its own direction . The dimensional formula of force is [MLT–2]
Solved example
2. A body of mass 1 kg is moving with velocity 30 ms–1 due north. It is acted on by a force of 10 N due east for 4 seconds. Find the velocity of the body after the force ceases to act.
Sol. M = 1 kg, F = 10 N due east; t = 4 s; u=0.
102 10ms 1 F a M === due east.
Final velocity due east at the end of 4 seconds,
v = u + at = 0 +10(4) = 40 ms–1
The body has initial velocity 30 ms–1 due north, which is not affected by applied force. At the end of 4 seconds, its velocity is () + ˆˆ 4030ij
221 (40)(30)50ms V =+=
Direction of resultant velocity:
5.3.2 Impulse
■ Impulse: The product of force and the duration of its action, which produces a finite change in the body's momentum.
■ Impulsive force is generally variable.
■ If the force varies with time, according to Newton’s second law of motion, the change in momentum is the integral of the force over time.
■ The change in momentum is the measure of impulse J . As the force F is variable, on integrating both sides
The resultant velocity makes an angle
tan13 4 North of east
Solved example
3. A rubber ball of mass 0.5 kg is moving along negative x -axis with a velocity of 2 m/s. It collides with a rigid wall and after the impact it starts moving along positive y -axis with same speed. If duration of collision is 0.01 second, find magnitude of force exerted by the wall on the ball.
Sol. () 0.52kgm/skgm/s i pii =×−=−
() 0.52kgm/skgm/s f pjj =×=
Try yourself:
2. A Velocity of a particle of mass 1 kg moving in xy-plane is given by () 2 23m/s.vtitj =+
Find the magnitude of force acting on the particle at t = 1 second. Ans: 5 N
■ Where 1P is the momentum at time t1, 2P is the momentum at time t2.
diagram
∴ Impulse = Change in mo mentum = Area under F-t diagram This is true for one dimensional motion.
Force
Time t2 t1
■ If the average force during the time (t2 –t1) is
av F , we can rewrite the above expression as () 2 []121 t avt FtPP =−
()() 2121 av FttPP ⇒−=−
()() Impulse21 av JFtt ∴=−
■ Impulse of a force is positive, negative or zero according to whether momentum of
the body increases, decreases or remains unchanged under the effect of force.
Applications of Impulse
■ A cricket player lowers or draws his hands back while catching the ball to increase the time over which the ball is stopped.
■ This results in a smaller force applied to stop the ball, reducing the force exerted on the player's hands and preventing injury.
■ If the player catches the ball abruptly, the same change in momentum occurs in a shorter time, requiring a greater force, which could lead to injury.
Solved example
4. A body of mass 2 kg has an initial speed 5 m/s. A force acts on it for some time in the direction of motion. The force-time graph is shown in figure. Find the final speed of the body.
Try yourself:
3 Force acting on a particle is varying with time according to the semi-circular graph shown in figure. Find the average impulsive force acting on the particle.
Area
Total
TEST YOURSELF
1. A balloon with its contents weighing 160 N is moving down with an acceleration of g/2 ms–2. The mass to be removed from it so that the balloon moves up with an acceleration of g/3 ms–2 is (g = 10 ms–2) (1) 5 kg (2) 10 kg (3) 6 kg (4) 3 kg
2. A body is acted on by a force given by F = (10 + 2t)N. The impulse received by the body during the first four seconds is (1) 40 N s (2) 56 N s (3) 72 N s (4) 32 N s
Answer Key
(1) 2 (2) 2
5.4 NEWTON’S THIRD LAW OF MOTION
■ Third Law of Motion: To every action, there is always an equal and opposite reaction.
■ Action: The force exerted by one body on another.
■ Reaction: The force exerted by the second body on the first body.
■ If ABF is the force exerted on body A by body B (i.e., action) and BAF is the force exerted on body B by body A (i.e, reaction) then according to third law =− ABBAFF
■ From the above, the third law can be stated as follows
■ Forces always occur in pairs . Force exerted on body by another body is equal and opposite to the force exerted on second body by first one.
■ Examples of Newton’s Third Law of Motion:
Book on a table: The book exerts a downward force on the table (action), and the table exerts an equal upward force on the book (reaction).
Walking: A person presses the ground backward with their feet (action), and the ground pushes the person forward with an equal force (reaction). The horizontal component of the reaction helps the person move forward.
Solved example
5. A cricket ball of mass 60 gm moving with a velocity of 20 m/s hits a bat and just after the collision starts moving with a velocity of 152 m/s making an angle of 45° with its original direction of motion. What is the magnitude of force exerted by the ball on the bat if the bat remains in contact with the ball for 0.1 second?
Sol. 20m/sui
0.0620kgm/s1.2kgm/s i pii
0.0615kgm/s0.9kgm/s f pijij =×+=+
By Newton’s third law, force exerted by the ball on the bat is 9.487 N.
Try yourself:
4. Explain why a horse cannot pull a cart and run in empty space
Ans: While trying to pull a cart, a horse pushes the ground backwards with a certain force at an angle. The ground offers an equal reaction in the opposite direction on the feet of the horse. The forward component of this reaction is responsible for the motion of the cart. In empty space, there is no reaction and hence a horse cannot pull the cart and run.
TEST YOURSELF
1. A car accelerates on horizontal road due to force exerted by (1) the driver of the car (2) the engine of the car
(3) the earth
(4) the road
2. The book is lying on the table. What is the angle between the action of the book on the table and the reaction of the table on the book?
(1) 0° (2) 30°
(3) 45° (4) 180°
Answer Key
(1) 4 (2) 4
5.5 CONSERVATION OF LINEAR MOMENTUM
5.5.1 Law of Conservation of Linear Momentum
■ Law of Conservation of Momentum: "The total momentum of an isolated system remains constant if no external force acts on it."
■ In an isolated system, the vector sum of the linear momenta of all bodies is conserved and unaffected by their mutual action and reaction.
5.5.2 Practical Applications of the Principle of Conservation of Linear Momentum
Collision
■ Consider two bodies A and B of masses m1 and m2 moving with velocities 12 and uu respectively (u1 > u2).
N System –N
During period of contact
■ Here N is the impulsive force whose magnitude is a function of time. N is action and N is reaction.
Impulse on B ==−
2 022 t NdtmVmu ...(1)
Impulse on A 2 011 (N) t dtmVmu=−=−
...(2)
From equations (1) and (2),
2211 mVmumVmu
2112 mVmVmumu
⇒= fi pp
∴ Final momentum of the system = Initial momentum of the system
Solved example
6. Two ice skaters A and B approach each other at right angles. A has mass 30 kg and velocity 1 m/s and B has a mass of 20 kg and velocity 2 m/s. They meet and stick together. Find the final velocity of couple.
Sol. Let v = Composite velocity after collision, θ = angle with the direction of 30 kg mass
Then, from momentum conservation, 2m/s 20 kg 1m/s 30 kg v
Along x axis, 50v cos θ = 30 × 1
Along y axis, 50v sin θ = 20 × 2
From above equations, ()=+ 22 509001600 v
50v = 50 v = 1 m/s
Try yourself:
5. A 2 kg block, moving on a frictionless horizontal surface with velocity 4 m/s, collides with a stationary 6 kg block. After the collision the blocks stick together. Find velocity of combined mass after the collision.
Ans: 1 m/s
Recoiling of a Gun
■ The recoil velocity of the gun can be calculated from the principle of conservation of linear momentum.
Suppose m1 = mass of bullet,
m2 = mass of gun,
1v = velocity of the bullet, 2 v = velocity of the gun.
■ Before firing, the gun and the bullet both, are a t rest. Therefore, total linear momentum before firing = 0. According to the principle of conservation of linear momentum, total linear momentum after firing should also be zero.
Try yourself:
6. A man of mass 70 kg, is standing on a frictionless horizontal surface with a 5 kg stone in hand. Find the separation between the man and the stone 1 s after the man throws the stone horizontally with a velocity of 2 m/s relative to ground.
Ans: 15/7 m
Balloon–Man System:
■ The negativ e sign shows that direction of 2 v is opposite to the direction of 1v i.e. the gun recoils. Further, as m2 >> m1 therefore, << 21vv i.e. velocity of recoil of the gun is much smaller than the vel ocity of the bullet .
■ If the barrel of the gun is inclined, momentum of the gun-bullet system is conserved in horizontal direction only.
■ In vertical direction, due to impulsive reaction force from hand, momentum of gun-bullet system is not conserved.
■ A man of mass m climbs a rope of length L suspended below a balloon of mass M. The balloon is stationary with respect to ground. (a) If the man begins to climb up the rope at a speed v rel (relative to rope), in what direction and with what speed (relative to ground) will the balloon move? (b) How much has the balloon descended when the man reached the balloon by climbing the rope?
(a) Given that initially the system is at rest, so initial momentum of the system is zero. In the climbing of the rope, there is no external force.
Therefore, final momentum of the system should also be zero
7. A bullet of mass m is fired horizontally with a muzzle velocity u from a stationary gun of mass M kept on a smooth horizontal surface. Find the velocity of the gun after the bullet is fired.
Sol. Let v be the velocity of the gun relative to ground.
Conserving momentum, m(u – v) + M(–v) = 0
Solved example
Furthermore, here it is given that =−
rel vvV ........... (2)
Substituting the value of v from Eqn. (2) in(1), we get ()=−+ rel MVmvV or () =− + rel mv V mM ..........(3)
This is the desired res ult and from this it is clear that the direction of motion of balloon is opposite to that of climbing () rel v , i.e., vertically down.
(b) If s be the displacement of (balloon + ladder) system (downward displacement) and be the upward displacement of the man relative to the ladder, then = ds V dt and = . rel d v dt
From eqn. (3), () = + dsmd dtMmdt
sL dsmmL ds MmMm
Key Insights:
■ In case of two body system, if m is displaced relative to M by d rel, the displacement of individual bodies relative to the ground are: and relrel mM Mdmd dd mMmM == ++ respectively, and they are always in pposite directions to each other.
Solved example
8. All surfaces are smooth. Find the horizontal displacements of the block and wedge when the block slides down from top to bottom.
θ M l m h
CHAPTER 5: Laws of Motion
Sol. When the block slides down on the smooth wedge, the wedge moves backwards. In the horizontal direction there is no external force.
0 x F ∴
x P = constant
fi PP (along x–axis) += 0 muMV
x1 = forward distance moved by the block along x–axis.
x2 = backward distance moved by the wedge along x–axis.
=−= 12 ; xx muMVmM tt mx1 =Mx2 m(L–x2)=Mx2
cosand MLM x MmMm ∴== ++
MmMm
L can also be written as cos θ
Try yourself:
7. A man of mass M standing at a height h from the floor in a gravity free space throws a ball of mass m straight down with a speed u. When the ball reaches the floor, find the distance of the man above the floor. Ans:
Explosion of Stationary or Moving Body:
■ When a stationary or moving body explodes into a number of pieces, resultant external force acting on the body during the explosion is zero, hence momentum of the system remains conserved.
Solved example
9. A bomb moving with velocity ˆˆˆ 405025m/s ijk ++ explodes into pieces of mass ratio 1 : 4. If the small piece goes out with velocity +− ˆˆˆ 2007015ijk m/s, find the velocity of the larger piece after explosion.
Sol. Let 5m be mass of bomb. The largest piece will have mass 4m. Using principle of momentum conservation,
ˆˆˆ 5405025 mijk ++
=+−+ ˆˆˆ 20070154 mijkmv
Which gives, + = ˆˆ 180140 4 vjk
ˆˆ 4535m/s jk=+
Solved example
10. A shell fired from a cannon with speed v m/s at angle θ with horizontal explodes into three pieces of equal masses at the highest point of trajectory. One piece falls down vertically while the other retraces its path. What is speed of the third piece?
Sol. If ‘m’ is mass of shell, at highest point, initial momentum = m ( v cos θ ), along horizontal. Given, momentum of one fragment
() =−θ cos 3 m v momentum of second fragment along horizontal = 0
If ‘v1’ is speed of third fragment, from momentum conservation,
5.5.3
Variable Mass System
System at time = t + ∆t V (V+∆V)
m (m–∆m) ∆m
Time = t Time = t + ∆t
■ Let a system of mass m be moving with velocity V at time = t. Under the action of resultant external force F ext.
■ Also let in the next time interval ∆ t , a quantity of mass ∆ m be ejected from the main body of the system with a velocity u relative to the main body of the system.
■ As a result velocity of the main body of the system becomes (V + ∆ V) relative to the ground and velocity of the ejected mass relative to the ground is ( V + ∆ V – u).
■ Therefore, change of momentum of the system in time interval ∆ t is given by
∆ p = [(m – ∆ m) (V + ∆ V) + ∆ m (V + ∆ V – u)] – mV
⇒∆ p = m ∆ V – u ∆ m
So, average external force acting on the system is given by, ()∆∆∆ ==⋅−⋅ ∆∆∆ extavg pVmFmu ttt
v1 = 4v cos θ
Try yourself:
8. A body at rest explodes and breaks up into 3 pieces. Two pieces having equal mass fly off perpendicular to each other with the same speed of 30 m/s. The 3 rd pieces has 3 times the mass of each of the other pieces. Find the magnitude and direction of its velocity immediately after the explosion.
Ans: 102 m/s, making 135° relative to either piece
■ Instantaneous external force acting on the system at time = t is given by, ∆→ ∆ =⋅=− ∆ 0 limextt pdVdmFmu tdtdt
⇒=+ ext dVdm mFu dtdt
Key Insights:
■ When direction of F ext is same as that of V, external force is positive, otherwise it is negative.
■ When direction of ‘ u ’ is same as the direction of ‘ V ’, it is written as ‘+ u ’
CHAPTER 5: Laws of Motion
otherwise (–u).
■ When the system is ejecting mass at a constant rate ( µ ), =−µ; dm dt if the system is collecting mass of a constant rate ‘ µ ’, =+µ. dm dt
Rocket Propulsion (Gravity Considered)
■ Let m0 be the mass of the rocket at time t = 0. Initially, let us suppose that the velocity of the rocket is u. Suppose, m is mass at any time t and v, is its velocity at that moment.
t=0, m=m0 m0 v u W
0 mt
mmt t=t, v=u, At At Exhaust velocity = vr dm Where dt
■ Let - (dm/dt) be the mass of the gas ejected per unit time and v r the exhaust velocity of the gases with respect to rocket. Usually(dm/dt) and v r are kept constant throughout the journey of the rocket. Now, let us write few equations which can be used in the problems of rocket propulsion At time t
Thrust force on the rocket, F t = v r(–dm/ dt)(upwards)
Weight of the rocket, W = mg (downwards)
Net force on the rocket, Fnet = Ft – W (upwards)
netr dm Fvmg dt
Net acceleration of the rocket,
Hence,
ln0 r m vugtv m ............ (i)
Key Insights:
■ If gravity is ignored and initial velocity of the rocket u = 0, Eq, (i) reduces to v = v r
0m m
■ If gravity is ignored and initial velocity of rocket is u then equation (i) reduces to V = u + v r ln 0m m
Solved example
11. Fuel is consumed at the rate of 50 kgs–1 in a rocket. Find the thrust on the rocket if the velocity of the exhaust gases is 2 kms–1. Also calculate the velocity of the rocket at the instant, when its mass is reduced to 1/10th of its initial mass if its initial velocity is zero. (neglect gravity)
(i) The thrust on the rocket, 35 21050110N dm Fu dt ==××=×
(ii) The velocity of the rocket, =+ 0 0log e m vvu m
300 0 0210log1010 e mm m m
= 2 × 103 loge 10
= 2 × 103 × 2.303 log10 10
= 4.606 × 103 m/s
Try yourself:
9. The first and second stage of two stage rocket separately weigh 100 kg and 10 kg and contain 800 kg and 90 kg fuel respectively. If the exhaust velocity of gases is 2 km/s then find velocity of rocket (nearly) (log105=0.6990) (neglect gravity)
Ans: 7.8 × 310 m/s
TEST YOURSELF
1. A man fires a bullet of mass 200 g at a speed of 5 m/s. The gun is of one kg mass. By what velocity the gun rebounds backwards?
(1) 0.1 m/s (2) 10 m/s
(3) 1 m/s (4) 0.01 m/s
2. An object flying in air with velocity () k ˆˆ 20i5 ˆ 2j12+− suddenly breaks in two pieces whose masses are in the ratio 1 : 5. The smaller mass flies off with a velocity () ++100i35 ˆ j ˆ 8 ˆ k. The velocity of the larger piece will be:
5.6 EQUILIBRIUM OF A PARTICLE
■ If two forces 1F and 2F , act on a particle, equilibrium requires
i.e. the two forces on the particle must be equal and opposite.
■ Equilibrium under three concurrent forces 1 , F 2F and 3F requires that the vector sum of the forces is zero.
From the above equation, it implies that F1x + F2x + F 3x = 0 F1y + F2y + F 3y = 0 F1z + F2z + F 3z = 0
■ where F1x, F1y and F1z are the components of 1F along x, y and z directions respectively.
(1) k
4i23j16 +−
(2) 100i35j8k
(3) k
20i5 ˆ 1j80+− (4) 20i15j80k
3. A rocket of initial mass 6000 kg ejects mass at constant rate of 200 kg/s with constant relative speed of 800 m/s. The acceleration of the rocket after 5 s is (neglect gravity)
(1) 50 m/s2 (2) 16 m/s2 (3) 60 m/s2 (4) 32 m/s2
Answer Key (1) 3 (2) 1 (3) 4
■ In other words, the resultant of any t wo forces say 1F and 2F , obtained by the parallelogram law of forces must be equal and opposite to the third force 3F
■ A particle is in equilibrium under forces F1, F2, ..., Fn if they can be represented by the sides of a closed n-sided polygon with arrows directed in the same sense.
Applications
■ If ‘ n ’ equal magnitude coplanar forces acting at a point simultaneously with the angle between any two adjacent forces is θ and kept in equilibrium then = θ 360 n
■ The above forces can be re presented b y side of a closed regular polygon taken in an order.
CHAPTER 5: Laws of Motion
Solved example
12. A mass of 3 kg is suspended by a rope of length 2 m from the ceiling. A force of 40 N in the horizontal direction is applied at midpoint P of the rope as shown. What is the angle the rope makes with the vertical in equilibrium and the tension in part of string attached to the ceiling? (Neglect the mass of the rope, g = 10 m/s2)
TEST YOURSELF
1. Two bodies of masses m1 = 5 kg and m2 = 3 kg are connected by a light string going over a smooth light pulley on a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass m1 will be : [Take g = 10ms–2]
Sol. In equilibrium, T2 = W = 30 N
Resolving the tension T 1 into two mutually perpendicular components, we have
T1 cos θ = T2 = 30 N, T1 sin θ = 40 N =⇒=°θθ 4 tan53 3
The tension in part of string attached to the ceiling 2222 13040N50N TWF=+=+=
Try yourself:
10. A block of mass 2 kg rests on a frictionless horizontal surface. It is in equilibrium under the action of a 10 N force and a force F as shown in figure. Find the value of normal reaction of the surface on the block.
(1) 30 N (2) 40 N (3) 50 N (4) 60 N
2. Two wooden blocks of masses M and m are placed on a smooth horizontal surface as shown in figure. If a force P is applied to the system as shown in figure such that the mass m remains stationary with respect to block of mass M, then the magnitude of the force P is
(1) (M + m)g tan β (2) g tan β (3) mg cos β (4) (M + m)g cosec β
Answer Key (1) 2 (2) 1
5.7 COMMON FORCES IN MECHANICS
■ In mechanics we come across many types of forces like gravitational force, contact force, tension in string, spring force etc.
■ Non-contact Force: Acts on an object without direct contact.
■ Contact Forces: Arise due to contact between objects.
Contact force between solids:
■ Example: Book on a table, normal force and friction
Contact force between solids and fluids: Examples: Viscous force, air resistance.
Tension in String: The force exerted by the string molecules on the object, constant in a light, inextensible string.
Spring force: The restoring force developed in a spring when compressed or elongated by an external force.
F = –Kx. Where x is compression or elongation.
■ There are four fundamental forces in nature:
Gravitational force and electromagnetic force are relevant in mechanics.
Weak nuclear force and strong nuclear force operate in domains outside the scope of mechanics.
5.7.1 Contact Force Between Two Surfaces
■ Blocks are placed in contact with each other on a frictionless horizontal surface.
towards left, called contact force. The block moves with an acceleration ‘ a’.
equation of motion: F – f = M1 a --- (1)
Forces acting on B
1) The reaction force ‘f’ of A is acting on B
∴ equation of motion f = M2 a ----- (2)
Adding (1) & (2)
From
(2)
Solved example
13. Two blocks of masses 5 kg and 2 kg are kept in contact with each other on a frictionless horizontal surface. If a force of 14 N is applied on the larger block, what is the acceleration of the system? What is the contact force between the two blocks?
Sol. M1 = 5 kg; M2 = 2 kg; F = 14 N.
■ The two blocks are considered as a system. Forces acting on A are:
external force F acting towards right
the force on A due to B ‘ f’ acting
Try yourself:
11. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. Force F is applied at one end of the rope. Find the force which the rope exerts on the block. Ans: )( F M Mm +
5.7.2 Frames of Reference
■ A frame of reference is any coordinate system used to describe the motion or position of a body.
■ Types of frames of reference:
Inertial (unaccelerated) frames
Non-inertial (accelerated) frames
■ Inertial frames of reference:
Frames where Newton’s laws of motion apply.
Frames that are either at rest or move with uniform velocity relative to a distant star.
In these frames, acceleration of a body is caused by real forces.
Examples: Normal force, tension, weight, spring force, muscular force, etc.
Equation of motion of mass ‘m’ moving with acceleration ‘ a ’ relative to an observer in an inertial frame is
→ ∑= real Fma
Inertial frames of reference:
Observers in all inertial frames measure the same net force and acceleration for a given object but may measure different velocities.
Basic laws of physics are identical in all inertial frames.
Inertial frames are also called Newtonian or Galilean frames.
Examples:
A lift at rest
A lift moving with constant velocity
A car moving with constant velocity on a straight road
■ Real force: Force acting on an object due to its interaction with another object.
■ Non-inertial frames of reference:
Frames moving with acceleration relative to a distant star.
Newton’s laws of motion do not apply correctly in non-inertial frames.
Acceleration is caused by pseudo forces or a combination of pseudo and real forces.
Examples:
An accelerating vehicle
A car moving in a circular path
A frame attached to the surface of Earth is a non-inertial frame due to Earth's rotation.
5.7.3 Pseudo Force or Inertia Force
■ In the diagram shown frame-(1) is inertial frame and frame-(2) is moving with acceleration a relative to frame-(1), so frame-(2) is non-inertial frame.
■ Suppose a body B of mass m is moving with acceleration f relative-to frame-(1) under the action of a real force P
■ Observer-(1) can write Newton’s second law of motion as
■ If r f be the acceleration of the body relative to observer-(2) who is observing the body from frame-(2).
■ Therefore observer-(2) can write Newton’s second law of motion as
+−= ....(3)
■ In the above eqs. (3), the term ‘ ma ’ is called pseudo force which appears when a body is observed from non-inertial frame of reference, but this force does not exist when a body is observed from inertial frame.
■ Examples of pseudo force: Centrifugal force, deflection of pendulum relative to accelerating car, gain or loss of weight experienced in an accelerating elevator
■ Apparent weight of a man in elevator cage: Suppose a person of mass m is standing on a weighing machine kept on the floor of an elevator cage.
■ Case-I: Elevator cage is moving up with acceleration ‘ a ’. From the frame of the elevator cage, forces acting on the man are shown in the diagram. ma
Weighing machine
mg = Weight of man, real force
R1 = Reaction of weighing machine which is the reading of the machine (apparent weight), it is real force
ma = Pseudo force acting on the man
■ Case-II: Elevator cage is moving downward with acceleration ‘ a’.
ma
mg R2
Weighing machine
ma = Upward pseudo force
mg = Weight, real force
R1 = Reaction of weighing machine which is the reading of the machine (apparent weight), it is real force
R2 + ma = mg
⇒ R2 = m(g – a)
■ Case-III: Elevator cage is falling freely.
In this case a = g
∴ From case-II,
10 a a WW g
So the person feels weightlessness.
■ Case-IV: Elevator cage is moving down with an acceleration a > g
From case-II,
10 a a WW g
So in this case the person’s feet looses contact with the weighing machinie and falls freely with acceleration g. While acceleration of elevator cage is a > g
Key Insights:
■ A simple pendulum hangs from the roof of a moving train. The string is inclined towards the rear end of the train. This implies that the train must be accelerating forward.
Solved example
14. A pendulum is hanging from the ceiling of a car having an acceleration a0, with respect to the road. Find the angle made by the string with vertical at equilibrium. Also find the tension in the sting in this position.
0a
Sol. A) T sin θ = ma0 ....(i) T cos θ = mg ....(ii)
∴= θ tan0 a g
∴ The string is making an angle
tan10 a g with vertical at equilibrium
B) squaring and adding (i) and (ii)
+=+θθ 2222222 sincos0 TTmag
=+22 0 Tmag
Solved example
15. A block slides down from top of a smooth inclined plane of elevation θ fixed in an elevator going up with an acceleration a 0. The base of incline has length L.
Find the time taken by the block to reach the bottom.
Sol. Let us solve the problem in the elevator frame. The free body force diagram is shown. The forces are
m N mg ma0 ma0
(i) N normal reaction to the plane,
(ii) mg acting vertically down,
(iii) ma0 (pseudo force) acting vertically down
If a is the acceleration of the body with respect to incline, taking components of forces parallel to the incline mg sin θ + ma0 sin θ = ma
∴ a = (g + a0) sin θ
This is the acceleration with respect to the elevator
The distance travelled is θ cos L . If t is the time for reaching the bottom of incline,
2 0 1 0sin. cos2 L gat
θθ 1 2 0 2 sincos L t ga
+ θ 1 2 0 4 sin2 L ga
Try yourself:
12. A 75 kg man stands in a lift. What force does the floor exert on him when the elevator starts moving upward with an acceleration of 2 ms–2. Given: g=10 ms–2
Ans: 900 kg-wt
Try yourself:
13. For what value of ‘a’ the block falls freely? h x a θ
Ans: g cot θ
5.7.4 Springs
■ A spring is fixed at one end. The other end of the spring is pulled by applying a force. At any instant, restoring force is directly proportional to the elongation. F
F s ∝ x
F = –kx
k is called as stiffness factor, spring constant or force constant. Negative sign indicates that elongation and restoring force are oppositely directed.
From the above equation, we have F k x = If x = 1, k = |F|
■ “Spring constant is numerically equal to the force require to deform a spring by unit length.”
Springs Connected in Parallel
■ If external force acts on a combination of springs such that they have same deformaton then we say that they are connected in parallel. If keq is the equivalent force constant of the combined spring, then F1 = –k1x ; F2 = –k2x F = –k eq x and F=F1 + F2
Spring in Series
■ If same force is transmitted in all springs then we say that springs are connected in series.
Solved example
16. The systen shown in the figure is in equilibrium at rest. The spring and string are massless. Now, the string is cut. Find the acceleration of the blocks just after the string is cut.
Sol. Under equilibrium of mass m
= mg
Now, the string is cut, therefore, T = mg
Force is decreased on mass m upwards and downwards on mass 2m
Accelerationof=
Acceleration32 of2= 22 mg mg m mmgmgg m ∴= =
Try yourself:
14. Figure shows a block A on a smooth surface attached with a spring of force constant k to the ceiling. In this state spring is in its natural length ℓ. The block A is connected with a massless and frictionless strng to another identical mass B hanging over a light and smooth pulley. Find the distance moved by A before if leaves contact with the ground.
TEST YOURSELF
1. Two masses 5 kg and 3 kg are suspended from the ends of an unstretchable light string passing over a frictionless pulley. When the masses are released, the thrust on the pulley is (g = 10ms–2)
(1) 80 N
(2) 37.5 N
(3) 150 N
(4) 75 N
2. In the Figure given below two masses m and m’ are tied with a thread passing over a pulley, m’ is on a frictionless horizontal surface. If acceleration due to gravity is g,
CHAPTER 5: Laws of Motion
the acceleration of m’ in this arrangement will be:
(1) g
(2) mg/(m + m')
(3) mg/m '
(4) mg/(m – m')
3. A small block of mass m rests on a smooth wedge of angle θ. With what horizontal acceleration ‘a’ should the wedge be pulled, as shown in Fig., so that the block falls freely?
(1) g cos θ (2) g sin θ (3) g cot θ (4) g tan θ
Answer Key (1) 4 (2) 2 (3) 3
5.8 SOLVING PROBLEMS IN MECHANICS
Solving problems from F.B.D. in various cases.
5.8.1
Free Body
Diagram : (F.B.D)
Newtons laws of motion - Applications:
■ Free Body Diagram (F.B.D) technique for solving problems using Newton’s laws of motion:
Step 1: Decide the system to apply the laws, ensuring all parts move with the same acceleration.
Step 2: List all the forces acting on the system, excluding internal forces.
Step 3: Draw the free body diagram representing the system as a point mass, with forces as vectors originating from a common point.
Step 4: Choose axes and write the equation of motion:
For straight-line motion, choose X, Y, Z axes.
For motion in a plane, resolve forces along X and Y-axes and find their corresponding accelerations.
Obtain separate equations of motion.
■ Single Object Suspended by a String
Suppose an object of mass M is suspended by a string from the ceiling as shown figure. The string is in a state of tension. The molecules of the string near the lower end exert force on the molecules of the object. The resultant of these electromagnetic forces is the force exerted by the string on the object which is called tension denoted by T. This supports the object and prevents it from falling. It is directed away from the object. If the string is weightless then the tension in the string is same at each and every point of the string. If we consider a point A of the string the tension in the part below A and above A are each equal to T and directed away from A. Similarly the tension at the upper end pulls the ceiling downward and the ceiling pulls the string upward. Here also the tension in the string is directed away from the ceiling
(a) pull of the string, T , upward, (b) gravitational force of the earth on the object, Mg, downward.
As the acceleration of the object is zero, the resultant force is zero by Newton’s second law.
T – Mg = 0; T = Mg Mg mg T' A
If the string is very light so that we can neglect its mass compared with the mass of the object suspended, then the tension in the string is same every where.
If the string is of mass m below A, the free body diagram at A is shown in Fig. The tension in the string T' is such that
T' – Mg – mg = 0; T' = (M + m) g
■ Single object suspended by a string and accelerated
A bucket of mass M pulled upward with an acceleration ‘a’ from a well by means of a rope of mass m, the tension in the string at the upper end can be found by taking the resultant force equal to the product of the mass of the system and its acceleration.
T – (M + m) g = (M + m) a
T = (M + m)(g + a).
To find the tension at the lower end, the free body diagram for the object is shown in figure. The forces on it are
This tension is equal to the force on our hands through which we pull the bucket up.
Key Insights:
■ If the bucket and rope system is allowed to move downward with an acceleration ‘ a’, the tension at the upper end is =+−()(). TMmga
Solved example
17. A mass of 1 kg attached to one end of a string is first lifted up with acceleration 4.9 m/s2 and then lowered with same acceleration. What is the ratio of tension in string in two cases?
Sol. When mass is lifted up with acceleration 4.9 m/s2 , T1 = m(9.8 + 4.9) When mass is lowered with same acceleration, T2 = m(9.8 – 4.9)
1 2 14.7 3:1 4.9 T T
Try yourself:
15. In the arrangement shown, find the tension in the string PQ.
A 2 kg 100 N P Q B 4 kg
Tension in the cable:
Ans: 66.67 N
■ If the lift is accelerating upwards or retarding downwards.
a
M = Mass of the lift + passengers
Key Insights:
■ If breaking force of cable is given then minimum distance in which lift can be stopped can be obtained from Tbreaking = M(g + a max); = 2 min max ; 2 u S a
The minimum time in which the lift can be stopped is tmin = max u a
u = initial velocity of the lift
If the lift is moving downward, with acceleration a or moving upward with retardation a a
Key Insights:
■ A body is kept on the floor of a lift at rest. The lift starts descending at acceleration a
If a > g the displacement of the body in time t is 12 2 gt
If a < g the displacement of the body in time t is 12 2 at
Solved example
18. A lift is going up. The total mass of the lift and the passenger is 1500 kg. The variation in the speed of the lift is as shown in the figure. Find the tension in the rope pulling the lift at t = 11th second.
Sol. Acceleration of lift at 11th second = slope of the line CD
3.6 2 10 12 A B C D = 11 03.6 1210 a = –1.8 m/s2
T = m(g + a) = 1500(9.8 – 1.8) = 1500 × 8 = 12000 N
Solved example
19. In the above problem, find the height to which the lift takes the passenger.
Sol. S = Area of the graph () 1 8123.636m 2 =+×=
Try yourself:
16. A lift of mass 1000 kg moves up with an acceleration ‘ a ’ in upward direction. If breaking tension in the string is 15000 N, find the maximum value of ‘ a’.
Ans: 5 2m/s
F = (m1 + m2 + m3) a = ++ 123 ....(1) F a mmm
Let T1 be the tension in the string between the blocks m1 and m2. T1 = m1a a m1 T1 = ++ 1 1 123 ....(2) mF T mmm
Let T2 be the tension in the string connected between the blocks m2 and m3 m2 T2
a
5.8.2
Connected Bodies
■ Case (i):
A force F is applied, as shown, All the bodies move with same acceleration ‘ a’.
T2 = m1
+
2a ()() + =+= ++ 12 2122 123 mmF TmmaT mmm
Solved example
20. Three blocks connected together by strings are pulled along a horizontal surface by applying force F. If T 3 = 36 N, what is tension T2? 1kg 8kg T1 T2 T 3 F 27kg
Sol. Suppose system slides with acceleration ‘ a ’. Then, for motion of blocks, T 3 = F = (m1 + m2 + m3) a
T 3 = 36a; i.e., a = 3 36 T = 1 m/s2,
T 3 – T2 = 27a
From the above equation,
36 – T2 = 27 × 1 ∴ T2 = 9 N
Try yourself:
17. Two identical blocks A and B, each of mass M , are connected to each other through a light string. The system is placed on a smooth horizontal floor. When a constant force F is applied horizontally on the block A, find the tension in the string.
Ans: 2 F
■ Case (ii): Atwood’s Machine:
■ It consists of two masses, m 1 and m 2 , connected by a light string passing over a smooth, frictionless pulley fixed to the ceiling.
■ When released, both masses move with the same acceleration, denoted as a.
■ As the pulley is frictionless, the tension in the string is the same on both sides of the pulley.
Let m1 be greater than m2 T T m1 m2
■ The mass m1 moves down and m2 moves up with same acceleration a.
Free body diagram of m1 T a m1g
CHAPTER 5: Laws of Motion
■ The forces acting on m1 are
its weight m1g acting down and
the tension T acting upward
∴ m1g – T = m1a ------ (1)
Free body diagram of m2 T a m2g
■ The forces acting on m2 are
its weight acting downward and
the tension T acting upward
∴ T – m2g = m2a ------- (2)
Adding (1) & (2) m1g – m2 g = (m1 + m2) a
∴ Acceleration () = + 12 12 mm ag mm ------- (3)
Substituting the value of a in equation-(1)
m1g – T = m1 () () + 12 12 mm g mm ; T = m1g – m1g () () + 12 12 mm mm
∴= + 12 12 2mm Tg mm -------------------- (4)
■ Determination of (g): The time of fall of m1 through a distance s is measured. Using the acceleration from equation (3), the kinematic equation of motion () =+=() + 12 22 12 11 0; 22 mm satsgt mm
g is calculated knowing s and t.
Key Insights:
■ If force on the clamp/string holding the pulley is T' from the free body diagram of pulley
T' – 2T= 0
( pulley is massless and at rest)
■ Case (iii):
■ Two bodies of masses m 1 and m 2 are connected by a light, inextensible string passing over a frictionless pulley.
■ One body m1 is on a horizontal surface, and the other body m2 is hanging vertically.
■ The surface on which m1 slides is smooth.
■ Both bodies move with the same acceleration. m1 m2
■ The forces on ‘m1’ are
the weight ‘m1g’ acting downwards.
Tension acting upwards. a
As the body is moving down m1g – T = m1a --------(1)
Since there is no acceleration in vertical direction, N – m2g = 0 (or) N = m2g
As the body moves towards right with an acceleration a, T = m2a --------(2)
From (1) and (2) m2g T N a
and tension in the string T = m2a
Key Insights:
■ Force on the pulley due to the string 22TT=+ == + 12 12 2 2 Tmmg mm
■ Case (iv): m1 m 2 T θ m2gsinθ T a a
CHAPTER 5: Laws of Motion
The inclined plane shown in the figure is smooth. Two bodies of masses m1 and m2 are connected by a light and inextensible string, as shown in the figure.
Free body diagrams: m1 a T m1g a
Equations of motion.
−= 11 mgTma ......(1)
−= θ 22 sin Tmgma ...... (2)
Along the normal to the inclined plane there is no acceleration.
∴ N – m2 g cos θ = 0 (or) N = m2g cos θ
Adding (1) and (2)
(m1 – m2 sin θ )g = (m1 + m2)a () = + θ 12 12 sin mmg a mm
substituting a in equation (1) =+θ 22sinTmamg
12 22 12 sin sin mm Tmgmg mm () −++ = + θθθ22 122122 12 sinsinsin mmmmmmg T mm () + = + 12 12 1sin mm T mm θ
Key Insights:
Force on the pulley due to the string is 90 2cos 2 FT
90 T T Pulley
■ Case (v):
Two bodies of masses m 1 and m 2 are connected by a light and inextensible string. The inclined plane is smooth
■ Free body diagram for m1:
m1gsinα T a N1 α α m1 gcosα m1g m1 N1 T x y α m1 gcosα m1gsinα m1g
11 sin mgTma α −= ---------- (1)
N1 – m1g cos α = 0
∴ N1 = m1g cos α
Free body diagram for m2: N2 T m2g m 2 m2 gcosβ m2gsin ββ β N2 T
m2 gcosβ m2g m2 gsinβ β
22 sin Tmgma β −= ......... (2)
N2 – m2g cos β = 0
∴ N2 = m2g cos β
Adding (1) and (2)
1212 sinsin mgmgmama αβ−=+
() 12 12 sinsin mmg a mm αβ = +
1111 sin;sin mgTmaTmgma αα −==−
1 (sin) Tmga ⇒=− α
12 1 12 sinsin sin Tmgmmg mm ⇒=−
α 1 12 Tmg mm ⇒= +
1212 sinsinsinsin mmmmαααβ +−+
() 12 12 sinsin Tmmg mm + ⇒= + αβ
Solved example
21. The pulley arrangements of figures (I) and (II) are identical. The mass of the rope is negligible. In figure (I), the mass m is lifted up by attaching a mass 2m to the other end of the rope. In figure (II), m is lifted up by pulling the other end of the rope with a constant downward force F = 2mg. Calculate the accelerations in the two cases.
m m 2m 2mg (I) (II)
Sol. Case I:
In figure (a), for the motion of mass m, T – mg = ma ..........(1) For motion of mass 2m figure (b)
2mg – T = (2m)a ..........(2)
Adding equation (1) and (2), we get 2mg – mg = 2ma + ma
mg = 3ma 3 ag⇒=
Case II : In figure (c), T' – mg = ma'
But T' = 2mg
∴ 2mg – mg = a' ∴ a' = g
Solved example
22. A man of mass 60 kg is standing on a weighing machine kept in a box of mass 30 kg, as shown in the diagram.
T T
If the man manages to keep the box stationary find the reading of the weighing machine.
Sol. We know that normal reaction = scale reading
For man T – Mg + R = M(0)
For box : T – R – mg = m(0)
TMgRT −+= Rmg 2R = (M – m)g
Ans: 12 42 mgmg,33 TT
Try yourself:
19. Two blocks of masses 2 kg and 5 kg are at rest on the ground. The masses are connected by a string passing over a frictionless pulley, which is under the influence of a constant upward force F = 50 N. Find the accelerations of 5 kg and 2 kg masses.
2.5 2m/s
Ans: Acceleration of 5 kg is zero and 2 kg is
5.8.3 Constraint Equations
■ These equations establish the relation between the accelerations of different masses connected by a string (s).
■ The number of constraint equations is generally equal to the number of strings in the system.
Solved example
23. Using constaint method find the relation between acceleration of 1 and 2.
Try yourself:
18. Figure shows three blocks of mass m , each hanging on a string passing over a pulley. Calculate the tension in the string connecting A to B and B to C.
Sol. At any instant of time let x 1 and x 2 be the displacement of 1 and 2 from a fixed line
Then, x1 + x2 = constant (or) x1 + x2 = l (length of string)
Differentiating with respect to time, we have v1 + v2 = 0 (or) v1 =–v2
Again differentiating with respect to time, we get a1 + a2 = 0 (or) a1 = –a2
This is the required relation between a1 and a2, i.e., accelerations of 1 and 2 are equal but in opposite directions.
Solved example
24. Find the constraint relation between a1, a2 and a3. 1 1 2 2 3 3 x3 x2 x4 x1
Sol. Points 1, 2, 3 and 4 are movable. Let their displacements from a fixed line be x 1, x 2, x 3, and x4
we have x1+ x4 = l1 (length of first string) . . .(i) and (x2– x4) + (x3 – x4) = l2 (length of second string) (or) x2+ x3 – 2x4 = l2 . . . (ii)
On double differentiating with respect to time, we get a1 + a4 = 0 . . . (iii) and a2 + a3 – 2a4 = 0 . . . (iv)
But since a4 = –a1 [From equation(iii)] we have, a2 + a3 + 2a1 = 0
This is the required constraint relation between a1, a2 and a3
Try yourself:
20. Using constraint equations find the relation between a1 and a2
Ans: a2 = –7a1
TEST YOURSELF
1. Consider the system shown in figure. The pulley and the string are light and all the surface are frictionless. The tension is the string is (Take g = 10 m/s2)
kg (1) 0 N (2) 1 N (3) 2 N (4) 5 N
kg
2. The acceleration of block B in the figure will be M1 A M2 B (1) () 2 12 mg 4mm + (2) () 2 12 2mg 4mm + (3) () 1 12 2mg m4m + (4) () 1 12 2mg mm +
3. Setup is hanging vertically. Take g = 10 m/s2
5.9.1 Contact force and Normal Reaction
■ When two bodies are in contact, electromagnetic forces act between the charged particles at their surfaces.
■ Each body exerts a force on the other, resulting in a contact force.
N
kg
A. Acceleration of block 10 kg I) 400 N
B. Tension in the string attached to block 10 kg II) 10 m/s2
C. Net force on block 10 kg III) 200 N
D. Pulling force on ceiling due to pulley IV) 100 N (A) (B) (C) (D) (1) I II III IV (2) III I II IV (3) II III IV I (4) II IV I III
(1) 4 (2) 1 (3) 3
5.9 FRICTION
■ Friction: The opposing force that acts tangentially between two surfaces to prevent relative motion.
■ It is a contact force along the common tangent of the contacting surfaces.
■ Friction is a non-conservative force.
■ Contact forces form an action-reaction pair.
■ Contact force can be resolved into two components:
Friction: Parallel to the contact surface.
Normal reaction force: Perpendicular to the contact surface.
■ Both normal reaction and friction are electromagnetic in nature.
■ Normal reaction: Remains unchanged during motion. It does not depend on the area of contact.
When a body of mass m is on an inclined plane of inclination θ , the weight of the body mg acts vertically downwards, and a component mg cos θ perpendicular to the inclined plane is exerting a contact pressure, then N = mg cos θ
5.9.2 Origin of Friction
■ A surface may appear smooth to the naked eye, but under a microscope, it reveals numerous surface irregularities.
■ When two bodies are in contact, these irregularities interlock and resist any relative motion between the bodies.
■ At the contact points, cold welds form due to intermolecular forces, essentially "welding" the surfaces together.
■ When a large force is applied, these cold welds are sheared off, allowing the body to slide on the surface.
■ Friction originates from electromagnetic forces.
5.9.3 Methods of Reducing Friction
■ Polishing: Reduces friction by smoothing surfaces, but excessive polishing can bring molecules too close, increasing friction due to intermolecular forces.
■ Using lubricants: A thin layer of oil or fluid between surfaces reduces friction, prevents overheating, and depends on the required reduction level and operating conditions.
■ Using ball bearings: Reduces friction by replacing sliding motion with rolling motion in vehicles, motors, and other machinery, as rolling friction is less than sliding friction.
■ Streamlining: Curved surfaces in automobiles and aeroplanes allow air to flow smoothly, reducing friction during motion.
5.9.4 Types of Friction
■ Friction is classified into three types. They are
■ Static friction (fs): The resistance encountered by a body at rest when an external force tries to move it.
■ Static friction is equal and opposite to the applied force parallel to the contacting surfaces.
■ When a small force is applied to a body at rest on a rough surface, it stays at rest due to the frictional force opposing the applied force.
■ As the applied force increases, the frictional force increases equally until the body begins to move.
■ At the point of sliding, static friction reaches its maximum value, equal and opposite to the applied force.
■ Any further increase in the applied force causes the body to slide.
■ The maximum value of static friction is called Limiting friction. ( fL) fsmax = fL
Key Insights:
■ Static friction opposes impending motion. The term impending motion means motion that would take place (but does not actually take place under the applied force), if friction were absent.
Kinetic Friction or Dynamic Friction
■ Kinetic friction (fk): The resistance encountered by a body sliding on a surface.
■ A sliding body experiences a force opposing its motion, eventually causing it to come to rest.
■ If the body moves with uniform velocity (a = 0), the resultant force on the body is zero, meaning the applied force is equal to the kinetic friction.
■ To keep the body moving with uniform velocity, the applied force must be numerically equal to the kinetic friction.
Rolling Friction
■ Rolling friction: The resistance encountered by a body rolling on a surface.
■ Ideally, a body like a ring or spherical ball rolling without slipping on a horizontal plane would experience no friction, as only the point of contact has no motion relative to the plane.
■ In practice, resistance occurs due to deformation at the point of contact, where both the body and surface deform. The flattened area at the contact point tends to slide against the surface, creating rolling friction.
fR
■ Because of the surface deformations, a rolling ball has to climb a hill as long as it is rolling. Thus sliding friction arises
■ For the same weight, rolling friction is much smaller than static or sliding friction.
Key Insights:
■ Rolling friction: Smaller when surfaces in contact are harder, as less deformation occurs, and the area of contact is smaller.
■ Rolling friction is a special case of dynamic friction, where the velocity at the point of contact is zero.
5:
■ Friction on bicycle wheels: When pedaled, the back wheel receives a torque, causing it to rotate.
If the surface is frictionless, the wheel slips.
If the surface is rough, friction prevents slipping, enabling translatory motion of the bicycle.
■ Since the back wheel moves forward, we can conclude that friction on the back wheel is forwards.
■ The front wheel which is in contact with the back wheel slides forward. If the surface on which the front wheel is present is smooth, it slides forward. If it is a rough surface, friction prevents it from sliding.
Smooth surface front wheel slides rough surface friction sliding
■ Friction on the front wheel provides the necessary torque to rotate it, acting backwards. If the bicycle is not pedaled, friction on both wheels acts backwards, opposite to the direction of motion.
5.9.5 Variation of Friction with Applied Force
■ A block B is at rest on a horizontal table, with a pan attached by a thread over a frictionless pulley.
Back wheel slips back
surface
■ As the weight in the pan increases, both the applied force and static friction increase equally.
■ When the static friction reaches its maximum value, equal to limiting friction (fL), the block is in equilibrium, and the net force is zero.
■ If the applied force exceeds fL, the block starts sliding.
■ The force required to keep the block moving at a constant velocity under friction (dynamic friction) is less than the force needed to start the motion.
■ Dynamic friction is always less than limiting friction for the same body on the same surface, and remains constant even if the applied force increases.
fs fmax fk Smooth sliding Fext (at rest)
Slope of the line ‘m’ = tan θ = tan 45° = 1.
■ The static friction is always equal to the force or component of force parallel to the two surfaces in contact in magnitude.
■ The static friction is always opposite to the applied force in direction. The minimum tangential force needed to move the body is equal to the limiting friction
■ The kinetic frictional force is always opposite to the direction of the velocity of sliding. The dynamic friction is equal to the applied force in magnitude only when it is sliding with constant velocity.
5.9.6
Laws of Friction
■ The lasw of friction describe the relationships between the force of friction, the normal force, and the coefficient of friction.
Laws of Friction–Static Friction
■ Static friction always acts in a direction opposite to that in which the body tends to move.
■ Static friction is a self adjusting force.
■ The magnitude of static friction is always equal to the force which tends to move the body.
■ The force of static friction is independent apparent of area of contact.
■ The value of limiting frictional force is directly proportional to the normal reaction.
LLs fNfN ∝⇒=µ
where ‘ µ S’ is a proportionality constant called “coefficient of static friction”.
L s f N µ=
The ratio of maximum static friction to n ormal reaction is called “coefficient of static friction”.
Laws of Friction–Kinetic Friction
■ Kinetic friction always acts in a direction opposite to the direction in which the body is moving.
■ Kinetic friction is independent of area of contact.
■ For small velocities kinetic friction is independent of velocity, whereas at high velocities, owing to heat energy produced,.
■ Kinetic friction is directly proportional to normal reaction.
KKK fNfN ∝⇒=µ
where ‘ µ k’ is a proportionality constant called “coefficient of kinetic friction”.
K k f N µ=
CHAPTER 5: Laws of Motion
The ratio of kinetic friction to normal reaction is called “coefficient of kinetic friction”.
Laws of Friction - Rolling Friction
■ Rolling friction increases with increase in area of contact.
■ Rolling friction is inversely proportional to radius.
■ Rolling friction is directly proportional to normal reaction.
RRR fNfN ∝⇒=µ
Where ‘ µ R’ is a proportionality constant called “coefficient of rolling friction”.
R R f N µ=
The ratio of rolling friction to normal reaction is called “coefficient of rolling friction”.
SKR µµµ >>
The value of coefficient of friction depends on the nature of the materials in contact and surface finish and is independent of area of contact and mass of the body. It is pure ratio and hence has no units or dimensions.
■ In general coefficient of friction is less than one but in some cases like in the case of highly polished copper plates it is greater than one..
5.9.7
Angle of Friction
■ The angle made by the resultant of normal reaction and the limiting friction with the normal reaction is called “Angle of Friction ( φ )”
■ In fig. OA represent N, OB represents fL. Completing the parallelogram OACB, OC gives the resultant of N and fL tan s ⇒=µφ
■ Where angle of friction is φ . The greater angle of friction the greater is the value of coefficient of friction.
Key Insights:
■ Cone of friction: When a body is subjected to a horizontal force P, the frictional force equals limiting friction, and the angle with the normal is the angle of friction.Varying P through 360° forms a cone of friction, with a semi-vertex angle equal to the angle of friction.
■ The forces exerted by the surface on the body are friction ‘ f L ’ and normal reaction ‘N’. As these two are mutually perpendicular, the net contact force acting on the body is given by () 222221L Rss FfNNNNµµ =+=+=+
21RsFmg µ ∴=+
2 Tan1(tan)RsFmg=+= φµφ
5.9.8 Motion of a body on Rough Horizontal Surface
■ Horizontal Force on a block: Frictional force is static friction and it is equal to force that tends to move the body. For the body to come in to motion the applied force ‘ F ’ must be at least equal to the limiting friction ‘fL’.
Key Insights:
Thus, when F = fL,
■ Consider a body of mass ‘m’ placed on a rough horizontal surface of coefficient of kinetic friction ‘ µ k’. A horizontal force ‘F’ greater than limiting friction is applied on it so that the body comes in to motion. If ‘N’ is the normal reaction, then the kinetic friction is given by fk = µ kN = µ kmg
■ If ‘ a ’ is the acceleration produced in the body,then
Solved example
25. A body of mass 60 kg is pushed up with just enough force to start it moving on a rough surface with µ s = 0.5 and µ k = 0.4 and the force continues to act afterwards. What is the acceleration of the body?
Sol. m = 60 kg; µ s = 0.5 ; µ k = 0.4
Solved example
26. A block of mass 4 kg is placed on a rough horizontal plane. A time dependent horizontal force F = kt acts on the block (k = 2 N/s). Find the frictional force between the block and the plane at t = 2 seconds and t = 5 seconds.
( µ = 0.2)
Sol. When t = 2 s, F = 4 N
fL = µ smg = 0.2 × 4 × 10 = 8 N
∴ F < fL ∴ friction = 4 N When t = 5 s, F = 2 (5) = 10 N
F > fL ∴ friction = 8 N
Try yourself:
21. Two blocks A and B attached to each other by a massless spring or kept on a rough horizontal surface ( µ = 0.1) and pulled by a force of 200 N as shown in figure. If at some instant, the 10 kg mass has an acceleration of 12 ms–2, what is the acceleration of 20 kg mass (in ms–2).
Ans: 2.5 –2ms
Pushing and Pulling of a Lawn Roller
■ A roller on horizontal surface pushed by an inclined force:
(sin) rrr fNmgFµµθ==+
■ Where µr is the coefficient of rolling friction between the roller and horizontal surface.
∴ The net pushing force is given by 1coscos(sin) rr PFfFmgFθθµθ=−=−+ or 1(cossin)rr PFmg θµθµ=−− ......... (1)
■ A lawn roller on a horizontal surface pulled by an inclined force:
CHAPTER 5: Laws of Motion
■ Then, the frictional force acting towards left is given by fr = µ r N = µ r (mg – F sin θ )
Where µ r is the coefficient of rolling friction between the roller and the horizontal surface.
∴ The net pulling force is given by 2coscos(sin) rr PFfFmgFθθµθ=−=−− or () 2cossin. rr PFmg θµθµ=+− ......... (2)
From (1) and (2), it is clear that P1 > P2 and concluded that pushing is difficult than pulling. or pulling is easier than pushing.
Applications
■ Applying an Inclined Pulling Force : The applied force ‘F’ can be resolved into two components ‘ F cos θ ’ and ‘ F sin θ ’. The body is in contact with the surface, there by
N + F sin θ = mg ⇒ N = mg – F sin θ … (1)
Force that tends to move the body, F tm = F cos θ
Let a lawn roller be pulled on a horizontal road by a force ‘F’, which makes an angle θ with the horizontal, to the right. The normal reaction is given by N = mg – F sin θ
If F tm < f, then body doesn’t slide
⇒ Frictional force = F tm = F cos θ
For the body to be pulled F tm = fl
⇒ F cos θ = fL
F cos θ = µS N ⇒ F cos θ = µS (mg – F sin θ)
F cos θ = µ Smg – µ SF sin θ
F cos θ + µ SF sin θ = µ Smg
F (cos θ + µ S sin θ ) = µ Smg
F = +
() cossin s s Fmg = + µ θµθ sin cos sin cossin cos mg
() sin coscossinsin Fmg = + φ θφθφ () sin cos Fmg ⇒=φ
For F to be minimum cos ( θ – φ ) should be maximum
From the figure,
This is the least force required to move the body on a horizontal force.
Applications
■ Applying an inclined pushing force: N mg F fL θ θ F sinθ F cosθ
The applied force ‘F’ can be resolved into two components ‘ F cos θ ’ and ‘ F sin θ ’. The body is in contact with the surface, there by
N = mg + F sin θ –––– (1)
Force that tends to move the body, F tm = F cos θ
If F tm < f, then body does not slide
⇒ Frictional force = F tm = F cos θ
For the body to be pulled F tm = fL
⇒ F cos θ = fL
F cos θ = µ SN
⇒ F cos θ = µ S (mg + F sin θ )
F (cos θ – µ S sin θ ) = µ Smg (cossin) s s Fmg = µ θµθ sinsin (coscossinsin)cos Fmg s
sin cos() Fmg = + φ θφ
(coscossinsincos()) θφθφθφ −=+
For F to be minimum θ = 0 min sin tan cos Fmgmg
From the figure, tan φ = µ s
Key Insights:
■ When pushing force F makes an angle θ w ith vertical, then the minimum value of F for which the body moves is gi ven by sin'cos' s s Fmg µ θµθ = Here, F must be positive for which sin'cos'0tan' s θµθθµ −>⇒> (or)1'tan()' θµθφ >⇒> where φ = tan–1 ( µ )
■ For angle φ ' < φ no motion will take place, however large the force F is.
Applications
■ Sliding of a chain on a horizontal table: Consider a uniform chain of mass ‘m’ and length ‘ L ’ lying on a horizontal table of coefficient of friction ‘ µ s ’. When 1/n th of its length is hanging from the edge of the table, the chain is found to be about to slide from the table.
Weight of the hanging part of the chain = mg n fs
CHAPTER 5: Laws of Motion
Weight of the chain lying on the table = 1 1 mg mgmg nn −=−
When the chain is about to slide from edge of the table, the weight of the hanging part of the chain is equal to frictional force between the chain on the table and the table surface. Thus,
1 ss mgmgn mgmg nnnn µµ
1(1)1(1) ss n n µµ =−⇒∴=
If ‘ l ’ is the length of the hanging part, then L n l = Substituting this in the above expression, we get, 1 (or) s s s lL n Lll µ µ µ + ===
∴ The maximum fractional length of chain hanging from the edge of the table in equilibrium is 1 s s L = + µ µ
■ Connected bodies: Two objects of masses m 1 and m 2 are connected by a very light string passing over a clamped light smooth pulley. The object of mass m2 is on rough horizontal table and the object of mass m 1 is hanging vertically. The coefficient of friction between m2 and the table is µ . T T a N
kk fN m2g m1g
Let the acceleration of the system of two objects be a . Considering the forces on
m 1 , we can write the equation for its acceleration ‘a’ as
m1a = m1g – T ---- (i)
Considering the forces on m2,
m2a = T – µ kN
⇒ m2a = T – µ km2g ---- (ii)
Adding Eqs (i) and (ii), we get (m1 + m2)a = (m1 – µ km2)g
■ Vehicle Coming To Rest On A Horizontal surface: A vehicle of mass ‘m’ is moving on a horizontal rough road with a velocity ‘V’. The coefficient of friction between the tyres and the road is ‘ µ ’. On applying brakes it is brought to rest owing to friction. Here, the loss of kinetic energy of the vehicle is equal to work done against friction. If ‘ S’ is the distance the vehicle travels before coming to rest,
W = ∆ KE 2211 0 22kk
One can say that the safe maximum speed of the vehicle for which it comes to rest within a distance of ‘ S’ is given by
2 k VgS =µ
If ‘ t ’ is the minimum time in which the body is to be brought to rest, then impulse
Solved example
27. When a car of mass 1000 kg is moving with a velocity of 20 ms–1 on a rough horizontal road, its engine is switched off. How far does the car move before it comes to rest if the coefficient of kinetic friction between the road and tyres of the car is 0.75?
Sol. Here, v = 20 ms–1, µ K = 0.75, g = 10 ms–2
Stopping distance
2 K v S g µ = = 26.67 m
Try yourself:
22. A block of mass 4 kg is sliding down a fixed rough inclined surface having angle of inclination 30° with horizontal with constant velocity of 1 m/s. Find the frictional force acting on the block.
Ans: 20 N
Applications
■ Body placed on an accelerating truck: Consider a block of mass ‘ m ’ placed on an open-ended truck, as shown in the figure. Let µs be the coefficient of friction between the surfaces in contact. Let the truck start from rest and accelerate at the rate ‘aT’. Now, the block in the non-inertial frame experiences a pseudo force ‘ maT’ in a direction opposite to the direction of motion of the truck.
The frictional force acting on the block = µsmg
For the body to be under equilibrium, the pseudo force must be less than or equal to the frictional force. Thus,
pfLTs Ts Ffmamg ag µ µ ≤⇒≤ ∴≤
Hence, the maximum acceleration with which the truck can move without sliding the block is given by aT(max) = µ sg
If the pseudo force ( F p ) acting on the block is less that limiting friction ( F l ), then block does not slide ⇒ frictional force = F tm , i.e., (F p)
⇒ f = F p ⇒ f = maT
If the pseudo force acting on the block exceeds limiting friction (Fpf>fL), then block starts sliding towards the open rear end of the truck. Now, kinetic friction is present between the block and truck, which opposes the sliding of the block. The acceleration of the block (aB) is given by Rpfk B FFf a mm == , Tk B mamg a m µ = BTk aag µ ∴=−
If ‘x’ is the distance of the block from the rear end, the time after which it
CHAPTER 5: Laws of Motion
leaves the truck is given by
() 2 Tk x t ag µ =
The velocity of the block at the instant it leaves the truck is given by () 2 Tk vagx µ =−
By the time the block leaves the truck, the truck travels a distance
() T Tk ax s ag µ =
Solved example
28. The rear side of a truck is open. A box of 40 kg mass is placed 5 m away from the open end, as shown in figure. The coefficient of friction between the box and the surface it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m/s 2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box). 2 2/ ams
Sol. Because of the acceleration of the truck the pseudo force on the box = m × a = 40 × 2 = 80 N.
This force acts opposite to the acceleration of the truck.
The frictional force on the truck which acts in forward direction f k = µ N = 0.15 × 40g = 58.8 N
Since pseudo force is greater than frictional force, so block will accelerate in backward direction relative to truck with a magni tude
8058.82 0.53m/s
40 a ==
The time taken by box to cover the distance 5 m 2 12 04.34s 2 s satt a =+⇒==
The distance travelled by truck in this duration 12 02(4.34)18.87m 2 s =+×=
Try yourself:
23. Two bodies A and B, of masses
5 kg and 10 kg, in contact with each other rest on a table against a rigid wall, as shown in fig. The coefficient of friction between the bodies and the table is 0.15. A force 200 N is applied horizontally to A. What are
(a) the reaction of the partition (b) the action-reaction forces between A and B?
(c) Find acceleration when the wall is removed?
(d) Does the answer to (b) change, when the bodies are in motion?
Ignore the difference between µ s and µ k.
Applications
■ Block on block: Consider a block of mass M at rest on a frictionless horizontal surface. A block of mass m is placed on top of it. Let µ s and µ k be the coefficients of static and kinetic friction between the two blocks.
■ Case-I:
A horizontal force F is applied on the lower block.
Suppose the blocks are moving together without any relative motion between them with acceleration a c . ...(1) c F a Mm ∴= + m A f a c
Now co nsider the motion of the upper block A only. Frictional force of acting on it is given by ...(2) c fmamFMm == +
Limiting frictio nal force between A and B is ...(3) Ls fmg =⋅ µ
The blocks will move together without any relative motion between the m if f ≤ fL
i.e.if s mF Mmmg ≤µ + i.e.if1 s M Fmg m µ ≤+
Therefore the magnitud e of maximum force that can be applied on the lower block for which the two blocks will move together is given by max 1 s M Fmg m µ =+
If the applied force F is greater than F max, slipping between the blocks takes place and in this case acceleration the blocks will be different.
■ Case-II:
Let a horizontal external force F be applied on the upper block.
CHAPTER 5: Laws of Motion
■ Case-III:
In this case co-efficient of static friction between the blocks is µ 1 and co-efficient of kinetic friction between the lower block and the floor be µ 2.
a) A horizontal external force F is applied on the lower block.
If the blocks move together without any relative motion between them with acceleration a c, then () ...(1) c F a Mm = +
Consider the motion of the lower block. m B f a c
∴ Frictional force acting on the block is given by,
...(2) c fmaMFMm == +
Limiting frictional force is
fL = µ s . mg … (3)
The blocks will move together without any relative motion between them if f ≤ fL
i.e.if mF Mmmg ≤µ +
i.e.if1 m FmgM µ
If the applied force F is greater than F max, then slipping between the blocks takes place and as a result acceleration between the blocks will be different.
Suppose the two block s are moving together without any relative motion between them with acceleration a c .
()2...(1) c FMmg a Mm µ −+ ∴= + m A f a c
Considering motion of the upper block A, frictional force acting on the block
()2...(2) c fmamFMmg Mm µ −+ == +
Limiting frictional force between A and B is fL = µ 1 mg ...(3)
No slipping takes place between A and B if f ≤ fL () () 2 i.e.1 if mFMmg Mmmg µ µ
−+
+
µµ µµ
i.e. if F ≤ ( µ 1 + µ 2) (M + m) g () () () max12 121 FMmg M mg m =++
b) A horizontal force F is applied on the upper block A. M
Suppose the blocks are moving together without any relative motion between them with acceleration a c .
Then () () 2 c FMmg a Mm µ −+ = + ...(1) M A µ2(M+m)g f a c
Considering motion of the lower block,
f – µ 2 (M + m) g = Ma c ...(2)
Also limiting friction between the blocks
fL = µ 1 mg ...(3)
The blocks will move together without any relative motion between them if f ≤ fL ()() () 2 i.e.21 if MmgMFMmgmg Mm µ µµ
()i.e.12if1 m FmgM µµ
If applied force F > F max, slipping between the blocks takes place and in this case acceleration of the blocks will be different.
Solved example
29. Find the direction of friction forces on each block and the ground. (Assume all surfaces are rough and all velocities are with respect to ground.)
2 m/s E D C B A 1 m/s
Solved example
30. A 2 kg block is placed over a 5 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.10. Find the acceleration of the two blocks if a horizontal force of 14 N is applied to the upper block (g = 10 ms –2).
Sol. Consider the motion of 2 kg block. The forces on it are
(ii) normal reaction N by the 5 kg block, vertically upwards,
(iii) force of friction f = µ N to the left and (iv) applied force 14 N.
In the vertical direction, there is no acceleration.
∴ N = 20 N
In the horizontal direction, the acceleration of the 2 kg block is a.
0.10
0.10)
14 – 2 = 2a; a = 6 ms–2.
Consider the motion of 5 kg block. The forces on it are
(i) gravitational force 5 g = 5 × 10 = 50 N, vertically downwards,
(ii) normal reaction N of the horizo ntal surface, vertically upwards
(iii) force of friction f = µ N to the right by Newton’s third law of motion and
(iv) normal reaction, N downwards by 2 kg block.
In the vertical direction, there is no acceleration.
∴ N = 50 N
In the horizontal direction, the acceleration of the 5 kg block is ‘a’.
µ N = 5a' ⇒ 0.10×20 =5a'; a' =0.4 ms–2.
Try yourself:
24. If force F is increasing with time and at t = 0, F = 0 where will slipping first start?
Ans: Between 1 kg and ground
Try yourself:
25. Two blocks of masses ‘ m ’ and ‘ M ’ are arranged as shown in the figure. The coefficient of friction between the two blocks is ‘µ’, whereas between the lower block and the horizontal surface is zero. Find the force ‘F’ to be applied on the upper block, for the system to be under equilibrium?
CHAPTER 5: Laws of
Key Insights:
■ In previous question, if F = 4 µ mg, for the upper block
4 µ mg – T – µ mg = ma .......... (1) for the lower block T – µmg = Ma ....... (2)
Adding (1) and (2), we get,
2 µ mg = (m + M)a () 2 amg mM
This is the acceleration with which the two blocks move in opposite directions.
