LAWS OF MOTION

1. A lift accelerated downward with acceleration 'a'. A man in the lift throws a ball upward with acceleration

a0 (a0 a). Then acceleration of ball observed by observer, which is on earth, is

(a) (a + a0 ) upward (b) (a − a 0 ) upward
(c) (a + a0 ) downward (d) (a − a 0 ) downward

2. A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball
as observed by the man in the lift and a man standing stationary on the ground are respectively
(a) g, g (b) g − a, g − a
(c) g − a, g (d) a, g

3. A man weighs 80kg. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration
of 5 m / s 2 . What would be the reading on the scale. (g = 10 m / s 2 )
(a) 400 N (b) 800 N
(c) 1200 N (d) Zero

4. A monkey of mass 20kg is holding a vertical rope. The rope will not break when a mass of 25 kg is suspended
from it but will break if the mass exceeds 25 kg. What is the maximum acceleration with which the monkey can
climb up along the rope (g = 10 m / s 2 )

(a) 10 m / s 2 (b) 25 m / s 2

(c) 2 .5 m / s 2 (d) 5 m / s 2

5. A rocket with a lift- off mass 3 .5 10 4 kg is blasted upwards with an initial acceleration of 10 m / s 2 . Then the
initial thrust of the blast is
(a) 1 .75 10 5 N (b) 3 .5 10 5 N
(c) 7 .0 10 5 N (d) 14 .0 10 5 N

6. A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N,
when the lift is stationary. If the lift moves downward with an acceleration of 5 m / s 2 , the reading of the spring
balance will be
(a) 49 N (b) 24 N
(c) 74 N (d) 15 N

2
7. With what minimum acceleration can a fireman slides down a rope while breaking strength of the rope is of
3
his weight
2
(a) g (b) g
3
1
(c) g (d) Zero
3
8. A ball of mass m moves with speed v and it strikes normally with a wall and reflected back normally, if its time
of contact with wall is t then find force exerted by ball on wall

2mv mv
(a) (b)
t t
mv
(c) mvt (d)
2t

9. A machine gun fires 20 bullets per second into a target. Each bullet weighs 150 gms and has a speed of 800 m/sec.
Find the force necessary to hold the gun in position

(a) 800 N (b) 1000 N (c)1200 N (d) 2400 N

10. The tension in the spring is

5N 5N

(a) Zero (b) 2 .5 N (c) 5 N (d) 10 N

11. A body of mass 5kg is suspended by a spring balance on an inclined plane as shown in figure. The spring balance
measure

(a) 50 N
(b) 25 N
(c) 500 N
M
(d) 10 N
30°

12. A lift is going up. The total mass of the lift and the passenger is 1500 kg. The variation in the speed of the
lift is as given in the graph. The tension in the rope pulling the lift at t = 11th sec will be
speed in m/sec

3.6
(a) 17400 N
(b) 14700 N
(c) 12000 N
(d) Zero 2 10 12

13. A rocket has a mass of 100 kg. 90% of this is fuel. It ejects fuel vapours at the rate of 1 kg/sec with a velocity of
500 m/sec relative to the rocket. It is supposed that the rocket is outside the gravitational field. The initial
upthrust on the rocket when it just starts moving upwards is
(a) Zero (b) 500 N
14. A man fires a bullet of mass 200 g at a speed of 5 m/s. The gun is of one kg mass. by what velocity the gun
rebounds backwards
(a) 0.1 m/s (b) 10 m/s
(c) 1 m/s (d) 0.01 m/s

15. A bullet of mass 5 g is shot from a gun of mass 5 kg. The muzzle velocity of the bullet is 500 m/s. The recoil
velocity of the gun is
(a) 0.5 m/s (b) 0.25 m/s
(c) 1 m/s (d) Data is insufficient

