PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION THEORY OF LAWS OF MOTION

REFERENCE FRAME An axis is a number line. Origin is the point where zeros of the axes coincide. Position of an object at a time is described by drawing one axis for motion along a straighline, two intersecting axes for motion in a plane and three perppendicular axes for motion in space. More than one axis is a ‘system of axes’. At any instant of time, the position of a particle is described by noting co-ordinates on axes. To describe motion, we need an axis system (frame) that is fixed on some body, called reference body. This axis system is called ‘reference frame’. If we sit in an aeroplane and look at a car moving on road, the aeroplane in the reference body and the car is the object in motion. To locate the car, we set up xyz-axes in the aeroplane. This frame is here the reference frame. A reference frame is also attached with a clock to measure time. Example : A car is moving on road. Mile stones give us the positions of the car. We can describe it by choosing a point as origin and laying down number line along the road. (Observer is on the road.) This number line is called an axis and is a reference frame. mile stones

6

5

4

3

2

1

–3 –2 –1

7

Curved Axis Position (at time ‘t’)

O (Origin)

Example : The coint on a carrom board may be located at any time by drawing two axes in the plane of the board. Two numbers x and y locate it at a moment ‘t’. Y

(x, y)

Origin x

y X

Example : An insect (P) flying in a room may be located at a time by drawing three axes along the edges of the room, meeting at a corner. Insect ‘P’ has three co-ordinates (x, y, z) at time ‘t’. Z P(x, y, z)

Origin x y

Y

X

In all the examples above, axis systems are reference frames as they are fixed in the body at which rests the observer. The frame fixed to the ground is called ‘ground frame’ that fixed to a car is called ‘car frame’, that fixed to the sun is called ‘heliocentric frame’ etc.

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PHYSICS : SHEET LAWS OF MOTION LAWS OF MOTION CONCEPT OF FORCE Force is familiar word in science. From your own experience, you know that forces can produce changes in motion. If a force is applied to a stationary body, the body comes in motion. It can speed up and slow down a moving body or change the direction of its motion. In nut shell, the force is cause of change in velocity of the body. In other words, force is the cause of acceleration of the body. Force is a vector quantity and if more than one forces act on a particle, we can find the resultant force using the laws of vector addition. When a heavy block is suspended by a rope, the rope exerts a force on the block to hold it and the block exerts a force on the rope to make it tight and stretched. This is according to Newton’s third law of motion which may be stated as followed :   “If a body A exerts a force F on another body B, then B exerts a force – F on A, the two forces acting along the same line.” Some Particular Forces (i) Gravitational force : The force of attraction between bodies by virtue of their masses is known as gravitational force. Let two blocks of mass m1 and m2 are separated by a distance ‘r’.  The force on block 1 by block 2 is F12 acting towards m2 along line joining m1 and m2. Similarly, the force on  block 2 by block 1 is F21 acting towards m1 along line joining m2 and m1 (as shown in figure) From the concept of Newton’s third law, F12 F21   m1 m2 F12  F21 r In the sense of magnitude, Gm1m 2 F12  F21  F  r2 Here, G = gravitational constant = 6.7 × 10–11 Nm2/kg2. (ii) Weight of body (mg) : It is defined as the force by which earth attracts a body m towards its centre. If body is situated either on the surface of earth or near the surface of earth, then gravitational acceleration is nearly constant and is equal to mg g = 9.8 m/s2. The force of gravity (weight) on a block of mass m is w = mg acting centre of earth (shown in figure). Caution : The weight of a body is not the mass of the body. Weight is the magnitude of force and is related to the mass by w = mg. (iii) The normal reaction force : When a body presses against a surface, the surface (even a seemingly rigid  surface) deforms and pushes on the body with a normal reaction force N that is perpendiuclar to the surface. For a simple introduction to the normal reaction force, consider a situation in which you put a book on your head and continue your stationary position. In this case, the pain you feel is due to the force that the book exerts on your head. In the language of physics, the book exerts a force on your head normal to the surface of contact in downward direction. According to Newton’s third law of motion, the head exerts a force of same magnitude on the book normal to the surface of contact in upward direction. These forces are known as normal reaction forces. Normal reaction forces in difference situations are shown below : N N

(a)

Direction of normal reaction on the blocks

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Direction of normal reaction on the surface

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

N

m

N m

N1

(b) Direction of normal reaction on the blocks

Inclined plane

Direction of normal reaction on the inclined plane

Normal reaction on horizontal surface

The number of normal reaction pairs is equal to number of contact surfaces. m2 B

(c)

A

NB

NB

NA

A

B

m1

NA

The normal reaction on upper block is in upward direction and normal reaction on lower block is in downward direction. N2

N1

Wall

(d)

N1

Sphere

N2

N2

N1 C

C

(e)

N2

N1

C

N2

C

N2

(f)

N1 N1

N2

(g)

N2 N1

N1

(iv) Friction : If we slide or attempt to slide over a surface, the moton is resisted by a bonding between the body and the surface. The resistance Direction of  attempted slide is considered to be single force f , called the frictional force or simply f friction. This force is directed along the surface, opposite the direction of the intended motion. Some times to simplify a situation, friction is assumed to be negligible (the surface is frictionless). (v) Tension : When a string or cord is attached to a body and pulled taut, the  cord pulls on the body with a force T directed away from the body and T T along the cord Fig. The force is often called a tension force because the cord Force is said to be in a state of tension (or to be under tension), which means that it is being pulled taut. The tension in the cord is the magnitude T of the force on the body.

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

If the cord (string) is massless, the tension in cord everywhere remains the same. The cord then exists only as a connection between two bodies. If the cord (string) is massive, the tension in it varies point to point. NOTE : â&#x20AC;˘ If string, slacks, tension in string becomes zero. â&#x20AC;˘ The direction of tension on a body or pulley is always away from the body or pulley. The direction of tensions in some cases are shown below : (a)

(b)

m2

m2

m1

(c)

T

m1

T T m2

T T m2

m2 String is massless and pulley is m1 light and smooth

m1

String is massless and pulley is light and smooth

T

m2

T

T

T T

m2

m2

T1

Pulleys are light and smooth and string is massless

T1

T1 m T1 1

T1

m1

(d)

T1

T2

m1

T2 m2

T

T T

(e)

m1

m2

m1

T m2

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

T2

T1 T1

m1 m2

(f)

m1

String is massless and there is friction between string and pulley

T2 m2

T1 m1 C

(g)

T1

m1

String is massless and there is friction between pulley and string

C

T2 m2

T2 m2

If m2 tends to move downwards, then T2 > T1 and T2 = T1eµ where µ = coefficient of friction  = angle subtended by string at the centre of pulley

T1

T2 T3

m1 m2

(h)

m1

String is not massless and there is friction between pulley and string

T4 m2

(vi) Spring force : Coiled metallic wire is known as spring. The distance between two successive collisions in a spring remains the same. If a spring is placed on L a smooth surface, the length between ends of spring is known as natural length (shown in figure). As you may have discovered itself, springs resist attempts to change their length. If the length of spring is greater than its natural length, the spring is in the condition of elongation (shown in figure ahead). If the length of spring is lesser than its natural length, the spring is in the condition of compression (shown in figure ahead). In fact, the more you alter a spring’s length, the harder it resists. From this point of view, spring force increases, when elongation or compression increases. For small elongation or compression of spring, spring force is proportional to its elongation or compression. i.e., Fx F = kx where ‘k’ is proportionality constant known as spring constant or stiffness constant. Its unit is N/m. The direction of spring force is always towards the natural length of spring.

L

Spring in natural length (a)

L+ x

Spring in the condition of elongation (b)

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L– x

Spring in the condition of compression (c)

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION L F=0

L+x

Lâ&#x20AC;&#x201C; x

F = kx

F = kx

m No spring force on body (d)

m

m

Spring force on the body is leftward (e)

Spring force on the body is rightward (f)

FREE BODY DIAGRAM A free body diagram is helpful in solving problems like several bodies connected by string, springs, surfaces of contact, with the Newtonâ&#x20AC;&#x2122;s second law. It is a striped-down diagram in which one body is considered. That body is represented by a sketch or simply a dot. The external forces on the body are drawn, and a co-ordinate system is superimposed, oriented so as to simplify the solution. While sketching a free body diagram, the following points should be kept in mind. 1. Normal reaction (N) always acts normal to the surface on which body is kept [Fig.]. N Vertical surface

N Block

Block N Horizontal surface Normal reaction = N Weight (action) = w W

W

Inclined plane

(a)

2.

Block

(b)

(c)

When two blocks A and B are connected by a string the tension for block A is towards B and for block B, it is towards A. T

T

A

B

NOTE : Always remember that while drawing a FBD, tension in each branch must form an action-reaction pair. 3. 4. 5.

If the pulley is light and frictionless and a string passes over it without any links, then tension on either side of the string is the same. While drawing FBD of a body, always take into account the forces which are acting on the body not those which the body exerts on others. The friction is a tangential force acting tangentially to the surfaces in contact.

Example 1. Three blocks A, B and C are placed one over the other as shown in figure. Draw free body diagarms of all the three blocks. Sol. Free body diagrams of A, B and C are shown below :

wA N1 FBD of A

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N1

N2

wB

wC

N2 FBD of B

A B C

N3 FBD of C

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PHYSICS : SHEET LAWS OF MOTION Here,

LAWS OF MOTION

N1 = normal reaction between A and B N2 = normal reaction between B and C N3 = normal reaction between C and ground

Example 2. A block of mass ‘m’ is attached with two strings as shown in figure. Draw the free body diagram of the block.

Sol. The free body diagram of the block is as shown in fig. T1 T2 mg

NEWTON’S LAWS OF MOTION First Law “If no force acts on a body, then the body’s velocity cannot change; that is the body cannot accelerate.” In other words, if the body is at rest it stays at rest and if it is moving, it will continue to move with the same velocity (same magnitude and same direction) unless it experiences a net external force. In simpler terms, we can say that the net force on Q body is zero, its acceleration is zero. That is, where F = 0, then a = 0. From the first law. We conclude that an isolated body (a body that does not interact with its environment) is either at rest or moving with constant velocity. Second Law The change of motion is proportional to the magnitude of force applied and is made in the direction of the straight line in which that force is applied. According to this law, the net force on a body is equal to the product of the body’s mass and the acceleration of the body. In equation form     Fnet Fnet  m a or a  m Third Law To every action there is always an equal and opposite reaction or the mutual actions of two bodies upon each other are always directed to contrary parts. According to this law, when two bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction.   2 F12 = – F21 This is, F12  F21 This law, which is illustrated in fig., is equivalent to stating that forces always F12 F21 occurs in pairs, or that a single isolated force cannot exist. 1

Problems Solving Steps : Normally problem based on Newton’s law can be solved in following steps : 1. Concentrate your mind on the system (body) which is considered by you. The considered body may be a single particle, a block or a combination of two or more blocks, two blocks connected by a string. etc.

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PHYSICS : SHEET LAWS OF MOTION 2. 3. 4.

LAWS OF MOTION

Show all forces acting on the system but it is restricted that the forces that the system exerts onother should not be shown here. Thus, we draw, a free body diagram of the system and indicate the magnitude and directions of all the forces as discussed in step 2. Choose a co-ordinate system including mutually perpendicular ‘x’ and ‘y’ axes in the plane of the forces. Write the components of all the forcs acting along the acceleration (assumed as x-axis) and perpendicular to acceleration (assumed as y-axis). That is Fx = ma ...(i) and for y-axis, Fy = 0 ...(ii) NOTE : Above Eqs (i) and (ii) hold in case of coplanar forces. If the forces are collinear, the Eq. (ii) is not needed.