Applications
■ Pile of blocks: Let ‘N’ identical blocks each of mass ‘m’ are placed one above the other. Let ‘µk’ be the coefficient of kinetic friction between any successive blocks in contact. A horizontal force ‘F’ is applied to pull the ‘ nth’ block from the top without dislodging the other blocks. While pulling this block, frictional forces act between ‘(n – 1)th’ and ‘ nth’ blocks and also between ‘nth’ and ‘(n +1)th’ blocks. Let the frictional forces be ‘f1’ and ‘f2’ respectively.
Ans: 2 µmg
f1 = µ k(n – 1) mg and f2 = µ k nmg.
Thereby to pull the block F = f1 + f2
= µ k (n – 1 + n)mg (21) k Fnmg µ ⇒=−
Key Insights:
■ If the total number of blocks is ‘N’, in order to pull the ‘ n th ’ block from the bottom,
the minimum horizontal force required is () 221 k FNnmg µ =−+
5.9.9 Bodies in Contact with Vertical Surface
■ The limiting friction fL = µ SN = µ SF
■ If mg < f L , then the block is at rest and friction is static and is equal to weight i.e.,
fs = mg.
If mg > fL, then the block slides down .
mg – f = ma
For the body to be under equilibrium
fL = mg
µ s N = mg
µ s F = mg ( N = F) N F mg fs
Key Insights:
■ If a book is held between two hands and pressed using a force ‘F’ by each hand, then weight is balanced by the total frictional force fs F mg F fs
W = 2fL
mg = 2 µ s F
∴=
2 s Fmg µ
Applications
■ A block of mass ‘ m ’ is pressed against a vertical rough wall of coefficient of friction ‘ µ ’ with a horizontal force ‘F’.
The minimum force to be applied parallel to the wall so as to move the body upwards is given by ()()() min FmgfmgNmgF µµ=+=+=+
31. A block of mass m kg is pushed up against a wall by a force P that makes an angle ‘ θ ’ with the horizontal as shown in figure. The coefficient of static friction between the block and the wall is µ . The minimum value of P that allows the block to remain stationary is Sol:
Try yourself:
26. Find the minimum force required to hold 2 kg block in equilibrium against a rough vertical wall 1 . 3 s µ= Take g = 10 m/s2
Ans: 17.32 N
Try yourself:
Applications
■ Body placed in contact with the front surface of Accelerating Truck:
The magnitude of pseudo force FPf = ma = N For the body to be under equilibrium, a mg fs N ma
the necessary condition is mg = µ s FPf mg = µ s ma
The minimum acceleration of the vehicle for which the block will be under equilibrium is given by s ag µ = Solved example
32. A block of mass m is placed behind in contact with vertical side of M as shown in the figure. The coefficient of static friction between m and M is µ . Find the least horizontal force with which m can be pushed so that the two blocks move together is (neglect friction between M and ground)
Acceleration of be system F a mM = +
CHAPTER 5: Laws of Motion
Normal reaction between th e blo cks is MF N Mm = + When ‘m’ is in equilibrium mg < µ N () () MFmgMmmgFMmM µ µ + ≤⇒≥ + The minimum force () FmgMm Mµ + =
27. In the arrangement shown the floor is frictionless. Coefficient of friction between A and B is 0.3. Find the minimum value of F for which blocks A and B will move together without any relative motion between them.
4 kg 1 kg B Ans: 167 N
5.9.10
Motion of a Body on an Inclined Plane
■ Angle of Repose: • The angle of repose (α) is the angle of inclination of a plane with respect to the horizontal at which a body is in limiting equilibrium.
■ As the angle of inclination (θ) increases, the body will begin to slide at a specific angle, α, called the angle of repose. mg fL N α α mg cosα mg sinα
■ If θ<α ; then sin(cos) Ls mgfmgθµθ <= and body is at rest. Now friction is static. The static frictional force ( f) = mg sin θ
■ If angle of inclination θ=α ; the body is ready to slide. The body is in limiting equilibrium, the net force on it should be equal to zero. That means mg sin α becomes equal to the limiting friction ( fL).
i.e.,(or)cossin LS fmgmgµαα = 1
tan(or)tan() SS∴==µααµ
The angle of repose is equal to angle of friction
■ If angle of inclination θ>α ; then sin(cos) Ls mgfmgθµθ >= and the body slides down on a rough inclined plane under kinetic friction.
The kinetic friction cos kkk fNmgµµθ==
u = 0 v The resultant force acting on the body down the plane is (sin) Rk Fmgf θ =−
sinsincosRkk FmgNmgmg ⇒=−=−θµθµθ
(sincos)RkFmg ⇒=−θµθ
∴ The acceleration of the body (sincos)RkFmg a mm θµθ ==
() sincosagkθµθ=−
If ‘v’ is the velocity of the body at point ‘B’ and the body starts from rest at A then by the equation v2 – u2 = 2as, we get 2202(sincos) k vgs θµθ−=−
Where ‘s’ is the distance travelled along the inclined plane
Then 2(sincos)vgskθµθ=−
The time taken by the body to slide down is given by 12 2 sutat =+ Here, u= 0, s = l
12
0()(sincos) 2 k ltgt θµθ∴=+− 2 (sincos) k l t g θµθ ∴=
Key Insights:
■ Let ‘ F ’ is the f orce to be applied parallel to the inclined plane to prevent the body from sliding down.
0(sincos)0 (sincos) Rs s FFmg Fmg =⇒−−= ∴=− θµθ θµθ 0 mgCos mgSin N h F mg fs
Force ‘ F ’ is required only if angle of incination ( θ ) is greater than angle of repose ( α ).
Solved example
33. A body is sliding down an inclined plane have coefficient of friction 0.5. If the normal reaction is twice that of resultant downward force along the incline. Find the angle between the inclined plane and the horizontal.
Sol. µ = 0.5.
()2sincosNmg=θ−µθ
()cos2sincosmgmg ⇒θ=θ−µθ cos2(sincos) ⇒θ=θ−µθ
11 tantan145 22 θθθ =−⇒=⇒=°
Try yourself:
28. IA block is placed on a rough inclined plane of inclination θ = 30°. If the force to drag it up along the plane is to be smaller than to lift it vertically. The coefficient of friction µ should be less than
Ans: 1 3 µ<
Applications
■ Body pr ojected up a Rough Inclined Plane: If a body is projected with an initial velocity ‘u’ to slide up the plane, the kinetic frictional force acts down the plane and the body suffers retardation due to a resultant force (sin) Rk Fmgf θ =+ down the plane as shown in the figure.
∴ The retardation on the body is given by (sin)sincos kk mgNmgmg a mm θµθµθ ++ ==
(cos) Nmg=θ
(sincos)agkθµθ∴=+
If the body com es to rest travelling a distance ‘s’ up the plane,
from v2 – u2 = 2as we get 2 02(sincos) k ugs −=−θ+µθ 2 2(sincos) k u s g ∴= θ+µθ
If ‘ h ’ is the maximum vertical height to which the body can rise, then from the figure sin h s =θ
CHAPTER 5: Laws of Motion
() 2 sin2sincos k hu s g ⇒== θθ+µθ () 2sin 2sincos k hu g θ = θ+µθ () 2sin 2sin1cot k u g θ = θ+µθ () 2 21cot k hu g
∴= +µθ
If ‘t’ is the time taken by the body before coming to rest final velocity v = 0. We have from
v = u + at 0(sincos) ugkt⇒=−+θµθ
()() 2 sincossincos a kk ul t gg ⇒== θ+µθθ+µθ
The time of ascent is less than the time of descent in this case .
Solved example
34. An insect crawls up a hemispherical surface, as shown (see the figure). The coefficient of friction between the insect and the surface is 1/3. If the line joining the centre of the hemisperical surface to the insect makes an angle θ with the vertical, the maximum possible value of θ is given by
Sol. cos Nmg=θ ......(1) sin Nmg µ=θ ......(2) h ↑ ↑ N P O r sin θ mg sin θ mg mg cos θ fS µSN θ θ r
From (1) and (2) we get, 11 cot3cot(3) θ==⇒θ= µ
Try yourself:
29. A 30 kg block is to be moved up an inclined plane at an angle of 30° to the horizontal, with a velocity of 5 ms –1. If the frictional force retarding the motion is 150 N, find the horizontal force required to move the block up the plane.
(g = 10 ms–2)
Applications
Ans: 346 N
■ A body is projected along a rough inclined plane of coefficient of kinetic friction
µ k with an initial velocity ‘ u ’. It travels a distance ‘ l ’ up the plane and retraces its path and reaches the same point of projection. Let its time of ascent is ‘ t a ’ and time of descent is ‘td’ and time of descent is ‘N’ times time of ascent. Then
Solved example
35. A body is released from rest from the top of an inclined plane of length ‘L’ and angle of inclination ‘θ’. The top of plane of length 1(1) Ln n > is smooth and the remaining part is rough. If the body comes to rest on reaching the bottom of the plane then find the value of coefficient of friction of rough plane.
Sol.
■ A body is projected along a rough inclined plane of coefficient of kinetic friction
µ K with an initial velocity ‘ u ’. It travels a distance ‘ l ’ up the plane and retraces its path and reaches the same point of projection with velocity ‘ v’.
During its ascent it moves with a retardation of (sincos)gkθµθ + , while coming down it moves with an acceleration of (sincos)gkθµθ 22 2(sincos)2(sincos) kk uv s gg == θ+µθθ−µθ
2 (sincos)tan (sincos)tan kk kk uu vv θ+µθθ+µ ⇒=⇒=
For smooth part : 2 21 L Va n =
For rough part : 2 2 1 02 n VaL n −=
12 1 22 Ln aaL nn
sin(sincos)(1) ggn ⇒=−−− θθµθ
sin[11]cos(1) gngn ⇒+−=− θµθ tan 1 n n
Try yourself:
30. A block is projected up a rough inclined plane having inclination 45° with horizontal, with velocity 2 m/s. It returns to the point of projection with velocity 1 m/s. Find the coefficient of kinetic friction between the block and the inclined surface.
Ans: 0.6
TEST YOURSELF
1. Two blocks A and B of masses 6 kg and 3 kg rest on a smooth horizontal surface as shown in the fig. If coefficient of friction between A and B is 0.4, the maximum horizontal force which can move them without separation is
(1) 72 N (2) 40 N (3) 36 N (4) 20 N
2. A person holds a block weighting 2 kg between his hands & keeps it from falling down by pressing it with his hands. If the force exerted by each hand horizontally is
50 N, the co-efficient of friction between the hand & the block is ( g = 10 ms–2)
(1) 0.2 (2) 0.4
(3) 0.1 (4) 0.5
3. If μ be the co-efficient of friction between the block and the cart, horizontal acceleration of the cart that is required to prevent block B from falling is:
# Exercises
JEE MAIN LEVEL
Level-I
Newton’s Second Law
Single Option Correct MCQs
1. The velocity of a body of mass 20 kg decreases from 20 m/s to 5 m/s in a distance of 100 m. Force on the body is
(1) –27.5 m (2) –47.5 m
(3) –37.5 m (4) –67.5 m
2. A force acts for 20 s on a body of mass 20 kg starting from rest, after which the force ceases and then body describes 50 m in the next 10 s. The value of force will be:
(1) 5 N (2) 40 N
(3) 20 N (4) 10 N
3. The momentum of a body in two perpendicular directions at any time ‘ t‘ are given by P x = 2t2 + 6 and =+ 32 3 2 y t P . The force acting on the body at t = 2 s is
(1) 5 units (2) 2 units
(3) 10 units (4) 15 units
4. Three forces 2 N, 3 N, 4 N are applied on a body of mass 2 kg for times 2 s, 2 s, 3 s respectively. The impulse imparted to the body is ____ Ns
(1) 11 (2) 22 (3) 33 (4) 44
5. A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to:
(1) 150 N (2) 3 N
(3) 30 N (4) 300 N
6. A particle of mass 0.5 kg is at rest. A force F starts acting on the particle which varies according to the F - t graph shown in figure. Then kinetic energy of the particle at the end of 3 s is
t(s) 3 (1) 530 J (2) 625 J (3) 67 J (4) 250 J
7. A certain force applied to mass m1 gives it an acceleration of 6 ms–2. The same force applied to mass m2 gives it an acceleration of 3 ms–2. If the two masses are fixed together and the same force is applied to the combination, acceleration of it would be
(1) 9 ms–2 (2) 4.5 ms–2 (3) 3 ms–2 (4) 2 ms–2
8. Two bodies of masses 4 kg and 16 kg at rest to acted upon by same force. The ratio of times required to attain the same speed is, (1) 1 : 1 (2) 4 : 1 (3) 1 : 4 (4) 1 : 2
9. Gravel is dropped on a conveyor belt at the rate of 2 kg/s. The extra force required to keep the belt moving at 3ms –1 is (1) 1 N (2) 3 N (3) 4 N (4) 6 N
10. A rigid ball of mass m strikes a rigid wall at 60° and gets reflected without loss of speed as shown in the figure below. The value of impulse imparted by the wall on the ball will be m 60° 60° v v
(1) mV (2) 2mV (3) 2 mV (4) 3 mV
Equilibrium of Particle
Single Option Correct MCQs
11. A mass ‘M’ is suspended by a rope from a rigid support. It is pulled horizontally with a force F. If the rope makes an angle ‘θ’ with vertical in equilibrium, then the tension in the string is (1) F sin θ (2) F/sin θ (3) F cos θ (4) F/cos θ
Numerical Value Questions
12. The weight of the suspended body, if the tension in the diagonal string is 20 N, is n ×10–1 N. Then, the value of n is ______.
Common Forces in Mechanics
Single Option Correct MCQs
13. A rope of length 10 m and linear density 0.5 kg/m is lying lengthwise on a smooth horizontal floor. It is pulled by a force of 25 N. The tension in the rope at a point 8 m away from the point of application of force is (1) 20 N (2) 15 N (3) 10 N (4) 5 N
14. Two blocks are in contact on a frictionless table one has a mass m and the other 2m as shown in figure. Force F is applied on mass 2m then system moves towards right. Now the same force F is applied on m. The ratio of force of contact between the two blocks will be in the two cases respectively.
2m m F F
(1) 1 : 1 (2) 1 : 2
(3) 1 : 3 (4) 1 : 4
15. Ten coins are placed on top of each other on a horizontal table. If the mass of each coin is 10 g and acceleration due to gravity is 10 ms–2, what is the magnitude and direction of the force on the 7th coin (counted from the bottom) due to all the coins above it?
(1) 0.3 N downwards
(2) 0.3 N upwards
(3) 0.7 N downwards
(4) 0.7 N upwards
16. A body of mass 2 kg is hung on a spring balance mounted vertically in a lift. If the lift descends with an acceleration equal to the acceleration due to gravity ‘g’, the reading on the spring balance will be
(1) (2 × g)kg (2) zero
(3) (4 × g)kg (4) (6 × g)kg
17. Figure shows a weight of 30 kg suspended at one end of cord and a weight of 70 kg applied at other end of the cord passing over a pulley. t weight of rope and pulley find the tension in the cord and acceleration of the system (g = 10ms–2) 70 kg
30 kg T T
(1) 2 m/s2; 120 N (2) 4 m/s2; 420 N
(3) 6 m/s2; 180 N (4) 4 m/s2; 240 N
18. A dynamometer D is attached to two blocks of masses 6 kg and 4 kg. Forces of 20 N and 10N are applied on the blocks as shown in Fig. The dynamometer reads
F = 20 N
F = 10 N 6 kg 4 kg D
(1) 10 N (2) 20 N
(3) 6 N (4) 14 N
19. A lift is going up, the total mass of the lift and the passengers is 1500 kg. The variation in the speed of lift is shown in fig. Then the tension in the rope at t = 1 s will be
V(m/s)
3.6 2 4 6 8 10 12 t(s)
(1) 17400 N (2) 14700 N
(3) 12000 N (4) 10000 N
20. With how much acceleration can you lower a 100 kg body from the roof of a house using a cord with a breaking strength of 80 kg weight without breaking the rope?
(1) 1.96 m/s2 downwards
(2) 9.8 m/s2 downwards
(3) 4.9 m/s2 downwards
(4) 19.6 m/s2 downwards
21. An object of mass 2 kg is placed in a frame (s1) moving with velocity 105m/s ij + and having acceleration + 2 510m/s ij This object is also seen by an observer standing in frame (s2) moving with velocity + 510m/s ij
The net force acting on object with respect to s2 frame is
(1) 1020N ij
(2) + 1020N ij
(3) −+
510N ij
(4) zero
22. The apparent weight of a person inside a lift is w1 when lift moves up with a certain acceleration and is w2 when lift moves down with same acceleration. The weight of the person when lift moves up with constant speed is
(1) 12 2 ww + (2) 12 2 ww
(3) 2w1 (4) 2w2
Solving Problems in Mechanics
Single Option Correct MCQs
23. Six forces lying in a plane and forming angles of 60° relative to one another are applied to the center of a homogeneous sphere with a mass m = 6 kg. These forces are radially outward and consecutively 1 N, 2 N, 3 N, 4 N, 5 N and 6 N. The acceleration of the sphere is
(1) zero (2) 1/2 m/s2 (3) 1 m/s2 (4) 2 m/s2
24. A block A of mass 7 kg is placed on a frictionless table. A thread tied to it passes over a frictionless pulley and carries a body B of mass 3 kg at the other end. If the system is released from rest, the velocity acquired by the blocks after 2 s is (given g = 10 ms–2)
(1) 3 ms–1 (2) 6 ms–1 (3) 5 ms–1 (4) 1.5 ms–1
25. Three blocks of masses 10 kg, 6 kg and 4 kg are connected by mass less strings and placed on a smooth horizontal surface. If forces of 10 N and 40 N are applied as shown, then tension at T1 is
(1) 10 N (2) 20 N (3) 25 N (4) 35 N
26. On the floor of an elevator, a block of mass 50 kg is placed on which another block of mass 20 kg is also placed. The elevator is moving up with a constant acceleration 1.5 m/s2. Force exerted by 20 kg block on the 50 kg block is nearly: 50 kg 20 kg
(1) 250 N (2) 230 N (3) 170 N (4) 150 N
27. A block of mass m is placed on the floor of lift which is moving with velocity v = 4 t 2 , where t is time in second and velocity in m/s. Find the time at which normal force on the block is four times of its weight m
(1) (3g/8)s (2) g s (3) 4g s (4) 3g s
28. Two blocks, each having a mass M, rest on frictionless surface as shown in the figure. If the pulleys are light and frictionless, and M on the incline is allowed to move down, then the tension in the string will be
(1) 2 sin 3 Mg θ (2) 3 sin 2 Mg θ (3) sin 2 Mg θ (4) 2Mg sin θ
29. A chain of mass ‘m’ is attached at two points A and B of two fixed walls, as shown in the figure. The tension in the chain at A is
(1) 1/2 mg cos θ (2) mg sin θ (3) 1/2 mg cosec θ (4) mg tan θ
30. A railway engine of mass 50 tons is pulling a wagon of mass 40 tons with a force of 4500 N. The resistance force acting is 1 N per ton. The tension in the coupling between the engine and the wagon is
(1) 1600 N (2) 2000 N
(3) 5000 N (4) 1500 N
31. Two blocks of masses ‘3m’ and ‘2m’ are in contact on a smooth table. A force P is first applied horizontally on block of mass ‘3 m’ and then on mass ‘2 m ’. The contact force between the two blocks in the two cases are in the ratio 3m P P 2m
(1) 1 : 2 (2) 2 : 3
(3) 3 : 2 (4) 5 : 3
32. Three equal weights of mass m each are hanging on a string passing over a fixed pulley, as shown in figure. What are the tensions in the string connecting weights A to B and B to C?
(1) 73 ; 44mgmg (2) 74 ; 33mgmg
(3) 42 ; 33mgmg (4) 3 ; 44 mg mg
Numerical Value Questions
33. Two masses of 5 kg and 2 kg, connected by a string, are at rest on a smooth horizontal surface. A constant force F=1N is applied on 5 kg mass, as shown in the figure. Two seconds later, the string connecting the masses is cut but ‘F’ continues to act. After two more seconds, velocity of lighter body is_____ m/s.
F= 1 N 2kg 5kg
Friction
Single Option Correct MCQs
34. A block of mass 6 kg is placed on a horizontal surface. Coefficient of friction between the surfaces in contact is 0.8. If a horizontal force of 30 N acts on the block then the frictional force on the block is ( g = 10 ms–2)
(1) 48 N (2) 66 N
(3) 60 N (4) 30 N
35. A body of mass M is resting on a rough horizontal plane surface, the coefficient of friction being equal to µ. At t = 0, a horizontal force F = F 0t starts acting on it.
Where F0 is a constant. Find the time T at which the motion starts?
(1) Mg/F 0
(2) Mg/ F 0
(3) F0/Mg
(4) None of these
36. A block of weight 100 N is pushed by a force F on a horizontal rough plane moves with an acceleration 1 m/s2,when force is doubled its acceleration becomes 10 m/s2. The coefficient of friction is ______ ( g = 10 ms–2)
(1) 0.2
(2) 0.4
(3) 0.6
(4) 0.8
37. A homogeneous chain lies in limiting equilibrium on a horizontal table of coefficient of friction 0.5 with part of it hanging over the edge of the table. The fractional length of the chain hanging down the edge of the table is
(1) 1/2 (2) 1/5
(3) 1/3 (4) 2/3
38. A coin is kept at distance of 10 cm from the centre of a circular turn table. If µ = 0.8 , the frequency of rotation at which the coin just begins to slip is
(1) 62.8 rpm
(2) 84.54 rpm
(3) 54.6 rpm
(4) 32.4 rpm
39. A body of weight 20 N is on a horizontal surface, minimum force applied to pull it when applied force makes an angle 60° with horizontal (angle of friction (ϕ = 30°)is
(1) 20 N
(2) 203 N
(3) 20 3 N
(4) 10 N
40. A block of mass 1kg is kept on a rough inclined plane at θ = 30° with horizontal. The block is connected with a string as shown. Between the block and inclined plane 3 tan37 4 s µ== Then tension in the string is (g = 10ms–2) θ
(1) < 5N but not equal to zero (2) > 5N
(3) 53N
(4) zero
41. A body of mass 2 kg slides down with an acceleration of 2 m/s2 on a rough inclined plane having a slope of 30°. The external force required to take the same body up the plane with the same acceleration will be (g = 10 m/s2)
(1) 4 N (2) 14 N (3) 6 N (4) 20 N
42. The force required just to move a body up an inclined plane is double the force required just to prevent the body sliding down. If the coefficient of friction is 0.25, the angle of inclination of the plane is (1) 37° (2) 45° (3) 30° (4) 42.6°
43. A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The coefficient kinetic friction between the block and the slab is 0.40. A horizontal force of 100 N is applied on the 10 kg block Find the resulting acceleration of the slab. (g = 10 ms–2) Block 100 N Slab 10 kg 40 kg
(1) 3 ms–2 (2) 4 ms–2 (3) 1 ms–2 (4) 2 ms–2
44. A horizontal force applie d on a body on a rough horizontal surface produces an acceleration ‘ a’ . If coefficient of friction between the body and surface, which is μ, is reduced to μ/3, the acceleration increases by 2 units. The value of μ is
(1) 2 3 g (2) 3 2 g (3) 3 g (4) 1 g
45. A block is placed on a rough horizontal plane. A time-dependent horizontal force F = Kt acts on the block. Here, K is a positive constant. Acceleration time graph of the block is: (1) a t
Level-II
Newton’s Second Law
Single Option Correct MCQs
1. A machine gun fires a bullet of mass 40 g with a velocity of 1200 ms –1 . The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second, at the most?
(1) One
(2) Four
(3) Two
(4) Three
2. A force F is applied to the initially stationary cart. The variation of force with time is shown in the figure. The speed of cart at t = 5 s is
(2) () 132 or FFF mm +
(3) () 123 or FFF mm
(4) () 223 or mm FFF +
4. A bullet is fired from a gun. The force on bullet is F = 600 – 2 × 10 5t newtons. The force reduces to zero just when bullet leaves barrel. Find the impulse imparted to bullet.
(1) 0.6 NS
(2) 0.9 NS
(3) 1.2 NS
(4) 1.5 NS
Equilibrium of Particle
Single Option Correct MCQs
(1) 10 m/s
(2)16.3 m/s
(3) 2 m/s
(4) zero
3. Three forces are acting on a particle of mass m , initially in equilibrium. If at the first instance, two forces
12 and FF are perpendicular to each other and suddenly the third force 3F is removed, then find the acceleration of the particle.
(1) () 312 or FFF mm +
5. A horizontal uniform beam AB of length 4 m and a mass of 20 kg is supported at the end B by means of a string that passes over a fixed, smooth pulley supporting a counterbalancing weight of 8 kg on the other side. What force, F, when applied at the point A in a suitable direction, will hold the beam in static equilibrium? A 20 kg
8 kg 30° B
(1) F = 160 N
(2) F4019N =
(3) () F200803N =−
(4) no force F, as specified, can hold the beam in static equilibrium
6. In the arrangement shown, m1 = 300 kg, m2 = 400 kg. If the system is in equilibrium, then the tension in the rope AB is
m2
(1) 50 kg wt (2) 500 kg wt
(3) 100 kg wt (4) 1000 kg wt
Common Forces in Mechanics
Single Option Correct MCQs
7. In the arrangement shown, a1 and a2 are the acceleration of the two blocks (1 and 2) with respect to ground (in upward direction of each block with respect to lift). Regarding the tension in the string, choose the incorrect statement. (m1 and m2) 1 2 m2 Lift is accelerating up with acceleration a0 m1
(1) T = m1g + m1a1
(2) T = m2g + m2a2
(3) T = m1(g – a1)
(4) () 12 0 12 2 Tga mm mm =+ +
8. A pendulum is hanging from the ceiling o f a cage. When the cage is moving up with certain acceleration and when it is moving down with the same acceleration,
the tensions in the string are T 1 and T 2 , respectively. When the cage moves horizontally with the same acceleration, the tension in the string is
(1) () 22 12 2TT + (2) 22 12 TT 2 +
(3) () 22 12 2TT (4) 22 12 TT +
9. The system shown in figure is released from rest when the spring was in its normal length. On releasing, the spring starts elongating if m m M
(1) M > m
(2) M > 2m
(3) m M 2 <
(4) For any value of M
Numerical Value Questions
10. A smooth block fitted with an ideal spring is pulled with an acceleration a = 5 m/s2 as shown. The plank is suddenly stopped. The acceleration of the block at that instant is () ˆ xi + . Then, the value of ‘ x’ is______. m a i j M
Solving Problems in Mechanics
Single Option Correct MCQs
11. In the arrangement shown, by what acceleration the boy must go up so that 100 kg block remains stationary on the wedge? The wedge is fixed and friction is absent everywhere. Take g = 10 m/s2
100 kg m = 50 kg
(1) 2 m/s2
(2) 4 m/s2
(3) 6 m/s2
(4) 8 m/s2
12. In the figure shown, the relation between the accelerations of A, B, and C is (here, a = acceleration of A relative to C)
14. Wh at is the relation between speeds of A and B in the given figure? Slender rod is constrained to move vertically down and wedge is constrained to move along horizontal.
(1) a + b + c = 0
(2) a = b + c
(3) a + b = c
(4) none of these
13. A bob of mass m1 hangs by a light inextensible string, which passes over a fixed smooth pulley P and connects a ring of mass m2. The ring is constrained to move along a smooth, rigid, horizontal rod. The instantaneous velocities and accelerations of the bodies are v1, a1, and v2, a2 respectively. Then,
(1) aB = aAcot θ (2)aB = aAtan
(3) aA = aBcos θ (4) aB = aAsec θ
Numerical Value Questions
15. A uniform rod of lengt h 60 cm and mass 6 kg is acted upon by two forces as shown in the diagram. The force exerted by 45 cm part of the rod on 15 cm part of the rod is ____ N
16. Two blocks A and B of masses m and 2m, respectively, are held at rest such that the spring is in natural length. The accelerations of both the blocks just after release are a and b. Then, (a + b) is n × g. Then, n is____.
Friction
Single Option Correct MCQs
17. A person of mass 72 kg, sitting on ice, pushes a block of mass of 30 kg on ice horizontally with a speed of 12 ms –1 . The coefficient of friction between the man and ice and between block and ice is 0.02. If g = 10 ms–2, the distance bet ween man and the block, when they come to rest, is
(1) 360 m (2) 10 m
(3) 350 m (4)422.5 m
18. A block of mass m is pressed against a vertical wall with a horizontal force F = mg. Another force F1 = mg/2 is acting vertically upon the block. If the coefficient of friction between the block and wall is 1/2, the force of friction between them is
= m 2 g
(1) 2,upmg (2) ,down 2 mg
(3) mg, up (4) zero
19. A body of mass 10 kg is on a rough inclined plane having an inclination of 30° with the horizontal. If coefficient of friction between the surfaces of contact of the body and the plane is 0.5, find the least force required to pull the body up the plane.
(1) 80.5 N (2) 91.4 N
(3) 85.4 N (4) 78.4 N
20. The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If coefficient of friction between the body and in clined
plane is 1 23 , the angle of inclined plane is (1) 60° (2) 45°
(3) 30° (4) 15°
21. A body sliding down an inclined plane has coefficient of friction 0.5. If the normal reaction is twice that of the resultant downward force along the incline, find the angle between the inclined plane and the horizontal.
(1) 90° (2) 30° (3) 60° (4) 45°
22. A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20. The accelerations of the two blocks, if a horizontal force of 12 N is applied to the lower block, are (g = 10 ms–2) (1) 2 ms–2 , 2 ms–2 (2) 2 ms–2 , 1 ms–2 (3) 3 ms–2 , 1 ms–2 (4) 4 ms–2 , 1 ms–2
Numerical Value Questions
23. The system is pushed by a force F, as shown in figure. All surfaces are smooth, except between B and C. Friction coefficient between B and C is μ. The minimum value of F to prevent block B from downward slipping is . 2 xmg µ The value of x is 2m m 2m
24. In the figure, a block of weight 60 N is placed on a rough surface. The coefficient of friction between the block and the surface is 0.5. The minimum weight of the block that does not slip on the surface is ( x × 10N). The value of x is .
Level-III
Single Option Correct MCQs
1. A rod of length L and mass M is acted on by two unequal forces F1 and F2(< F1), as shown in the following figure. The tension in the rod at a distance y from the end A is given by
25. A block placed on a rough inclined plane of inclination ( θ = 30°) can just be pushed upwards by applying a force F as shown. If the angle of inclination of the inclined plane is increased to ( θ = 60°), of the same block can just be prevented from sliding down by application of a force of same magnitude. The coefficient of friction between the block and the inclined plane is 1 1 n n + , where n is
(1) 12 F1F LL yY
(2) 21 F1F LL yY
(3) () 12 FF L y
(4) none of these
2. Three blocks m 1, m 2 and m 3 are arranged as shown. Find the force ( F ) that must be applied on m 3, so that the blocks are relatively at rest w.r.t. m3 m1 m2 m 3
26. In the figure shown, block B suddenly starts moving at an acceleration of 6 m/s2. The mass of each block is the same. The coefficients of friction are as shown. The displacement of block B (in metres) relative to A in 2s is n × 10, where n is . (g = 10 m/s2)
(1) () ++ = 1232 1 Fmmmmg m
(2) () = 1232 1 Fmmmmg m
(3) () +− = 1232 1 Fmmmmg m
(4) () −+ = 1232 1 Fmmmmg m
3. In the arrangement shown in figure, mA = m, mB = 2m and mC = M. Assume all surfaes to be smooth and pulleys to be light. The acceleration of block C is
A B
(1) 3 38 + Mg Mm (2) 3M8m + Mg (3) 2 38 + Mg Mm (4) 3 Mg m
4. A chain of length ‘L’ and mass ‘M’ is hanging by fixing its upper end to a rigid support. The tension in the chain at a distance ‘ x ’ from the rigid support is
(1) zero (2) Mg (3) ()MgLx L (4) ()MgLx x
5. A heavy uniform chain lies on horizontal table top. If the co-efficient of friction between the chain and the table surface is 0.25, the maximum percentage of the length of the chain that can hang over one edge of the table is
(1) 20% (2) 25% (3) 35% (4) 15%
6. A block of mass M1 = 10 kg is placed on a slab of mass M2 = 30 kg. The slab lies on a frictionless horizontal surface, as shown in figure. The coefficient of static friction between the block and slab is μ1 = 0.4 and that of dynamic friction is μ2 = 0.15. The acceleration with which the slab will move is (g = 10 ms–2)
F = 50 N
(1) 5 ms–2 (2) 2 ms–2
7. A block P of mass m is placed on a horizonal frictionless plane. A second block of same mass m is placed on it and is connected to a spring of spring consant K. The two blocks are pulled by distance A. Block Q oscillates without slipping. What is the maximum value of frictional force between the blocks?
(g = 10 ms–2) Smooth Q P µs
(1) 2 KA (2) KA
(3) smg (4) zero
8. A lift of total mass M is raised by cables from rest to rest through a height h. The greatest tension that the cables can safely bear is nMg. The maximum speed of lift during its journey if the ascent is to be made in shortest time is
(1) 1 2 ghn n + (2) 2 ghn
(3) 2 1 ghn n + (4) n1 2 n gh
9. A light, smooth, inextensible string connected with two identical particles, each of mass m, passes through a light ring R. If a force F pulls the ring, the relative acceleration between the particles has magnitude m
(1) 2 tan F m θ (2) Fsec m θ
(3) tan F m θ (4) zero
10. A body of mass m resting on a smooth horizontal plane starts moving under a constant force F . During its rectilinear motion, the angle θ between the direction of force and horizontal plane varies as θ = kx, where k is a constant and x is the distance travelled by the body from the initial position. What is the velocity of the body?
(1) sin F mk θ (2) 2sin F mk θ
(3) s Fco mk θ (4) 2s Fco mk θ
11. A bead of mass m is attached to one end of a spring of natural length R and spring constant () + = 31 . mg k R The other end of the spring is fixed at a point A on a smooth vertical ring of radius R. The normal reaction at B, t after it is released to move, is 30° A B
(1) 2 mg (2) 3mg
(3) 33mg (4) 33 2 mg
12. A particle of mass ‘m’ is at rest at the origin at time t = 0. It is subjected to a force F(t) = F0e–bt in the x-direction.
(1) The velocity of m at 101is1.
(2) The velocity of m at 0 2 21 is1. F t bmbe
(3) The velocity of m at time t is () 01.Fbt e mb
(4) None of the above
13. A mass M is suspended, as shown, in string, spring, pulley system. Assume ideal conditions. The system is in equilibrium. The net force acting on the lower support is xMg. The value of ‘x’ is ____.
14. A wedge of mass 2m and a cube of mass m are shown in figure. Between cube and wedge, there is no friction. The minimum coefficient of friction between wedge and ground, so that wedge does not move, is 2m 450 m
15. The system is in equilibrium. There is no friction anywhere. Spring, string, and pulley are massless. Spring compression is x cm. Find x.
k = 1000 N/m
14 kg
300
16. A batsman hits back a ball of mass 0.4 kg straight in the direction of the bowler without changing its initial speed of 15 ms–1. The impulse imparted to the ball is ______Ns.
17. Two blocks of masses M1 = 1 kg and M2 = 2 kg, kept on smooth surface, are connected to each other through a light spring (K = 100 N/m), as shown in the figure. When we push mass M1 with a force F = 10N and M1 is seen to move with acceleration α1 = 2 m/s2, what will be the acceleration of M 2?
2. A train is moving with acceleration along a straight line with respect to ground. A person in the train finds that
(1) Newton’s 2nd law is false but Newton's 3rd law is true.
(2) Newtons 3rd law is false but Newton's 2nd law is true.
(3) All the three Newton's laws are false but 2nd law can be applied by considering a pseudo force.
(4) All the three Newton's laws are true.
3. A particle is found to be at rest when seen from a frame S1 and moving with a constant velocity when seen from another frame S2
18. Force F is applied on upper pulley. If F = 30 t, where t is time in second. Find the time (in s) when m1 loses contact with floor. (Take g = 10 m/s2)
F = 30t N
m1 = 4kg, m2 = 1 kg
THEORY-BASED QUESTIONS
Single Option Correct MCQs
1. A lift is going up with uniform velocity. When brakes are applied, it slows down. A person in that lift experiences
(1) more weight
(2) less weight
(3) normal weight
(4) zero weight
a) Both the frames are inertial
b) Both the frames are non-inertial.
c) S1 is inertial and S2 is non-inertial.
d) S1 is non-inertial and S2 is inertial.
(1) a, b are true.
(2) c, d are true.
(3) b, c are true.
(4) a, d are true.
4. For a stationary railway platform on earth, choose the correct statement.
(A) It is an inertial frame of reference for an observer on earth.
(B) It is a non-inertial frame of reference for an observer on the moon.
(1) Only A
(2) Only B
(3) Both A and B are true.
(4) Both A and B are false.
5. A cork and a metal bob are connected by a string, as shown in the figure. If the beaker is given an acceleration towards left, then the cork will be thrown towards
(1) right (2) left
(3) upwards (4) downwards
6. A block of weight W is suspended from the mid-point of a rope whose ends are at the same horizontal level. The force required to straighten the rope is
(1) W (2) 2 W
(3) W/2(4) infinitely large
7. Fuel is consumed at the rate of 50 kg/s in a rocket. The velocity of exhaust gases is 2 km/s. The thrust on rocket is
(1) 2 × 104 kg-wt
(2) 1 × 104 kg-wt
(3) 1 × 105 kg-wt
(4) 2 × 105 kg-wt
8. A ball falls towards the earth. Which of the following is correct?
(1) If the system contains the ball, the momentum is conserved.
(2) If the system contains the earth, the momentum is conserved.
(3) If the system contains the ball and the earth, the momentum is conserved.
(4) If the system contains the ball and the earth and the sun, the momentum is conserved.
9. A lift is moving down with an acceleration equal to the acceleration due to gravity. A body of mass M kept on the floor of the lift is pulled horizontally. If the coefficient of friction is μ, then the frictional resistance offered by the body is (1) Mg (2) Mg (3) zero (4) Mg/2
10. A block of mass 5 kg is lying on a rough horizontal surface. The coefficient of sta tic
and kinetic friction are 0.3 and 0.1 and g = 10 ms–2. The frictional force on the block is (1) 25 N (2) 15 N (3) 10 N (4) zero
11. A rectangular block is placed on a rough horizontal surface in two different ways, as shown. Then,
(a) (b)
(1) friction will be more in case ( a) (2) friction will be more in case ( b) (3) friction will be equal in both the cases (4) friction depends on the relations among its dimensions
12. A block is placed on a rough floor and a horizontal force F is applied on it. The force of friction f by the floor on the block is measured for different values of and a graph is plotted between them.
a) The graph is a straight line of slope 45°.
b) The graph is a straight line parallel to the F-axis.
c) The graph is a straight line of slope 45° for small F and a straight line parallel to the F-axis for F.
d) There is a small kink on the graph.
(1) a and c are true. (2) a, b and d are true. (3) c and d are true. (4) all are true.
13. The limiting friction between two bodies in contact is independent of (1) nature of surfaces in contact
(2) the area of surfaces in contact
(3) normal reaction between the surfaces (4) all the above
14. If we imagine two ideally smooth surfaces, and if they are kept in contact, the frictional force acting between them is
(1) zero
(2) a small finite value but not zero
(3) very high
(4) we can’t predict
15. With increase in temperature, the frictional force acting between two surfaces
(1) increases
(2) decreases
(3) remains the same
(4) none of the above
16. If a man is walking, the direction of friction is
(1) opposite to direction of motion
(2) same as that of direction of motion
(3) perpendicular to that of direction of motion
(4) 45° to the direction of motion
17. A boy of mass M is applying a horizontal force to slide a box of mass M´ on a rough horizontal surface. Coefficient of friction between the shoes of the boy and the floor is μ and that between the box and the floor is μ'. In which of the following cases, it is certainly not possible to slide the box?
(1) < ´, M < M´ (2) > ´, M < M´
(3) < ´, M > M´ (4) > ´, M > M´
18. Consider the following statements and identify the correct statements.
A) Frictional force always opposes motion
B) Static frictional force is always greater than kinetic frictional force for a given pair of surfaces.
C) Frictional force is a non-conservative force.
(1) All statements are true.
(2) Only B and C are true.
(3) Only C is true.
(4) Only B is true.
19. While walking on ice, one should take small steps to avoid slipping. This is because smaller steps ensure
(1) larger friction
(2) smaller friction
(3) larger normal force
(4) Both A and C
20. A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping, throughout these motions). The directions of frictional force acting on the cylinder are
(1) up the incline while ascending and down the incline while descending (2) up the incline while ascending as well as descending
(3) down the incline while ascending and up the incline while descending (4) down the incline while ascending as well as descending
21. Newton’s first law of motion describes which of the following?
(1) Energy (2) Work
(3) Inertia (4) Moment of inertia
22. A passenger in a moving train tosses a coin. If the coin falls behind him, the train must be moving with
(1) an acceleration (2) a deceleration (3) a uniform speed (4) any of the above
23. Two trains A and B are running in the same direction on parallel tracks such that A is faster than B, and packets of equal weight are exchanged between them. Then,
(1) A will be retarded and B will be accelerated
(2) B will be retarded and A will be accelerated
(3) there will be no change in A but B will be retarded
(4) there will be no change in B but A will be retarded
24. The area under force-time curve gives (1) work (2) power
(3) displacement (4) impluse
25. A man of mass m is on the floor of a lift. The lift is moving up with acceleration 'a'. Then,
a) the net unbalanced force on him is ‘ma ’
b) the normal reaction exerted by the floor on the man is m (g + a)
c) the apparent weight is greater than his true weight
(1) a, b, and c are correct.
(2) a, b, and c are incorrect.
(3) a and c are correct.
(4) b and c are correct.
26. A block of mass 10 kg is suspended through two light spring balances, as shown in the figure. Then, choose the correct option.
27. S-I : If three forces F 1 , F 2 , and F 3 are represented by three sides of a triangle, and F 1 + F 2 = – F 3 , then these three forces are concurrent forces and satisfy the condition for equilibrium.
S-II : A triangle made up of three forces F1, F2, and F 3 as its sides, taken in the same order, satisfy the condition for translatory equilibrium.
28. A body is kept on the floor of a topless lift at rest. The lift starts descending at an acceleration ‘a’.
S-I : If a < g, the displacement of the body in a time ‘t’ is 12 2 at w.r.t ground.
S-II : If a > g, the displacement of the body in a time ‘t’ is 12 2 gt w.r.t ground.
29. S-I : The coefficient of friction between two surfaces will increase if the surfaces are made rough.
10 kg
(1) The upper scale will read 10 kg and lower will read zero.
(2) Both scales will read 5 kg.
(3) The readings of individual scales may be anything but their sum will be 10 kg.
(4) Both scales will read 10 kg.
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect
(3) if statement I is correct but statement II is incorrect,
(4) if statement I is incorrect but statement II is correct.
S-II : Rolling friction is more than sliding friction.
30. S-I : Shock absorbers reduce the magnitude of change in momentum.
S-II : Shock absorbers increase the time of action of impulsive force.
31. S-I : If the weight of the lift is equal to the tension force of the cable wire, then it moves with uniform velocity.
S-II : If the lift moves downward with an acceleration, then the constant force between the boy’s feet and lift floor is more than the weight of boy.
Assertion and Reason Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A), (2) if both (A) and (R) are true but (R) is not the correct explanation of (A), (3) if (A) is true but (R) is false, (4) if both (A) and (R) are false.
32. (A) : When a firecracker (rocket) explodes mid-air, its fragments fly in such a way that centre of mass continue moving in the same path, which the firecracker would have followed, had it not exploded.
(R) : Explosion of cracker (rocket) occurs due to internal forces only and no external force acts for this explosion.
33. (A) : Kinetic friction depends upon normal reaction of the body.
(R) : Static friction is a self-adjusting force.
34. (A) : Newton’s first law is not universally valid.
(R) : Newton’s first law defines an inertial frame.
35. (A) : A cricketer moves his hands backwards while holding a catch.
(R) : While taking a catch, when the time of action increases, the impulse decreases.
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. A body of mass 5 kg starts from origin with a velocity () 3041 ˆˆ 0msuij =+ . If a constant force () N ˆ 5 ˆ Fij =−+ acts on the body, then pick up the correct statements.
(1) y-component of the velocity becomes zero at t = 40.
(2) x-component of the velocity becomes zero at t = 150.
(3) x-component of the velocity becomes zero at t = 50.
CHAPTER 5: Laws of Motion
(4) The direction of motion changes only once along x and y-directions.
2. A force represented as shown in figure, acts on a body of mass 16 kg. The body starts from rest.Then, 10 t (s) F (newton) 5 0 50 75
(1) the change in momentum during the first 5 seconds of its motion is 250 N-S
(2) the change in momentum of the body during the interval 2s < t < 7s is 300 N-S
(3) the velocity of the body at the end of 7 s is 25 m/s
(4) the velocity-time curve of the motion can be represented as
v
3. In the figure, the blocks A, B, and C of mass m each have acceleration a 1, a 2, and a3, respectively. F1 and F2 are external forces of magnitudes 2 mg and mg, respectively. Then,
(1) a1 = a2 = a3 (2) a1 > a2 > a3
(3) a1 = a2, a2 > a3
(4) a1 > a2, a2 = a3
4. A particle moves in XY plane under the influence of a force such that its linear momentum is ()() () ˆ con ˆ ssi PAktiktj =+−
,
where A and k are positive constants.
(1) The magnitude of the force is Ak.
(2) The angle between force and momentum is π /2.
(3) The angle between force and momentum is π.
(4) The angle between acceleration and velocity vectors is always π /2.
5. A block of mass ‘m’ is in equilibrium being supported by light strings AC, BC, and CD, as shown. AB = 5 a ; CB = 3 a ; AC = 4 a . A horizontal force F = mg in the plane of ABC acts parallel to AB, as shown. Then,
7. A particle of mass m, initially at rest, is acted on by a force 2 0 2 1 tT FF
during the interval 0 < t < T . The velocity of the particle at the end of the interval is (1) 50 6 FT m (2)
(1) tension in the string AC is 7 5 mg
(2) tension in the string AC is mg/5
(3) tension in the strong BC is mg/5
(4) tension in the strong BC is 7 5 mg
6. Consider the Atwood machine, as shown in the figure. The larger mass is stopped for a moment 2.0 s after the system is set into motion. Find the time elapsed before the string is tight again.
8. A vertical force of magnitude F acts at the top of a string of mass ‘ m’ and length ‘l’. A body of mass M hangs at the bottom of the string. Then, F M y = y = 0
(1) acceleration of the system is F Mm +
(2) acceleration of the system is F mMg
+
(3) tension in the string at y = 0 is Mm MF +
(4) tension in the string at y = is F
9. A block of mass ‘m’, is raised by a man of mass m 2 in two different ways as shown.
(m1 = 25 kg, m2 = 50 kg)
Fig (1) Fig (2)
(1) The action of the man on the floor in first case is 750 N.
(2) The action of the man on the floor in second case is 250 N.
(3) The action of the man on the floor in first case is 250 N.
(4) The action of the man on the floor in second case is 750 N.
10. Find the minimum acceleration of the wedge, so that the block is under free fall (assume all surfaces are smooth).
(1) 2 2 kx a m =
(2) 1 1 Fkx a m =
(3) F = m1a1 + m2a2
(4) 12 12 F aa mm == + at the time of maximum elongation of the spring
12. Two blocks A and B, each of same mass, are attached by a thin inextensible string through an ideal pulley. Initially, block B is held in position, as shown in figure. Now, the block B is released. Block A will slide to right and hit the pulley in time tA. Block B will swing and hit the surface in time tB Assume the surface as frictionless. Then,
ℓ ℓ
(1) amin = g cot θ
(2) amin = g tan θ
(3) Force of contact between the block and wedge is zero.
(4) Force of contact between the wedge and horizontal surface is equal to the weight of the wedge.
11. A spring connects two particles of masses m1 and m2. A horizontal force F acts on m1. Ignoring friction, when the elongation of the spring is x, then
(1) tA = tB
(2) tA < tB
(3) tA > tB
(4) data is not sufficient to get relationship between tA and tB
13. A particle is found to be at rest when seen from a frame S1 and moving with a constant velocity when seen from another frame S 2 Mark the possible options.
(1) Both the frames are inertial.
(2) Both the frames are non-inertial.
(3) S1 is inertial and S2 is no-ninertial.
(4) S1 is non-inertial and S2 is inertial.
14. In the following mass-pulley systems shown in figure-1 and figure-2, two blocks A and B are connected, as shown. Find the possible correct relation(s) between their velocities.
16. Figure shows two blocks A and B connected to an ideal pulley string system. I n this system, when bodies are released, then (neglect friction and take g = 10 m/s2)
(1) In figure-1 and figure-2, B A V 3 V =
(2) In figure-1 and figure-2 , BB AA VV 3and2 VV ==
(3) Constraint equation in figure-1 is
VB – 3VA = 0
(4) Constraint equation in figure-1 is VB – 2VA = 0
15. Two moving points A and B are as shown in figure-1 and figure-2, respectively. The relations between the speeds of A and B are VA and VB. In figure-1, , B A V x V = and in figure-2 B A V Vy. = then,
(1) Acceleration of block A is 1 m/s 2
(2) Acceleration of block A is 2 m/s 2
(3) Tension is string connected to block B is 40 N
(4) Tension in string connected to block B is 80 N
17. Two blocks A and B of equal mass m are connected through a massless string and arranged as shown in figure. Friction is absent everywhere. When the system is released from rest,
B
(1) tension in string is 2 mg
(2) tension in string is 4 mg
(3) acceleration of A is 2 g
(4) acceleration of A is 3 4 g
CHAPTER 5: Laws of Motion
Numerical Value Questions
18. A block of mass 5 kg is at rest on a smooth horizontal surface. Water coming out of a pipe horizontally at the rate of 2 kg s –1 hits the block with a velocity of 6 ms–1. The initial acceleration of the block is _______ms –2
19. A disc of mass 5 kg is kept floating horizontally in mid-air by firing 10 bullets per second, vertically up. If the mass of each bullet is 50 g and bullets rebound with same speed, the speed of each bullet is _______ m/s. (g = 10 ms–2)
20. A man of mass m = 60 g is standing on a weighing machine fixed on a triangular wedge of angle θ = 60°, as shown in the figure. The wedge is moving up with an upward acceleration a = 2 m/s2. The weight registered by machine is ____ N.
22. The acceleration of the block B in the figure, assuming the surfaces and the pulleys P1 and P2 are all smooth, is ; F nn
23. The acceleration of movable pulley P, if acceleration of block A = 1 m/s 2↓ , then a p
21. In the two systems, as shown, what is the value of mass M, so that extensions in two springs (of same spring constant k) is same (in kg)?
Integer Value Questions
24. A force FVA =×
is exerted on a particle in addition to the force of gravity, where V is the velocity of the particle and A is a constant vector in the horizontal direction. The minimum speed of a particle of mass 3 kg to be projected, so that it continues to move undeflected with a constant velocity, is xg/A. Then, ‘x’ is (consider the plane of the ground as xy plane) ______.
25. A chain consists of five links, each with mass 100 g, as shown. It is lifted vertically upward with a constant acceleration of 2 ms–2. Then, the force F exerted by the agent in lifting the chain is ‘ x’ N. Then, x is N.
26. A loop placed around a plug is used to pull it with a force F. Then, the tension of the ropes forming the loop depend on the magnitude of the angle ‘ θ ’ is cos 2 F T x = θ
28. As shown in figure the cart mass is (m1) and block mass is (m2). Assume ideal conditions, whereever necessary. Acceleration of the cart is 2 12 g 4mm xm + Then, ‘x’ is ____ cart m2
. Find the value of ‘x’.
27. A mass M is suspended as shown in string, spring, pulley system. Assume ideal conditions. The system is in equilibrium. If the extension produced in the spring is , xMg k then the value of ‘x’ is ______.
29. In the below diagram, m 2 >> m 1 , the acceleration of the cart is xg . Then, x is cart
m2
30. A block of mass m1 is placed on a smooth horizontal surface, as shown, and connected to two other blocks, through pulleystring system. Assume ideal conditions wherever necessary. Here, block m1 moves towards left due to constant external force, F = 2 N.
3
Given m 1 = 2 kg, m 2 = 0.5 kg, m 3 = 1 kg, g = 10 ms–2. If the system stays in equilibrium, the value of F is x × 3 N, where ‘x’ is _____.
31. A horizontal force F acts on a point P of an inextensible string passing over a smooth light pulley, which is attached to a large block of mass M, as shown. Another small block of mass ‘m’ is placed on the larger block. The acceleration of M is xF / M . Find the value of ‘x’.
34. The speed of the ball at the instant the force acting on the ball is maximum is (1) v0 (2) 2v0 (3) 3v0 (4) 4v
Passage II : Three masses m 1 = m , m 2 = 2 m, and m3 =3m are hung on a string passing over a frictionless pulley, as shown in figure. The mass of the string is negligible. The system is released from rest.
Passage-based Questions
Passage I: A ball of mass ‘m’ is thrown with a speed v0 towards a bat. It rebounds along the same direction with speed 2 v0. The variation of interaction force between the bat and ball is shown.
32. Maximum force exerted by the bat on the ball is
35. If a1, a2, and a3 are the accelerations of masses m1, m2 and m3, respectively, then (1) a1 < a2 < a
(3)
36. The tension in the string between masses m2 and m3 is (1) mg (2) 3 mg (3) 4 mg (4) 5 3 mg
Passage III: In the arrangement shown, all strings are light and inextensible, and pulleys are light and smooth. All surfaces are smooth. The block A moves on smooth horizontal surface towards left at a = 12 ms–2. The block B always remain in horizontal pos ition.
33. Average force exerted by the bat on the ball is
37. Acceleration block B is (1) 6 ms–2 (2) 2 ms–2 (3) 4 ms–2 (4) 1ms–2
38. The tension in the string connecting the pulley P 4 and block B is (consider mass of the block A as mA)
(1) 4mA
(2) 8mA
(3) 16mA
(4) cannot be expressed in terms of mA
Matrix Matching Questions
39. In the arrangement shown, strings A, B, and C are connected with two blocks of same mass M. Then , match the following and choose the correct option.
M M C A
Column I
Column II
A. String C is pulled with a sudden jerk I) String A will break
B. String B is pulled with a sudden jerk II) String B will break
C. String C is pulled steadily III) String C will break
D. String A is pulled with a sudden jerk IV) None of these
(A) (B) (C) (D)
(1) IV II I I
(2) III IV I I
(3) III II IV I (4) III II I I
40. The entires in column-II show some arrangements of two or more blocks connected by means of light inextensible ropes and smooth pulleys. The tension in
the string PQ is T = ηmg. Column-I lists, some value of η. Match appropriate options (neglect friction)
Column I Column II
(A) (B) (C) (D) (1) III V II IV
(2) V II V IV
(3) I III V IV
(4) II III IV I
CHAPTER 5: Laws of Motion
FLASHBACK (Previous JEE Questions)
JEE Main
Laws of Motion
1. Consider a block and trolley system as shown in figure. If the coefficient of kinetic friction between the trolley and the surface is 0.04, the acceleration of the system in ms –2 is: (Consider that the string is massless and unstretchable and the pulley is also massless and frictionless): (2024)
kg
N fk
kg (1) 3 (2) 4 (3) 2 (4) 1.2
2. A body of mass 4 kg experiences two forces 158 ˆˆˆ 7 Fijk =++ and 234 ˆˆˆ 3 Fijk =−− The acceleration acting on the body is: (2024)
5. Given below are two statements:
Statement (I) : The limiting force of static friction depends on the area of contact and independent of materials.
Statement (II) : The limiting force of kinetic friction is independent of the area of contact and depends on materials.
In the light of the above statements, choose the most appropriate answer from the options given below: (2024)
(1) Both statement I and statement II are correct.
(2) Both statement I and statement II are incorrect.
(3) Statement I is correct but statement II is incorrect.
(4) Statement I is incorrect but statement II is correct.
6. All surfaces shown in figure are assumed to be frictionless and the pulleys and the string are light. The acceleration of the block of mass 2 kg is: (2024)
3. A cricket player catches a ball of mass 120 g moving with 25 m/s speed. If the catching process is completed in 0.1 s then the magnitude of force exerted by the ball on the hand of player will be (in SI unit): (2024)
(1) 24 (2) 12 (3) 25 (4) 30
4. A heavy iron bar of weight 12 kg is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle 60° with the horizontal, the weight experienced by the man is: (2024)
(1) 6 kg (2) 12 kg
(3) 3 kg (4) 63 kg
7. A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5m. The impulse of force imparted by the ground to the body is given by: (given g = 9.8 m/s2) (2024)
(1) 4.32 kg ms–1 (2) 43.2 kg ms–1 (3) 23.9 kg ms–1 (4) 2.39
8. Three blocks A , B and C are pulled on a horizontal smooth surface by a force of 80 N as shown in figure
A B C F = 80 N 2 kg 3 kg 5 kg T1 T2
The tensions T 1 and T 2 in the string are respectively: (2024)
(1) 40N, 64N
(2) 60N, 80N
(3) 88N, 96N
(4) 80N, 100N
9. A block of mass m is placed on a surface having vertical cross section given by y = x2/4. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is: (2024)
(1) 1/4 m (2) 1/2 m
(3) 1/6 m (4) 1/3 m
10. In the given arrangement of a doubly inclined plane two blocks of masses M and m are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25. The value of m, for which M = 10 kg will move down with an acceleration of 2m/s2 is: (take g =10 m/s2 and tan37°=3/4) (2024)
(1) 9 7 (2) 8 1 (3) 4 3 (4) 5 3
12. A block of mass m slides down the plane inclined at angle 30° with an acceleration g 4 . The value of coefficient of kinetic friction will be (2023) (1) 1 23 (2) 3 2 (3) 231 2 + (4) 231 2
13. The time taken by an object to slide down 45° rough inclined plane is n times as it takes to slide down a perfectly smooth 45° incline plane. The coefficient of kinetic friction between the object and the inclined plane is (2023) (1) 2 1 1 n (2) 2 1 1 n + (3) 2 1 1 n (4) 2 1 1 n
11. A light string passing over a smooth light fixed pulley connects two blocks of masses m1 and m2. If the acceleration of the system is g/8, then the ratio of masses is (2024)
14. The figure represents the momentum time (p-t) curve for a particle moving along an axis under the influence of the force. Identify the regions on the graph where the magnitude of the force is maximum and minimum, respectively, if ( t3 – t2) < t1 (2023)
(1) c and b (2) b and c (3) c and a (4) a and b
15. A machine gun of mass 10 kg fires 20 g bullets at the rate of 180 bullets per minute with a speed of 100 ms –1 each. The recoil velocity of the gun is (2023)
(1) 0.02 m/s (2) 2.5 m/s
(3) 0.6 m/s (4) 1.5 m/s
16. A b ody of mass 10 kg is moving with an initial speed of 20 m/s. The body stops after 5 due to friction between body and the floor. The value of the coefficient of friction is. (take acceleration due to gravity g = 10 ms–2) (2023)
(1) 0.2 (2) 0.4 (3) 0.5 (4) 0.3
17. A block ‘ A ’ takes 2 s to slide down a frictionless incline of 30° and length ‘l’, kept inside a lift going up with uniform velocity ‘ v’. If the incline is changed to 45°, the time taken by the block to slide down the incline will be approximately (2022)
(1) 2.66 s (2) 0.83 s (3) 1.68 s (4) 0.70 s
18. A uniform metal chain of mass m and length ‘ L ’ passes over a massless and frictionless pulley. It is released from rest with a part of its length ‘l’ hanging on one side and rest of its length ‘L - l’ hanging on the other side of the pulley. At a certain point of time, when lL x = , the acceleration of the chain is 2 g .
The value of x is (2022)
(1) 6 (2) 2 (3) 1.5 (4) 4
19. A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient of friction between the conveyor belt and bag is 0.4. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion is [Take g = 10 m/s–2] (2022) (1) 2 m (2) 0.5 m (3) 3.2 m (4) 0.8 m
20. A block of metal weighing 2 kg is resting on a frictionless plane (as shown in figure). It is struck by a jet releasing water at a rate of 1 kgs 1 and at a speed of 10 ms–1. Then, the initial acceleration of the block, in ms–2, will be: (2022) a = ? Plane 2 kg
(1) 3 (2) 6 (3) 5 (4) 4
21. A block of mass 40 kg slides over a surface, when a mass of 4 kg is suspended through an inextensible massless string passing over frictionless pulley, as shown below. The coefficient of kinetic friction between the surface and block is 0.02. The acceleration of block is. (Given g = 10 ms–2) (2022)
40 kg 4 kg
(1) 1 ms–2 (2) 1/5 ms–2
(3) 4/5 ms–2 (4) 8/11 ms–2
22. A block of mass 2 kg moving on a horizontal surface with speed of 4 ms –1 enters a rough surface ranging from x = 0.5 m to x = 1.5 m. The retarding force in this range of rough surface is related to distance by F = – kx, where k = 12 Nm–1. The speed of the block as it just crosses the rough surface will be (2022)
(1) zero (2) 1.5 ms–1 (3) 2.0 ms–1 (4) 2.5 ms–1
23. A system of two blocks of masses m = 2 kg and M = 8 kg is placed on a smooth table, as shown in figure. The coefficient of static friction between two blocks is 0.5. The maximum horizontal force F that can be applied to the block of mass M so that the blocks move together will be (2022) F M m
(1) 9.8 N (2) 39.2 N (3) 49 N (4) 78.4 N
24. A block of mass 200 g is kept stationary on a smooth inclined plane by applying a minimum horizontal force N, Fx = as shown in figure. The value of x =_____________. (2022)
25. A system to 10 balls, each of mass 2 kg, is connected via massless and unstretchable string. The system is allowed to slip over the edge of a smooth table, as shown in figure. Tension on the string between the 7th and 8th ball is_____________N, when the 6th ball just leaves the table. (2022)
10th ball
1st ball
26. Two inclined planes are placed as shown in figure. A block is projected from the point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top point B at a height 10 m. After reaching the point B, the block slides down on inclined plane BC. Time it takes to reach the point C from point A is () + 21s t . The value of t is (use g = 10 m/s2) (2022)
10 m
27. A mass of 10 kg is suspended vertically by a rope of length 5 m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is θ = tan–1(x × 10–1). The value of x is .
28. A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The coefficient of static friction between the chain and the surface of the table is 0.5.
The maximum length of the chain hanging from the table is m.
29. The initial mass of a rocket is 1000 kg. Calculate at what rate the fuel should be burnt, so that the rocket is given an acceleration of 20 ms−2. The gases come out at a relative speed of 500 ms −1 with respect to the rocket. [Use g = 10 m/s 2] (2022)
(1) 6.0 × 102 kg s–1 (2) 100 kg s–1
(3) 10 kg s–1 (4) 60 kg s–1
30. A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle θ, as shown in figure. The coefficient of kinetic friction is μ k. Then, the block’s acceleration ‘a’ is given by (g is acceleration due to gravity) (2022)
(4) cossin K FF g mm
JEE Advanced
31. A particle of mass m is moving in the xy -plane such that its velocity at a point (x, y) is given as () ˆ2ˆ vyxxy =α+ where α is a non-zero constant. What is the force F acting on the particle? (2022)
CHAPTER 5: Laws of Motion
(1) ()22ˆˆ Fmxxyy =α+
(2) ()22ˆˆ Fmyxxy =α+
(3) ()22ˆˆ Fmyxxy =α+
(4) ()22ˆˆ Fmxxyy =α+
32. A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m = 0.4 kg is at rest on this surface. An impulse of 1.0 Ns is applied to the block at time t = 0 so that it starts moving along the x-axis with a velocity v(t) = v0e–t/T, where v0 is a constant and τ = 4 s. The displacement of the block, in metres, at t = T is Take e–1 = 0.37 (2022)
CHAPTER TEST-JEE MAIN
Section-A
1. A ship of mass 3 × 107 kg, initially at rest, is pulled by a force of 5 × 104 N through a distance of 3 m. Assume that the resistance due to water is negligible. The speed of the ship is
(1) 1.5 m/s (2) 60 m/s
(3) 0.1 m/s (4) 5 m/s
2. A physics textbook of mass m rests flat on a horizontal table of mass M placed on the ground. Let N a–b be the contact force exerted by body ‘a’ on body ‘b’. According to Newton’s 3rd Law, which of the following is an action-reaction pair of forces?
(1) mg and Ntable→book
(2) (m+M)g and Ntable→book
(3) Nground → table and mg + Ntable→book
(4) Nground → table and Ntable→book
3. The system shown in the figure is in equilibrium at rest and the spring and string are massless. Now, the string is cut. The acceleration of masses 2m and m, just after the string is cut, will be
(1) g/2, g
(2) g, g
(3) g, g/3
(4) g/4, g/4
4. Three blocks of masses m1, m2, and m3 are connected by a massless string, as shown in figure, on a frictionless table. They are pulled with a force T 3 = 40 N. If m1 =10 kg , m2 = 6 kg, and m3 = 4 kg , then tension T2 will be
3
(1) 10 N (2) 20 N
(3) 32 N (4) 40 N
5. A balloon with mass m is descending with an acceleration a (where a < g). How much mass should be removed from it so that it starts moving up with an acceleration a?
(1) 2ma ga + (2) 2ma ga (3) ma ga + (4) ma ga
6. A block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and, from its other end, another block B of mass m 2 is suspended. The coefficient of kinetic friction between the block and the table is μk. When the block A is sliding on the table, the tension in the string is
(1) () () 12k 12 mm1g mm +µ + (2) () () 12k 12 mm1g mm −µ + (3) () () 2k1 12 mmg mm +µ + (4) () () 2k1 12 mmg mm −µ +
7. Two blocks A and B of masses 1.5 kg and 0.5 kg, respectively, are connected by a massless inextensible string passing over a frictionless pulley, as shown in the figure. Block A is lifted until block B touches the ground and then block A is released. The initial height of block A is 80 cm when block B just touches the ground. The max height reached by block B from the ground after block A falls on the ground i s B A 1.5 kg 0.5 kg
(1) 80 cm (2) 120 cm (3) 140 cm (4) 160 cm
8. Calculate the acceleration of the block and trolley system shown in the figure. The coefficient of kinetic friction between the trolley and the surface is 0.05. ( g = 10 m/s2 , mass of the string is negligible and no other friction exists).