16. A force of 50 dynes is acted on a body of mass 5 g which is at rest for an interval of 3 seconds, then impulse is

(a) 0.15 10 −3 Ns (b) 0.98 10 −3 Ns

(c) 1.5 10 −3 Ns (d) 2 .5 10 −3 Ns

17. Two blocks are connected by a string as shown in the diagram. The upper block is hung by another string. A force
F applied on the upper string produces an acceleration of 2m / s 2 in the upward direction in both the blocks. If
T and T be the tensions in the two parts of the string, then

F
T
(a) T = 70 .8 N and T = 47 .2 N 2 kg
(b) T = 58 . 8 N and T = 47 .2 N
(c) T = 70 .8 N and T = 58 . 8 N T'

(d) T = 70 .8 N and T = 0 4 kg

18. Three equal weights A, B and C of mass 2 kg each are hanging on a string passing over a fixed frictionless pulley
as shown in the figure The tension in the string connecting weights B and C is

(a) Zero
(b) 13 N
A
(c) 3.3 N B
(d) 19.6 N
C
19. A body of weight 2kg is suspended as shown in the figure. The tension T1 in the horizontal string (in kg wt) is

30°
(a) 2 / 3 T1
(b) 3 /2

(c) 2 3
(d) 2 2 kg-wt

20. One end of a massless rope, which passes over a massless and frictionless pulley P is tied to a hook C while the
other end is free. Maximum tension that the rope can bear is 360 N. with what value of minimum safe
acceleration (in ms −2 ) can a monkey of 60 kg move down on the rope
P

(a) 16
(b) 6
(c) 4
C
(d) 8

21. A block of mass 4 kg is suspended through two light spring balances A and B. Then A and B will read respectively

(a) 4 kg and zero kg

(b) Zero kg and 4 kg B
(c) 4 kg and 4 kg
(d) 2 kg and 2 kg
4kg

22. In the arrangement shown in figure the ends P and Q of an unstretchable string move downwards with uniform
speed U. Pulleys A and B are fixed. Mass M moves upwards with a speed
(a) 2U cos
A B
(b) U cos
2U
(c) M
cos P Q
U
(d)
cos
23. The pulleys and strings shown in the figure are smooth and of negligible mass.
For the system to remain in equilibrium, the angle should be

(a) 0 o
(b) 30 o
(c) 45 o
m m
(d) 60

24. Two skaters have weight in the ratio 4 : 5 and ate 9 m apart, on a smooth frictionlesss
surface. They pull on a rope stretched between them. The ratio of the distance covered
by them when they meet each other will be
(a) 5 : 4 (b) 4 : 5 (c) 25 : 16 (d) 16 : 25

25. Three forces are acting on a particle of mass m initially in equilibrium. If the first two
forces (R1 and R2) are perpendicular to each other and suddenly the third force (R3) is
removed, then the acceleration of the particle is
R R + R2 R − R2 R
(a) 3 (b) 1 (c) 1 (d) 1
m m m m

26. n balls each of mass m impinge elastically each second on a surface with velocity u.
The average force experienced by the surface will be
(a) mnu (b) 2 mnu (c) 4 mnu (d) mnu/2

27. A ball of mass m moving with a velocity u rebounds from a wall. The collision is
assumed to be elastic and the force of interaction between the ball and wall varies as
shown in Fig. Then the value of F0 is

(a) mu/T (b) 2 mu/T (c) 4mu/T (d) mu/2 T

28. In order to raise a mass of 100 kg, a man of mass 60 kg fastens a rope to it and passes
the rope over a smooth pulley. He climbs the rope with acceleration 5g/4 relative to
the rope. The tension in the rope is (take g = 10ms-2)

4875 4875 4875 4875

(a) N (b) N (c) N (d) N
8 2 4 6
29. A plumb bob is hung from the ceiling of a train compartment. The train moves on an
inclined track of inclination 30° with horizontal. The acceleration of train up the plane
is a = g/2. The angle which the string supporting the bob makes with normal to the
ceiling in equilibrium is
(a) 30 (b) tan −1 (2 / 3) (c) tan −1 ( 3 / 2) (d) tan-1 (2)

30. The system shown in Fig. is released from rest. The spring gets elongated

(a) If M > m (b) If M > 2m (c) If M > m/2 (d) For any value of M
(Neglect the friction and masses of pulley, string, and spring.)