Example 3. In the arrangement shown in figure, the strings are light F = 28 N 4kg and inextensible. The surface over which blocks are placed is 2kg 1kg smooth. Find : (a) the acceleration of each block ; (b) the tension in each string. Sol. (a) Let ‘a’ be the acceleration of each block and T1 and T2 be the tensions, in the wo strings as shown in figure. y 4kg

x

T2

T1 2kg

F = 28 N

1kg

Taking the three blocks and the two strings as the system. 4kg

Using or

2kg

1kg

F = 28 N

Fx = max 28 = (4 + 2 + 1)a

28  4 m / s2 7 (b) Free body diagram (showing the forces in x-direction only) of 4 kg block and 1 kg block are shown in figure.

or

a

a = 4 m/s 2 4kg

Using For 1 kg block, or  For 4 kg blocks, 

T2

y T1

1kg

F = 28 N

x

Fx = max F – T1 = (1) (a) 28 – T1 = (1) (4) = 4 T1 = 28 – 4 = 24 N T2 = (4) (a) T2 = (4) (4) = 16 N

Example 4. Two blocks of mass 4 kg and 2 kg are placed side by side on a smooth horizontal surface as shown in the figure. A horizontal force of 20 N is applied on 4 kg block. Find the acceleration of each block.

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4kg 2kg

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

Sol. Since, both the blocks will move with same acceleration (say a) in horizontal direction. a 20 N

4kg 2kg

y x

Let us take both the blocks as a system. Net external force on the system is 20 N in horizontal direction. Using Fx = max 20 = (4 + 2) a = 6a or

a

10 m / s2 3

Example 5. Find the acceleration of 3 kg mass when acceleration of 2 kg 3 kg 2 kg mass is 2 ms–2 as shown in figure. 2 ms –2 (A) 3 ms–2 (B) 2 ms–2 (C) 0.5 ms–2 (D) Zero Sol. From force diagram of 2 kg block : 10 N 3 kg 2 kg 10 – kx = 2a = 4  kx = 6 2 ms –2 From force diagram of 3kg block, 2 kg 10 N kx = 3a kx a or 6 = 3a 3 kg

kx

6 a   2m / s 2 3

Example 6. The system shown adjacent is in equilibrium. Find the acceleration of the blocks A, B & C all of equal masses m at the instant when (Assume springs to be ideal) (a) The spring between ceiling & A is cut. (b) The string (inextensible) between A and B is cut. (c) The spring between B & C is cut. Also find the tension in the string when the system is at rest and in the above 3 cases. Sol. The system is in equilibrium : kx3 = mg ...(1) 2mg + kx3 = kx1 ...(2)  3mg = kx1

K A B K C

kx1

from fig.(1) from fig.(2)

A+B kx3 Fig(2)

2mg

10 N

kx3 C mg Fig(1)

(a) when spring between ceiling and block is cut. then, elongation of spring between B and C remains same just after cutting. 2m  ac = 0 ( Q kx3 = mg) For (A + B) A+B kx3 + 2mg = 3mg  3mg = 2ma 3 2  a = g = 15 m / s 2 T  a A = a B = 15 m / s 2 B For tension, kx3 mg + kx3 – T = maB mg

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

3mg 2 mg T=  2 (b) when string between A and B is cut. the elongation in springs do not change just after cutting the string. mg – kx1 = maA mg – 3mg = maA { Q kx1 = 3mg} – 2mg = maA aA = – 2g  aA = 2g (upward) For B mg + kx3 = m aB Q mg or mg + mg = m aB { Q kx3 = mg}

mg + mg – T =

 For C,

a A = 2g

kx1 A mg

B kx3

(downward) kx3

mg – kx3 = maC mg – mg = maC { Q kx3 = mg} aC = 0 T = 0 (c)when spring between B and C is cut. mg = maC aC = g  (downward)

C

or Q

2mg – kx1 = 2maB 2mg – 3mg = 2maB g aB =  2 g aA = aB =  2 T – (mg) = maB mg T – (mg) = 2 mg T = mg + 2 3mg T= 2

mg

C mg

{ Q aA = aB} { Q kx1 = 3mg}

kx1 A+B 2mg

(upward)

T B

PULLEYS Using pulling force method, the problems based on pulleys can be made simple. We can understand this method using an example. Let two enequal masses ‘m’ and ‘2m’ are attached to the ends of a light inextensible string passing over a smooth massless pulley. We have to find the acceleration of the system.

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kx3

mg

a

a m

2m

mg

2mg

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

Net pulling force on the system is 2mg – mg = mg and total mass being pulled is 2m + m = 3m Therefore, acceleration of the system

a

Net pulling force Total mass to be pulled

a

mg g  3m 3

Motion of Connected Bodies (i) Unequal masses (m1 > m2) suspended from a pulley : m2g T

T

m2

m1

m1g

m2g

m1 + m2

m1g

m2 m2g

m1 m1g

Acceleration  a 

 m1  m2  g  m1  m2 

 2m1m 2  Tension  T   g  m1  m 2  (ii) Bodies accelerated on a horizontal surface by a falling body : T = m2a ...(i) m1g – T = m1a ...(ii)

N

a

 m1  Acceleration  a   g  m1  m 2 

T

Smooth

T

 mm  Tension  T   1 2  g  m1  m 2  (iii) Motion on a smooth inclined plane : m1g – T = m1a T – m2g sin  = m2a 

and

m2g m1

a

...(i) ...(ii)

m1g

T

N

T

m2

 m  m 2 sin   a  1 g  m1  m 2  T

m2

m1

m2g sin

a

m2gcos m1g

m1m 2 1  sin  g  m1  m 2 

m2g

Example 7. A rope of mass 5 kg is moving vertically in vertical position with an upwards force of 100 N acting at the upper end and a downwards force of 70 N acting at the lower end. The tension at midpoint of the rope is (A) 100 N (B) 85 N (C) 75 N (D) 105 N

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION 100N

Sol. From force diagram of rope, mg + 70 – 100 = ma or 50 + 70 – 100 = 5a or 20 = 5a  a = 4 m/s2

m=5kg a T

Let tension at middle point is T. From force diagram of half lower portion : 25 + 70 – T = 2.5a or 95 – T = 10  T = 95 – 10 = 85 N.

mg

70N

2.5g a 2.5g

70N

P

Example 8. What should be the minimum force ‘P’ to be applied to the string so that block of mass ‘m’ just begins to move up the frictionless plane. Mg tan  Mg cot  (A) (B) 2 2 Sol. From force diagram of pulley, P + P cos  = T From force diagram of block : T > mg sin  or P + P cos > mg sin 

or

mg sin  P 1  cos 

Pmin 

Mg cos  (C) 1  sin 

(D) None P P

N T

T mgsin

mgcos mg

mg sin  1  cos 

B

C

A

Example 9. In the fig., mass of A, B and C are 1 kg, 3 kg and 2 kg respectively. Find : (a) the acceleration of the system and (b) tension in the string. Neglect friction. (g = 10 m/s2) Sol. (a) In this case net pulling force = mAg sin 60º + mBg sin 60º – mCg sin 30º  110

M

60º

30º

 3 3  1   310     210    24.64 N  2 2  2 

Total mass being pulled = 1 + 3 + 2 = 6 kg  Acceleration of the system 24.64  4.11 m / s 2 6 (b) For the tension in the string between A and B. FBD of A mAg sin 60º – T1 = mAa  T1 = mAg sin 60º – mAa T1 = mA (g sin 60º – a) a

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T1 A

a mAg sin 60º

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PHYSICS : SHEET LAWS OF MOTION 

LAWS OF MOTION

  3 T1  1  10   4.11  4.55 N 2  

For the tension in the string between B and C. FBD of C T2 – mCg sin 30º = mCa  T2 = mC (a + g sin 30º) 

T2 C

  1  T2  2  4.11  10     18.22 N  2  

mCg sin 30º

Example 10. Two masses ‘m’ and ‘M’ are attached to the strings as shown in the figure. If the system is in equilibrium, then 2M 2m (A) tan   1  (B) tan   1  m M 2M 2m (C) cot   1  (D) cot   1  m M Sol. At point A, 2T sin 45º = mg 

M 45º

T1

mg mg T  2sin 45º 2

T T

T

T1 sin  = Mg + T cos 45º

45º

Also, or

T1 cos 

mg T1 sin  Mg  2 2M   1 mg T1 cos  m 2

B M

45º A m

mg T1 sin   Mg  cos 45º 2 mg 2 T1 cos  = T sin 45º

or

45º m

At point B,

or

a

Mg

mg

T1 sin   Mg 

mg mg sin 45º  2 2

A

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V1 P1 6m/s

66 v1  0 Here 2 Since, pulley P1 is in rest. so, pulley P1 is also in rest. So, velocity of C is 4 m/s in upward direction.

B C

D

4m/s

Sol.

6m/s

6m/s

Example 11. In the figure shown the velocity of different blocks is shown. The velocity of C is (A) 6 m/s (B) 4 m/s (C) 0 m/s (D) none of these

P2 B

A C

D

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

Example 12. At what value of m1 will 8 kg mass be at rest. Sol. Here T0 = 8g = 80 N Also, T0 = 2T  T = 40 N From force diagram of 5 kg block, 50 – T = 5a 50 – 40 = 5a 10 = 5a a = 2 m/s2 From force diagram of m1, T – m1g = m1a or 40 – m1g = m1 × 2 or 40 = 10m1 + 2m1 = 12 m1 

m1 

8kg

5kg m1

T0 8kg

T0

40 10  kg . 12 3

T T 5kg a

T T m1 a

m1g

5g

Example 13. Two masses ‘A’ and ‘B’, lie on a frictionless table. They are attached to either end of a light rope which passes around a horizontal movable pulley of negligible mass. Find the acceleration of each mass MA = 1 kg, MB = 2 kg, MC = 4 kg. The pulley P2 is vertical.

B

P1 P2

A

C

aB aC

Sol. aA

From contstant relation, aA  aB 2 From force diagram of pulley, T3 = 2T Here T = mBaB and T = mAaA and mCg – T3 = mCaC After solving, 4g 2g aA  , aB  , 5 5 aC 

...(i)

T T3

...(ii) ...(iii) ...(iv) ...(v) aC 

T

3g 5

Example 14. Block ‘A’ of mass m/2 is connected to one end of light rope which passes over a pulley as shown in the Fig. Man of mass ‘m’ climbs the other end of rope with a relative acceleration of g/6 with respect to rope find acceleration of block ‘A’ m/2 and tension in the rope. A Sol. The acceleration of man with respect to ground is ‘a’. g  a  a0  6

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g/6 m

14 14

PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

g  a0 6 From force diagram of man, T – mg = ma

or

a

a0 m 2 A

g  T  mg  m   a 0  6 

m

...(i)

From force diagram of block A, m m g  a0 2 2 Subtracting eqn (i) and (ii), T

or or or

...(ii)

m m mg g  a0   ma 0 2 2 6 3 mg mg ma 0   2 2 6 3 4mg ma 0  2 6 4mg 2 a0   6 3m 4 a0  g 9

From eqn (ii) m m a0  g 2 2 m 4 m T  g g 2 9 2 13mg T 18 T

Example 15. Inclined plane is moved towards right with an acceleration of 5 ms–2 as shown in figure. Find force in newton which block of mass 5 kg exerts on the incline plane. (All surfaces are smooth) Sol. For discussion of the block inclined plane is taken as reference frame. ma 0s

in5

º 0

5 m/s2

ma sin 37º 37º

37

mgco s37º+ ma

37º

N

53º

ma0

5kg

cos53 º

mg

From the force diagram, N = mg cos 37º + ma0 cos 53º N = 5 × 10 × 0.8 + 5 × 5 × 0.6 N = 40 + 15 = 55 N

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF2 MOTION .5k g

spring balance

Example 16. Find the reading of spring balance as shown in figure. Assume that mass ‘M’ is in equilibrium. (All surfaces are smooth)

M 37º

kx0

Sol.