9. A block of mass 1 kg lies on a horizontal surface in the truck. The coefficient of friction between the block and the surface is 0.6. If the acceleration of the truck is 5 ms –2 , the frictional force acting on the block is (g=10 m/s2) (1) 2 N (2) 5 N (3) 3 N (4) 6 N
10. Two point size bodies of same mass are knotted to a horizontal string, one at the end, and the other at the midpoint of it. The string is rotated in horizontal plane with the other end as centre. If T is tension in the string between centre of circle and first body then the tension in the string between the two bodies is (1) T/2 (2) 2T (3) 2T/3 (4)3T/2
11. A body of mass m at rest explodes into three pieces, in the ratio of masses 1:1:2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms–1 and 40 ms–1, respectively. The velocity of the third piece will be
(1) 15 ms–1 (2) 25 ms–1 (3) 35 ms–1 (4) 50 ms–1
12. An object flying in air with velocity () k ˆˆˆ 4i5j6 +− suddenly breaks into two pieces, whose masses are in the ratio 1 : 5. The smaller mass flies off with a velocity () k ˆˆ 20i ˆ 7j4++ . The velocity of the larger piece will be (1) k ˆˆ 4i3 ˆ j40
CHAPTER
(2) 0.8i4.6j8k
+− (3) 20i7j4k
(4) 4i3j
40k +−
13. Water (density ρ) is flowing through the uniform horizontal tube of cross-sectional area A with a constant speed v, as shown in the figure. The magnitude of force exerted by the water on the curved corner of the tube is (neglect viscou s forces) 600
(1) 32 Avρ (2) 2 ρ Av2 (3) 22 Avρ (4) 2 2 Avρ
14. A uniform sphere of weight W and radius 5 cm is being held by a string, as shown in the figure. The tension in the string will be 8cm
(1) 12 5 W (2) 5 12 W
(3) 13 5 W (4) 13 12 W
15. A man of mass 60 kg is standing on a weighing machine kept in a box of mass 30 kg, as shown in the diagram. If the man
manages to keep the box stationary, the reading in the weighing machine is
18. A small block of mass 1 kg is placed over a plank of mass 5 kg. The length of the plank is 8 m. Coefficient of friction between the block and the plank is 0.5 and the ground over which plank is placed is smooth. A constant force F = 50 N is applied on the plank in horizontal direction. The time after which the block will separate from the plank is 1kg
kg F = 50 N
(1) 150 N (2) 600 N
(3) 300 N (4) 400 N
16. In the situation shown below, block A, having mass 3 m, and block B, having mass m , are held at rest on a horizontal, frictionless surface with a massless spring compressed between them. After the blocks are released from rest, what is the ratio of the magnitude of B momentum to the magnitude of A momentum?
(1) 9/1 (2) 3/1
(3) 1/1 (4) 1/3
17. A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is 10 N
(1) 2 N
(2) 20 N
(3) 50 N
(4) 100 N
19. A mass m rests on a horizontal surface. The coefficient of friction between the mass and the surface is μ. If the mass is pulled by a force F, as shown in figure, the limiting friction between the mass and the surface will be 300
(1) mg
(2) ()3/2F mg µ−
(3) [mg – (F/2)]
(4) [mg + (F/2)]
20. A box weighing 20 kg is pushed along the floor at a constant speed by applying a horizontal force. If the coefficient of friction is 0.25, then force applied is (g = 10 ms–2)
(1) 5 N (2) 10 N(3) 50 N (4) 200N
21. A block is kept on the rough horizontal surface and force acting on it as a function of time is shown by a graph. Find velocity (in m/s) of the particle at the end of 10 s. F(N) µk = 0.15 µs = 0.2 t(s) 2kg F 0 6
22. The coefficient of static friction between two blocks is 0.5 and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is ___ N. (Take g = 10 ms–2)
Table 2 kg 1 kg µ= 0.5 F
23. Two blocks of mass 10 kg each are placed on a smooth horizontal surface, as shown in the figure. The coefficient of friction between the blocks is 0.2. Initially, the blocks are at rest. A time varying horizontal force F = (20t) N is applied on the upper block. Find the value of t (in s) when slipping takes place between the two blocks.
10 kg 10 kg F
24. The figure shows the position-time ( xt ) graph of one-dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is___ Ns (in magnitude).
2 0 2 4 6 t(s) x(m) 8 10 12 14 16
25. In the figure shown, pulley and spring are ideal and strings are light and inextensible. Initially all the bodies are at rest when string connecting A and B is cut. Find the initial acceleration of pulley (in m/s 2).
CHAPTER TEST-JEE ADVANCED
2022 P1 Model
Section-A
[Integer Value Questions]
1. A block of 7 kg is placed on a rough horizontal surface and is pulled through a variable force F (in N) = 5t, where ‘t’ is time in seconds at an angle of 37° with the horizontal, as shown in figure. The coefficient of static friction of the block with the surface is one. If the force starts acting at t = 0 s, then the block starts to slide at time t = 10(x) s.
Find x. [Take g = 10 ms–2]
2. For the arrangement shown in the figure, the ratio of tension in the string to the frictional force on 10 kg is________ (Use g = 10 ms–2 and sin3703 5 = )
3. The wedge shown can slide without friction on a horizontal floor. Mass of the wedge is M and its angle of inclination is θ = 30°. A block of mass ‘m’ slides down the wedge without friction when released on its inclined face. If path of the block relative to the ground makes an angle of ϕ = 60° with the horizontal, find the ratio of mass of the block to that of the wedge
4. A block B of mass M B = π kg placed on a rough horizontal surface is connected with block A with the help of string (ideal). The string passes over a fixed horizontal cylinder. The coefficient of friction between block B and ground as well as between string and cylinder is 7 11 µ= . Find the minimum mass of block A in kg so that block B starts sliding. (Take value of 22 7 π= and e = 2.72)
A5. Three identical smooth cylinders, each of mass m and radius r are resting in equilibrium within a fixed smooth cylinder of radius R (only a part of this cylinder has been shown in the figure). Find the largest value of R in terms of r, for the small cylinders to remain in equilibrium. If R is expressed as () 12 RrK =+ , what is value of K?
6. The varying force horizontal F = K 0 t is applied on a block of mass ‘ m ’ placed at rest on a rough horizontal surface (μ is coefficient of friction). At a time , 0 3 , tmg K µ = the acceleration of the block is given by a = p g where p = __( K 0 is a positive constant).
7. The coefficient of friction between the block P and the surface is 0.25. For the system to be in equilibrium, the value of M is ___ kg.
8. The velocity of point A on the rod is 2ms –1 (leftwards) at the instant shown in figure. The velocity of the point B on the rod at this instant is ______ ms–1
VA = 2ms–1 B 5 m 1 m
Section-B
[Multiple Correct Option MCQs]
9. One end of the rigid rod (point A) is moving with a constant velocity v in a horizontal direction. Other end of the rod (point B) is moving vertically. Which of the following statements is/are correct, when the rod is inclined at angle α?
(1) Magnitude of acceleration of point B is
3sin v l α
(2) Magnitude of angular acceleration of the rod is 2 23 sin v l α
(3) Magnitude of angular velocity of the rod is sin v l α
(4) Magnitude of angular velocity of the rod is sin v l α
10. The ring on a fixed rod, as shown in figure, is given a constant horizontal acceleration () /3 g ag = . If the Maximum deflection of the string from the verti cal is θ0, then m m (ring) smooth horizontal rail ℓ
(1) θ0 = 30°
(2) θ0 = 60°
(3) at maximum deflection, tension in string is equal to mg
(4) at maximum deflection, tension in string is equal to 2 3 mg
11. The figure shows a block of mass m placed on a smooth wedge of mass M. The minimum value of M and tension in the string are calculated, so that the block of mass m will move vertically downwards (take g = 10 m/ s2). Then,
(1) 'cot 1cot M M θ = −θ
(2) 'tan 1tan M M θ = −θ
(3) tension in string is tan Mg θ
(4) tension in string is t Mg co θ
12. My weight is w = mg , but when I weigh myself in an elevator with a normal weighing machine, I am surprised to see that it appears to be a different amount W ap . If the coordinate y is measured vertically upward as positive, which of the following statement(s) are true, if acceleration of elevator is ? v a
(1) If W ap is greater than w, then the elevator is certainly moving upwards.
(2) If a v is positive, then W ap will be smaller than w.
(3) If v ag = , then W ap = 0
(4) If the elevator is moving upwards, then W ap could be greater than w, but it could also be smaller.
13. A block of mass m is placed on a fixed inclined plane, making an angle θ = 37° with horizontal. The block is moving down the incline with constant velocity. The only forces acting on block are mg downwards and net reaction exerted by inclined plane (g is acceleration due to gravity). Choose the correct statement(s).
15. In the given arrangement, strings and pulleys are light and all surfaces are frictionless. Assuming, at t = 0, system is released from rest, find the speed of block A at t = 2 s. (Assume horizontal surface to be sufficiently long)
(1) The horizontal component of net reaction exerted by plane on block is zero.
(2) The magnitude of net reaction exerted by plane on block is not less than mg.
(3) The magnitude of net reaction exerted by plane on block is not greater than mg.
(4) If mass of block is somehow doubled, the net force on block still remains the same.
Section-C
[Single Option Correct MCQs]
14. A body of mass 8 kg, at rest, exploded into 3 pieces of masses 1 kg, 2 kg and 5 kg. The first two pieces fly off perpendicular with velocities in the ratio 3 : 2, and 5 kg piece flies off at 10 m/s. Then,
(1) velocity of 1 kg part is 30 m/s
(2) velocity of 2 kg part is 20 m/s
(3) velocity of 1 kg part is 15 m/s
(4) velocity of 2 kg part is 10 m/s
(1) 8 m/s (2) 9 m/s (3) 10 m/s (4) 12 m/s
16. A block of mass m , lying on a rough horizontal plane, is acted upon by a horizontal force P and another force Q , inclined at an angle θ to vertical. The block will remain in equilibrium, if coefficient of friction between block and surface is
17. Determine the magnitude and direction of the net force with respect to ground acting on a stone of mass 0.2 kg, just after it is dropped from the window of a train, which is accelerating at 5 m/s 2. Take g = 10 m/s2
(1) 2N downward
(2) 5N, making an angle of tan–1 (2) with trains motion
(3) 2N upwards
(4) 5N, making an angle of tan–1 (2) with vertical
18. A machine gun fires a bullet of mass 40 g with a velocity of 1200 ms–1. The man
ANSWER KEY
JEE Main
Theory-based Questions
JEE Advanced Level
CHAPTER 5: Laws of Motion
holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most?
(1) One (2) Four (3) Two (4) Three
(36) 1 (37) 2 (38) 2 (39) 4 (40) 3
Flashback
Chapter Test-JEE Main
Chapter Test-JEE Advanced
CIRCULAR MOTION CHAPTER 6
Chapter Outline
6.1 Kinematics of Circular Motion
6.2 Dynamics of Circular Motion
6.1 KINEMATICS OF CIRCULAR MOTION
■ In translatory motion, all particles of a rigid body have the same linear displacement and velocity.
■ In rotational motion, particles move in circles around an axis of rotation, with the same angular displacement and angular velocity.
■ Examples:
Fan blades in rotary motion.
Earth revolving around the sun.
Car wheels performing rolling motion (combination of rotational and translatory motion).
■ Force causes translatory motion, while torque causes rotational motion.
■ Torque is the rotational analogue of force in rotational motion.
6.1.1 Fundamentals of Circular Motion
■ Let us discuss a few important terms related to circular motion.
Radius Vector ( r )
Angular Displacement (q)
■ The angle described by the radius vector in a given time interval is called angular displacement ( q ).
■ As shown in the figure, O represents the centre of the circle on which the particle is rotating, and P0 is the initial position of the particle at time t = 0. The position of the particle at an instant of time ‘t’ is P. So, the angular displacement during time ‘t’ is ‘q’.
In case of particle moving in a circle
In case of rigid body rotating about the axis
■ SI unit of angular displacement is radian. It has no dimensions.
Key Insights:
■ Small angular displacements are vectors because they obey the laws of vector addition.
Large angular displacements are not vectors as they do not obey the laws of vector addition.
CHAPTER 6: Circular Motion
r Angular displacement is analogous to linear displacement in translatory motion. The direction of angular displacement is along the axis of rotation, as given by the right hand screw rule. In vector form, =θ× dsdr
When a particle completes one revolution, q = 2π radian. When a particle completes N revolutions in a circle, the angular displacement q = 2 π N.
If a body rotates about a fixed axis, then all the particles will have the same angular displacement in a given time but linear displacement increases with increase in distance of the particle from the axis of rotation.
Angular Velocity (w)
■ “The rate of angular displacement of a particle is called angular velocity.”
Average Angular Velocity
■ Suppose a particle is revolving in a circular path. At time t = 0, it is at P0. In time t1, radius vector describes an angle q 1, and in time t2, the radius vector describes an angle q 2 . The average angular velocity w avg for this time interval is defined as
Solved example
1. When a motorcyclist takes a U-turn in 4 s, what is the average angular velocity of the motorcyclist?
Sol. When the motorcyclist takes a U-turn, angular displacement q = π rad and t = 4 s. The average angular velocity,
Try yourself:
1. When a car takes a turn through an angle of 45° in 2 seconds, find its average angular velocity.
Ans: 0.3925 seconds
Instantaneous Angular Velocity
■ The angular velocity of the particle at a particular instant of time is called instantaneous angular velocity.
■ SI unit of angular velocity is radian/s. Its dimensional formula is [ T–1].
■ Angular velocity is a pseudo vector. Its direction is given by the right hand screw rule. The direction of the angular velocity will be along the axis of rotation.
Key Insights:
■ If the particle moves with a constant speed in a circular path, its instantaneous and average angular velocities are equal.
■ When a body performs ‘N’ rotations in a time interval ‘t’, then its average angular velocity is 2 N t π ω=
■ If a particle makes ‘n’ rotations per second (rps), its angular velocity is given by 2. ω=π=
■ If T is the time period of revolution of a particle, its angular velocity is 2 π ω= T , where 1 = n T
■ Angular velocity of a particle depends on about which point it is determined or about which axis of rotation it is determined.
Example: As shown in the figure below, the angular velocity of the particle about ‘O’ is 0 α ω= t , and the angular velocity of the particle P about ‘A’ is 0 β ω=⋅ω≠ω
■ If a particle is moving on a circle from P to P’ in time t , as shown in fig., the angular velocity with respect to O will be ()0/ω=β t , while with respect to A, it will be () / ω=αAt But from geometry, b = 2 a , so w 0 = 2 w A.
■ In figure below the angular velocity of B relative to A will be
w rel = w B – w A
So, the time taken by one to complete one revolution around O with respect to the other:
In the above case, if the two particles are revolving in opposite directions in a circular path, then w rel = w B + w A
Applications
■ If two particles are moving on two different concentric circles with different velocities,
then angular velocity of B relative to A as observed by A will depend on their positions and velocities. When A and B are closest to each other and moving in the same direction,
relBABA rrrrr
ω==relBA rel relBA vvv rrr
■ When a body is in pure rotation about an axis, all the particles of the body will have the same angular velocity, irrespective of their distance from the axis of rotation.
■ Angular velocity of the hands of a clock:
Angular velocity of the seconds hand: The seconds hand completes one revolution in 60 s.
221 ;rads 6030 πππ ω=ω== T
Angular velocity of the minute hand: The minute hand completes one revolution in one hour (3600 s).
21 rads 36001800 ππ ω==
Angular velocity of the hour hand. The hour hand completes one revolution in 12 hours.
21 rads 12360021,600 ππ ω== ×
■ Spin angular velocity of earth about its own axis.
21 rads 24360043,200 ππ ω== ×
Angular Acceleration (a)
■ “The rate of change of angular velocity of a particle is called angular acceleration.”
CHAPTER 6: Circular Motion
Average Angular Acceleration (aavg)
■ Average angular acceleration is given by ()2121 2121 2 ∆ωπ− ω−ω α=== ∆−−
Instantaneous Angular Acceleration (ainst)
■ The angular acceleration of a particle at a particular instant of time is called instantaneous angular acceleration.
SI unit is rad s–2 Its dimensional formula is [T –2].
■ Angular acceleration is a pseudo vector whose direction is in the direction of change in angular velocity.
Key Insights:
■ When angular velocity increases, angular acceleration is in the same direction; when angular velocity decreases, angular acceleration is in the opposite direction.
■ If the particle is moving with constant angular velocity, its angular acceleration is zero.
6.1.2 Relation between Linear Velocity and Angular Velocity
■ The instantaneous angular velocity w at the instant of time
is
■ At that instant of time, the linear velocity of the particle is given by
but ∆ S = ∆ q .r
■ Linear velocity = Radius of the circle × angular velocity
■ If a particle has velocity V while it is rotating in the plane of the paper with its radius vector r , then its angular velocity
ω will be in a plane perpendicular to the plane of the paper.
Solved example
2. A car is moving in a circular path with a uniform speed v. Find the magnitude of change in its velocity when the car rotates through an angle q .
Sol. Change in velocity ∆=−
The magnitude of change in velocity 222cos ∆=+−θ vvuuv
As the speed is uniform,
Try yourself:
2. A particle is rotating along a circular path with constant speed of 2 m/s. Find the magnitude of change of velocity when it rotates through an angle of 90°. Ans: m/s22
6.1.3 Equations of Circular Motion and Comparison with Equations of Translational
6.1.4 Uniform Circular Motion (UCM)
■ When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion.
■ Suppose, an object is moving with uniform speed v in a circle of radius r , as shown in the figure (a). Since the velocity of the object is changing continuously in terms of direction, the object undergoes acceleration.
(a)
Fig. (b)
Fig. (c)
The magnitude of the acceleration is
■ If ∆t is small, ∆q will also be small and then arc PP’ can be approximately taken to b e
Therefore, the centripetal acceleration a c is: 2
■ Thus, the acceleration of an object moving with speed v in a circle of radius
r has a magnitude 2 , v r and it is always directed towards the centre. This is why this acceleration is called centripetal acceleration (a term proposed by Newton, because ‘centripetal’ comes from a Greek term which means ‘centre-seeking’). Since v and r are constant in magnitude, the magnitude of the centripetal acceleration is also constant. However, the direction changes but it always pointing towards the centre. Therefore, a centripetal acceleration is not a constant vector.
CHAPTER 6: Circular Motion
■ We can express centripetal acceleration a c in terms of angular speed: 222 2 ω ===ω c vr ar rr
■ The distance moved by the object during one time period is s = r q = 2 π r
In terms of frequency u , we have 222224 ω=πυ⇒=ω=πυ c arr
Solved example
3. An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 s.
(a) What is the angular speed and the linear speed of the motion?
(b) Is the acceleration vector a constant vector? What is its magnitude?
Sol. This is an example of uniform circular motion.
Here, R = 12 cm.
The angular speed w is given by 2720.44rad/s 100 π ω==π×= T
The linear speed v is: 0.44125.3cms1 vR=ω=×=
The direction of velocity v is along the tangent to the circle at every point.
The acceleration is directed towards the centre of the circle.
Since this direction changes continuously, acceleration here is not a constant vector.
However, the magnitude of acceleration is constant
Try yourself:
3. A ball of 200 g is at one end of a string of length 20 cm. It is revolved in a horizontal circle at an angular frequency of 6 rpm. Find (i) the angular velocity
(ii) the linear velocity
(iii) the centripetal acceleration Ans: (i) 0.628 rad/s (ii) 0.1256 m/s (iii) 0.079 2m/s
6.1.5 Non-uniform Circular Motion
■ Non-uniform circular motion: When a particle moves in a circular path and its radius vector makes unequal angular displacements in equal time intervals.
■ Alternatively, if the speed of the particle changes with respect to time, the motion is considered non-uniform circular motion.
We know v = r w
The linear acceleration is
dvd ar dtdt ==ω×
a Xr W X V
■ In non-uniform circular motion, the velocity of the particle changes in both magnitude and direction.
■ The acceleration has two components:
Radial component: Due to the change in the direction of velocity (centripetal acceleration).
Tangential component: Due to the change in the magnitude of velocity.
■ The radial acceleration (also known as centripetal acceleration) is directed towards the center of the circular path. 2 2 ==ω C v ar
The tangential component of acceleration is due to the change in the magnitude of velocity of the particle known as tangential acceleration
The magnitude of net acceleration of a particle in non-uniform circular motion is given by
Here, r a is the tangential component of acceleration () ta and v 2 / r is the radial component () c a
Hence, the resultant acceleration in nonuniform circular motion is
.......... (1)
The direction of the net acceleration with radial component is given by
Key Insights:
■ In uniform circular motion, speed ( v) of the particle is constant i.e., 0 = dv dt
Thus, a t = 0 a t =0 V a r a t V a r and a = a r = 2 v r
■ In accelerated circular motion dv dt = p ositive. i.e., tangential accelerati on of particle is parallel to velocity v
■ In decelerated circular motion, a t u a r
dv dt = negative, i.e., tangential acceleration is anti-parallel to velocity v
■ In non-uniform circular motion, there are linear acceleration and angular acceleration.
Linear acceleration ( a ): It has two components. One is radial towards the centre of the circle and the other is tangential to the circular path.
Angular acceleration (a): It is along the axis of rotation.
Solved example
4. A stone is thrown horizontally from a height with a velocity v x = 15 m/s. Determine the
CHAPTER 6: Circular Motion
normal and tangential acceleration of the stone in 1 second after it begins to move. a t V V x qq X y a n a v y
Sol. The horizontal component of acceleration is zero. The net acceleration of the stone is directed vertically downward and is equal to the acceleration due to gravity g. Thus, 22==+tnagaa
From figure, we can see that cos xnn vaa vag θ=== and sin θ===ytt vaa vag Hence 2 222 == + y t x vgt ag vvgt and 222 xx n x vgv ag vvgt == + on substituting numerical values, v x = 15 m/s, g = 9.8 m/s2, we get a t = 5.4 m/s2 and a n = 8.3 m/s2
Try yourself:
4. A particle starts moving from rest along a circular path of radius 1 m. Its speed is increasing at a rate of 1 m/s2. Find its total acceleration at t = 1 s. Ans: 1.414 m/s2
TEST YOURSELF
1. The length of minutes hand in a pendulum clock is 10 cm. The speed of the tip of the hand is (1) 1 ms 6000 π
(2) 1 ms 18000 π (3) 1 ms 3600 π (4) 1 ms 1200 π
2. A car is moving with a speed of 30 m/s on a circular path of radius 500 m. If the speed is increasing at the rate of 2 m/s 2, the net acceleration of the car is (1) 3.6 m/s2 (2) 2.7 m/s2 (3) 1.8 m/s2 (4) 2 m/s2
3. Find the linear velocity, if 2and22. ˆˆ ˆ krijω==+
(1) 4 ˆ 4 ˆ ij + (2) 4 ˆ 4 ˆ ki + (3) 4 ˆ 4 ˆ ji (4) ˆ 4 ˆ 4 ij
(1) 2 (2) 2 (3) 3
6.2 DYNAMICS OF CIRCULAR MOTION
6.2.1
Centripetal Force
■ In uniform circular motion, the velocity of a particle changes in direction but not in magnitude, leading to centripetal acceleration, which is always directed towards the center of the circle.
■ Centripetal force: The net force required for uniform circular motion, directed towards the center of the circle, to deviate the body from its tangential path into a circular path.
■ The direction of centripetal force is along the radius, towards the center of the circle.
■ This force is supplied by external sources like gravitation, tension, friction, or Coulomb force.
■ If the centripetal force ceases, the body will fly off tangentially at that point due to its inertia of direction.
From Newton’s seco nd law of motion, cc Fma = 2 c
Examples:
■ When a ball at the end of string is rotated about an axis, the required centripetal force is provided by the tension in the string.
Solved examples
1. Moon’s motion: The gravitational force between the Earth and the Moon provides the necessary centripetal force for the Moon's circular motion.
2. Car turning: The static frictional force between the car tyres and the road provides the necessary centripetal force.
3. Aeroplane turning: The lift provided by the banked wings of the aeroplane provides the necessary centripetal force.
4. Electron orbiting nucleus: The Coulombic force between the nucleus and electron provides the necessary centripetal force for the electron’s orbit.
5. One end of a massless spring of spring constant 100 N/m and natural length 0.5 m is fixed and the other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The spring remains horizontal. If the mass is made to rotate at an angular velocity of 2 rad/s, find the elongation of the spring.
Sol. The restoring force in the spring provides the centripetal force to the mass. If ‘x’ is the extension in the spring, then
()2KxmLx=+ω
Putting the values, we get
1cm x ≈
Try yourself:
5. A small ball of mass 100 g is tied to one end of a string of length 1.5 m. The block is made to rotate on a smooth horizontal surface in a circular path about a vertical axis passing through the other end of the string. Find the centripetal force acting on the block if its speed is 2 m/s.
Ans: 4 N 15
6.2.2 Centrifugal Force
■ Suppose a block of mass m is tied to one end of a string and the other end of the string is attached to a fixed point O. The block is moving along a circular path as viewed by an observer standing on the ground as shown in Fig. (1). In this case acceleration of the block (centripetal acceleration) is directed towards the centre O and a c = w2r where r = radius of the circle.
Therefore, observer (1) can write T = ma c = m w 2r
CHAPTER 6: Circular Motion
■ Now, suppose another observer (2) is standing on the rotating block. Relative to observer (2), the block is stationary and since observer (2) is in non-inertial frame of reference, a pseudo force, called centrifugal force, whose direction is radially outward, acts on the block. Since the block is in equilibrium relative to observer (2), we can write T – m w 2r = 0 ⇒ T = m w 2r.
Solved example
6. In the fig. shown below, with what angular speed ‘ w ’ must ‘ m ’ with a radius ‘ r ’ rotate on a frictionless table so that ‘M’ does not move? If m = 1.0 kg, M = 10.0 kg, and r = 0.5 m, find w .
Sol. a) Tension in the string = Weight of the hanging mass M
The tension in the string supplies the required centripetal force to the body of mass m to revolve in a circular orbit of radius r
Try yourself:
6. A stone of mass 2.0 kg is tied to the end of a string of 2 m length. It is whirled in a horizontal circle. If the breaking tension of the string is 400 N, calculate the maximum velocity of the stone.
Ans: 20 –1ms
6.2.3 Dynamics of Non-uniform Circular Motion
■ If a particle moves with varying speed in a circular path, it has centripetal (radial) acceleration and a tangential acceleration.
■ Therefore, the force acting on the particle also have a radial and a tangential component. Because the total acceleration
; Ct aaa =+
the total force acting on the particle is ct FFF =+ ; as shown in figure.
CF is directed towards the centre of the circle, it is responsible for the centripetal acceleration.
2 2
CC mv Fmamr r ===ω
tF is tangential to the circle and is responsible for the tangential acceleration. Due to this, the speed changes. The magnitude of tangential force, tt dv Fmammr dt ===α
■ The magnitude of net force acting on the particle in non-uniform circular motion (F) is given by
■ If the net force makes acute angle with velocity ⇒ The motion is accelerated circular motion F c q V F t
■ If the net force makes obtuse angle with velocity ⇒ The motion is retarded circular motion. F c V F F t
Solved example
The direction of net force () F makes an angle ‘ a ’ with radial component. It is given by tan t c F F α=
Key Insights:
■ If the net force is always perpendicular to velocity FV⊥⇒ The motion is uniform circular motion. F c
7. A particle of mass 100 g is at rest on a smooth horizontal surface. It starts moving in a horizontal circular path of radius 1 m. If speed of the particle increases at a rate of 1 m/s, find the centrifugal force acting on it after 1 s. Sol. 1 00 2 1 1m/s 0.11 F0.1N 1 v c dvdvdt dt v mv r =⇒= ⇒= ×
Try yourself:
7. Starting from rest a particle of mass 200 g continues to move along a circular path of radius 50 cm. Length of path traversed by the particle is given by S = 2t2 m. Find force acting on the particle at t = 1 s.
Ans: 6.45 N
Applications
■ A particle is revolving in horizontal plane on the inner surface of smooth inverted cone at a height ‘ h’ above the vertex.
N sin q N N cos q h q mg
On resolving normal reaction in to comments, N sin q is balanced by weight N sin q = mg
N sin q provides required centripetal force
2 cos mv N r θ= 22 sin tan cos Nmggr Nmvv r θ ∴=⇒θ= θ
() 2 or rrgvgh hv =⇒=
• A conical pendulum occurs when a simple pendulum is displaced at an angle θ from the vertical and the bob moves in a circular path in the horizontal plane, causing the string to sweep out a cone.
Consider a conical pendulum of length ‘l’ and mass ‘m’ which is making an angle q with vertical. Let the bob is revolving in circular path in horizontal plane with an angular velocity w
T sin q T l q
T cos q Mg
CHAPTER 6: Circular Motion
On resolving tension into components the vertical component is balanced by weight.
cos Tmg θ=
The horizontal co mponent of tension provides the required centripetal force.
sin2Tmrθω = sin2but cos Tmr Tmg ∴= θω
sinsin r r θθ =⇒= 2 2 sinsin cos coscos g gg ll θθ×ω ⇒= θ
⇒ω=⇒ω= θθ
The time period 2cos 2 T g πθ π ω ==
In terms of speed of bob, 2 tan v Rg θ=
tan vrv rgrg lr ⇒θ=⇒=
22221/4 22() rgrgvvrg lrlrlr ⇒=⇒==
Solved example
8. A cone rotates with a period of 4 s as shown in figure. What is the relation between linear velocities of the points P and Q?
P 3 cm 15 cm Q
Using similar triangles,
Try yourself:
8. A mass m is connected to a vertical revolving axle by two string of length l, each making an angle of 45° with the axle. Both axle and mass are revolving with angular velocity w . Gravity is directed downward. Find the tension in the upper string, T up, and lower string, Tlow
■ To prevent toppling outward on a horizontal track, the cyclist leans inward at an angle θ.
■ The total reaction force (N) from the ground acts along a line making angle θ with the vertical.
■ Horizontal component ( N sin q ) of N is equal to the force of friction.
■ The vertical component N cos q of normal reaction N will balance the weight of the cyclist, while the horizontal component N sin q will provide the necessary centripetal force to the cyclist.
2 sin fNmV r θ == and N cos q = mg
Dividing equations, Vertical N cos θ N sin θ θ Mg r
We must have sin2 cos NmVr Nmg θ θ =
2 ortan V rg θ=
Therefore, the cyclist should bend through an angle
2 tan1 V rg θ = to get the necessary centripetal force.
6.2.4 Turning of Cyclist on the Road
■ When a cyclist takes a turn, the frictional force (f) acting on the tyres provides the necessary centripetal force.
■ It follows that the angle through which cyclist should bend will be greater, if:
The radius of the curve is small, i.e., the curve is sharper
The velocity of the cyclist is large
Solved example
9. When a cyclist is moving along a horizontal circular track of radius 10 m on a rough surface, its angle of inclination with vertical is 45°. Find the speed of the cyclist. [g = 10 m/s2]
Sol. 2 tantan v vgr rg =θ⇒=θ
1010tan45m/s10m/s=××°=
Try yourself:
9. A cyclist is moving along a horizontal circular track of radius 10 m with a speed of 10 m/s. Mass of the bicycle with the cycle is 100 kg. The cyclist makes an angle q with vertical. Find the reaction of the road on the cycle.
Ans: 1414 N
Applications
■ Body placed on a rotating disc: Consider a body of mass ‘m’ placed at a distance ‘x’ from the centre of a circular disc of radius ‘ R ’. The disc is rotating in a horizontal plane with a constant angular velocity ‘ w ’ about a vertical axis passing through its centre. The two forces acting on the body are the centrifugal force acting radially outwards, trying to push the body outwards, and the frictional force between the surfaces in contact, trying to prevent the body from sliding. For the body to be under equilibrium,
2 s s g mxmg x µ ωµω=⇒=
The minimum time period of rotation for which the body can be under equilibrium is given by
2 s Tx g π µ =
CHAPTER 6: Circular Motion
Solved example
10. A disc revolves with a speed of 1 33 3 rev/min, and has a radius of 15 cm . Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the coefficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?
Sol. For the coin to revolve with the disc, the force of friction should be enough to provide the necessary centripetal force, i.e., 2 mv mg r ≤µ
Now, v = r w , where 2 T π ω= is the ang ular frequency of the disc. For a given µ and w , the condition is 2 rg µ ω ≤
The condition is satisfied by the nearer coin (4 cm from the centre).
Try yourself:
10. A coin is placed at a radial distance of 20 cm. The record is spinning with 1 revolution per second. Find the minimum coefficient of friction between the coin and the record for which the coin will not skid.
Ans: 0.788
Applications
■ Body placed in contact with inner wall of rotating hollow cylinder: Consider a hollow cylinder of radius ‘R ’ rotating in a horizontal plane about a vertical axis passing through its centre and parallel to its length with a constant angular velocity ‘ w ’. A block of mass ‘ m’ is in contact with the inner wall of the cylinder. The forces involved in the system are as shown in the figure.
Here, normal reaction N = mR w 2
For the body to be under equilibrium,
friction (fs) = weight (mg)
µ s N = mg w mr w 2 f s mg N
Try yourself:
As per no slip condition,
2 2/ s s mRmg gR µω ωµ ≥ ⇒≥ Thus the minimum angular velocity
min s g R ω= µ
The maximum time period of revolution of the cylinder 2 s R T g µ =π
Solved example
11. A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall without falling when the floor is suddenly removed ?
Sol. The frictional force f (vertically upwards) opposes the weight mg. The man remains stuck to the wall after the floor is removed if mg ≤ fL i.e., mg < µ mR w 2
The minimum angular speed of rotation of the cylinder is min g R ω µ = 1 9.8 0.153 4.67rads. = × = .
11. A vertical hollow cylinder of inner radius 20 cm is spinning about its axis with 2 revolution per second. A block is kept in contact with the vertical inner wall of the cylinder. Find the minimum coefficient of friction between the block and the cylinder wall for which the block will rotate with the cylinder without slipping.
Ans: 0.31
Applications
■ Vehicle taking a turn on a horizontal road: Consider a vehicle of mass ‘m’ moving on a horizontal rough road with a velocity ‘V’. The coefficient of friction between the tyres and the road is ‘ µ s ’. When it takes a turn of radius of curvature ‘r’, the friction between the tyres and the road supplies the necessary centripetal force.
2 ss mv mgvrg r µµ=⇒=
This is the safe maximum speed with which the vehicle can negotiate the turn, to avoid skidding of the vehicle. The maximum angular velocity is
Sg r µ ω=
Solved example
12. A car is driven round a curved path of radius 18 m without the danger of skidding. The coefficient of friction between the tyres of the car and the surface of the curved path is 0.2. What is the maximum speed in kmph of the car for safe driving? (g = 10 ms–2)
So1. Maximum speed, s vgr =µ
Try yourself:
12. A car is moving along a circular road of mean radius 20 m. The car is on the verge of skidding in radial direction when its speed is 8 m/s. Find the minimum coefficient of friction between tyres and road surface for safe drive.
Ans: 0.32
6.2.5 Motion of a Vehicle at a Turn
■ When vehicles travel along a horizontal circular path, the centripetal force is provided in the following ways:
By friction only.
By banking of roads only.
By both friction and banking of roads.
■ Motion of a vehicle on an unbanked rough road:
r f V O f N mv2 r [[ mg centrifugal force
■ Consider a car of mass ‘ m ’ moving on a rough horizontal surface. The force exerted by the engine brings forward motion. If this force is just sufficient to overcome the opposing frictional force between the tyres and the road, the car moves with uniform velocity along a straight line path.
■ Let the driver steer the vehicle to take a turn. While taking this turn, the vehicle has a tendency to move away from the centre of the curvature due to inertial force (centrifugal force). Simultaneously, a frictional force fs (static frictional) acts towards the centre. Let ‘µs ’ is the coefficient of static friction and v max is the maximum safe velocity of the vehicle.
CHAPTER 6: Circular Motion
Then,maxmax or s s mvv mgr rg == µ µ
■ If the velocity ‘v’ increases or µs decreases, the vehicle tends to follow a circular path of greater radius of curvature.
■ For a given radius of curvature and coefficient of friction, the safe maximum velocity of the vehicle is given by
max s vrg =µ .
Key Insights:
■ If µ = 0 then r = ∞ or the body skids tangentially
■ When a vehicle takes a turn on an unbanked rough surface, as friction provides the centripetal force, this results in wear and tear of the tyres.
■ When a car takes a turn, friction between the tyres and the road helps it follow the desired curved path. However, if friction is insufficient or absent, the car skids.
■ To avoid relying on friction, the outer edge of the road is raised relative to the inner edge, a process known as banking of the road. The angle through which the outer edge is raised is called the angle of banking (θ).
N cos q = mg .... (1)
The other component N sin q is directed towards the centre of the circular path. Which provides the centripetal force for correct angle of banking or optimum angle of banking.
2 sin mv N r θ= .... (2)
Dividing equation (2) by (1) we get, 2 tan(or)tan v vrg rg θθ ==
For a given 2 , tan v r g θ θ =
■ If ‘ v ’ increase s the body will follo w a circular path of larger radius of curvature.
■ Motion of a vehicle on a rough banked road:
■ When a vehicle takes a turn on a smooth banked road, of given radius and given banking angle there is a limit for the speed and it is equal to tan. vrg=θ
■ When the velocity of the vehicle exceeds this value, the vehicle will follow a circular path of larger radius and this may cause an accident.
■ If friction is present between the road and the tyres, the components of friction and normal reaction provide the centripetal force.
■ Case–I: If
2 sin mv N r θ < , the vehicle possesses the tendency to skid up the plane. The safe maximum speed for avoiding skidding can be obtained by taking friction acting down the plane.
2 max sincos.Nfmv r θθ+= cossin Nfmg θθ−= N N cos q mg f sin q f cos q
(centrifugal force)
tan 1tan rg v θµ µθ + =
or,maxtan Vgr θλ=+ where λ = tan–1 ( µ s) = Angle of friction.
Key Insights:
■ If q = 0° (on a horizontal curved path) vrg =µ
■ If µ = 0 (on a smooth banked road) v = tan rg θ
■ Case–II: If 2 sin mv N r θ> , the vehicle possesses the tendency to slip down the plane.
The minimum speed for avoiding slipping down the plane can be obtained by taking friction up the plane
2 sincosmin cossin NNvfN NNrg
==
min tan tan 1tan rg vrg
µθ ==− +
Solved example
13. A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race - car and the road is 0.2, what is the (a) optimum speed of the race - car to avoid wear and tear on its tyres, and (b) maximum permissible speed to avoid slipping?
Sol. On a banked road, the horizontal component of the normal force and the frictional force contribute to provide centripetal force to keep the car moving on a circular turn without slipping. At the optimum speed, the normal reaction’s component is enough to provide the needed centripetal force, and the frictional force is not needed. The optimum speed u 0 is given by
1/2 0tan Rg υθ =
Here R = 300 m, q = 15°, g = 9.8 ms-2
The maximum permissible speed u max is given by
Try yourself:
CHAPTER 6: Circular Motion
14. A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 106 kg. What provides the centripetal force required for this purposethe engine or the rails? What is the angle of banking required to prevent wearing out of the rail?
Sol. The centripetal force is provided by the lateral thrust by the rail on the flanges of the wheels. By the third law, the train exerts an equal and opposite thrust on the rail causing its wear and tear. Angle of banking
13. An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?
Ans: 14.8 km
Key Insights:
■ If the speed maintained is such that 2 sin mv N r θ= then the frictional force will be zero even on rough banked road.
TEST YOURSELF
1. The whole set up shown in the figure where particles ( m 1 and m 2 are) rotating with constant angular velocity w on a horizontal frictionless table. If T1 and T2 are tension in given strings then ratio of tensions T1/T2 is
m2
(1) M1/M2
(2) M1 + 2M2/2M2
(3) M2/M1
(4) M2 + 2M1/M1
2. A body of mass m is tied to one end of a spring and whirled round in a horizontal plane with constant angular velocity and elongation in the spring is 1 cm. If the angular velocity is doubled, the elongation in the spring becomes 5 cm. The original length of spring is
(1) 13 cm
(2) 14 cm
(3) 15 cm
(4) 16 cm
3. A motorcyclist moving with a velocity of 144 km h–1 on a flat road takes a turn on the road at a point, where the radius of curvature of the road is 40 m. The acceleration due to gravity is 10 ms–2. In order to avoid sliding, he must bend with respect to the vertical plane by an angle
(1) q = tan–1(4)
(2) q = 45°
(3) q = tan–1(2)
(4) q = tan–1(6)
4. A curved road of 50 m radius is banked at correct angle for a given speed. If the speed is to be doubled keeping the same banking angle, the radius of curvature of the road should be changed to
(1) 25 m
(2) 100 m
(3) 150 m
(4) 200 m
CHAPTER 6: Circular Motion
# Exercises
JEE MAIN LEVEL
Level-I
Kinematics of Circular Motion
Single Option Correct MCQs
1. The ratio of angular speed of a second hand to that of hour-hand of a clock is
(1) 3600 : 1 (2) 720 : 1
(3) 72 : 1 (4) 60 : 1
2. A particle moves in a circle of radius 20 cm. Its linear speed is given by v = 2t, where t is in second and v in m/s. The radial and tangential acceleration at t = 3 s respectively are
(1) 180 m/s2, 2 m/s2 (2) 160 m/s2, 1 m/s2
(3) 190 m/s2, 3 m/s2 (4) 170 m/s2, 4 m/s2
3. A body is moving in a circular path with a constant speed. It has
(1) A constant velocity
(2) A constant acceleration
(3) An acceleration of constant magnitude
(4) An acceleration which varies with time in magnitude
4. The angular speed of a fly wheel making 120 revolutions/minute is
(1) 2rad/s π (2) 2 4rad/s π
(3) rad/s π (4) 4rad/s π
5. Suppose, a disc is rotating counter-clockwise in the plane of the paper. Then,
(1) its angular velocity vector will be perpendicular to the page pointing up out of the page
(2) its angular velocity vector will be perpendicular to the page pointing inwards
(3) its angular velocity vector acts along the tangent to the disc
(4) None of the above is correct
6. A body is in pure rotation. The linear speed ν of the particle, the distance r of the particle from the axis, and the angular velocity ω of the body are related as r . Then,
(1) 1 r
(2) w∝ r
(3) w= 0
(4) w is independent of r
7. The position vector of a particle is ratiatj cossin . The velocity of the particle is (1) directed towards the origin (2) directed away from the origin (3) parallel to the position vector (4) perpendicular to the position vector
8. A point on the rim of a wheel 3 m in diameter has linear velocity of 18 ms–1. The angular velocity of the wheel is (1) 4 rads–1 (2) 12 rads–1
(3) 6 rads–1 (4) 18 rads–1
9. The speed of a motor increases from 1200 rpm to 1800 rpm in 20 s. How many revolutions does it make in this period of time?
(1) 400 (2) 200
(3) 500 (4) 800
10. Starting from rest, a wheel rotates with uniform angular acceleration 2π rads−2. After 4 s, if the angular acceleration ceases to act, its angular displacement in the next 4 s is (1) 8 π rad (2) 16 π rad
(3) 24 π rad (4) 32 π rad
11. Two particles A and B are moving on two different concentric circles with different velocities V A and V B . Then, the angular velocity of B relative to A, as observed by A, is given by
(1) VV rr BA BA (2) VV rr BA BA (3) vv rr AB AB (4) vv rr BA BA + +
12. A particle is moving in a circular path. The acceleration and momentum vectors at an instant of time are aij23m/s 2 and
64kgm/sPij Then, the motion of the particle is
(1) uniform circular motion
(2) circular motion with tangential acceleration
(3) circular motion with tangential retardation
(4) we cannot say anything from ap and only
13. A table fan rotating at a speed of 2400 rpm is switched off and the resulting variation of the rpm with time is shown in the figure. The total number of revolutions of the fan before it come to rest i s
2400 t(s) Rev/min 600 0 8 16 24
(1) 160 (2) 280 (3) 380 (4) 420
Dynamics of Circular Motion
Single Option Correct MCQs
14. A particle of mass ‘m’ describes a circle of radius ( r ). The c entripetal acceleration of the particle is 2 4 r . The momentum of the particle (1)
(3)
(2)
(4)
15 The centripetal force required by a 1000 kg car that takes a turn of radius 50 m at a speed of 36 kmph is
(1) 1000 N (2) 3500 N
(3) 1600 N (4) 2000 N
16. A stone of mass 0.5 kg is attached to a string of length 2 m and is whirled in a horizontal circle. If the string can withstand a tension of 9 N, the maximum velocity with which the stone can be whirled is
(1) 6 ms–1 (2) 8 ms–1
(3) 4 ms–1 (4) 12 ms–1
17. A particle does uniform circular motion in a horizontal plane. The centripetal force acting on the particle is 10 N. It’s kinetic energy is 1 joule. The radius of the circular path is (1) 0.1 m (2) 0.2 m
(3) 2 m (4) 0.02 m
18. A particle of mass m is tied to a light string and rotated with a speed v along a circular path of radius r. If T is tension in the string and mg is gravitational force on the particle, then the actual forces acting on the particle are
(1) mg and T only
(2) mg, T, and an additional force of mv r 2 directed inwards
(3) mg, T, and an additional force of mv r 2 directed outwards
(4) only a force mv r 2 directed outwards
19. When a car takes a sudden turn, it is likely to fall
(1) away from the centre of curvature
(2) towards the centre of curvature
(3) towards forward direction
(4) towards backward direction
20. A car moves along a horizontal circular road of radius r with velocity V. The coefficient of friction between the wheels and the road is μ. Which of the following statements is not true?
(1) The car will slip if vrg
(2) The car will slip if v rg 2
(3) The car will slip if v rg 2
(4) The car will slip at lower speed, if it moves with some tangential acceleration, than if it moves at constant speed.
21 A curved section of a road is banked for a speed V. If there is no friction between the road and the tyres, then
(1) a car moving with speed υ will not slip on the road
(2) a car is more likely to slip on the road at speeds higher than υ, than at speeds lower than υ
(3) a car is more likely to slip on the road at speeds lower than υ, than at speeds higher than υ
(4) a car can remain stationary on the road without slipping
22 The centripetal force required by a 1000 kg car that takes a turn of radius 50 m at a speed of 36 kmph is
(1) 1000 N (2) 3500 N (3) 1600 N (4) 2000 N
23. One end of a massless spring of spring constant 100 Nm –1 and natural lengt h 0.5
CHAPTER 6: Circular Motion
m is fixed and the other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. If the mass rotates at an angular velocity of 2 rad s–1, the elongation of the spring is approximately (1) 4 cm (2) 3 cm (3) 1 cm (4) 2 cm
Numerical Value Questions
24. What is the smallest radius of a circle (in m)at which a cyclist can travel if its speed is 36 km h –1 , angle of inclination 45° and g = 10 ms–2?
Level-II
Kinematics of Circular Motion
Single Option Correct MCQs
1. A car is moving with a speed of 30 ms –1 on a circular path of radius 500 m. If its speed is increasing at the rate of 2 ms –2, the net acceleration of the car is
(1) 3.6 ms–2 (2) 2.7 ms–2
(3) 1.8 ms–2 (4) 2 ms–2
2. The angular frequency of a fan increases from 30 rpm to 60 rpm in π s. A dust particle is present at a distance of 20 cm from axis of rotation. The tangential acceleration of the particle in π s is
(1) 0.8 ms–2 (2) 0.34 ms–2
(3) 0.2 ms–2 (4) 1.2 ms–2
3. Consider the motion of a particle described by x = a cos t , y = a sin t, and z = t . The trajectory traced by the particle as a function of time is
(1) helix (2) circular (3) elliptical (4) straight line
4. A point situated on a wheel decelerates, obeying the relation w = w 0 – a q , where q is angular displacement counted from t = 0. Which of the following graphs represents the angular acceleration of the wheel?
Numerical Value Questions
5. A stone tied to a 180 cm long string at its end is making 28 revolutions in a horizontal circle in every minute. If the magnitude of acceleration of stone is 1936 x ms–2, then the value of x is __________.
6. A rod is rotating with an angular velocity of 3πt3 rad/s. The angle rotated by the rod from t = 0 to the time when it has an angular velocity of 24π rad/s is given by απ radian. Find 2 α
7. A boy whirls a stone of small mass in a horizontal circle of radius 2.5 m and at height 20 m above level ground. The string breaks and the stone flies off horizontally and strikes the ground after travelling a horizontal distance of 10 m. The magnitude of the centripetal acceleration (in m/s2) of the stone, while in circular motion, is 2 n . Find the value of n. (Take g = 10 m/s2)
Dynamics of Circular Motion
Single Option Correct MCQs
8. The centripetal force required for a 1000 kg car travelling at 36 kmph to take a turn by 90° in travelling along an arc of length 628 m is
(1) 250 N (2) 500 N (3) 1000 N (4) 125 N
9. A cyclist is moving on a smooth horizontal curved surface with a speed of 5 ms–1. If angle of leaning is 30°, radius of curvature of road should be
(1) 53 (2) 253 . (3) 5 3 (4) 25 3
10. A particle of mass m is suspended from a ceiling through a string of length L. If the particle moves in a horizontal circle of radius r, as shown in the figure, then the speed of the particle is θ L
rg Lr22 (4) g r Lr22
11. A car is negotiating a curved road of radius R . The road is banked at an angle θ. The coefficient of friction between the types of the car and the road is μ s . The maximum safe velocity on this road is
(1) gRs s tan ta 1n (2) g R s s tan ta 1n (3) g R s s tan tan 21
(4) gRs s 2 1 tan tan
12. A hemispherical bowl of radius R is set rotating about its axis of symmetry, which is kept vertical. A small block kept in the bowl rotates with the bowl without slipping on its surface. If the surface of bowl is smooth, and the angle made by the radius through the block with the vertical is θ. The angular speed at which the bowl is rotating is (1) g R cos q (2) g R cos q (3) g R sin q (4) g R sin q
13. Two particles of masses m 1 and m 2 are connected to a string and the system is rotated in a horizontal plane with ‘P’ as centre. The ratio of tension in the two parts of string is
(1) m mm 1 12 + (2) mm m 12 1 +
(3) mm m 12 2 2 2 + (4) 21 12 m mm +
14. A conical pendulum of length 1 m makes an angle θ = 45° with respect to z-axis and moves in a circle in the x-y plane. The radius of the circle is 0.4 m and its centre is vertically below O. The speed of the pendulum, in its circular path, will be (Take g = 10 ms–2)
two concentric circles of radii R 1 and R 2 with equal angular speed ω. At t = 0, their positions and direction of motion are shown in the figure. The relative velocity vvt AB at 2 is given by
(1) 0.4 m/s
(2) 2 m/s
(3) 0.2 m/s (4) 4 m/s
15. Two particles A and B are moving on
Numerical Value Questions
16. A curved in a level road has a radius of 75 m. The maximum speed of a car turning this curved road can be 30 m/s without skidding. If radius of curved road is changed to 48 m and the coefficient of friction between the tyres and the road remains same, then maximum allowed speed would be __________m/s.
17. A circular race track is banked at 45° and has a radius of 40 m. If the coefficient of friction between the wheels and the track is 1 2 , find the maximum speed (in m/s) at which the car can travel around the track without skidding. (Round off to the nearest integer)
18. A ball of mass m = 0.5 kg is attached to the end of a string having length L = 0.5 m. The ball is rotated on a horizontal circular path about a vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in radian/s) is__________.
19. A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 4 m, rotating about its vertical axis. The coefficient of friction between the wall and his clothing is 0.4. The minimum rotational speed of cylinder (in rad/s) to enable the man to remain stuck to the wall (without falling), when the floor is suddenly removed, is 10 n . Find the value of n. (Take g = 10 m/s2)
20. A curve on a level road has a radius of 75 m. The maximum speed of a car turning this curved road can be 15 m/s without skidding. If radius of curved road is changed to 48 m and the coefficient of friction between the tyres and the road remains same, then maximum allowed speed would be_____m/s.
21. A hemispherical bowl of radius R = 10 m is set rotating about its axis of symmetry, which is kept vertical. A small block kept in bowl rotates with bowl without slipping on its surface. The surface is smooth. The angle made by radius through the block with vertical is 60°. Angular speed at which the bowl is rotating is h rad/s. The value of h is________. [g = 10 m/s2]
Level-III
Single Option Correct MCQs
1. Three point masses, each of mass m , are joined together using a string to form an equilateral triangle of side a. The system is placed on a smooth horizontal surface and rotated with a constant angular velocity ω about a vertical axis passing through the centroid. Then, the tension in each string is
2. Two spheres of equal masses are attached to a string of length 2 m, as shown. The string and the spheres are then whirled in a horizontal circle about O at a constant rate. If T1 is the tension in the string between P and Q, and T 2 is the tension in the string between P and O, then T1/T2 equals _______. P is connected to the midpoint of string.
3. A large mass M hangs stationary at the end of a light string that passes through a smooth fixed tube to a small mass m that moves around in a horizontal circular path. If l is the length of the string from m to the top end of the tube and θ is the angle between this part and vertical part of the string, as shown in the figure, then time taken by m to complete one circle is equal to
4. A particle of mass m moves along the internal smooth surface of a vertical cylinder of radius R, as shown. The force that acts on the wall of the cylinder, if, initially, the velocity v0 of the particle makes an angle α with the horizontal (aAssume that particle does not leave contact with the curved surface of the cylinder), is a
at P. If particle also moves with bowl without slipping, then angular velocity is (g = 10 m/s2 and R = 2m) w 45º
5. An undeformed spring with spring constant k has length l0. When the system rotates at an angular velocity ω, the weight of mass m causes an extension in the spring. If l is the length of the rotating spring, then w l m
(1) l ml k 0 2 (2) ll ml k 0 0 2 (3) l kl km 0 2 (4) ll ml k 0 0 2
6. A hemispherical bowl is rotated about the axis shown with constant angular velocity ω. A particle is kept, as shown,
(1) 2rad/s
(2) 5rad/s
(3) 3 rad/s
(4) 10rad/s
7. A motorist moves along a circular track at 144 kmh –1. The angle he should make with the vertical if the track is 880 m long is (g = 10 ms–2)
(1) 45° (2) tan 18 7
(3) tan 17 8 (4) tan 116 7
8. The radius of curvature of a railway line at a place, when the train is moving with a speed of 36 kmh–1, is 1000 m, the distance between the two rails being 1.5 m. The elevation of the outer rail above the inner rail is
(1) 15 98 . (2) 3 98
(3) 45 98 (4) 1 98
9. A particle moves in a circular path of radius (R). If centripetal force ( F) is kept constant but the angular velocity is doubled, the new radius of the path will be
(1) 2R (2) R 2
(3) R 4 (4) 4R
10. A rotor of radius ( r ) is rotating about its own vertical axis and a person in contact with inner wall of rotor remains in equilibrium without slipping down. If ω is angular velocity of rotor and μ is minimum coefficient of friction between person and the wall of rotor, then which of the following is correct?
A)B)C)D) 2 2 11 r r
(1) A and B are true. (2) A and D are true. (3) C and B are true.
(4) C and D are true.
Numerical Value Questions
11. A particle whose velocity is given as vitj ˘˘ 6m/s is moving in xy plane. At t = 0, particle is at origin. If the radius of curvature of path at point 2 3 2 3 mm , is R m, find 4R.
12. A particle is moving along a circle and its linear velocity is 0.5 ms–1 when its angular velocity is 2.5 rad s –1 . If its tangential acceleration is 0.2 ms –2 , then its angular acceleration is ______ rad/s 2 .
13. A car moves with a constant tangential acceleration a = 0.62 m/s2 along a horizontal surface circumscribing a circle of radius R = 40 m. The coefficient of sliding friction between the wheels of the car and the surface is μ = 0.2. The car can ride without sliding up to a distance 12x metres, if, at the initial moment of time, its velocity is equal to zero. The value of x is ___. (Round off to nearest integer)
14. A car moves uniformly along a horizontal sine curve ya x sin , where a and α are certain constants. The coefficient of friction between the wheels and the road is equal to μ. Velocity of the car riding without sliding is Vgx a Then, the value of x is_____.
15. A coin is kept at distance of 10 cm from the centre of a circular turn table. If = 0.8 the frequency of rotation at which the coin just begins to slip is ___ rpm. (Take g = π2 ms–2) ( Round off to the nearest integer)
16. A car is moving along a circular track of radius 103m with a constant speed of 36 kmph. A plumb bob is suspended from the roof of the car by a light rigid rod of length 1 m. The angle made by the rod with the track is_________degrees. (g = 10 ms–2)
17. A particle describes a horizontal circle on the smooth inner surface of a conical funnel, whose vertex angle is 90°. If the height of the plane of the circle above the vertex is 9.8 cm, the speed of the particle is _______ m/s. [g = 9.8 m/s2]
18. A person is in contact with the inner wall of a vertical hollow cylinder of radius 2 m, without floor under his feet. He remains in equilibrium without slipping down when the cylinder is rotated about its own vertical axis. If the coefficient of static friction between person and the wall of cylinder is 0.4, the minimum angular velocity of the cylinder should be ______ m/s. [g = 9.8 m/s 2]
19. When the road is dry and coefficient of friction is μ, the maximum speed of a car in a circular path is 10 m/s. If the road becomes wet and ’ = /2, then the maximum speed permitted is ____m/s.
THEORY-BASED QUESTIONS
Single Option Correct MCQs
1. Given below are two statements:
Statement I : The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.
Statement II : The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector
In light of the above statements, choose the correct answer from the options given below.
(1) Both statement I and statement II are correct.
(2) Both statement I and statement II are incorrect.
(3) Statement I is correct but statement II is incorrect.
(4) Statement I is incorrect but statement II is correct.
2. Which of the following statements are true about circular motion?
(A) If a particle is moving with constant speed on a circle, then its total acceleration is zero.
(B) If a particle moves along a circle with increasing speed, then the angle between total acceleration vector and velocity vector is acute angle.
(C) If a particle moves along a circle with constant speed, then tangential acceleration and angular accelerations are zero but centripetal acceleration is not zero.
(D) If a particle moves along a circle with speed, which is directly proportional to time t, then tangential acceleration is constant but centripetal acceleration is not constant.
(1) B, C, and D (2) A and B (3) B and C (4) All are correct
3. Average acceleration of a particle in uniform circular motion in one revolution is (1) zero (2) 2 u r
(3) 22 u r (4) 2 2 u r
4. The direction of angular acceleration of a body moving in a circle in the plane of the paper is (1) along the tangent (2) along the radius inward (3) along the radius outward (4) along the perpendicular to the plane of the paper
5. Velocity vector and acceleration vector in a uniform circular motion are related as (1) both in the same direction (2) perpendicular to each other (3) both in opposite directions (4) not related to each other
6. Centripetal acceleration is (1) a constant vector (2) a constant scalar (3) a magnitude changing vector (4) not a constant vector
7. A cyclist bends while taking turn to (1) reduce friction
8. A car is moving up with uniform speed along a flyover bridge, which is part of a vertical
circle. Choose the correct statement from the following.
(1) Normal reaction on the car gradually decreases and becomes minimum at highest position of bridge.
(2) Normal reaction on the car gradually increases and becomes maximum at highest position.
(3) Normal reaction on car does not change.
(4) Normal reaction on the car gradually decreases and becomes zero at highest position.
9. For a particle moving along a circular path, which of the following relations are correct?
(a)
v (b)
(c) va c ⊥ (d) a c
(1) a, b, d (2) b, c, d
(3) a, b, c (4) a, c, d
10. A car takes a circular turn with a uniform speed u. If the reaction at inner and outer wheels be denoted by R1 and R2, then
(1) R1 = R2 (2) R1 < R2
(3) R1 > R2 (4) none of these
11. A car driver going at speed suddenly finds a wide wall at a distance r. To avoid hitting the wall, he should
(1) apply the brakes
(2) turn the car in a circle of radius r
(3) turn the car in a circle of radius 2r
(4) jump on the back seat
12. When a car takes a sudden turn, it is likely to fall
(1) away from the centre of curvature
(2) towards the centre of curvature
(3) towards forward direction (4) towards backward direction
Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect,
(3) if statement I is correct but statement II is incorrect,
(4) if statement I is incorrect but statement II is correct.
13. Statement-I : If a body moving in a circular path has constant speed, then there is no force acting on it.
Statement-II : The direction of the velocity vector of a body moving in a circular path is changing:
Assertion and Reason Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) if (A) is true but (R) is false,
(4) if both (A) and (R) are false.
14. (A) : When a particle moves in circle with a uniform speed, its velocity and acceleration both change.
(R) : The centripetal acceleration in circular motion is dependent on angular velocity of the body.
15. (A) : Centripetal acceleration causes change of direction of velocity.
(R) : Tangential acceleration causes change in the magnitude of velocity.
CHAPTER 6: Circular Motion
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. A particle is acted upon by constant magnitude force perpendicular to it, which is always perpendicular to velocity of particle. The motion is taking place in a plane. It follows that
(1) velocity is constant
(2) acceleration is constant
(3) kinetic energy is constant
(4) it moves in a circular path
2. In circular motion of a particle,
(1) particle cannot have uniform motion
(2) particle cannot have uniformly accelerated motion
(3) particle cannot have net force equal to zero
(4) particle cannot have any force in tangential direction
3. A car moves on a circular road describing equal angles about the centre in equal intervals of time. Which of the following statements about the velocity of car is/are not true?
(1) Velocity is constant.
(2) Magnitude of velocity is constant but the direction changes.
(3) Both magnitude and direction of velocity change.
(4) Velocity is directed towards the centre of circle.
4. A particle of mass m describes path of radius ‘ r ’ and is radial or normal or centripetal acceleration depends on time ‘ t’ as aR = kt2, here k is +ve constant. Then,
(1) at any time t, force acting on particle is (2) power developed at any time t is mkrt
(3) power developed at any time t is mk3/2/ r3/2t
(4) tangential acceleration is also varying.
5. A particle P of mass m is attached to a vertical axis by two strings AP and BP of length L each. The separation AB = L . P rotates around the axis with an angular velocity ω. The tensions in the strings AP and BP are T1 and T2, respectively. Then,
(1) T1 = T2
(2) T1 + T2= m w 2L
(3) T1 – T2= 2mg
(4) BP will remain taut only if 2g/ L
6. A particle is describing circular motion in a horizontal plane in contact with the smooth inside surface of a fixed right circular cone, with its axis vertical and vertex down. The height of the plane of motion above the vertex is h and the semi-vertical angle of cone is α. The period o f revolution of the particle h a
(1) increases as h increases
(2) decreases as h increases
(3) increases as α increases
(4) decreases as α increases
7. The upper end of the string of a simple pendulum is fixed to a vertical z-axis, and set in motion such that the bob moves along a horizontal circular path of radius 2 m, parallel to the x – y plane, 5 m above the origin. The bob has a speed of 3 m/s. The string breaks when the bob is vertically above the x-axis and it lands on the x–y plane at a point (x, y). Then, Z
10. A smooth circular road of radius r is banked for a speed v = 40 km/h. A car of mass m attempts to go on the circular road. The friction coefficient between the tyre and the road is negligible. Choose the correct statements.
(1) The car cannot make a turn without skidding.
(2) If the car turns at a speed less than 40 km/h, it slips down.
(3) If the car turns at the correct speed of 40 km/h, the force by the road on the car is equal to mv2/r
(4) If the car turns at the correct speed of 40 km/h, the force by the road on the car is greater than mg as well as greater than mv2/r.
(1) x = 2 m (2) x > 2 m
(3) y = 3 m (4) y = 2 m
8. A particle moves in a circle of radius 20 m. Its linear speed is given by v = 2t, where t is in seconds and v is in ms–1. Then,
(1) the radial acceleration at t = 2 s is 80 ms–2
(2) tangential acceleration at t = 2 s is 2 ms–2
(3) net acceleration at t = 2 s is greater than 80 ms–2
(4) tangential acceleration remains constant in magnitude
9. A simple pendulum of length L with bob of mass M is oscillating in a plane about a vertical line between angular limits -ϕ and ϕ. For an angular displacement θ, [|θ| < ϕ] the tension in the string and velocity of the bob are T and v, respectively. Which of the following relations hold good under the above conditions?
(1) T cos q = Mg
(2) TMg Mv L cos 2
(3) Tangential acceleration = g sin q
(4) T = Mg cos q
Numerical Value Questions
11. The speed of a motor increases from 1200 rpm to 1800 rpm in 20 s. How many revolutions does it make in this period of time?
12. A particle P is moving in a circle of radius r with uniform speed v. AB is the diameter of circle and C is the centre. The ratio of angular velocity of P about A and C is ___
13. A circular disc is rotating about its own axis at uniform angular velocity ω. The disc is subjected to uniform angular retardation by which its angular velocity is decreased to ω/2 during 120 rotations. The number of rotations further made by it before coming to rest is ___
14. If the angular displacement of a particle is given by q = t3 + t2 + t + 1, then its angular velocity at t = 2 s is ______ rad s–1 .
15. A fly wheel is rotating about its own axis at an angular velocity 11 rad s–1. Its angular velocity in revolutions per minute is ___.
16. A stationary wheel starts rotating about its own axis at uniform angular acceleration of 8 rad/s2. The time taken by it to complete 77 rotations is ____ s.
17. A car is moving with a speed of 30 ms –1 on a circular path of radius 500 m. If its speed is increasing at the rate of 2 ms –2, the net acceleration of the car is ________ ms –2
181135 .
18. A body of mass m is made to revolve along a circle of radius r in a horizontal plane at a speed V m/s, with the help of a light horizontal string, such that the tension in the string is 10 N. Keeping speed constant, if mass is increased to 2 m and radius is decreased to r/2, then tension in the string is ___ N.
19. An object of mass 10 kg is whirled around a horizontal circle of radius 4 m by a revolving string inclined 30° to vertical. If the uniform speed of the object is 5 m/s, the tension in the string (approximately) is ___ N.
Integer Value Questions
20. Two point size dense particles of same mass, knotted to a single massless string at different points, are whirled along concentric circles in a horizontal plane. The ratio of distances of particles from centre of circles is 1 : 2. If the tension in the string between two particles is 4 N, the tension in the remaining string is___ N.
21. In a circular motion of a particle, the tangential acceleration of the particle is given by a t = 2t m/s2. The radius of the circle described is 4 m. The particle is initially at rest. Time after which total acceleration of the particle makes 45° with radial acceleration is ___ s.
22. A block of mass ‘m’ is kept on a horizontal ruler. The friction coefficient between the ruler and the block is ‘μ’. The ruler is fixed at one end and the block is at a distance (L) from the fixed end. The ruler is rotated about the fixed end in the horizontal plane. Angular
speed of the ruler is uniformly increased from zero at an angular acceleration ‘α’ and, when the angular speed becomes
, the block will slip.
Then, n = ___.