31. A balloon of mass M is descending at a constant acceleration . When a mass m is

released from the balloon, it starts rising with the same acceleration . Assuming that
its volume does not change, what is the value of m ?
2 +g +g
(a) M (b) M (c) M (d) M
+g +g 2

32. Which figure represents the correct F.B.D. of rod of mass m as shown in figure :

(1) (2) (3) (4) None of these

33. Two forces of 6N and 3N are acting on the two blocks of 2kg and 1kg
kept on frictionless floor. What is the force exerted on 2kg block by 1kg block ?

6N
2kg 1kg 3N
///////////////////////////////////////////////
(1) 1N (2) 2N (3) 4N (4) 5N

34. The 50 kg homogeneous smooth sphere rests on the 30º incline A and bears against the smooth
vertical wall B. The contact forces at A and B.

1000 500 1000 1000

(1) NA = N, NB = N (2) NA = N, NB = N
3 3 3 3
500 500 500 400
(3) NA = N, NB = N (4) NA = N, NB = N
3 3 3 3

35. Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A 15
kg weight is attached to the rope at the mid point which now no longer remains horizontal. The
minimum tension required to completely straighten the rope is :

15
(1) 15 kg kg (2)
2
(3) 5 kg (4) nfinitely large

36. A block is dragged on smooth plane with the help of a rope which moves with velocity v. The
horizontal velocity of the block is :
V

//////////////////
////////////////////////////

v v
(1) v (2) (3) v sin (4)
sin cos

37. A rod AB is slipping on a frictionless surface as shown. It acceleration of A is 10 m/s 2 downward,

the acceleration of B will be :

(1) 10 cot (2) 10 tan (3) 10 sin (4) 10 cos

38. For the arraangement shown in figure, pulleys and strings are ideal Find out the acceleration of 2
kg block

g g 2 2g 2g
(1) (2) (3) (4)
3 3 3 2

39. The velocity of end ‘A’ of rigid rod placed between two smooth vertical walls moves with velocity
‘u’ along vertical direction. The velocity of end ‘B’ of that rod, rod always remains in contact with
the vertical walls.

(1) u cot (2) u tan (3) u sin (4) u cos

40. Two bodies of 5 kg and 4 kg are tied to a string as shown in the fig. If the table and pulley both are
smooth, acceleration of 5 kg body will be equal to-

g 4g 5g
(1) g (2) (3) (4)
4 9 9

41. An object of mass 3 kg is at rest. If a force F = (6t 2 ˆi + 4t ˆj) N is applied on the object, then the
velocity of the object at t = 3 s is :

(1) 18 ˆi + 3 ˆj (2) 18 ˆi + 6 ˆj (3) 3 ˆi + 18 ˆj (4) 18 ˆi + 4 ˆj

42. Three masses of 1 kg , 6 kg and 3 kg are connected to each other with threads and are placed on
table as shown in figure. What is the acceleration with which the system is moving ? Take g = 10m
s–2 .
 (1) Zero (2) 1 m s–2 (3) 2 m s–2 (4) 3 m s–2

43. A balloon of gross weight w newton is falling vertically downward with a constant acceleration
a(<g). The magnitude of the air resistance is :

a a a
(1) w (2) w 1 + (3) w 1 − (4) w
g g g

44. A body of 2 kg has an initial speed 5ms–1. A force acts on it for some time in the direction of motion.
The force time graph is shown in figure. The final speed of the body.

