Kinematics Important Questions and Solutions for JEE Main and Advanced

2

KINEMATICS SINGLE CORRECT QUESTION

1.

Particles A and B are moving with constant velocities along x and y axis respectively, the graph of separation between them with time is Class 11 chemistry chapter 2 notes

(A)

s

(B) t

2.

3.

s

(C) t

s

(D) t

t

A particle starting from rest covers a distance S with uniform acceleration and then a distance of 2S withuniform velocity. Finally it stops after covering further a distance 3S with uniform deceleration. If the entirepath travelled by the particle is a straight line, then the ratio of its avearge velocity and the maximum velocitywill be: (A) 3/5 (B) 4/5 (C) 2/5 (D) 1/5 A particle is thrown horizontally with velocity 10 m/s from an inclined plane, which is also moving with acceleration 10 m/s2 vertically upward. The time after which it lands on the plane (Take g = 10 m/s2) (A) 1 sec 2 v = 10m/s A = 10m/s (B) 2 sec (C)

1 sec 2 30°

1 sec (D) 3 4.

5.

s

A particle P is projected from a point on the surface of smooth inclined plane (see figure).

Simultaneouslyanother particle Q is released on the smooth inclined plane from the same position.

P and Q collide on theinclined plane after t = 4 second. The speed of projection of P is : (A) 5 m/s (B) 10 m/s (C) 15 m/s (D) 20 m/s P is a point moving with constant speed 10 m/s such that its velocity vector always maintains an angle 60° with line OP as shown in figure (O is a fixed point in space). The initial distance between O and P is 100 m.

After what time shall P reach O. (A) 10 sec.

(B) 15 sec.

(C) 20 sec.

6.

7.

(D) 20 3 sec Velocity versus displacement graph of a particle moving in a straight line is as shown in figure. The acceleration of the particle is (A) constant (B) increases linearly with x (C) increases parabolically with x (D) none A particle moves in space along the path z = ax3 + by2 in such a way that

dy dx =c= . Where a, b and c are dt dt

constants. The acceleration of the particle is (A) (6ac2x + 2bc2) k̂

(B) (2ax2 + 6by2) k̂

(C) (4bc2x + 6ac2) k̂

(D) (bc2x + 2by) k̂ (4)

8.

9.

A projectile is thrown from the origin. The vertical velocity and the horizontal velocity vary with respect to time according to the graphs shown. Answer the following questions: (g = 10 m s/2) &  is initial angle of projection. Value of t1 is ...... & initial angle of projection is .............

(A) 1,600 (B) 2, 600 (C) 1, 450 (D) 2,450 A car 2m long and 3m wide is moving at 10 m/s when a bullet hits it in a direction making an angle of tan-1

3   with the car as seen from the ground. The bullet enters one edge of the car at the corner and passes out 4

10.

11.

12.

13.

at diagonally opposite corner. Neglecting gravity, the time for the bullet to cross the car is (A) 1.0 s (B) 0.4 s (C) 0.2 s (D) 0.6 s Two trains are moving towards each other with a velocity Vm/s and 3 Vm/s. The trains are at a distance L from each other. If the trains are of length L and 2L then find out the time in which they will cross each other.

(A) L/V sec.

(B) 2L/V sec.

(C) 3L/2V sec.

(D) L/2V sec.

In the above question. the total path length used for the crossing (A) 3L (B) 4L (C) 5L (D) 6L Two trains, which are moving along different tracks in opposite directions towards each other, are put on the same track due to a mistake. Their drivers, on noticing the mistake, start slowing down the trains when the trains are 300 m apart. Graphs given below show their velocities as function of time as the trains slow down. The separation between the trains when both have stopped, is:

(A) 120 m (B) 280 m (C) 60 m (D) 20 m A particle of mass m is projected at an angle of 60° with a velocity of 20 m/s relative to the ground from a plank of same mass m which is placed on smooth surface. Initially plank was at rest. The minimum length of the plank for which the ball will fall on the plank itself is (g = 10 m/s2) NCERT Solutions for Class 11 Maths Chapter 9

(A) 40 3 m

(B) 20 3 m

(C) 10 3 m

(D) 60 3 m (5)

14.

A river of width 500 m has flow velocity 3m/s. A man starts swimming perpendicular to river flow at a speed of 4m/s relative to water. Simultaneously a car starts moving on a road, that runs along the opposite side of the river, with constant speed. The car starts from a point on the other side that is exactly opposite to the point from where the man starts swimming. The speed of the car if both, the car and the man, meet at the same point at the moment man has crossed the river, is: (A) 5 m/s

15.

16.

17.

(B) 4 m/s

(C) 3 m/s

(D) 7 m/s A particle is thrown from a platform (which is in the state of rest) with a velocity V at an angle 60° with the horizontal. The range is obtained to be R. If the platform is moving horizontally with a velocity V/2 and the particle is thrown in the same direction (with the same velocity and angle with respect to the platform as earlier), then new range of the projectile will be (A) 3R/2 (B) 5R/2 (C) 2R (D) 3R A projectile is thrown with velocity U = 20m/s ± 5% at an angle 60°. If the projectile falls back on the ground at the same level then which of following can not be a possible answer for range.

Consider g = 10m/s2.