23. A small wedge, whose base is horizontal, is fixed to a vertical rod, as shown in the figure. The sloping side of the wedge is frictionless and the wedge is spun with constant angular velocity ω about vertical axis, as shown in figure. The value of angular speed ω, for which the block of mass m just does not slide down the wedge, is xg l sin cos 2 . Then, the value of x is ____________.
24. A solid body starts rotating about a stationary axis with an angular acceleration β = at , where a = 2 × 10 –2 rad/s 2 . How soon from beginning of rotation will the total acceleration vector of an arbitrary point of the body form an angle α = 60° with its velocity vector? (Round off to nearest integer)
25. A small body of mass m can slide without friction along a bent trough, which is in the form of a semicircular arc of radius ( R ). At height h , the body will be at rest with respect to the trough, if the trough rotates with uniform angular velocity ω about a vertical axis. If hRg n , the value of n is ______.
26. The system shown in figure is rotated in a horizontal circle with angular velocity ‘ω’.
Passage-based Questions
Passage I: A particle undergoes uniform circular motion. The velocity and angular velocity of the particle at an instant of time are vijx ij346 m/sandrad/s .
28. The value of x (in rad/s) is (1) 8
(2) –8
(3) 6
(4) can’t be calculated
29. The radius of circle (in metres) is
(1) 1/2 m
(2) 1 m
(3) 2 m
(4) can’t be calculated
Tension in the string connecting m1 and m2, when slipping just starts between the blocks, is 6 × N. The coefficient of friction between the two masses is 0.5, and there is no friction between m2 and the ground. Dimension of masses can be neglected. Take R = 0.5 m, m 1 = 2 kg, m 2 = 1 kg. Find the value of x (g = 10 ms–2)
27. The acceleration of body A, which slides without initial velocity down a helical groove with a pitch h and a radius R, at the end of the
nth turn is aghhRxnR hR 2222422 222 4 4 .
Find the value of x. (Disregard friction) A R h
30. The acceleration of the particle at the given instant is
(1) 50k
(2) 42k
(3) 23ij + (4) can’t be calculated
Passage II: A smooth hemispherical bowl of radius R = 0.1 m is rotating about its own axis (which is vertical), with an angular velocity ω. A particle of mass 10–2 kg on the inner surface of the bowl, at an angled position θ (with vertical), is also rotating with same velocity ω. The particle is at a height h from the bottom of the bowl. [Take g = 9.8 ms–2]
31. The value of θ in terms of R and h is
(1) sin 1 Rh R (2) cos 1 Rh R
(3) tan 1 Rh R (4) sec 1 Rh R
32. The relation between h and ω is
(1) hRg 2 (2) h = R – g w 2
(3) Rhg 2 (4) R = h – g w 2
33. What is the minimum value of ω needed for non-zero value of h?
(1) 7 rad/s (2) 7 2 rad/s
(3) 2 7 rad/s (4) 72rad/s
Matrix Matching Questions
34. A particle is moving in a uniform circular motion on a horizontal surface. Particle position and velocity at time t = 0 are in shown in the figure in the coordinate system. Which of the indicated variable in column-I is/are correctly matched by the variables on the vertical axis in the graphs shown in column-II, for the particle’s motion?
vatt=0 y x
Column-I
Column-II
A. x-component of velocity I) t
B. y-component of force keeping particle moving in a circle II) t
C. Angular velocity of the particle III) t
D. x-coordinate of the particle IV) t
(A) (B) (C) (D)
(1) II II III IV
(2) II II III III
(3) I II III IV
(4) II II I IV
35. A particle is moving with speed V = 2 t 2 on the circumference of a circle of radius (R). Match the questions in Column-I with corresponding results in Column-II.
Column-I Column-II
A. Magnitude of tangential acceleration of particle I) Decrease with time
B. Magnitude of centripetal acceleration of particle II) Increase with time
C. Magnitude of angular speed of particle with respect to centre of circle III) Remains constant
D. Value of tan θ when ‘θ’ is angle between the total acceleration vector and centripetal acceleration vector of particle IV) Proportional to R V) Inversely proportional to R
(A) (B) (C) (D)
(1) II,V I,IV I,IV II,V
(2) II I,IV II,V II,V
(3) II II,V II,V I,IV
(4) II II,V I,IV II,V
FLASHBACK (P revious JEE Q uestions )
JEE Main
1. A ball of mass 0.5 kg is attached to a string of length 50 cm. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is 400 N. The maximum possible value of angular velocity of the ball in rad/s is,: (2024)
(1) 1600 (2) 40
(3) 1000 (4) 20
2. A train is moving with a speed of 12 m/s on rails which are 1.5 m apart. To negotiate a curve radius 400 m, the height by which the outer rail should be raised with respect to the inner rail is (Given, g = 10 m/s2): (2024)
(1) 6.0 cm (2) 5.4 cm
(3) 4.8 cm (4) 4.2 cm
3. If the radius of curvature of the path of two particles of same mass are in the ratio 3:4, then in order to have constant centripetal force, their velocities will be in the ratio of: (2024)
(1) 3:2
(2) 1:3
(3) 3:1
(4) 2:3
4. A stone of mass 900g is tied to a string and moved in a vertical circle of radius 1m making 10 rpm. The tension in the string, when the stone is at the lowest point is (if π2 = 9.8 and g = 9.8 m/s2) (2024)
(1) 97 N (2) 9.8 N (3) 8.82 N (4) 17.8 N
5. A coin is placed on a disc. The coefficient of friction between the coin and the disc is μ. If the distance of the coin from the center of the disc is r, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is: (2024)
(1) g r µ (2) r gµ
(3) g r µ (4) rg µ
6. A car is moving with a constant speed of 20 m/s in a circular horizontal track of radius 40 m. A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be (Take g = 10 m/s2) (2023)
(1) π/6 (2) π/2 (3) π/4 (4) π/3
7. A car is moving on a horizontal curved road with radius of 50 m. The approximate maximum speed of car, if friction between tyres and road is 0.34, will be [take g = 10 ms–2] (2023) (1) 17 ms–1 (2) 3.4 ms–1 (3) 22.4 ms–1 (4) 13 ms–1
8. An object moves at a constant speed along a circular path in a horizontal plane with centre at the origin. When the object is at x = + 2 m, its velocity is 4 j m/s . The object’s velocity (v) and acceleration (a) at x = – 2 m will be (2023)
(1) vms,ams ==482 ji //
(2) vms,ams 482 ji //
(3) vms,ams 482 ij //
(4) vms,ams ==482 ij //
9. A stone of mass 1 kg is tied to one end of a massless string of length 1 m. If the breaking tension of the string is 400 N, then maximum linear velocity the stone can have without breaking the string, while rotating in horizontal plane, is (2023) (1) 40 ms–1 (2) 20 ms–1 (3) 400 ms–1 (4) 10 ms–1
10. A small block of mass 100 g is tied to a spring of spring constant 7.5 N/m and length 20 cm. The other end of spring is fixed at a particular point A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity 5 rad/s about point A, then tension in the spring is (2023)
(1) 0.25 N (2) 0.75 N (3) 1.5 N (4) 0.50 N
11. A child of mass 5 kg is going round a merrygo-round that makes 1 rotation in 3.14 s. The radius of the merry-go-round is 2 m. The centrifugal force on the child will be (2023)
(1) 40 N (2) 80 N (3) 50 N (4) 100 N
12. A car is moving on a circular path of radius 600 m such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete first quarter of revolution, if it is moving with an initial speed of 54 km/h, is t(1 – e–π/2) s. Then, the value of t is ____.
(2023)
13. A stone tied to 180 cm long string at its end is making 28 revolutions in horizontal circle in every minute. If the magnitude of acceleration of stone is 1936/ x ms –2, then the value of x is _______. (Take π = 22/7) (2023)
14. A smooth circular groove has a smooth vertical wall, as shown in figure. A block of mass m moves against the wall with a speed v. Which of the following curves represents the correct relation betwee n the normal reaction on the block by the wall ( N ) and speed of the block (v)? (2022)
(1)
(4) N
15. A ball is spun with angular acceleration a = 6 t 2 – 2 t, where t is in seconds and α is in rads –2 . At t = 0, the ball has angular velocity of 10 rads –1 and angular position of 4 rad. The most appropriate expression for the angular position of the ball is (2022)
(1) 3 2 4210 ttt
(2) tt t 43 23 104 (3) 2 36 1012 43tt t (4) 2 2 454 3 t t t
16. One end of a massless spring of spring constant k and natural length l0 is fixed while the other end is connected to a small object of mass m, lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity ω about an axis passing through
fixed end, then the elongation of the spring will be (2022)
(1) kml m 0 0 2 (2) ml km 2 0 2
(3) ml km 2 0 2 (4) kml m 2 0 2
17. A fly wheel is accelerated uniformly from rest and rotates through 5 rad in the first second. The angle rotated by the fly wheel in the next second will be (2022)
(1) 7.5 rad (2) 15 rad
(3) 20 rad (4) 30 rad
18. A boy ties a stone of mass 100 g to the end of a 2 m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80 N. If the maximum speed with which the stone can revolve is K π rev/min , then the value of K is (Assume the string is massless and unstretchable) (2022)
(1) 400 (2) 300
(3) 600 (4) 800
19. A person moved from A to B on a circular path, as shown in figure. If the distance travelled by him is 60 m, then the magnitude of displacement would be (Given, cos 135° = – 0.7) (2022)
(1) 42 m (2) 47 m (3) 19 m (4) 40 m
20. A disc with a flat small bottom beaker placed on it at a distance R from its centre is revolving about an axis passing through the centre and perpendicular to its plane with an angular velocity ω. The coefficient of static friction between the bottom of the beaker and the surface of the disc is μ. The
beaker will revolve with the disc if (2022)
(1) R g 22 (2) R g 2
(3) R g 22 (4) R g 2
21. For a particle in uniform circular motion, the acceleration a at any point P(R, θ) on the circular path of radius R is (when θ is measured from the positive x-axis and v is uniform speed) (2022)
(1) v R i v Rj 22 sin cos
(2) v R i v Rj 22 cossin
(3) v R i v Rj 22 cossin
(4) v R i v Rj 22
22. A tube of length 50 cm is filled completely with an incompressible liquid of mass 250 g and closed at both ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity xF rads–1. If F is the force exerted by the liquid at the other end, then the value of x will be ______. (2022)
JEE Advanced
23. At time t = 0, a disc of radius 1 m starts to roll without slipping on a horizontal plane with an angular acceleration of 2 3 rads–2 A small stone is stuck to the disc. At t = 0, it is at the contact point of the disc and the plane. Later, at time t s , the stone detaches itself and flies off tangentially from the disc. The maximum height (in m) reached by the stone measured from the plane is 1 210 + x . The value of x is _____. [Take g = 10 ms –2] (2022)
CHAPTER 6: Circular Motion
CHAPTER TEST – JEE MAIN
Section–A
1. A motorcyclist takes a U-turn in 4 s. His angular velocity will be _____rad s –1
(1) 0.7855 (2) 0.7955
(3) 0.8955 (4) 0.9955
2. A stationary wheel starts rotating about its own axis at constant angular acceleration. If the wheel completes 50 rotations in first 2 seconds, then the number of rotations made by it in next two seconds is (1) 75 (2) 100
(3) 125 (4) 150
3. Three particles describe circular path of radii r1, r2, and r3 with constant speed, such that all the particles take the same time to complete the revolution. If w 1, w 2, and w 3 are the angular velocities v1, v2, and v3 are the linear velocities, and a 1, a 2, and a 3 are the linear accelerations, then which of the following is not correct?
(1) w 1: w 2 : w 3 = 1 : 1 : 1
(2) v1: v2 : v3 = r1: r2 : r3
(3) a1: a2 : a3 = 1 : 1 : 1
(4) a1: a2 : a3 = r1: r2 : r3
4. A ceiling fan is rotating about its own axis with uniform angular velocity ω. The electric current is switched off. Then, due to constant opposing torque, its angular velocity is reduced to 2ω/3, as it completes 30 rotations. The numb er of rotations it further makes before coming to rest is
(1) 18 (2) 12
(3) 9 (4) 24
5. A particle in a rotating disc covers an angle q = 2t – t3. Then,
(1) average angular velocity during first
second is 2 rad/s
(2) average angular acceleration during first second is –4 rad/s2
(3) angular acceleration at the end of first second is –6 rad/s2
(4) average angular acceleration during first second is –6 rad/s2
6. In uniform circular motion, assuming V = velocity, r radius vector, angular velocity relative to centre of the circle, and a = acceleration, which of the following is incorrect?
(1) vv00 but
(2) 00 but
(3) rr00 but
(4) aa00 but
7. A wheel is subjected to uniform angular acceleration about its axis. Initially, its angular velocity is zero. In the first 2 s, it rotates through an angle ‘θ 1’. In the next 2 s, it rotates through an additional angle ‘θ2’. The ratio of q q 2 1 is
(1) 1 (2) 2
(3) 3 (4) 5
8. The linear velocity of a point on the surface of earth at a latitude of 60° is
(1) 800 3 m/s (2) 800 3 π m/s
(3) 800 5 18 × m/s (4) 2000 27 π m/s
9. A merry-go-round has a radius of 4 m and
completes a revolution in 2 seconds. Then, acceleration of a point on its rim is________ rad/s2.
(1) 4 π 2 (2) 2 π 2
(3) π 2 (4) zero
10. A car is travelling with linear velocity (V) on a circular road of radius (r). If it is increasing its speed at the rate of a (m/s 2 ), then the resultant acceleration will be
(1)
(4)
11. Two point-sized bodies of the same mass are knotted to a horizontal string, one at the end, and the other at the midpoint of it. The string is rotated in a horizontal plane with the other end as centre. If T is tension in the string between centre of circle and first body, then the tension in the string between the two bodies is
(1) T/2 (2) 2T (3) 2T/3 (4) 3T/2
12. A body of mass m is tied to one end of a spring and whirled a round in a horizontal plane with constant angular velocity. Elongation in the spring is 1 cm. If the angular velocity is doubled, the elongation in the spring becomes 5 cm. The original length of spring is
(1) 13 cm (2) 14 cm
(3) 15 cm (4) 16 cm
13. Four point-sized steel spheres, each of mass 1 kg, are placed on a turn-table and are connected by 4 strings, each of length a, to form a square. If the spheres are rotated with an angular velocity of 1 2 π rev/s , the tension in the connecting strings (in newtons) is
(1) a (2) a 2 (3) a 2 (4) 2a
14. A circular disc is made to rotate in a horizontal plane about its centre at the rate of 2 rps. The greatest distance of a coin placed on the disc from its centre, so that it does not skid, is (μ is coefficient of friction)
(1) µ 62 m (2) 6.2 cm
(3) µ 62 . cm (4) µgcm 62
15. A constant power is supplied to a rotating disc. The relationship between the angular velocity (ω) of the disc and the number of rotations (n) made by the disc is governed by (1) w∝ n1/3 (2) w∝ n2/3 (3) w∝ n3/2 (4) w∝ n2
16. A cyclist moves around a circular path of radius 3923 metres with a speed of 19.6 ms –1 . He must lean inwards at an angle θ with the vertical, such that tan θ is (1) 1 (2) 3 (3) 1 3 (4) 2
17. A uniform metal rod of length L and mass M is rotating about an axis passing through one of the ends perpendicular to the rod with angular speed ω. If the temperature increases by t °C, then the change in its angular velocity is proportional to which of the following? (Coefficient of linear expansion of rod = α)
(1) w (2) w
(3) w 2 (4) 1 w
18. A vehicle is moving on a banked road, θ is angle of banking, and μ is coefficient of friction between tyre of vehicle and road. The vehicle is moving along a horizontal
circular path of radius (r), and velocity of vehicle for which it does not slip is v. Choose the correct option.
(1) v should be between 0 and µgr .
(2) v should be between 0 and rgtan ta 1n
(3) V should lie between rgtan ta 1n
(4) V should lie between rgtan ta 1n and rgtan ta 1n
19. Three point masses, each of mass m , are joined together using a string to form an equilateral triangle of side a. The system is placed on a smooth horizontal surface and rotated with a constant angular velocity ω about a vertical axis passing through the centroid. Then, the tension in each string is
(1) ma w 2 (2) 3ma w 2
(3) maw 2 3 (4) maw 2 3
20. A chain of 100 links is 1 m long and has a mass of 2 kg. With the ends fastened together, it is set rotating at 3000 rpm, in a horizontal plane. The centripetal force on each link is
(1) 3.14 N
(2) 31.4 N
(3) 314 N
(4) 3140 N
Section–B
21. A boy is sitting on a horizontal platform in the shape of a disc, at a distance of 5 m from
CHAPTER 6: Circular Motion
its centre. The boy begins to slip when the speed of the wheel exceeds 10 rpm. If the coefficient of friction between the boy and platform is π 2 x , then the value of x is ___.
(g = 10 ms–2)
22. A particle is resting on an inverted cone, as shown. It is attached to the cone by a thread of length 20 cm. String remains parallel to slope of cone. If the cone is given angular acceleration of 0.5 rad/s 2 , then at what time (in s) does mass loose contact with the surface? (Assuming sufficient friction) (g = 10 ms–2)
m=20 gm
23. A point-sized body is moving along a circle at an angular velocity of 2.8 rad s–1. If centripetal acceleration of body is 7 ms–2, then its speed is ____ ms –1
24. A sphere of mass 300 grams is rotated at a uniform angular velocity of 2 rad s–1 along a circle of radius 1.25 m in a horizontal plane with the help of a light horizontal string. The tension in the string is ___N.
25. A wheel having radius 10 cm is coupled by a belt to another wheel of radius 30 cm. The first wheel increases its angular speed from rest at a uniform rate of 1.57 rad s–2 . The time for the second wheel to reach a rotational speed of 100 rev/min is ____ s. (Assume that the belt does not slip) [Take π = 3.14]
CHAPTER TEST – JEE ADVANCED
2022 P2 Model
Section-A
[Integer Value Questions]
1. The small mass m and its supporting wire become a simple pendulum when the horizontal cord is cut. The ratio of tension in the supporting wire immediately after the cord is cut to the tension in the wire before the cord is cut is equal to 1/ x. The value of x is ______.
600 wire cord
2. Two particles having masses M and m are moving in circular paths having radii R and r . If their time periods are same, then the ratio of their angular velocities will be ___.
3. A toy car of mass m can travel at a fixed speed. It moves in a circle on a fixed horizontal table. A string is connected to the car and attached to a block of mass M that hangs, as shown in fig. The portion of string below the table is always vertical. The coefficient of friction between the surface of the table and the tyres of the toy car is μ. The ratio of maximum radius to minimum radius, for which the toy car can move in a circular path with centre ‘ O’ on table, is r r Mxm Mm max min . Then, the value of x is___________. m M
4. Water of density ‘ρ’ flows with a linear speed ν through a horizontal tube in the form of a ring of radius R. If the diameter of the tube is d << R. If the tension in the rubber tube is T n vd 122 , then the value of n is _____.
R
5. A metal ring of mass m and radius R is placed on a smooth horizontal table and is set rotating about its own axis, such that the ring moves with constant speed. If the tension in the ring is mv nR 2 π , then the value of n is ___.
6. A vehicle whose wheel track is 1.7 m wide and whose centre of gravity is 1 m above the road and central between the wheels, taken a curve whose radius 50 m, on a level road. If the speed at which the inner wheel would leave the road is close to n × 5 m/s, then the value of n is ______. (Round off to nearest integer)
7. A particle is moving with constant speed in a circle, as shown. If the angular velocity of A with respect to O is ω, then the angular velocity of particle A with respect to C at the given instant is x 3 , where x = ___.
8. The ring that can slide along the rod is kept at mid-point of a smooth rod of length L. The rod is rotated with constant angular velocity ω about vertical axis passing through its one end. The ring is released from mid-point. If the velocity of the ring, when it just leaves the rod, is x L 2 , then x is equal to_____.
Section – B
[Multiple Option Correct MCQs]
9. A rod of length 1 m rotates about the z-axis, passing through the point O in the xy-plane, with an angular velocity of w = 10(rad/s) + 5(rad/s2)t in the counter-clockwise direction, and O is at rest. The velocity and acceleration of point A, at t = 0, are
(1) g 2 (2) 3 2 g
(3) g 2 (4) g 5
11. A particle is moving in a circular path, whose position vector is given by ratiatj cossin . Then, the tangential component of acceleration at t 2 s is not equal to
(1) 0 (2) a m/s2 (3) a/2 m/s2 (4) a/4 m/s2
(1) vj 10 ^ m/s
(2) vj = 10 ^ m/s
(3) aji () ^^ 5100m/s 2
(4) aji () ^^ 5100m/s 2
10. A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the origin, as shown in figure and whose equation is x2 = 4ay. The wire frame is fixed, and the bead can slide on it without friction. The bead is released from the point y = 4a on the wire frame from rest. The tangential acceleration of the bead, when it reaches the position given by y = a, is not equal to
12. A long horizontal rod has a bead that can slide along its length and is initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A, with a constant angular acceleration α. If the coefficient of friction between the rod and bead is μ, and gravity is neglected, then the time after which the bead starts slipping may not be:
(1) (2)
(3) 1 (4) infinitesimal
13. The kinetic energy k of a particle moving along circle of radius R depends on the distance covered s as k = as 2 , where a is a positive constant. FT, FC, and total force acting on the particle are (1) Fa s CR = 2 2
(2) Fas s R 21 2 2 1 2
(3) FT = 2as
(4) Fa R s C = 2 2
14. A heavy particle is tied to the end A of a string of length 1.6 m. Its other end O is fixed. It revolves as a conical pendulum with the string making 60° with the vertical. Then, (Take g = 9.8 m/s2)
(1) its period of revolution is 4 7 π s
(2) the tension in the string is double the weight of the particle
(3) the speed of the particle = 283.m/s
(4) the centripetal acceleration of the particle is 9832 .m/s
Section – C
[Single Option Correct MCQs]
15. A particle moving along the plane trajectory has the form of parabola y = ax2, where a is
ANSWER KEY JEE Main
a positive constant. Radius of curvature at x = 0 is
(1) 1 (2) 2a
(3) 2a2 (4) 1 2a
16. A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A bob is suspended from the roof of the car by a light wire of length 1.0 m. The angle made by the wire with the vertical is
(1) 0° (2) π 3
(3) π 6 (4) π 4
17. A motorist moves along a circular track at 144 km h–1. The angle he should make with the vertical, if the track is 880 m long, is (g = 10 ms–2)
(1) 45° (2) tan 18 7
(3) tan 17 8 (4) tan 116 7
18. What is the smallest radius of a circle at which a bicyclist can travel, if his speed is 7 m/s and = 0.25 between road and tyres? (1) 10 m (2) 5 m (3) 15 m (4) 20 m
■ W ork is done when a body undergoes displacement parallel to the line of action of the force.
■ Work done = magnitude of displacement × component of force along the displacement, or magnitude of force × component of displacement along the force.
7.1.1 Work Done by a Constant Force
■ Work done WFSFSFS coscos Similarly, as shown in fig (2), the component of displacement along force is S cos q Hence WFSFSFS coscos
B S F x Fy=Fsin q F x =Fsin qq y Fig. 1
Fig. 2
Key Insights:
■ When body gets displaced along the direction of force, then work done is equal to the product of force and displacement. W = FS
■ If FFiFjFk xyz and SSiSjSk xyz then WFSFSFSFS xxyyzz
■ If the force FFiFjFk xyz displaces the object from the point ( x1, y1, z1) to (x2, y2, z2) then work, WFxxFyy xy2121 Fzz z 21
Work is a Scalar Quantity
■ Dimensions and units of work: The dimensional formula of work is [ML2T–2]. Its unit in SI system is joule and is denoted by J.
■ Joule: Work done is said to be one ‘joule’ if a force of 1 newton displaces a body through
CHAPTER 7: Work, Energy, and Power
a distance of 1 m along the direction of force.
1 J = 1 N × 1 m
In CGS system, the unit of work is erg.
■ erg: Work done is said to be one ‘erg’ if a force of one dyne displaces the body through a distance of 1 cm along the direction of force.
1 erg = 1 dyne × 1 cm
Relation between joule and erg
■ 1 J = 1 N × m = 105 dyne × 100 cm =107 dyne cm
1 J = 107 erg
7.1.2 Nature of Work
■ Work done by a force may be positive, negative or zero.
Positive work
■ Positive work means that force (or one of its components) is parallel to displacement. If ‘ q ’ is the angle between force and displacement, then 0° < q < 90°.
Examples:
■ When a person lifts a body from the ground, the work done by the lifting force (upward) is positive.
Negative Work
■ Negative work means that force (or one of its components) is opposite to displacement. If ‘ q ’ is the angle between force and displacement, then 90° < q < 180°.
Examples:
■ When a person lifts a body from the ground, the work done by the force of gravity (downward) is negative.
■ When a body is made to slide over a rough surface, the work done by the frictional force is negative.
Zero Work
■ Under three conditions, work done becomes zero.
■ The condition are listed below:
If the force is perpendicular to the displacement ( q = 900)
If there is no displacement [S = 0]
If resultant force acting on the body becomes zero. (F = 0).
■ work done can be calculated from force–displacement graph. The area enclosed by the graph on displacement axis gives the amount of work done by the force.
■ Force–displacement graph of a constant force is a straight line PQ, as shown below. F O Q R S P
Work = Area OPQR = OP × OR = (F)(S)
7.1.3
The Work Done by a Force in Various Situations
■ A body of mass ‘m’ is lifted from ground to a height ‘h’ by using a minimum force F For minimum force F = mg
■ The minimum work done by the lifting force is WFhFhm Fgh o === .cos0 F mg m h
Work done by the gravitational force is WFhFghmgh gg o cos. 180
■ One end of a string is fixed to a support and a body of mass ‘m’ is attached to the other end of the string.
■ If the point of support moves with acceleration ‘a ’ in the upward direction, work done by tension in the string when the body moves upward through a distance ‘h’ is WThThmgah
Tmgma o ∵ cos0 T a mg
■ If the body moves downward with acceleration ‘ a ’ then work done by the tension in the string is W= ∵ ThThmgah mgTma o cos 180
Here, work done is negative as force and displacement are opposite to each other.
■ A body of mass ‘m’ is placed on a rough horizontal surface. A force F acts on the body parallel to the surface.
■ If the body moves with uniform velocity, the work done on the body is,
WFSF F
WFSFSF netnet F ∵ 00 0
Workdoneby, cos S f WfSfS mgS k fkk k
Workdoneby Here,|W , cos | 180 |W f |
F
In the above case, if the body moves with uniform acceleration a work done by the force is
■ A pendulum of length ‘ L’ carries a bob of mass ‘m’ . The bob is slowly pulled aside through an angle with the vertical by a horizontal force F. mg B F y M A y x
Displacement, rABAMMB
LjLi 1cossin
Applied force ==FFai ;
Gravitational force = Fmggj
Work done by the gravitational force
WFrmgggL .c1os
Work done by the horizontal force F is WFrF FaL sin
■ A ladder of mass ‘m’ and length ‘ℓ’ resting on a level floor is slowly lifted and held against a wall at an angle ‘q’ with the floor.
l/2 h G q
■ Work done by the gravitational force is W g = –mgh
■ Where ‘ h ’ is the height gained by centre of gravity.
hl 2 sin
Since,sin h l 2
Wmg l g -sin 2
■ A uniform chain of mass ‘ m ’ and length ‘ℓ’ is suspended vertically. The lower end of the chain is slowly lifted upto the point of suspension.
Work done by gravitational force is
W g = –mgh
■ Where ‘ h ’ is the height gained by center of gravity of the lower half of chain, with respect to its initial position G.
h = ℓ/4+ℓ/4=ℓ/2
■ Work done by the gravitational force is
W g = m g 22
Wmg g 4
■ A uniform chain of mass ‘ m ’ and length
‘L’ rests on a table, having 1 n th part of its length hanging down from the edge of table. The work done by the pulling force to slowly bring the hanging part of chain on to the table is
W = m n gh , where ‘h’ is the height gained by CG of hanging chain with respect to the
CHAPTER 7: Work, Energy, and Power
edge of table. F G h=L/2n
W = m n g L n mgL n 222 =
■ A body of mass ‘m’ is on a smooth inclined plane of inclination ‘ q ’ to horizontal. If ‘l’ is the length of incline, q mgsin q mg N mgcos q
Work done by gravitational force in pulling the block down from top to bottom of the incline is W=mgsin g q
Here, work done by normal reaction is zero.
■ A body of mass ‘m’ is slowly moved up the smooth inclined plane of inclination q and length ℓ by a constant horizontal force F.
Fcos q q q q Fsin q N F mgcos q mgsinq mg
Work done by horizontal force, WF = + Fcos q × ℓ
Work done by gravitational force, W g = –mgsin q × ℓ
Net work done by the resultant force on
the body = W g + WF = ( Fcos q – mgsin q )ℓ
■ A body of mass ‘ m ’ is sliding down a rough inclined plane of inclination q to horizontal. If ‘ ℓ ’ is the length of incline and µK is the coefficient of kinetic friction
Work done by gravitational force in pulling the block down the incline is
W g = m(g sin q ) ℓ.
Work done by frictional force is
Wf = – fkℓ = –µkm(g cos q ) ℓ. q mg fk mgcosq mgsinq N
∴ Net work done by the resultant force on the body, W = FR.ℓ
W=(mg sin q – µk mg cos q )ℓ
■ Consider two blocks of mass m 1 and m 2 ( m 1 < m 2) connected by an inextensible string passing over a smooth frictionless pulley. Two blocks are released from same level. At any instant ‘t’, ‘x’ is the respective displacement of blocks.
Work done by gravitational force on block m1, W1= –m1gx
Work done by gravitational force on block m, W2= + m2gx
Work done by gravitational force,
W g = –m1gx + m2 gx
⇒ W g = (m2 – m1)g 1 2 2 at
a mmg mm () 21 12
Solved example
1. A force of 10 N acts on a body of mass 1.0 kg at rest. Find the work done in 4 seconds.
Sol. F = 10 N; m = 1.0 kg ; u = 0, t = 4 s;
acceleration, a = F m = 10 1 = 10 ms–2
displacement, S = ut + 1 2 at2
S = 0 + 1 2 (10) 42 = 80 m
W mmgt gmm () () 21 222 212 x x m1 m2
work done, W = FS = maS = 1.0 × 10 × 80 = 800 J
Try yourself:
1. A massive box is dragged along a horizontal floor by a rope. The rope makes an angle of 600 with the horizontal. Find the work done if the tension in the rope is 200 N and the box is moved through a distance of 20 m.
dwFdxF idxi
7.1.4 Work Done by a Variable Force
Ans: 2000 J
dwFdx .. Areaoftheshadedstrip.
Tot talworkdone=W= AreaunderF -graph. dw Fdxx a 0
In terms of rectangular components
FFiFjFk
drdx idyjdzk
WFiFjFkdx xyz xyz ().( i idyjdzk
WFdxFdyFdz A B xy A B Az B A B )
Solved example
2. A particle is displaced by 2m from point (0, 0) along the straight line x – y = 0 under the influence of a force Fy ixjN . Find the work done by the applied force.
y O A(1,1) x
Sol. Displacement, OA = 2m
Equation of straight line, x – y = 0
∴ x = y = dx = dy.
Position vector of the particle at any time is rx iyj
drdx idyj
dwFdry ixjdxidyj
CHAPTER 7: Work, Energy, and Power
2. A force () N ˆˆ 437 ˆ =−− Fijk displaces a particle from point P(0,1,4) to point Q(2,3,2).
Then the work done by the force is (1) 16 N (2) 14 N (3) 12 N (4) 10 N
3. A long rod ABC of mass m and length L has two particles of masses m and 2m attached to it as shown in the figure. The system is initially in the horizontal position. The work to be done to keep it vertical with A at the bottom is (g = acceleration due to gravity)
L/2 2 mm L/2 C B A
(1) 2mgL (2) 3mgL (3) 5mgL/2 (4) 3mgL/2
Answer Key
(1) 2 (2) 1 (3) 2
ydxxdydwxxdxydyxy
wxdxydy As Jou 0 1 0 1 122 2 10 1 2 101lle
Try yourself:
2. Find the work done by the force Fx23N in moving a particle along x-axis from x = 1 m to x = 2 m.
Ans: 6 J
TEST YOURSELF
1. A person draws water from a 5 m deep well in a bucket of mass 2 kg of capacity 8 litres by a rope of mass 1 kg. The total work done by the person is? (Assume g = 10 m/s 2)
(1) 550 J
(2) 525 J
(3) 125 J
(4) 500 J
7.2 KINETIC ENERGY
■ Mechanical energy is of two types, namely (1) kinetic energy and (2) potential energy. Kinetic energy is the energy possessed by a body by virtue of its motion.
■ It is measured by the amount of work that the body can do against an opposing force acting on it before coming to rest.
Expression for kinetic energy:
Kinetic energy of the body, Kmv = 1 2 2 .
7.2.1 Relation between Kinetic Energy and Linear Momentum
■ When a body of mass m moves with a velocity ‘v’, the relation between its kinetic energy and linear momentum is given by KE 1 222 2 222 mv mv m p m pmv
Also, KE = 1 2 pv..
Solved examples
3. If linear momentum of a body of mass 0.5 kg is pijj3411kgm/s, find its kinetic energy.
Sol. p 34116 222kgm/skgm/s
KEjoulejoule p m 22 2 6 205 36
4. A car and truck have same momentum. Which will have more kinetic energy?
Sol. From the expression, Kp m = 2 2 , if ‘p’ is constant, ‘K’ is inversely proportional to the mass of the body. Hence, the kinetic energy of the car is more than that of the truck.
Try yourself:
3. A man standing on the edge of a roof of a 20 m tall building projects a ball of mass 100 g vertically up with a speed of 10 ms–1 and simultaneously throws another ball of same mass vertically down with the same speed. Find the kinetic energy of each ball when they reach the ground ( g = 10 ms–2 )
Ans: 25 J, 25 J
7.2.2 Work Enery Theorem
■ “The work done on a particle by the net force is equal to the change in its kinetic energy.”
■ Consider a one dimensional motion along x-axis of body of mass m, which is acted upon by variable force. If at any instant, the speed of the body is v, then at that instant,
dwFdxmv dv dx idxi
mv dv dx dxmvdv
wmvdv v v i f 1 2 1 2 22 mvmv wKK fi fi
Thisisworkenergytheorem
If the applied force is FFiFjFk xyz , instantaneous velocity of the particle is vvivjvk xyz , and instantaneous
position is rx iyjzk , then
dwFdr
FiFjFkdx id xyyjdzk z . .
dwFdFdFd mvdvmvdvmvdv
wmvdvmv xxyyzz xxyyzz xx v v xi xf y vyy v vzz v xfyfzfxiyi dvmvdv mvvvmvv yi yf zi zf 1 2 1 2 2222222 v zi
Solved example
5. A block of mass 1 kg rests on a horizontal surface. A vertically upward force F = (20 + x) N is applied on the block where x is the displacement of the block in meter. Find the velocity of the block when its displacement is 2 m.
Sol. Net vertically upward force = F net = (20 + x)–10 N = (10 + x)N wFdxdxxd nex tnet 0 2 0 2 0 2 10 22J
By work-energy theorem, 1 2 1222211vv m/s.
Try yourself:
4. A bullet of mass 25 g moving with a velocity of 500 ms–1 enters into a wooden block and comes out of it with a velocity of 100 ms –1 . Find the work done by the bullet while passing through the wooden block.
Ans: +3000 J
7.2.3
Work Energy Theorem for a System of Interacting Particles
■ For a system of particles interacting through gravitational, electromagnetic, spring, or frictional forces, the net work done by external and internal forces is equal to the change in kinetic energy of the system.
6. A plank of mass 2 kg rests on a smooth surface. A block of mass 1 kg rests on the plank at end A. The block is projected along the plank with velocity 2 m/s. There is friction between the block and the plank. After some time the plank and the block start moving togather without any relative motion between them. Find the net work done by the frictional force.
Block (1 kg)
V0=2 m/s Plank (2 kg) Sol. Conserving
ww
Try yourself:
5. Two particles of masses 1 kg and 2 kg are released from rest when they are at a sepraration r under the influence of their mutual force of attraction they start moving towards each other. Find the work done
by the force of attraction when velocity of lighter particle is 2 m/s.
Ans: 3 J
7.2.4 Applications of Work Energy Theorem in Various Situations
■ A particle of mass ‘ m’ is thrown vertically up with a speed ‘ u ’. Neglecting the air friction, the work done by gravitational force, as particle reaches maximum height is
■ A pariticle of mass ‘m’ falls freely from a height ‘h’ in air medium onto the ground. If ‘v’ is the velocity with which it reaches the ground, the work done by air friction is Wf and work done by gravitational force W g = mgh WWkmghWm gfv f , 1 2 20 Wmvm fgh 1 2 2
■ A block of mass ‘ m ’ slides down a frictionless smooth incline of inclination ‘ q ’ to the horizontal. If ‘h’ is the height of incline, the velocity with which the body reaches the bottom of incline is WKmghm gvvgh 1 2 202 q u=0 h
■ A body of mass ‘m’ starts from rest from the top of a rough inclined plane of inclination ‘q’ and length ‘l’. The velocity ‘v’ with which it reaches the bottom of incline if µκ is the coefficient of kinetic friction is,
WWK gf
(sin)(cos) mgmgmv k 1 2 20
vgk 2 sincos
■ Bob of pendulum of length ‘l’ is projected horizontally with a speed ‘ u’ at its lowest position then the speed of the bob when it makes an angle q with the lower vertical is,
WKWKK ggfi , thus, mglmvv (c1os)() 1 21 22
vvgl 1 221cos
■ A bullet of mass ‘m’ moving with velocity ‘ v’ stops in a wooden block after penetrating through a distance x. If ‘F’ is the resistance offered by the block to the bullet,
WKK Ffi
fk
or, fmv x k = 1 2 2
■ A pile driver of mass ‘ m’ is dropped from a height ‘h’ above the ground. On reaching the ground it pierces through a distance ‘s’ and then stops finally. If ‘R’ is the resistance offered by ground, then
u1=0
air (soil) S h
ϑ1=ϑ=√ 2g ϑ2=0
WWKK gRfi
mg(h + s)+(–Rs)=0, or, Rmg h s 1
Here, time of penetration is given by impulse equation, Rmgtmgh 02 . t s gh = 2 2
■ A bullet of mass ‘m’ is fired horizontally with a velocity ‘v’ which strikes a wooden block of mass ‘M’ suspended vertically . If the bullet is embedded in the block , the height ‘h’ to which the system would rise is hmv Mmg 2 1 2
■ By the principle of conservation of linear momentum, mv = (M + m)v1, or, v1 = mv/ (M + m), Where V1 is the common velocity of (bullet + block) system.
m+M h V1 m+M V m
By work energy theorem W g = Kf - Ki MmghMmv 0 1 21 2
∴ hmv Mmg 2 1 2
■ A body of mass ‘ m ’ is initially at rest. By the application of a constant force , its velocity changes to v 0 in time t 0. The kinetic energy of the body at any time ‘t’ is Kmv t ft 1 20 2 0 2 W = ∆ K, or, W
TEST YOURSELF
1. A force of 10 N acts on a body at rest for 5 s and then force of 8 N acts in opposite direction to the earlier for 4 s. If the kinetic energy gained by the body is 18 J, then mass of the body is
(1) 2 kg (2) 18 kg (3) 9 kg (4) 8 kg
2. A force of 1 N acts on a 1 kg mass at rest for 1 s. In another case 1 N force acts on 1 kg mass at rest and moves it through 1 m. The ratio of kinetic energies in the two cases is (1) 1 : 1 (2) 1 : 2
(3) 1 : 4 (4) 1:2
Solved example
7. Starting from origin a particle of mass 1 kg moves along x -axis under the action of a force and at time = t, its position is given by x = (2t2 –8t) m. Find the work done by the force when the particle momentarily sto ps. Sol. xttv dx
3. Velocity-time graph of a particle of mass 2 kg moving along a straight line path is as shown in the figure. Work done by all forces acting on the particle is v(m/s) t(s) 10 2
(1) 100 J (2) –100 J
(3) –200 J (4) 200 J
4. Position of a particle of mass 30 g varies as x = 3t – 4t2 + t3, where x is in metre and t is in seconds. Work done by the net force acting on the particle during the first 4 seconds is (1) 5.28 J (2) 7.5 J (3) 8.45 J (4) 9.32 J
Answer Key
Try yourself:
6. A block of mass m rests on the surface of earth. A vertically upward constant force F = 3mg starts acting on the block. Find the velocity of the block when its vertical displacement is h.
Ans: 2 gh
(1) 3 (2) 2 (3) 2 (4) 1
7.3 POTENTIAL ENERGY
■ Potential energy is the energy possessed by a body due to its position or configuration.
■ It is the stored energy that can be released as kinetic energy when the body is allowed to move.
( ∴ F = mg; s=h)
F h mg
U = mgh
■ The above expression actually represents the increase in the stored energy from the reference position (earth surface) to the final position at a height ‘h’
Solved example
8. Consider n number of identical cubes each of mass “m” and side “a” lying on a level ground. Find the work done in piling them one above the other. a na 2 2
Sol. The initial total potential energy of all the blocks lying on the ground is Unmg a 12
The potential energy, when the blocks are piled to form a single column, is
Unmg na nmg a 2 2 22 ==
Work done in arranging the blocks one above the other is W = U2 – U1
Wnmg a nmg a 2 22 mg a nn 2 2 or, Wmg a nn 2 1
7.3.1 Conservative and Non–conservative Forces
■ onservative forces (e.g., gravity, elastic forces) are path-independent and conserve mechanical energy.
■ Non-conservative forces (e.g., friction, air resistance) dissipate energy and depend on the path taken by the object.
Conservative Force
■ A force is conservative if the work done around a closed path is zero and pathindependent.
■ Work done by a conservative force is stored as potential energy.
■ Examples: Gravitational force, electrostatic force, and spring force.
■ For a conservative force, change in potential energy = negative of the work done by the force.
conservative force is given by F dU dr
Key Insights:
■ By definition, the negative of the work done by gravitational force is defined as the change in gravitational potential energy of the body.
dUFdr .
dUFdrmgdr U Uhh i f 00 Uf – Ui = mg (h – 0) = mgh.
■ Work done, potential energy and kinetic energy depend on frame of reference selected to observe the motion of the body.
Examples:
In the absence of air resistance, work done by gravity on a vertical trip is Wg = -mgh upwards and Wg = +mgh downwards, with net work = 0 for a round trip.
For lifting a body of mass m to height h, the work done by gravity is the same (W = mgh) regardless of the path. Work depends on initial and final points, not the path.
In the absence of friction, a body is projected up the incline plane, then work done by the gravitational force in moving the body through a distance L is W g = –mg sinq.L and in return journey the work done by the gravitational force is W g = +mg sin q L. On reaching the ground the net work done by the gravitational force in a round trip is zero.
Non–conservative Force
■ A force is non-conservative if the work done around a closed path is not zero and depends on the path.
■ Examples: Frictional force, viscous force.
■ Work done by non-conservative forces is not stored as potential energy. .
Solved example
9. Magnitude of force of attraction between two particles at a separation r is given by FK r = 2 Find the change in potential energy
CHAPTER 7: Work, Energy, and Power
of the system when the separation is increased from R to 2R.
Try yourself:
7. A particle of mass 100 g can move along x -axis under the action of a conservative force and at a distance x measured from the origin, its potential energy is given by U = (2x – 4x2)J. Find the acceleration of the particle after it is released at x = 1 m.
Ans: –60 2m/s
7.3.2 The Potential Energy of A Spring
■ The spring force is an example of a variable force which is conservative. Figure shows a block attached to a spring and resting on a smooth horizontal surface. The other end of the spring is attached to a rigid wall. The spring is light and may be treated as massless. In an ideal spring, the spring force F s is proportional to x, where x is the displacement of the block from the equilibrium position. The displacement could be either positive figure (b) or negative figure (c). This force law for the spring is called Hooke’s law and is mathematically stated as F s = –kx
■ The constant k is called the spring constant. Its unit is Nm-1 . The spring is said to be stiff if k is large and soft if k is small. Suppose that we pull the block outwards as in figure (b). If the extension is xm, the work done by the spring force i s
■ This expression may also be obtained by considering the area of the triangle as in figure (d). Note that the work done by the external pulling force F is positive since it overcomes the spring force. W kx m 2 2
(a)
(b)
F s = 0
= 0
(c) X = 0 F s is negative x is positive F s is positive x is negative
■ The same is true when the spring is compressed with a displacement x c (< 0). The spring force does work W s = - kx c 2/2 while the external force F does work + kx c 2/2. If the block is moved from an initial displacement xi to a final displacement xf, the work done by the spring force W s is
Wkxdx kxkx s x x if i f 22 22
■ Thus, the work done by the spring force depends only on the end points. Specifically, if the block is pulled from xi and allowed to return to xi , then
Wkxdx kxkx s x x ii i i 22 22 0
■ The work done by the spring force in a cyclic process is zero.
■ The spring force is position-dependent (Hooke's Law: Fs = -kx) and the work done depends only on the initial and final positions, making it a conservative force.
■ The potential energy (U(x)) of the spring is defined as zero when the block-spring system is at the equilibrium position.
■ For an extension (or compression) x, the above analysis suggests that Ux Kx () = 2 2
■ You may easily verify that – dU/dx = –kx, the spring force. If the block of mass m in Fig. is extended to x m and released from rest, then its total mechanical energy at any arbitrary point x, where x lies between –x m and + x m will be given by
1 2 1 2 1 2 222 kxkxmv m
■ where we have invoked the conservation of mechanical energy. This suggests that the speed and the kinetic energy will be maximum at the equilibrium position, x = 0,
i.e., 1 2 1 2 22mvkxmm = where v m is the maximum speed. or, v k m x mm =
■ Note the k/m has the dimensions of [ T–2] and our equation is dimensionally correct. The kinetic energy gets converted to potential energy and vice-versa. However, the total mechanical energy remains constant. This is graphically depicted in Fig.
K(KE)
-x m x m
E = K+U V(PE) x 0
Fig. Parabolic plots of the potential energy U and kinetic energy K of a block attached to a spring obeying Hooke’s law.
■ The two plots are complementary one decreasing as the other increases the total mechanical energy E = K + U remains constant.
Application
■ A block of mass ‘m’ attached to a spring of spring constant ‘k’ oscillates on a smooth horizontal table. The other end of the spring is fixed to a wall. It has a speed ‘ V’ when the spring is at natural length. The distance it moves on table before it comes to instantaneous rest is
E=-kx mg x
WS.F + W g + WN = ∆ K
Let the mass be oscillating with amplitude x on compressing the spring.
WkSFx 1 2 2 ; Wg = mgx cos 900 = 0
WN = Nx cos 900= 0, since no work is done by gravity and normal reaction ( N).
WS.F = Kf – Ki
CHAPTER 7: Work, Energy, and Power
Solved example
10. When elongation produced in a spring is x, potential energy stored in it is 4 J, when elongation produced is y, potential energy stored in it is 9 J. What will be the potential energy stored in the spring when elongation produced in it is (x + y)? Sol. 1 2 1 2
Solved example
11. A 2 kg block is connected with two springs of force constants K 1 = 100 N/m and K 2 = 300 N/m as shown in figure. The block is released from rest with the springs unstretched. Find the acceleration of the block in its lowest position ( g = 10 m/s2)
K1 K2
2kg
Sol. Let ‘x’ be the maximum displacement of block downwards. Then, from conservation of mechanical energy: decrease in potential energy of 2 kg block = increase in elastic potential energy of both the springs.
mgxkkx 1 212()2 or, xmg kk 22210 100300 01 12 .m
Acceleration of block in this position is a kkxmg m 12 (upwards)
40001210 2 =10 m/s2 upwards
Try yourself:
8. To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed 18.0 km/h on a smooth road and colliding with a horizontally mounted spring of spring constant 6.25 × 10 3 Nm –1 . What is the maximum compression of the spring?
Ans: 2.00 m
7.3.3 Potential Energy Curve
■ The plot of potential energy of a system as a function of distance [i.e., U(x)] is called potential energy curve. Following is the potential energy curve for a particle moving in one dimension. U(x)
If the particle is moving left, it is accelerated, and its velocity increases.
If moving right, it is retarded and slows down, as the potential energy curve is "uphill."
■ When the slope of the potential energy curve is negative, the force is positive and directed right.
In the F-x graph, force is directed to the right between x1 and x2 and x3 and x4
■ At points where the slope of the potential energy curve is zero, the force is also zero.
In the F-x graph, force is zero at points like x2, x3, x4, and in regions like 0 < x < x1 and x > x5
At these points, the particle is in equilibrium.
■ A particle can have three types of equilibrium.
■ Stable equilibrium: When a particle is displaced from equilibrium, and the force acts towards the equilibrium point to restore it, the particle returns to equilibrium.
In F-x graph, at points like x 2 or x4, if displaced, the force will bring the particle back to equilibrium, making it stable.
■ As conservative force is negative derivative of the potential energy [i.e., F(x) = dU dx ], the corresponding force-displacement graph is obtained by determing slope of U(x) curve at various points, as shown in the above figure.
■ When the slope of the potential energy curve is positive, the force is negative and directed left.
■ Unstable equilibrium: When a particle is displaced, and the force acts away from the equilibrium point, accelerating the particle further, it does not return to equilibrium.
In F-x graph, at x3, if displaced, the force pushes the particle further away, making it unstable.
■ Neutral equilibrium: When a particle is displaced and no force acts on it, it remains at rest or moves with uniform velocity.
In F-x graph, at regions like 0 < x < x1 or x > x5, where force is zero, the particle is in neutral equilibrium.
Solved example
12. The force acting on a particle moving along x-axis varies with the position of the particle shown in the figure then body is in stable equilibrium at
1) x = x1
2) x = x2
3) both x1 and x2
4) neither x1 nor x2
Sol. If particle at x2 is displaced to right, the force on it is negative and is brought back to x2. Similarly if particle is displaced to left, the force on it is positive and is again brought back to x2. Hence the particle is in stable equilibrium at x2
Try yourself:
9. The given plot shows the variation of U, the potential energy of interaction between two particles with the distance r separating them.
a) B and D are equilibrium points
b) C is a point of stable equilibrium
c) The force of interaction between the two particles is attractive between C and D and repulsive between D and E
d) The force of interaction between particles is repulsive between points E and F.
Which of the above statements are correct
1) a and b
2) a and d
3) b and d
4) b and c
CHAPTER 7: Work, Energy, and Power
Ans: If particle at C is displaced to either of the side,s the force will bring it back to C. Hence ‘C’ is a point of stable equilibrium. Force on a particle at D is attractive, it is directed towards C. Similarly force of interaction in the area between C and D is attractive. D and E
Turning Points
■ A plot of potential energy function U(x) of a particle moving in one dimension is shown below. U(x) x E 6 J 5 J 4 J 3 J 2 J 1 J x1 x2 x 3 x 4 x 5
■ In the absence of non conservative force, total mechanical energy of the system is constant i.e. E = K ( x ) + U ( x ). If total mechanical energy which is represented by a straight line, is given as shown in the above diagram then kinetic energy of particle at any position can be found by the equation K(x) = E – U(x).
■ In the region x < x 1 , U ( x ) > E ⇒ K(x) = –ve, which is not possible. Hence particle cannot be present in this region. If particle is moved from x 2 towards x 1 When the particle reaches x 1 , the force on th e particle is positive because slope dU dx is negative. The particle does not remain at x 1 but instead begins to move to right. Hence x1 is a turning point, where KE = 0 and TE = PE = 4 J.
K(x4) = 4 –1 = 3 J and in the region x > x5, U(x) = E
∴ K(x) = 0
TEST YOURSELF
1. 5 J of work is required to stretch a spring through 5 cm beyond its unstretched length. The extra work required to stretch it further by an additional 5 cm shall be
(1) 5 J (2) 10 J
(3) 15 J (4) 25 J
2. In the figure shown, the ball strikes the block and sticks to it. The maximum compression of spring is ___ ( L = Natural length, K = spring constant) m v0k L 2m
(1) 0 2 3 m V K = (2) 0 3 2 m V K
(3) 02 m V K (4) 03 m V K
3 The elastic potential energy of a stretched spring is given by E = 50x2, where x is the displacement in metre and E is in joule, then the force constant of the spring is
(1) 50 Nm (2) 100 Nm-1
(3) 100 N/m2 (4) 100 Nm
Answer Key
(1) 3 (2) 4 (3) 2
7.4 LAW OF CONSERVATION OF MECHANICAL ENERGY
■ “The total mechanical energy of a system is constant if the internal forces are conservative and external forces do no work.”
Mechanical energy, ( E) = PE + KE
E = U + K = constant
■ Under only conservative forces mechanical energy of system is constant.
E2 = E1 ⇒ K2 + U2 = K1 + U1 K + U = constant
∆Κ + ∆U = 0
∆(Κ + U) = 0
■ Mechanical energy, E = constant (or) ∆Ε = 0
■ If kinetic energy of the body increases its potential energy will decrease by an equal amount and vice-versa K K=E U 135o E=U ∆ k ∆p
K + U = const.(c); K = –U + C; y = mx + c
Hence, slope m = tanθ = −1;
Slope = K U KU 1
∴ Gain in KE = loss in PE
Key Insights:
■ Under external and non–conservative forces, mechanical energy of system is not conserved.
■ According to work energy theorem, work done by all forces external, and internal (conservative and non – conservative) forces acting on the body is equal to change in kinetic energy.
W ext + Wci + Wnci = ∆K = K2 – K1
■ We know work done by conservative force is equal to negative of change in potential energy
■ For example, some non–conservative force like friction is also acting on the particle, the total mechanical energy is no more constant, it changes by the amount of work done by the frictional force, i.e, ∆( Κ + U) = ∆Ε = Wf
■ Where Wf is the work done against friction. The lost energy is transformed into heat
■ The heat energy developed is exactly equal to mechanical energy disspated. We can therefore, write ∆ Ε + Q = 0; where Q is heat produced.
7.4.1 Law of Conservation of Energy in Case of Freely Falling Body
■ A body of mass ‘m’ is at a height ‘h’ from the ground. As the body falls freely under gravity its potential energy decreases and kinetic energy increases
CHAPTER 7: Work, Energy, and Power
= mgh + 0 = mgh . . . (1)
At point C: As the body falls freely, consider any point ‘C’ in its path, where AC = x.
PE of the body = mg (h – x) = mgh – mgx
Let v1 be the velocity of the body at ‘C’.
Initial velocity u = 0 (at A)
Velocity at C = v = v1 (Say)
Acceleration a = g
Distance travelled = s = x
Using the formula v2 – u2 = 2 as,v1 2 = 2gx
KE of the body at C
Kmvmgxmg Cx === 1 2 1 2 12 2 .
Total energy at C = PE + KE
= mgh – mgx + mgx = mgh . . . (2)
At point B: The body touches the ground at ‘B’ with a velocity ‘ v2’.
PE at B = mg× 0 = 0
Again, u = 0; v = v2; a = g and s = h
v2 2 - 0 = 2gh
KE atBmvmghmgh 1 2 1 2 22 2
At ‘B’ the total energy of the body = PE + KE = 0 + mgh = mgh............. (3)
Thus, at any point on the path of a freely falling body, the total energy is ‘ mgh’.
Thus, the law of conservation of energy is verified.
At point A: The highest point
PE of the body = mgh
Velocity of the body, ‘ v’ = 0
KE of the body = 1 2 2 mv = 0
Total energy at A = PE + KE
7.4.2
Law of Conservation of Energy in Case of Vertically Projected Body
■ Consider a body of mass ‘m’ is projected vertically upwards from the ground with velocity ‘u’ so that it reaches a maximum height ‘ h ’ from the ground. As the body
moves up against gravity its potential energy increases and kinetic energy decreases. v=0 V1 u B A h x C
At ground B, as the height of the body is zero, the PE of the body = mg× 0 = 0
The KE of the body = 1 2 mu2 TE = PE + KE 0 1 2 1 2 22 mumu
At point C: During its motion, consider a point C, where the velocity of the body is v1.
Let BC = x
At C, the P.E. of the body = mgx
But, vugxvugx 1 22 1 22 22 or,
The KE of the body 1 2 1 2 2 1 12 222 mvmugxmumgx ()
The TE of the body = PE + KE = mgx + 1 2 mu2 –mgx = 1 2 mu2 . . . . (2)
At point A: When the body reaches the maximum heigh ‘h’ above the ground, its final velocity becomes zero.
In the equation, v2 – u2 = 2asv = 0, u = u, a = –g and s = h
Then, 0 – u2 = –2gh hu g = 2 2
The PE of the body = mghmg u g mu 2 2 2 1 2 . . . (3)
Thus the total energy at all points A, B and C remains constant and equal to 1 2 mu2.
Hence the law of conservation of energy is verified in the case of a body projected vertically upwards.
Solved example
13. A uniform chain of length ‘L’ and mass ‘m’ is on a smooth horizontal table, with 1 n th part of its length hanging from the edge of the table. Find the kinetic energy of the chain as it completely slips off the table.
Sol. With respect to the top of the table, the initial potential energy of the chain, U1 = PE of the chain lying on the table + PE of the hanging part of the chain L n mg m n g L n mgL n 1 1 0 222
PE of the chain, when it just slips off the table,
Umg LmgL 222
From law of conservation of energy
∆ K =−∆ U and Kf –Ki = – (Uf – Ui)
KKmgLmgL n if 0 ;222 KmgL n f 2 1 1 2
If ‘v’ is the velocity of the chain, then,
Try yourself:
10. Two blocks are connected by a string as shown in figure. They are released from rest. After they moved a distance L, find their common speed. ( µ is the coefficient of friction)
TEST YOURSELF
1. A body of mass 2 kg is projected vertically up with a velocity of 100 ms–1. If it rises to a height of 400 m, the energy utilized to overcome friction is (g = 10 ms –2)
(1) 10 kJ (2) 8 kJ
(3) 1 kJ (4) 2 kJ
2. A block sliding along a horizontal frictionless surface with a velocity of 2 ms –1 comes to the bottom of a frictionless inclined plane making an angle of 30° with the horizontal and comes to a stop after ascending the inclined plane as indicated. If g = 10 ms –2 , the value of h is stops here 2 m/s h 30°
(1) 5 m (2) 10 m (3) 0.2 m (4) 2 m
3. A body projected obliquely with velocity 19.6 ms–1 has its kinetic energy at the maximum height equal to 3 times its potential energy
CHAPTER 7: Work, Energy, and Power
there. Its position after 1 second of projection from the ground is (h = maximum height)
(1) h/2 (2) h/4 (3) h/3 (4) h
Answer Key
(1) 4 (2) 3 (3) 4
7.5 VERTICAL CIRCULAR MOTION
Velocity of the Body at Any Point on the Vertical Circle
■ Let V q be the velocity of the body when the string makes an angle q with verticle as shown in figure.
■ According to law of conservation of mechanical energy, EA = EP KA + UA = KP + UP KP = KA – (UP – UA) O P p h V1 V2 A r-h r q
But, KP = , KA = 1 21 2 mV and UP – UA = mgh 1 2 1 2 2 1 2 mVmV – mgh. VVgh 2 122 from the triangle OP’ P, h = r (1-cos q ) VVgr 2 1221cos
VVgr 12211 cos........()
■ If V2 is the velocity of the body at highest point ( q = 180°)
VVgr 212211
VVgr 2 2 1242........()
Tension in the String at any Point
■ Let T q be the tension in the string and V q be the velocity of the body when the string makes an angle q with verticle.
At the highest point, q=180°, Vq = V2, the tension in the string is
2 2 (minimum) ..... (5)
When the string is horizantal q = 90°, V q = V hor tension in the string at this position is T mV r hor horz 2 .............(6)
Key Insights:
■ The difference in maximum and minimum tension in the string is
42426
■ Ratio of maximum tension to minimum tension in the string is
■ At point ‘P’, the net force towards the centre is T q− Mg cos q , this force provides the centripetal force at this point.
Hence, Tmg mV r cos 2 . ...... (3)
At the lowest point, q = 0°
V q= V1, the tension in the string is T mV r Lmg 1 2 (maximum) ...... (4)
When the body is at a point ‘P’, as shown in the above figure:
i) Tangential force acting on the particle is F t = mg sin q
Tangential acceleration, a t = g sin q
ii) Centripetal force acting on the particle is F mV r Tm cg 2 cos Centripetal acceleration, a
iii) The net force acting on the particle at
point ‘P’ is FFF tc 22
Net acceleration of the particle at the
point ‘P’ is aaatc 22 .
Angle made by net force or net acceleration with centripetal
component is φ and tan F F a a t c t c
■ In a vertical circular motion, if velocity at the highest point is ngr , then velocity and tensions at different positions will be V2 = √ ngr
V1 = √ (n+4) gr T q T 1 T 2 q Vq=√ngr+2gr (1+cos q) T q= mg ( n+2+3 cos q )
T 1= mg ( n+5 )
The ratio of velocity at the lowest and highest position is V V n n 1 2 4
The ratio of tensions at the lowest and highest position is T T n n max min 5 1
Applications
■ Condition for looping a loop: Consider a body of mass m tied at the end of a string and whirled in a vertical circle of radius ‘ r’. Let v1 and v2 be the velocities of body and T1, T2 be the tensions in the string at the lowest point A and highest point B respectively. The velocities of the body at
CHAPTER 7: Work, Energy, and Power
points A and B will be directed along the tangents to the circular path at the points while tensions in the string will always be directed towards the centre of rotation ‘O’ as shown in the figure.
in vertical plane
■ The forces acting on the body when it is at A are
Tension ‘T1’ in the upward direction
Weight ‘mg’ of the body acting in the downward direction.
The net force on the body supplies the necessary centripetal force,
■ The forces acting on the body when it is at B are
Tension T 2 acts in the downward direction.
Weight ‘mg’ of the body acting in the downward direction.
So, the net force acting towards centre (centripetal force) is T2 + mg, Hence,
■ The body will complete the vertical circular path only and only if minimum tension at highest point T20 min
mv r mg mv r mgvgr 2 2 2 2 02 ;
So, the minimum speed at the highest point B to just complete the vertical circle is vgr 2min . This is also called critical velocity.
■ If the velocity of the body at the highest point, is less than rg , the string becomes loose (T2 < 0) and the body falls down without continuing along the circumference. According to the law of conservation of energy, EA = EB
KA+UA=KB+UB ⇒ KA=KB+UB–UA
but, KmvKmvAB== 1 2 1 12 2 2 2 ; , UB–UA=2mgr
Hence, 1 2 1 2 12 2 2 2 mvmvmgr
vvrg 1 2 224 ..............(3)
■ To complete the verticle circle, there is a minimum velocity of the body at A so that its velocity at B will not be less than rg .
■ To find the minimum velocity of the body at A to just complete the vertical circle, we substitute vgr 2 = in equation (3).
VgrgrVrg 1 22 1 mi45 nmin.
So, for a body to be in vertical circular path its minimum velocity ( v1)min at its lowest point must be vrg 15 min
■ The minimum te nsion to be maintained in the string at the lowest point to just complete the vertical circle is
T1(min) = 6mg
Key Insights:
■ The cr itical velocity of the body at any point ‘P’ in vertical circle to just looping the loop is
VVgr minminco 1s 221
Vgrgr min co 52s 1
Vgr min co 32s
■ The minimum tension to be maintained in the string when the string makes an angle q with vertical to just looping the loop is
T mV r mg min min cos 2 .
But, Vgr min co 32s
Tmgr r mg min cos cos 32
Tmg min co 31s
■ When a body is moving in a vertical circle with critical speed, then velocities and tensions in the string at different positions are as shown in below figure.
90o 120o 180o 60o T T1 T2 T V1 V2 = √ gr, T2 = 0
V1 = √5 gr T1= 6 gm
E
V = √4 gr, T= 3mg 9mg 2 2 V = √2 gr, T=
V3 = √3 gr, T3 = 3mg
■ In vertical circle with critical speed, the ratio of minimum speeds to be maintained at highest, horizontal and lowest positions are in the ratio of 135 :: . The respective kinetic energy ratio at these points is 1 : 3 : 5.
Applications
■ In case of non uniform circular motion in a vertical plane if velocity of the body at the lowest point is lesser than 5 gr , the particle will not complete the circle in vertical plane. Now, it can either oscillate about the lowest point or it leaves the circle without looping.
■ Case ( i): Condition for oscillating about the lowest position: Let V L is the minimum velocity given to a body at the lowest position to reach the horizontal position. According to law of conservation of mechanical energy.
TEA = TEB KEA + UA = KEB + UB 1 2 200 mVmg Lr
VgLr = 2
■ If Vgr 12 = it oscillates with angular amplitude of 90 0. And the tension in the string when the string is in horizontal position is T = 0.
CHAPTER 7: Work, Energy, and Power
P (Vq = 0) VL r q
Key Insights:
■ If velocity of the body at the lowest point VL < 2 gr , then the maximum height reached by the body just before its velocity becomes zero is given by h V g L 1 2 2 = .
■ The angle made by the string with the vertical when its velocity becomes zero is given by cos1 2 2 V gr L
■ Case (ii): Condition for leaving the circular path without looping: When the velocity given to the body at the lowest point is equal to 5 gr then it just completes vertical circle. If the velocity given to the body at the lowest position is > 2 gr and < 5 gr then the body process horizontal position but cannot reach the highest position. At a certain point the tension in the string becomes zero and the body posscess some velocity. Now the only real force acting on the body at that position is gravitational force. So the body follows its own parabolic path from that point.
■ If velocity of the body at the lowest position is < 2 gr it oscillates with angular amplitude of q <90°. At position ‘P’ the tension in the string is Tmg mV r cos 2 0
Let V L is the velocity of the body at the lowest position. We know the tension at any point is given by T mV r Pmg 2 cos
If TP = 0 then P q vq TP m g 2gh<v<5gh mV r mg 2 co0 s m r Vghmg rh r L220
Vghgrg Lh 22 = 0
3gh = VL 2 + gr hVgr g L 2 3
And the angle made by the string with the vertical at this position is q then h = r (1 – cos q ) r Vgr g V gr
1 3 2 33 22 cos,cos
■ Uniform vertical circular motion:
Consider a body of mass ‘ m ’ tied at the one end of a string and whirled in a vertical circle of radius ‘r’ with uniform speed ‘ v’ by fixing the other end at ‘O’.
Tmg mv r Tmv r mg coscos 22
When the body is at lowest point ( q = 0) and the tension in the string is maximum T r Lmg mv 2 maximum
When the body is at higest point(q=180°) then tension in the string is mini mum.
Tmv r Hmg 2 minimum
Key Insights:
■ The ratio of tensions at the lowest and highest positions is T T mv r mg
r mg max min 2 2
T vgr vgr max min 2 2
■ The difference of tensions at the lowest and highest position of the body is given by TTmv r mg mv r mg maxmin 22
T max – Tmin = 2mg.