F(N)

2.5

0 2 4 4.5 6.5 t(s)

(1) 9.25 ms –1
(2) 5 ms–1 (3) 14.25 ms–1 (4) 4.25 ms–1

45. Two small masses m1 and m2 are at rest on a frictionless, fixed triangular wedge whose angles are
30° and 60° as shown. They are connected by a light inextensible string. The side BC of wedge is
horizontal and both the masses are 1 metre vertically above the horizontal side BC of wedge. There
is no friction between the wedge and both the masses. If the string is cut, which mass reaches the
bottom of the wedge first? (Take g = 10m/s2)

m1 m2

B 60° 30° C
////////////////////////////////////////////////////////////
(1) Mass m1 reaches the bottom of the wedge first.
(2) Mass m2 reaches the bottom of the wedge first.
(3) Both reach the bottom of the wedge at the same time.
(4) It’s impossible to determine from the given information.
46. In the shown mass pulley system, pulleys and string are massless. The one
end of the string is pulled by the force F = 2mg. The acceleration of the block will be

(1) g/2 (2) 0 (3) g (4*) 3g

47. A particle of mass m, initially at rest, is acted upon by a variable force F for a brief interval of time T. It begins to move
with a velocity u after the force stops acting. F is shown in the graph as a function of time. The curve is a semicircle.
F02
(a) u =
2m
F0
Force

T2
(b) u =
8m
FT
(c) u = 0
4m O Time T
FT
(d) u = 0
2m

48. A particle of mass m moving with velocity u makes an elastic one dimensional collision with a stationary particle of
mass m. They are in contact for a very short time T. Their force of interaction increases from zero to F0 linearly in
time T/2, and decreases linearly to zero in further time T/2. The magnitude of F0 is
(a) mu / T
(b) 2mu / T F
(c) mu / 2T
(d) None of these F0

t
T/2 T

49. Figures I, II, III and IV depict variation of force with time
F (N) F (N)

0.3
0.25
(I) (II)
t (10–3s) t (10–3s)

0 1.0 0 1.0 2.0

F (N) F (N)

1.0 1.0
(III) (IV)

t (10–3s) t (10–3s)

0 1.0 0 1.0
The impulse is highest in the case of situations depicted. Figure
(a) I and II (b) III and I
(c) III and IV (d) IV only
50. A ball of mass 50 g is dropped from a height of 20 m. A boy on the ground hits the ball vertically
upwards with a bat with an average force of 200 N, so that it attains a vertical height of 45 m. The time
for which the ball remains in contact with the bat is [Take g = 10 𝑚/𝑠 2 ]

(1) 1/20th of a second

(2) 1/40th of a second
(3) 1/80th of a second
(4) 1/120th of a second

51. In which of the following graphs, the total change in momentum is zero?

52. A weight Mg is suspended from the middle of a rope whose ends are at the same level. The rope is no
longer horizontal. The minimum tension required to completely straighten the rope is

𝑀𝑔
(1)
2
(2) Mg cosθ
(3) 2Mg cosθ
(4) Infinitely large

53. Tension in the rope at the rigid support is (g = 10 𝑚/𝑠 2 )

(1) 760 N
(2) 1360 N
(3) 1580 N
(4) 1620 N
54. In the figure given below, with what acceleration does the block of mass m
will move? (Pulley and strings are massless and frictionless)

𝑔
(1)
3
2𝑔
(2)
5
2𝑔
(3)
3
𝑔
(4)
2

55. 𝑇1 and 𝑇2 in the given figure are

(1) 28 N, 48 N
(2) 48 N, 28 N
(3) 96 N, 56 N
(4) 56 N, 96 N

56. A uniform rope of mass M and length L is fixed at its upper end vertically from a rigid support. Then
the tension in the rope at the distance l from the rigid support is
𝐿
(1) Mg
𝐿+1
𝑀𝑔
(2) (𝐿 − 𝐼)
𝐿
(3) Mg
𝐼
(4) Mg
𝐿
57. In the arrangement as shown, tension 𝑇2 is (g = 10 𝑚/𝑠 2 )