37º

53º

53º

N0 mg

N

From force diagram of wedge, N cos 53º = kx0

...(i) N mg sin

37 º

mgcos37º

From force diagram of smaller block : N = mg cos 37º  kx0 = N cos 53º kx0 = mg cos 37º cos 53º kx0 = 2.5 × 10 × 0.8 × 0.6 = 12 N Hence, reading is kx0 = 12 N. Example 17. A flexible chain of weight ‘W’ hangs between two fixed points ‘A’ & ‘B’ which are at he same horizontal level. The inclination of the chain with the horizontal at both the points of support is . What is the tension of the chain at the mid point ? (A)

W cosec θ 2

Sol.  Also, 

(B)

W tan θ 2

(C) T

W Tsin   2 W T 2 sin  T cos  = T0 T0 = T cos  W W T0  cos   cot  2sin  2

W cot θ 2

(D) none T

T

T sin

T cos

T0 W 2

Comprehension Type — [18 to 20] A paticle of mass ‘m’ is constrained to move on x-axis. A force ‘F’ acts on the particle. ‘F’ always points toward the position labeled ‘E’. For example, A +ve when the particle is to the left of E’, ‘F’ points to the right. The magnitude of x m E ‘F’ is a constant ‘F’ except at point ‘E’ where it is zero. The system is horizontal. ‘F’ is the net force acting on the particle. The particle is displaced a distance ‘A’ towards left from the equilibrium position ‘E’ and released from rest at t = 0.

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PHYSICS : SHEET LAWS OF MOTION Example 18. What is the period of the motion ?  2Am  (A) 4  F  

LAWS OF MOTION

 2Am  (B) 2  F  

 2Am  (C)  F  

(D) None

Sol. The time period is four times the time taken by particle to cover the distance A. Here the acceleration of the particle is a  

A

t0 

F . m

1 2 at 0 2

2A 2mA  a F

 The time period is

T  4t 0  4

2mA F

Example 19. Velocity-time graph of the particle is v

v

(A)

(B)

t

v

t v

(C)

(D)

t

t

Sol. Since, acceleration is constant in magnitude. So, graph is straight line. Since, acceleration is always directed towards E. If the particle is in leftward acceleration is 

F . So, slope of v – t graph is positive. m

But when the particle is rightward, the acceleration is a  

F i.e. negative. m

Hence, option (A) is correct.

Example 20. Find minimum time it will take to reach from x   (A)

3 mA 2 F

Sol. Here

2 1

A

(B)

mA F

2 1

(C) 2

A to 0. 2

mA F

2 1

(D) None

1 2 Ft 02 at  2 0 2m

2mA F But the time taken x = –A to x = –A/2 is t1 

t0 

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PHYSICS : SHEET LAWS OF MOTION Here or 

LAWS OF MOTION

A 1 F 2    t1 2 2  m mA t1  F

t  t 0  t1 

mA F

2 1

Example 21. Two blocks are connected by a spring. The combination is suspended, at rest, from a string attatched to the ceiling, as shown in the figure. The string breaks suddenly. Immediately after the string breaks, what is the initial downward acceleration of the upper block of mass 2m ? (A) 0 (B) 3g/2 (C) g (D) 2g kx0 Sol. Step-I : Discuss the problem before cutting the string : From force diagram of lower block, B kx0 = mg From force diagram of upper block : T = 2mg + kx0 mg Step-II : Discuss the problem after cutting the string, T 2mg + kx0 = 2ma or 2mg + mg = 2ma 2m or 3mg = 2ma kx0 3 2mg a g  2

m

m1 m2 30º

y

x

Example 22. A block of mass m1 = 3.70 kg on a frictionless inclined plane of angle 30º is connected by a cord over a massless, frictionless pulley to a second block of mass m2 = 2.30 kg hanging vertically (fig.). what are (a) the magnitude of the acceleration of each block and (b) the direction of the acceleration of the hanging block? (c) What is the tension in the cord ? Sol. (a) The free body diagram for each block is shown in fig.

2m

T

m

1

gs

in

m1gcos m1g

T

For block 1, Fx = m1a to T – m1g sin  = m1a and Fy = 0 or N – m1g cos  = 0 For block 2,

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a

...(i) ...(ii)

m2g

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PHYSICS : SHEET LAWS OF MOTION m2g – T = m2a Eliminating T from Eqs. (i) and (iii), we have a

a

LAWS OF MOTION ...(iii)

 m 2  m1 sin  g m1  m 2

 2.30  3.70sin 30º  9.8   0.735 m / s

2

3.70  2.30 (b) the result is positive, indicating that the acceleration of block 1 is up the incline and the acceleration of block 2 is downward. (c) The tension force of the cord is T = m1a + m1g sin  T = 3.70 × 0.735 + 3.70 × 9.8 sin 30º F = 30t N T = 20.8 N Example 23. Force ‘F’ is applied on upper pulley. If F = 30t where ‘t’ is time in second. Find the time when m1 loses contact with floor. Sol. From force diagram of pulley : F = 3T 30t = 3T  T = 10 t For lossing contact, 2T  m1g or 20t  40 40 t  2 sec. or 20

m1

m2

m1=4kg m2=1kg

IMPORTANT FEATURES 1. Newton’s first law is not true in all reference frames, but we can always find reference frames in which it is true. such frames are called inertial reference frames, or simply inertial frames. Thus, an inertial reference frame is one in which Newton’s laws hold. 2. Newton’s laws are not valid in the non-inertial frames. They are to be modified by introducing the concept of pseudo force. 3. The pseudo force is always directed in a direction opposite to the direction of the acceleration of the noninertial frame. 4. While drawing free body diagrams (FBD) in which pseudo force is involved, we must first see the acceleration of the non-inertial frame and then in the FBD, plot the pseudo force with a value ma in a direction opposite to the acceleration of non-inertial frame. 5. If a pulley is massless, net force on it is zero even if it is accelerated. For example, in the following figure : T1 T1 T1 P T2 T2

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T2

T2

FBD of pulley P

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

T1 = 2T2 whether the pulley is accelerated or not provided the pulley is massless. This because   Fnet  mass  acceleration and Fnet will be zero if pulley is massless. FRICTION  When a force f tends to slide a body along a surface, a frictional force from the surface acts on the body. The frictional force is parallel to the surface and directed so as to oppose the sliding. It is due to bonding between  the body and the surface. If the body does not slide the frictional force is a Static Frictional Force f s . If there  is sliding the frictional force is kinetic frictional force f k . Three Properties of Friction   1. If the body does not move, then the static frictional force f s and the component of F that is parallel to the  surface are equal in magnitude, and f s is directed opposite that component. If that parallel component increases,

2. 3.

magnitude fs also increases. If the applied parallel component exceeds a certain (maximum) value, the body slides on the surface.  The magnitude of f s has a maximum value called limiting value fs,max or fL that is given by fL = µsN where µs is the coefficient of static friction and N is the magnitude of the normal force. If the body begins to slide on the surface, the magnitude of the frictional force rapidly decreases to a constant value fk given by fk = µkN where µk is coefficient of kinetic friction. The value of µs is greater or equal to µk. NOTE : In problems if µs and µk are separately not given. But only µ is given. Then µ = µs = µk

Example 24. A baseball player with mass m = 79 kg, sliding into second base, is retarded by a frictional force of magnitude 470 N. What is the coefficient of kinetic friction µk between the player and the ground ? Sol. The free body diagram from the player is shown in the right. The force of friction is related to the normal force N by f = µkN. The vertical component of acceleration is zero, so the vertical component of Newton’s second law is f N – mg = 0  N = mg f 470 mg µk    0.61 Thus, N (79) (9.8) Angle of Friction As shown in fig, a body A is in contact with surface B. The forces acting on A is shown. Surface B applies two contact forces on body A. P

N A fL

Angle of friction O

F B

w =M g

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PHYSICS : SHEET LAWS OF MOTION 1. 2.

LAWS OF MOTION

Normal reaction force N, whose magnitude is equal to weight w( = mg) of body and is directed upwards. Maximum static frictional force fL (= µN), which is tangential to surface of A and directed opposite to the  direction of applied force F .   The resultant force of these two forces is represented by the vector OP . The angle between vector OP and normal reaction force N is the angle of friction (). f µN tan   L  µ  N N or a = tan–1(µ) The magnitude of net tangential force applied by surface B on A,  | OP | N 2  f L2  | OP | N 2  (µN) 2  | OP | N 1  µ 2

Angle of Repose Suppose inclined plane BC, makes an angle  with the horizontal and a body A is placed on it is position of rest [Fig.]. D N = mg cos

µN A mg cos

mg sin mg

B

A

C

H

Angle is gradually increased, till the body placed on its surface just begins to slide down. If  is the inclination at which the body just begins to slide down, then it is called angle of repose. In the figure is the angle of repose. From figure, mg sin  = µN = µ mg cos  or µ = tan  But µ = tan , wher is angle of friction.  tan  = tan  or = or angle of repose is equal to angle of friction. Example 25. A particle of mass 1 kg rests on rough conact with a plane inclined at 30º to the horizontal and is just about to slip. Find the coefficient of friction between the plane and the particle. Sol. The given angle 30º is really the angle of repose . Hence, 1 µ  tan 30º  3

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PHYSICS : SHEET LAWS OF MOTION  Example 26. A force F  ˆi  4 ˆj acts on block shown. The force of friction y acting on the block is : (A)  ˆi (B)  1.8 iˆ (C)  2.4 ˆi (D)  3 ˆi

LAWS OF MOTION F x

1 Kg µ = 0.3

Sol. The maximum value of friction is fmax = µN Here N + 4 = mg = 10  N=6N  fmax = 0.4 × 6 = 2.4 N But applied force in horizontal direction is Fh = 1 N. Fh < fmax. So, static friction comes into play.   Fh = f  f=1N   f   ˆi Some Particular Cases (i) If force is applied on lower block : The maximum friction acting on m2 is f2 = µ2m2g. If the system moves with the common acceleration, then µ2 m2 F – µ1(m1 + m2)g = (m1 + m2)a µ1 F and f2 = m2a m1  µ2m2g = m2a  a = µ2g (ii) If force is applied on upper block : f2 = limiting friction between m1 and m2. f1 = limiting friction between the surface and m1. µ2 F m2 If F > f2, then both blocks move with different acceleration and the µ1 maximum friction acts between the blocks. m1 F – f2 = m2a2  F – µ2m2g = m2a2 N2 and N2 = m2g N1 = N2 + m1g N1 = (m1 + m2)g f2 m2 F  f1 = µ1N1 f1 = µ1(m1 + m2)g m2g f2 = µ2m2g If f2 < f1, then m1 remains at rest. If f2 > f1, then m1 moves in the direction of f2. f2 – f1 = m1a1 N1

N2

f2

m1 f1 m1 g

If F < f2, then no relation is found between m1 and m2. i..e, m1 and m2 move together.