■ A bucket containing water is made to revolve in a vertical circle, such that water does not fall even in the inverted position at the highest point of the path. If ‘ r ’ is the radius of the circle then the maximum time period of revolution of the bucket is Tr g 2 :
At the highest point of the path if the velocity of the bucket Vgr ≥ then water does not fall. rgrg r
Let T q be the tension in the string when it makes an angle q with the vertical. At point ‘P’, the net force towards the centre is T q mg cos q . This net force provids the centripetal force at this point ‘ P’.
■ Reaction of Road on Car:
1) When car moves on a concave bridge then
Bridge
force = Nm
2) When car moves on a convex bridge
i) If it starts from rest at P , what is the resultant force acting on it at Q?
ii) At what height above the bottom of the loop should the block be released so that the force it exerts against the track at the top of the loop equals its weight?
Sol. By conservation of mechanical energy between points P and Q
mgRmgRmv 5 1 2 2
i.e., vgR = 8
Now in case of circular motion, NorTmvRmg 2 cos
And as at Q, q =90° N mv R mgR Rmg 28 8
So resultant force on m at Q.
Fmgmgmg 865 22
ii) At highest point N mv Rmg 2 (as q = 1800)
But according to given problem N = mg so mv Rmgmg 2 , i.e., vgR = 2
14. A small block of mass m slides along a smooth track as shown in the fig.
For achieving this, we apply conservation of mechanical energy mghmvmgR ’1 2 22 or, hRv g RgR g R ’2 2 2 2 2 3 2
Solved example
15. A mass m is released from the top of a vertical circular track of radius r with a horizontal speed v0. Calculate the angle θ with respect to the vertical where it leaves contact with the track.
Sol. The forces acting on the body are its weight mg and reaction N as shown in Fig.
So for circular motion of the body at any position q
Try yourself:
11. A nail is located below the point of suspension of a simple pendulum of length ‘ l ’. The bob is released from horizontal position. If the bob loops a verticle circle with nail as centre, find the distance of nail from point of suspension
Ans: lxxll 5 2 3 5,
Try yourself:
12. Find the angle ‘ q ’ with the lower vertical at which the resultant acceleration of the bob is along the horizontal, when the pendulum is released from horizontal position.
Ans: tan 21
Key Insights:
The body will leave contact where N = 0
i.e., mg mv r cos 2 0 i.e.,
cos v rg 2 ------ (2)
Now applying conservation of mechanical energy between H and P, we get
1 2 1 2 21 0 2 mvmvmgr cos
where ‘v’ is the velocity at ‘P’
[ as y = r (1-cos q )] or, vvgr 2 02213 cos......()
substituting the value of v2 in Equation (2)
coscos v rg 0 2 21 cos v rg 0 2 3 2 3
∴ cos 10 2 3 2 3 v rg
■ If V0 = 0 then cos 2 3
■ If the body is to lose contact at ‘ H ’ itself then q=0 0 vgr 0
■ The height of the point where the body leaves contact with the track is hR = 2 3 (if v0 = 0)
■ The vertical distance of the point p from the top most point of hemisphere where the body leaves contact with the track is y R = 3 (if v0 = 0)
TEST YOURSELF
1. A small glass marble of mass m oscillates between the two edges, inside a hemispherical glass bowl of radius r . If v is the speed of marble at the lowest position, the normal reaction at that position is (1) 2 mv r (2) 22 mv r (3) 32 mv r (4) 32 2 mv r
2. A 5 kg body is rotated in a vertical circle with a constant speed of 4 ms–1 using a string of length 1 m, when the tension in the string is 31 N, then the body will be (1) at the lowest point (2) making an angle 30° with vertical (3) at the highest point (4) at horizontal position
3. A body is tied at the end of a string of length l and rotated in a vertical circle. The string is just taut when the body is at the highest point. Velocity of the body when the string is in horizontal position is (1) 3
7.6.1 Relation between Average Power and Instantaneous Power
■ A particle starts from rest and moving with uniform acceleretion, gains a velocity ‘ V’ in time ‘t’
7.6 POWER
■ Power is the rate at which work is done or the rate at which energy is transferred.
■ If ‘W’ is the total work done by a force in a time interval ‘t’, then the average power is P W avt == Totalwork Totaltime
■ Instantaneous power is the dot product of force and velocity, assuming constant force. The instantaneous power is
■ we know instantaneous power is PFV inst = .
From the above it can be concluded that PPavinst = 1 2
7.6.2 Dimensions and Units of Power
■ Power is a scalar quantity with dimensions [ML2T–3].
SI unit of power is J/s or watt (W).
CGS unit of power is erg/s.
Practical unit of power is horse power (HP)
1 HP = 746 W.
Watt: The power of an agent is said to be one watt, if one joule of work is done in one second.
1 watt = 1 1 joule s = 1 J s–1
■ The slope of W–t curve gives instantaneous power, (as shown in fig.(a).
P = dW dt tan
Work Time q
Fig. (a)
The area under P–t graph gives work done, as shown in fig. (b).
P= dW dt WPdt . Power Time P dt
Fig. (b)
7.6.3 Efficiency of Crane or Motor
■ The ratio of output power to the input power is called efficiency.
Outputpower
Inputpower 100100 0P Pi Efficiency can also be defined as the ratio of useful work done to total energy spent. If η is the efficiency of motor, then total input power, Pmgh it 100
8.6.4 Applications of Power in Various Situations
■ If a machine gun fires ‘ n ’ bullets per second such that mass of each bullet is ‘ m ’ and coming out with a velocity ‘ v ’
then the power of the machine gun is
P = Nmv t 1 2 2 (where N bullets are fired in time ‘t’ then n = N/t)
P av = 1 2 2 mnv
■ A conveyor belt moves horizontally with a constant speed ‘v’. Gravel is falling on it at a rate of ‘dm/dt’ then,
a) Extra force required to drive the belt is F dm dt v
b) Extra power required to drive the belt is PFV dm dt vv
P dm dt v 2
■ If pump lifts w ater from depth h and delivers at a rate dm dt with a velocity v then power delivered, P dmdtghv 2 2
■ The instantaneous power of the engine is P=(R+ma)⋅VP , where R is resistance, m is mass, a is acceleration, and V is velocity.
Net force on the car is, F – R = ma driving force of the eng ine is, F = R + ma a F R
Instantaneous power PFV
P = (R + ma)V
CHAPTER 7: Work, Energy, and Power 162
■ The car moves on a rough horizontal r oad with a constant speed ‘ V ’ then the instantaneous power of engine is
PFV (V = constant) But F = f
P = fV (Here, f = frictional force on rough horizontal surface) P = µmgV
■ A body of mass ‘m’ is initially at rest. By the application of constant force its velocity changes to “v0” in time ‘t o ’ then
v = u + atv0 = at0
Acceleration of the body is a v t = 0 0
a) Instantaneous power at an instant of time ‘t’ is PFV = (ma) (at) = ma2 t
Pm v t ta v stint 0 0 2 0 0
b) Average power during the time ‘ t’ is P av = 1 2 1 2 0 0 2 .;pPm t int st av v
■ A motor pump is used to deliver water at a certain rate from a given pipe. To obtain ‘ n’ times water from the same pipe in the same time, the amount by which (a) force and (b) power of the motor should be increased, are —
If a liquid of density ‘ρ’ is flowing through a pipe of cross-section ‘A’ at speed ‘V’, the mass coming out per second will be dm dt AV
To set ‘n’ times water in the same time dm dt n dm dt 1 AVnAV
As the pipe and liquid are same ρ ’ = ρ AAvnV ,
a) Now as F = V dm dt F F V dm dt V dm dt nVn dm dt V dm dt n ’’(). ’ 2 FnF 12
b) PFV P P FV FV nFnV FV n PnP ’’’()() ’. 2 3 3
To get ‘n’ times of water, force must be increased n2 times, while power must be increased n3 times.
■ An automobile of mass ‘ m ’ accelerates, starting from rest, while the engine supplies constant power, its position and velocity change w.r.t time.
i) Velocity: FV = P = constant
i.e., m dV dt VPFm dV dt
or, VdVP m dt
On integrating both sides, we get VP m tc 2 21
As initially, the body is at rest, i.e., V = 0 at t = 0 ⇒ c1 = 0, V Pt m Vt 2 1 2 1 2
ii) Position: From the above expression, V Pt m 2 1 2
By integrating both sides, we get s P m tc 22 3
at
=
Solved examples
16. An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ms –1 . The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power.
Sol. The downward force on the elevator is,
F = mg + Ff = (1800 × 10) + 4000 = 22000 N
The motor must supply enough power to balance this force.
Hence, P = Fv = 22000 × 2 = 44000 W = 59 hp
17. An electric pump on the ground floor of a building taken 15 minutes to fill a tank of volume 30 m 3 with water. If the tank is 40 m above the ground and the efficiency of the pump is 30%, find the electric power consumed by the pump in filling the tank. [Density of water = 103 kg/m3]
Sol. mass of the water to be filled = (volume of the tank) × density of water m = 30 × 103 kg h = 40 m, t = 15 × 60 =900 sec.
Efficiency, η = 30%
Input power, Pi = 100 mgh t 10030109840 30900
Try yourself:
13. A machine gun fires 240 bullets per minute with certain velocity. If the mass of each bullet is 10–2 kg and the power of the gun is 7.2 kW, find the velocity with which each bullet is fired.
Ans:: 600 m/s
Try yourself:
14. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration a c is varying with time t as a c = k 2 rt 2 where k is a constant. What is the power delivered to the particle by the forces acting on it?
Ans:: mk2r2t
TEST YOURSELF
1. A body of mass m is accelerated uniformly from rest to a speed v in a time T . The instantaneous power delivered to the body as a function time is given by
2. A bus of mass 6 metric tons is pulled at a speed of 18 kmph on a smooth incline of inclination 1 in 20. Power of the engine is, (g = 10 ms –2) (1) 15 kW (2) 1500 W (3) 30 kW (4) 1.2 W
3. A train is moving with a uniform velocity of () 241 ˆ ijk ˆ s ˆ m −+ If the force required to overcome friction is () N ˆˆˆ 32−+ ijk the power of the engine is (1) 7 W (2) 10 W (3) 13 W (4) 15 W
Answer Key
(1) 1 (2) 1 (3) 3
# Exercises
JEE MAIN LEVEL
Level-I
Work: Single Option Correct MCQs
1. Work done against gravity in slowly taking a 5 kg block to 2 m height in 2 s is
(1) 49 J (2) 98 J
(3) – 49 J (4) –98 J
2. A body constrained to move in z direction is subjected to a force by () . ˆˆ 3105N ˆ =−+ Fijk
What is the work done by this force in moving the body through a distance of 5 m along z axis?
(1) 15 J (2) –15 J
(3) –50 J (4) 25 J
3. A string is used to pull a block of mass m vertically up by a distance h at a constant acceleration g /3. The work done by the tension in the string is
(1) 2 3 mgh (2) 3 mgh
(3) mgh (4) 4 3 mgh
4. A uniform chain has mass M and length L respectively. It is lying on a smooth horizontal table with half of its length hanging vertically down. The work done in pulling the chain completely on to the table is
(1) MgL/2 (2) MgL/4
(3) MgL/8 (4) MgL/16
5. A force F acting on an object varies with distance x as shown here. 1 2 3 4
Work done by the force in moving the object from x = 0 to x = 6 m is
(1) 18 J (2) 27 J
(3) 36 J (4) 40 J
6. A force F = (2 + x ) N acts on a particle in x -direction, where x is in metres. Find the work done by this force during a displacement from x = 1 m to x = 2 m.
(1) 3.5 J (2) 4.2 J
(3) 6.8 J (4) 8.0 J
7. A block of mass m is lowered with the help of a rope of negligible mass through a distance d with an acceleration of 3 g . Work done by the rope on the block is
(1) 2 3 Mgd
(2) 2 3 Mgd
(3) 3 Mgd
(4) 3 Mgd
8. A particle of mass 0.1 kg is subjected to a force which varies with distance as shown in figure. If it starts its journey from rest at x= 0, then its velocity at x = 12 m is 10 4 8 displacement (m) 12 O Force (N)
(1) 0
(2) 202m/s
(3) 203m/s
(4) 40 m/s
Numerical Value Questions
9. A block of mass 1 kg is suspended by a light thread from an elevator. The elevator, starting from rest, is accelerating upward with uniform acceleration 0.2 m/s 2 . The work done by tension on the block in 1 s is_________J.[g = 9.8 m/s 2]
Kinetic Energy
Single Option Correct MCQs
10. Two bodies of masses of 1 g and 4 g are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is
(1) 4:1 (2) 2:1
(3) 1:2 (4) 1:16
11. If the kinetic energy of a body increases by 2%, then the momentum
(1) increases by 2% (2) increases by 1% (3) decreases by 1 % (4) decreases by 2%
12. If the speed of a vehicle increases by 2 ms–1, its kinetic energy is doubled, then original speed of the vehicle is
(1) (21)1 ms + (2) 2(21)1 ms
(3) 2(21)1 ms + (4) (221)1 ms +
13. A body of mass 5 kg has a linear momentum of 20 kgm/s. If a constant force of 5 N acts on it in the direction of its motion for 10 s. The change in kinetic energy of the body is (1) 320 J (2) 600 J
(3) 200 J (4) 450 J
14. A particle does uniform circular motion in a horizontal plane. The centripetal force acting on the particle is 10 N. It's kinetic energy is 1 joule. The radius of the circular path is (1) 0.1 m (2) 0.2 m (3) 2 m (4) 0.02 m
15. A particle of mass 2 kg is moving in one dimension under a force that performs work
of 50 J on the particle. If the initial speed (in m/s) of the particle is zero, the final speed of particle is (1) 0 m/s (2) 5 m/s (3) 52m/s (4) 25m/s
16. A particle experiencing a force =− ˆ 3 ˆ 4 Fij undergoes a displacement of 2. ˆ di = If the particle has a kinetic energy of 4 J at the beginning of the displacement, what is the kinetic energy at the end of the displacement? (1) 9 J (2) 15 J (3) 12 J (4) 10 J
Numerical Value Questions
17. A car accelerates from rest to u m/s. The energy spent in this process is E J. The energy required to accelerate the car from u m/s to 2u m/s is nE J. The value of n is _______.
18. A bomb of mass 20 kg, initially at rest, explodes into two pieces of masses 8 kg and 12 kg. If the kinetic energy of the 8 kg piece is 576 J, find the kinetic energy (in joules) of the 12 kg piece.
19. A block of mass m sliding on the smooth horizontal surface with the velocity V meets a long horizontal spring fixed at one end and having spring constant k as shown in the figure. The maximum compression of the spring is Vm nk where n is equal to . v
20. A block of mass 10 kg starts sliding on a surface with initial velocity of 9.8 m/s. The distance covered by the block before coming to rest is____× 10–1m [µ=0.5].
Potential Energy
Single Option Correct MCQs
21. A body of mass 0.5 kg falls from a height of 10 m to a height of 6 m above the ground. Find the loss in potential energy taking place in the body (g = 10 m/s 2)
(1) 20 J (2) 4 J (3) 6 J (4) 8 J
22. A simple pendulum is swinging in a vertical plane. The ratio of its potential energies when it is making angles 30° and 60° with the vertical is
(1) ()1:23 (2) ()23:1
(3) 1 : 2 (4) 2 : 1
23. A rectangular plank of mass m1 and height a is on a horizontal surface. On the top of it another rectangular plank of mass m2 and height b is placed. The potential energy of the system is,
(1) ()()12g 2 + + ab mm
(2) 12 2g22 +
(3) 1 22g22
mb mam
(4) 1 21g22
24 The potential energy of a body is given by, U = A–Bx2 (Where x is the displacement). The magnitude of force acting on the particle is
(1) Constant
(2) Proportional to x
(3) Proportional to x2
(4) Inversely proportional to x
25. A uniform metre scale of mass 2 kg is suspended from one end. If it is displaced through an angle 60° from the vertical, the increase in its potential energy is,
(1) 4.9 J (2) 9.8 J
CHAPTER
7: Work, Energy, and Power
(3) 9.8 3 J (4) 4.9 (2 – 3 ) J
26. The work done in carrying a body from A to B is 20 J and from B to C is 15 J. The work done in moving the body from A to C, assuming the force involved in all the cases is conservative force, is
(1) 35 J (2) 25 J
(3) 5 J (4) 175J
27. When you lower a book of mass m down slowly through a distance h from yourself, the work done by you on the book is
(1) mgh (2) – mgh
(3) zero (4) > mgh
Numerical Value Questions
28. A 0.4 kg mass takes 8 s to reach the ground when dropped from a certain height P above the surface of the earth. The loss of potential energy in the last second of fall is ________ J. (Take g = 10 m/s2)
Conservation of Mechanical Energy
Single Option Correct MCQs
29. A body of mass 200 g begins to fall from a height where its potential energy is 80 J. Its velocity at a point where kinetic and potential energies are equal is
(1) 108m/s (2) 4 m/s
(3) 400 m/s (4) 20 m/s
30. A stone projected vertically upwards from the ground reaches a maximum height h . When it is at a height h/4, the ratio of kinetic and potential energy is
(1) 3 : 4 (2) 3 : 1
(3) 4 : 3 (4) 1 : 3
31. A body is projected vertically up with certain velocity. At a point 'P' in its path, the ratio of its potential to kinetic energy is 9 : 16. The ratio of velocity of projection to velocity at 'P' is
(1) 3 : 4 (2) 5 : 4
(3) 9 : 25 (4) 25 : 16
Numerical Value Questions
32. A ball of mass 4 kg, moving with a velocity of 10 ms–1, collides with a spring of length 8 m and force constant 100 Nm–1. The length of the compressed spring is x m. The value of x, to the nearest integer, is _____.
Vertical Circular Motion
Single Option Correct MCQs
33. A 2 kg stone is swung in a vertical circle by attaching it at the end of a string of length 2 m. If the string can withstand a tension 140.6 N, the maximum speed with which the stone can be rotated is
(1) 22 ms–1 (2) 44 ms–1 (3) 33 ms–1 (4) 11 ms–1
34. A ball of mass m is rotated in a vertical circle with constant speed. The difference in tension at the top and bottom would be
(1) 6 mg (2) 5 mg
(3) 2 mg (4) mg
35. A 1 kg ball is rotated in a vertical circle by using a string of length 0.1 m. If the tension in the string at the lowest point is 29.4 N, its angular velocity at that position is
36. A body is revolving in a vertical circle with constant mechanical energy. The speed of the body at the highest point is 2rg . The speed of the body at the lowest point is
(1) 7rg (2) 6rg
(3) 8rg (4) 9rg
37. A body is moving in a vertical circle of radius r by a string . If the ratio of maximum to minimum speed is 3:1 , the ratio of maximum to minimum tensions in the string is
(1) 3 : 1 (2) 5 : 1
(3) 7 : 1 (4) 9 : 1
38. The velocity of a body revolving in a vertical circle of radius r at the lowest point is 7. gr The ratio of maximum to minimum tensions in the string is
(1) 8 : 1 (2) 4 : 1
(3) 7:1 (4) 1:7
39. A simple pendulum of length l carries a bob of mass m . When the bob is at its lowest position, it is given the minimum horizontal speed necessary for it to move in a vertical circle about the point of suspension. When the string is horizontal, the net force on the bob is
(1) mg (2) 3 mg
(3) 10g m (4) 4 mg
40. A 1 kg ball is rotated in a vertical circle by using a string of length 0.1 m. If the tension in the string at the lowest point is 29.4 N, its angular velocity at that position is
(1) 7 rads–1 (2) 14 rads–1
(3) 3.5 rads–1 (4) 25.6 rads–1
Numerical Value Questions
41. A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 50 cm long. The speed of the bob, when the length makes an angle of 60° to the vertical, will be (g=10 m/s2) ______m/s.
Power
Single Option Correct MCQs
42. A force of 234N ˆˆˆ ++ ijk acts on a body for 4 seconds, produces a displacement of () 345m. ˆˆˆ ++ ijk The power used is
(1) 9.5 W (2) 7.5 W
(3) 6.5 W (4) 4.5 W
43. An engine pumps up 100 kg of water through a height of 10 m in 5 s. Given that the efficiency of the engine is 60%, what is the power of the engine? (g = 10 ms –2) (1) 33 kW (2) 3.3 kW (3) 0. 33 kW (4) 0.033 kW
44. For a projectile of mass m projected with initial velocity u at an angle of projection 'θ' with the horizontal, the instantaneous power delivered by gravity at the point of maximum height is (1) zero (2) mgu cos θ (3) mgu sin θ (4) –mgu cosθ
45. A man of mass 60 kg lifts a 15 kg mass to the top of a building of height 10 m in 5 minutes. His efficiency is (1) 10% (2) 20% (3) 30% (4) 40%
Numerical Value Questions
46. If force () 60153N ˆˆˆ =+− Fijk and velocity () 2451 ms ˆˆˆ =−+ Vijk , then if the instantaneous power is x × 5 watts, find the value of x Level-II
Work
Single Option Correct MCQs
1. A uniform rod of mass 2 kg and length l is lying on a horizontal surface. If the work done in raising one end of the rod through an angle 45° is W , then the work done in raising it further 45° is (1) W (2) 2W (3) 2 W (4) () 21 W
2. A uniform rod of length 2 m and mass 5 kg is lying on a horizontal surface. The work done in raising one end of the rod with the other end in contact with the surface until the rod makes an angle 30° with the horizontal, is (g = 10 ms–2)
(1) 25 J (2) 50 J
(3) 253J (4) 503J
3. A force F = ax2 +bx + c acts on a block along positive x-direction. In the expression of force, a, b, and c are constant. Find the work done by force on the block, if block displaces from x=0 to x=d
(1) 32 32 adbdcd ++
(2) 3 2 2 adbd +
(3) 32 64 adbdcd ++
(4) None of these
4. A force () N ˆˆ =−+ Fkyixj (where k is a positive constant) acts on a particle moving in the XY-plane. Starting from the origin, the particle is taken along the positive X-axis to the point (a, 0)and then parallel to the Y-axis to the point (a, a). The total work done by the force F on the particle is
(1) –2ka2 (2) 2ka2
(3) –ka2 (4) ka2
Numerical Value Questions
5. The work done (in J) by an applied variable force F = α x + b x3 from x = 0 and x = 2 m, where x is displacement and constants α=|1 Nm–1| and b =|1 Nm–3|, is ____.
Kinetic Energy
Single Option Correct MCQs
6. A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to (1) x (2) ex (3) x2 (4) loge x
7. A vehicle of mass M is moving on a rough horizontal road with a momentum p. If the coefficient of friction between the tyres and the road is μ, then the st opping distance is
8. A juggler keeps four balls in air. He throws each ball vertically upwards with the same speed at equal intervals of time. The maximum height attained by each ball is 20 m. Find the kinetic energy of the first ball when the fourth ball is in hand. Assume that the mass of each ball is 10 g.
(1) 0.5 J (2) 1 J (3) 1.5 J (4) 2.5 J
Numerical Value Questions
9. A point mass of 0.5 kg is moving along x-axis as x=t2+2t, where x is in metres and t is in seconds. Find the work done (in J) by all the forces acting on the body during the time interval [0, 2 s].
10. Block A of mass 1 kg is placed on the rough surface of block B of mass 3 kg. Block B is placed on smooth horizontal surface. Blocks are given the velocities as shown. Find net work done by the frictiona l force.[in –veJ]
11. An open knife edge of mass M is dropped from a height h on a wooden floor. If the blade penetrates a distance 5 m into the wood, the average resistance offered by the wood to the blade is 1 h Mg s
+
. Then, s = ____m.
Potential Energy
Single Option Correct MCQs
13. If potential energy of a pair of nucleons is plotted as a function the separation between them, choose the correct option.
(1) For r > r 0, force between nucleons is repulsive
(2) For r < r 0, force between nucleons is attractive
(3) For r > r 0, force between nucleons is attractive
(4) Nuclear force is much weaker than coulumbic force for short range of separation.
14. The potential energy of a particle in a force field is 2 AB U rr =− , where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is
(1) A/B
(2) B/A
(3) B/2A
(4) 2A/B
12. A block of mass m, moving with a speed v, compresses a spring through a distance x before its speed reduces by 25%. The spring constant of the spring is (1) 2
15. The given plot shows the variation of U, the potential energy of interaction between two particles with the separation r between them.
Then,
A) ‘B’ and ‘D’ are equilibrium points
B) ‘C’ is a point of stable equilibrium
C) ‘E’ is also in equilibrium but it is unstable
(1) A only is true.
(2) A and B are true.
(3) B and C are true.
(4) A, B, and C are true.
16. A ball of mass 10 g is projected upward with an initial velocity 10 m/s and it comes back, finally achieving a velocity of 5 m/s at the point of projection. Find the work done by air resistance. (Neglect buoyancy force due to air)
(1) 0.375 J (2) 0.275 J
(3) –0.375 J (4) zero
17. A uniform metre scale of mass 2 kg is suspended from one end. If it is displaced through an angle 60° from the vertical, the increase in its potential energy is
(1) 4.9 J (2) 9.8 J
(3) 9.83J (4) () 4.923J
18. A spring of constant 100 N/m is stretched by applying equal forces, each of magnitude F at the two ends. The energy stored in the spring is 200 J. Now, spring is cut into two equal parts, and one of the parts is stretched by applying equal forces, each of magnitude F at the two ends. The energy stored is (1) 200 J (2) 100 J (3) 400 J (4) 50 J
19. A point mass m (neglect volume of the point mass) is floating on the surface of water
contained by a vertical cylinder. The water is up to height H. Due to leakage at the bottom of the cylinder, total water is spread out near the bottom of the cylinder. The total mass of water is M. The work done by gravity is (1) g2 H M (2) (m+M)gH
(3) 2g + M mH (4) () g2 + H mM
20. A spring of force constant k is stretched by a small length x. The work done in stretching it further by a small length y is
21. The radius of a well is 7 m. Water in it is at a depth of 20 m and depth of water column is 10 m. Work done in pumping out water completely from the well is ________. (g = 10 m–2) (in MJ)
22. A rectangular block of dimensions 6 m × 4 m × 2 m and of density 1.5 g/cc is lying on a horizontal ground with the face of largest area in contact with the ground. The work done in arranging it with its smallest area in contact with the ground is _____.(g = 10 ms–2)(in kJ)
23. A particle of mass m is moving in a horizontal circle of radius r under centripetal force equal to 2 K r , where K is a constant. If the total energy of the particle is K Pr , then P is _________.
24. A small block of mass 0.1 kg is pressed against a horizontal spring fixed at one end to compress the spring through 5.0 cm, as shown. The spring constant is 100 N/m. When released, the block moves horizontally till it leaves the spring. It will hit the ground 2 m below the spring at a horizontal distance of ________.
Conservation of Mechanical Energy
Single Option Correct MCQs
25. Figure shows a particle sliding on a frictionless track, which terminates in a straight horizontal section. If the particle starts slipping from the point A, how far away from the track will the particle hit the ground?
1 m
(1) 1 m (2) 0.1 m
(3) 10 m (4) 0.01 m
26. A particle is subjected to a force F =–kx with k = 15 N/m. If it is released from a point 20 cm away from the origin, its kinetic energy, when it is 10 cm from the origin, is
(1) 0.225 J (2) 0.45 J
(3) 0.75 J (4) 0.075 J
27. A ball at rest is dropped from a height of 12 m. It loses 25% of its kinetic energy on striking the ground and bounces back to a height h. The value of h is
(1) 3 m (2) 6 m
(3) 9 m (4) 12 m
28. The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?
(1) 3.14 ms–1
(2) 5.28 ms–1
(3) 1.54 ms–1
(4) 8.26 ms–1
29. When a body is projected vertically up, at a point 'P' in its path, the ratio of potential to kinetic energies is 3 : 4. If the same body were to be projected with two times the initial velocity, the ratio of potential to kinetic energies at the same point is
(1) 3 : 25 (2) 3 : 28
(3) 1 : 20 (4) 1 : 25
30. A block of mass M, moving on a fricitionless horizontal surface, collides with the spring of spring constant K and compresses it by length L . Then, maximum momentum of the block after collision is M K
(1) L Mk (2) zero (3)
(4)
31 In the ideal pulley-particle system shown, the mass m 2 is connected with a vertical spring of spring constant K ( m 2 > m 1 ). If the mass m2 is released from rest when the spring is underformed, find the maximum compression of the spring. m m
(1) () 21 mmg K (2) () 12 mmg K +
(3) 1 mg K (4) ()221mmg K
Numerical Value Questions
32. A particle is released from height H . At a certain height from the ground, its kinetic energy is twice its gravitational potential energy. The velocity of the particle at that height is 3 VngH = . The value of n is ____.
33. A pendulum is suspended by a string of length 250 m. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at 60° with vertical, as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be ______ms–1(if g = 10 m/s–2).
600 I = 250 cm
Power
Single Option Correct MCQs
34. An air pump is pumping air into a large vessel against a pressure of 1 atmosphere. In each stroke, 1000 cc of air enters into the vessel. If the number of strokes made by the pump per minute is 30, the power of the pump is (1 atm =105 Pa) (1) 5 kW (2) 50 kW (3) 100 kW (4) 50 W
35. Sand is falling on a conveyor belt at the rate of of 5 kgs–1. The extra power required to move the belt with a velocity of 6 kgs –1 is (1) 30 W (2) 180 W (3) 150 W (4) 1.2 W
36. A uniform rope of mass 1 kg and length 1 m is lying on the ground. One end of the rope is pulled up by a worker with the constant velocity of 1 m/s. The average power supplied by the worker in pulling the entire rope just off the ground, such that the rope becomes vertical, is (take g=10 ms/ 2)
(1) 5.5 W (2) 6 W (3) 10.5 W (4) none of these
Numerical Value Questions
37. A body of mass 1 kg begins to move under the action of a time depends force
32Nˆˆ =+ Ftitj , where ˆ i and ˆ j are the unit vectors along x and y axis. The power developed by the above force, at the time t=2 s, will be __________ W.
38. Water falls from a height of 60 m at the rate of 15 kgs–1 to operate a turbine. The losses due to frictional force are 10% of energy. If the power generated by the turbine is n×0.1 kW, then find the value of n
Vertical Circular Motion
Single Option Correct MCQs
39. An inclined track ends in a circular loop of diameter D . From what height on the track shoud a particle be released so that it completes that loop in the vertical plane?
(1) 5 2 D (2) 2 5 D
(3) 5 4 D (4) 4 5 D
40. A coin is sliding down on a smooth hemispherical surface of radius R . The height from the bottom, where it loses contact with the surface is
(1) R/3 (2) 2R/3
(3) R (4) 4R/3
Numerical Value Questions
41. The minimum speed of a bucket full of water whirled in a vertical circle of radius 10 m at the highest point, so that the water may not fall, is (g = 10 ms–2)_____(ms–1).
42. A small block slides with velocity 0.5 gr on the horizontal frictionless surface, as shown in the figure. The block leaves the surface at point C. If the value of cos 1 n n θ= + , find the value of n
43. Figure shows a smooth track, a part of which is a circle of radius R. A block of mass m is pushed against a spring of spring constant k fixed at the left end, and is then released. The initial compression of the spring, so that the block presses the track with a force mg when it reaches the point P, where the radius of the track is horizontal, is pmgR k Find the value of p m
Level-III
Single Option Correct MCQs
1. A particle of mass m is projected with a velocity u at an angle α with the horizontal. Work done by gravity during its descent from its highest point to the position where its velocity vector makes an angle 2 α with the horizontal is
(1) 122 tan 2 mu α
(2) 122 tan 22 mu α
(3) 1222 costan 22 mu α α
(4) 1222 cossin 22 mu α α
2. A block of mass m is attached to two unstretched springs of constant k1 and k2 The other ends are free. There is no friction anywhere. The block is displaced towards left by small distance x1 and released. The maximum compression in spring of force constant k2 is x2. Then, ratio of 1 2 x x is
3. A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. The work done by the force of gravity during the time the particle goes up is (1) 0.5 J (2) –0.5 J (3) −1.25 J (4) 1.25 J
4. A particle of mass m moves with a variable velocity v , which changes with distance covered x along a straight line as , vkx = where k is a positive constant. The work done by all the forces acting on the particle during the first second is
5. A uniform chain of length L is placed on a smooth table of height h ( h > L ) with a length l hanging from the edge of the table. The chain begins to slide down the table. When the end of the chain is about to leave the edge of the table, its velocity is
6. A block of mass 5 kg is sliding down a smooth inclined plane, as shown. The spring arranged near the bottom of the inclined plane has a force constant 600 N/m. Find the compression in the spring at the moment the velocity of the block is maximum. 4m 5m 3m
(1) 5 cm (2) 10 cm
(3) 15 cm (4) 25 cm
7. A block of mass m is attached with a massless spring of force constant k . The block is placed over a rough inclined surface for which the coefficient of friction is 3 4 µ= The minimum value of M required to move the block up the plane is (Neglect mass of pulley, string, and friction in pulley) m M
370
(1) 3 5 m (2) 4 5 m
(3) 6 5 m (4) 3 2 m
8. A particle is released from a height S . At a certain height, its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are, respectively,
(1) 3 , 42 SgS (2) 3 , 42 SgS
(3) 2 , 22 SgS (4) 3 , 42 SgS
9. Calculate the work done for following F-d curve?
70 –30 F(N) d(m) 5 3 0
(1) 30 J (2) 40 J (3) 45 J (4) 50 J
10. A body of mass m =10–2 kg is moving in a medium and experiences a frictional force F=–Kv2. Its initial speed is v0=10 m/s. If, after 10 s, its energy is 2 0 1 , 8 mv the value of K will be
11. An object of mass m initially at rest on a smooth horizontal plane starts moving under the action of force F = 2 N. In the process of its linear motion, the angle θ (as shown in figure) between the direction of force and horizontal varies as θ= kx , where k is a constant and x is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be sin =θ n E k . The value of n is _____.
m q
Smooth horizontal Surface
12. Two blocks are connected to an ideal spring of stiffness 200 N/m. At a certain moment, the two blocks are moving in opposite directions with speeds 4 ms−1 and 6 ms−1, and the instantaneous elongation of the spring is 10 cm. The rate at which the spring energy
2 2 kx
is increasing (in joules per second) is 100 n. Find the value of n.
13. When a 2 kg car, driven at 20 m/s on a level road, is suddenly put into neutral gear (i.e. allowed to coast), the velocity decreases in the following manner: 20 m/s 1 20 = +
V t , where t is the time in seconds. The power (in watts) required to drive this car at speed 10 m/s on the same road i s _____.
14. A force acting on a body depends on its displacement S , is given by 1 3 FS ∝ . The power delivered by force is directly proportional to Sx. Then, the value of x is _______.
15. A block moving horizontally on a smooth surface with speed of 40 m/s splits into two equal parts. If one of the parts moves at 60 m/s in the same direction, then the fractional change in the kinetic energy will be x : 4, where x=______.
16. A ball is projected vertically down with an initial velocity from a height of 20 m onto a horizontal floor. During the impact, it loses 50% of its energy and rebounds to the same height. The initial velocity (in m/s) of its projection is _______.
17. The potential energy of a 1 kg particle free to move along the x -axis is given by
The total mechanical energy of the particle is 2 J. If the maximum speed is m/s 2 x , then x = ________.
18. A light spring of length 1 m is fixed at point A on a fixed cylinder of radius = π 2 l Rm . The other end of the string is attached to a
ball of mass 1 kg. The ball is imparted with a velocity (v0) of 10 m/s horizontally. The tension in the string, when the ball is at the lowest point, is 30x N. Find the value of x.
19. A spring of force constant 50 N/m is stretched by small length of 1 cm. The work done in stretching it further by a small length of 1 cm is x ×10–4 J. Find the value of x
20. Two blocks A and B of equal mass ( m = 10 kg) are connected by a light spring of spring constant k = 150 N/m. The system is in equilibrium. The minimum value of initial downward velocity V 0 of the block B, for which the block A bounces up, is 20 m/s. 3n Find the value of n
21. The potential energy of a particle of mass m, free to move along x-axis, is given by 12 2 Ukx = for x <0 and U =0 for x ≥ 0 ( x denotes the x-coordinate of the particle and k is a positive constant). If the total mechanical energy of the particle is E, then its speed at 2 E x k =− is _____.
22. A 60 hp electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to _____m/s. (Take 1 hp = 746 W, g = 10 ms –2 ) (Round off to the nearest integer)
23. A particle of mass m is suspended by a string of length l from a fixed rigid support. A sufficient horizontal velocity 03 vgl = is imparted to it suddenly. Calculate the angle (in degree) made by the string with the vertical when the acceleration of the particle is inclined to the string by 45°.
THEORY-BASED QUESTIONS
Single Option Correct MCQs
1. The area under a 'force-displacement' curve gives (1) impulse (2) power (3) work (4) time
2. When a body is falling freely from some height, work done by gravity is (1) negative
(2) positive
(3) zero
(4) may be positive or negative
3. A boy carrying a box on his head, walking on a straight level road from one place to another, is doing no work. The statement is (1) partly correct (2) correct (3) incorrect (4) insufficient data
4. No work is done by a force on an object if a) the force is always perpendicular to its velocity
b) the force is always perpendicular to its acceleration
c) the object is stationary but the point of application of the force moves on the object
d) the object moves in such a way that the point of application of the force remains fixed.
(1) b, c, and d are correct.
(2) a, b, and d are correct.
(3) a, c, and d are correct.
(4) All are correct.
5. A) Work done by frictional force is always negative.
B) A body at rest can have mechanical energy.
C) Mechanical energy of freely falling body decreases gradually.
(1) Only A is true.
(2) Only B is true.
(3) Only C is true.
(4) All the three are true.
6. A lorry and a car moving with same momentum are brought to rest by applying the same retarding force. Then, (1) lorry will come to rest in a shorter distance
(2) car will come to rest in a shorter distance
(3) both come to rest in the same distance
(4) none
7. If the force acting on a body is inversely proportional to its speed, then the kinetic energy of the body is (1) constant
(2) directly proportional to time
(3) inversely proportional to time
(4) directly proportional to square of time
8. A body can have (1) zero momentum and finite kinetic energy
(2) zero kinetic energy and finite momentum
(3) zero acceleration and increasing kinetic energy
(4) finite acceleration and zero kinetic energy
9. Same force acts on two bodies of masses m1 and m2 (m2 > m1). After travelling the same distance,
(1) kinetic energy of m2 is greater than kinetic energy of m1
(2) kinetic energy of m1 is greater than kinetic energy of m2
(3) ratio of kinetic energies of those two is equal to ratio of their momentum
(4) kinetic energy of m1 is equal to kinetic energy of m2
10. A constant force acts on a body, which is initially at rest. If a graph is drawn by plotting its kinetic energy along y-axis and displacement along x-axis, shape of the graph is
(1) straight line passing through origin
(2) parabola
(3) straight line parallel to x-axis
(4) straight line parallel to y-axis
11. A curve is drawn expressing the kinetic energy of a particle as a function of the distance traversed (on x-axis). The slope of this curve represents the instantaneous
(1) velocity
(2) acceleration
(3) force
(4) power
12. The kinetic energy of a particle continuously increases with time.
a) The resultant force on the particle must be parallel to the velocity at all instants.
b) The resultant force on the particle must be at angle less than 90° all the time
c) Its height above the ground level must continuously decrease
d) The magnitude of its linear momentum is increasing continuously.
(1) a and b are correct.
(2) b and c are correct.
(3) a and d are correct.
(4) b and d are correct.
13. The work done by all the forces (external and internal) on a system equals the change in
(1) total energy
(2) kinetic energy
(3) potential energy
(4) none of these
14. Identify the false statement from the following.
(1) Work - energy theorem is not independent of Newton's second law.
(2) Work-energy theorem holds in all inertial frames.
(3) Work done by friction over a closed path is zero.
(4) Work done is a scalar quantity.
15. Two springs have their force constants k 1 and k 2 and they are stretched to the same extension. If k2 > k1, work done is
(1) same in both the springs
(2) more in spring k1
(3) more in spring k2
(4) none
16. Two bodies of masses m1 and m2 have equal momentum. Their KE are in the ratio
(1) 21 : mm
(2) m1 : m2
(3) m2 : m1
(4) 22 12 : mm
17. A particle of mass m is being circulated on a vertical circle of radius r. If the speed of particle at the highest point is v, then
(1) 2 mv mg r =
(2) 2 mv mg r >
(3) 2 mv mg r <
(4) 2 mv mg r ≥
18. The tube AC from a quarter circle in a vertical plane. The ball B has an area of cross-section slightly smaller than that of the tube, and can moves without friction through it. B is placed at A and displaced slightly. It will
(1) always be contact with the inner wall of the tube
(2) always be in contact with the outer wall of the tube
(3) initially be in contact with the inner wall and later with the outer wall
(4) initially be in contact with the outer wall and later with the inner wall
19. A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to (1) t1/2 (2) t (3) t3/2 (4) t2
20. Internal force can change (1) kinetic energy
(2) total momentum of system (3) total energy of the system (4) all of the above
21. In the case of conservative force, (1) work done is indepenent of the path (2) work done in a closed loop is zero (3) work done against conservative force is stored in the form of potential energy (4) all the above
22. A body of mass M , moving with velocity V , explodes into two equal parts. If one comes to rest and the other body moves with velocity V, what would be the value of V? (1) V (2) 2 V (3) 4V (4) 2V
23. A) When a body slides on rough surface, its momentum is not conserved.
B) When a ball falls from a height, momentum of earth ball system is conserved.
C) In case of explosion of flying bomb, momentum is conserved.
(1) A and B are true.
(2) A, B, and C are true.
(3) A and C are false.
(4) A is false.
Assertion and Reason Type Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) if (A) is true but (R) is false,
(4) if both (A) and (R) are false.
24. (A) : Water at the foot of the waterfall is always at a different temperature from that at the top.
(R) : The potential energy of water at the top is converted into heat energy during falling.
25. (A) : A spring has potential energy, both when it is compressed or stretched.
(R) : In compressing or stretching, work is done on the spring against the restoring force.
26. (A) : Mass and energy are not conserved separately but are conserved as a single entity called mass–energy.
(R) : Mass and energy conservation can be obtained by Einstein equation for energy.
27. (A) : Work done in moving a body over a closed loop is zero for conservative force in nature.
(R) : Work done is independent of nature of force.
28. (A) : A body can have energy without having momentum, but it cannot have momentum without having energy.
(R) : Momentum and energy have same dimensions.
29. (A) : In one dimensional motion with constant acceleration, the power delivered is proportional to time.
(R) : In general, power is equal to product of force applied and its velocity.
30. (A) : The work done by the tension in the string of a simple pendulum in one complete oscillation is zero.
(R) : No work is done by the tension in the string since tension is always at right angles to the motion of bob.
31. (A) : The work done by the spring force in a cyclic process is zero.
(R) : The spring force is a conservative force.
32. (A) : Total energy is negative for a bound system.
(R) : Potential energy of a bound system is negative and more than kinetic energy.
JEE ADVANCED LEVEL
Multi Option Correct MCQs
1. A constant force F acts on a moving body along a smooth quarter circular arc of radius R fixed on horizontal surface. Calculate the
work done by force F from A to B. B R O A
(1) If force F always points towards B, then work is 2 FR
(2) If force F always acts tangentially, then work is 2 π FR .
(3) If force F always acts parallel to AB, then work is 2 FR .
(4) If force F always acts parallel to AB, then work is ‘zero’.
2. Work done by a force on an object is zero, if
(1) the force is always perpendicular to its acceleration
(2) the object is stationary but the point of application of the force moves on the smooth object
(3) the force is always perpendicular to its velocity
(4) the object moves in such a way that the point of application of the force remains fixed
3. A particle of mass 1 kg is moving in the xyplane, whose position at an instant t is r = 3 ()23s2 ˆˆ 1co sintitj +− , where r is in metres and t is in seconds. The kinetic energy of the particle at an instant t is
(1) 4.5 J
(2) 10 J
(3) 18 J
(4) does not depend on the time
4. A block of mass 10 kg is pulled along a smooth surface in the form of arc of radius 10 m. The applied force F is 200 N, as shown in figure. If the block starts from point A, then
(1) the speed at point B is 15.7 m/s
(2) the speed at point B is 27 m/s
(3) the work done by gravity in moving from A to B is negative
(4) the work done by tension on block is zero
5. A small ball of mass 1 kg is given vertical velocity of magnitude v0=5 m/s and it swings in a vertical plane by the help of a light string of length 2 m, fixed at one end. The breaking strength of string is 20 N. l=2m
(1) String does not break before reaching the lowest point.
(2) String breaks before reaching lowest point.
(3) the acceleration of the ball just after angle 11 cosisg 4 θ= .
(4) None of the above
6. Two identical blocks each of mass m , are connected by a light spring of constant k = 100 N/m. Initially, spring is compressed up to 0.2 m and then released. Which of the following statements is/are correct?
(1) The work done by the spring on the block A, when the spring comes in natural length, is 1 J.
(2) The work done by the spring on the block B, when the spring comes in natural length, is 1 J.
(3) The work done by the spring on the system (block A+block B) is 2 J.
(4) The work done by the spring on the system (block A+block B) is –2 J.
7. An elevator is going upward with constant velocity v0. One end of light spring of constant k is fixed to the ceiling of the elevator. A naughty boy inside the elevator compresses the spring through a compression x 0 and, in the mean time, the elevator moves by a distance s. The work done by the naughty boy with respect to the ground on the spring is
(1) 2 0 1 2 kx
(2) ()2 0 1 2 kxs +
(3) 0 1 2 kxs
(4) same as that in elevator frame
8. A body of mass m is moving along a circular path of radius R such that, at any instant, the kinetic energy 2 0 0 t KK t = , where t0
and K 0 are appropriate constants. Then, (1) the magnitude of tangential component of force acting on it must be constant
(2) the magnitude of centripetal force acting on the body will be directly propotional to t2
(3) after a long time, the resultant force will make a very small angle with the radius (4) speed of the particle remains constant
9. A particle slides down from rest on an inclined plane of angle ‘θ’ with horizontal and the distances are as shown. The particle slides down from the top to the position ‘A’ , where velocity is v. Then, choose the correct options. P S A H h q
(1) (V2–2gh) will remain constant.
(2) (V2–2gS sinθ) will remain constant.
(3) 22 VgSHh ps
(4) 22 VgSH p
will remain constant.
will remain constant.
10. A 0.1 kg block is pressed against a horizontal spring fixed at one end to compress the spring through 5 cm. The spring constant is 100 Nm–1. The ground is 2 metres below the spring. Which of the following are correct?
(Take g=10 ms–2) 2 m
(1) When released, the block shall have a kinetic energy of 1 J 8 .
(2) The initial horizontal velocity of the block is 51 ms 2 .
(3) The block shall reach the ground in 2 . 5 × S
(4) The block shall hit the ground at a horizontal distance of 1 metre from the free end of the spring.
11. A body of mass 1 kg is taken from infinity to a point P. When the body reaches that point, it has a speed of 2 ms –1. The work done by conservative force is –5 J. Which of the following is true (assuming non–conservative and pseudo-forces to be absent)?
(1) Work done by applied force is + 7 J.
(2) The total energy possessed by the body at P is + 7 J.
(3) The potential energy possessed by the body at P is + 5 J.
(4) Work done by all forces together is equal to the change in kinetic energy.
12. Which of the following is/are conservative force(s)?
(1) 23 Frr = (2) 5 F r =−
(3) ()
13. Figure shows a block of mass m and moves from position A to position B in the presence of force F through three different paths. The work done by the force F on the block is W1, W2, and W3 for paths 1, 2, and 3, respectively. If
(1) W 1 = W 2 = W 3 , then F is certainly conservative
(2) W 1≠ W 2≠ W 3, then F is certainly nonconservative
(3) W 1 = W 2 = W 3 , then F can never be conservative
(4) W 1 ≠ W 2 ≠ W 3 , then F is certainly conservative
14. A block of mass m is pushed up on smooth irregular inclined plane by applying force F 0 along the tangent on the path of the block. During moving of block from A to B, B
A m h
(1) the work done by applied force F is mgh if block is moving slowly
(2) the work done by applied force F is more than mgh if the speed of block at B is greater than that of point A
(3) the work done by applied force is lesser than mgh if the speed of block at B is less than that of point A
(4) the applied force F is non-conservative in nature
15. The potential energy v (in joules) of a particle of mass 1 kg, moving in xy plane, obeys the law v=3x+4y, when (x,y) are the coordinates of the particle in metres. If the particle is at rest at (6,4) at time t=0, then
(1) the particle has constant acceleration.
(2) the work done by the external force from the position of rest of the particle and the instant at which the particle is crossing x-axis is 25 joules
(3) the speed of the particle when it crosses the y-axis is 10 m/s
(4) the coordinates of the particle at time t = 4 s are (–18 m, –28 m)
16. The velocity of a particle of mass 10 –2 kg is given as () s ˆˆ 234m ˆ/=++ vxuyjzk (where x , y , and z are in metres). The amount of work done by applied force in displacing a particle from (1 m, 2 m, 0) to (–1 m, 3 m, 1 m) is
(1) 0.31 joules
(2) 0.24 joules
(3) 0.48 joules
(4) 0.72 joules
17. A particle starts to move along the path, which is given as x2=(8y)(x and y are in m), due to applied force () N ˆˆ =+ Faxibyj (a>0 and b>0; x and y are in metres). The amount of work done in displacing particle from x = 0 to x = 1 metre is
(1) +
64joules 128 ab
(2) +
8joules24 ba
(3) +
247joules 18 ab
(4) +
34joules 5 ab
18. A particle is dropped from height H . At a point, its kinetic energy is x times of its potential energy. Find the speed of the particle at that point. (Reference of PE is ground)
(1) [2gxH]1/2
(2) () 1/2 21gxH x +
(3) () 1/2 2 1 gH x
+
(4) () 1/2 2 1 gxH x
+
19. Two blocks of masses M and 2 M are connected to a light spring of spring constant K that has one end fixed, as shown in figure.
The blocks are released from the spring at its natural length condition. 2 m M K
(1) Maximum extension in the spring is 422 Mg K
(2) Maximum kinetic energy of the system is 222 Mg K
(3) Maximum energy stored in the spring is four times that of maximum kinetic energy of the system.
(4) When the kinetic energy of the system is maximum, energy stored in the spring is 422 Mg K .
20. One end of a light spring of force constant K is fixed to the wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is 12 2 Kx . Choose the possible cases:
(1) The spring was initially stretched by a distance x and finally was in its natural length.
(2) The spring was initially in its natural length and finally compressed by a distance x.
(3) The spring was initially compressed by a distance x and finally was in its natural length.
(4) The spring was initialy in its natural length and finally stretched by a distance x.
21. A block of mass m is pulled by an external agent by an inextensible string at point P, as shown in the figure. m a v p
(1) Kinetic energy of the block is 2 50 mv .
(2) Power delivered by the external agent is 25 mav
(3) Total power delivered by the string is zero.
(4) Work done by the string on the block is non-zero.
22. Two blocks are displaced through same distance along a rough horizontal surface by applying the forces, as shown in the given figure. m m
A B F qq
(1) Block A takes more time than block B.
(2) The net work done on both blocks is the same.
(3) The work done by applied force F on block A is more than that on block B.
(4) The average power supplied by the force F on the block A is more than that on block B.
23. A particle of potential energy shown in graph is moving towards right. The initial kinetic energy of particle is 21 J at x = 0. Mark the correct option(s).
(1) The particle accelerates for 0 ≤ x < 1 m.
(2) At x = 1 m, the particle has minimum KE energy.
(3) The kinetic energy is 41 J at x = 3 m.
(4) The particle is trapped within 0 < x ≤ 4.
24. A mass m is released with a horizontal speed v from the top of a smooth and fixed hemispherical bowl of radius r. The angle θ w.r.t. the vertical, where it leaves contact with the bowl, is
(1)
25. A simple pendulum is vibrating with an angular amplitude of 90°. For what values of θ with vertical is the acceleration directed?
1) Vertically upwards
2) Horizontally
3) Vertically downwards (1)
11 cos,0,90 3 (3)
90,11cos,0 3 (4)
11 cos,90,0 3
26. A small block slides with velocity 0.5 gr on the horizontal frictionless surface as shown in figure. The block leaves the surface, at point C. The angle θ in the figure is
(1) 14 cos 9
(2) 13 cos 4
(3) 11 cos 2
(4) None of the above
27. A particle initially at rest starts moving from point A on the surface of a fixed smooth hemisphere of radius r , as shown. The particle loses its contact with hemisphere at point B. C is the centre of the hemisphere. The equaion relating α and β is A r B b α C
(1) 3sin α=2cosβ
(2) 2sin α=3cosβ
(3) 3sin β=2cosα
(4) 2sin β=3cosα
Numerical Value Questions
28. A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain onto the table? (g = 10 m/s2)(in J)
29. A bucket filled with water weighing 20 kg is raised from a well of depth 20 m. If the linear density of the rope is 0.2 kg per metre, the amount of work done is _______. (g = 10 ms–2)(in J)
30. Force acting on a particle is () . ˆ 23 ˆ N ij + Work done by this force is zero, when a particle is moved along the line 3y + kx =5. Here, the value of k is ____.
31. Acceleration-time graph of a particle is shown. Work done by all the forces acting on the particle of mass m in time interval t1 and t2, while a1 and a2 are the accelerations at time t1 and t2. The the value of work done is
Then, the value of n is____.
length 200 m, whose axis is horizontal and 10 m above the surface of water of the swimming pool. The work done in filling half of the vessel is 6 1010.25 3 n J ππ
+
. Find the value of n.
33. A ladder 'AB' of weight 300 N and length 5 m is lying on a horizontal surface. Its centre of gravity is at a distance of 2 m from end A. A weight of 80 N is attached at end B. The work done in raising the ladder to the vertical position with end 'A' in contact with the ground is (in J) _______.
34. If the potential energy of a particle of mass 1 kg in a conservative field is 22 Uxy = ; all quantities are in SI. If, at a point defined by position vector () m ˆˆ =+ rij , then particle is observed moving perpendicular to r, find the magnitude of tangential acceleration in m/s2.
35. A body of mass 1 kg is dropped from a height of 5 m onto the ground. If the body penetrates 2 cm into the ground, the average resistance offered by the ground onto the body is__________. (g = 10 ms–2) (in N)
32. Water of density 1000 kg/m3 is to be lifted from a rectangular swimming pool of cross-sectional area 200 m 2 and stored in a cylindrical vessel of radius 1 m π and
36. A block of mass 2 kg is moving with speed v 0 towards a massless unstretched spring (K=10 N/m). It is found that, for maximum speed 06.4m/s, = v the block compresses the spring, stops at that position, and does not return. Friction coefficient at surface is 1 x µ= . The value of x is________.
37. A block A of mass 10 kg is placed over a long plank B of mass 20 kg. The system rests on a smooth horizontal surface. The coefficient of friction between blocks is 0.5. A horizontal constant force of 60 N is applied on the plank B. (Take, g = 10 m/s2) 10 kg A
B F 20 kg
If the block A and the plank B are taken as a system, find the work done by the friction on the system in 2 s.
38. A rope of mass m and length l = 0.9 m is shown in the given figure. The pulley is smooth and its radius is negligible. The initial distance of end A from the pulley is 3 l . Find the speed of the rope (in m/s) when the end A reaches the pulley.
Integer Value Questions
41. The free end of a flexible rope of length L and mass λ per unit length is released from rest in the position shown in the (a) part of the figure. If the velocity v of the moving portion of the rope in terms of y is 2
then the value of k is __________.
(b) (a) v L 2
39. A 2 kg stone is swung in a vertical circle by attaching it at the end of a string of length 2 m. If the string can withstand a tension of 140.6 N, the maximum speed with which the stone can be rotated is (in ms –1)_____.
40. The figure shows a smooth track, a part of which is a circle of radius R. A block of mass m is pushed against a spring of spring constant K fixed at the left end and is then released. The initial compression of the spring so that the block presses the track with a force mg when it reaches the point P, where the radius of the track is horizontal, is nmgR K . The value of n is _______.
42. A wedge with a rough circular track AB is fixed on the plane XX 1. The radius of the track is ‘R’ and the coefficient of friction of the track varies as µ= µ 0 x . The work done on mass m in moving from A to B is 01 mgR n µ
+
. Here, C is the centre of the circular track AB and x is the distance along the positive X-axis from origin ‘O’. Then, find the value of n.
43. A particle is projected along a horizontal field whose coefficient of friction varies as 2, A r µ= where r is the distance from the origin in metres and A is a positive constant. The initial distance of the particle is 1 m from the origin and its velocity is radially outwards. The minimum initial velocity at this point, so the particle never stops, is . kgA Find the value of k
44. The potential energy for field F is given by U ( x , y )=sin( x + y ). Find the force acting on the particle of mass m at a position given by coordinates 0,. 4 π (Consider all values in SI system)
45. A uniform chain of length L and mass M overhangs a horizontal table with its two-third part on the table. The friction coefficient between the table and the chain is μ. The work done by the friction during the period the chain slips off the table is () 2 MgL x −µ . Find the value of x
46. Two blocks m1 and m2 of 1 kg and 2 kg are connected by a spring of spring constant K on a rough horizontal surface of coefficient of friction 0.4. A horizontal force F is applied on m1 so that it could just move the block of mass m2. Find the value of F. F m1 m2
47. A block of mass 1 kg is attached to three unstretched springs of spring constants k 1 =40 N/m, k 2 =15 N, and k 3 =45 N/m, as shown in the figure. The block is displaced towards left on the horizontal smooth surface through a distance of 10 cm and is released. Find the maximum velocity during its motion, in m/s.
48. A small ball of mass 1 kg is suspeneded from a fixed point by a light string of length 5 m. It is thrown from the lowest point horizontally with a speed of 102m . The speed of ball, just before the string is slacking, is 10 m/s. n
Find the value of n.
49. A ball of mass m is rotated in a vertical circle with constant speed. The difference in tension at the top and bottom would be (in multiples of mg) ________.
50. A machine gun fires 360 bullets per minute. Each bullet moves with a velocity of 600 ms–1. If the power of the gun is 5.4 kW, the mass of each bullet is (in g) _______.
Passage-based Questions
Passage 1:
A car of mass 1000 kg, moving with constant acceleration, covers the distance between two points 60 m apart in 6 s. Its speed as it passes the second point is 15 m/s.
51. What was the kinetic energy of the car at the first point?
(1) 12500 J
(2) 6250 J
(3) Zero
(4) None of these
52. What was the kinetic energy of the car at 7.5 m prior distance from first point?
(1) Zero
(2) 12500 J
(3) 6250 J
(4) None of these
188
Passage II:
The potential energy U [in J] of a particle is given by (ax+by), where a and b are constants. The mass of the particle is 1 kg and x- and y-are the co-ordinates of the particle, in metres. The particle at t=0 is at rest at (4a, 2b).
53. Find the speed of the particle when it crosses y-axis.
(1) 422 ab + (2) ()2222 ab +
(3) () 22ab + (4) ()222 ab +
54. Find the coordinates of the particle at t = 1 s.
(1) (3.5a, 1.5b) (2) (3a, 2b)
(3) (3a, 3b) (4) (3a, 4b)
Passage III:
A block of mass m starts from rest with an acceleration. The acceleration–time graph of the block is shown in figure below.
55. The kinetic energy versus time graph is (1)
KE
(2) t KE
CHAPTER 7: Work, Energy, and Power
(3) t KE
(4) t KE
56. The power versus time graph is (1)
Matrix Matching Questions
57. An observer is situated in a car moving with constant acceleration a0=5 m/s2. A block A of mass m = 1 kg is moving on rough horizontal surface of coefficient of friction µ = 0.5. The car and block both start from rest. At t = 1 s, the change in kinetic energy of block A in the frame of car is 50 J .
(2)
(3)
(4)
F = 20 N a0=5 m/s2 µ=0.5
W2 = Work done by gravity in the frame of car in time t = 1 s
W4 = Work done by friction on block in the frame of car in time t = 1 s
Match Column-I with Column-II and select the correct option from the codes given below.
Column–I
A. W1=Work done by normal reaction in the frame of car in time t = 1 s
B. W3 = Work done by applied force F in the frame of car in time t = 1 s
Column–II
I) –25 J
II) zero
C. W5 = Work done by pseudo force on the block in the frame of car in time t = 1 s III) 50 J
D. W1 + W2 + W3 + W4 + W5 IV) 100 J
(A) (B) (C) (D)
(1) II IV I III
(2) III IV I II
(3) III IV II I
(4) IV III II I
58. In the given figure, the velocity versus time graph is shown for a particle of mass 1 kg.
Match column-I with column-II and choose the correct option.
Column–I
Column–II
A. Work done is maximum I) t = 1 s to t = 2 s
B. Power is maximum II) t = 3 s to t = 5 s
C. Work done is zero III) t = 2 s to t = 3 s
D. Force is maximum IV) t = 0 to t = 1 s
(A) (B) (C) (D)
(1) I III II,IV II
(2) III IV I II
(3) I III,IV II III
(4) II III I III,IV
FLASHBACK (Previous JEE Questions)
JEE Main
1. A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 60° by a force of 10 N parallel to the inclined surface as shown in figure. When the block is pushed up by 10 m along inclined surface, the work done against frictional force is : [g=10m/s2] (2024)
(1) 53 J (2) 5 J
(3) 5 × 103 J (4) 10 J
2. A block of mass 100 kg slides over a distance of 10 m on a horizontal surface. If the coefficient of friction between the surfaces is 0.4, then the work done against friction (in J) is: (2024)
CHAPTER 7: Work, Energy, and Power 190
(1) 4200 (2) 3900
(3) 4000 (4) 4500
3. A bob of mass ‘m’ is suspended by a light string of length ‘L’. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes half circle reaching the top most position B. The ratio of kinetic energies (..) (..) A B KE KE is: (2024)
(1) 3 : 2 (2) 5 : 1
(3) 2 : 5 (4) 1 : 5
4. A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is struck by a bullet of mass 10–2 kg moving with a speed of 2×102 ms–1. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use g=10 m/s2) (2024)
(1) 0.30 m
(2) 0.20 m
(3) 0.35 m
(4) 0.40 m
5. The potential energy function (in J) of a particle in a region of space is given as U = (2x2+3y3+2z) Here x, y and z are in meter. The magnitude of x-component of force (in N) acting on the particle at point P (1, 2, 3) m is: (2024)
(1) 2 (2) 6 (3) 4 (4) 8
6. A body of mass 2 kg begins to move under the action of a time dependent force given by () ˆ2ˆ66 FtitjN =+ . The power developed by the force at the tim e t is given by: (2024)
(1) (6t4+9t5)W (2) (3t3+6t5)W
(3) (9t5+6t3)W (4) (9t3+6t5)W
7. Two bodies of mass 4 g and 25 g are moving with equal kinetic energies. The ratio of magnitude of their linear momentum is: (2024)
(1) 3 : 5 (2) 5 : 4
(3) 2 : 5 (4) 4 : 5
8. A body of mass 1000 kg is moving horizontally with a velocity 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: (2024)
(1) 6 (2) 2
(3) 3 (4) 5
9. An artillery piece of mass M1 fires a shell of mass M2 horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is: (2024)
(1) M1/(M1+M2)
(2) 2 1 M M
(3) M2/(M1+M2)
(4) 1 2 M M
10. Two bodies are having kinetic energies in the ratio 16 : 9. If they have same linear momentum, the ratio of their masses, respectively, is (2023)
(1) 16 : 9 (2) 3 : 4
(3) 4 : 3 (4) 9 : 16
11. The ratio of powers of two motors is 3 , 1 x x + that are capable of raising 300 kg water in 5 minutes and 50 kg water in 2 minutes, respectively, from a well of 100 m deep. The value of x will be (2023)
(1) 2.4 (2) 16 (3) 2 (4) 4
12. Identify the correct statements from the following. (2023)
A. Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket is negative.
B. Work done by gravitational force in lifting a bucket out of a well by a rope tied to the bucket is negative.
C. Work done by friction on a body sliding down an inclined plane is positive.
D. Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity is zero. Work done by the air resistance on an oscillating pendulum is negative.
Choose the correct answer from the options given below.
(1) B and D only (2) B, D, and E only (3) B and E only (4) A and C only
13. A particle of mass 10 g moves in a straight line with retardation 2 x , where x is the displacement in SI units. Its loss of kinetic energy for above displacement is 10 J. n x
The value of n will be____. (2023)
14. The momentum of a body is increased by 50%. The percentage increase in the kinetic energy of the body is ________%. (2023)
15. A closed circular tube of average radius 15 cm, whose inner walls are rough, is kept in vertical plane. A block of mass 1 kg just fits inside the tube. The speed of block is 22 m/s, when it is introduced at the top of tube. After completing five oscillations, the block stops at the bottom region of tube. The magnitude of work done by the tube on the block is ___________ J. (Given: g = 10 m/ s2) (2023)
16. If the maximum load carried by an elevator is 1400 kg (600 kg-Passengers + 800 kgElevator), which is moving up with a uniform speed of 3 ms –1, and the frictional force acting on it is 2000 N, then the maximum power used by the motor is ____kW. (g = 10 m/s2) (2023)
17. If momentum of a body is increased by 20%, then its kinetic energy increases by (2022)
(1) 36% (2) 40% (3) 44% (4) 48%
18. A bullet of mass 200 g having initial kinetic energy 90 J is shot inside a long swimming pool, as shown in the figure. If its kinetic energy reduces to 40 J within 1 s, the minimum length of the pool the bullet has to a travel so that it completely comes to rest is (2022)
Water Pool
(1) 45 m (2) 90 m (3) 125 m (4) 25 m
19. Sand is being dropped from a stationary dropper at a rate of 0.5 kgs–1 on a conveyor belt moving with a velocity of 5 ms –1. The power needed to keep the belt moving with the same velocity will be (2022)
(1) 1.25 W (2) 2.5 W
(3) 6.25 W (4) 12.5 W
20. Two cylindrical vessels of equal crosssectional area 16 cm2 contain water up to heights 100 cm and 150 cm, respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process is [Take density of water = 10 3 kg/m3 and g = 10 ms–2] (2022)
(1) 0.25 J (2) 1 J
(3) 8 J (4) 12 J
21. A particle of mass 500 g is moving in a straight line with velocity v=bx5/2. The work done by the net force during its displacement from x = 0 to x = 4 m is (Take b = 0.25 m –3/2s–1) (2022)
(1) 2 J (2) 4 J (3) 8 J (4) 16 J
22. A block of mass m (as shown in figure) moving with kinetic energy E compresses a spring through a distance of 25 cm when its speed is halved. The value of spring constant of used spring will be nE Nm –1 , where n = _______. (2022)
Smooth surface
23. A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at 60° with vertical, as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be ________ms–1.(If g=10 m/s–1) (2022)
600 1 = 250 cm
24. A pendulum of length 2 m consists of a wooden bob of mass 50 g. A bullet of mass 75 g is fired towards the stationary bob with a speed v. The bullet emerges out of the bob with a speed 3 v and the bob just completes the vertical circle. The value of v is ______ms–1.