(1) 50 N
(2) 100 N
(3) 50√3 𝑁
(4) 100√3N

58. If pulleys shown in the diagram are smooth and massless and a1 and a2 are acceleration of blocks of
mass 4 kg and 8 kg respectively, then

(1) 𝑎1 = 𝑎2
(2) 𝑎1 = 2𝑎2
(3) 2𝑎1 = 𝑎2
(4) 𝑎1 = 4𝑎2

59. Figure shows a uniform rod of length 30 cm having a mass 3.0 kg. The rod is pulled by constant forces
of 20 N and 32 N as shown. Find the force exerted by 20 cm part of the rod on the 10 cm part (all
surfaces are smooth) is

(1) 36 N
(2) 12 N
(3) 64 N
(4) 24 N
60. A particle of mass m strikes a wall with speed v at an angle 30° with the wall
elastically as shown in the figure. The magnitude of impulse imparted to the ball by the wall is

(1) mv
𝑚𝑣
(2)
2
(3) 2mv
(4) √3 mv

61. In accordance with Newton’s third law of motion

(1) Action and reaction never balance each other
(2) For appearance of action and reaction, physical contact is not necessary
(3) This law is applicable whether the bodies are at rest or they are in motion
(4) All of these

62. A 6 kg bomb at rest explodes into three equal pieces P, Q and R. If P flies with speed 30 m/s and Q with
speed 40 m/s making an angle 90° with the direction of P. The angle between the direction of motion
of P and R is about
(1) 143°
(2) 127°
(3) 120°
(4) 150°

63. A block of mass m is released on a smooth inclined plane of inclination θ with the horizontal. The force
exerted by the plane on the block has a magnitude
(1) mg
mg
(2)
cosθ
(3) mg tanθ
(4) mg cosθ

64. A vehicle is moving on a track with constant speed as shown in figure. The apparent weight of the
vehicle is

(1) Maximum at A
(2) Maximum at B
(3) Maximum at C
(4) Same at A, B and C
65. The figure shows the position - time (x – t) graph of one-dimensional motion of a body of mass 0.4
kg. The magnitude of each impulse is

(1) 0.4 Ns (2) 0.8 Ns (3) 1.6 Ns (4) 0.2 Ns

66. Two fixed frictionless inclined planes making an angle 30º and 60º with the vertical are shown in
the figure. Two blocks A and B are placed on the two planes. What is the relative vertical
acceleration of A with respect to B?

(1) 4.9 ms–2 in horizontal direction (2) 9.8 ms–2 in vertical direction
(3) Zero (4) 4.9 ms–2 in vertical direction

67. A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F(t) = F 0e–bt in the
x direction. Its speed v(t) is depicted by which of the following curves ?

(1) (2)

(3) (4)

68. A mass of 10 kg is suspended vertically by a rope form the roof. When a horizontal force is applied
on the rope at some point, the rope deviated at an angle of 45º at the roof point. If the suspended
mass is at equilibrium, the magnitude of the force applied is (g = 10 ms –2)
(1) 200 N (2) 100 N (3) 140 N (4) 70 N
69. A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F
= kt acts in the same direction on the moving particle during time interval T so that its momentum
changes from p to 3p. Here k is a constant. The value of T is:

2k k p 2p
(1) (2) 2 (3) 2 (4)
p p k k

70. A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied
horizontally at the mid-point of the rope such that the top half of the rope makes an angle of 45°
with the vertical. Then F equals : (Take g = 10 ms–2 and the rope to be massless)

(1) 75 N (2) 90 N (3) 100 N (4) 70 N

71. A particle moving with velocity V is acted by the three forces shown by the vector triangle PQR.
The velocity of the particle will :