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

If F < f1 then the system is in rest. If F > f1, the system moves with the common acceleration a. In this case, F – f1 = (m1 + m2)a or F – µ1(m1 + m2)g = (m1 + m2)a (iii) The ratio of masses on an inclined plane : The coefficient of friction = µ. N T

m2

sin mg

m1

T

T m2

m1

2

m2gcos m1 g

(a) When m1 starts moving downwards, then m1  sin   µ cos  m2

(b) When m2 starts moving downwards, m1  sin   µ cos  m2

(c) When no motion takes place, m1  sin  m2

(iv) Blocks in contact on an inclined plane. m1

m2

µ2 µ1

In this type of problem, find the accelerations of blocks without contact. N1

µ 1m

a1 in gs

m1

os gc

N2

1

m1 m1gcos

µ 2m

a2 in gs

m2

os gc

2

m2 m2gcos

(a) If a1 > a2, then both blocks move separately with respective accelerations a1 and a2. (b) If a1 < a2, then both blocks move together with a common acceleration a. In this case, both blocks are treated as a system of mass (m1 + m2) (m1 + m2)g sin  – µ1m1g cos – µ2m2g cos  = (m1 + m2)a Example 27. Figure shown two blocks in contact sliding down an inclined surface of kg inclination 30º. The friction coefficient between the block of mass 2.0 kg and the g 2 k 4 incline is µ1 = 0.20 and that between the block of mass 4.0 kg and the incline is 30º µ2 = 0.30. Find he acceleration of 2.0 kg block. g = 10 m/s2. Sol. Since, µ1 < µ2, acceleration of 2 kg block down the plane will be more than the acceleration of 4 kg block if allowed to move separately.

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

In this case both blocks are treated as a system of mass (4 + 2) = 6 kg and will move down with the same acceleration. Net force down the plane is F = (m1 + m2)g sin  – µ1m1g cos  – µ2m2g cos  F = (4 + 2)g sin 30º – (0.2) (2) g cos 30º – (0.3) (4) g cos 30º  3  3  1 F  (6) (10)    (0.4) (10)    (1.2) (10)    2  2   2 

F = 30 – 13.76 = 16.24 N Therefore, acceleration of both the blocks down the plane will be a

F 16.24   2.7 m / s 2 m1  m 2 4  2

Comprehension Type — 28 to 32 refer figure Example 28. When F = 2N, the frictional force between 5 kg block and µ = 0.1 ground is (A) 2N (B) 0 (C) 8 N (D) 10 N µ = 0.3 Sol. f1 max = 0.1 × 10 × 10 = 10 N f2 max = 0.3 × 15 × 10 = 45 N For F = 2N, static friction come in to play at both surfaces.  F = f2  f2 = 2N

10 kg

F

5 kg

Example 29. When F = 2N, the frictional force between 10 kg block and 5 kg block is (A) 2 N (B) 15 N (C) 10 N (D) None Sol. From force diagram of 10 kg block, F = f1  f1 = 2N Example 30. The maximum ‘F’ which will not cause motion of any of the blocks. (A) 10 N (B) 15 N (C) data insufficient Sol.  F = f1  F = f1max = 10 N Example 31. The maximum acceleration of 5 kg block (A) 1 m/s2 (B) 3 m/s2 Sol. 5 kg

(C) 0

(D) None

(D) None

10N = f 1 max

45N = f 2 max

f1max < f2max  Hence, static friction comes into play between ground and 5 kg block.  Acceleration of 5 kg block is zero.

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

Example 32. The acceleration of 10 kg block when F = 30 N (A) 2 m/s2 (B) 3 m/s2 (C) 1 m/s2 Sol.26. F – f1 max = 10 a F – 10 = 10 a or 30 – 10 = 10 a 

a

(D) None

20  2 m / s2 10

Example 33. The coefficient of static and kinetic friction between the two blocks and also between the lower block and the ground are µs = 0.6 and µk = 0.4. Find the value of tenstion ‘T’ applied on the lower block at which the upper block begins to slip relative to lower block. Sol. The maximum static friction on upper block and lower block is f1 max = µSm1g = 0.6 × 2 × 10 = 12 N The maximum static friction between ground and lower block is F2 max = µS(m1 + m2)g = 0.6 × 4 × 10 = 24 N Assume that both blocks move together. T – µk(m1 + m2)g = (m1 + m2)a T – 0.4 × 4 × 10 = 4a or T – 16 = 4a ...(1) For upper block :

M = 2kg M = 2kg

(µ s =0.6, µk =0.4) T

 T  16  f1  2a  2   4 

or

f1 

T  16 2

For just sliding, f1 = f1 max or or

T  16  12 2 T – 16 = 24 T = 24 + 16 = 40 N

Example 34. A block of mass 1 kg is horizontally thrown with a velocity of 10 m/s on a stationary long plank of mass 2 kg whose surface has a µ = 0.5. Plank rests on frictionless surface. Find the time when m1 comes to rest w.r.t. plank. m1 V0

Sol.

m2

µm1g  µg m1 µm1g a2  m2 a1  

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25 25

PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION m1g m2

a rel  a1  a 2  µg  µ

 or

1 a rel  5  0.5  10 2 5 a rel  5   7.5 m / s 2 2 vrel = 0, urel = v0 vrel = urel + arel t 0 = v0 – 7.5 t

t

v0 10 4   sec. 7.5 7.5 3

Example 35. A thin rod of length 1 m is fixed in a vertical position inside a train, which is moving horizontally with constant acceleration 4 m/s2. A bead can slide on the rod, and friction coefficient between them is 1/2. If the bead is released from rest at the top of the rod, find the time when it will reach at the bottom. Sol. From force diagram of bead, N = ma0 = 4m µN N mg – µN = ma or or 

mg

1 mg  4m  ma 2 10 m – 2m = ma a = 8 m/s2 L

a0 = 4m/s2

1 2 at 2

1  8  t2 2

or

1

or

t2 

1 4

t=

1 second 2

Example 36. A car begins to move at time t = 0 and then accelerates along a straight track with a speed given by V(t) = 2t2 ms–1 for 0  t  2. After the end of acceleration, the car continues to move at a constant speed. A small block initially at rest on the floor of the car begins to slip at t = 1 sec. and stops slipping at t = 3 sec. Find the coefficient of static and kinetic friction between the block and the floor. Sol.  v = 2t2 

a

dv  4t dt

at t = 1, a = 4 m/s2 Since, block moves with the car upto t = 1 sec. Hence, static friction comes into play between block and car.  fs = ma But fs =  µsmg or a = µsg

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

or

4 = µsg

µs 

4 4   0.4 g 10

Let the velocity of block at t = 3 second is v0. The velocity of block at t = 1 second is u0 = 2 × 12 = 2 m/s The velocity of car at t = 2 second is uC = 2 × 22 = 8 m/s The acceleration of car is µkg at 1<t3  v0 = u0 + µkg (3 – 1) or v0 = u0 + 2µkg or vC = u0 + 2µkg or 8 = 2 + 2 µk × 10 or 6 = 20 µk 

µk 

6  0.3 20

Example 37. A body of mass 2 kg rests on a horizontal plane having coefficient of friction µ = 0.5. At t = 0 a  horizontal force F is applied that varies with time F = 2t. The time constant t0 at whcih motion starts and distance moved in t = 2t0 second will be ........... and .............. respectively. Sol. Here fmax = µ mg = 0.5 × 2 × 10 = 10 N for sliding starts, F > µ mg F > 10 or 2t > 10 or t>5  t0 = 5 second. Here or or or

a

2t  µmg 2t  0.5  2  10  m 2

2t  10 2 a=t–5 a

dv   t  5 dt t

or

v

0

 t2  dv    t  5 dt    5t  t0  5 2 5 t

 t 2 25  v   5t  25 2 2 

 t2 25  v    5t   2 2

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

or

ds  t 2 25     5t   dt  2 2

or

s   ds  

2t 0 10

t 0 5

 t2 25   2  5t  2  dt 10

 t 3 5t 2 25     t 2 2 5 6  1000   125 125 125  125   250  125     m   6   6 2 2  6

Example 38. With what minimum velocity should block be projected from left end ‘A’ towards end ‘B’ such that it reaches the other end ‘B’ of conveyer belt moving with constant velocity ‘v’. Friction coefficient between block and belt is µ. (A) µgh Sol.

(B)

2µgL

or

a = µg 02 = v02 – 2 µgL v02 = 2 µgL

v 0  2µgL

(C)

3µgL

A

m µ

V0 V

B

L

(D) 2 µgL

m

Example 39. Block M slides down on frictionless incline as shown. Find the minimum M friction coefficient so that ‘m’ does not slide with respect to ‘M’. Sol. Since, m does not slide on the block M. So, both block move together. 37º From force diagram of combined mass (m + M), (M + m)g sin 37º = (m + M)a0  M  m g sin 37º a0    m  M 3 a 0  10   6m / s 2 along inclined surface.  N+ma0 sin37º 5 ma0 From force diagram of smaller block in the frame of triangular block, 37º  N + ma0 sin 37º = mg ma0 cos37º f mg and ma0 cos 37º = f But f  µN 37º or ma0 cos 37º µ(mg – ma0 sin 37º) or a0 cos 37º µ (10 – 6 × 0.6) or 6 × 0.8 µ × 6.4 or 4.8  6.4 µ 4.8 µ or 6.4 6 µ or 8 6 3 µ min    0.75 or 8 4

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IMPORTANT FEATURES 1. The direction of friction force on each of the blocks in Fig.(a) is such as it either stops the relative motion or attempts to do so. For example, if a force F is applied on block A of a two block system, the direction of frictional forces at different contacts on different bodies will be as shown : f1

A B

f2

(a)

A

F f1

B f2 (b)

Here,

2.

3.

f1 = force of friction between A and B f2 = force of friction between B and ground Force of friction f = 0, if no driving force is applied. f  fL (= µsN) If driving force is applied but no relative motion is there. and f = µkN if relative motion is there. A common mistake which the students do in hurry is that they always write fL = µmg (in case of horizontal ground) or fL = µmg cos (in inclined surface). The actual formula is fL = µN. Here, N is equal to mg or mg cos upto when no force is acting at some angle (  0º) with the plane. F F N = mg fL = µmg

F

F

F

or N = mg fL = µmg

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N = mg cos fL = µmg cos

N = mg cos fL = µmg cos

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION THINKING PROBLEMS

1.

Explain why a man getting out of a moving train must run in the same direction for a certain distance.

2.

During a high jump event, it hurts less when an athlete lands on a heap of sand. Explain.

3.

An athlete runs over a certain distance before taking a long jump. Explain.

4.

Ripe mangoes fall down when the three is shaken. Why?

5.

A heavy picture frame is suspended by a string passing over a peg in the vertical wall and attached at its two ends to two points in the upper horizontal edge of the frame. Explain why the picture is likely to fall if the string is too short.

6.

Out of the coefficient of static friction, coefficient of kinetic friction and coefficient of rolling friction, the last is the least. True or false ?