(if g = 10 m/s2) (2022)
25. A rolling wheel of mass 12 kg is on an inclined plane at position P and connected to
a mass of 3 kg through a string of fixed length and pulley, as shown in figure. Consider PR as friction-free surface.
The velocity of centre of mass of the wheel, when it reaches at the bottom Q of the inclined plane PQ, will be 1 m/s 2 xgh .
The value of x is __________. (2022)
26. If the kinetic energy of a moving body becomes four times of its initial kinetic energy, then the percentage change in its momentum will be (2021)
(1) 100% (2) 200%
(3) 300% (4) 400%
27. A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time t is proportional to (2021)
(1) 3 2 t (2) 1 2 t
(3) 1 4 t (4) 3 4 t
28. A porter lifts a heavy suitcase of mass 80 kg and, at the destination, lowers it down by a distance of 80 cm with a constant velocity. Calculate the work done by the porter in lowering the suitcase. (Take g=9.8 ms –2) (2021)
(1) –62720.0 J (2) –627.2 J
(3) +627.2 J (4) 784.0 J
29. Two persons A and B perform same amount of work in moving a body through a certain distance d with application of forces acting at angle 45º and 60º with the direction of displacement, respectively. The ratio of force applied by person A to the force applied by person B is 1 x . Then, the value of x is _________. (2021)
30. A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5:1. The velocity of the bob at the highest position is_________ m/s. (Take g = 10 m/s2) (2021)
31. The potential energy ( U ) of a diatomic molecule is a function dependent on r (inter-atomic distance) as αβ =−−1053, U rr where α and β are positive constants. The equilibrium distance between two atoms will be 2 , a b α
where a = ____. (2021)
32. Two solids A and B of mass 1 kg and 2 kg, respectively, are moving with equal linear momentum. The ratio of their kinetic energy(KE)A:(KE)B will be A/1, so the value of A will be _____. (2021)
JEE Advanced
33. A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height 3h from the ground, as shown in the figure. A spherical ball of mass m is released on the slide from rest at a height h from the top of the terrace. The ball leaves the slide with a velocity u0 =u0x and falls on the ground at a distance d from the building, making an angle θ with the horizontal. It bounces off with a velocity v and reaches a maximum height h 1. The acceleration due to gravity is g and the coefficient of restitution of the ground is 1 3 . Which of the following statement(s) is(are) correct? (2023)
(1) 02ˆ ughx =
(2) ()2ˆˆ vghxz =−
(3) θ=60°
(4) 1 23 d h =
34. A student skates up a ramp that makes an angle 30° with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed v0 and wants to turn around over a semicircular path xyz of radius R , during which he/she reaches a maximum height h (at point y) from the ground, as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then, (g is the acceleration due to gravity) (2020)
R z y x h
(1) 2 0 1 2 2 −= vghgR
(2) −= 2 0 3 2 2 vghgR
(3) the centripetal force required at points x and z is zero
(4) the centripetal force required is maximum at points x and z
35. A block of mass 2M is attached to a massless spring with spring constant k . This block is connected to two other blocks of masses M and 2M using two massless pulleys and strings. The accelerations of the blocks are a1, a2, and a3, as shown in the figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is x0. Which of the following option(s) is/are correct? [g is the acceleration due to gravity. Neglect friction] (2019)
(1) At an extension of 0 4 x of the spring, the magnitude of acceleration of the block connected to the spring is 3 10 g
(2) 0 4 xMg k =
(3) When spring achieves an 0 2 x extension of for the first time, the speed of the block connected to the spring is 3 5 M gk . (4) a2 –a1 =a1 –a3
36. A particle is moved along a path AB-BC-CDDE-EF-FA, as shown in figure, in presence of a force () N ˆ 2 ˆ =α+α Fyixj , where x and y are in metres and α=–1 Nm–1. The work done on the particle by this force F will be _______ joules. (2019)
37. A particle of mass m is initially at rest at the origin. It is subjected to a force and starts moving along the x-axis. Its kinetic energy K changes with time as dK t dt =γ , where γ is a positive constant of appropriate dimensions. Which of the following statements is (are) true? (2018)
CHAPTER 7: Work, Energy, and Power
(1) The force applied on the particle is constant.
(2) The speed of the particle is proportional to time.
(3) The distance of the particle from the origin increases linearly with time.
(4) The force is conservative.
CHAPTER TEST–JEE MAIN
Section-A
1. A force of 5 N is applied on a 20 kg mass at rest. The work done in the third second is
(1) 25 J 8 (2) 25 J 4
(3) 12 J (4) 25 J
2. A uniform chain of mass m and length l is kept on a horizontal table with half of its length hanging from the edge of the table. Work done in pulling the chain onto the table so that only
th 1 5 of its length now hangs from the edge is
(1) 8 mgl (2) 50 mgl
(3) 18 mgl (4) 21 200 mgl
3. A uniform rope of length l and mass m hangs over a horizontal table with two-third part on the table. The coefficient of friction between the table and the chain is μ. The work done by the friction during the period the chain slips completely off the table is
(1) 2 9 mglµ (2) 2 3 mglµ
(3) 1 3 mglµ (4) 1 9 mglµ
4. Two spheres, whose radii are in the ratio 1 : 2, are moving with their velocities in the ratio 3 : 4. If their densities are in the ratio 3 : 2, the ratio of their kinetic energies is
(1) 27 : 64 (2) 1 : 1
(3) 27 : 256 (4) 9 : 64
5. The kinetic energy of a body is K . If onefourth of its mass is removed and velocity is doubled, its new kinetic energy is (1) K (2) 3K
(3) 4K (4) 9K/4
6. A particle of mass m moves with a variable velocity v , which changes with distance covered x along a straight line as , vkx = where k is a positive constant. The work done by all the forces acting on the particle during the first t seconds is
7. A lorry and a car of mass ratio 4 : 1 are moving with KE in the ratio 3 : 2 on a horizontal road. Now, brakes are applied and braking forces produced are in the ratio 1 : 2. Then, the ratio of stopping timings of lorry and car is
(1) 1 : 1 (2) 1 : 3
(3) 26:1 (4) 62:1
8. n identical cubes, each of mass m and side l, are on the horizontal surface. Then, the minimum amount of work done to arrange them one on the other is
(1) nmgl (2) 2 ln 2 mg
(3)
9. A block is projected upward along a smooth inclined plane of inclination ‘α’. Initial gravitational potential energy of the block is zero. Draw the gravitational potential energy of the block versus time graph during total time of motion of the block on the inclined plane.
10. In case of a freely falling body, the ratio of kinetic energy at the end of the third second, to increase kinetic energy in the next three seconds, is (1) 1 : 1 (2) 1 : 2 (3) 1 : 3 (4) 1 : 9
11. One end of a spring of natural length h and spring constant k is fixed at the ground and the other is fitted with a smooth ring of mass m, which is allowed to slide on a horizontal rod fixed at a height h, as shown in figure. Initially, the spring makes an angle of 37° with the vertical when the system is released from rest. Find the speed of the ring when the spring becomes vertical. 4 cos 37 5
=
(1) hk m (2) 2 hk m
(3) 2 hk m (4) 4 hk m
12. A ring of mass m can slide over a smooth vertical rod, as shown in figure. The ring is connected to a spring of force constant 4 kmg R = , where 2 R is the natural length of the spring. The other end of spring is fixed to the ground at a horizontal distance 2R from the base of the rod. If the mass is released at a height 1.5 R, then the velocity of the ring as it reaches the ground is
slowly lowered to its equilibrium position. This stretches the spring by x. If the same object is attached to the same vertical spring but permitted to fall suddenly, the spring stretched by
(1) x/2 (2) x
(3) 2x (4) 4x
15. A block of mass m is moving with a constant acceleration on a rough plane. If the coefficient of friction between the block and the ground is μ, the power delivered by the external agent after a time t from the beginning is equal to
(1) ma2t
(2) μmgat (3) μm(a+ μg)gt (4) m(a+μg)at
16. A motor is used to lift water from a well of depth 10 m and to fill a water tank of volume 30 m3 in 10 minutes. The tank is at a height of 20 m above the ground. If 20% of energy is wasted, the power of the motor is
(1) 18.75 kW
(1) gR (2) 2 gR
(3) 2 gR (4) 3 gR
13. Two blocks A and B are released on the inclined plane of angle 30° and a circular track of radius R from different heights h1 and h2, respectively. The mass of each block is m. If F1 and F2 are the respective forces experienced by two blocks at the bottommost point of the tracks and F 1 = F 2, then find the value of h2 for R = 8 m.
(1) 1
(2) 2 m (3) 3 m (4) 4 m
14. An object is attached to a vertical unstretched light spring suspended from a ceiling and
(2) 20 kW
(3) 22.5 kW
(4) 37.5 kW
17. A body is revolving in a vertical circle with constant mechanical energy. The speed of the body at the highest point is 2rg . The speed of the body at the lowest point is (1) 7 gr (2) 6 gr
(3) 8 gr (4) 9 gr
18. A body is moving in a vertical circle of radius r by a string. If the ratio of maximum to minimum speed is 3:1 , the ratio of maximum to minimum tensions in the string is (1) 3 : 1 (2) 5 : 1
(3) 7 : 1 (4) 9 : 1
19. A mass of 0.1 kg is rotated in a vertical circle using a string of length 20 cm. When the string makes an angle 30° with the vertical, the speed of the mass is 1.5 ms –1 . The tangential acceleration of the mass at that instant is
20. The velocity of a body revolving in a vertical circle of radius r at the lowest point is 7 gr . The ratio of maximum to minimum tensions in the string is
(1) 8 : 1 (2) 4 : 1
(3) 7:1 (4) 1:7
Section-B
21. A particle is displaced from (2,0,2) to (5,4,2) under influence of force 8 ˆ 6 ˆ Fij =+ N. If W is work done (in J), then what is the value of W/10?
22. The momentum of a body is increased by 50%. The percentage increase in the kinetic energy of the body is _______ %.
23. A body of mass 1 kg travels on straight line path with velocity v=(2x2+3) m/s. The net work done by force during its displacement from x = 0 to x = 2 m is_____J.
24. A block of mass 2 kg, moving at a speed of 10 m/s, accelerates at 3 m/s2 for 5 s. Find its final kinetic energy in joules.
25. A ladder ‘AB’ of weight 300 N and length 5 m is lying on a horizontal surface. Its centre of gravity is at a distance of 2 m from end A. A weight of 80 N is attached at end B. The work done in raising the ladder to the vertical position with end ‘A’ in contact with the ground is _______ (in J).
CHAPTER TEST – JEE ADVANCED
2022 P2 Model
Section-A
[Integer Value Questions]
Interger Value Questions
1. The masses m 1 = 10 kg and m 2 = 5 kg are connected by an ideal string, as shown in figure. The coefficients of friction between m1 and the surface is µ=0.2. Assuming that the system is released from rest, calculate the velocity of blocks when m2 has descended by 4 m. [g = 10 ms–2] (in cm–1) m1 m2
2. The potential energy function for a particle executing linear simple harmonic motion is given by () 12 2 Uxkx = , where k is the force constant of the oscillator. For k = 0.5 Nm–1, the graph of U(x) on y-axis is shown figure. The particle has total energy of 1 J. Moving under this potential, it turns back when it reaches x=±____m.
U(x) Y X –2m–m x=0 + 1m + 2m
3. A circular track of radius 1 m is shown in the given figure. A small block is moving on the horizontal part of the track with speed 044m/s = v . The value of sin θ at which it leaves the track is x/5. Find the value of x. O q
4. A smooth hollow fixed hemisphere has a radius R of 23 m. A vertical smooth circular track of centre O is as such OC=r. The block is released from rest from point A. The minimum value of r, so that block competes its turn about O, is 3n m. Find the value of n. (Take θ=37°)
5. A pump motor is used to deliver water at a certain rate from a given pipe. To obtain twice as much water from the same pipe, in the same time, the power of motor has to be increased to n times. Find the value of n.
6. A system consists of two cubes of masses m 1 and m 2, respectively, connected by a spring of force constant k. Find the force (F), in newtons that should be applied to the upper cube for which the lower one just lifts after the force is removed.
(Take m1=0.1 kg, m2=0.2 kg, g=10 m/s)2
8. A wind power generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by blades into electrical energy. If wind speed is v, power delivered is proportional to v x. Then, the value of x is _________.
Multiple Choice MCQ
9. The kinetic energy of a body is given by equation K=(2t3) joules (t is in s). The correct graph shown between applied force (F) and time (t) is ___________.
7. One end of a light spring of natural length d and spring constant k = 64 N/m is fixed on a rigid wall and the other is attached to a smooth ring of mass m =1 kg, which can slide without friction on a vertical rod fixed at a distance d = 3 m from the wall. Initially, the spring makes an angle of 37° with the horizontal, as shown in fig. When the system is released from rest, find the speed of the ring when the spring becomes horizontal.
3 sin 37 5
10. An ideal spring supports a disc of mass M.
A body of mass m is released from a certain height, from where it falls to hit M. The two masses stick together at the moment they touch and move together from then on. The oscllations reach a height a above the original level of the disc and depth b below it. The constant of the force of the spring is
maximum angle that the string makes after hitting the pin, find cos θ.
(1) cos Ll Ll α+ +
(2) Lcosl Ll α+
(3) cos Ll Ll α−
(4) cos Ll Ll α− +
(1) 2mg ba
(2) mg ba (3) 2mg ab
(4) () 2 mg ab
11. A machine delivers power, given by
Ppt tt = + , where p0 are constants. The machine starts at t = 0 and runs forever. What is the maximum work that the machine can perform?
(1) Infinite
(2) Zero
(3) p0t0
(4) Cannot be predicted, data insufficient
12. A simple pendulum consisting of a mass M is attached to a string of length L. It is released from an angle α. A pin is located at a distance l < L below the suspension. If θ is the
13. Two blocks of masses M and 2 M are connected to a light spring of spring constant K that has one end fixed, as shown in figure. The blocks are released with the spring at its natural length condit ion.
2 m M K
(1) Maximum extension in the spring is 422 Mg K
(2) Maximum kinetic energy of the system is 222 Mg K
(3) Maximum energy stored in the spring is four times that of maximum kinetic energy of the system.
(4) When the kinetic energy of the system is maximum, energy stored in the spring is 422 Mg K .
14. A block m is kept stationary relative to cage on a rough inclined surface of an accelerating cage, as shown in the figure. At the given instant, m a=gtanq q
(1) normal reaction performs positive work on the block in a given time interval
(2) power delivered by normal reaction relative to cage is zero
(3) work done by frictional force on the block is negative in any time interval
(4) net work done by normal reaction and friction between the block and cage is zero in any time interval
Single Option Correct MCQs
15. A block of mass m is pulled along a horizontal surface by applying a force at an angle ‘θ’ with the horizontal. If the block travels with a uniform velocity and has a displacement d, and the coefficient of friction is μ, then find the work done by the applied force.
16. Two identical cylindrical vessels, each of area of cross section A , are on a level ground. Each contains a liquid of density ρ. The heights of liquid columns are h 1 and h 2 . If the two vessels are connected by means of a narrow pipe at the bottom, the work done by gravity in equalising the liquid levels is
■ Centre of Mass (COM) is a point where the entire mass of a system of particles is supposed to be concentrated.
■ The mass moment of a particle is the product of its mass and position vector about a reference point.
Fig. 1
In the figure, mr is the mass moment of a particle of mass m at P about ‘O’. (The direction of r is from point to the position of mass.)
Key Insights:
■ Algebraic sum of moments of all masses about centre of mass is equal to zero.
C represents equivalent position of total mass of the system of particles as a point mass.
■ Let F be the sum of all the external forces acting on different point masses of a system. Then acceleration of COM is F a M =
■ It means that centre of mass behaves as a point where entire mass of system is concentrated as a point mass and all the external forces applied on the system are exerted on it.
■ Important points about centre of mass: Mass is not actually concentrated at centre of Mass. It is a mathematical concept. It is just a point only.
■ For symmetric bodies with uniform mass distribution, centre of mass lies at the geometric centre.
Examples of centre of mass for different shapes are given in the following table.
of the Body
Plane
or Solid sphere
ring
or Solid cylinder
plate
C = Centroid
■ The center of mass may lie outside the material of the body, depending on mass distribution (e.g., rings, hollow spheres).
■ It depends on the system's mass distribution and is closer to denser regions.
■ For complex systems, like a classroom with students and benches, the center of mass is at a point based on mass concentration.
■ The center of mass follows the simplest path, even if individual particles move along complex trajectories.
CHAPTER 8: System of Particles and Rotational Motion
■ Rolling wheel
Example: In a rolling wheel or an Indian club's motion, the center of mass moves in a simple path, while particles move in more complicated paths.
The motion of the single point represents the motion of the entire body. This point within the boundary of the body is called the centre of mass or centre of inertia.
■ The sum of mass moments of all particles of a system about its centre of mass is always equal to zero.
Here, C is the centre of mass of system.
Fig. 4 Indian club motion
■ Centre of mass always lies within the boundary of the system.
Coordinates of COM of a System of Particles
■ Position of centre of mass of a system of particles does not depend upon position, orientation of rectangular co-ordinate system with respect to which the coordinates of centre of mass is given.
Solved example
1. Find the position of centre of mass of the system of 3 objects of masses 1 kg, 2 kg and 3 kg, located at the corner of an equilateral triangle of side 1 m. Take 1 kg mass object at the origin and 2 kg along the x-axis.
Sol.
\ Co-ordinates of centre of mass ( x cm, ycm)
Try yourself:
1. Four particles of masses 1 kg, 2 kg, 3 kg and 4 kg are placed at the four vertices A, B, C and D of the square of side 1 m. Find the coordinates of centre of mass of the particles.
Ans: (0.5 m, 0.7 m)
Centre of Mass of a Two Particle System:
■ Let us consider the origin of co-ordinate system to be at centre of mass and particles are on x-axis. Now xcm= 0, coordinates of m1, m2 are –x1, x2.
Fig. 7
■ Masses of particles are at inverse proportion to their distances from centre of mass.
Solved example
2. If the separation between carbon atom and oxygen atom in CO molecule is 1.4 Å, find the distance of its centre of mass from the C atom.
Sol. mC = 12 amu, m0 = 16 amu
Try yourself:
2. Two blocks of masses 10 kg and 30 kg are placed on x-axis. The first mass is moved on the axis by a distance of 2 cm towards right. By what distance should the second mass be moved to keep the position of centre of mass unchanged?
Ans: 2/3 cm, towards left
■ Let 'x' be the distance between two masses
12 xxx += and 12 21 mx mx =
■ For two particle system, 1122 12 cm mxmx x mm + = +
When m1, m2 are moved by distances Dx1, Dx2 then centre of mass moves by a distance
CHAPTER 8: System of Particles and Rotational Motion
D x cm here D x 1 , D x 2 and D x cm represent changes in x1, x2 and x cm respectively. Then, ()() 1122 12 cm mxmx x mm ∆+∆
∆= +
For many particle system, in general, ()()() 1122 12 . nn cm n mrmrmr r mmm ∆+∆+…+∆ ∆= ++…+
()ii i mr m ∑∆ = ∑
If the centre of mass remains at the same position even after displacements of m1, m2, then D x cm =0.
()() 11220mxmx ∴∆+∆=
()() 1122 mxmx ∆=−∆
–ve sign indicates that the two masses are displaced in opposite directions.
■ Consider two groups of particles containing N1 and N2 number of particles. M1 and M2 are masses of two groups and their centres of masses are at points C 1 and C 2 such that their co-ordinates are C 1( X 1,Y 1,Z 1) and C 2( X 2, Y 2, Z 2) respectively. Then, coordinates of COM of system of two groups are
1122 12 cm MXMX X MM + = +
1122 12 cm MYMY Y MM + = + 1122 12 cm MZMZ Z MM + = +
Solved example
3. When ‘n’ number of particles, each of mass ‘m’, are at distances x1 = 1, x2 = 2, x3 = 3, ...... x n = n units from origin on the x-axis, then find the distance of their centre of mass from origin.
Try yourself:
3. When ‘n’ number of particles of masses m, 2m, 3m,..... nm are at distances x1 = 1, x2 = 2, x3 = 3, ...... x n = n units, respectively, from origin on the x-axis, then find the distance of centre of mass of the system from origin. Ans: 1 3 21 n
8.1.2 Centre of Mass of Bodies with Cavity or Hole
■ Let M is the mass of the body before making a hole or cavity. Its centre of mass is 'C'. m1 is the mass that is removed in making the cavity and C1 is COM of m1 when it fills the cavity. m2 is mass of the remaining part of the body.Then ‘C ‘ will become COM of masses m1 and m2. Shift of COM is CC2 = x2
\ Shift in COM due to making of cavity is 11 22 2 mx xC m =⋅ COM always lies on the other side of the cavity w.r.t. ‘C’.
Example-1
Here, M = m1 + m2 For two particle system, m1x1 = m2 x2
■ From a uniform circular disc of radius R, centre C, a circular hole of radius 'b' is made with its centre C1 at a distance 'a' from 'C'. Shift of CM is CC2= S
If t, ρ represent thickness, density of plate, 22 1, mbtMRt =πρ=πρ and () 22 21 mMmtRb =−=πρ−
x1 = a and x2 = S m1x1 = m2x2 ()()222() btaRbtS ⇒πρ=π−ρ 2 22 ba S Rb ∴= If ()() 111222(1) muvmvu
Cavity
Example-2
■ From a uniform solid sphere of radius R1, and centre C, a spherical cavity of radius b with centre at C1 is made and C C1= a [Figure is similar the above case]
33 1 44 , 33 mbMR=πρ=πρ
Example-3
■ From a uniform solid cylinder AB of radius R , length 4 R , a hemispherical cavity of radius R is made at one edge.
CHAPTER 8: System of Particles and Rotational Motion
We know that BC1=3 R/8 11 313 2 88 RR xCCR ∴==−= C C2 = S = ? 3 1122 213 38 R
() 340 RsSR=πρ⇒=
Solved example
4. A thin disc of radius R/2 is placed on another disc of radius R. The two discs are made of same material and have same thickness. Find the distance of centre of mass of the system from the centre of the larger disc. Assume that the discs are very thin so that vertical shift of COM can be ignored.
Sol. Let s be the mass per unit area of each disc.
Solved example
5. A circular hole of radius R /2 is punched in a thin circular disc of radius R as shown in figure. Find the coordinates of COM of the remaining disc.
6andy0cc R x ==
So the coordinates of COM are (R/6, 0)
Try yourself:
4. Identical blocks, each of mass M and length L, are placed one above the other, such that each extends out by maximum length, as shown in figure. Find the maximum extension of the nth block from the top, so that the blocks will not fall.
Ans: x L n n = 2
8.1.3
Centre of Gravity
Sol. Let us consider two objects 1 and 2.
Object 1 is a solid disc of radius R having surface density of mass + s . Object 2 is a solid disc of radius R/2, having same thickness as object 1 but surface density of ‘– s’. Object 2 is made of a hypothetical material.
When object 2 is placed on object 1, the given object is formed.
■ The center of gravity (CG) is the point where the body’s total weight acts, and it coincides with the center of mass (COM) for small bodies. For large bodies like mountains and oceans, CG and COM do not coincide due to varying gravitational acceleration, with CG shifting toward regions of higher gravity.
Table 2: Distinction between centre of mass and centre of gravity
Centre of mass
(A) It is the point in the body where the total mass is supposed to be concentrated and at which, if an external force is applied, the body undergoes translatory motion.
Centre of gravity
(I) It is the point in the body through which the weight of the body is supposed to be acting.
Centre of mass Centre of gravity
(B) It refers to mass of the body.
(C) It depends on masses of the particles and their relative positions.
(II) It refers to weight of the body.
(III) It depends on the masses of the particles, their relative positions, and also the gravitational field.
8.1.4 Centre of Mass of Bodies of Continuous Mass Distribution
■ Co-ordinates of COM as
CM xdmxdm x dmM ∫∫ ==
CM ydmydm yM dm ∫∫ ==
and CM zydmydm dmM ∫∫ == ∫
■ The integration is to be performed under proper limits so that, as the integration variable goes through the elements covering the entire body.
Solved example
6. If the linear density of a rod of length L varies as ABx , find the position of its centre of mass.
Sol. Let the x -axis be along the length of the rod and origin at one of its ends. As rod is along x–axis, for all points on it, y and z coordinates are zero.
CHAPTER 8: System of Particles and Rotational Motion
Centre of mass will be on the rod. Now consider an element of rod of length dx at a distance x from the origin. Then, dmdxxdx =A +B X dm dm xA +B d A+ Bd CM 0 L 0 L 0 L 0 L
2BL 32A+BL L3A+ 2BL 32A CM 23 2 2 + +BL
Try yourself:
5. Find the distance of centre of mass of a uniform solid cone of height ‘ h ’ and base radius R , from the vertex on the line of symmetry is
Ans: 3h/4.
■ Proceeding in the similar manner, we can find the COM of certain rigid bodies. Centre of mass of some well known rigid bodies are given below:
■ Centre of mass of a uniform semicircular ring COM
Fig.15
■ Centre of mass of a uniform semicircular disc
■ Centre of mass of a hemispherical shell
■ COM of a uniform conical shell
Fig.20
■ COM of a uniform solid cone (or pyramid) of height h is at a height 4 h from its base.
8. h
■ Centre of mass of a solid hemisphere
■ Centre of mass of a uniform circular arc of radius R subtending an angle '2α' at the centre
Fig.21
Height of COM of a uniform triangular plate a bove the midpoint of the base is 3 h where h is the altitude of the triangle above this base. This is true from any one of the three bases.
Solved example
7. L-shap ed piece is cut from a thin metallic sheet as shown in figure. Find the coordinates of its COM.
Sol. Let us divide the object into two parts 1 and 2, as shown in the figure. If s be the surface density of mass,
CHAPTER 8: System of Particles and Rotational Motion
8.1.5 Motion of the Centre of Mass
Velocity of centre of mass is
Further, mv = momentum of a particle p . Therefore, equation (i) can be written as momentum of centre of mass
Acceleration of centre of mass is
eforethecoordinatesofCMare L L , 3 2
Try yourself:
6. A piece of sheet, whose shape is as shown in figure, is cut from a large thin sheet. Find the coordinates of its centre of mass.
Further, in accordance with Newton’s second law of motion, Fma = . Hence, equation (ii) can be written as
■ In the addition of the forces, all the internal forces will cancel each other as they exist in action - reaction pair.
■ Thus, the cent re of mass of a system of particles moves as it was a particle of mass equal to the whole system with all the external forces acting directly on it.
Y Explosion
Parabolic path of the projectile
Path of the COM of fragments
O x x1
Fig.2
■ The center of mass of projectile fragments continues along the same parabolic path as before the explosion, as the internal explosion forces do not affect it. The external force (gravity) remains unchanged, so the center of mass follows the same trajectory.
Solved example
8. A thin rod AB of length 5 m and mass m inclined to a vertical wall is slipping. At an instant, end B is moving on the floor with a velocity of 2 m/s. Find the magnitude of velocity of COM of the rod at this instant.
Velocity of particle A is given by Vsinq = 4cosq
[ \ The massless rod AB is rigid]
Sol. Since the rod is uniform, its COM must lie at its centre. In this case, the rod can be replaced by two identical particles A and B, each of mass m/2, connected by a massless rigid rod AB of length 5 m.
Try yourself:
7. An ob ject A is dropped from rest, from the top of a 30 m high building, and at the same moment, another object B is projected vertically upwards with an initial speed of 15 m/s from the base of the building. Mass of the object A is 2 kg while mass of the object B is 4 kg. Find the maximum height above the ground level attained by the centre of mass of the A and B system (take g = 10 m/s2).
Ans: 15 m.
Motion of a Man in a Boat
■ The mutual forces between the man and the boat are internal forces which do not move the centre of mass.
■ If x1, x2 are the distances of M, m from the origin (shore) then distance of their COM is ()() 12 cm Mxmx x....(1) Mm + = +
Fig.23
■ Suppose the man has moved a distance l w.r.t. the boat towards the shore. To make the position of COM constant, the boat moves away from the shore through a distance s w.r.t. the shore. Their final distances will become 11 xxs ′ =+ and 22 xxls ′ =−+ and ()() 12 12 (2) cm Mxsmxls x mm ++−+ = … +
On equating (1) and (2) and simplifying, we get ml s mM = +
■ Final distance of man from the shore is 22 xxls ′ =−+ 22 mlMlxlx mMmM =−+=− ++
Similarly final distance of the centre of boat from the shore is 111 ml xxsx mM ′ =+=+ +
CHAPTER 8: System of Particles and Rotational Motion
■ If the man moves with velocity of magnitude V relative to the boat, then magnitude of velocity of boat w.r.t. the shore is Vb. dl V dt = and b ds V dt = mldsmdl s mMdtmMdt =⇒= ++ bm m VV mM ⇒= +
Directions of V b and V m are always opposite.
■ If there are more-than one person on the boat, algebraic sum of the displacements will be equal to the net distance moved by the boat.
■ Example: Let m 1 ,m 2 be masses of two persons standing on the boat. Suppose m 1 moves towards the shore through a distance l 1 and m 2 moves away from the shore through a distance l2 w.r.t. the boat, 11 1 12 ml s mmM = ++ away form the shore 22 2 12 ml s mmM = ++ towards the shore
Net distance moved by the boat s=s1 –s2.
■ If s 1>s 2, the boat moves away from the shore
If s1 >s2, the boat moves towards the shore. If s1=s2, the boat remains at same position.
■ In this example, if both the persons move in the same direction, then s = s1+ s2 and the boat moves opposite to the direction of motion of persons.
Solved example
9. A block of mass m is placed on the top of a wedge of mass M and the system is placed on a horizontal surface. Base length of the wedge is L. If all the surfaces are smooth, by the time the block reaches the bottom, Find the displacement of the wedge.
Sol. Here, resultant horizontal external force is zero. Hence, horizontal shift of centre of mass is zero.
Consider a wedge of mass M and base length L resting on a horizontal surface.
A small block of mass ‘m’ is on the top of the wedge. All surfaces are smooth.
As the block slides down the incline plane, the wedge displaces backwards.
Here, centre of mass displaces only in vertical direction due to external gravitational force. However, since there is no external force along horizontal direction, centre of mass will not be displaced in that direction. So, by principle of moments,
m1x1 = m2x2 m[L – x] = Mxt
where x and (L – x) are horizontal displacements of block and wedge, respectively.
\ Displacement of wedge is x mL Mm
If L is the length of inclined plane and q is angle of inclination, then x mL Mm cos
Try yourself:
8. Find the acc eleration of centre of mass of the blocks of masses m1 and m2 (m1 > m2) in Atwood’s machine.
■ Center of mass is reference frame independent.
■ Without external force, it is at rest or moves with uniform velocity.
■ Internal forces don’t affect its position.
■ Moves in translatory motion, not rotatory.
■ Newton’s second law governs its motion.
Laboratory and Centre of Mass Frames of References
■ Centre of mass frame of reference:
The center of mass frame (C-frame) is when the origin is fixed at the center of mass of an isolated system.
In this frame, the position vector of CM is zero, so its velocity and linear momentum are also zero.
This frame is also called the zero momentum frame.
In the absence of any external force, the centre of mass frame is an inertial frame.
■ Relation between Velocities in Laboratory and C-frames: In laboratory frame, position vector and velocity of COM for two particle system are
■ In C-frame, let the position vectors of two masses be ′′
12 and rr respectively.
Characteriestics of Centre of Mass
■ Center of mass depends on masses, positions, and for continuous mass, shape and distribution.
■ Sum of moments about it is zero.
■ No mass particle at center of mass (e.g., uniform ring).
1122212 1 1212 mrmrmrr r mmmm
Similarly,
Therefore, velocities of particles in the centre of mass frame of reference are 1212 1 12 drmdrdr v dtmmdtdt
CHAPTER 8: System of Particles and Rotational Motion
Key Insights:
■ Linear momenta of the particles in C-frame are
Momentum of system in C-frame is
■ Total KE of system of two particles in C-frame: If 12 , vv
and cm V are velocities of two particles of masses m 1, m 2 and their centre of mass, respectively, in laboratory frame, velocity of the particles in C-frame are 1122cmcm vvvandvvv ′′ =−=−
KE of system in C-frame 22 1122 11 ' 22 mvmv ′ =+
■ KE of system 22 1122 11 22 mvmv=+ ()=+ 2 12 KE1 ofCOM 2 cm mmv
These two are not equal.
■ In the C-frame, colliding particles have equal and opposite momenta with collinear velocities.
■ In head-on collisions, the lines of motion are the same, while in oblique collisions, they are different.
■ Block of mass m sliding on smooth wedge of mass M:
■ Displacement of the wedge by the time m comes down is mL x mM = +
■ A balloon of mass M , a light rope, and a monkey of mass m are at rest in air as shown in figure. If the monkey reaches the top of the rope of length L then the balloon descends by a distance mL Mm +
Solved example
10. Particle 1 of mass 1 kg is moving along positive x-axis with velocity 1 m/s and particle 2 of mass 2 kg is moving along positive y-axis with velocity 2 m/s. Using the COM as frame of reference, find the velocity of COM.
Try yourself:
9. Particle A of mass 2 kg is moving along negative x -axis with velocity 2 m/s and particle B of mass 1 kg is moving along positive x -axis with velocity 1 m/s. Using the COM of the system as frame of reference, find velocity of COM.
Ans: 1 m/s
TEST YOURSELF
1. P is the centre of mass of a system of four point masses A, B, C and D, which are coplanar but not collinear.
a) P may or may not coincide with one of the point masses.
b) P must lie within or on the edge of at least one of the triangles formed by taking A, B, C, and D, three at a time.
c) P must lie on a line joining two of the points A, B, C, D.
d) P lies outside the quadrangle ABCD.
(1) a and b are correct.
(2) c and d are correct.
(3) a and c are correct.
(4) b and d are correct.
2. A thin uniform rod of length L is bent at its midpoint, as shown in figure. The distance of centre of mass from the point ‘O’ is
(1) πL
(2) L/8
(3) 2L
(4) πL/2
3. A uniform wire of length L is bent in the form of a circle. The shift in its centre of mass is
(1) L/π (2) 2L/π (3) L/2π (4) L/3π
Answer Key
(1) 1 (2) 2 (3) 3
8.2 COLLISION
■ A collision involves strong interaction, potentially without physical contact, causing significant deformation.
■ There are three stages: before, during, and after, with large forces during and zero forces before/after.
■ Conservation of momentum helps relate initial and final velocities, as forces are often unknown.
8.2.1 Types of Collision
Types of Collisions on the Basis of Law of Conservation of Kinetic Energy
■ Collisions are classified into elastic (kinetic energy remains constant) and inelastic (kinetic energy changes) based on the change in kinetic energy.
Perfectly Elastic Collisions
■ A perfectly elastic collision occurs when both linear momentum and kinetic energy are conserved, with no loss of kinetic energy.
■ Examples: Collision between atomic particles
Inelastic Collision
■ Inelastic collision: Kinetic energy is not conserved, but linear momentum is conserved; total energy is conserved.
■ Perfectly inelastic collision: The bodies stick together after the collision, with momentum conserved but kinetic energy not conserved. Example: a bullet embedding in wood.
■ Classification by direction:
■ Head-on (one-dimensional): Motion along the same straight line before and after the collision.
■ Oblique: Motion not along the initial line of motion, can be two-dimensional or three-dimensional.
■ Line of impact: The line perpendicular to the common tangent at the point of contact between colliding bodies.
■ Analysis of collision:
■ Force and momentum: Bodies experience strong forces, changing momentum, but total momentum is conserved.
■ Energy loss: Mechanical energy is mostly not conserved; sound, light, and heat are generated.
■ Perfectly elastic collision: Kinetic energy remains constant before and after impact.
■ Non-perfectly elastic collision: Kinetic energy changes.
■ Perfectly inelastic collision: Bodies move together with the same velocity, often losing kinetic energy.
■ Relative velocity: Velocity of separation after collision may differ from velocity of approach.
■ System of particles: All particles involved in a collision form a system, with deformation and recovery impulses during impact.
Deformation period Recovery period
■ The period of impact is the time interval during which two colliding bodies remain in contact before separating.
■ During the collision, bodies with unequal velocities press against each other, causing local deformation, and the period of compression continues until both bodies attain the same velocity.
u1
u2
Oblique impact
■ The period of restitution is the time between when two bodies attain equal velocities and when they fully separate, after transferring momentum and "pushing" against each other.
Consideration of Line of Impact
■ The line of impact is found by drawing a tangent at the point of contact, then a normal to the tangent, which indicates the direction of the impulsive force during collision.
■ Identify the point of contact just before the collision.
■ Draw the common tangent at the contact point and then a normal to the tangent.
■ The line of impact is the normal, and the relative velocity components along this line are used to calculate the coefficient of restitution (e).
() () 21 12 n n v e u = , where (u12)n and (v21)n are the magnitudes of com ponents of relative velocity between them along the line joining them just before and just after the impact, respectively. ( u 12 ) n is known as velocity of approach and ( v 21)n is known as velo city of seperation.
8.2.3 Perfectly Elastic Head-on Collision between Two Bodies u1 m1
Before collision m2 u2 υ1 υ2
After collision m1 m2
■ Let us consider an elastic one-dimensional (head –on) collision between two particles. Let m1 and m2 be the masses of two particles. Suppose u1, u2 and v1, v2 be their respective velocities before and after the collision.
■ Applying the law of conservation of linear momentum, we have 11221122 mumumvmv +=+ or ()() 111222...(1) muvmvu −=−
■ According to law of conservation of kinetic energy, we have 2222 11221122 1111 2222 mumumvmv +=+ or ()()() 2222 111222....2 muvmvu −=−
■ Th us, in an elastic collision, in one dimension, the relative velocity of approach before collision is equal to the relative velocity of separation after collision. From eq. =+− (),1221 avvuu
Substituting this value of v1 in eq. (1), we get ()() 11221222 muvuumvu −+−=−
■ When m 1= m 2 = m , then from eq. (b) and (c) v1 = u2 and v2 = u1
■ When u2 = 0, i.e., the second body is at rest, we have () 11 2 12 2mu v mm = + and () () 121 1 12 mmu v mm = +
Now, in this case, we consider the following conditions:
■ If m1 = m2 = m, then v2 = u1 and v1= 0
If m2>>m1, then v1= –u1 and 20 v If m1>>m2, then v1= u1 and v2 = 2u1
CHAPTER 8: System of Particles and Rotational Motion
Solved example
11. A sphere of mass 2 kg moving with a velocity collides elastically with another sphere which is at rest. If after the collision, the first sphere moves in the same direction with its velocity reduced to one third of its initial velocity, find the mass of the second sphere.
Sol. Here m1 = 2 kg, u1 = u1, u2 = 0, vu 11 1 3 =
In elastic collisions
12. T wo perfectly elastic spheres of masses 2 kg and 3 kg moving in opposite directions with velocities 8 ms–1 and 6ms–1 respectively collide with each other. Find their velocities after the impact.
Sol. Here m1 = 2 kg, m2 = 3kg, u1 = 8ms–1, u2 = –6 ms–1
According to the law of conservation of linear momentum.
m1u1 +m2u2 =m1v1 +m2v2
i.e.,(2)(8) + (3) (–6) = 2 v1 + 3 v2
⇒ 2v1 + 3v2 = –2 .... (i)
Since the collision is elastic, u1 – u2 = v2 – v1
i.e., 8 – (–6) = v2 – v1
⇒ v2 – v1 = 14 .... (ii)
On solving the equations (i) and (ii) we get v1 = –8.8 ms–1 , v2 = 5.2 ms–1
Try yourself:
10. A particle of mass m moving with a speed v hits elastically another stationary particle of mass 2m on a smooth horizontal circular tube of radius r. The time in which the next collision will take place is equal to
■ The value of e depends on the nature of the colliding bodies.
■ e is a number. It has no units and it is dimensionless
Determination of Coefficient of Restituion
■ To determine the coefficient of restitution between two materials ,one of them is taken in the form of a very heavy plate and the other in the form of a small sphere. The small sphere is dropped onto the plate from a height 'h'.
Try yourself:
11. A small bl ock of mass M moves with velocity 5 m/s towards another block of same mass M placed at a distance of 2 m on a rough horizontal surface. Coefficient of friction between the blocks and ground is 0.25. Collision between the two blocks is elastic, the separation between the blocks, when both of them come to rest, is (g = 10 m/s2)
Ans: 3 m
8.2.4 Coefficient of Restitution = relativevelocityofseperationaftercollision relativevelocityofapproachbeforecollision e 21 12 vv e uu =
Key Insights:
■ Coefficient of restitution is termed degree of elasticity.
■ Practically, e lies between 0 and 1 .
■ For a perfectly elastic collision, e = 1
■ For a completely inelastic collision, e = 0
■ The equation of coefficient of restitution is applied in the direction of line of impact of the bodies.
■ It hits the plate with a velocity v1 (as shown in Fig. 4). Let the sphere rebound to a height ' h 2 ' after collision with the plate. Assume the velocity of the plate before and after collision to be zero (u2 = v2 = 0) since the plate is very heavy.
Fig.4
direction to u1
■ Equation for the height attained by a freely falling body after number of rebounds on the floor
A body is dropped from a height h0. The body strikes the ground and it rebounds to a height h1. It again falls from that height and rebounds to a height h2 and so on. h1 h2 h 3 h 4 h 5
Fig. 5
222 vuas −= 22020 vgh⇒−= 20vgh =
The velocity with which it strikes the ground just before 1st impact is
20vgh =
21 12 vv e uu =
1 0 0 20 v e gh =
10 2 vegh =−
Let the ball rebound to a height h1
222 vuas −=
2 2 01022() eghgh ⇒−=−
2 0122 eghgh⇒⋅=
2 10 heh⇒=
Similarly, after 2nd rebounce, the body rises to a height h2, which is given by 42(2) 200 heheh ==
CHAPTER 8: System of Particles and Rotational Motion
Similarly, 62(3) 300 heheh ==
2
0 n n heh =
■ Let V, V1 be the velocities of ball just before and just after 1st impact with the ground.
11 1 0 0 VV eVeV VV ==⇒=
If V2 is the velocity after 2nd impact, V2 = eV1 = e(eV)=e2V
Similarly, velocity of the ball just after nth impact if V n =enV
■ Total distance travelled by the body before it stops bouncing is 2 02 1 1 Hhe e
■ Total time taken by the ball to stop bouncing is
Solved example
13. A ba ll is dropped from a height h on to a floor. If, in each collision, its speed becomes e times of its striking value (a) Find the total change in momentum of the ball (b) Find the average force exerted by the ball on the floor.
Sol. (a) Change in momentum in I collision mvmvmvv 1010
Change in momentum in II collision, = m (v2 + v1)
Change in momentum in nth collision, = m (v n + vn–1)
Adding these all, total change in momentum of the ball is
pmvvvv
s or, v vevvev
pmvee a nn011 0 2 22 122 ... ......
pmve e p 102 2 0 012 1 1 , m mghe e 2 1 1
(b) Nowasso, Weknow, Fdp dt Fp T T h g e e av 21 1 f from,12 1 1 2 1 12 1 1 pmghe e Fmghe e g h e av e mg
Solved example
14. A body ‘ A ’ with a momentum ‘ P ’ collides with another identical stationary body ‘ B ’ one dimensionally. During the collision, ‘ B’ gives an impulse ‘J’ to the body ‘A’ . Then the coefficient of restitution is
Sol. According to law of conservation of linear momentum, m1u1 + m2u2 = m1v1 + m2v2
i.e., mu + m(0) = mv1 + mv2
⇒ P – P1 = P2 where P2 = J, given \ Coefficient of restitution, e vv u mvmv mu PP P PPP P PP P e JP P J P 212121 2 22 1 222
Try yourself:
12. A steel ball is allowed to fall freely from a height of 36 cm onto a smooth floor. Find the height to which it rises after rebounding from the floor. Coefficient of restitution between the steel ball and the floor = 0.60.
Ans: 0.13 m
8.2.5 Inelastic Head-on Collision
■ Two bodies of masses m1 and m2 are moving with velocities u 1 and u 2 ( u 1 > u 2). They collide as a result of this collision, their velocities change to v1 and v2 respectively. u1 v1 v2
u2
Before collision During collision After collision m1 m1 m
Fig.6
11221122...(1) mumumvmv +=+
21 12 ; vv e uu = ()2112 vveuu −=−
2112....(2) vveueu =+−
1212....(3) vveueu =−+
Substitute v2 and v1 from (2) and (3) in (1) and on simplifying we have, 1222 11 2112 (1) memmeu vu mmmm
−+ =+ ++ 1121 22 1221 (1) meumem vu mmmm
+− =+ ++
Key Insights:
■ If m1= m2 = m and u2 = 0 11 12(1);(1) 22 uu veve =−=+ 1 2 1 1 ve ve = +
8.2.6 Perfectly Inelastic Head-on Collision
■ In a collision, if the bodies stick together or one gets embedded into other and move with common velocity after collision, it is called a perfect inelastic collision.
u1 u2
Before collision After collision m1 m2 V
■ Applying the law of conservation of linear momentum, we have
m1 u1+ m2 u2 = (m1 + m2)v 22 1122 11 . 22 KEmumu =+ common velocity
Initial KE 22 1122 11 22 mumu=+
Final KE () 2 12 1 2 mmv=+
⇒∆= +
Loss of KE = KEi – KEf ()() 122 12 12 KE 2 mm uu mm
Key Insights:
■ In the above case, if the two bodies move in opposite directions, then ()()
∆=+ + 122 12 12 KE 2 mm uu mm
■ If 'e' is the co-efficient of restitution in a collision, then loss in KE is ()()()
∆=−− + 1222 12 12 KE1 2 mm uue mm when they move in same direction initially.
■ For any collision, general formula for loss in KE is
∆=−− + 1222 12 12 KE1 2 mm uue mm
Solved examples
15. Ball 1 collides with an another identical ball 2 at rest as shown in figure. For what value of coefficient of restitution e, the velocity of
CHAPTER 8: System of Particles and Rotational Motion
second ball becomes two times that of 1 after collision? 1 2
Sol. Here, m1 = m2 and u2 = 0.
After collision, v e u 2 1 2 and v e u 1 1 2 Given , v2 = 2 v1 on solving we get, e = 1 3
16. 6.65 A sphere of mass 50 × 10–3 kg moving with a velocity of 2 ms–1 hits another sphere which is at rest. Assuming the collision to be head–on collision and if they stick together after collision and move in the same direction with a velocity of 0.5 ms –1, find the mass of the second sphere
Sol. m1 = 50 × 10–3 kg, u1 = 2 ms–1, u2 = 0, v = 0.5 ms–1, v = v1 = v2 = 0.5 ms–1. m1u1 + m2u2 = (m1 + m2) v 501020 501005 3 2 3 2 m m . m2 5033 102501005 05 10025 05 101501033 . kg
Try yourself:
13. After perfectly inelastic collision between two identical particles moving with same speed in different directions, the speed of the combined particle becomes half the initial speed of either particle. The angle between the velocities of the two before collision is Ans: q = 120°
Try yourself:
14. A particle of mass 0.1 kg moving with an initial speed v collides with another particle of same mass kept at rest. If after collision total energy becomes 0.2 J, then the minimum and maximum values of ‘ v’ are
Minimum value of v is 2 m/s in case of perfectly elastic collision and maximum value of v is 22 m/s in case of perfectly inelastic collision.
Ans: q = 2 m/s
8.2.7 Ballistic Pendulum
■ A ballistic pendulum determines the velocity of a bullet by embedding it into a suspended wooden block and measuring the height it rises, using the conservation of linear momentum.
The system rises to a height h, as shown in the figure. Applying the law of conservation of linear momentum, we have
Using conservation of mechaical energy, KE at the lowest point = PE at highest point
KE of the system after collision is given by 12KE() 2 mMv=+
PE at highest point =(m + M) gh
Loss of KE = KEi – KEf
Solved example
17. A bullet of mass 2 g travelling with a velocity of 500 ms–1 is fired into a block of wood of mass 1kg suspended from a string of 1 m length. If the bullet penetrates the block of wood and comes out with a velocity of 100 ms–1, find the vertical height through which the block of wood will rise (assuming the value of g to be 10 ms–2).
Sol. Let the masses of the bullet and the block be ‘ m ’ and ‘ M ’ respectively. Let their velocities after the impact be v and V respectively. Let the initial velocity of the bullet be ‘u’.
According to the law of conservation of linear momentum. mu = mv + MV
Here, m = 2 × 10–3 kg, u = 500 ms–1, v = 100 ms–1
210500210100133 V
V = 0.8 ms–1
When the block rises to a height of ‘h’, according to the law of conservation of energy.
Try yourself:
15. A wooden block A of mass m is suspended by a string of length ℓ. Another wooden block B moving horizontally strikes block A and sticks to it. If the combined mass rises to a height h above the initial position of A, find initial velocity of B.
Ans: 22 gh
8.2.8 Oblique Collision
■ Let 'u' and 'v' be the velocities of the sphere before and after the impact, making angles ‘q‘ and 'α' respectively with the normal at C.
u Normal at the point of contact C v A C B qα
■ The components of u , along and perpendicular to the normal at C are ucosq and usin q respectively.
■ Similarly, the components of v along and perpendicular to the normal at C are v cos α and v sin α respectively.
■ Now, the components of velocities along the plane AB remain unchanged, as there is no friction.
Thus u sin q = v sin α ...(1)
CHAPTER 8: System of Particles and Rotational Motion
■ However, the components of velocities normal to plane AB, change according the Newton’s experimental laws.
Thus, (cos0)(0cos) veuα−=−θ cos(cos)veu ⇒α=−θ or v cos α = eu cos q (numerically) ... (2)
From equations (1) and (2)
222 sincos....(3)ue =θ+θ and tan tan....(4) e θ α=
■ Equations (3) and (4) give the magnitude and direction of the velocity after impact.
■ Special cases:
■ If the impact is direct (q = 0), then v = eu and a = 0.
■ For an elastic collision (e = 1), v = u and a = q, meaning velocities and angles of approach and separation are the same.
■ For a perfectly inelastic collision, e = 0.
Equations (3) and (4) yield that v = u sin q and α = 90°.
Loss in K.E. Due to the Impact
2 221cos 2 u me =−θ
■ Special cases:
■ For a perfectly elastic collision e = 1
\ Loss in KE = 0
■ For a perfectly inelastic collision e = 0 θ ∴= 22 cos LossinKE 2 mu
Impulse Due to Impact
■ The impulse of the impact, normal to the plane AB = Change in momentum along the normal to the plane AB
= mv cos α – m(–u cos q ) [in the upward direction]
= mv cos α + u cos q ]
= m[eu cos q + u cos q] = m(1 + e) u cos q
Solved example
18. Two billiard balls of the same size (radius r) and same mass are in contact on a billiard table. A third ball also of the same size and mass strikes them symmetrically and remains at rest after the impact. The coefficient of restitution between the balls is
From conservation of linear momentum
mumvv u e == = 230 3 0 cosor
Now relativevelocityofseparation relat tivevelocityofapproach in common normal direction
Hence, e v u u u === cos / 30/ 3 32 2 03
Try yourself:
16. A ball of mass m hits a floor with a speed v making an angle of incidence q with the normal . The coefficient of restitution is ‘ e’. Find the speed of the reflected ball and the angle of reflection of the ball.
Ans: evv e ’scoins, ta n22 21 ’=tan
Applications of Oblique Impact
■ Application-I: Consider the collision between block A and ball B. Block A is constrained to move in horizontal direction and ball B is free to move in the plane of the figure. All the surfaces have been considered smooth. At the instant of collision, action and reaction force F (internal force) is exerted on block A in vertical direction by ground. Following equations can be set up: [t = tangential, n =normal]
Fig.10
VA, VB are initial velocities of block and ball respectively and V' A, V' B are final velocities.
■ The component along t -axis of the momentum of ball B is conserved since friction is absent. Hence the t component of the velocity of ball B remains unchanged, i.e., (VB)t = (V'B)t
■ The component along the horizontal x-axis of the total momentum of block A and ball B is conserved,
i.e., mAVA+ mB (VB)x= mAVA'+ mB(VB')x
■ The component along the n -axis of the relative velocity of block A and ball B after impact is obtained by definition of coefficient of restitution,
i.e., (V'B)n –(V'A)n = e [(VA)n–(VB)n]
■ Application-II: Suppose a ball is projected with speed u at an angle q with horizontal. It collides at some distance with a wall parallel to y-axis as shown in Fig. Let V x and V y be the components of its velocity along x and y directions at the time of impact with wall.
■ Co-efficient of restitution between the ball the wall is e. Component of its velocity along y direction (common tangent) V y will remain unchanged while component of its velocity along x-direction (common normal) V x will become eV x in opposite direction.
CHAPTER 8: System of Particles and Rotational Motion
Solved example
19. Prove that in case of oblique elastic collision of two particles of equal mass if one is at rest, the recoiling particles always move off at right angles to each other.
Sol.
■ Further, since V y does not change due to collision, the time of flight (time taken by the ball to return to the same level) and maximum height attained by the ball will remain same as it would had been in the absence of collision with the wall.
Thus, tOAB = tCD+ tDEF and HA= HE
In elastic collision momentum is conserved. So, conservation of momentum along x - axis yields mumvmv uvv 1122 11221 coscos ,cosco i.e.s andalongg-axisyields y vv 02 1122 sinsin
Squaring and adding Eqns. (1) and (2), we get
21212 cos .... (3)
As the collision is elastic
i.e.,
.... (4) Put
in Eqn (3) , we get 20 1212vv
As it is given that vv 1200≠≠ and So,i.e co. s,,1212090 0
Try yourself:
17. A steel ball A of mass 2 kg is rest on a smooth horizontal floor. Another steel ball B of mass 1 kg, moving with a velocity of 2 m/s striks A obliquely and clasifically. After the collision, direction of velocity of B makes an angle of 60° with its original direction of motion. Find velocity of A after the inpact.
1Ans: m/s
TEST YOURSELF
1. Choose t he correct statements.
A) For any collision, momentum transfer takes place.
B) In any collision, energy transfer takes place.
C) Impulsive forces are involved in collisions.
(1) A and C are true.
(2) A and B are true.
(3) A, B and C are true.
(4) A, B, and C are false.
2. During a collision, a relatively large force acts on each colliding particle for a relatively _____time
(1) short
(2) moderately short (3) large
(4) depends on a particular case
3. A heav y sheet of wood and a light ball is moving towards each other, as shown in the figure. What will be the speed of the ball after col lision?
(1) 3v (2) 2v (3) v (4) v/2
Answer Key
(1) 1 (2) 1 (3) 1
8.3 ROTATIONAL MECHANICS
■ A rigid body is an ideal object that maintains its shape regardless of external forces, and in its motion, the distance between any two parts remains constant.
■ For planar motion, a rigid body can undergo pure translation, rotation, or rolling, with all points rotating by the same angle and instantaneous angular velocity.
8.3.1 Kinematic Analysis of a Rigid Body
■ Any kind of motion of a rigid body can be visualized in a two stage process (we can procede in any convenient order):
■ Stage-I: Find complete motion of centre of mass for a given time period.
■ Stage-II: Find instantaneous state of the body as a whole at any given time.
■ Kinematic eq u ations of a rigid body in rotation are same as that of an isolated particle in circular motion
Angular
Velocity and Angular Acceleration
■ In a small time dt, the particle moves along the arc of a circle by a distance dsi, where, dsi = vi dt ....(1) where vi is the sp eed of the particle. The angle swept out by the line from the centre to the particle, in radians, is this distance divided by the radius r i: ....(2)
■ The rate of change of angle with respect to time, dq/dt, is the same for all the particles of the wheel. It is called the angular velocity w of the wheel,
■ From equ ation (2) the speed of the i th particle is related to its radius r i and the angular velocity of the wheel by ...(4)
In vector notation Vr =ω×
■ If a par ticle has velocity V while it is rotating in the plane of the paper with its radius vector r , then its angular velocity ω will be in a plane perpendicular to the plane of the paper along the axis of rotation.
Solved example
20. A particle is moving anticlockwise in a circular path of radius 2 m with an angular velocity w = 5 rad/s. Find the vector expression of its velocity when the radius vector of the particle makes an angle of 60° with x-axis.
CHAPTER 8: System of Particles and Rotational Motion
Vjims P P 53 553 k/ /.
Angular acceleration ( α ): We define angular acceleration a as the time rate of change of angular velocity. Thus, d dt
Try yourself:
18. A particle is moving in a circular path, about z-axis with angular velocity 10 rad/s. Plane of the circular path is parallel to xy-plane. Find the velocity of the particle when its co-ordinates are (1 m, 1 m, 1 m)
Ans: 10(–î + ĵ) m/s
■ The rate of change of angular velocity with respect to time is called the angular acceleration α. For rotation about a fixed axis,
■ The units of angular acceleration are radians per second squared. The relation between the tangential linear acceleration of the ith particle of the wheel and the angular acceleration is obtained by taking the
derivative of the speed (not velocity) vi in equation (4) with respect to t:
i iti dvd ar dtdt ω ==
Thus ait = r iα
■ Each particle of the wheel also has a radial linear acceleration or the centripetal acceleration which points inward along the radial line, and has the magnitude.
2 2 i ici i v ar r ==ω
0tan d const dt ω α==α= then w = w 0 + α0 t where w0 is the angular velocity at t = 0, and 2 000 1 2 ttθ=θ+ω+α
Angular acceleration: () 22 000 2 ω=ω+αθ−θ
8.3.2 Kinetic Energy of a Body about a fixed Axis-Moment of Inertia
Speed of the particle (say v) is: v = r w
v = rw r Fig.9
Its kinetic energy becomes 22211 22 dKEdmvdmr==ω
Then total kinetic energy of the body is 122 2 KEdKEdmr=∫=∫ω
But ω is same for all the particles in a rigid body. ()222) 11......(15 22 KErdmI ∴=∫ω=ω where 2....(16)Irdm =∫ and for a system of descrete particles, 2...(17)ii Irm =∑
■ The new physical quantity I is called second ordered moment of mass or moment of inertia.
■ If we look at the Eq (15) for kinetic energy of a rotating body it reminds us the similar expression for kinetic energy in “Particle Dynamics”: 12 2 KEmv =
■ Moment of inertia depends on the body's mass, axis of rotation, and mass distribution about the axis.
■ A heavier body or a hollow body has a greater moment of inertia, and a person’s moment of inertia decreases when folding hands and increases when stretching them.
Solved example
21. A rigid body is rotating about a fixed axis and its kinetic energy of rotaion is 100 joule. If its angular momentum changes by 50%, what will be its final kinetic energy?
Sol. Initially,joule Finally L I k L I L fI 2 22 2 100 15 2 225 2 ,. 2 22.5100joule=225joule.
Try yourself:
19. If angular m omentum of a body spinning about a fixed axis changes by 1%, find the corresponding change in its kinetic energy. Ans: 2%
8.3.3 Radius of Gyration
■ The radius of gyration is the distance from the axis of rotation at which the entire mass of a body could be concentrated to have the same moment of inertia denoted by K.
Solved example
22. Four particles, each of mass 100 g, are placed at the corners of a square of side 10 cm. Find the moment of inertia of the system about an axis passing through the centre of the square and perpendicular to its plane. Also find the radius of gyration of the system.
41 2 0 0052210 210 04 0077 232 3 () . kg cm. m kI M m
Try yourself:
20. The radius of gyration of a system of particles about z -axis is 0.5 m. If total mass of the system is 2 kg, find momentum of inertia of the system about z-axis.
0.5Ans: kg 2m
8.3.4 Theorems on Moment of Inertia
Perpendicular Axes Theorem:
■ ‘P is located at a distance y from X-axis, x from Y-axis and 22 xy + from Z-axis.
CHAPTER 8: System of Particles and Rotational Motion
■ Moment of inertia of the body about x-axis is 2 x Iydm =∫
■ Moment of inertia of the body about y-axis is 2 y Ixdm =∫ and moment of inertia of the body about Z-axis is 22 zyx IxdmydmII =∫+∫=+ zxy III∴=+
■ This is known as perpendicular axis theorem. Note that this theorem is applicable only for planar bodies.
Solved example
23. Radius of gyration of a disc of mass 5 kg about a transverse axis passing through its centre is 14.14 cm. Find its radius of gyration about its diameter and hence, calculate its moment of inertia about its diameter.
Sol. Radius of gyration of a disc about a transverse axis passing through its centre K I M MR M R I MR 2 2 22 1414 2 ;. cm
Radiusofthedisc,R1414220 .cm
Radiusofgyrationofthediscaboutits
diame eter, cm Momentofinertiaaboutitsd K I M MR M R 2 42 20 2 10 i iameter kgm
Radiusofthedisc,R1414220 .cm
Radiusofgyrationofthediscaboutits
diame eter, cm Momentofinertiaaboutitsd K I M MR M R 2 42 20 2 10 i iameter
Try yourself:
21. Radius of gyration of a thin ring of mass m passing through its centre and perpendicular to the plane of the ring is R . Then find its moment of inertia about a tangent lying in the plane of the ring.
1.5Ans: 2mR
Parallel Axes Theorem
■ The parallel axis theorem states that the moment of inertia about any axis parallel to the center of mass axis is equal to the moment of inertia about the center of mass axis plus the product of the mass and the square of the distance between the two axes.
Solved example
24. The radius of gyration of a body about an axis at a distance of 12 cm from its centre of mass is 13 cm. Find its radius of gyration about a parallel axis through its centre of mass.
Sol. By parallel axes theorem, M (13)2 = I 0 + M (12)2, I 0 = M (132 – 122) = M(25) Its radius of gyration about a parallel axis through its centre of mass
K I M ===0255cm
Try yourself:
22. ____ of mass 5 kg moment of inertia of a rigid body about an axis is 12 kg m2. Distance of the COM of the body from the given axis is 0.5 m. Find moment of inertia of the body about an axis passing through its COM and parallel to the given axis.
10.75Ans: kg 2m/s
Application of Theorems of Moment of Inertia
Moment of Inertia of Thin Rod
■ The body we have taken for deriving this theorem is planar or laminar in xy-plane. This is only for the sake of simplicity. But parallel axis theorem is true for a general three dimensional body.
■ Conside r an element of length dx at a distance x from C. Mass of that element is dm=(m/l) dx.
■ Now moment of inertia of that element about AB is dI= (dm)x2
■ Moment of inertia of the entire rod about AB is given by 2 /22 0 mmIdIxdx 12 =∫=∫=
■ If we consider an axis PQ parallel to AB and passing through one end of the rod, moment of inertia about PQ can be found using parallel axes theorem
■ In the first case, radius of gyration is 12 and in the second case it is 3
■ About an axis passing through an end and inclined to the rod at an angle α : In the diagram, AB is the rod of length and mass M. r=sin
CHAPTER 8: System of Particles and Rotational Motion
■ About the same axis moment of inertia of the entire ring is 22IdIdmRmR =∫=∫=
■ Radius of gyration K=R
■ If we consider an axis along its diameter we can find moment of inertia of the ring using perpendicular axis theroem. Here XX' and YY ' be two axes along the diameters of the ring.
■ These two axes are perpendicular to each other and intersect at the centre O.
Moment of Inertia of Circular Ring
■ About the given axis, moment of inertia of that element is dI = (dm)R2
Radius of gyration /2 KR =
■ Consider a tangent in the perpendicular plane to the circular ring. The moment of inertia of the ring about a parallel axis through the centre of mass is IC=MR2
■ By parallel axis theorem, moment of inertia about a tangent perpendicular to the plane of the ring is 222 IMRandKR ==
■ Consider a tangent in the plane of the circular ring parallel to the diameter. We
know that moment of inertia of the circular ring about a diameter is 2 2 MR By parallel axis theorem, moment of inertia about a tangent in the plane of the ring is I = IC + Md2 2 2() 2 MRMRdR =+= 2 33 and 22 MR IKR⇒==
Moment of Inertia of a Circular Disc
■ Moment of inertia of the disc about a tangent perpendicular to its plane is 32 2 mR I = (use parallel axis theorem). If that tangent is in the plane of the disc, moment of inertia about that tangent is 52 4 mR
■ About a diameter moment of inertia of the disc is 2 mR 4
Solved example
25. A hollow thin disc has mass m inner radius R, and outer radius 2R. Find its moment of inertia about a tangent lying in its plane.
Sol. Moment of inertia of the disc about an axis passing through its centre C and perpendicular to the plane of the disc is,
C D 1 2 2 5 2 222
Byperpendicularaxistheorem,
1 2 5 4 2
Byparallelaxistheorem, d
2 5222 4 2 21 4
Try yourself:
23. A thin disc of mass 1 kg has inner radius 2 cm and outer radius 4 cm. Find its radius of gyration about an axis passing through its COM and perpendicular to the plane of the disc.
Ans: c10m.
Moment of Inertia of a Rectangular Lamina
■ Consider a rectangular lamina of length l and breadth b. Mass of that lamina is m.
IZ IY
■ Moment of inertia of the rectangular lamina about an axis of rotation passing through its centre and perpendicular to its plane can be obtained by using perpendicular axis theorem.
Where IX =mb2/12 and IY =ml2/12
+
Solved example
26. A bar magnet of length 5.0 cm and breadth 1.2 cm is rotated about an axis passing through its centre and perpendicular to its plane. Find its moment of inertia if the mass of the magnet is 200 g.
Sol. Momentofinertia I
Try yourself:
24. A squar thin sheet has moment of inertia I about one side. What is its moment of inertia about on axis passing through its centre and perpendicular to its plane?
Ans: I/2
Moment of Inertia of a Uniform Solid Sphere
■ 22 5 IMR =
CHAPTER 8: System of Particles and Rotational Motion
Moment of Inertia of a Hollow Sphere (or) Spherical Shell
■ Hollow sphere of mass M and radius R is 22 3 IMR =
■ About an axis passing through its tangent 72 5 IMR =
■ Moment of inertia about an axis passing through its tangent can be obtained by applying parallel axes theorem.
I = IC + MR2
3 IMR =
Solved example
27. Four spheres, each of diameter 2a and mass m, are placed with their centres on the four corners of a square of the side b. Calculate the moment of inertia of the system about any side of the square.
Sol.
Moment of Inertia of a Uniform Hollow Cylinder
■ Its moment of inertia about its own axis is I = MR2 .