(A) 34.6 m (B) 37.5 m (C) 32.0 m (D) 39.0 m A bomb of mass 3m is kept inside a smooth closed box of mass 3m and length 4L at it's centre. It explodes in two parts of mass m & 2m. The two parts move in opposite direction and stick to the apposite side of the walls of box. Download NCERT Maths Book Class 11 Solutions PDF What is the distance moved by the box during this time interval.

18.

4L L L L (A) 0 (B) (C) (D) 6 12 3 Figure shows path followed by a particle and position of a particle at any instant. Four different students have represented the velocity vectors and acceleration vectors at the given instant. Which student has made the wrong vector diagram?

P

 v

trajectory of particle

90°

particle at a given instant

19.

 v

Sita

 v

>90°

 a

 a Gita

° 90    a

 v >90°  a

Shyam

Ram

(A) Sita (B) Gita (C) Ram (D) Shyam A girl is standing on ground near a horizontally moving conveyor belt. She throws her suitcase horizontally.

Such that the initial velocity of suitcase relative to stationary girl is along dotted line shown in top view of conveyor belt. (given Vbelt > Vsuitcase) Belt Which of the following arrows represents direction of friction at initial moment.

Initial velocity of suitcase

vbelt

Initial velocity of suitcase

position of girl C

B

D A

(A) A

(B) B

(C) C

(D) D (6)

20.

21.

y Two projectiles are projected with velocity vA, vB at angles A (from horizontal) and B vA (from vertical) as shown in the figure below, such that vA > vB but having same horizontal component of velocity. Which of the following can not be correct ? B vB (A) TA > TB (B) HA > HB A (C) RB > RA (D) RA > RB x Two boats start from mid point of river in case (i) and in mid point of pond of still water in case (ii).

If all velocity shown in figure are relative to water then mark the correct option about time taken to reach the nearest bank. JEE Main previous year question paper in PDF

v A d 

v C



d River Flow

B



v

Case (ii)

(B) 

24.

D

(A) Boat A takes least time (B) Boat B takes maximum time.

(C) All the four boats A,B,C and D take same time. (D) Time of reaching bank can not be compared.

A cannon is mounted on a trolley. A cannon ball is fired at an angle  and velocity vrel relative to trolley which is free to recoil frictionlessly on a smooth surface. Which of the following trejectory (represented by dark curves) of projectile for a ground is not possible? vrel 

(A) 23.

Still water

v Case (i)

22.





(C)



(D) 

A man can swim in still water at a speed of 5 km/hr. He wants to cross a river 6 km wide, flowing at the rate of 4 km/hr. If he heads in a direction making an angle of 127° with stream direction, then he will reach a point on the other bank (approx) (A) upstream at a distance of 1.5 km (B) downstream at a distance of 1.5 km (C) directly on the other side of the bank (D) downstream at a distance of 2 km The velocity-time graph of a body moving in a straight line is shown in the figure. The ratio of displacement to the distance travelled by the body in first 6 seconds is

(A) 1 : 1

(B) 1 : 2

(C) 1 : 3

(D) 1 : 4 (7)

ASSERTION AND REASON 25.

Statement-1: Any body projected from earth at an oblique angle with vertical must follow a parabolic trajectory.

Statement-2: Any body that experiences uniform force and having initial velocity making oblique angle with the force will follow parabolic trajectory.

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

(C) Statement-1 is true, statement-2 is false.

(D) Statement-1 is false, statement-2 is true. Read Gravitation class 11 notes COMPREHENSION

26.

27.

28.

Passage for Q. no. (26 to 28) A spaceship launched vertically from Mars has reached a height of 320 m and a velocity of 80 m/s at time t = 0. At this instant its controls are switched off. If continues to move upward under the influence of Martian gravity, approxiately equal to 3.72 m/s2. At the same instant another spaceship at height 1500 m is moving downward at 25 m/s and slowing down at a rate of 0.8 m/s2.

When are the two at the same height from Mars for the first time (A) 2.5 sec.

(B) 2 sec.

(C) 19 sec.

(D) 50 sec.

How high above the planets surface will the first meeting take place?

(A) 3 km (B) 18 km (C) 1.14 km (D) 5.14 km Which of the graph best explanins the motion

(A)

(B)

(C)

(D)

Passage for Q. no. (29 to 31) Raindrops are falling with a velocity 10 2 m/s making an angle of 45° with the vertical. The drops appear to be falling vertically to a man running with constant velocity. The velocity of rain drops change such that the

29.

rain drops now appear to be falling vertically with 3 times the velocity it appeared earlier to the same person running with same velocity.

The magnitude of velocity of man with respect to ground is :

30.

(A) 10 2 m/s (B) 10 3 m/s (C) 20 m/s (D) 10 m/s After the velocity of rain drops change, the magnitude of velocity of raindrops with respect to ground is: (A) 20 m/s

31.

(B) 20 3 (C) 20 m/s (D) 10 3 m/s The angle (in degrees) between the initial and the final velocity vectors of the raindrops with respect to the ground is : (A) 8º (B) 15º (C) 22.5º (D) 37º

(8)

Passage for Q. no. (32 to 33) Consider the projectile motion shown in the figure. The projectile is thrown from the origin and vector R represents the instantaneous displacement vector. The angle made by it with the horizontal is . Now consider the graph of tanversus time t. Get Magnetic Effects of Electric Current Notes here.