R Q

(1) change according to the smallest force QR

(2) increase
(3) decrease
(4) remain constant

72. Same spring is attached with 2 kg, 3 kg and 1 kg blocks in three different cases as shown in figure.
If x1, x2 and x3 be the extensions in the spring in these cases then (Assume all the blocks to move
with uniform acceleration

(A) x1 = 0, x3 > x2 (B) x2 > x1 > x3 (C) x3 > x1 > x2 (D) x1 > x2 > x3
73. System shown in figure is in equilibrium and at rest. The spring and string are massless Now the
string is cut. The acceleration of mass 2m and m just after the string is cut will be :

(1) g/2 upwards , g downwards (2) g upwards, g/ 2 downwards

(3) g upwards , 2g downwards (4) 2g upwards , g downwards

74. The system starts from rest and A attains a velocity of 5 m/s after it has moved 5 m towards right.
Assuming the arrangement to be frictionless every where and pulley & strings to be light, if the
F
constant force F applied on A then find the value of . [M.Bank_NLM_3.54]
15

(1) 1 (2) 3 (3) 5 (4) 7

75. A massless spring balance is attached to 2 kg trolley and is used to pull the trolley along a flat
surface as shown in the fig. The reading on the spring balance remains at 10 kg during the motion.
The acceleration of the trolley is (Use g = 9.8 ms–2)

(1) 4.9 ms–2 (2) 9.8 ms–2 (3) 49 ms–2 (4) 98 ms–2
 Answer
1. Sol. (d) The effective acceleration of ball observed by observer on earth = (a – a0)

As a0 a, hence net acceleration is in downward direction.

2. Sol. (c) Due to relative motion, acceleration of ball observed by observer in lift = (g – a) and for man
on earth the acceleration remains g.

3. Sol. (c) For accelerated upward motion

R = m (g + a) = 80 (10 + 5) = 1200 N
4. Sol. (c) Tension the string = m (g + a) = Breaking force
20(g + a) = 25 g a = g / 4 = 2 .5 m / s 2

5. Sol. (c) Initial thrust must be

m [g + a] = 3 .5 10 4 (10 + 10 ) = 7 10 5 N

6. Sol. (b) When the lift is stationary W = mg

49 = m 9 . 8 m = 5 kg.

7. Sol. (c) If man slides down with some acceleration then its apparent weight decreases. For critical
condition rope can bear only 2/3 of his weight. If a is the minimum acceleration then,
Tension in the rope = m(g − a) = Breaking strength
2 2g g
m (g − a) = mg a= g− =
3 3 3
8. Sol. (a) For exerted by ball on wall
= rate of change in momentum of ball
mv − (−mv ) 2mv
= =
t t

m Bv B 0 . 2 5
9. Sol. (c) vG = = =1m /s
mG 1
By the conservation of linear momentum m Bv B = m ava
m B v B 5 10 −3 500
vG = = = 0 .5 m / s
mG 5

10. Sol. (a) Zero

11. Sol. (b) Since downward force along the inclined plane
= mg sin = 5 10 sin 30 = 25 N

12. Sol. (c) At 11th second lift is moving upward with acceleration
0 − 3 .6
a= = −1 .8 m /s 2
2
Tension in rope, T = m(g − a)
= 1500 (9 .8 − 1 .8 ) = 12000 N
 dm
13. Sol. (b) F =u = 500 1 = 500 N
dt

14. Sol. (d) F = mnv = 150 10 −3 20 800 = 2400 N .

15. Sol. (c) 5N force will not produce any tension in spring without support of other
5N force. So here the tension in the spring will be 5N only.