Try yourself:
25. Moment of inertia of a solid sphere about a diameter is I. What is its moment of inertia about an axis whose distance from its centre is 2R?
Ans: 11 I
Moment of Inertia of a Uniform Solid Cylinder
■ Moment of inertia of a solid cylinder about its own axis 22 22 RR mm=∑=∑ 2 2 MR I ∴= I = 2 MR2
It is same as that of a circular disc.
■ Moment of inertia about an axis perpendicular to its geometrical axis and passing through its centre of mass is
22 124 lR IM =+
It is same as that of a circular ring.
■ Moment of inertia about an axis perpendicular to its geometrical axis and passing through its centre of mass is
22 122 lR IM =+
Solved example
28. A solid cylinder has mass m , radius R and length R. Find its moment of inertia about an axis passing through its centre and perpendicular to its length.
Sol. Im RR mR 22 2 124 1 3
Try yourself:
26. A thin cylindrical shall has mass m, radius R and length 2R. Find its moment of inertia about an axis passing through one of its ends and perpendicular to its own geometrical axis.
Ans: 11 6 2 mR
8.3.5 Torque
■ Mathematically, torque is called as first order moment of force or simply Moment of force.
CHAPTER 8: System of Particles and Rotational Motion
■ Torque of the force F about a point O in vector form is given by rFτ=×
■ The direction of torque is perpendicular to the plane containing both r and F . Its direction can be determined by using right hand thumb rule.
■ Moment of force has dimensions [ML2T-2].
■ The SI unit of moment of force is newton meter (Nm).
■ Let 12 and FF be the forces acting on the body at points of position vectors 1r and 2r respectively, then 121122 netrFrF τ=τ+τ=×+×
Solved example
29. When a force of 10.0 N is exerted on the handle of a door, the door can be just opened. If the handle is at a distance of 50 cm from the hinges, find the torque applied on the door.
Sol. rF sin = 0.50 × 10.0 sin 900 = 5 Nm.
Try yourself:
■ Let a force F acts on a single particle at a point P in XOY plane whose position vector with respect to the origin O is r
■ The magnitude of moment of the force or torque of the force F about the point O is given by t = Fd
From figure, sinsin ddr r θ=⇒=θ t = rFsin q r is magnitude of position vector r F is magnitude of the force F q is the angle between r and F
27. A thin rod AB of length 1.5 m is hinged at end A. A 10 N force is applied at end B as shown in figure. Find the torque of the applied force about end A. A B 30° 10 N Ans: 7.5 N-m
8.3.6 Couple
■ Let 1F and 2F be the forces in a couple, then 12 FF =− . If these forces act on a body simultaneously and the lines of action of these forces do not coincide, we say that a couple is acting on the body. Net force
12 net FFFzero =+= . Therefore, net force exerted by a couple is zero. Let 1r and 2r be the position vectors of points of application of 1F and 2F respectively, then net torque:
netrFrFrFrF
121(-) rrFmaybenonzero=−×
■ Therefore net torque exerted by a couple may be non-zero. The couple has another interesting property; let us look at the expression for net torque due to a couple. What does 12rr refer to? It indicates the separation between the lines of action of
F2
12 FandF , which is independent of the choice of the fulcrum. Therefore, “net torque or the moment of the couple will be equal to product of magnitude of single force with the separation between the lines of force and it is independent of choice of fulcrum or of reference frame”.
Solved example
30. A thin rod Ab of length 1 m is at rest on a smooth horizontal surface. AC is its midpoint. Forces 20 N, 15 N and 5 N are applied at points A, C and B. Find the moment of couple acting on the rod.
= Moment of 15 N and 5 N forces about point A = (15 × 0.5 + 5 × 1) N–m = 12.5 N–m.
Try yourself:
28. Two equal and opposite forces, each of magnitude 10 N are applied at the ends of a diameter of a disc of radius 10 cm. Find the torque acting on the disc.
Ans: 2 N -m.
8.3.7 Rigid Body in Equilibrium
■ A rigid body is in equilibrium if:
■ The net force acting on the body is zero (∑F = 0).
■ The net torque acting on the body is zero (∑τ= 0).
Solved example
31. A roller of ma ss 300 kg rests against a step of height 20 cm. If the radius of the roller is 50 cm, find the minimum horizontal force to be applied passing through its centre of mass to take the roller on to the step.
Sol. Taking the moments of forces about ‘O’, h = 0.2 m, R = 0.5 m, xRRh 222
xRRh2204.m
To climb up the step, moment due to horizontal force ≥ moment due to weight.
Sol. Moment of couple = Smooth of moment of forces about any point
FRhmgx Fmgx Rh F 3009804 03 ..3920 . N
Try yourself:
29. A force Fijk234N is applied at point P(1 m, 1 m, 1 m). Find the torque of the force about x-axis.
Ans: 1 N-m.
CHAPTER 8: System of Particles and Rotational Motion
F 1 represents some weight or load to be lifted and d1 is called the load arm. Force F2 is the effort to lift the load, and the distance d2 is called effort arm.
■ The ratio F 1 /F 2 is called mechanical advantage (M.A).
Principle of Moments
■ An ideal lever is essentially a light rod pivoted at a point called fulcrum along its length. Two forces F1 and F2, parallel to each other, and usually perpendicular to the lever, as shown here, act on the lever at distances d 1 and d 2 respectively from the fulcrum.
■ The lever is a system in mechanical equilibrium. If R be the reaction of support at the fulcrum. Then R=F1+F2 for translational equilibrium. For rotational equilibrium F1d1=F2d2. In the case of lever,
■ If d2 > d1, then M.A > 1 means small effort can be used to lift a large load.
Table 3 Translatory and Rotatory Equilibrium
Solved example
32. PQR is a rigid equilateral triangular frame of side length ‘L’. Forces F 1, F 2 and F 3 are acting along PQ, QR and PR. Find the relation between the forces if system is in equilibrium.
Sol. If the system is in rotational equilibrium, find the relation between the forces.
Perpendicular distance of any force shown in the figure from centroid C of triangle is L 23
The forces F 1 and F 2 produce anti-clockwise turning effect whereas F 3 produces clockwise turning effect about ‘C’.
Since the system is in rotational equilibrium the total torque acting on the system about the centroid is zero.
F L F L F L 123 232323 0
Hence, F1 +F2 –F 3 = 0
\ F 3 = F1 + F2.
Try yourself:
30. The rod shown in figure has mass m and length L . Determine the value of F so that rod remains in equilibrium.
■ skidding and
■ overturning
■ The cause of skidding and the speed limit imposed for its avoidance have already been discussed in topic laws of motion. Now we study why a vehicle overturns at a curved path and how that can be prevented.
■ Suppose the radius of the curved path is 'r' the speed of the vehicle is 'v' and the mass of the vehicle is 'm'. The centre of mass 'C' of the system is at a height 'h' above the ground. The distance between the inner and outer wheels of vehicle is '2 l'
■ The forces acting on the vehicle are:
■ The force of gravity 'mg' acting vertically downwards through 'C'
■ The normal reactions N1 and N2 acting vertically upward through the inner and outer wheels respectively.
■ The forces of static friction f 1 and f 2 acting horizontally towards the centre of the curved path at the bottom of inner and outer wheels respectively. N1 f1 f2 C mg h 2l N2
8.3.8 Overturning of Vechicle on an Unbanked Path
■ When a vehicle negotiates a sharp turn on a level road, its driver should be careful about two possible mishaps:
Here, the two frictional forces provide the centripetal force. Then 2 12 ffmv r +=
Since the vehicle has no vertical acceleration, we have N1+N2=mg
■ The various torques acting on the vehicle about the centre of mass is
■ Torque due to N1 is t1= N1l (clockwise)
■ Torque due toN 2 is t 2 = N 2 l (anticlockwise)
■ Torque due to weight mg is t 3 = 0. (weight acts through centre of mass)
■ Torque due to f1 = t 4 = f1h. (clockwise)
■ Torque due to f2= t 5 = f2h. (clockwise)
For the vehicle not to overturn the net torque about the centre of mass ‘C‘ is zero.
i.e. the sum of anticlockwise torques = the sum of clockwise torques.
\ t 2 = t 1+ t 4 + t 5
N2l = N1l +(f1 + f2) h or () 12 21 NNffh +
On solving the equations, we get
■ Hence N 2 is always positive, but N 1 may become zero (or negative if ‘v‘ is sufficiently large). In that event, the inner wheels leave the ground and the car tends to overturn towards the outer edge of road. Then 2 mvhgr g0orv 2rh
which should be the speed for the limiting case.
Solved example
33. A car is moving along a circular path of mean radius 100 m. Separation between inner and outer wheels of the car is 2 m and distance of centre of mass of the car with driver above the ground is 1 m. Assuming that there is sufficient friction between the wheels and the road surface, find the maximum speed of the car to avoid the risk of overturning.
CHAPTER 8: System of Particles and Rotational Motion
Sol. VgR h max . 101002 1 1020 m/s m/s
Try yourself:
31. A car is moving along a circular path of mean radius 150 m. Separation between inner and outer wheels of the car is 2 m. The car can move with a maximum speed of 1510m/s. There is sufficient friction between the wheels and the road surface to prevent skidding. Find the maximum height of the COM of the car with the driver to prevent over tur ning. Ans: 1.33 m.
8.3.9 Work Done by Torque
■ Let us consider a body rotating about a fixed axis under an external torque τ
■ rFτ=× where F is force applied on the body. Let the body be rotated by a small angle d q . Then displacement of point of application is
() dldrvr =θ×=ω×
work done is: ()() dWFdlFdrrFd =⋅=⋅θ×=×⋅θ
....(20) Wd ∴=∫τ⋅θ
And power developed by the torque is ().....(21)dWrFd dtdt ×⋅θ ==τ⋅ω
■ In a perfectly rigid body there is no internal motion. The work done by external torques is therefore, not dissipated and goes on to increase the kinetic energy of the body. The rate at which work is done on the body is to be equated to the rate at which kinetic energy increases. The rate of increase of kinetic energy is
■ We assume that the moment of inertia does not change with time. This means that the mass of the body does not change, the body remains rigid and also the axis does not change its position with respect to the body.
Since α= dω/dt.
■ Equating rates of work done and of increase in kinetic energy.
tw = i w α ; t =I α
■ Just as force produces acceleration, torque produces angular acceleration in a body. The angular acceleration is directly proportional to the applied torque and is inversely proportional to the moment of inertia of the body.
Solved example
34. A disc of mass 5 kg and radius 1 m can rotate about a spindle passing through a point on its circuference and, initially, the disc is at rest. A torque is applied on the disc such that finally the disc starts rotating at 60 rpm about the spindle. Find the work done by the torque.
Work Energy Theorem
■ The work done by a constant torque acting on a body is equal to change in rotational kinetic energy.
Solved example
35. What is the work done in increasing the angular frequency of a circular ring of mass 2 kg and radius 25 cm from 10 rpm to 20 rpm about its axis?
Sol. Work done = Increase in rotational kinetic energy
36. A solid cylinder of mass M and radius ‘ R ’ is mounted on a frictionless horizontal axle so that it can freely rotate about this axis. A string of negligible mass is wrapped round the cylinder and a body of mass ‘ m’ is hung from the string. The mass is released from rest. Find the tension in the string and the angular speed of the cylinder as the mass falls a distance h.
Sol.
Sol. WI
=KineticenergyKE
Try yourself:
32. A solid sphere can rotate about fixed spindle passing through its centre. Its moment of inertia about that spindle is 2 kg m2. A torque starts acting on the sphere and after some time its kinetic energy becomes 16 joule. Find the angular velocity of the sphere at that instant.
Ans: 4 rad/s.
The acceleration ‘a’ of the falling body is given by mg – T = ma... (1) Torque on the cylinder is TRI T I R MRa R a R 2 22. or T Ma = 2 ...(2)
from (1) and (2), T mMg Mm 2
From conservation of energy, we have mg
Onsolving,4 22 1 2
Try yourself:
33. A thin rod AB of mass m and length ℓ is hinged at point O such that it can rotate in a vertical plane. End of A the rod is held in horizontal position. Now, the rod is released. Find the velocity of end B when the rod is vertical.
Hint: Find I 0 conserve energy.
Ans: 3 3 7 g
Key Insights:
■ Two rigid bodies of moment of inertia I1 and I2 are rotating with angular velocities ww 12 and in the same direction about the common axes. If they couple, then:
■ The common angular velocity of the system LL12 = ;
CHAPTER 8: System of Particles and Rotational Motion
■ Loss of energy of the system,
■ In the above case, if the two bodies are rotating in opposite directions before coupling, then w2 should be taken as negative,
Table 2 Comparison of Translational and Rotational Motion Linear motion Rotational motion about a fixed axis
7. Kinetic energy KMv = 22 / Kinetic energy KI 22 /
8. Power P = Fv Power P =
9. Linear momentum pMv = Angular momentum LI
8.3.10 ANGULAR MOMENTUM
■ Angular momentum is to rotational mechanics what linear momentum is to particle mechanics, with its change being equal to the net external torque.
Angular Momentum of an Isolated Particle
■ Angular momentum of the particle about origin is defined as moment of momentum or mathematically:
0.....(22)Lrp =×
0L
is perpendicular to the plane containing randp and its direction can be identified using right hand thumb rule.
0sin90Lrp = 2()prmrvr ==ω=ω
0 I =ω
■ In a time interval ‘dt’, r rotates through an angle ‘dq’, as shown in Fig. (b). Now, we are interested in finding out area of the region (shaded) swept by r . 12 2 dArd=θ (area of a sector) and 2 11 222 dAdL rvr dtdtm θ === Vectorially ....(23) 2 dAL dtm =
■ Consider a particle of mass m moving with a velocity v , whose angular momentum is to be calculated about two fixed points O and O1. 1randr are the position vectors of the particle with respect to O and O1 .
and 1 , OO r , O is the position vectors of O1 with respect to O.
So, () o Lmrv =×
and ()11O Lmrv =×
()()11 1 OOOO LLmrrvmrv ∴−=−×=×
(or) 1....(24)OOOO LLrP ′ −=×
■ This equation is true for any two fixed points O and O1 .
■ In Fig. (a), we considered a special case of time independent angular momentum. If L changes with time, we can find its rate of change. () dLddrdP rPPr dtdtdtdt =×=×+×
Here, dr v dt =
and v
is parallel to P ; and net dP F dt =
....(25) netnet dL rF dt ∴=×=τ
■ Rate of change of angular momentum of an isolated particle is equal to net torque acting on the particle. We started off this section with a predetermined notion to establish the above equation and so far we succeded for the case of an isolated particle.
Solved example
37. A particle of mass 0.1 kg is moving along a straight line 3x - 4y + 4 = 0 with velocity 10 m/s. Find the magnitude of angular momentum of the particle about the origin.
Here x and y are measured in metre.
Try yourself:
34. A particle is projected at time t = 0 from a point ‘O’ with a speed ‘u’ at an angle 45o to horizontal. Find the angular momentum of particle at time t = u/g. Ans: gmu 223 /
Angular Momentum of System of Particles
■ Sum of angular momenta of the individual particles is the net angular momentum of the system of particles.
■ The same principle can be applied even for continuous mass distribution.
CHAPTER 8: System of Particles and Rotational Motion
() 1....(28)Occ O LLMrv ∴=+×
■ Similarly, we can also apply the eq. 25 for system of particles.
() ,,int....(29)
■ This equation says that internal forces do not change angular momentum.
Angular Momentum of a Rigid Body(Pure Translation)
■ A rigid body in planar motion can either be in instantaneous translation or in instantaneous rotation. When a rigid body is in pure translation all the particles in the body must have identical instantaneous velocities. In this case angular velocity of the body is zero. Pure translation does not mean that the body must move in a straight line path. In the case of pure translation. () () Lrvdmrdmv =∫×=∫×
() ....(30) c LMrv∴=×
■ Hence to calculate angular momentum of a body in pure translation : “the whole body can be treated as a point mass located at its centre of mass”. V V V
From the concept of centre of mass, c vdmMv∫= ; M is total mass of the system and c v
is velocity of centre of mass of the system. Therefore,
()11 , ....(27) Oc OOO LLMrv −=×
Angular Momentum of a Rigid Body (Rotation about a Fixed Axis)
■ Consider a rigid body which is rotating about z -axis with an angular velocity ω. We want to calculate component of angular
momentum of the body along the axis of rotation(Lz). If mass distribution of the body is symmetric about Z-axis, Lx and L y will vanish; then, net angular momentum about origin, L o =L z and, & zzzz LII=ωτ=α
■ In general () O Lrvdm =∫× ; where r is position vector of elementary mass ‘dm’. Its velocity , vr=ω×ω is angular velocity of the body and is directed along z-axis.
And distance of the element from z-axis is 22 xy + .
From the given data, it can be shown that
■ Net torque about axis of rotation is equal to product of moment of inertia of the object and its angular acceleration.
Solved example
38. A thin spherical shell of mass 2 kg and radius 20 cm is spinning about its diameter with angular velocity of 30 rad/s. Find its angular momentum about the axis of rotation. Sol.
Try yourself:
35. A thin rod of mass 1 kg and length 2 m is spinning with angular momentum 2 kg m2/s about an axis passing through its centre and perpendicular to its length. Find its kinetic energy.
Ans: 6 joules
Impulse–Momentum Equation
■ cm PorPFdt∆∆=∫
So, PFdt∆=∫ is known impulsemomentum equation.
■ If iP and fP denote initial and final linear momenta we can write if PFdtP +∫=
■ Similarly if an external torque τ changes the angular momentum L of the rigid body, we can write cm LorLdt ∆∆=∫τ
■ Here dt∫τ is called angular impulse during time interval Dt about CM. It is also a vector quantity. Here τ is net torque if a number of torques act.
■ So Ldt∆=∫τ is known as impulse angular momentum equation.
So, if LdtL +∫τ= w here iL and Lf respectively. Denote initial and final angular momenta
Solved example
39. A disc of mass 2 kg and radius 0.4 m is rotating about a spindle passing through its centre and perpendicular to its plane with angular velocity 5 rad/s. Due to an external torque acting on the disc for a short interval of time, its angular velocity is increased to 10 rad/s. calculate the angular impulse acted on the disc.
36. A solid disc is rotating about a spindle passing through its centre and perpendicular to its plane with kinetic energy 9 joules. An external torque acts on the disc for a short interval of time and, as a result, its kinetic energy increases to 16 joules. If moment of inertia of the disc about the axis of rotation is 2 kg m2, find the angular impulse acted on the disc.
Ans: 2 J/s
Conservation of Angular Momentum
If t z = zero, Lz = I z ω= constant
■ This principle is very powerful and is valid even in relativistic mechanics and in quantum mechanics.
■ There are instances, for example nuclear reactions, where conservation of mass is not valid but conservation of linear momentum and conservation of angular momentum are valid.
Solved example
40. If the radius of the earth is suddenly halved, keeping its mass constant, find its time period of rotation around its own axis.
Sol. When the radius of the earth gets reduced suddenly, keeping its mass constant, the angular momentum of the earth is constant.
IMR T constantconstant 2 5 22
Since mass ‘M’ is constant, RT 2
The radius changes from R to R/2 R RT T 2 2 4 24 6 hours hours
41. A horizontal disc is freely rotating about a vertical axis passing through its centre at the rate of 100 rpm. A bob of wax of mass 20 g falls on the disc and sticks to it at a distance of 5 cm from the axis. If the moment of inertia of the disc about the given axis is 2×10 –4 kg m 2 , find the new frequency of rotation of the disc.
Sol. I1 = Moment of inertia of the disc
= 2 × 10–4 kg m2
I2 = Moment of inertia of the disc + Moment of inertia of the bob of wax on the disc 4243 2102102010 mr =×+=×+×
()0.052244 kg-m2100.510 =×+×
CHAPTER 8: System of Particles and Rotational Motion
37. A uniform rod AB of mass ‘m’ and length ‘2a’ is allowed to fall under gravity with AB in horizontal. When the speed of the rod is ‘v’, suddenly the end ‘A’ is fixed. Find the angular velocity with which it begins to rotate.
8.3.11 Rolling Motion
■ A rolling wheel combines pure translation and pure rotation, with a point on the rim tracing a cycloid.
■ Types of rolling:
■ Rolling without slipping: The point of contact remains at rest.
■ Rolling with slipping: The point of contact slips forward or backward.
Rolling without Slipping
Fig.29
■ At P 0 , the linear velocity v r, due to rotation is directed exactly opposite to the translational velocity vcm. Further the magnitude of vr here is R w , where R is the radius of the disc. The condition that P0 is instantaneously at rest requires v cm =Rω. Thus for the disc the condition for rolling without slipping is Vcm=Rω.
Some Important Points Related to Rolling
■ Velocity at any instantaneous point on the rolling body: Consider a point x on a rolling body, the velocity of point x is the vector sum of velocity due to translation and rotation. Thus
■ In one complete rotation of wheel, the centre of mass of the wheel moves a displacement s =2αR. In this motion there is no relative motion between the point of contact of the moving body and the surface. If T is the time to complete a rotation, then we have
■ Here q is the angle made by the radius vector with upward verticle in clockwise direction.
■ If V cm= Rω, the wheel is in pure rolling. The velocity of the point of contact is zero.
■ If V cm > Rω, wheel slips forward. The wheel moves through a distance greater than 2πR in one full rotation.
■ If Vcm< Rω, wheel slips backward. The wheel moves through a distance less than 2πR in one full rotation.
Rolling motion = Pure translatory motion + Pure rotatory motion.
■ Velocities at different points on the rolling wheel:
At 0 ,,02 R BvV θ==
At ,,902 R RvV θ==
At ,,603 R PvV θ==
Here, RV is resultant velocity at a given point on wheel.
Solved example
42. A disc is rolling without slipping with velocity V on a horizontal surface. P is a point on its rim whose position is defined by angle q Find the speed of P.
Sol. Since the disc is rolling without slipping, velocity of point of contact O on the disc is zero. At the instant shown in figure, the whole disc is rotating with angular velocity w about an axis passing through O and perpendicular to the plane of figure.
engt
CHAPTER 8: System of Particles and Rotational Motion
Try yourself:
38. A disc is rolling without slipping on a horizontal surface. AB is a horizontal diameter and speed of point A is V . What is the velocity of its centre at that instant?
Ans: V 2
Rolling bodies over moving platform:
■ Case-1: If points of contact of surface is moving with velocity u with respect to ground, then Vcm– w R = u
■ Case-2 : For no sliding on the moving platform, u = ωR-Vcm.
■ Case-3: For accelerated surface, acm–αR =
Solved example
43. A disc is rolling without slipping on a plank with angular velocity w = 10 rad/s. The plank is moving horizontally with velocity 2 m/s as
shown in figure. Find the velocity of centre of disc relative to the ground. Radius of the disc is 200 cm.
w =10 rad/s
Sol. V cm – VP = w R ⇒ V cm = VP + w R
= (2+10×02) m/s = 4 m/s.
Try yourself:
2 m/s
39. A disc is rolling without slipping on a plank with a velocity v relative to the plank. The plank itself is moving with velocity v relative to ground. Find the speed of A relative to ground.
Hint: VVVCgAAC .//
Ans: 5 V
■ Direction of friction in case of translation and rotation combined
r FR aRRx II τ =α==
net acceleration of point p is ptr aaa =+
p FFRx a MI =− (Towards right) ....(i)
■ From the above equation it is clear that motion tendency at point P depends upon both x and I
Eq(i) can be written as, 2 1 p FRx a MK
=−
■ To find the direction of friction at P, let us consider the surface to be frictionless.
Acceleration of point P due to translation
t F a M = (towards right)
Acceleration of point P duet rotation only,
....(ii)
■ If K2 > Rx : friction will act in backward direction.
If K2 = Rx : no friction will act
■ If K 2 < Rx : friction will act in forward direction
Solved example
44. A force F is applied at the top of a ring of mass M and radius R placed on a rough horizontal surface. Friction is sufficient to prevent slipping. Find the frictional force acting on the ring and also find the acceleration of centre of mass.
Sol. fFR R mR mR
FfMa 1 2 0 2 ;So,
FrictionandaccelerationfaFM == 0.
Try yourself:
40. A horizontal force F is applied to the centre c of a solid disc of mass m and radius R. If the disc starts rolling without slipping, calculate its acceleration.
Ans: 2 F/3 m.
■ Kinetic energy of rolling body: A body rolling without slipping posses both translational KE and rotational K.E.
KE = KE t + KE r () 2211 22 tcmcmr KEMVMVVVRV ====ω=
22 2 11 22rcm VV KEIMK RR
22 22 22 11 22 KK MVMV RR
2211 22 tr KEKEKEMVMV ⇒=+=+β
12(1) 2 MV =+β
hence, total kinetic energy of the rolling body without slipping: 12 .(1) 2 KEMV=+β
Where a R α= is a Dimensionless quantity. Its value vearies from body to body and axis to axis.
■ For example:
■ For ring
22 2 1 K IMRMK R ==⇒==β
■ For Solid sphere
22 2 22 55 K IMRMK R ==⇒==β
■ For disc 2 2 2 MR IMK ==
2 1 2 K R ∴==β
■ For hollow sphere 222 3 IMRMK == 2 2 2 3 K R ∴==β
CHAPTER 8: System of Particles and Rotational Motion
K.E of the rolling body without slipping can also be written as: KErolling = KE t +KE r () 2211 22cmcmcm MVIVR =+ω=ω 22211 22 cm MRI =ω+ω 122 2 cm KEMRI =+ω
■ As we know MR 2 +I cm =I P is the moment of inertia of the wheel about the point of contact 'P' then 12 2 rollingP KEI=⋅ω
■ Rolling motion of a body may be treated as pure rotation about an axis normal to the plane passing through the point of contact with same angualar velocity.
■ Fraction of translational KE in the total kinetic energy 1 1 = +β
■ Fraction of rotational KE in the total kinetic energy 1 β = +β
■ Ratio of translational KE to rotational 1 KE = β
Solved example
45. A uniform spherical shell rolls down a fixed inclined plane without slipping. Find the ratio of rotational kinetic energy to translational kinetic energy as it reaches the lowest point of the incline.
Sol. As there is no slip, vR is valid and rotational kinetic energy KImk R 1 2 1 2 222 cm and translational kinetic energy , KmvmR
Therefore
Try yourself:
41. A solid disc of mass 1 kg is rolling without slipping with kinetic energy 3 joule. Find the velocity of its centre.
Ans: 2 m/s.
■ Angular momentum of rolling wheel: V cm
■ Rolling of a body on an inclined plane: Consider a rigid round body of mass 'm' and radius ' R' is placed on a rough inclined plane making an angle q with the horizontal as shown in figure. mg N f mgcosq mgsinq q q
The forces acting on the body are:
cmcm mRvI =×+ω
(or) 2 cm LmRI=ω+ω
(or) () 2 cmP LImRI =+ω=ω
Solved example
46. A solid cylinder of mass m and radius R is rolling without slipping with vlocity V on a horizontal plane (x-axis). P is a point whose coordinates are (0, h ), such that angular momentum of the cylinder about P is zero.
Sol. LmVhRkIk
mVhRmRV R k PC 1 2 2
Since ewecanwrite, L
mVhRmRVhR P 0 1 2 0 3 2 ,
Try yourself:
42. A thin ring of mass 2 kg and radius 0.5 m is rolling slipping with velocity of 2 m/s. Find its angular momentum about the point of contact.
Ans: 2 kg /s.2m
■ Weight 'mg' acting vertically downwards. The component of the weight along the inclined plane is 'mg sin q ' and perpendicular to the inclined plane is 'mg cos q '.
The torque produced by the weight about the centre is t 1 = mg(0)=0. (Weight acts along the centre).
■ Normal reaction N. The torque produced by the normal reaction N is t 2 =N(0)= 0. (Normal reaction passes through the centre).
■ The static frictional force 'fs' acts along the inclined plane upwards.
The torque produced by the frictional force 'f s ' is t 3 =f s R
\The resultant force normal to the inclined plane is zero because N balances mg cos q
\ N=mg cos q
■ The net force acting down the inclined plane is FR = mg sin q – fS
■ If α is the acceleration of the body down the inclined plane
∴=
sin .....(1) S mgf a m θ−
Net torque acting on the body is
123 τ=τ+τ+τ
∴τ=++
00 S fR
■ Due to this torque the angular acceleration of the body is given by S fR I α=
But a R α= and I = mK2 2 22....(2)SS afRfR a RmKmK ∴=⇒=
[K is radius of gyration]
Substitute equation (2) in (1)
CHAPTER 8: System of Particles and Rotational Motion
Coefficient of static friction is tan 1 s
Ex: For solid spere 2 5 β= 2 tan 7 s ⇒µ≥θ
■ Velocity of the rolling body (without slipping) on reaching the bottom of the inclined plane is PEA= KEB 121 2 mghmv=+β
sin f ag m =θ− 2 2 sin mKa a m gR=θ− 2 22 2 sin sin 1 Kag agaRK R
⇒=θ−∴=θ +
2 2 sin 1 agK where R θ ⇒==β +β
Static frictional force
=θβ
sin....(4) 1 S fmg
■ Condition for pure rolling without slipping is () cos SLLss fffNmg ≤=µ=µθ sincos 1 mgs mg
sin h hl l θθ =⇒=
■ time taken by the rolling body to reach the bottom of the plane is 2 u0 12 2agsin 1 l utatt a =
2(1) sinsin lhtl g +β
or 12(1) sin h t g =+β θ
Hence 1 t ∝+β
For sliding bodies β = 0, 212 sinsin lh t gg ⇒== θθ
Smaller the β , smaller the time taken by the rolling body to reach the bottom of the inclined plane.
Combination of Pure Translation and Rotation
■ Consider the motion of a rigid body from position 1 to position 2. Here particle A translates by a distance S A. The straight line AB rotates through an angle q simultaneously.
B q A SA A
B
■ Displacement of B relative to the ground is given by BBAA SSS =+
BBAA dSdSdS dtdtdt ⇒=+
(or)
BBAA VVV =+
But BABABAVr =ω×
So, () BBABAA VrV =ω×+ Similarly, we can obtain () BBABAA ara =α×+
■ Fraction of translational KE in the total K.E. is
■ Fraction of rotational KE in the total K.E. is
8.3.12 Toppling and Sliding
■ Consider a block of mass m resting on a rough horizontal surface. Let F be the force applied on it at a height h above the base as shown.
■ If friction f is sufficient to prevent sliding, the block is in translational equilibrium.
So, we have F=f and N=mg where N is normal reaction. If N also passes through can unbalanced torque
(F and f form a couple) acts which has a tendency to topple the block about point P.
■ To cancel this effect, N shifts towards right say a distance x. Such that, net torque becomes zero.
FhNxmgxandxFh
■ If F or h (or both) increases, distance x also increases. But x can not go beyond right edge at P. So, the extreme case is N passes through P as shown.
CHAPTER 8: System of Particles and Rotational Motion
From equations (1) and (2), we see that the cube will topple before sliding if 1 2 .
48. A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point directly above the centre of the face, at a height 3a/4 above the base. What is the minimum value of F for which the cube begins to topple about an edge?
Sol. In the limiting case, normal reaction will pass through O. The cube will topple about O if torque of F exceeds the torque of mg. In this case, the normal force shifts to the edge O. F a mg a Fmg 3 42 2 3 ;
Solved examples
47. A force F is applied on the top of a cube, as shown in figure. The coefficient of friction between the cube and the ground is m . If F is gradually increased, find the value of m for which the cube will topple before sliding.
Sol. Let m be the mass of cube and ‘a’ be the side of cube.
The cube will slide if Fmg ... (1) and it will topple, if torque of F about P is greater than torque of mg about P, i.e.,
Fa a mg . 2 ; or Fmg > 1 2 ... (2)
So, the minimum value of F is 2 3 mg .
Try yourself:
43. A uniform cylinder of height h and radius r is placed with its circular face on a rough inclined plane and the inclination of the plane to the horizontal is gradually increased. If m is the coefficeint of friction, then under what conditions will the cylinder a) slide before toppling b) topple before sliding?
Ans: (n4 – 1)x
8.3.13 Instantaneous Axis of Rotation
■ A rigid body undergoing combined translation and rotation has an instantaneous axis of rotation, where the velocity is zero.
■ In a rolling uniform disc, the center of mass and points on the axis move in pure translation, while other points rotate about the axis.
■ Pure rolling occurs when the instantaneous point of contact remains stationary. The instantaneous center of rotation (IC) can be found where the position vector from IC to a point is perpendicular to its velocity.
■ There are three possibilities.
■ If v and w are known, there IC is located along the line drawn perpendicular to V at P such that v r = ω
TEST YOURSELF
1. A thin rod of mass m and length 2L is made to rotate about a normal axis through centre of rod. If its angular velocity changes from ‘0’ to ‘ω’ in time t, the torque acting on it is
(1) mL2 w /12t
(2) 4mL2 w /3t
(3) mL2 w /t
(4) mL2 w /3t
■ If AV and BV are velocities of two points A and B of the rigid body, draw perpendiculars at A and B to the lines of action of AB VandV . The point of intersection of these perpendiculars gives the location of IC as shown,
Here AAVr=ω and BBVr=ω
■ If VA and BV are parallel, then IC can be located by proportional triangles as shown.
2. A child is standing with folded hands at the centre of a platform rotating about its central axis. The KE of the system is K The child now stretches his arms so that the MI of the system is doubled. The KE of the system now is
(1) 2K
(2) K/2
(3) 4K
(4) K/4
3. A motor rotates a pulley of radius 25 cm at 20 rpm. A rope around the pulley lifts a 50 kg block. The power output of the motor is
(1) 220 W
(2) 520 W
(3) 350 W
(4) 261 W
Answer Key
(1) 4 (2) 2 (3) 4
CHAPTER 8: System of Particles and Rotational Motion
# Exercises
JEE MAIN LEVEL
Level-I
Centre of Mass
Single Option Correct MCQs
1. Mass of a ring is non-uniformly distributed around its geometric centre. If R is radius of the ring, then
a) centre of mass does not coincide with geometric centre
b) position of centre of mass from the geometric centre will be x(0 < x < R)
c) centre of mass will be nearer to the greater mass distribution
d) centre of mass may lie outside the periphery
(1) Only a and b are correct.
(2) Only b and c are correct.
(3) a, b, and c are correct.
(4) a, b, c, and d are correct.
2. Two particles, each of the same mass, move due north and due east, respectively, with the same velocity ‘V’. The magnitude and direction of the velocity of centre of mas s is
(1) NE 2 V (2) 2NE V
(3) 2V SW (4) SW 2 V
3. Consider the two following statements.
A) Linear momentum of the system remains constant.
B) Centre of mass of the system remains at rest.
Choose the correct statement from the options given below.
(1) A implies B and B implies A.
(2) A does not imply B and B does not imply A.
(3) A implies B but B does not imply A. (4) B implies A but A does not imply B.
4. Two particles of equal mass have velocities –1 1 ˆ 4ims =
v and –1 2 ˆ 4j. ms =
v First particle has an acceleration () –2 1 ˆ 5i ˆ 5jms =+ a , while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of (1) straight line
(2) parabola
(3) circle
(4) ellipse
5. A shell of mass m in flight explodes into three equal fragments. One of the fragments reaches the ground earlier than the other fragments. The acceleration of the centre of mass of the shell system just after the first one reaches the ground is (1) g (2) g/3
(3) 3g/4 (4) 2g/3
6. A 50 kg man is standing at one end of a 25 m long boat. He starts running towards the other end. On reaching the other end of the boat, his velocity is 4 ms–1. If the mass of the boat is 250 kg, final velocity of the boat is (in ms–1)
(1) 2/5 (2) 2/3
(3) 8/5 (4) 8/3
7. Particles of masses 1 g, 2 g, 3 g, and 4 g are placed at x =1 cm, x = 2 cm, x = 3 cm, and x = 4 cm, respectively. Then, x cm =
(1) 1 cm (2) 2 cm (3) 3 cm (4) 4 cm
8. Four cubes of side a each, of mass 40 g, 20 g, 10 g, and 20 g, are arranged in x-y plane as shown in the figure. The coordinates of centre of mass of the combination with respect to point O is
(1) (19a/18, 17a/18)
(2) (17a/18, 11a/18)
(3) (17a/18, 13a/18)
(4) (13a/18, 17a/18)
9. Three particles of masses 1 kg, 2 kg, and 3 kg are placed at the vertices A, B, and C of an equilateral triangle ABC. If A and B lie at (0, 0) and (1, 0) m, the coordinates of their centre of mass are
(1) 37mandm 26
(2) 73mandm 64
(3) 73mandm 124
(4) 77mandm 1212
11. Four particles A, B, C and D with masses m A = m , m B = 2 m , m C = 3 m, and m D = 4 m are at the corners of a square. They have accelerations as shown. The acceleration of centre of mass of the particles is
(1) () ˆ 5 ˆ a ij (2) ˆ 5 a i (3) ˆ 5 a j (4) () ˆ 5 ˆ a ij +
12. In the system shown in the figure m1 > m2 If another block is attached below the block of mass m1, then the acceleration of COM will
(1) decrease
10. The centre of mass of a non-uniform rod of length L, whose mass per unit length 2 , λ= kx L where k is a constant and x is the distance from one end, is (1) 3 4 L (2) 3 L (3) 4 L (4) 2 3 L
(2) increase (3) remains the same (4) may increase or decrease
Numerical Value Questions
13. A uniform wire of length 2L is bent to form an angle of 60°, as shown. The distance of the centre of mass from the vertex O is 3 , L n where n is O 60°
14. Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is m/s.
15. A shell is fired from the ground at an angle θ with the horizontal, with a velocity V. At its highest point, it breaks into two equal fragments. If one fragment comes back through its initial line of motion with the same speed, then the speed of the second fragment will be xvcos θ. Here, x =
16. Two rods of lengths l 1 = 1.5 m and l 2 = 1 m, and masses m1 = 1 kg and m2 = 2 kg, are joined end to end. If the centre of mass of the rod from the end of the first rod is 19 m, 6 y then y is
17. Masses 1 kg, 1.5 kg, 2 kg, and M kg are situated at (2, 1, 1), (1, 2, 1), (2, -2, 1), and (-1, 4, 3). If their centre of mass is situated at (1, 1, 3/2), the value of M is x/2 kg. Here, x =
CHAPTER 8: System of Particles and Rotational Motion
18. The position of the centre of mass of a uniform semicircular wire of radius R, placed in x-y plane with its centre at the origin and the line joining its ends as x-axis, is given by 0,.
Then, the value of |x| is .
19. A boat of mass 40 kg is floating on water. A boy of mass 10 kg on the boat moves by 8 m (w.r.t. the boat) towards the shore. The distance by which the boat moves away from the shore is x/10. Then, x is m.
20. Two particles of masses 4 kg and 6 kg are at rest separated by 20 m. If they move towards each other under mutual force of attraction, the position of the point where they meet is (from 4 kg body) m
21. Three particles of masses 8 kg, 4 kg, and 4 kg, situated at (4, 1), (–2, 2), and (1, –3), are acted upon by external forces 6N,6N,and14N.
jii If the acceleration of centre of mass of the system is k/8, then k is ____m/s2
22. Two particles of mass 2 m are at a distance d apart. Under their mutual gravitational force, they start moving towards each other. The acceleration of their centre of mass, when they are d/2 apart, is
Collisions
Single Option Correct MCQs
23. A particle of mass m and momentum P moves on a smooth horizontal table and collides directly and elastically with a similar particle (of mass m) having momentum –2p. The loss (–) or gain (+) in the kinetic energy of the first particle in the collision is
(1) + 2 2 p m (2) 2 4 p m
(3) + 32 2 p m (4) zero
24. A heavy ball moving with speed v collides with a tiny ball at rest. The collision is elastic. Then, immediately after the impact, the second ball will move with a speed approximately equal to (1) v (2) 2v (3) v/2 (4) v/3
25. A body of mass 6 kg travelling with a velocity of 10 m/s collides head-on and elastically with a body of mass 4 kg travelling at a speed 5 m/s in the opposite direction. The velocity of the second body after the collision is (1) 0 m/s (2) 6 m/s(3) 8 m/s (4) 13m/s
26. A body of mass 5 kg makes an elastic collision with another body at rest and continues to move in the original direction after collision, with a velocity equal to 1/10th of its original velocity. Then, the mass of the second body is
(1) 4.09 kg (2) 0.5 kg
(3) 5 kg (4) 5.09 kg
27. A body of mass m moving with a velocity of 8 m/s, collides with another body of mass 15 kg moving with 4 m/s in the opposite direction. After collision, they stick and move together. The loss of KE is 270 J. Then, m is .
(1) 5 kg (2) 10 kg
(3) 15 kg (4) 20 kg
28. Given below are two statements.
Statement I : In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision.
Statement II : In an elastic collision, the linear momentum of the system is conserved.
In light of the above statements, choose the correct answer from the options given below.
(1) Both statement I and statement II are correct.
(2) Both statement I and statement II are incorrect.
(3) Statement I is correct but statement II is incorrect.
(4) Statement I is incorrect but statement II is correct.
29. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
Assertion (A) : Collision between two particles is not necessarily associated with physical contact between them.
Reason (R) : Only in physical contact, momentum transfer takes place.
In light of the above statements, choose the correct answer from the options given below.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A).
(2) Both (A) and (R) true but (R) is not the correct explanation of (A).
(3) (A) is true but (R) is false.
(4) (A) is false but (R) is true.
30. A ball of mass 0.4 kg, moving with a uniform speed of 2 ms–1 , strikes a wall normally and rebounds. Assuming the collision to be elastic and the time of contact of the wall as 0.4 s, find the force exerted on the ball.
(1) 1 N (2) 2 N (3) 3 N (4) 4 N
31. A body of mass 5 kg moving with a velocity of 10 m/s collides with another body of mass 20 kg, at rest, and comes to rest. The velocity of the second body after collision is (1) 2.5 m/s (2) 5 m/s
(3) 7.5 m/s (4) 10 m/s
Numerical Value Questions
32. In the arrangement shown, the pendulum on the left is pulled aside. It is then released and allowed to collide with the other pendulum which is at rest. A perfectly inelastic collision occurs and the system rises to a height of 1/4 h . The ratio of the masses of the pendulums is . h
33. Two balls collide and bounce off each other, as shown in the figure. The 1 kg ball has a speed of 10 cm/s after collision. The velocity of the 0.5 kg ball will be____.
34. An object of mass 40 kg, having velocity 4, ˆ m/s i collides with another object of mass 40 kg, having velocity 3jm/s . If the collision is perfectly inelastic, then the loss of mechanical energy (in SI unit) is .
35. A body of mass 1 kg collides head on elastically with a stationary body of mass 3 kg. After collision, the smaller body reverses its direction of motion and moves with a speed of 2 ms –1 . The initial speed of the smaller body before collision is ms–1
36. Two bodies of the same mass are moving with the same speed but in different directions in a plane. They have a completely inelastic collision and move together after the collision, with a final velocity which is half of the initial velocity. The angle between the initial velocities of the two bodies (in degrees) is
Rotational Mechanics
Single Option Correct MCQs
37. A thin uniform rod of mass M and length L is hinged at an end and released from rest in the horizontal position. The tension at a point located at a distance L /3 from the hinge point, when the rod becomes vertical, will be
CHAPTER 8: System of Particles and Rotational Motion
(1) 11 13 Mg (2) 6 13 Mg (3) 22 27 Mg (4) 2 Mg
38. A uniform solid cylinder of mass m and radius R is placed on a rough horizontal surface. A horizontal constant force F is applied at the top point P of the cylinder so that it starts pure rolling. The acceleration of the cylinder is
(1) F/m (2) 4F/3m (3) 2F/3m (4) F/2m
39. A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of frictional force acting on the cylinder are
(1) up the incline while ascending and down the incline while descending (2) up the incline while ascending as well as while descending (3) down the incline while ascending and up the incline while descending (4) down the incline while ascending as well as while descending
40. The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height h, from rest, without sliding, is
(1) gh (2) g 5 gh
(3) 4 3 gh
(4) 10 7 gh
41 An equilateral prism of mass m rests on a rough horizontal surface with coefficient of friction μ. A horizontal force F is applied on the prism, as shown in figure. If the coefficient of friction is sufficiently high so that the prism does not slide before toppling, the minimum force required to topple the prism is
(1) 3 mg (2) 4 mg
(3) 2 3 mg (4) 7 4 mg
42. In pure rolling motion of a ring,
a) it rotates about instantaneous point of contact of ring and ground
b) its centre of mass moves in translatory motion only
c) its centre of mass will have translatory as well as rotatory motion
(1) Only a is correct.
(2) a and c are correct.
(3) a and b are correct.
(4) a, b, and c are correct.
43. A disc of radius 10 cm is rotating about its axis at an angular speed of 20 rad/s. Find the linear speed of a point on the rim.
(1) 1 m/s (2) 4 m/s (3) 2 m/s (4) 6 m/s
44. Three thin rods, each of length L and mass M, are placed along x, y and z axes in such a way that one end of each rod is at the origin, as shown. The moment of inertia of this system about z-axis is
45. Two discs of same mass and different radii are made of different materials such that their thickness is 1 cm and 0.5 cm, respectively. Densities of materials are in the ratio 5 : 3. The moment of inertia of these discs, respectively, about their diameters will be in the ratio
(1) 2 : 3 (2) 5 : 6
(3) 3 : 10 (4) 4 : 10
46. The radius of gyration of a uniform rod of a uniform length ‘l’ , about an axis passing through a point 3 l away from the centre of the rod, and perpendicular to it is
47. A ring of mass M = 10 kg and radius R = 2 m is taken as shown in the figure. Calculate the moment of inertia of ring about the diametric axis
(1) 25 kg m2 (2) 30 kg m2 (3) 40 kg m2 (4) 20 kg m2
48. The moment of inertia of a uniform thin rod of length L and mass M about an axis passing through a point at a distance of L/3 from one of its ends and perpendicular to the rod is
(1) 72 48 ML (2) ML2
(3) 2 9 ML (4) 2 3 ML
49. A small pulley of radius 20 cm and moment of inertia 0.32 kg–m 2 is used to hang a 2 kg mass with the help of a massless string. If the block is released, for no slipping condition, acceleration of the block will be (Take g = 10 ms–2) 2 kg
(1) 2 m/s2 (2) 4 m/s2 (3) 1 m/s2 (4) 3 m/s2
50. A ring is kept on a rough inclined surface. But the coefficient of friction is less than the
CHAPTER 8: System of Particles and Rotational Motion
minimum value required for pure rolling. At any instant of time, let KT and KR be the translational and rotational kinetic energies of the ring. Then,
(1) KR = KT (2) KR > KT (3) KT > KR (4) KR = 0
51. A solid sphere of mass 1.5 kg rolls on a frictionless horizontal surface. Its centre of mass is moving at a speed of 5 ms −1. Then, it rolls up an incline of 30° to horizontal. The height attained by the sphere before it stops is (g =10 ms−2)
(1) 5 m (2) 1.5 m (3) 1.75 m (4) 3.65 m
Numerical Value Questions
52. A body rotating with an angular speed of 600 rpm is uniformly accelerated to 1800 rpm in 10 s. The number of rotations made in the process is
53. Moment of inertia of a body about a given axis is 1.5 kg m 2. Initially, the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular acceleration of 20 rad/s2 must be applied about the axis for a duration of s.
54. For the figure shown, a rod of mass 10 kg (of length 100 cm) has some point masses tied to it at different positions. Find the distance of a point (from A) at which, if the rod is picked over a knife edge, it will be in equilibrium about that knife edge (in cm).
20 kg 15 kg 5 kg 7.5 kg 15 cm
B A
50 cm 20 cm
100 cm
55. A uniform solid cylindrical roller of mass m is being pulled on a horizontal surface with force F parallel to the surface and applied at its centre. If the acceleration of the cylinder is a and it is rolling without slipping, then the value of = 3 F x ma. Find the value of x.
56. A solid sphere of mass 1 kg rolls without slipping on a plane surface. Its kinetic energy is 7 ×10–5 J. The speed of the centre of mass of the sphere is cms–1
57. A wheel, which is initially at rest, is subjected to a constant angular acceleration about its axis. It rotates through an angle of 15° in time t seconds. The angle (in degrees) through which it rotates in the next 2 t seconds is
58. Two thin uniform circular rings each, of radius 10 cm and mass 0.1 kg, are arranged such that they have a common centre and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through their common centre and perpendicular to the plane at one of the rings is × 10–4 kg m2.
59. The radius of gyration of a body about an axis at a distance of 3 cm from its centre of mass is 5 cm. The radius of gyration about a parallel axis through centre of mass is cm.
60. Particles, each of mass 1 kg, are placed at 1 m, 2 m, and 4 m on x -axis with respect to origin. Then, moment of inertia of the system about y-axis is 3n kgm2. Then, the value of n is equal to .
61. A light rope is wound around a hollow cylinder of mass 6 kg and radius 60 cm, which is free to rotate about its geometrical axis. The rope is pulled with a force of 50.4 N. The angular acceleration of the cylinder will be___ rad s–2 .
Level-II
Centre of Mass
Single Option Correct MCQs
1. Three particles of masses 1.0 kg, 1.5 kg, and 2.5 kg are placed at three corners of a right angle of triangle of sides 4.0 cm, 3.0 cm and 5.0 cm, as shown in the figure. The centre of mass of the system is at a point
(1) 0.6 cm right and 2.0 cm above 1 kg mass
(2) 0.9 cm right and 2.0 cm above 1 kg mass
(3) 2.0 cm right and 0.9 cm above 1 kg mass
(4) 1.5 cm right and 1.2 cm above 1 kg mass
2. A rod of length L has non-uniform linear mass density given by () 2 ,
where a and b are constants and 0
L. The value of x for the centre of mass of the rod is at
(1) 4 323 ab L ab + +
(2) 32 43 ab L ab + +
(3) 32 23 ab L ab + +
(4) 3 22 ab L ab + +
3. The coord inat es of centre of mass of a uniform flag-shaped lamina (thin flat plate) of mass 4 kg (the coordinates of the same in SI system are shown in figure) is
(1) (1.25 m, 1.50 m)
(2) (0.75 m, 0.75 m)
(3) (0.75 m, 1.75 m)
(4) (1 m, 1.75 m)
4. As shown in fig, when a spherical cavity (centred at O ) of radius 1 is cut out of a uniform sphere of radius R (centred at C), the centre of mass of remaining (shaded) part of the following sphere is at G, i.e., on the surface of the cavity. R can be determined by ehich of the equation? 1
(1) (R2+R−1)(2−R)=1
(2) (R2+R+1)(2−R)=1
(3) (R2 –R+1)(2−R)=1
(4) (R2 –R−1)(2−R)=1
5. Two bodies of mass 1 kg and 3 kg have position vectors ˆˆˆ 2 ijk ++ and , ˆˆˆ 32 −−+ijk respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector
(1) ˆˆˆ 2 ijk −+
(2) 32 ˆˆˆ ijk−−+
(3) ˆ 2 ˆ 2 ik−+
(4) ˆ 2 ˆ 2 ˆ ijk−−+
6. Two blocks of masses 10 kg and 30 kg are placed on the same straight line with coordinates (0, 0) cm and ( x , 0)cm, respectively. The block of 10 kg is moved on the same line through a distance of 6 cm towards the other block. The distance through which the block of 30 kg must be moved to keep the position of centre of mass of the system unchanged is (1) 4 cm towards the 10 kg block (2) 2 cm away from the 10 kg block
CHAPTER 8: System of Particles and Rotational Motion
(3) 2 cm towards the 10 kg block (4) 4 cm away from the 10 kg block
7. A circular hole of radius 2 a is cut out of a circular disc of radius a as shown in fig. The centroid of the remaining circular portion with respect to point ‘ O’ will be Y axis x- axis 0 a/2 a
(1) 2 3 a (2) 1 6 a (3) 10 11 a (4) 5 6 a
Numerical Value Questions
8. The distance of centre of mass from end A of a one-dimensional rod (AB), having mass density 2 02 1 x L ρ=ρ− kg/m and length L (in metres) is 3L m α . The value of α is where x is the distance from end A.
9. A man of 60 kg is running on the road and suddenly jumps into a stationary trolly car of mass 120 kg. Then, the trolly car starts moving with velocity 2 ms–1. The velocity of the running man was__________ms–1 when he jumps into the car.
(1) 6 (2) 7
(3) 8 (4) 10
10. The position of the centre of mass of a uniform semicircular wire of radius ' R ' placed in x-y plane, with its centre at the origin and the line joining its ends as x-axis, is given by 0, xR π . Then, the value of |x| is
Collisions
Single Option Correct MCQs
11. It is found that, if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc . The values of pd and pc are, respectively, (1) (–89, –28) (2) (–28, –89) (3) (0, 0) (4) (0, 1)
12. A particle of mass m moving with velocity v, collides with a stationary particle of mass 2 m. After collision, they stick together and continue to move together with velocity
(1) v (2) 2 v
(3) 3 v (4) 4 v
13. A ball of mass 200 g rests on a vertical post of height 20 m. A bullet of mass 10 g, travelling in horizontal direction, hits the centre of the ball. After collision, both travel independently. The ball hits the ground at a distance of 30 m and the bullet at a distance of 120 m from the foot of the post. The value of initial velocity of the bullet will be (if g =10 m/s2)
(1) 120 m/s (2) 60 m/s (3) 360 m/s (4) 400 m/s
14. As per the given figure, a small ball P slides down the quadrant of a circle and hits the other ball Q of equal mass, which is initially at rest. Neglecting the effect of friction and assuming the collision to be elastic, the velocity of ball Q after collision will be:
(g = 10 m/s2)
(1) 0 (2) 4 m/s
(3) 0.25 m/s (4) 2 m/s
15. A particle of mass m with an initial velocity ˆ ui collides perfectly elastically with a mass 3m at rest. It moves with a velocity ˆ vj after collision. Then, v is given by (1) 2 3 vu = (2) 1 6 vu = (3) 3 u v = (4) 2 u v =
16. A mass ‘ m ’ moves with a velocity v and collides inelastically with another identical mass. After collision, the first mass moves with velocity 3 v in a direction perpendicu lar to the initial direction of motion. Find the speed of the second mass after collision. m
√3 (1) v (2) 3v (3) 2/3 v (4) /3 v
17. A body of mass M, moving at speed V 0 , collides elastically with a mass 'm ' at rest. After the collision, the two masses move at angles θ1 and θ2 with respect to the initial direction of motion of the body of mass M The largest possible value of the ratio M/m, for which the angles θ1 and θ2 will be equal, is (1) 4 (2) 1 (3) 3 (4) 2
Numerical Value Questions
18. A ball with a speed of 9 m/s collides with another identical ball at rest. After the collision, the direction of each ball makes an angle of 30º with the original direction. The ratio of velocities of the balls after collision is x : y, where x is .
19. A ball of mass 10 kg, moving with a velocity of 103 m/s along the x-axis, hits another ball of mass 20 kg which is at rest. After the collision, the first ball comes to rest while the second ball disintegrates into two equal pieces. One piece starts moving along y-axis with a speed of 10 m/s. The second piece starts moving at an angle of 30° with respect to the x-axis. The velocity of the ball moving at 30° with the x-axis is x m/s. The value of x to the nearest integer is . The configuration of pieces after collision is shown in the figure below.
pieces. One of the pieces starts moving along the y-axis at a speed of 10 m/s. The second piece starts moving at a speed of 20 m/s at an angle θ (degrees) with respect to the x-axis. The configuration of pieces after collision is shown in the figure. The value of θ to the nearest integer is After Collision
Rotational Mechanics
Single Option Correct MCQs
21. Seven identical circular planar disks, each of mass M and radius R, are welded symmetrically, as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is P
20. A ball of mass 10 kg, moving with a velocity 1031 ms along the x-axis, hits another ball of mass 20kg, which is at rest. After the collision, the first ball comes to rest and the second one disintegrates into two equal
(3) 732 2 MR (4) 1812 2 MR
22. From a uniform circular disc of radius R and mass 9 M, a small disc of radius R/3 is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is:
(3) 7.5 × 102 kg/m3 (4) 1.49 × 102 kg/m3
24. Moment of inertia of a square plate of side l about the axis passing through one of the corner, and perpendicular to the plane of square plate is given by (1) 2 6 Ml (2) Ml2 (3) 2 12 Ml (4) 22 3 Ml
(1) 10 MR2 (2) 372 9 MR
(3) 4 MR2 (4) 402 9 MR
23. The solid cylinder of length 80 cm and mass M has a radius of 20 cm. Calculate the density of the material used, if the moment of inertia of the cylinder about an axis CD parallel to AB, as shown in figure, is 2.7 kg m2
25. The radius of gyration of a uniform rod of length , about an axis passing through a point L/4 away from the centre of the rod and perpendicular to it is (1) 7 48 L (2) 3 8 L (3) 1 8 L (4) 1 4 L
26. An object of mass 8 kg is hanging from one end of a uniform rod CD of mass 2 kg and length 1 m pivoted at its end C on a vertical wall as shown in figure. It is supported by a cable AB such that the system is in equilibrium. The tension in the cable is (Take g = 10 m/s2) B
cm wall C A
° 30° D 8 kg
(1) 14 9. kg/m3
(2) 7.5 × 101 kg/m3
(1) 240 N (2) 30 N (3) 300 N (4) 90 N
27. Shown in the figure is rigid and uniform one metre long rod AB held in horizontal position by two strings tied to its ends and attached to the ceiling. The rod is of mass ‘m’ and has another weight of mass 2 m hung at a distance of 75 cm from A. The tension in the string at A is
(1) 2 mg (2) 1 mg (3) 0.5 mg (4) 0.75 mg
28. A slender uniform rod of mass M and length is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod, when it makes an angle θ with the vertical, is
CHAPTER 8: System of Particles and Rotational Motion
29. A cord is wound round the circumference of a wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance ‘h’, the square of angular velocity of wheel will be
30. Consider a uniform ro d of mass M = 4 m and length L , pivoted about its centre. A mass m, moving with velocity ν and making angle 4 θ=π to the rod’s long axis, collides with one end of the rod and sticks to it. The angular speed of the rod-mass system, just after the collision, is
(1) 32 7 L ν (2) 3 7 L ν (3) 3 72 L ν (4) 4 7 L ν
31. A thin circular ring of mass M and radius R is rotating with a constant angular velocity of 2 rads −1 in a horizontal plane about an axis vertical to its plane and passing through the centre of the ring. If two objects, each of mass m, are attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity (in rads−1) of
(1) () M Mm + (2) () 2 2 Mm M + (3) () 2 2 M Mm + (4) ()22Mm M +
32. Consider a situation in which a ring, a solid cylinder, and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and they have identical diameter. Choose the correct statement for this situation.
(1) The sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.
(2) The ring has the greatest and the cylinder has the least velocity of the centre of mass at the bottom of the inclined plane. (3) All of them will have same velocity.
(4) The cylinder has the greatest and the sphere has the least velocity of the centre of mass at the bottom of the inclined plane.
33. A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm/s. Its kinetic energy is
34. A solid cylinder and a solid sphere, having same mass M and radius R, roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be (1) 5 3 (2) 4 5 (3) 3 5 (4) 14 15
35. Solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is
37. I CM is the moment of inertia of a circular disc about an axis (CM) passing through its centre and perpendicular to the plane of disc. IAB is its moment of inertia about an axis AB perpendicular to plane and parallel to axis CM at a distance 2 3 R from centre, where R is the radius of the disc. The ratio of IAB and ICM is x : 9. The value of ‘x’ is .
2 R M B 3 C R A
38. Two discs of the same mass and different radii are made of different materials such that their thickness is 1 cm and 0.5 cm, respectively. The densities of materials are in the ratio 3 : 5 . The moment of inertia of these discs, respectively, about their diameters will be in the ratio of 6 x . The ratio of x is .
Numerical Value Questions
36. A uniform solid cylinder with radius R and le ngth L has moment of inertia I 1, about the axis of the cylinder. A concentric solid cylinder of radius 2 R R ′ = and length 2 L L ′ = is carved out of the original cylinder. If I2 is the moment of inertia of the carved out portion of the cylinder, then 1 2 = I I (Both I 1 and I 2 are about the axis of the cylinder)
39. Two identical solid spheres, each of mass 2 kg and radii 10 cm, are fixed at the ends of a light rod. The separation between the centres of the spheres is 40 cm. The moment of inertia of the system about an axis perpendicular to rod passing through its middle point is × 10–3 kg-m2
40. If a solid sphere of mass 5 kg and a disc of mass 4 kg have the same radius, then the ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be 7 x . Then, the value of x is .
41. A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be ms–1
42. A thin uniform rod of length 2 m, cross-sectional area ‘ A ’, and density ‘ d ’ is rotated about an axis passing through the centre and perpendicular to its length, with angular velocity ω. If value of ω, in terms of its rotational kinetic energy E, is , αE Ad then the value of α is
43. A pulley of radius 1.5 m is rotated about its axis by a force F = 12 t –3 t 2 N applied tangentially (while t is measured in seconds). If moment of inertia of the pulley about its axis of rotation is 4.5 kg m 2, the number of rotations made by the pulley before its direction of motion is reversed will be K π The value of K is .
44. A solid disc of radius 20 cm and mass 10 kg is rotating with an angular velocity of 600 rpm, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc to rest in 10 s is _____ π × 10 –1 Nm.
CHAPTER 8: System of Particles and Rotational Motion
47. The position vector of 1 kg object is () ˆ 3 ˆ rijm =− and its velocity is
v3j1 ˆ ms ˆ k =+ . The magnitude of its angular momentum about the origin is x Nm , where x is
48. A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity of 3 m/s (as shown in figure). Maximum height with respect to the initial position covered by it will be cm. (take g = 10 m/s2)
m 3 S
49. For a rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is 5 x . The value of x is
50. A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is π : 22, then the value of its angular speed will be _________ rad/s.
is 1 x . The value of x will be .
45. A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface, hits one end of the rod with a velocity u, in a direction perpendicular to the rod. The collision is completely elastic. After collision, the particle comes to rest. The ratio of masses m M
46. A circular plate is rotating in horizontal plane, about an axis passing through its centre and perpendicular to the plate, with an angular velocity ω. A person sits at the centre having two dumbbells in his hands. When he stretches out his hands, the moment of inertia of the system becomes triple. If E is the initial kinetic energy of the system, then the final kinetic energy will be E x . The value of x is
51. A uniform disc of mass 0.5 kg and radius r is projected with velocity 18 m/s at t = 0 s on a rough horizontal surface. It starts off with a purely sliding motion at t = 0 s. After 2 s it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after 2 s will be . J (given: coefficient of friction is 0.3 and g = 10 m/s2).
18 m/s t=0 t=2 s
Level-III
Single Option Correct MCQs
1. Four point masses m , 2 m , M 1 and M 2 are placed on a plane, as shown in figure. If the coordinates of the centre of mass of the system are 67 ,, 510
then the value of M1 and M2 are, respectively,
(1) 3m and 4m (2) 4m and 3m (3) 2943 and 44 mm (4) 29and6 4 mm
2. A circular plate of uniform thickness has a diameter of 28 cm. A circular portion of diameter 21 cm is removed from the plate, as shown. O is the centre of mass of complete plate. The position of centre of mass of the remaining portion will shift towards left from O by (in cm)
(1) h/m (2) h/4 (3) 2h/3 (4) 2h/4
(1) 2.5 (2) 3.5 (3) 4.5 (4) 5.5
3. A paraboloid shaped solid object has equation y = Kx2 about y-axis, as shown in figure. If the height of the body is h, then the position of the centre of mass from the origin (assume density to be uniform throughout) is X Y O h
4. Shown in the figure is a system of three particles of mass 1 kg, 2 kg, and 4 kg, connected by two springs. The acceleration of A, B, and C at any instant is 1 ms–2, 2 ms–2, and 1/2 ms–2, respectively, directed as shown in the figure. External force acting on the system is
a2 a1 (1) 1 N (2) 7 N (3) 3 N (4) 10 N
3
5. Six identical balls are lined in a straight groove made on a horizontal frictionless surface, as shown. Two similar balls, each moving with a velocity v, collide elastically with the row of 6 balls from left. What will happen?
(1) One ball from the right rolls out with a speed 2v and the remaining balls will remain at rest.
(2) Two balls from the right roll out with speed v each and the remaining balls will remain stationary.
(3) All the six balls in the row will roll out with speed v/6 each and the two colliding balls will come to rest.
(4) The colliding balls will come to rest and no ball rolls out from right.
6. A body of mass m, having an initial velocity v , makes head on elastic collision with a stationary body of mass M . After the collision, the body of mass m comes to rest and only the body having mass M moves. This will happen only when (1) m >> M
(2) m << M
(3) m = M
(4) 1 2 mM =
7. A ball is dropped onto a floor. If the coefficient of restitution is 3 , 2 the percentage loss of energy of the ball on rebounding from the floor is
(1) 12.5 % (2) 50% (3) 25% (4) 75%
8. A sphere of mass m, moving with a constant velocity u, hits another stationary sphere of the same mass. If e is the coefficient of restitution, then the ratio of the velocity of two spheres after collision will be
(1) 1 1 e e + (2) 1 1 e e +
(1) e = 0.8 (2) e = 0 (3) e = 1 (4) e = 1/2
11. A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with centre at O, as shown. The moment of inertia of the loop about the axis xx' is
12. Particles of masses 1 g, 2 g, 3 g, ……., 100 g are kept at the marks 1 cm, 2 cm, 3 cm,……, 100 cm, respectively, on a meter scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the meter scale.
(1) 0.34 kg –m2
(2) 0.43 kg –m2
(3) 4 kg –m2
(4) 2 kg –m2
13. Which of the following graphs represents the relation between log K and log I, where K is the radius of gyration and I is the moment of inertia?
9. A body falling from a height of 10 m rebounds from hard floor. If it loses 20% energy in the impact, then coefficient of restitution is
(1) 0.89 (2) 0.56
(3) 0.23 (4) 0.18
10. A ball is given velocity 3 ugd = at an angle of 45° with the horizontal. It strikes a wall at a distance d and returns to its original position. Find the coefficient of resistitution e.
(1) log I
(2) log I
14. A solid cylinder is suspended symmetrically, through the massless strings, as shown in the figure. The distance of centre of mass from the initial rest position, for the cylinder to fall by unwinding the strings to achieve a speed of 6 ms–1, is cm.
(1) 120 (2) 210 (3) 160 (4) 270
15. A ‘T’ shaped object, with dimensions shown in the figure, is lying on a smooth floor. A force F is applied at the point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of P with respect to C.