32.

33.

34.

what is the angle of projection?

(A) tan–1k (B) cot–1k (C) sin–1k (D) None What is the initial speed ?

(A) gt (B) gT{1+k2}1/2k (C) gT{1+k2}1/2/k (D) None Passage for Q. no. (34 to 36) A particle is thrown with speed u from A above the ground at an angle  with the vertical as shown in the figure. A strong horizontal wind gives the particle a constant acceleration a in the +x direction. If the particle strikes the ground at a point B directly below A.

The time taken by the particle to reach to B is (A)

35.

2u sin  g

2u cos  g

(C)

2u sin  a

(D)

2u cos  a

The maximum distance of the particle from the line AB is

u 2 sin 2  (A) 2a 36.

(B)

u 2 sin 2  (B) 2g

u2

(C)

2 g2  a2

u 2 sin 2 

(D)

2 g2  a2

The distance AB is

u 2 sin 2 (A) a

u 2 sin 2 (B) g

  2u 2 a sin   cos   sin   (C) g g  

2u 2 g   sin   cos   sin   (D) a a  

(9)

Passage for Q. no. (37 to 39) 16 m. The circle lies in  horizontal plane. At time t = 0 train is at point P and moving in anticlockwise direction. At this instant a stone

A small toy train is moving with a constant speed of 10 m/s in a circle of radius

ˆ m/s relative to train. Answer following questions based is projected from the train with velocity ( 6iˆ  8k)

on above situation. Current Electricity class 12 notes y

origin

x

P at t = 0

37.

What is velocity vector of stone relative to train when the stone is at highest point of its trajectory?

38.

(A) 4iˆ  10ˆj (B)  4î  10k̂ (C) 4î  10k̂ What is angular displacement of train during this time interval?

(D)  4î  10k̂

  (B) (C)  (D) 2 2 4 What is position vector of point where stone finally lands relative to origin of coordinate system,

(A) 39.

take

16 =5 

(A)  9.2î  8 ĵ

(B) 9.2î  8ˆj

(C) 4.6î  16 ĵ

(D)  4.6î  16 ĵ

MULTIPLE CORRECT QUESTION 40.

41.

42.

The displacement-time graph of a body acted upon by some forces is shown in figure. Select the correct alternative (s) (A) work done by the forces during BC is zero (B) work done by the forces during AB is zero (C) work done by the forces during AB is positive (D) work done by the forces during OA is positive (OA is a part of a parabola) The coordinates of a particle moving in a plane are given by x(t) = a cos (pt) and y (t) = b sin (pt), where a,b(<a) and p are positive constants of appropriate dimensions. Then (A) the path of the particle is an ellipse (B) the velocity and acceleration of the particle are normal to each other at t = /(2p) (C) the acceleration of the particle is always directed towards a focus (D) the distance travelled by the particle in time interval t = 0 to t = /(2p) is a Find the correct statements out of the following (A) If acc. = 0, motion is uniform (B) If acc. = constant, acc. is uniform but motion not (C) If acc.  const., both acc. and motion are not uniform (D) If acc. = const., both acc. and motion are uniform (10)

43.

44.

45.

A spy after dropping through a distance of 196 m from a helicoptre opens the parachute and decelerates at the rate of 1 ms–2. He reaches the ground with a speed of 4.6 ms–1.

(A) Time before the opening of parachute is 2 s (B) Time after opening the parachute is 15 s (C) Velocity attained after falling through 196 m is 196 ms–1.

(D) Velocity attained after falling through 196 m can not be 196 ms–1.

Choose the correct alternative(s) (A) If the greatest height to which a man can throw a stone is h, then the greatest horizontal distance upto which he can throw the stone is 2h (B) The angle of projection for a projectile motion whose range R is n times the maximum height H is tan–1(4/n) (C) The time of flight T and the horizontal range R of a projectile are connected by the equation gT2 = 2Rtan where  is the angle of projection (D) A ball is thrown vertically up. Another ball is thrown at an angle  with the vertical. Both of them remain in air for the same period of time. Then the ratio of heights attained by the two balls is 1 : 1.

Pick the correct statements : (A) Average speed of a particle in a given time is never less than the magnitude of the average velocity.

(B) It is possible to have a situation in which

46.

47.

48.

dv d  0 but v 0 dt dt

(C) The average velocity of a particle is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval.

(D) The average velocity of a particle moving on a straight line is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval. (Infinite acceleration is not allowed) A man is standing on a road and observes that rain is falling at angle 45° with the vertical. The man starts running on the road with constant acceleration 0.5 m/s2. After a certain time from the startof the motion, it appear to him that rain is still falling at angle 45° with the vertical, with speed 2 2 m/s. Motion of the man is in the same vertical plane in wich the rain is falling. Then which of the following statement(s) are true : (A) It is not possible JEE Main Previous Year Question paper with solution PDF download (B) Speed of the rain relative to the ground is 2 m/s (C) Speed of the man when he finds rain to be falling at angle 45° with the vertical, is 4 m/s.

(D) The man has travelled a distance 16 m on the road by the time he again finds rain to be falling at angle 45°.