7.

When we walk we rub our footwear opposite to the direction of motion. True or false ?

8.

Action and reaction are equal and opposite and so they balance each other. True or false ?

9.

The motion of the cm of a system is determined by external forces only and not the internal forces of the system. True or false ?

10. The momentum of a system of particles is always conserved. True or false? 11. What will be the trajectory of the bob of a pendulum if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position. 12. Figure shows the position-time graph of a particle of mass m = 0.5 kg. Suggest a suitable example to fit the curve. What is the interval between ten consecutive impulses ? What is the magnitude of each impulse ? 13. A rod not reaching the rough floor is inserted between two identical blocks. A horizontal force F is applied to the upper end of the rod. Which of the blocks will move first ?

14. A massive homogeneous cylinder that can revolve without friction around a horizontal axis is secured on a flat car. A bullet flying horizontally with a velocity v strikes the cylinder and drops on to the car. Does the speed acquired by the car after the impact depend on the point where the bullet strikes the cylinder ? 15. The application of brakes on the driving wheels of a car means the reversal of the direction of the force of friction between the wheel and the ground. True or false ?

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LAWS OF MOTION SOLUTION OF THINKING PROBLEMS

1. A man runs for a certain distance after alighting from a moving train to maintain the same velocity of all parts of his body relative to the train so that he may not fall forward due to the inertia of rest of the lower part of his body and inertia of motion of the upper of his body. 2. Because of greater distance and hence greater time over which him motion is stopped, he experiences less force : áFñ = change in momentum = D p / D t. For a given D p, D t µ 1/áFñ 3. An athlete runs over a certain distance before taking a long jump to gain inertia of motion so that he may take a longer jump. 4. Due to inertia of rest ripe mangoes get detached from their weak stem. 5. Suppose 2l is the length of the string and 2a is the distance between the points on the upper edge of the frame where the ends of the string are attached. Then the inclination of each part of the string to the - 1 a . If T is the tension of the string, then for the equilibrium of the picture, horizontal is cos l W = 2T sin q = 2T 1- a 2 / l 2

Thus it is seen that a shorter length means a greater tension. So the picture is likely to fall if the length is too short. 6. True 10.

7. True

8. False

9. True

False, momentum is conserved only when the external force on the system is zero.

11.(a) The bob will fall vertically downward as the velocity of the bob is zero. 12.

The motion of a particle between two parallel walls; 2s; 8 kg ms–1.

13.The right block will move first. To justify the answer let us suppose that F is the force applied at the end of the rod. Let F1 be the magnitude of the action and reaction between the rod and the light block and F2 that of the rod and and left block. Considering translational equilibrium of the rod. F1 – F2 = F Therefore, F1 > F2 Therefore, the right block will tend to move first. 14. No, the speed of the cart will not depend on the point of impact. The momentum of the revolving cylinder is zero because one half has as much negative linear momentum as the other half has positive linear momentum. For this reason the bullet will impart to the ‘car + cylinder’ system the same momentum as it would be a cylinder rigidly fixed to the cart. 15.Tru. In a car the driving force and the frictional force are in the same direction. This is not in violation of the law of friction. The application of brakes means the application of force on the axle in the opposite direction. Consequently the frictional force is also reversed.

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ASSERTION & REASON THE NEXT QUESTIONS REFER TO THE FOLLOWING INSTRUCTIONS A statement of Statement-1 is given and a Corresponding statement of Statement-2 is given just below it of the statements, mark the correct answer as – (A) If both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1. (B) If both Statement-1 and Statement-2 are true and Statement-2 is NOT correct explanation of Statement-1. (C) If Statement-1 is true but Statement-2 is false. (D) If both Statement-1 and Statement-2 are false. (E) If Statement-1 is false but Statement-2 is true. 1. 2. 3.

4. 5. 6. 7. 8.

Statement-1 : Statement-2 : Statement-1 : Statement-2 : Statement-1 : Statement-2 : Statement-1 : Statement-2 : Statement-1 : Statement-2 : Statement-1 : Statement-2 : Statement-1 : Statement-2 : Statement-1 : Statement-2 :

9.

Statement-1 : Statement-2 : 10. Statement-1 : Statement-2 : 11. Statement-1 : Statement-2 : 12. Statement-1 : Statement-2 : 13. Statement-1 : Statement-2 : 14. Statement-1 : Statement-2 : 15. Statement-1 : Statement-2 :

Aeroplanes always fly at low altitudes. According to Newton’s third law of motion. The slope of momentum versus time curve give us the acceleration. Acceleration is given by the rate of change of momentum. The apparent weight of a body in an elevator moving with some downward acceleration is less than the actual weight of body. The part of the weight is spent in producing downward acceleration, when body is in elevator. When the lift moves with uniform velocity the man in the lift will feel weightlessness. In downward accelerated motion of lift, apparent weight of a body decreases. A player lowers his hands while catching a cricket ball and suffers less reaction force. The time of catch increases when cricketer lowers its hand while catching a ball. A reference frame attached to earth is an inertial frame of reference. The reference frame which has zero acceleration is called a non inertial frame of reference. A table cloth can be pulled from a table without dislodging the dishes. To every action there is an equal and opposite reaction. A body subjected to three concurrent forces cannot be in equilibrium. If large number of concurrent forces acting on the same point, then the point will be in equilibrium, if sum of all the forces is equal to zero. Impulse and momentum have different dimensions. From Newton’s second law of motion, impulse is equal to change in momentum. On a rainy day, it is difficult to drive a car or bus at high speed. The value of coefficient of friction is lowered due to wetting of the surface. When a bicycle is in motion, the force of friction exerted by the ground on the two wheels is always in forward direction. The frictional force acts only when the bodies are in contact. Pulling a lawn roller is easier than pushing it. Pushing increases the apparent weight and hence the force of friction. Two bodies of masses M and m (M > m) are allowed to fall from the same height if the air resistance for each be the same then both the bodies will reach the earth simultaneously. For same air resistance, acceleration of both the bodies will be same. The acceleration of a body down a rough inclined plane is greater than the acceleration due to gravity. The body is able to slide on a inclined plane only when its acceleration is greater than acceleration due to gravity. A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table. For every there is an equal and opposite reaction.

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Level # 1 1.

The upper half on an inclined plane of inclination  is perfectly smooth while the lower half is rough. A block starting from rest at the top of the plane will come to rest at the bottom if the coefficient of friction between the block and the lower half of the plane is given by : 2 1 (A)   2 tan  (B)   tan  (C)   (D)   tan  tan 

2.

A rope which can withstand a maximum tension of 400 N is hanging from a tree. If a monkey of mass 30 kg climbs on the rope, in which of the following cases will the rope brake? Take g = 10 m/s 2 and neglect the mass of the rope (A) the monkey climbs up with a uniform speed of 5 m/s (B) the monkey climbs up with a uniform acceleration of 2 m/s2 (C) the monkey climbs up with a uniform acceleration of 5 m/s2 (D) the monkey climbs down with a uniform acceleration of 5 m/s2

3.

A block of mass m is projected up an inclined plane of inclination

 with an initial velocity u. If the coefficient

of kinetic friction between the block and the plane is  , the distance up to which the block will rise up the plane , before coming to rest, is given by : (A) 4.

u2  2g sin 

(B)

u2  2g cos 

(C)

u2 4g sin 

A flat car given an acceleration a0 = 2 m/s2 starting from rest. A cable is connected to a crate A of weight 50 kg as shown. Neglect friction between the floor and the car wheels and also the mass of the pulley.

(D)

u2 4g cos 

A

a0

Calculate corresponding tension in the cable if   0.30 between the crate and the floor of the car. (A) 350 N (B) 300 N (C) 450 N (D) none of these 5.

A 40 kg trunk sliding across a floor slows down from 5 to 2 m/s in 6 seconds. Assuming that the force acting on the trunk is constant, find is magnitude and its direction relative to the velocity vector of the trunk (A) 20 N in the direction opposite to the velocity (B) 30 N in the direction opposite to the velocity (C) 20 N in the same direction to the velocity (D) none of these

6.

If the coefficient of friction between a car’s wheels and the roadway is 0.70, what is the least distance in which a car can accelerate from rest to a speed of 15 m/s? (A) 20.8 m (B) 14.6 m (C) 16.4 m (D) none of these

7.

In figure , when m is 3 kg, the acceleration of block m is 0.60 m/s2, while a = 1.6 m/s2 If m = 4 kg. Find the frictional force on the block M as well as its mass .(neglect mass & friction of pulley) (A) 12.2 N (B) 10.11 N (C) 8.66 N (D) none of these

8.

Find the frictional force on block 30 kg. (A) 20 N (C) 40 N

9.

(B) 30 N (D) 45 N

A heavy homogeneous sphere is suspended by a light string, one end of which is attached to a vertical wall and the other, to a point on the vertical line through the centre of sphere. What should be the coefficient of friction between the sphere and the wall for the sphere to remain in equilibrium. (A)  < 0.5 (B)  > 1 (C)  = 0 (D) none of these

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10. A short, right circular cylinder of weight W rest in a horizontal, V shaped notch of angle 2 . If the coefficicent of friction is  , Find the horizontal force P parallel the axis necessary for slipping to occur.

W (A) cos 

W (B) tan 

W (C) sin 

2

(D) none of these

11. When a cube is in limiting equilibrium on an inclined plane, it is also about to topple. The coefficient of friction between the cube and the plane is (A) 1

(B) 1/2

(C)1 /

(D) 1 / 3

2

12. A scooter is moving on a straight horizontal surface with a velocity u, calculate the shortest distance in which a scooter can be stopped, if the coefficient of friction between tires and the road is 

u2 (A) 2g

u 2g (B) 2

u2 (C) 4g

(D) none of these

13. In the figure shown the force with which the man should pull the rope to hold the plank in position is F. If weight of the man is 60 kgf, the plank and the pulleys have negligible masses,then (A) F = 150 N (B) F = 300 N (C) F = 600 N (D) F = 1200 N 14. A given object takes n times as much time to slide down a 45 0 rough inclined as a takes to slide down the perfectly smooth 450 incline. The coefficient of kinetic friction between the object and the incline is given by (A)  k 

1 (1  n 2 )

(B)  k  1

1 n2

(C)  k 

1 (1  n 2 )

(D)

1

1 n2

15. For the system in figure, the pulleys are light and frictionless. The tension in the string will be (A)

2 mg sin  3

(C) ½ mg sin

(B)

3 mg sin  2

(D) 2 mg sin

16. A particle is lying at rest on a rough horizontal plane. The plane is tilted till its starts moving & then kept fixed. If the static and the dynamic coefficient of friction are  s and

 k , Find the velocity of particle after it

has travelled a distance d.