17. A uniform disc of mass m and radius R is projected horizontally with velocity V0 on a rough horizontal floor, so that it starts off with a purely sliding motion at t = 0. After t0 seconds, it acquires a purely rolling motion as shown in figure. The velocity of the centre of mass of the disc at t0 is
16. A uniform rod of length ‘ ’ is free to rotate about a fixed horizontal axis through O. The rod begins rotating from rest from its unstable equilibrium position. When it is turned through an angle θ , its angular velocity ω is given as
18 A solid cylinder and a hollow cylinder, both of the same mass and same external diameter, are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?
(1) Both reach together when the angle of inclination is 45°
(2) Both reachtogether
(3) Hollow cylinder
(4) Solid cylinder
19. A bullet of mass m, moving with a speed v, hits a stationary block of mass M at the topmost point and gets embedded in it. If friction is sufficient to prevent slipping, then the block
(1) must topple down
(2) must not topple down
(3) may topple down (4) will remain at rest
Numerical Value Questions
20. A rod AB of is such that its linear density (mass per unit length) μ varies as =, µ a bx where x is the distance of the section from end A (and b > l). The distance of the centre of mass from end A is ,
where z is
21. Two blocks of masses m = 2 kg and M = 4 kg are inter connected by an ideal spring of stiffness 4/3 Nm–1. If m is pushed with a velocity v0 = 3 ms–1, Then the velocity of the centre of mass of the system will be ms–1.
2 kg v0 4 kg
22. A system consists of two particles. At t = 0, one particle is at the origin and the other particle, having a mass of 600 kg, is on the y-axis at y = 80 m. At t = 0, the centre of mass of the system is on the y -axis at y = 24 m and has a velocity given by 2 6, ktj where k =1 ms –3. The acceleration of the centre of mass, in ms −2, at time t = 2 s is
23. A particle of mass m, moving with speed π m/s, hits elastically another stationary particle of mass 2 m on a smooth horizontal circular tube of radius 5 m. The time in which the next collision will take place is s.
24. A body of mass m1, moving with uniform velocity of 40 m/s, collides with another mass m2, at rest, and then, the two together begin to move with uniform velocity of 30 m/s. The ratio of their masses 1 2 m m is
CHAPTER 8: System of Particles and Rotational Motion
25. A metal ball falls from a height of 32 m on a steel plate. If the coefficient of restitution is 0.5, to what height (in m) will the ball rise after the second bounce?
26. A ball of mass 10 kg, moving with a velocity of 103m/s along the x–axis, hits another ball of mass 20 kg, which is at rest. After the collision, the first ball comes to rest while the second ball disintegrates into two equal pieces. One piece starts moving along the y-axis with a speed of 10 m/s. The second piece starts moving at an angle of 30° with respect to the x -axis. The velocity of the ball moving at 30° with x-axis is x m/s. The value of x to the nearest integer is The configuration of pieces after collision is shown in the figure below.
y–axis
V1 = 10 m/s
x–axis 300 Piece-1
27. Consider two uniform discs of the same thickness and different radii, R 1 = R and R2 = a R made of the same material. If the ratio of their moments of inertia, I1 and I2, respectively, about their axes is I1 : I2 = 1 : 81 then the value of ' a ' is
28. I1 is the moment of inertia of a solid cylinder of radius R and length l about its geometrical axis. A concentric cylinder having radius R/2 and length l /2 is carved from the original cylinder and I 2 is the MOI of the carved cylinder. Find 1 2 . I I
29. Three spheres, each of mass m and radius R are kept in touch with each other such that their centres form an equilateral triangle. The moment of inertia of the system about
a median of triangle is 2 . 5 x MR The value of x is .
30. A shell of mass m is projected with a velocity v at an angle 60° to horizontal. When it reaches the maximum height, its angular momentum with respect to point the of projection, is 33 mv xg The value of x is
31. A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits at one end of the rod with a velocity u, in a direction perpendicular to the rod. The collision is completely elastic. After the collision, the particle comes to rest. The ratio of masses 1 is m Mx
The value of x will be
32. A shaft rotating at 6000 rpm is transmitting a power of 2π kW. The magnitude of the driving torque is equal to Nm.
33. A rod of length L is held vertically on a smooth horizontal surface. The top end of the rod is given a gentle push. At a certain instant of time, when the rod makes an angle of 37° with the horizontal the velocity of COM of the rod is 4 ms–1. The velocity of the end of the rod in contact with the surface at that instant is m/s.
34. A thin rod of mass 12 m and length 6 L is bent into a regular hexagon. The MI of the hexagon about a normal axis to its plane and through centre of system is KmL2. Then, the value of K is
35. Moment of inertia of a body about a given axis is 1 kgm2. Initially, the body is at rest. In order to produce a rotational kinetic energy of 800 J, the constant angular acceleration
of 10 rad/s2 must be applied about the axis for a duration of s.
36. Moment of inertia of a thin rod of mass M and length L, about an axis passing through its centre, is 2 12 ML Its moment of inertia about a parallel axis at a distance of L/4 from this axis is given by 2 48 KML . The value of K is .
37. A particle of mass m is moving in time t on a trajectory given by () 2 , ˆ 1055 ˆ rtitj =α+β− where a and b are dimensional constants. The angular momentum of the particle becomes the same as it was for t = 0 at time t equal to s.
38. A force 34 ˆ 4 ˆˆ Fijk =++ N is applied on an intersection point of x = 2 m plane and x-axis. The magnitude of torque of this force about a point (2 m, 3 m, 4 m) is _____Nm.
THEORY-BASED QUESTIONS
Single Option Correct MCQs
1. Among the following, correct one in the case of two particle system is
(1) CM lies nearer to lighter body
(2) CM always lies at midpoint of line joining the bodies
(3) CM lies nearer to heavier body
(4) CM always lies outside the system
2. A cylinder is completely filled with water. If 1/4 of the volume of water leaks out, its centre of mass
(1) moves up
(2) moves down
(3) does not change
(4) moves towards vertical surface
3. Three identical spheres, each of radius R, are placed touching each other on a horizontal table. The centre of mass of the system is located
(1) at one of the centres of the spheres
(2) at the midpoint joining the centres of any two spheres
(3) at the point of intersection of the medians of the triangle formed by the centres of the three spheres
(4) at the midpoint of a median of the triangle formed by the centres of the three spheres
4. Choose the wrong statement.
(1) In the case of small bodies [uniform gravitational field], centre of mass and centre of gravity coincide with each other.
(2) In non-uniform gravitational field, centre of mass and centre of gravity will not coincide with each other.
(3) Centre of mass is independent of acceleration due to gravity, but centre of gravity depends on acceleration due to gravity.
(4) Centre of mass describes the stability of the body whereas centre of gravity describes the nature of motion of the body.
5. A shell is thrown vertically up. The shell at the highest point explodes into two equal fragments. The centre of mass of the two fragments
(1) goes further up and then falls
(2) falls down with an initial speed
(3) falls down with zero inital velocity
(4) comes to rest
6. A uniform metre stick is placed vertically on a horizontal frictionless surface and released. As the stick is in motion, the centre of mass moves
(1) vertically up
(2) vertically down
(3) in a parabolic path
(4) horizontally
CHAPTER 8: System of Particles and Rotational Motion
7. When an external force is applied at the centre of mass of a system of particles, then it undergoes
(1) Only translatory motion
(2) Only rotatory motion
(3) Both translatory and rotatory motion
(4) An oscillatory motion
8. If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that
(1) linear momentum of the system does not change in time
(2) kinetic energy of the system does not change in time
(3) angular momentum of the system does not change in time
(4) potential energy of the system does not change in time
9. A thin rod has a non-uniform density. It is mounted on an axle passing perpendicular to it, through its centre of mass, as shown, and is then rotated about the axle.
The axle divides the rod into two parts, one on each side of it. Which of the following must be true, no matter how the mass in the rod is distributed?
(1) The two parts have the same mass.
(2) The magnitudes of the momentum of the two parts are equal.
(3) The magnitudes of the angular momentum of the two parts, about the centre of mass, are equal.
(4) The kinetic energies of the two parts are equal.
10. Choose the incorrect statement.
(1) In the process of explosion, some changes may occur in momentum of individual fragments due to internal forces but the motion of the centre of mass is unaltered.
(2) Motion of centre of mass depends upon the external force.
(3) The location of centre of mass relative to the body depends on the reference frame used to locate it.
(4) The position of centre of mass depends upon the shape of the body and the distribution of mass.
11. A body of mass M1 collides elastically with another mass M2 at rest. There is maximum transfer of energy when
(1) M1 > M2 (2) M1 < M2
(3) M1 = M2
(4) same for all values of M1 and M2
12. Which of the following statements is wrong?
(1) KE of a body is independent of the direction of motion.
(2) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(3) If two protons are brought towards each other, the PE of the system increases.
(4) A body can have energy without momentum.
13. A ball hits the floor and rebounds after inelastic collision. In this case,
(1) the momentum of the ball just after the collision is the same as that just before the collision
(2) the mechanical energy of the ball remains the same in the collision
(3) the total momentum of the ball and the earth is conserved
(4) the total energy of the ball and the earth is conserved
14. Choose the incorrect options.
(1) Kinetic energy is not conserved in an elastic collision.
(2) Momentum of the system is conserved during a collision.
(3) Kinetic energy of the system is conserved during an elastic collision.
(4) Coefficient of restitution for perfectly inelastic collision = 0.
15. If two balls collide in air while moving vertically, then momentum of the system is conserved because
(1) gravity does not affect the momentum of the system
(2) force of gravity is very less compared to the impulsive force
(3) impulsive fore is very less than the gravity
(4) gravity is not acting during collision
16. During elastic collision, (1) law of conservation of momentum is valid
(2) law of conservation of kinetic energy is valid
(3) both (1) and (2)
(4) neither (1) nor (2)
17. Which of the following is not a perfectly inelastic collision?
(1) Striking of two glass balls
(2) A bullet striking a bag of sand
(3) An electron captured by proton
(4) A man jumping on to a moving cart
18. Coefficient of restitution depends upon (1) the relative velocities of approach
(2) the masses of the colliding bodies
(3) the nature of the colliding bodies
(4) all of the above
19. Moment of inertia of a body depends upon (1) distribution of mass of the body
(2) position of axis of rotation
(3) temperature of the body
(4) all of the above
20. In a rectangle ABCD, AB = 2l and BC = l. The centre of rectangle is at the origin of co-ordinate system. AB and CD are parallel to x-axis. The MI is least about
(1) BD (2) BC
(3) x-axis (4) y-axis
21. The graph between rotational energy E r and angular velocity (ω) is represented by which of the following curve?
(1) Y X E r 0 ω (2) Y X E r 0 ω
(3) Y X E r 0 ω (4) Y X E r 0 ω
22. When temperature of a body made of brass increases, its moment of inertia (1) increases (2) decreses (3) remains the same (4) becomes zero
23. If I is the moment of inertia of a solid sphere about an axis parallel to a diameter of the sphere and at a distance x from it, which of the following graphs represents the variation of I with x?
(1) I X (2) I X
(3) I X (4) I X
24. Choose the correct statement (1) The total torque on a system is independent of the origin if the total external force is zero.
(2) The angular momentum L and angular velocity ω are always anti-parallel vectors.
CHAPTER 8: System of Particles and Rotational Motion
(3) We cannot have a situation in which total external torque is non-zero while net external force is zero.
(4) The centre of gravity always coincides with centre of mass.
25. When a constant torque is applied on a rigid body, then
(1) the body moves with linear accelaration
(2) the body rotates with constant angular velocity
(3) the body rotates with constant angular accelaration
(4) the body undergoes equal angular displacement in equal intervals of time
26. A solid uniform sphere rotating about its axis with kinetic energy E1 is gently placed on a rough horizontal plane at time t = 0. Assume that, at time t = t 1, it starts pure rolling and, at that instant, total KE of the sphere is E2. After some time t = t2, KE of sphere is E3. Then,
(1) E1 = E2 = E 3 (2) E1 > E2 = E 3
(3) E1 > E2 > E 3 (4) E1 < E2 = E 3
27. In both the figures, all other factors are same, except that in figure (i), AB is rough and BC is smooth while in figure (ii), AB is smooth, and BC is rough.
Kinetic energy of the ball on reaching the bottom:
(1) is same in both the cases
(2) is greater in case (i)
(3) is greater in case (ii)
(4) information insufficient
28. A sphere is rolled on a rough horizontal surface. It gradually slows down and stops. The force of friction tries to
(1) increase the linear velocity
(2) increase the angular velocity
(3) increase the linear momentum
(4) decrease the angular velocity
29. A disc is rolling (without slipping) on a rough surface. C is its centre and Q and P are two points equidistant from C. Let Vp, VQ, and VC be the magnitudes of velocities of points P, Q, and C, respectively. Then,
S-II : Centre of gravity of a body is the point through which the whole weight of the body acts.
32. S-I : In a collision between two objects, the centre of mass of the system always remains stationary if there are no external forces acting on it.
S-II : The centre of mass of a system of particles will remain at rest or move with a constant velocity unless acted upon by an external force. This is a consequence of the law of inertia and is consistent with the concept of conservation of momentum.
(1)
(2)
(3)
(4) VQ < VC > VP Statement Type Questions
Each question has two statements: statement I (S-I) and statement II (S-II). Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect
(3) if statement I is correct but statement II is incorrect,
(4) if statement I is incorrect but statement II is correct.
30. S-I : The centre of mass of a body may lie where there is no mass.
S-II : The centre of mass has nothing to do with the mass.
31. S-I : Centre of mass is that fixed point of a system of particles or a rigid body within the boundaries of the system where the entire mass is concentrated.
33. S-I : Two bodies A and B, initially at rest, of masses 2m and m, respectively, move towards each other under mutual gravitational force of attraction. At the instant when the speed of A is v and that of B is 2v, the speed of the centre of mass of the system is 4v/3.
S-II : The speed of the centre of mass of a system changes if an external force acts on the system.
34. S-I : The initial kinetic energy is equal to the final kinetic energy but the kinetic energy during the collision time is not constant. Such a collision is called an elastic collision.
S-II : A collision between two particles in which the two particles move together after the collision is called a completely inelastic collision.
35. S-I : If a uniform metal disc is remoulded into a solid sphere, then the moment of inertia about the axis of symmetry increases than that before.
S-II : For a given body and for a given plane, the moment of inertia is minimum about an axis passing through the centre of mass.
36. S-I : A rigid body executes both translatory and rotatory motions.
S-II : A rigid body executes pure rotational motion when acted upon by unbalanced torque.
37. S-I : If there is no external torque on a body about its centre of mass, then the velocity of the centre of mass remains constant.
S-II : The linear momentum of an isolated system(Fext= 0) remains constant.
Assertion and Reason Type Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R).
Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A), (2) if both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) if (A) is true but (R) is false, (4) if both (A) and (R) are false.
38. (A) : In a perfectly inelastic one-dimensional collision between two spheres, velocity of both spheres just after collision, is equal.
(R) : For any collision between finite bodies, momentum is always conserved.
39. (A) : Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
(R) : Principle of conservation of momentum holds true for all kinds of collisions.
40. (A) : Moment of inertia of a body is same, whatever be the axis of rotation.
(R) : Moment of inertia does not depends on mass but not the size of the body.
41. (A) : If a rod is thrown up vertically in the absence of air drag, it returns back to the ground without any change in its angular velocity.
(R) : In the absence of any torque, the angular momentum of system is conserved.
CHAPTER 8: System of Particles and Rotational Motion
42. (A) : Two cylinders, one hollow (metal) and the other solid (wood), with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.
(R) : By the principle of conservation of energy, the rotational kinetic energies of both the cylinders are different when they reach the bottom of the incline.
43. (A) : A ring moving down on a smooth inclined plane will be in slipping motion.
(R) : Work done by friction in pure rolling motion is zero.
44. (A) : A solid sphere is rolling on a rough horizontal surface. Acceleration of contact point is zero.
(R) : A solid sphere cannot roll on the smooth surface.
45. (A) : Heavy boxes are loaded along with empty boxes on a cart. Heavy boxes should be kept at the centre of gravity of the cart, for more stability.
(R) : The vertical line passing through the centre of gravity should fall within the base of cart for stability.
46. (A) : The centre of mass of an electron and proton, when released, moves faster towards proton.
(R) : This is because proton is lighter.
47. (A) : A shell moving in a parabolic path explodes in mid air. The centre of mass of the fragments will follow the same parabolic path.
(R) : Explosion is due to internal forces, which cannot alter the state of motion of centre of mass.
48. (A) : The centre of mass of system of n particles is the weighted average of the position vector of the n particles making up the system.
(R) : The acceleration of the centre of mass of a system is independent of internal forces.
49. (A) : In an elastic collision of two billiard balls, both kinetic energy and linear momentum remain conserved.
(R) : During the collision of the balls, as the collision is elastic, there is no exchange of energy. Therefore, both energy and momentum are conserved.
50. (A) : Moment of inertia of a circular disc of mass M and radius R about x , y-axes (passing through its plane) and z -axis, which is perpendicular to its plane, were found to be I x, I y , and I z, respectively. The respective radii of gyration about all the three axes will be different.
(R) : A rigid body making rotational motion does not have fixed mass and shape.
51. (A) : If there is no external torque on a body about its centre of mass, then the velocity of centre of mass remains constant.
(R) : The linear momentum of an isolated system may not remains constant.
52. (A) : If polar ice melts, then day will be longer.
(R) : Moment of inertia of the earth increases and, thus, angular velocity decreases.
53. (A) : A body is moving along a circle with a constant speed. Its angular momentum about the centre of the circle remains constant.
(R) : In this situation, a constant non-zero torque acts on the body.
54. (A) : For a particle moving on a straight line with a uniform velocity, its angular momentum is constant.
(R) : The angular momentum is zero when the particle moves with a uniform velocity.
55. (A) : The velocity of a body at the bottom of an inclined plane of given height is more when it slides down the plane compared to when it is rolling down the same plane.
(R) : In rolling down, a body acquires both kinetic energy of translation and rotation.
56. (A) : A solid sphere and hollow sphere, when released from the top of a smooth and fixed inclined plane, reach the ground simultaneously.
(R) : Acceleration while sliding is independent of mass and dimension of the body.
JEE ADVANCED LEVEL
Multiple Option Correct MCQ
1. A n object comprises of a uniform ring of radius R and a uniform chord AB (not necessarily made of the same material), as shown. Which of the following cannot be the centre of mass of object? A B X Y
(1) , 33
RR (2) , 34
RR
(3) , 22
RR (4) , 22
RR
2. If a disc of radius r is removed from the disc of radius R, then which of the following are true?
(1) The minimum shift in centre of mass is zero
(2) The maximum shift in centre of mass cannot be greater than 2 . r Rr +
(3) Centre of mass must lie where mass exists.
(4) The shift in centre of mass is 2 . r Rr +
3. A long, thin, inextensible, and very flexible uniform wire is lying on the rough horizontal floor. One end of the wire is bent back and then pulled backward with constant velocity V such that, at any instant of time, the moving part of the wire, always remains just above the part of the wire which is still at rest at that instant on the floor, as shown in diagram. If the wire has unit length and unit mass, then
floor
(1) speed of centre of mass of the moving part at all the times will remain constant
(2) speed of centre of mass of the moving part at all the times will be 3/4 V
(3) minimum force needed to pull the moving part will be V2/2
(4) minimum force needed to pull the moving part will be 3/8 V2
4. A block is kept on a wedge, which is kept on a horizontal floor, as shown in figure. All the surfaces are frictionless. Then, M
(1) centre of mass will not shift horizontally
(2) centre of mass will shift horizontally
(3) centre of mass will shift vertically
(4) centre of mass will shift neither horizontally nor vertically
5. Which of the following is/are correct?
(1) If centre of mass of three particles is at
CHAPTER 8: System of Particles and Rotational Motion
rest and two of them are moving along different lines, then the third particle must also be moving.
(2) If centre of mass remains at rest, then net work done by the forces acting on the system must be zero.
(3) If centre of mass remains at rest, then the net external force must be zero.
(4) If speed of centre of mass is changing then there must be some net work being done on the system from outside.
6. A block of mass m slides down on an inclined plane of a wedge of same mass m (see fig.). Friction is absent everywhere. Which of the following statements is/are correct?
(1) Vertical component of acceleration of block is 2 2 cos 1sin g θ +θ downwards.
(2) Acceleration of centre of mass of (block + wedge) is 2 2 sin 1sin g θ +θ downwards.
(3) Acceleration of centre of mass of (block + wedge) is 2 2 cos 1sin g θ +θ downwards.
(4) Vertical component of acceleration of block is 2 2 2sin 1sin g θ +θ downwards.
7. A cannon and a supply of cannon balls are inside a sealed rail road car at rest. The cannon fires to the right, the car recoils to the left. The balls remain in the car after hitting the farther wall. Assume friction to be absent. If L is the length of the rail road
car, D is the distance traversed by the rail road car relative to the ground, m is the total mass of cannon balls fired, and M is the total mass of remaining system (cannon + car), then
(1) D < L (2) D > L
(3) ML D Mm = + (4) mL D Mm = +
8. A block of mass m is placed at rest on a smooth wedge of mass M placed at rest on a smooth horizontal surface. As the system is released,
(2) After the collision, the centre of mass of ‘B’ plus ‘W’ moves with the velocity 2mV
mM +
(3) When ‘B’ reaches its highest position of W, the speed of W is + 2 . mV mM
(4) When B reaches its highest position of W, the speed of W is . mV
mM +
10. Ball A of mass m strikes a stationary ball B of mass M and undergoes an elastic collision. After collision, ball A has a speed one-third of its initial speed. The ratio of M/m is (1) 1/4 (2) 1/2
(3) 2 (4) 4
(1) The centre of mass of the system remains stationary
(2) The centre of the system has an acceleration g vertically downwards (3) momentum of the system is conserved along the horizontal direction (4) acceleration of centre of mass a cm < g is vertically downwards
9. In the figure, the block ‘B’ of mass m starts from rest at the top of a wedge 'W' of mass M. All surfaces are without friction. 'W' can slide on the ground. ‘B’ slides down on to the ground, moves along it with a speed ‘V’, has an elastic collision with the wall, and climbs back on to ‘W’.
B(m) W(M)
(1) From the beginning, till the collision with the wall, the centre of mass of ‘B’ plus ‘W’ does not move horizontally.
11. A particle moving with kinetic energy E makes a head-on elastic collision with an identical particle at rest. During the collision, (1) elastic potential energy of the system is always zero
(2) maximum elastic potential energy of the system is E/2
(3) minimum kinetic energy of the system is E/2
(4) kinetic energy of the system is constant
12. Two blocks A and B, each of mass m , are connected by a massless spring of natural length L and spring constant K. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length, as shown in the figure. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A and B and makes a perfectly elastic collision with A. Then,
(1) the kinetic energy of the A-B system, at maximum compression of the spring, is zero
(2) the kinetic energy of the A-B system, at maximum compression of the spring, is mv2/4
(3) the maximum compression of the spring is () /2 vmK
(4) all of these
13. A set of n identical cubical blocks lies at rest parallel to each other, along a line on a smooth horizontal surface. The separation between the near surfaces of any two adjacent blocks is L. The block at one end is given a speed v towards the next one at time t = 0. All collisions are completely inelastic. Then,
(1) the last block starts moving at () 1 nL t v =
(2) the last block starts moving at () 1 2 nnL t v =
(3) the centre of mass of the system will have a final speed v
(4) the centre of mass of the system will have a final speed v/n
14. In a one-dimensional collision between two identical particles, B is stationary and A has momentum p before impact. During impact, A gives impulse J to B. Then,
(1) the total momentum of A plus B system is p before and after the impact and (p–J) during the impact
(2) during the impact, B gives impulse J to A
(3) the coefficient of restitution is (2 J/p)–1
(4) the coefficient of restitution is (2 J/p)+1
15. A projectile is fired on a horizontal ground. Coefficient of restitution between the projectile and the ground is e . Let a , b and c be the ratio of time of flight 1 2 , T T
maximum height 1 2 , H H
and horizontal
CHAPTER 8: System of Particles and Rotational Motion
range 1 2 R R
in the first two collisions with the ground. Then,
(1) 1 a e = (2) 2 1 b e =
(3) 2 1 c e = (4) all of these
16. A ball strikes the ground at an angle α and rebounds at an angle β with the vertical, as shown in the figure. Then, ab
(1) coefficient of restitution is tan a /tan b (2) if α < β, the collision is inelastic (3) if α = β, the collision is elastic (4) if α > β, the collision is inelastic
17. Ball A of mass m moving with velocity V collides head on with a stationary ball B of mass m. If e is the coefficient of restitution, then which of the following statements are correct?
(1) The ratio of velocities of balls A and B after the collision is 1 . 1 e e +
(2) The ratio of the final and initial velocities of ball A is 1 . 2 e
(3) The ratio of the velocities of balls A and B after the collision is 1 . 1 e e +
(4) The ratio of the final and initial velocities of ball B is 1 2 e +
18. A ball of mass m moving horizontally at a speed v collides with the bob of a simple pendulum at rest. The mass of the bob is also m
(1) If the collision is perfectly inelastic, the height to which the two balls rise after the collision is v2/8g.
(2) If the collision is perfectly inelastic, the kinetic energy of the system immediately after the collision becomes half of that before collision.
(3) If the collision is perfectly elastic, the bob of the pendulum will rise to a height of v2/2g.
(4) If the collision is perfectly elastic, the kinetic energy of the system immediately after the collision is equal to that before collision.
19. A ball A collides elastically with an another identical ball B with velocity 10 m/s at an angle of from the line joining their centres C1and C2. Select the correct option(s).
(1) the angular velocity of rod ω = 10 rad/s, counter-clockwise
(2) the angular velocity of rod ω = 5 rad/s, counter-clockwise
(3) the velocity of centre of mass of rod, V cm = 2.5 m/s
(4) the velocity of centre of mass of rod, V cm = 5 m/s
21. A rod of length l is moving in a vertical plane(x-y plane) when the lowest point A of the rod is moved with a velocity v. Which of the following statements are true?
(1) Velocity of ball A after collision is 5 m/s.
(2) Velocity of ball B after collision is 53m/s .
(3) Both the balls move at right angles after collision.
(4) Kinetic energy will not be conserved here, because collision is not head on.
20. A uniform rod of mass m = 2 kg and length l = 0.5 m is sliding along two mutually perpendicular smooth walls with the two ends P and Q having velocities VP = 4 m/s and VQ = 3 m/s, as shown. Then,
(1) Angular velocity of the rod is
(2) Angular velocity of the rod is
(3) Velocity of the end B is 2 vtan θ .
(4) Velocity of the end B is 2 vcot θ
.
.
22. If a sphere is rotating about a diameter at uniform angular speed, then which of the following options is/are incorrect:
(1) The particles on the surface of the sphere do not have any linear acceleration.
(2) The particles on the diameter mentioned above do not have any linear acceleration.
(3) Different particles on the surface have different angular speeds.
(4) All the particles on the surface have the same linear speed.
23. A ring type flywheel of mass 100 kg and diameter 2 m is rotating at the rate of (300/π) revolutions per minute. Then,
(1) the moment of inertia of the flywheel is 100 kgm2
(2) the kinetic energy of rotation of the flywheel is 5 kJ
(3) the flywheel, if subjected to a rotating torque of 200 Nm, will come to rest in 5 s
(4) the moment of inertia of the flywheel is 143 kgm2
24. ABCD is a square plate with centre O. The moments of inertia of the plate about the perpendicular axis through O is I and about the axes 1, 2, 3, and 4 are I 1, I 2, I 3, and I 4, respectively. It follows that
CHAPTER 8: System of Particles and Rotational Motion
applied on the body, where α = 1.0 N–1 and β = 1.0 N. The torque acting on the body about the origin at time t =1.0 s is τ. Which of the following statements is (are) true?
(1) 1 N.m 3 τ=
(2) The torque τ is in the direction of the unit vector ˆ k +
(3) The velocity of the body at t = 1 s is () 1 2m/s. ˆˆ 2 vij =+
(4) The magnitude of displacement of the body at t = 1 s is 1 m. 6
27. A force ˆˆ FAiBj =+ acts at a point whose radius vector with respect to origin is ˆˆ raibj =+ , where a , b , A, and B are constants. Then,
(1) torque about ‘O’ is () ˆ aBbAk
(2) arm of force 22 aBbA AB + about origin
(1) I2 = I 3 (2) I = I1 + I 4
(3) I= I2 + I 4 (4) I1 = I 3
25. Four identical rods, each of mass m and length l, are joined to form a rigid square frame. The frame lies in the x-y plane, with its centre at the origin and the sides parallel to the x and y axes. The moment of inertia about
(1) the x-axis is 22 3 ml
(2) the z-axis is 42 3 ml
(3) an axis parallel to the z-axis and passing through a corner is 102 3 ml
(4) one side is 52 3 ml .
26. Consider a body of mass 1.0 kg at rest at the origin at time t = 0. A force ˆˆ Ftij =α+β is
(3) torque about ‘O’ is zero
(4) torque about ‘O’ is () ˆ aBbAk +
28. The torque of a force () 537N ˆˆˆ ijk +− about the origin is τ . If the force acts on a particle whose position vector is () ˆˆˆ 2mijk ++ , then the torque acting on the particle is
(1) () ˆˆ 17127.m ˆ N ijkτ=−+−
(2) 482N.m τ=
(3) ()17127N.m ˆˆˆ ijk
(4) () 17127N.m ˆˆˆ ijk ++
29. A small ball of mass m, suspended from the ceiling at a point O by a thread of length
l, moves along a horizontal circle with a constant angular velocity ω . Which of the following statements is/are correct?
(1) Angular momentum is constant about O.
(2) Angular momentum is constant about C.
(3) Vertical component of angular momentum about O is constant.
(4) Magnitude of angular momentum about O is constant.
30. A particle of mass m collides with the end of a spinning rod of mass 2m and length l, at the end of the rod. If the coefficient of restitution of collision is 2 3 e = , then
2m V0 l = V0 W0
(1) speed of particle after collision is 0 6 v
(2) speed of centre of rod after collision is 50 12 v
(3) angular speed of rod just after collision is 3 2 o v l ω=
(4) change in kinetic energy of the particle is 0 k ∆=
31. A particle of mass m is projected with a velocity v, making an angle of 45° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection, when the particle is at its maximum height h, is
(1) zero (2) 3 42g mv (3) 3 2g mv (4) m23 gh
32. The potential energy of a particle of mass m at a distance r from a fixed point O is given by () 2 4 kr Vr = , where k is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius R about the point O. If v is the speed of the particle and L is the magnitude of its angular momentum about O, which of the following statements is (are) true?
(1) k vR m =
(2) 2 k vR m = (3) 2 2 mk LR = (4) 2 LmkR =
33. A uniform meter stick of mass 200 g is suspended from the ceiling through two vertical strings of equal lengths fixed at the ends. A small object of mass 40 g is placed on the stick at a distance of 70 cm from the left end. Find the tension in the two strings.
(1) Tension in left end is 1.12 N.
(2) Tension in right end is 1.28 N.
(3) Tension in left end is 1.28 N.
(4) Tension in right end is 1.12 N.
34. A uniform circular disc of mass M and radius R rolls without slipping on a horizontal surface. If the velocity of its centre is v0, then the total angular momentum of the disc about fixed point ' p ' at a height 3 2 R above the centre C
CHAPTER 8: System of Particles and Rotational Motion
(2) The force of friction is to the right and equal in magnitude to 3 F
(3) The force of friction is to the left and equal in magnitude to . 3 F
(4) The force of friction is to the right and equal in magnitude to 4 . 3 F
(1) increases continuously as the disc moves away
(2) decreases continuously as the disc moves away
(3) is always equal to 2MRV0
(4) is always equal to MRV0
35. If a thin ring of mass 1 kg and radius 1 m is pure rolling at a speed of 1 m/s, then
(1) its total kinetic energy is 1 J
(2) its translational kinetic energy is 1/2 J
(3) its rotational kinetic energy is 1/2 J
(4) its total kinetic energy is 1/2 J
36. A spool of wire of mass M and radius 2R is unwound under a constant force F. Assume that the spool is a uniform, solid cylinder that does not slip.
37. Consider a cylinder of mass M and radius R lying on a rough horizontal plane. It has a plank lying on its top, as shown in figure. When a force F is applied on the plank, it moves and causes the cylinder to roll. The plank always remains horizontal. There is no slipping at any point of contact.
Then,
(1) The acceleration of the centre of mass is 4 . 3 F M
(1) the acceleration of the cylinder is 4cos 83 F mM θ +
(2) the acceleration of the cylinder is 4sin 83 F mM θ +
(3) the frictional force between plane and cylinder is cos 38 FM Mm θ +
(4) the frictional force between plane and cylinder is sin 38 FM Mm θ +
38. A wheel of radius 3R and mass m is placed on an inclined plane of inclination θ and a massless, inextensible string is wound around an axle of radius R , as shown in figure. If the inclined plane is frictionless
and string does not slip on the axle, then which of the following statements about the acceleration of the wheel and the tension in the string is/are true? (I = Moment of inertia of the wheel)
(1) Acceleration of the wheel is 2 I 1 gsin mR θ +
(2) Acceleration of the wheel is 2 g I 1 mR +
(3) Tension in the string is 2. 1 I mgsin mR θ +
(4) Tension in the string is 2 1 I mg mR +
39. The figure shows a system consisting of
(i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed ω and
(ii) an inner disc of radius 2R rotating anticlockwise with angular speed ω /2.
The ring and disc are separated by frictionless ball bearings. The system is in the x-z plane.
The point P on the inner disc is at a distance R from the origin, where OP makes an angle of 30° with the horizontal Then with respect to the horizontal surface,
(1) the point O has linear velocity 3 ˆ i Rω
(2) the point P has velocity 113ik 4 ˆˆ 4
(3) the point P has velocity
(4) the point P has velocity
Numerical Value Questions
40. A cylindrical rod of length 1.2 m is made of three materials. The first part of 0.5 m is made of iron, the second part of 0.3 m is of copper, and the last part of 0.4 m is of aluminium. If the densities of iron, copper, and aluminium are 7.9×103 kg/m3, 8.9 ×103 kg/m3, and 2.7 ×103 kg/m3, respecitively, the centre of mass of the rod from the right end of the aluminium rod is m.
41. Two rods of lengths l1 = 1.5 and l2 = 1 m, and masses m1 = 1 kg, and m2 = 2 kg, are joined end to end. The centre of mass of the rod from rod is 19/6y m. Then, y is
42. Four particles of masses 2, 2, 4, and 4 kg are arranged at the corners A, B, C, and D of a square ABCD of side 2 m, as shown in the figure. The perpendicular distance of their centre of mass from the side AB is C D 4kg 4kg 2kg 2kg A B (0, 0) (2, 0) (0, 2) (0, 2)
43. From a hemisphere of radius R, a cone of base radius R/2 and height R is removed as shown in figure. Calculate the height of centre of mass of the remaining object in cm, if R = 28 cm.
maximum height attained by their centre of mass, in metres. [Take g = 10 ms–2]
44. A small particle of mass m starts sliding down from rest along the smooth surface of a fixed hollow hemisphere of mass M (4 m ). The distance of centre of mass of (particle + hemisphere) from centre O of hemisphere, when the particle is separated from the surface of hemisphere, is 69 5 R α Find the value of 'α'.
47. A platform of mass m and counter weight of M + m (where, M = 3m) are connected by a light cord, which passes over two smooth pulleys. A man of mass M is standing on the platform, which is at rest. If the man leaps vertically upwards (and out of platform) with velocity 10 ms–1, find the distance (in m) through which the platform will descend.
45. Two particles A and B of mass 5 kg and 2 kg are projected simultaneously from same ground level, as shown in figure. Find the maximum height (in cm) reached by the centre of mass. Assume that only gravitational force of attraction by the earth is present, and take g = 10 ms–2 .
46. Two balls A and B of equal mass are projected upwards simultaneously, one from the ground with speed 50 ms −1 and the other from a tower of height 40 m above the first ball, with initial speed 30 ms−1. Calculate the
48. A body of mass 1 kg makes an elastic collision with another body at rest and continues to move in the original direction after collision with a velocity equal to 1/5th of its original velocity. Find the mass of the second body in kg.
49. A steel sphere of mass 100 g, moving with a velocity of 4 m/s, collides with a dust particle elastically moving in the same direction with a velocity of 1 m/s. The velocity of the dust particle after the collision is ____ m/s.
50. Ball 1 collides with another identical ball 2 at rest, as shown in figure. For what value of coefficient of restitution e , the velocity
of second ball becomes two times that of 1 after collision?
51. A ball strikes a horizontal floor at an angle θ=45°. The coefficient of restitution between the ball and the floor is 1 2 e = The fraction of its kinetic energy lost in collision is .
52. A ball of mass 0.2 kg is dropped down vertically from a height of 1 m above the ground. If it rebounds to a height of 0.64 m, find the coefficient of restitution between the surface of the ball and the ground.
53. A circular disc X of radius R is made from an iron plate of thickness t and another plate Y of radius 4R is made from an iron plate of thickness t/4. The ratio between moments of inertia y x I I is .
54. A ring and a solid sphere rotating about an axis passing through their centres have same radii of gyration. The axis of rotation is perpendicular to plane of ring. The ratio of radius of ring to that of sphere is 2 . x Then, the value of x is .
55. The length of a rod is 1012m. . The radius of gyration of a cylinder rod about an axis of rotation perpendicular to its length and passing through the centre will be m.
56. ABCDECA is a planar body of mass m of uniform thickness and same material. The dimensions are as shown in the figure. The moment of inertia of the body about an axis passing through point A and perpendicular to planar body is I1 and that of about an axis passing through C and perpendicular to planar body is I2. Then, 1 2 I I is
57. Three identical uniform rods, each of length 1 m and mass 2 kg, are arranged to form an equilateral triangle. The moment of inertia of the system about an axis passing through one corner and perpendicular to the plane of the triangle is ___kg–m 2.
58. A square plate of mass M and edge L is shown. The moment of inertia of the plate about the axis in the plane of plate and passing through one of its vertices, making an angle 15° from horizontal, is 112 , 8 ML x where x is
59. A light rod of length 2 m is pivoted at its centre and two masses of 5 kg and 2 kg are hung from the ends, as shown in figure. The initial angular acceleration of the rod, assuming that it was horizontal in the beginning, is α =____(rad/s2). Take g = 9.8 m/s2. 5 kg
60. A uniform circular disc of mass 1.6 kg and radius 0.5 m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude 0.6 N are applied simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc (in rad/s) is .
61. A horizontal platform of mass 200 kg in the form of a circular disc rotates at 10 rpm along a vertical axis passing through its centre. A man weighing 80 kg is standing on its edge. The angular velocity (in rpm) with which the platform rotates, if the man moves to its centre, is
62. A horizontal circular platform of radius 0.5 m and mass 0.50 kg is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass 0.1 kg, are attached to the platform at a distance 0.25 m from the centre on its either side, along its diameter (see figure). Each gun simultaneously fires the ball horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the ball has horizontal speed of 9 ms −1 with respect to the ground. The rotational speed of the platform (in rad/s) after the balls leave the platform is
63. A disc of mass 4 kg and radius R is free to rotate about a horizontal axis passing
8: System of Particles and Rotational Motion
through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now, the system is released. When the body comes to lowest position, its angular speed
will be 4 3 x R rad/s, where x = .
(g = 10 m/s2)
64. A uniform solid sphere of 1 kg rolls on a horizontal surface without slipping. If the velocity of centre of mass is 5 cm/s, the work done in stopping the sphere is N × 10–4 J. The value of N is .
65. A solid sphere of mass 10 kg rolls without sliding with uniform velocity 2 m/s along a horizontal table. The total energy of the sphere is J.
Integer Value Question
66. Two blocks A and B are connected by a massless string, as shown in figure. Force of F = 30 N is applied on block (assuming no friction) B. Then, distance travelled by the centre of mass in 2 seconds, starting from rest, is x metres. Find the value of x
B F A 10 kg 20 kg
67. A ball of mass m moving with a certain velocity collides against a stationary ball of mass m. The two balls stick together during collision. If E is the initial kinetic energy, then the loss of kinetic energy in the collision is E/A. Find the value of A.
68. A ring and a disc are initially at rest, side by side, at the top of an inclined plane, which makes an angle of 60° with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reachi ng the ground is () 23 s, 10 then the height of the top of the inclined plane, in metres, is x/4. The value of x is _____. (Take, g = 10 ms–2)
Passage-based Questions
Passage-I:
A (2n + 1) identical bricks, each of length L and height h, are arranged, with h are arranged,with n bricks above and n bricks below the central brick AB, as shown in figure, for n = 3. Each brick is displaced with respect to the one in contact by L/10. The x–y axes are defined with respect to the lowest brick, as shown.
(1) 0, 2 L (2) ,0 2 L
(3) ,0 23 L (4) L/3
72. The position coordinates of the centre of mass of triangular and circular plates with respect to point O are (L is side of triangles)
(1) () 31 23 π− π+ right of O
(2) 2 3 π+ right of O
(3) 1 3 π+ left of O
69. The y-coordinate of the centre of mass of the system is
(1) () 1 2 Lhn + (2) 1 2 nh +
(3) 2 nh (4) () 1 1 2 nh +
70. The x-coordinate of the centre of mass for the arrangement shown in the figure is
(1) 16 45 L (2) 32 45 L
(3) 22 35 L (4) 35 44 L
Passage-II:
Equilateral triangular, square, and circular plates of same thickness but made up of different materials are arranged as shown in figure.
(4) 3 3 π+ rigth of O
Passage-III:
Persons A, of mass 60 kg, and B, of mass 40 kg, are standing at rest on a horizontal platform of mass 50 kg. The platform is supported with wheels on a horizontal frictionless surface and is initially at rest. Both A and B jump from the platform simultaneously and in the same horizontal direction. Just after the jump, the persons A and B move away from the platform with a speed of 3 m/s relative to the platform and along the horizonta l.
A B
71. The coordinates of the centre of mass of the triangular plate of side L, (if O is the reference point) is
73. Just after both A and B jump from the platform, velocity of centre of mass of the system (A, B, and the platform) is ____m/s.
74. Final speed of the platform in situation (i), i.e., just after both A and B have jumped, will be nearly _____m/s.
Passage-IV:
A block of mass m moves with velocity v o towards a stationary block of mass M on a smooth horizontal surface.
m M k
75. Velocity of centre of mass of the system is
(1) 0 mv Mm + (2) 0 mv M
(3) 0 mv Mm (4) 0 2 v
76. Initial velocity of block of mass m w.r.t. centre of mass of the system is (1) o Mv mM + (2) 0 mv Mm +
(3) 0 2 v (4) 0 3 v
Passage-V:
In fig, block A of mass 2 kg is moving to the right with a speed of 5 m/s on a horizontal frictionless surface. Another block B of mass 3 kg, with a massless spring attached to it, is moving to the left on the same surface and with a speed of 2 m/s. Let us take the direction to the right as the positive x -direction. At some instant, block A collides with the spring attached to block B. At some other instant, the spring has maximum compression and then, finally, the blocks move with their final velocities. Assume that
(i) the spring force is conservative and, so, there is no conversion of kinetic energy to interval energy, and
(ii) no sound is made when block A hits the spring.
CHAPTER 8: System of Particles and Rotational Motion
78. Final speed of centre of mass of the system of blocks A and B will be _____m/s.
Passage-VI:
Two pendulum bobs of mass m and 2m collide elastically at the lowest point in their motion. If both the balls are released from height h above the lowest point, answer the following.
79. Speed of the bob of mass m just after collision is 5 3 xgh Then, the value of x is .
80. If the height to which the bob of mass m rises after the collision is 9 yh , then the value of y is .
Passage-VII:
Two smooth spheres, A and B, of equal radius lie on a horizontal table. A is of mass m and B is of mass 3 m . The spheres are projected towards each other with velocity vectors 52and2, ˆˆ
ijij +− respectively, and when they collide, the line joining their centres is parallel to the vector ˆ i . The coefficient of restitution between A and B is 1/3.
81. Their velocities after impact are (1) 22, ˆˆˆˆ 3 ijij +− (2) 22, ˆˆˆˆ 3 ijij (3) ˆˆˆˆ ,3 ijij −+ (4) ˆˆˆˆ ,3 ijij
82. The loss in kinetic energy caused by the collision is (1) m (2) 2m (3) 3m (4) 4m
Passage-VIII:
In the given figure, F = 10 N, R = 1 m, mass of the body is 2 kg, and moment of inertia of the body about an axis passing through O and perpendicular to plane of body is 4 kgm 2. O is the centre of mass of the body .
77. Final speed of block A will be ____m/s.
83. If ground is sufficiently rough, what is the kinetic energy of the body (in J) after 2 s?
84. If the ground is smooth, what is the total kinetic energy (in J) of the body after 2 s?
Passage-IX:
The minimum coefficient of static friction between an inclined plane and cylinder for the cylinder not slipping on the inclined plane is µ. The angle of inclination is 45° . Now the angle of inclination is changed to 30°, then the cylinder reaches the bottom with an acceleration a ms–2 (given g = 10 ms–2)
85. The value of µ is
86. The magnitude of a is
Matrix Matching Questions
87. A rectangular slab ABCD has dimensions a × 2 a , as shown in figure. Match the following two columns
Column I
Column II
(A) Radius of gyration about axis-1 (I) 12 a (B) Radius of gyration about axis-2 (II) 2 3 a (C) Radius of gyration about axis-3 (III) 3 a (D) Radius of gyration about axis-4 (IV) 4 3 a (V) 43 a
(A) (B) (C) (D)
(1) II III III I
(2) II III IV I
(3) I IV II III
(4) II IV I III
88. Square frame is made of four uniform rods. Mass of each rod is m and length is l.
Column I (Axis of rotation)
Column II (MI about axis of rotation)
(IV) 82 3 m
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) III IV II I
(2) IV II I III
(3) IV III I II
(4) II I IV III
89. A uniform disc can rotate about a fixed horizontal axis passing through its centre and perpendicular to its plane. A light chord is wrapped around the rim of the disc (mass 2 kg and radius 1 m) and mass of 1 kg is tied to the free end. If it is released from rest, then match the followin g.
Column–I Column–II
(A) The tension in the chord is ____ N. (I) 10
(B) In the first 4 seconds, the angular displacement of the disc is___rad.
(C) The work done by the torque on the disc in the first 4 seconds is _____J.
(II) 5
(III) 40
(D) Acceleration of mass is________m/s2 (IV) 200 (V) 50
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) III II IV I
(2) II IV III I
(3) I III IV II
(4) II III IV II
90. A ring type flywheel of mass 200 kg and diameter 2 m is rotating at the rate of 300/π revolutions per minute. Answer the following w.r.t. the situation .
List-I
List-II
(A) Moment of inertia of the flywheel is___kg.m2 (I) 10
(B) The kinetic energy of rotation of the flywheel is _____kJ. (II) 200
CHAPTER 8: System of Particles and Rotational Motion
(C) The flywheel, if subjected to a rotating torque of 200 N.m will come to rest in time t = s. (III) 5
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) III I III II
(2) II III I III
(3) I II III III
(4) II I I I
91. A disc of radius R rolls without slipping on a rough surface with constant angular velocity ω . If v is the linear speed of the centre of mass of the disc, then match the speed of points.
List-I List-II
(A) A (I) 2Rω
(B) B (II) v
(C) C (III) 2v
(D) O (IV) 0
(V) v/2
Choose the correct answer from the options given below.
(A) (B) (C) (D)
(1) IV II I, III II
(2) IV I II, III II
(3) IV II II, III I
(4) I II IV, III II
FLASHBACK (P revious JEE Q uestions )
JEE Main
Numerical Value Questions
1. A solid circular disc of mass 50 kg rolls along a horizontal floor so that its center of mass has a speed of 0.4 m/s. The absolute value of work done on the disc to stop it is ______ J. (2024)
2. The identical spheres each of mass 2 M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 4 m each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is 42 x where the value of x is (2024)
3. A disc of radius R and mass M is rolling horizontally without slipping with speed . It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is: (2024)
4. A uniform rod AB of mass 2 kg and Length 30 cm at rest on a smooth horizontal surface. An impulse of force 0.2 Ns is applied to end B (penpendicular to the rod). The time taken by the rod to turn through at right angles will be s x π , where x = ____. (2024)
5. Four particles each of mass 1 kg are placed at four corners of a square of side 2 m. Moment of inertia of system about an axis
perpendicular to its plane and passing through one of its vertex is _____ kgm 2 . (2024)
6. A ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is 7 x where x is ________. (2024)
7. A cylinder is rolling down on an inclined plane of inclination 60°. It's acceleration during rolling down will be /2 3 x ms , where x = _________. (use g=10m/s2). (2024)
8. A body of mass 5 kg moving with a uniform speed 32 ms−1 in X–Y plane along the line y = x + 4. The angular momentum of the particle about the origin will be ______ = kgm2s−1. (2024)
9. A particle of mass m projected with a velocity ‘ u ’ making an angle of 30º with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is : (2024)
(1) 33 16 mu g
(2) 32 2g mu
(3) 3 2 mu g
(4) zero
10. Consider a Disc of mass 5 kg, radius 2 m, rotating with angular velocity of 10rad/s
about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is _______ J. (2024)
11. Two discs of moment of inertia I 1=4kgm 2 and I 2 =2kgm 2 about their central axes & normal to their planes, rotating with angular speeds 10 rad/s & 4 rad/s respectively are brought into contact face to face with their axe of rotation coincident. The loss in kinetic energy of the system in the process is_____J. (2024)
12. Two identical spheres each of mass 2 kg and radius 50 cm are fixed at the ends of a light rod so that the separation between the centers is 150 cm. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is 2 20xkgm , where the value of x is_____ (2024)
13. A body of mass 'm' is projected with a speed ‘ u’ making an angle of 45° with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as 23 mu Xg . The value of ' X ' is _______. (2024)
14. In the figure shown, ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then BC/AB is close to (2022)
CHAPTER 8: System of Particles and Rotational Motion
(1) 1.85 (2) 1.37 (3) 1.5 (4) 3
15. A spherical shell of 1 kg mass and radius R is rolling with angular speed ω on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is 2 3 ω a R The value of a will be (2022)
(1) 2 (2) 3 (3) 5 (4) 4
16. Consider a badminton racket with length scales as shown in the figure.
The mass of the linear and circular portions of the badminton racket are same ( M) and the mass of the threads are negligible. The moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at r /2 distance from the end A of the handle will be Mr2 (2022)
17. Three identical spheres, each of mass M, are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 3 m each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be m. x The value of x is . (2021)
18. Three blocks A, B, and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal mass m while C has mass M. Block A is given an inital speed v towards B, due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically. 5/6th of the initial kinetic energy is lost in the whole process. What is the value of M/m? (2020)
(1) 3 (2) 4 (3) 2 (4) 5
19. A small particle of mass m is projected at an angle θ with the y x-axis with an initial velocity v0 in the x-y plane, as shown in the figure. At a time 0sin , θ < v t g the anguIar momentum of the particle is
(2) continuously increases (3) first increases and then decreases (4) remains unchanged
21. Two particles of equal mass m have respective initial velocities ˆ ui and 2
They collide completely inelastically. The energy lost in the process is (2019)
(1) 32 4 mu (2) 12 8 mu (3) 12 3 mu (4) 22 3 mu
22. Distance of the centre of mass of a solid uniform cone from its vertex is z 0. If the radius of its base is R and its height is h, then z0 is equal to (2016) (1) 2 4 h R (2) 3 4 h (3) 5h 8 (4) 32 8 h R
where i,j,andk ˆˆˆ are unit vectors along x, y, and z-axis, respectively. (2020)
(1) 2 0 ˆ cos mgvtj −θ (2) 0c ˆ os mgvtk θ (3) 2 0 1ˆ cos 2 mgvtk −θ
20. A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc (2020) (1) continuously decreases
23. A body A of mass M, while falling vertically downwards under gravity, breaks into two parts; a body B of mass 1/3 M and a body C of mass 2/3 M. The centre of mass of bodies B and C taken together shifts compared to that of body A towards (2015) (1) depends on height of breaking (2) does not shift (3) body C (4) body B
24. A body A of mass m = 0.1 kg has an initial velocity of 3–1 ms ∧ i It collides elastically with another body B of the same mass, which has an initial velocity of –1 5. ms j ∧ After collision, A moves with a velocity 4. ∧∧ =+
vij The energy of B after collision is written as 10 x J The value of x is . (2010)
25. Moment of inertia (MI) of four bodies having same mass M and radius 2R are as follows:
I1= MI of solid sphere about its diameter
I2 = MI of solid cylinder about its axis
I 3 = MI of solid circular disc about its diameter
I 4 = MI of thin circular ring about its diameter
If 2(I2 + I3) + I 4 = xI1, then the value of x will be . (2011)
26. A particle of mass m is dropped from a height h above the ground. At the same time, another particle of the same mass is thrown vertically upwards from the ground with a speed of 2. gh If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of , h g is (2005)
JEE Advanced
27. A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height 3h from the ground, as shown in the figure. A spherical ball of mass m is released on the slide from rest at a height h from the top of the terrace. The ball leaves the slide with a velocity = 00 ˆ uux and falls on the ground at a distance d from the building, making an angle θ with the horizontal. It bounces off with a velocity v and reaches a maximum height h1. The acceleration due to gravity is g and the coefficient of restitution of the ground is 1 . 3 Which of the following statements is(are) correct? (2023)
CHAPTER 8: System of Particles and Rotational Motion
(3) θ = 60° (4) = 1 23 d h
28. One end of a horizontal uniform beam of weight W and length L is hinged on a vertical wall at point O and its other end is supported by a light inextensible rope. The other end of the rope is fixed at point Q, at a height L above the hinge at point O. A block of weight a W is attached at the point P of the beam, as shown in the figure (not to scale). The rope can sustain a maximum tension of 22 W. Which of the following statements is(are) correct? (2023)
(1) The vertical component of reaction force at O does not depend on a.
(2) The horizontal component of reaction force at O is equal to W for a = 0.5.
(3) The tension in the rope is 2 W for a = 0.5.
(4) The rope breaks if a > 1.5.
29. A thin circular coin of mass 5 g and radius 4/3 cm is initially in a horizontal x-y plane. The coin is tossed vertically up (+z direction) by applying an impulse of π × 2 10N-s 2 at
a distance 2/3 cm from its centre. The coin spins about its diameter and moves along the +z direction. By the time the coin reaches back to its initial position, it completes n rotations. The value of n is __________.
[Given: The acceleration due to gravity g = 10 ms−2] (2022)
Passage-I:
A pendulum consists of a bob of mass m = 0.1 kg and a massless inextensible string of length L= 1.0 m. It is suspended from a fixed point at height H = 0.9 m above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse P = 0.2 kg-m/s is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is J kg–m2/s . The kinetic energy of the pendulum just after the lift-off is K joules. (2021)
30. The value of J is
31. The value of K is .
32. A solid sphere of mass 1 kg and radius 1 m rolls without slipping on a fixed inclined plane with an angle of inclination θ = 30° from the horizontal. Two forces of magnitude 1 N each, parallel to the incline, act on the sphere, both at a distance r = 0.5 m from the centre of the sphere, as shown in the figure. The acceleration of the sphere down the plane is ____ms–2. (Take g = 10 ms–2) (2021)
CHAPTER TEST – JEE MAIN
Section - A
1. From a uniform circular disc of radius a , an isosceles right angled triangle with the hypotenuse as the diameter of disc is removed. The distance of the centre of mass of the remaining portion from the centre of the circle is (1) 3(π–1)a (2) () π− 1 6 a (3) ()31 a π− (4) ()31 a π+
2. The centre of mass of the letter ‘F’, which is cut from a uniform metal sheet from point ‘A’, is
(1) 1533 , 77 (2) 1523 , 77
(3) 2233 , 77 (4) 3322 , 77
3. The coordinates of centre of mass of particles of mass 10, 20, and 30 g are (1, 1, 1) cm. The position coordinates of mass 40 g, which when added to the system, the position of combined centre of mass be at (0, 0, 0) are,
(1) (3/2, 3/2, 3/2)
(2) (–3/2, –3/2, –3/2)
(3) (3/4, 3/4, 3/4)
(4) (–3/4, –3/4, –3/4)
4. A projectile of mass 3m explodes at highest point of its path. It breaks into three equal parts. One part retraces its path, and the second one comes to rest. The range of the projectile was 100 m, if no explosion would have taken place. The distance of the third part from the point of projection, when it finally lands on the ground, is (1) 100 m (2) 150 m (3) 250 m (4) 300 m
5. The object at rest suddenly explodes into three parts with the mass ratio 2 : 1 : 1. The parts of equal masses move at right angles to each other with equal speed v. The speed of the third part after explosion will be (1) v (2) 2v (3) 2v (4) /2 v
6. Two billiard balls of same size and mass are in contact on a billiard table. A third ball of the same size and mass strikes symmetrically and then comes to rest right after the impact. The coefficient of restitution between the balls is (1) 1/2 (2) 3/2 (3) 2/3 (4) 9/4
7. Two particles of masses m 1 and m 2 in projectile motion have velocities 12,vandv respectively, at time t = 0. They collide at time. Their velocities become 12vandv at time 2t0, while still moving in air. The value of ()() '' 11221122 || mvmvmvmv +−+
8. A body of mass m, moving with velocity collides head-on elastically with another body of mass 2m, which is initially at rest. The ratio of KE of the colliding body before
CHAPTER 8: System of Particles and Rotational Motion
and after the collision will be
(1) 1 : 1 (2) 2 : 1
(3) 4 : 1 (4) 9 : 1
9. A particle P moving with speed v undergoes a head-on elastic collision with another particle Q of identical mass but at rest. After the collision,
(1) both P and Q move forward with speed v/2
(2) both P and Q move forward with speed 2 v
(3) P comes to rest and Q moves forward with speed v
(4) P and Q move in opposite directions with speed 2 v
10. A neutron travelling with a velocity v and kinetic energy E collide perfectly elastically head on with the nucleus of an atom of mass number A at rest. The fraction of total energy retained by neutron is
(1) 2 1 1 A A
+
(2) 2 1 1 A A +
(3) 2 1 A A
(4) 2 1 A A +
11. Mass per unit area of a circular disc of radius a depends on the distance r from its centre as σ(r) = A + Br2. The moment of inertia of the disc about the axis perpendicular to the plane and passing through its centre is
(1)
24 45 ABa a (2)
2 24 46 ABa a
2 24 46
12. From a uniform circular disc of radius R and mass 9 M , a small disc of radius R /3 is removed, as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is
2R 3
(1) 4MR2 (2) 10MR2
(3) 402 9 MR (4) 372 9 MR
13. A uniform cylinder of mass M and radius R is to be pulled over a step of height a(a < R) by applying a force F at its centre O, perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of F required is
(3) a MgR (4) 2 2 1 a MgR
14. A wheel having moment of inertia 2 kgm2 is rotating about its geometrical axis at the rate of 60 rpm about its axis. The average torque which can stop the wheel’s rotation in one minute would be
(1) 18 Nm π (2) 2 15 Nm π (3) 12 Nm π (4) 15 Nm π
15. If a particle moving horizontally hits a stationary ring, then about point of collision,
(1) angular momentum of the ring is conserved
(2) angular momentum of the particle is conserved
(3) angular momentum of particle-ring system is conserved
(4) all of these
16. Two rotating bodies A and B of masses m, and 2m, with moments of inertia IA and IB ( I B > I A ), have equal kinetic energy of rotation. If L A and L B be their angular momenta respectively, then
(1) 2 B A L L = (2) LA = 2LB
(3) LB > LA (4) LA > LB
17. A cubical block of side 30 cm is moving with velocity 2 ms–1 on a smooth horizontal surface. The surface has a bump at a point O, as shown in figure. The angular velocity (in rad/s) of the block immediately after it hits the bump is
(1) 13.3 (2) 5.0 (3) 9.4 (4) 6.7
18. The moment of inertia of a body about a given axis is 1.2 kgm2. Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of 25 rad/s2 must be applied about that axis for a duration of
(1) 4 s (2) 2 s
(3) 8 s (4) 10 s
19. As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a flywheel (disc) of radius r and mass m. When released from rest, the bob starts falling vertically. When it has covered a distance of h, the speed of the bob will be
CHAPTER 8: System of Particles and Rotational Motion
Section - B
21. The distance between the centres of carbon and oxygen atoms in the carbon monoxide gas molecule is 1.13 × 10–10 m. The distance of centre of mass of the molecule relative to carbon atom is______ × 10 –12 m.
22. Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 ms–1 to the heavier block in the direction of the lighter block. The velocity of the centre of mass is m/s–1.
23. A rubber ball is dropped from a height of 5 m on a planet, where it rises back to a height of 1.8 m. The ball loses its velocity on bouncing by a factor of x/5. Then, x is
24. A body of mass 2 kg, moving with a speed of 4 m/s, makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its initial speed. The speed of the two-body centre of mass x/10 m/s. Then, the value of x is .
(1) 3 2 gh (2) 3 4 gh (3) 4 3 gh (4) 2 3 gh
20. The angular position of a particle on the rim of a rotating wheel is given by θ = 4t – 3t2 + t3, where θ is in radians and t is in seconds.
(a) What is the angular velocity at t = 2 s?
(b) What is the average angular acceleration for the time interval that begins at t = 2 s and ends at t = 4 s?
(1) 4 rad/s, 12 rad/s2
(2) 6 rad/s, 14 rad/s2
(3) 4 rad/s, 6 rad/s2
(4) 10 rad/s, 12 rad/s2
25. A shaft rotating at 3000 rpm is transmitting a power of 3.14 kW. The magnitude of torque is . Nm. (Take π=3.14)
CHAPTER TEST – JEE ADVANCED
2022 P2 Model
Section-A
[Integer Value Questions]
1. Two particles A and B, initially at rest, move towards each other by mutual force of attraction. At the instant when the speed of A is n and speed of B is 3 n , the speed of the centre of mass of the system is .
2. A linear rod of length l has linear mass density 01, λ=λ+ X l where λ0 is a constant and X is distance from left end of the rod. The distance of centre of mass from its left end is (5/N)l. The value of N is___.
3. Particle A experiences a perfectly elastic collision with a stationary particle B. The particles fly apart symmetrically relative to the initial direction of motion of particle A with angle of divergence θ(0 < θ < 90°). Ratio between the masses of the two bodies B A m m is 1 , 12cos+θ n where n is
4. Three objects A, B, and C are kept in a straight line on a frictionless horizontal surface. These have masses m, 2m, and m, respectively. The object A moves towards B with a speed of 9 m/s and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in m/s) of object C.
6. A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of mass 0.1 kg moving in a straight line with velocity 80 m/s hits the rod at its bottom-most point and sticks to it (see figure). The angular speed (in rad/s) of the rod immediately after the collision will be 10x. The value of x is
5. Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force F at a point distance b above its centre of mass (see in figure). If the coefficient of friction is μ = 0.4, the maximum possible value of × 100 b a for box not to topple before moving is
7. A thin disc rotates about an axis passing through its centre and perpendicular to its plane with a constant angular velocity ω . I is the moment of inertia of that disc and L is its angular momentum about the given axis. Then, rotational kinetic energy of the disc E is proportional to I–n. The value of n is .
8. A spool, initially at rest, is kept on a frictionless fixed inclined plane, making an angle θ = 37° with the horizontal. The mass of the spool is 2 kg and it is pulled by a string, as shown, with a force of T = 10 N. The string connecting the spool and the
pulley is initially horizontal. Find the initial acceleration of the spool on the incline. Express your answer in m/s 2
Section – B
[Multiple Option Correct MCQs]
9. A ball of mass is thrown at an angle of 45° to the horizontal from the top of a 65 m high tower AB, as shown in fig, at t = 0. Another identical ball is thrown with velocity 20 m/s horizontally towards AB from the top of a 30 m high tower CD, after 1 s from the projection of the first ball. Both the balls move in the same vertical plane. If they collide in mid-air,
CHAPTER 8: System of Particles and Rotational Motion
(1) 0 4 v
(2) 20 v
(3) 0 2 v
(4) 0 2 v
11. In the figure shown, if the coefficient of restitution between A and B is 1 2 e = , then
(1) the two balls will collide at time t = 5 s
(2) the two balls witll collide at time t = 2 s
(3) distance AC is 40 m
(4) velocity of combined ball just after they strike together is ˆ–1520s ˆ m ij
10. In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles after collision, is
(1) velocity of B after collision is v/2
(2) impulse between two during collision is 3 4 mv
(3) loss of kinetic energy during the collision is 32 16 mv
(4) loss of kinetic energy during the collision is 12 4 mv
12. The potential energy of a particle of mass m at a distance r from a fixed point O is given by () 2 , 2 = Vrkr where k is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius R about the point O. If v is the speed of the particle and L is the magnitude of its angular momentum about O, which of the following statements is (are) true?
(1) 2 k vR m =
(2) k vR m =
(3) 2 LmkR =
(4) 2 2 mk LR =
13. A particle of mass m is moving in a horizontal circle inside a smooth inverted fixed vertical cone above height h from apex. Angle of cone is θ. Then, θ h
(1) normal force on particle by surface of cone is mg cos θ
(2) normal force on particle by surface of cone is mg cosecθ
(3) time period of revolution of particle increases if θ increases, keeping h constant
(4) time period of revolution increases if h increases, keeping θ fixed
14. Three solid cylinders A, B, and C, each of mass m and radius R, are allowed to roll down on three inclined planes A', B', and C' respectively (each of inclination θ). For A' and A, coefficient of friction is zero. For B' and B, coefficient of friction is greater than zero but less than tan 3 θ , and for C' and C,
coefficient of friction is more than θ tan 3 On reaching the bottom of the inclined plane,
(1) C has maximum angular speed
(2) B has minimum total kinetic energy
(3) A takes minimum time to reach the bottom
(4) B has minimum translational kinetic energy
Section – C [Single Option Correct MCQs]
15. A non-uniform rod of length L is lying along the x-axis with one end at origin O, as shown in figure. The linear mass density (i.e., mass per unit length) λ varies with x as λ = a+ bx , where a and b are constants. Find the distance of the centre of mass from origin O.
(1) L(3a+ 2bL)/4(2a+bL)
(2) L(2a + 3bL)/4(a + 2bL)
(3) L(3a + 2bL)/3(2a + bL)
(4) L(2a + 3bL)/3(a + 2bL)
16. A smooth ball is dropped from height h on a smooth fixed incline, as shown in figure. After collision, the velocity of the ball is directed horizontally. Then, the coefficient of restitution is
h θ
(1) cot2 θ
(2) sin2 θ
(3) tan2 θ
(4) cos2 θ
CHAPTER 8: System of Particles and Rotational Motion
17. Point masses m 1and m 2 are placed at the opposite ends of a rigid rod of length L and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod, through which the axis should pass so that the work required to set the rod rotating with angular velocity ω 0 is minimum, is given by
18. A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q, as shown in figure. When string is cut, the initial angular acceleration of the rod is