A train is running with uniform velocity in east direction. A car is running on a road parallel to the track with uniform speed. After some time, the road becomes perpendicular to railway track. The car driver notices that initial speed of train with respect to itself was 7 m/s and later on it became 13 m/s. What can be the true speed of the car ?

(A) 12 m/s (B) 5 m/s (C) 13 m/s (D) 7 m/s  A cubical box dimension L=5/4 metre starts moving with an acceleration a = 0.5 m/s2 î from the state of  rest. At the same time, a stone is thrown from the origin with velocity V = v î + v ˆj – v k̂ with respect to 1

2

3

 earth. Acceleration due to gravity g = 10m/s2(– ˆj ). The stone just touches the roof of box and finally falls at the diagonally opposite point then :

(A) v1 =

3 2

(B) v2 = 5

(C) v3 =

5 4

(D) v3 =

5 2 (11)

49.

A particle moves along x-axis such that its position varies with time as x = 50t– 5t2. Select the correct alternative(s). Class 11 Maths NCERT book PDF (A) The particle has travelled 130 m distance by t = 6s.

(B) At t = 4 and t = 12s, the particle is at 120 m distance from starting point.

(C) The particle will never be at distance of 130 m from starting point.

(D) During the whole motion, the particle does not have same velocity at 2 different instants.

MATCH THE COLUMN

50.

A particle is moving along a straight line. Its velocity varies with time as v = kt where k is a positive constant and t is the time. Match the graphs in Column-II with statements in Column-I. Column (I) Column (II) (A) Acceleration vs time curve

(p)

(B) Acceleration vs displacement curve

(q)

(C) Velocity vs time curve

(r)

(D) Displacement vs velocity curve

(s)

51.Column (I) (A) A particle is projected with certain velocity at an

Column (II) (p)

angle  to the horizontal. If y-axis represents kinetic energy of the ball and x-axis represents height h, then (B)A ball is projected vertically upwards with an

(q)

initial velocity. If x-axis represents height h and y-axis represents kinetic energy , then (C) A ball is thrown up with a certain velocity at an

(r)

angle  to the horizontal. The trajectory of the ball can be represented as.

(D) A ball is thrown up in an electrostatic atmosphere

(s)

where downward acceleration = (qE/m) which is here equal to (1/g). Then if y-axis denotes height traversed and x-axis denotes time, then (12)

52.

A boat is being rowed in a river. Air is also blowing. Direction of velocity vectors of boat, water and air in ground frame are as shown in diagram. Class 12 English book

Column I (A) Direction in which boat is being steered (B) Direction in which a flag on the boat may flutter

53.

Column II (Possible directions) (P) (Q)

(C) Direction of velocity of water

(R)

relative to boat (D) Direction of velocity of air relative to a piece of wood floating on river

(S)

Column – 

Column – 



dv (A) 0 dt

(p) velocity is constant



v (B)  0 t

(q) acceleration is constant



dr (C)  0 dt

(r) speed may be constant



dv  (D) . v  0 dt

(s) acceleration may be zero SUBJECTIVE QUESTION

54.

A large heavy box is sliding without friction down a smooth plane of inclination . From a point P on the bottom of the box, a particle is projected inside the box. The initial speed of the particle with respect to the box is u and the direction of projection makes an angle  with the bottom as show in the figure.

(A) Find the distance along the bottom of the box between the point of projection P and the point Q where the particle lands (Assume that the particle does not hit any other surface of the box. Neglect air resistance).

(B) If the horizontal displacement of the particle as seen by observer on the ground is zero, find the speed of the box with respect to the ground at the instant when the particle was projected.

(13)

55.

56.

A small cannon ‘A’ is mounted on a platform (which can be rotated so that the cannon can aim at any point) and is adjusted for maximum range R of shells. It can throw shells on any point on the shown circle (dotted) on ground. Sundenly a windstorm starts blowing in horizontal direction normal to AB with a speed 2 times the velocity of shell. At what least distance can be shell land from point B. (Assume that the velocity of the windstorm is imparted to the shell in addition to its velocity of projection. Also assume that the platform is kept stationary while projecting the shell.)

Hailstones falling vertically with a speed of 10 m/s, hit the wind screen (wind screen makes an angle 30° with the horizontal) of a moving car and rebound elastically. Find the velocity of the car if the driver finds the hailstones rebound vertically after striking. Maths NCERT exemplar class 10

57.

A stone is thrown vertically upward with an initial velocity v0. Find the distance travelled in time

58.

Two particles A & B are projected from the same point in different directions in such a manner that vertical components of their initial velocities are same. Which one of them will have more time of flight and which one of them will have more range?

3v 0 2g

.

y

A

B

A B

x

59.

To a man running upwards on the hill, the rain appears to fall vertically downwards with 4 m/s. The velocity vector of the man w.r.t. earth is (2i + 3j) m/s. If the man starts running down the hill with the same speed, then determine the relative speed of the rain w.r.t. man.

60.

Two cars A & B are at same point at rest. Car A accelerates at 2 m/s2. After 2 sec. of start of car A car B start moving with some constant acceleration in same direction. Car B catches car A 8 sec after the start of car A. Find the acceleration of car B.

61.

A passenger who just missed the train stands on the platform, sadly watching the last two boggies of the train.