 2( s   k )gd  (A)   2 1/ 2  (1   s ) 

1/ 2

(B)

2 s  k (1   s2 )1/ 2

(C)

2( s   k ) 1   s2

(D)

2( s   k ) (1  s2 )1/ 2

17. In the arrangement shown, the pulleys are fixed an ideal, the strings are light, m 1 > m 2, and S is a spring balance which is itself massless. The reading of S (in units of mass) is (A) m 1 - m 2 (B) ½ (m 1 + m 2) (C)

m1m2 m1  m2

(D)

2m1m2 m1  m2

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18. A uniform chain of length l is placed on a rough table, with length l/n, where n > 1, hanging over the edge. If the chain just begin to slide off the table by itself from this position, the coefficient of friction between the chain and the table is (A)

1 n

(B)

1 n 1

(C)

1 n 1

(D)

n 1 n 1

19. Two masses M, m (M > m) are joined by a light spring passing over a smooth light pulley. If the blocks are allowed to move for some time, after which M is stopped momentarily (brought to rest and released at once). After this (A) both blocks will move with different acceleration (B) the string will become taut (under tension) again when the blocks aquire the same speed (C) the string will become taut again when the blocks cover equal distances (D) none of these

m m M

20. A boy B drags a wedge A by an in extensible string passing over the pulleys 1,2,3 and 4 as shown in figure. If all the pulleys are smooth and boy walks with constant velocity of magnitude V, the magnitude of relative velocity between the boy and the wedge is = (A) V (B) 2V (C) 1.5 V (D) 1.25 V 21. In the given figure all the surfaces are smooth. Find the time taken by the block to reach from the free end to the pulley attached to the plank. Distance between free end and pulley is  (A)

2Mm (2m  M)F

(B)

Mm (2m  M)F

(C)

Mm (m  M)F

(D) none of these

MORE THAN ONE ANSWER IS CORRECT 22. A block of mass 0.1 is held against a wall applying a horizontal force of 5 N on the block. If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is: (A) 2.5 N (B) 0.98 N (C) 4.9 N (D) 0.49 N

23. The pulley and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle  should be (Shown in figure) [JEE (Scr.) 2001] (A) 0° (B) 30° (C) 45° (D) 60°

2m m

m

24. A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given by (A)

2 Mg

(C)

 M  m

2

 m2 g

(B)

2 mg

(D)

 M  m 

m

[JEE (Scr.) 2001] 2

M

M2 g F

25. What is the maximum value of the force F such that the block shown in the arrangement, does not move? (A) 20 N (B) 10 N (C) 12 N (D) 15 N [JEE (Scr.) 2003]

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 = 60°

1 2 3

m = 3 kg

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26. A block P of mass m is placed on a horizontal frictionless plane. A second block of same mass m is placed on it and is connected to a spring of spring constant k, the two blocks are pulled by distance A. Block Q oscillates without slipping. What is the maximum value of frictional force between the two blocks. (A)

kA 2

(B) KA

(C)

 s mg

k

Q

s

P

(D) Zero

27. A particle starts sliding down a frictionless inclined plane. If S n is the distance traveled by it from time

t  n  1 sec to t  n sec , the ratio S n S n 1 is (A)

2n  1 2n  1

(B)

2n  1 2n

(C)

2n 2n  1

(D)

2n  1 2n  1

28. A reference frame attached to the earth (A) is an inertial frame by definition. (B) cannot be an inertial frame because the earth is revolving round the sun. (C) is an inertial frame because Newton’s laws are applicable in this frame. (D) cannot be an inertial frame because the earth is rotating about its own axis.

29. A spring of force constant ‘k’ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of [JEE-1999] (A) (2/3) k (B) (3/2) k (C) 3 k (D) 6 k 30. The string between blocks of mass m and 2m is massless and inextensible. The system is suspended by a massless and inextensible. The system is suspended by a massless spring as shown. If the string is cut find the magnitudes of accelerations of mass 2m and m (immediately after cutting) [JEE-2006]

2m m

g g g g (A) g, g (B) g, (C) , g (D) , 2 2 2 2 31. Two particles of mass ‘m’ each are tied at the ends of a light string of length 2a. F The whole system is kept on a frictionless horizontal surface with the string held m m tight so that each mass is at a distance ‘a’ from the centre ‘P’ (as shown in the P a a figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force ‘F’. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes 2x, is [JEE-2007]

F a F x F x F a2  x2 (A) 2m 2 (B) 2m 2 (C) (D) 2m a a  x2 a  x2 2m x 32. A particle moves in the X–Y plane under the influence of a force such that its linear momentum is  pt  A  ˆi cos  kt   ˆjsin  kt   , where ‘A’and ‘k’ are constants. The angle between the force and the momentum is [JEE-2007] (A) 0º (B) 30º (C) 45º (D) 90º

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FILL IN THE BLANKS 1.

A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is 5 m/s 2, the frictional force acting on the block is ................ newtons.

2.

A uniform rod of length L and density  is being pulled along a smooth floor with a horizontal acceleration  (see Figure). The magnitude of the stress at the transverse cross-section through the mid-point of the rod is ...............

3.

L 

Imagine that you are holding a book weighing 4 N at rest on the palm of your hand. Complete the following sentences: (A) A downward force of magnitude 4 N is exerted on the book by __________ (B) An upward force of magnitude __________ is exerted on __________ by the hand. (C) Is the upward force in part (B) the reaction to the downward force in part (A)? (D) The reaction to the force in part (A) is a force of magnitude __________ exerted on __________ by __________ Its direction is __________ (E) The reaction to the force in part (B) is a force of magnitude __________ exerted on __________ by __________ Its direction is _____ (F) The forces in parts (A) and (B) are equal and opposite because of Newton’s __________ law. (G) The forces in parts (B) and (E) are equal and opposite because of Newton’s __________ law.

TRUE / FALSE 4. 5. 6.

7.

When a person walks on a rough surface, the frictional force exerted by the surface on the person is opposite to the direction of his motion. A simple pendulum with a bob of mass m swings with an angular amplitude of 40°. When its angular displacement is 20°, the tension in the string is greater than mg cos20°. The pulley arrangements of Figure (A) and (B) are identical. The mass of the rope is negligible. In (A) the mass m is lifted up by attaching a mass 2m to the other end of the rope. In (B), m is lifted up by pulling the other end of the rope with a constant downward force F = 2 mg. m 2m m F = 2mg The acceleration of m is the same in both cases. (A) (B) Two particles of mass 1 kg and 3 kg move towards each other under their mutual force of attraction. No other force acts on them. When the relative velocity of approach of the two particles is 2 m/s, their centre of mass has a velocity of 0.5 m/s. When the relative velocity of approach becomes 3 m/s, the velocity of the centre of mass is 0.75 m/s.

TABLE MATCH 8.

Match Table I and Table II and select the correct answer using the codes given below the lists:

Column-I A. B. C. D. 9.

Column-II

Normal reaction Frictional force Nuclear force Force of tension

P. Q. R. S.

Electromagnetic force [MLT–2] Super position principle Short range force

Match Table I and Table II and select the correct answer using the codes given below the lists:

Column-I A. B. C.

D.

Column-II

Thrust on a body of mass m floats inside a liquid Normal reaction on a body of mass m placed on rough horizontal surface stationary Friction on a body of mass m placed on a rough horizontal surface in the absence of pulling and pushing force Friction on a body of mass m moving on horizontal rough surface.

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P. mg Q. Zero R.

µmg

S.

greater than zero

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37 37

PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION PASSAGE TYPE QUESTIONS

PASSAGE # 1 A moving company uses the pulley system in figure 1 to lift heavy crates up a ramp. The ramp is coated with rollers that make the crate’s motion essentially frictionless. A worker piles cinder blocks onto the plate until the plate moves down, pulling the crate up the ramp. Each cinder block has mass 10 kg. The plate has mass 5 kg. The rope is nearly massless, and the pulley is essentially frictionless. The ramp makes a 30° angle with the ground. the crate has mass 100 kg.

Cr ate

30°

Plate

Let W1 denote the combined weight of the plate and the cinder blocks piled on the plate. Let T denote the tension in the rope. And let W2 denote the crate’s weight. 1.

What is the smallest number of cinder blocks that need to be placed on the plate in order to lift the crate up the ramp? (A) 3 (B) 5 (C) 7 (D) 10

2.

Ten cinder blocks are placed on the plate. As a result, the crate accelerates up the ramp. W hich of the following is true?

3.

(A) W1  T  W2 sin 30

(B) W1  T  W2 sin 30

(C) W1  T  W2 sin 30

(D) W1  T  W2 sin 30

The ramp exerts a “normal” force on the crate, directed perpendicular to the ramp’s surface. This normal force has magnitude: (B) W2 sin 30

(A) W2 4.

(D) W2  sin 30  cos 30 

(C) T  W2 sin 30

(D) T  W2

The net force on the crate has magnitude: (A) W1  W2 sin 30

5.

(C) W2 cos 30

(B) W1  W2

After the crate is already moving, the cinder blocks suddenly fall off the plate. Which of the following graphs best shows the subsequent velocity of the crate, after the cinder blocks have fallen off the plate? (Up theramp is the positive direction.)

+ v 0

(A) A

A time

+ v 0

B time

(B) B

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+ v 0

(C) C

C

+ v time

0

D

time

(D) D

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38 38

PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

PASSAGE # 2 A safety engineering firm is producing a film for high school driver education classes. The firm uses skilled test drivers driving both small cars and larger vans. The vans weigh three times as much as the cars and have larger tires with twice the tread width. In a demonstration that tests reaction times and skid-to-stop distances and shows them on the film, three guns that fire a yellow paint onto the road are mounted on the bumpers and fired electrically. When the driver hears the report of the first gun, he locks the breaks, and the touch of his foot on the brake pedal fires a second yellow pellet. The third pellet is fired when the car stops. The safety engineers also design several remote controlled cars and vans in order to film crash results. 6.

Drivers of a car and a van brake hard and skid to a stop from 50 mph. The skid marks are measured to be the same length for both. Why are the stopping distances the same length? (A) The mechanical work done by friction to stop both is the same (B) The frictional force between tires and road is three times greater for the heavier van so it slides the same distance as the car (C) The frictional force for the car and van are the same (D) The wider tires on the van require less friction force than the narrow tires on the car.

7.

Two drivers in identical cars skid to a stop from speeds of 20 mph and 40 mph. How do the lengths of the skid marks compare? (A) They are the same length (B) The 40 mph mark is twice as long (C) The 40 mph mark is four times as long (D) The 40 mph mark is eight times as long.

8.

A remote controlled car and van are crashed head on at the same speed. Why does the car suffer more damage in the collision? (A) The car and van had the same momentum (B) The forces during collision are equal and opposite, so the smaller and weaker car suffers more damage. (C) The van exerts a larger force on the car (D) The mechanical work down in stopping the car is greater.

9.

The van going at a speed of 15 mph collides head on with a car going at a speed of 45 mph. Because the van weighs three times as much, their vector momenta are equal and opposite. However, examination shows that the car suffers more damage than the van. Why? (A) The massive van exerts a larger force on the car. (B) The lighter car exerts a smaller force on the van (C) the forces exerted during the collision are equal and opposite, so the weaker car suffers more damage. (D) The mechanical work required to stop the van is smaller.

10. A driver in a van skids to a stop from 20 mph and a driver in a car skids to a stop from 60 mph. How do the length of the skid marks compare? (A) The van skids three times further than the car, because it is three times heavier. (B) The car skids three times further because it is lighter an si going three times faster. (C) The car will skid nine times as fast as the van (D) They skid the same length because the speed ratio is 1:3, whereas the weight ratio is 3:1. 11. For the reaction time test, one driver is tested at 20 mph and 60 mph. It is noted that the distance between the first two paint marks is three times farther for the 60 mph test than the 20 mph test. How do his reaction times compare at 20 mph and 60 mph? (A) His reaction time at 60 mph is three times faster. (B) His reaction times remain the same (C) His reaction time at 20 mph is three times faster. (D) His reaction time at 20 mph is one-third as long as at 60 mph.