The second last boggy takes time 3 sec. to pass by the passenger, and the last one takes time 2 sec. to pass by. How late is the passenger for the departure of the train? Assume that the train accelerates at constant rate.

A boy sitting at the rear end of a railway comparment of a train, running at a constant acceleration on horizontal rails, throws a ball towards the fore end of the compartment with velocity 20m/s (relative to train) at an angle 370 above the horizontal, when the train is running at speed of 10m/s. If same boy catches the ball without moving from seat & at the same height of projection. Then find the speed of train at the instant of his catching the ball.

(Here g = 10m/s2, sin370 = 3 / 5 ) A huge Diwali rocket is projected vertically upword so as to attain a maximum height of 160m. The rocket explodes just as it reaches the top of its trajectory sending out luminous particles in all possible directions all with same speed v. The display, consisting of the luminous particles, spreads out as an expanding, brilliant sphere. The bottom of this sphere just touches the ground when its radius is 80m. With what speed (in m/s) are the luminous particles ejected by the explosion?

A Diwali rocket explodes at the topmost point of its trajectory, 550 m from the point of projection. One of the fragments is found at a location 550 m east and 120m north of the launch point. Second fragment is found at a location 550m east and 65 m south of the launch point. First two fragments are of equal mass m and third fragment has mass 2m. If all the three fragments struck the ground simultaneously, what is the location of the third fragment? Express your answer in cm.

62.

63.

64.

(14)

65.

A bead slides along a frictionless wire lying on a horizontal plane. It makes an angle of 41° with x-axis as shown in figure. NCERT Maths class 10 exemplar In addition to any normal forces exerted by the wire, the bead is subject to an external y v 2

  x   y  force that depends on position according to formula F  F0   î  F0   ˆj  x0   y0 

bead

P v0

O

66.

wire

y0

2

45°

x0 x

If the bead has speed v0 at the origin O, find its speed as (in m/s) it leaves the end P of wire Given data : F0 = 1 N, x0 = 15 m, y0 = 15 m , mass of bead = 2 kg Initial kinetic energy of bead = 26 J A small squirrel jumps from pole 1 to pole 2 in horizontal direction. Squirrels is observed by a very small observer at origin. What is average velocity vector of squirrel ? If average velocity vector is expressed as v x î  v y ĵ  v z k̂ , express your answer as sum of magnitudes of its components | v x |  | v y |  | v z | in unit

m/s.

z

Pole 2 y

15m

Pole 1

50 2 m 50

m

20 m

observer

67.

68.

•

x

90°

50m A transparent lift A is going upwards with velocity 20 ms–1 and retarding at the rate of 8 ms–2. Second transparent lift B is located in front of it and is going down at 10 ms–1 with retardation of 2 ms–2. At the same instant a bolt from the ceiling of lift A drops inside lift A. If height of car of lift A is 16 m then find the distance travelled by bolt as observed by a person in lift B till the time it collides with floor of lift A. A high speed Jet starts from rest at s = 0 and is subjected to the acceleration shown in the figure. Determine the velocity of the after it has travelled 50 m. a 2

50 m/s

O

50m

s

(15)

SOLUTION [KINEMATICS]

KINEMATICS 1.

SOLUTION

(D)

a=v 2.

(A)

dv =  v0 + kx  .K dx

a – x is a linear equation.

7.

For first part acceleration = a1, u = 0 For second part acceleration = 0, v = a1t1  2a s 1

d2z = (6ac2x + 2bc2), dt 2

2S 2S 2S 2  3S 1  t3   S  a1t12 ; t 2  a1 ; v a1 and a2 2 a 2 

as 0 = v2 – 2a2 (3S)

Vav 

a1 3

 a = (6ac2x + 2bc2) k̂

8.

(C)

9.

(C) Let the speed of bullet be v. The component of relative velocity in the direction of the length of the car is v cos   10 and in the direction of the width of the car is v sin. Then from s = vt.

2  3S 2S 3 a1 / 3 a1

 t3 

dz dx dy = 3ax2 + 2by dt dt dt

z = ax3 + by2,

dz dx dy d2z = 3acx2 + 2bcy, 2 = 6acx + 2bc dt dt dt dt

For third part retardation = a2, u = a1t1, v = 0

t1 

(A)

S  2S  3S t1  t 2  t 3 ;

as t1 + t2 + t3  2S/ a1  2S/ a1  3 2S/ a1  5 2S/ a1

 Vav 

6S 3  2Sa ; 5 2S/ a 5

 Vav : Vmax =

3 :1  3: 5 5

2  (vcos   10)t and 3  (vsin )t Eleminating v, we get t = 0.2 s

Answer is (a).



3.





(D) a BP  a Bg  a Pg

    g  (  g )  2g 10.

2u y

2 10 sin 30 1 T   sec ay 2g cos 30 3 4.

(A)

train B should cover total distance 4L relative to A to cross it with relative velocity 4V.

(B) It can be observed from figure that P and Q shall collide if the initial component of velocity of P on inclined plane i.e. along incline. u|| = 0 that is particle is projected perpendicular to incline.

t 11.

4L  L/ v 4v A L

(C)

V

2L

B 4L, 0

B



2u  2u Time of flight T  g cos   g cos 

(–L, 0)

A (L, 0)

(2L, 0)

4L, 0

Total path length used 5L

 5.

gT cos  u = 10 m/s 2

12.

(C) Velocity of approach of P and O is





× 10 × 40 = 200 m; x2 = –

1 2

× 8 × 20

 xreletive = x1 + x2 = 280  separation = 300 – 280 = 20 m 13.

100 P reaches O after   20 sec.

5

dv =k dx

(A) Applying conservation of momentum along X 0 = m × 20 cos 60 + mvp 

NCERT Solutions for class 12 maths (B) v = v0 +kx.

1 2

= –80 m

dx  v cos 60  5 m / s dt

If can be seen that velocity of approach is always constant.

6.

(D) x1 =

vp = – 10 m/s vrel = 20 m/s ()

; Time of flight (T) =

2u sin  2  20  3 = = 23 sec g 2  10 (16)

SOLUTION [KINEMATICS] Minimum length of plank = vrel × T = 20 × 23 = 403 m 14.

15.

(C) Velocity of swimmer as seen from ground will be equal to velocity of river and for the car and swimmer to reach same point in same time car's speed should be same as river speed.

we get two solution t1 = 19.0 sec, t2 = 27.4 sec. Therefore for 1st time t1 = 19 sec.

27.

(C)

(C) As we know the two spaceships meet twice in their journey, we required y1 at t1 y1 = 320 + 80 × 19 – 1.86 × 192 = 1.14 Km.

28. 

(C) Spaceship 2 meets spaceship 1 when the latter is moving up and a second time when moving down as shown in graph (multiple). Few interpretations have been given to understand it better i.e. at a = 1.0 m/s2 and at a = 0.4 m/s2.

29. to 31. (D,A,B)

2V cos 60 . V sin 60 2V . V sin 60 R ; R'  g g

In the first case : From the figure it is clear that

2V 2 sin 60 . cos 60  ; g

2V 2 sin 60 R'  g

VRM is 10 m/s downwards and VM is 10 m/s towards right.

R '  2R 16.

(D) R =

202  sin120 = 20 3 g

=

R 2 U = R U

In the second case :

 R

Velocity of rain as observed by man becomes

2 5 = × 20 3 = 2 3 100

magnitude.

20 3 – 2 3 < R < 20 3 + 2 3  31.1m < R < 38.1m



17.

(B) Centre of mass will be at rest before and after explosion.

18.

(D) 19. (A) 20. (C)

VR '  VR 'M  VM

21. (C)

22. (C)

23. (B) 24.(B)



Assertion & Reason 25.

(D) It is an approximation valid for small speeds only.

26.

(C) The first spaceship wll trace a upward motion as well as downward motion after its velocity at the peak is zero.

The angle rain makes with vertical is

10 10 3

or   30

 Change in angle of rain = 45 - 30 = 15°.

32.

(a) at x = 0, y = 0 vector R is tangent to path i.e.

tan   ;

times in

New velocity of rain

tan  

Passage

3

33.

dy Vy  = angle of projection.  tan–1 (k) dx Vx

(d) T is the time of flight when particle again covers to ground.

2 2u sin  gT Tu  u  gT 1  K g 2sin  2.K

34. to 36. (C,A,D) Class 11 physics chapter 2 notes 37. to 39 (A, A, D) At t = 0

 vT  10jˆ m/s  ˆ m/s v S / T  ( 6iˆ  8k) We can denote the motions of the two spaceshil by 2 equations i.e.

1 y  y 0  ut  at 2 ; 2 2

y1  320  80t  1.86t , y 2  1500  25t  0.4t

2

   vS / T  vs  v T    vS  vS/ T  v T Time of flight of the stone =

When spaceships meet then y1 = y2 i.e. 320 + 80t – 1.86t2 = 1500 – 25t + 0.4 t2

ˆ m/s = ( 6iˆ  10ˆj  8k)

At highest point t =

2v y g

=

8 sec 5

4 sec 5 (17)

SOLUTION [KINEMATICS] or we can say that velocity is along negative x-axis and acceleration along y-axis.

v Angular displacement of train = t =   t r

 Hence, at t  2p , velocity and acceleration of the particle are

 10   =  16 /    4 / 5  = 2  

normal to each other. So, option (b) is also correct. At t = t,

At highest point class 12 chemistry chapter 1 notes

position of the particle r(t)  xi  yj  a cos ptiˆ  bsin ptjˆ and acceleration of the particle is

 vS  6iˆ  10ˆj  vT  10iˆ

ˆ  p2[xiˆ  yj] ˆ  p 2 r(t) a(t)  a x ˆi  a y ˆj  p2[a cos ptiˆ  bsin ptj]

   vST  vS  vT = 4iˆ  10ˆj

Therefore acceleration of the particle is always directed towards origin.

x - coordinates

Hence option (c) is also correct.

x = xi + vx T

P At t = 0, particle is at (a,0) and at t  2p , particle is at (0,b).

16 6 8 = – = 5 – 9.6 = – 4.6 m  5

Therefore, the distance covered is one fourth of the elliptical path not a.

y - coordinates y = 10 × (8 / 5) = 16 m

Hence option (d) is wrong.

 Thus r  4.6iˆ  16ˆj  0kˆ

42.

(A,B,C)

43.

(A,B,C) Let t1 be the time before opening of parachute.