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39 39

PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

PASSAGE # 3 A student performs a series of experiments to determine the coefficient of static friction and the coefficient of kinetic friction between a large crate and the floor. The magnitude of the force of static friction is always less than or equal to

 s N . where  s denotes the coefficient of static friction and N denotes the normal f s  s N

force exerted by the floor on the crate–

Static friction exists only when the crate is not sliding across the floor. The force of kinetic friction is given by

f k   K N , where  K denotes the coefficient of kinetic friction. Kinetic friction exists only when the crate is sliding across the floor. The crate has mass 100 kg. In this situation, the normal force points upward. Experiment 1– The student pushes horizontally (rightward) on the crate and gradually increases the strength of this push force. The crate does not begin to move until the push force reaches 400 N. Experiment 2– The student applies a constant horizontal (rightward) push force for 1.0 second and measures how far the crate moves during that time interval. In each trial the crate starts at rest, and the student stops pushing after the 1.0 second interval. The following table summarizes the results. Trial 1 2 3

Push force (N) 500 600 700

Distance (m) 1.5 1.5 2

12. The coefficient of static friction between the crate and floor is approximately– (A) 0.25 (B) 0.40 (C) 2.5 (D) 4.0 13. In experiment 1, when the rightward push force was 50 N the crate didn’t it move? (A) The push force was weaker than the frictional force on the crate. (B) The push force had the same strength as the gravitational force on the crate. (C) The push force was weaker than the frictional force on the crate. (D) The push force had the same strength as the frictional force on the crate. 14. The coefficient of kinetic friction between the crate and the floor is approximately– (A) 0.20 (B) 0.30 (C) 0.40 (D) 0.50 15. In trial 3, what is the crate’s speed at the moment the student stops pushing it? (A) 1.0 m/s (B) 2.0 m/s (C) 3.0 m/s (D) 4.0 m/s

1 time(s)

1 time(s)

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(D) 1 time(s)

position

(C)

position

(B)

position

(A)

position

16. For trial 3, which of the following graphs best shows the positions of the crate as a function of time? The student first starts pushing the crate at time t = 0.

1 time(s)

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40 40

PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

Level # 2

Section-A 1.

2.

In figure, block 1 is one-fourth the length of block 2 and weighs 1 m one-fourth as much. Assume that there is no friction between block 2 and the surface on which it moves and that the coefficient 2 4m of sliding friction between blocks 1 and 2 is k = 0.2. After the system is released, find the distance block 2 has moved when only one-fourth of block 1 is still on block 2. Block 1 and block 3 have the same mass.

A dinner plate rests on a tablecloth, with its center 0.3m from the edge of the table. The tablecloth is suddenly yanked horizontally with a constant acceleration of 9.2 m/s2. The coefficient of sliding friction between the tablecloth and the plate is

 k = 0.75. Find (a) the acceleration, (b) the velocity,,

and (c) the distance of the plate from the edge of the table, when the edge of the tablecloth passes under the center of the plate. Assume that the tablecloth just fits the tabletop.

3.

m 3

2

a = 9.2 m/s x

A research balloon of total mass M is descending vertically with downward acceleration a. How much balalst must be thrown from the car to give the balloon an upward acceleration a, presuming that the upward life of the air on the balloon does not change?

F 4.

5.

6.

Someone exerts a force F directly up on the axie of the pulley shown in figure. Consider the pulley and string to be massless and the bearing frictionless. Two objects, m 1 of mass 1.2 kg. and m 2 of mass 1.9 kg, are attached as shown to the opposite ends of the string, which passes over the pulley. The object m 2 is in contact with the floor. (a) What is the largest value the force F may have so that m 2 will remain at rest on the floor? (b) What is the tension in the string if the upward force F is 110 N? (c) With the tension determined in part (b), what is the acceleration of m 1? Two particles, each of mass m, are connected by a light string of length 2L, as shown in figure. A steady force F is applied at the midpoint of the string (x = 0) at a right angle to the initial position of the string. Show that the acceleration of each mass in the direction at 90 0 to F is F x given by a x  2m 2 2 1/ 2 in which x is the perpendicular distance of (L  x ) one of the particles from the line of action of F. Discuss the situation when x = L.

m1 m2 2L m

m F

A worker wishes to pile sand onto a circular area in his yard. The radius of the circle is R. No sand is to spill onto the surrounding area; Figure show that the h

greatest volume of sand that can be stored in this manner is  s R 3 / 3 where

 s is the cofficient of static friction of sand on sand. (The volume of a cone

R

is Ah/3, where A is the base area and h is the height)

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41 41

PHYSICS : SHEET LAWS OF MOTION 7.

8.

LAWS OF MOTION

A block of mass M is connected with a particle of mass m by a light inextensisble string as shown in figure. Assuming all contacting surfaces as smooth, find the acceleration of the wedge after releasing the system.

M m

If A and B each weight 170 N and

B

BAR

Blocks A and B in figure are connected with a bar of negligible weight.

 A = 0.2 and  B = 0.4, calculate the

A

acceleration of the system and the force in the bar.

8 15

9.

In the system of connected bodies in figure the coefficient of friction is 0.2 under bodies B and C. Determine the acceleration of each body and the tension in this cord.

10. A smooth ring of mass M, is threaded on a string whose ends are then threaded over two smooth fixed pulleys with masses m and m’ tied on-to them respectively, the various portions of the strings being vertical. The system being free to move, show that the ring will remain at rest if :-

4 1 1   M m m'

m

m´ M

11. Two men, of masses M and M + m, start simultaneously from the ground and climb with uniform accelerations up the free ends of a weightless inextensible rope which passes over a smooth pulley at a height h from the ground. If the lighter of the two men reaches the pulley in t second, show that the heavier cannot get nearer

m  gt 2   h . to it than  Mm  2  12. A 25 kg block A rests on an inclined surface and a 15 kg counter weight B is attached to a cable as shown. Neglecting friction, determine the acceleration of A, acceleration of B and tension in the cable after the system is released from rest. Cable is parallel to the plane Take g = 10 m/s2.

Section-B 1.

The 10 kg solid cylinder is resting in the inclined V-block. If the coefficient of static friction between the cylinder and the block is 0.5, determine (a) the frictional force F acting on the cylinder at each side before force P is applied (b) the value of P required to start sliding the cylinder up the incline (g = 9.8 m/s2)

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42 42

PHYSICS : SHEET LAWS OF MOTION 2.

LAWS OF MOTION

A block A of mass m and length l is placed on a horizontal floor. A rectangular box B is used to cover A. The distance between interior of the walls of B is L (>1) and the mass of B is also m. The coefficient of friction between A and floor is between B and floor is

1 and that

 2 ( 2  1 ) . Initially the left end of A

touches the left wall of B as shown in figure and both A and B moves with velocity  0 towards the right. All collisions between A ad B are elastic and contact time during each collision is very short. Find an expression for the period between two consecutive collisions.

3.

Consider the situation shown in figure the block B moves on a frictionless surface, while the coefficient of friction between A and the surface on which it moves is 0.2. Find the acceleration with which the masses move and also the tension in the strings. (Take = 10 m/s2)

4.

Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 ms-2. What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, upto what acceleration of the belt can the man continue to be stationary relative to the belt? Mass of the man = 65 kg. (g = 9.8 m/s2) The rear side of truck is open and 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 below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 ms-2 , find the time when box falls off the truck. (g = 9.8 m/s2)

5.

6.

A small disc A is placed on an inclined plane forming an angle  with the horizontal and is imparted an initial velocity v 0. Find how the velocity of the disc depends on the angle  if the friction cofficient   tan  and at the initial moment .  0   / 2

7.

A monkey of mass m clings to a rope slung over a fixed pulley. The opposite end of the rope is tied to a weight of mass M lying on a horizontal plate. The coefficient of friction between the weight and the plate is  . Find the acceleration of weight and the tension of the rope for three cases.(a) the monkey does not move with respect to the rope (b) the monkey moves upwards with respect to the rope with an acceleration b (c) the monkey moves downwards with respect to the rope with an acceleration b.

8.

–2

a = 2 ms

Each of the three plates has a mass of 10 kg. If the coefficients of

5m

M

m

18 N

static and kinetic friction at each surface of contact are  s = 0.3 and

 k  0.2 , respectively, determine the acceleration of each plate when the three horizontal forces are applied.

(Take g = 10 m/s2)

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D C

15 N

100 N

B A

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43 43

PHYSICS : SHEET LAWS OF MOTION 9.

LAWS OF MOTION

In the arrangement shown in figure, the rod of mass m held by two smooth walls, remians always perpendicular to the surface of the wedge of mass M. Assuming all the surfaces are frictionless, find the acceleration of the rod and that of the wedge.

m

10. A plank of mass M is placed on a rough horizontal surface and a constant horizontal force F is applied on it. A man of mass m runs on the plank. Find the acceleration of the man so that the plank does not move on the surface. Co-efficient of friction between the plank and the surface is  . Assume that the man does not slip on the plank.

M

F

11. A certain constant force starts acting on a body moving at a constant velocity v. After a time interval t , the velocity of the body is reduced by half and after the same time interval the velocity is again reduced by half. Determine the velocity of the body after a time interval 3  t from the moment when the constant force starts acting. M

12. A cart with a mass of M = 0.5 kg is connected by a string to a weight having a mass m = 0.2 kg. At the initial moment the cart moves to the right along a horizontal plane at a speed of v0 = 7 m/s. Find the magnitude and direction of the velocity of the cart, the place it will be at and the distance it covers in t = 5 seconds.

V0

m F(N)

13. A 20 kg block is originally at rest on a horizontal surface for which the coefficient of friction is 0.6. If a horizontal force F is applied such that it varies with time as shown. Determine the speed of the block in 10s.

F

200

0

5

14. For the system at rest shown in the figure, determine the accelerations of all the loads immediately after the lower thread keeping the system in equilibrium has been cut. Assume that the threads are weightless and inextensible, the strings are weightless, the mass of the pulley is negligibly small and there is no friction at the point of suspension.

t(s)

10

m1

m3

m2

m4

k 15. (a) In the arrangement shown in the figure the floor is smooth and the friction exists only between the blocks. The coefficient of static friction  s = 0.6 and coeffecient of kinetic friction

 k = 0.4, the

m1

m2 masses of the block are m 1 = 20 kg and m 2 = 30kg. Find the acceleration of each block if (1) F = 180 N (2) F = 200 N (b) Do the previous for F = 180 N if F is directed up as shown in the figure.

m1 F

k

F

m2

16. A body with zero initial slips from the top of an inclined plane forming an angle  with the horizontal.The coefficient of friction  between the body and the plane increases with the distance s from the top according to the law µ = bs. The body stops before it reaches the end of the plane. Determine the time t from the beginning of motion of the body to the moment when it comes to rest.

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44 44

PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

Level # 3 1.

A particle of mass 10–2 kg is moving along the positive x-axis under the influence of a force F(x) = – where K = 10–2 Nm 2. At time t = 0 it is at x = 1.0 m and its velocity is v = 0. (a) Find its velocity when it reaches x = 0.50 m. (b) Find the time at which it reaches x = 0.25 m (IIT 1998)

2.

3.

4.