Mulitple Choice Question 40.

(A,B,D) Slope of s-t curve gives the velocity of the particle.

In parts AB and BC slopes are constant

Using h  ut 

Therefore velocity is constant, Hence WA  B = KA  B = 0

19.6  0 

and WB  C = KB  C = 0 For part OA slope is increasing,

2 i.e, t1 

Hence WOA = KOA = + ve 41.

(A,B,C) x = a cos pt  cos(pt) 

y  bsin pt  sin(pt) 

we get, 

y ; b

x a

1 2 gt , we get 2

1  9.8  t12 2

19.6 4 4.9

i.e., t1 = 2 s using v2 – u2 = 2 gh, we get v2 = 2 × 9.8 × 19.6 = 19.6 ms–1.

Squaring and adding (1) and (2),

where v is the velocity attained after faling through a distance of 19.6 m.

x 2 y2  1 a 2 b2

Taking t2 as time taken after opening of parachute. using, v = u + at we have

Path of the particle is in ellipse.

4.6 = 19.6 – 1 × t2

Hence option (a) is correct.

i.e.,

4.6 = 19.6 – t2

then t2 = 19.6 – 4.6 = 15 s

From the given equations we can find,

44.

(A,B,C,D)

dx  v x  apsin pt dt

45.

(A,B,C) |Displacement|  Distance. A particle revolving in a circle has zero average velocity everytime it reaches the starting point.

d2x  a x  ap 2 cos pt ; dt 2

Average speed of a particle in a given time is never less than the magnitude of the average velocity.

dy d2 y  v y  bp cos pt and 2  a y   bp 2 sin pt dt dt

It is possible to have a situation in which

dv 0 dt

but

  At time t  2p or pt  2 ;

d v 0. dt

   ax and vy become zero  because cos  0  only vx and vy are 2   left,

The average velocity of a particle is zero in a time interval. It is possible that the instantaneoous velocity is never zero in the interval.

(18)

SOLUTION [KINEMATICS] 46.

(C, D)

(A)

Vrg  Vrm  Vmg

Vrm  Vrg  Vmg



(A)  (P)

Vrm cos 45  Vrg cos 45 Vrm  2 2 m / s  Vrg (B)





Vrm cos 45  Vmg  Vrg cos 45 1 1 2 2  4 m/s 2 2

Vmg  2 2

(B)  (Q)

using v2 = u2 + 2as for the motion of man. s = 16 m.

47.

(A,B) Say moving in opposite direction Vc = speed of driver

(C) Vt = speed of train. important questions for class 12 physics with answers



Vt + Vc = 7  Initially 

Vt2  Vc2 =132 Finally

(C)  (S)

(7–Vc)2 + Vc2 = 132

Vc2 – 7Vc – 60 = 0 Vc = 12 48.

(A,B,C) H =

T=

u 2y

53.

2u y g

= 1 sec

1 a × t2 2

(as

5 1 = 4x × 1 – × (0.5) × 12 4 2

 a





 uy = 5m/s

2g

2u sin  = g

=4×t–

(D)

Vc = – 5]

in × direction)

(D)  (R) ]

(A – p,q; B – r; C – r,s; D – s)  d    a  0 i.e.  = constant dt   |  | speed is also constant so a  p, q.  v  a avg  0  vf  vi t

ux = 3/2 m/sec

so velo change  acceleration is not zero. Now if only direction change then speed is constant But

So C  r, s  du  ·u  component of acceleration in the direction of velocity dt is zero i.e. tangential acceleration is zero, so speed must be constant so d  s

 v2 = 5/4 m/sec 49.

(A, B, D)

Match the column 50.

(A p; B p;C q;Dr)

51.

(A– q;B– r; C s; D s)

52.

(A p; B s, q; C s; D  p,s)

(Example. circular motion)

So, b  r  dr  0  u  0 it may be variable or constant but  0.  dt acceleration may or may not be zero but speed is non zero.

5 = u2 × 1 2

Along 2 axis

aavg  0

Subjective Type u 2 sin 2 u cos(  ) , (b) (down the plane) g cos  cos 

54.

(a)

55.

Let the speed of shell be u and the speed of wind be v.

The time of flight T remains unchanged due to windstorm (19)

SOLUTION [KINEMATICS] T

2u g

.............. (1)

From the fig (2),

BC  R 2  (2R) 2  5R 2  5R

Horizontal component of velocity of shell in absence of air

uH 

Hence the minimum required distance is

u 2

...............(2) BD = BC – DC =

Hence the net x and y component of velocity of shell (see figure) are

u x  2u 

uy 



56.

u cos  ............(3a) 2

5R–R=( 5

– 1)R

For the driver to observe the rain move vertically upward after the elastic collision, rain should come at an angle 30° with the horizontal (as clear from figure).

Let, velocity of rain w.r.t. car be Vm/c

VR / G  VR / C  VC / G

u sin  2

.............(3b)

VR / G( x)  VR / C(x)  VC / G(x)

The x and y coordinate of point P where shell lands is But VR / G(x )  0, since rain fall vertically down.

x  u x T  ( 2u 

u

2u cos )  2R  R cos  ...........(4a) g 2

 u  2u y  uyT   sin    R sin   2  g



 VC/ G( x)  VR / C(x)   ( Vcos 30)  VC / G(x)  V cos 30

..........(4b)

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

The distance S between B and P is given by

2

S = (x – 0)2 + (y – R)2 = (2R + R cos )2 + (R sin – R)2 = R2(6 + 4 cos – 2 sin)

 4cos  2sin    R 2 [6  20    20   20  s min imum  R 6  20  R 6  2 5 or R( 5  1)

Now, VR / G( y)  VR / C( y)

Alternate : Circle in fig. (1) represents locus of all points where shell lands on the ground in absence of windstorm.



[Since : VR/G(y) = - 10 m/s ; VC/G(y) = 0]

Let the speed of shell be ‘u’ and the speed of wind be v =



2 u. Let T

Then

V sin 30° = 10



V = 20 m/s.

Substituting V = 20 m/s in equation (i)

be the time of flight, which remains unaltered even when the windstorm blows. Since R is the maximum range angle of projection is 45° with the horizontal.

u R T 2

- 10 = - V sin 30° + 0

VC / G ( x)  V cos 30  20 x

 VC / G  10 3 ˆj ................(1)

As a result of flow of wind along x-axis, there is an additional shift (x) of the shell along x-axis in time of flight.

x  vT  2 uT  2R Hence locus of all points where shell lands on ground shifts along x-axis by 2R as shown in fig. (2).

57.

3  10 3 m/s.

2

10 3 m/s Ans.

5V02 8g

3V0 V0 V0 t = 2g = g + 2g

V0 Time to attain maximum height = g

chemistry class 11 important questions (20)

SOLUTION [KINEMATICS] 

Velocity of train when man enters = 0 + aT = aT

V02 2g

h=

Considering time taken to cross IInd last boggy

At B, u = 0,

h1 = 0 +

L = (aT) × 3 +

1  V0  g  2  2g 

2

V0 [Distance travelled in 2g ]

1 a(3)2 2

Considering time taken to cross last boggy L = a (T+3) × 2 +

V02  h1 = 8g

58.

V02 V02 + 2g 8g

=

5V02 8g

3T +

T=

Ans.

[Ans. Both will have same time of flight; B]

62.

9 = 2T + 6 + 2 2 7 sec.

2

Ans. ]

[Ans: 42 m/s] For ball

VAY = VBY

T=

....(2)

Let man catches ball at time t = T sec.

Vertical component equal 

1 a(2)2 2

L Equating  a  from equation (1) and (2)  

 Total distance travelled = h + h1 =

....(1)

2VAY 2VBY = g g

y - axis

Sy = Uyt +

1 2 ayt 2

0 = (20sin37)T + ½ (–g) T2 so time of flight is same for both RA = VAx T

VBx > VAx 59.

Ans.

RB = VBxT



RB > RA Ans. T=

20 m/s Case I : Man going up the incline

 Vm  2iˆ  3jˆ



40sin 37 12 = g 5

Displacement of ball w.r.t. train in horizontal direction is zero



 Vrm  4ˆj       Vrm  Vr  Vm  Vr  Vrm  Vm

1 2

  Sball / T x   U ball / T  t   a ball / T  t 2 

  Vr = 2iˆ  ˆj Case II: Man running down the plane with the same speed

  Vm '   Vm   2iˆ  3jˆ



0 = (20cos37) ×

  Vr '  Vr  2iˆ  ˆj



   Vrm '  Vr '  Vm '  4iˆ  2jˆ

aT =

 relative speed of rain w.r.t. man  2 2 = Vrm '  4  2 60.

20 m/s Ans.

8 sec after A start both will have same point.

1 1 × 2 × 82 = ×aB × 62 2 2

61.

= 10 +

[Ans. 32/9 m/s2]

64 = 18 aB,

aB =

40 3

64 32 = m/s2 18 9

2

For train

VT = UT + aTt =

12 1  12  + (–aT)   5 2  5 



40 12 × 3 5

VT = 42 m/s 63.

(0020)

64.

(2750)

65.

(0006)

66.

(0105)

physics important questions class 11

Let acceleration be a. Man is late by T and length of each boggy is L. (21)

SOLUTION [KINEMATICS] 67.

t = 4s

In the ref. frame of B :

(0051)

Vb/B = 30 m/s  ab/B

u = 20 m/s

= 12m/s2 T = 4s

h

V A = 20 m/s

SA/B

h1

Bolt (b) a=g

2

aA = 8 m/s

h = 16m aB = 2 m/s2 [A]

[B]

VA/B = 30 m/s 

VB = 10 m/s

T=

 2  16    = 4s  2 

SA/B = 30(4) –

ab/A = 2m/s2 

30 10 = = 2.5s 12 4



(30) 2 900 h1 = = 2(12) 24

1 (10) (4)2 = 120 – 80 = 40 2

distance = 2h1 + h – SA/B = 75 + 16 – 40 = 51 m 68.

(0050)

for Bolt (b) : 0 = 30 – 12(t1) a=

t1 =

t=0

for A :

In the ref. frame of A : Vb/A = 0 Sb/A = 16 m 

aA/B = 10m/s2 

75 h1 = 2

v dv ds

 a ds   v dv v2 1  × 50 × 50 2 2



v = 50 m/s

class 11 maths ncert solutions chapter 1

(22)