A smooth semicircular wire-track of radius R is fixed in a vertical plane. One end 3R of a massless spring of natural length is attached to the lowest point O of the 4 wire-track. A small ring is held stationary at point P such that the spring makes an angle of 60° with the vertical. The spring constant K = mg/R. Consider the instant when the ring is released, and (a) draw the free body diagram of the ring. O (b) determine the tangential acceleration of the ring and the normal reaction. (IIT 1996)

Two masses m and 2m are connected by a massless string which passes over a light frictionless pulley as shown in the figure. The masses are initially held with equal lengths of strings on either side of the pulley. Find the velocity of the masses at the instant the lighter mass moves up a distance of 6.54 m. The string is suddenly cut at that instant. Calculate the time taken by each mass to reach the ground. (g = 981 cm/s2) (IIT 1977)

m

P

2m 13.08 m

Ground

m2

Two cubes of masses m 1 and m 2 be on two frictionless slopes of block A which rests on a horizontal table. The cubes are connected by a string which passes over a pulley as shown in the figure. To what horizontal acceleration  should the whole system (that is blocks and cubes) be subjected so that the cubes do not slide down the planes. what is the tension of the string in this situation? (IIT 1978)

A lift is going up. The total mass of the lift and the passengers is 1500 Kg. The variation in speed of the lift is as given in the graph. (a) What will be the tension in the rope pulling the lift at t equal to (i) 1 sec (ii) 6 sec (iii) 11 sec. (b) What is the height to which the lift takes the passengers? (c) What will be the average velocity and the average acceleration during the course of the entire motion? ( g = 9.8 m/s2) (IIT 1976)

2

60°

 ƒ m1

A 5.

K

2 x 

 v(m/s) 3.6

O

2

10

12 t (sec)

6.

Two balls A and B of masses 100 gm and 250 gm respectively and connected by a stretched string of negligible mass, and placed on a smooth table. When the balls are released simultaneously, the initial acceleration of ball B is 10 cm/s2 westward. What is the magnitude and direction of the initial acceleration of the ball A? (IIT 1975)

7.

A spring weighing machine kept inside a stationary elevator reads 50 kg when a man stands on it. What would happen to the scale reading if the elevator is moving upward with (a) constant velocity (b) constant acceleration ? (IIT 1972)

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45 45

PHYSICS : SHEET LAWS OF MOTION 8.

LAWS OF MOTION

Two identical blocks A and B are placed on a rough inclined plane of inclination 45°. The coefficient of friction between block A and incline is 0.2 and that of between B and incline is 0.3. the initial

2m

9.

B

A

separation between the two blocks is 2m . The two blocks are released from rest, then find (a) the time after which front faces of both blocks come in same line and (b) the distance moved by each block for attaining above position. (IIT 2004)

B A

45°

In the figure masses m 1, m 2 and M are 20 kg, 5 kg and 50 kg P1 m1 respectively. The coefficient of friction between M and ground is zero. The coefficient of friction between m 1 and M and that between m 2 and ground is 0.3. The pulley and the string are massless. The string is perfectly horizontal between P 1 and F M m 1 and also between P2 and m 2. The string is perfectly vertical P2 m2 between P1 and P2. An external horizontal force F is applies to the mass M. Take g = 10 m/s2. (a) Draw a free body diagram for mass M, clearly showing the forces. (b) Let the magnitude of the force of friction between m 1 and M be  1 and that between m 2 and ground be  2. For a particular F it is found that  1 = 2  2. Find  1 and  2. Write down equations of motion of all the masses. Find F, tension in the string and accelerations of the masses. (IIT 2000)

10. Block A of mass m block B of mass 2m are placed on a fixed triangular wedge by means of a massless, in extensible string and a frictionless pulley as shown in Figure. The wedge is inclined at 45° to the horizontal on both sides. The coefficient of friction between block A and the wedge is 2/3 and that between block B and the wedge is 1/3. If the system A and B is released from rest, find (a) The acceleration of A, (b) tension in the string, (c) the magnitude and direction of friction acting on A.

A m 45°

(IIT 1997, May)

(g = 9.8 m/s2, sin 37° =

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2m 45°

11. Masses M1, M2 and M3 are connected by strings of negligible mass which pass over massless and frictionless pulleys P 1 and P2 as shown in the figure. Th masses move such that the portion of the string between P1 and P2 is parallel to the inclined plane and the portion of th string between and the coefficient of kinetic friction between the masses and the surfaces is 0.25. The inclined plane makes an angle of 37° with the horizontal. If the mass M1 moves downward with a uniform velocity, find: (a) the mass of M1 (b) tension in the horizontal portion of the string. 3 ) 5 12. In the Figure (a) and (b), AC, DG and GF are fixed inclined planes. BC = EF = x and AB = DE = y. A small block of mass M is released from the point A. It slides down AC and reaches C with a speed V C. The same block is released from rest from the point D. It slides down DGF and reaches the point F with speed V F . The coefficient of kinetic frictions between the block and both the surfaces AC and DGF are  . Calculate VC and VF . (IIT 1980)

B

P1 M2

P2

M1 M3

37°

(IIT 1981)

D

A

G

B

(a)

C

E

(b)

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46 46

PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION m

13. Two blocks connected by a massless sting slide down an inclined plane having an inclination of 37°. The masses of two blocks are m 1 = 4 kg m 2 = 2 kg respectively and the coefficient of friction of m 1 and m 2 with inclined plane are 0.75 and 0.25 respectively. Assuming the string to be taut, find (a) the common acceleration of two masses and (b) the tension in the string (sin 37° = 0.6, g = 9.8 m/s2) (IIT 1979) 14. In the diagram shown, the blocks A, B and C weight 3 kg, 4 kg and 5 kg respectively. The coefficient of sliding friction between any two surfaces is 0.25. A is held at rest by a massless rigid rod fixed to the wall while B and C are connected by a light flexible cord passing around a frictionless pulley. Find the force F necessary to drag C along the horizontal surface to the left at constant speed. Assume that the arrangement shown in the diagram, B on C and A on B, is maintained all through. (g = 9.8 m/s2) (IIT 1978)

1

m 2

37°

A B ƒ

C

15. An aeroplane requires for take off a speed of 80 km/h, the run on the ground being 100 metres. The mass of the plane is 10000 kg and the coefficient of friction between the plane and the ground is 0.2. Assume that the plane accelerates uniformly during the take off. What is the minimum force required by the engine of the plane for the take off? (g = 9.8 m/s2) (IIT 1977)

16. A circular disc with a groove along its diameter is placed horizontally on a rough surface. A block of mass 1 kg is placed as shown. The co-efficient of friction between the block and all surfaces of groove and horizontal surface in contact 25 m/s2 2 is µ  . The disc has an acceleration of 25 m/s2 towards left. Find the 5 4 3 acceleration of the block with respect to disc. Given cos   , sin   . [2006] 5 5

2. (D)

3. (C)

4. (E)

5. (A)

6. (D)

7. (B)

9. (E)

10. (A)

11. (E)

12. (A)

13. (D)

14. (D)

15. (B)

8. (E)

Level # 1 Q. A n s. Q. A n s. Q. A n s. Q. A n s.

1 A 10 C 19 C 28 BD

2 C 11 A 20 D 29 B

3 C 12 A 21 A 30 C

4 A 13 A 22 B 31 B

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5 A 14 B 23 C 32 D

6 C 15 C 24 D

7 A 16 A 25 A

8 D 17 D 26 A

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9 B 18 B 27 A

47 47

PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

Fill in the Blanks / True-False / Match Table 1. 5 2. 0.5 p 3. (a) Earth (b) 4N, Book (c) No (e) 4N, hand, book, downward, (f) Second law 4. False 5. False 6. False 8. [(A — P, Q, R); (B — P, Q, R); (C — Q, S); (D — P, Q, R)] 9. [(A — P, S); (B — P, S); (C — Q); (D — R, S)]

(d) 4N, Earth, book, upward (g) Third law. 7. False

Passage Type Question Q. Ans. Q. Ans.

1 B 10 C

2 D 11 B

3 C 12 B

4 C 13 D

5 C 14 B

6 B 15 D

7 C 16 C

8 B

9 C

Level # 2

Section-A 1.

 7.47

 a   a g 

3. 2M 

2. (a) 7.35 m/s2, (b) 4.26 m/s (c) 1.54 m

4. (a) 37 N, (b) 55 N, (c) 36 m/s2, Upward 8. 2.02 m/s2, P = 15 N

4 mg 17m  M

7.

9. T = 174.1 N , f A = 1.27 m/s2 , downwards , f B = 1.08 m/s2 , f C = 0.9 m/s2

12. 1.272 m/s2 (down the plane) , 1.8 m/s2 (at 150 with horizontal ) , 149 N

Section-B 1. (a) 24.5 N , (b) 109 N

4.

2.

2(L  ) ( 2  1 )g

Net force on the man = 65 × 1 = 65 N a0 = Mg = 0.2 ×9.8 = 1.96 m/s 2

6. v =

0 1  cos 

(b) a 

7. (a) a 

5.

2 5 = 4.34 s 0.53

( m  M )g mMg (1  ) , T m M Mm

m(g  b)  Mg 2m1m3 g , mM (m2 m3)(m1 m2)m2m3

8. aB = 0, aC = 4 m/s2 , aD = 0.2 m/s2

10.

3. a = 6 m/s2 T 2 = 48 N, T 1 = 32 N

(c) a 

m(g  b)  M g mM[g (1  )  b] , T mM mM

mg cos  sin  9.  m sin   M  sin   

F  ( M  m )g F  ( M  m )g  < a< + m m m m

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11.

mg cos  ,  m sin   M  sin   

7 v 4

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PHYSICS : SHEET LAWS OF MOTION

LAWS OF MOTION

12. The cart will be at the same place and will have a speed v = 7 m/s, directed to right & total covered distance is 17.5 m. 14. a1 = a2 = a3 = 0 a 4 

13. 24.8 m/s

(ii) a1 = 6.08 m/s2 , a2 = 2.6 m/s2

m3  m 4  m1  m 2 g m4

15. (a) (i) a1 = a2 = 3.6 m/s2 16. t 

(b) a1 = 5.08 m/s2 , a2 = 3.4 m/s2

 gb cos 

Level # 3 1. (a) – 1 m/s (b) 1.48 sec

5 3    2. (a) a T   8  g  

3 (b) N    mg 8

3. Velocity = 6.54 m/s, Time taken by m is 2.8 second, Time taken by 2 m is

2 second. 3

 m1 sin   m2 sin   m1 m 2 sin   4.    m cos   m cos   g , T   m 2 1 cos   m 2 cos    1 

5. (a) (i) 17400 N (ii) 14700 N (iii) 1200 N (b) 36 m (c) Average velocity = 3 m/s, Average acceleration = 0 6. 25 cm/s2 Eastward. 7. (a) ramains same (b) increase 8. (a) t = 2 sec (b) SB = 7 2 m 9.

(c) SS = 8 2 m

(b)  1 = 30 N,  2 = 15 N, F = 60 N Tension = 18 N (Acceleration of M) = (Acceleration of m 1) = 0.6 m/s2 (m 1 is at rest w.r.t. M).

10. (a) zero

2 2    (b)  3  mg  

mg

(c)

3 2

11. (a) M1 = 4.2 kg

(b) 9.8 N

12. VC  VF  2 g y   x  13. (a) 1.31 m/s2

(b) 5.2 N

14. 78.4 N

15. 4.43 x 104 N

16. 10 m/s2

—X—X—X—X—

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2.LAWS OF MOTION
2.LAWS OF MOTION

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