16. Sol. (d) Since action and reaction acts in opposite direction on same line, hence angle between
them is 180°.

17. Sol. (a) FBD of mass 2 kg FBD of mass 4kg

T T 8N
4N

2 kg 4 kg

T 20 N
40 N
T − T − 20 = 4 ….(i) T − 40 = 8 …(ii)
By solving (i) and (ii) T = 47 .23 N and T = 70 .8 N

18. Sol. (b) Tension between m 2 and m 3 is given by

2 m 1m 3
T= g
m1 + m 2 + m 3

2 2 2
= 9 .8 = 13 N m1
2+2+2 m2
T

m3

19. Sol. (c) T sin 30 = 2kg wt T sin 30

30o
T = 4 kg wt T
30o
T1 = T cos 30 °
T cos 30 T1
= 4 cos 30 °

=2 3 2 kg-wt

20. Sol. (c) If monkey move downward with acceleration a then its apparent
weight decreases. In that condition
Tension in string = m(g − a)
This should not be exceed over breaking strength of the rope i.e. 360 m(g − a) 360 60(10 − a)
2
a 4 m/s

21. Sol. (c) As the spring balances are massless therefore the reading of both balance should be equal.
22. Sol. (d) b b
A y B
l
O

M
P Q

As P and Q fall down, the length l decreases at the rate of U m/s.

From the figure, l 2 = b 2 + y 2
Differentiating with respect to time
dl db dy db dl
2l = 2b + 2y As = 0, = U
dt dt dt dt dt

dy l dl dy 1 U
= = U=
dt y dt dt cos cos

23. Sol. (c) From the figure for the equilibrium of the system
1
2T cos = 2 mg cos = = 45
2
T T

T T

m m

24 Sol. (a) Acceleration of the skaters will be in the ratio

F F
: or 5:4
4 5
1
Now according to the problem, s = 0 + at 2 , we get
2
s1 a1 5
= =
s2 a 2 4
R12 + R 22 R 3
25. Sol. (a) a = = R 3 = R12 + R 22
m m

26. Sol. (b) Change in momentum of one ball = 2 mu, time taken = 1 s
Total change in momentum n(2mu)
Fav = = = 2 mnu
Time taken 1

27. Sol. (c) Area under the force-time graph is impulse, and impulse is change in momentum.
Area of graph = change in momentum
1 4mu
TF0 = 2 mu F0 =
2 T
28. Sol. (c) Let T be the tension in the rope and a the acceleration of rope.
5g
The absolute acceleration of man is, therefore, −a .
4
Equations of motion for mass and man gives:
T - 100g = 100a (i)
5g
T − 60g = 60 −a (ii)
4
4875
Solving Eqs (i) and (ii), we get T = N
4

29. Sol. (b) T sin - mg sin 30° = ma

T sin = mg sin 30° + mg/2 (i)

T cos = mg cos 30° (ii)
Dividing Eqs. (i) by (ii), we get
2
tan =
3
30. Sol. (d) Let spring does not get elongated, then net pulling force on the system is Mg + mg
- mg or simply Mg. Total mass being pulled is M + 2 m. Hence, acceleration of the
system is
Mg
a=
M + 2m
Now since a < g, there should be an upward force on M so that its acceleration
becomes less than g. It means there is some tension developed in the string. Hence,
for any value of M spring will be elongated.

31. Sol. (b) Suppose F = upthrust due to buoyancy

Then while descending, we find
Mg - F = Ma (i)
When ascending, we have
F - (M - m)g = (M - m)a (ii)
2
Solving Eqs. (i) and (ii), we get m = M
+g

32. Sol. (3) Force exerted by string is always along the string and of pull type. When there is a contact
between a point and a surface the normal reaction is perpendicular to the surface and of push type.
33. Sol.(3)
Both blocks are constrained to move with same acceleration.

6 – N = 2a [Newtons II law for 2 kg block]

N – 3 = 1a [Newtons II law for 1 kg block]

34. Sol. (1)

mg – NA cos 30 = 0 [Equilibrium in vertical direction]

mg
NA =
cos30
1000
NA = N
3

NB – NA sin 30 = 0 [Equilibrium in horizontal]

NB = NA sin 30
1000 1 500
NB = NB = N
3 2 3

35. Sol. (4)

T cos + T cos – 150 = 0 [Equilibrium of point A]
75
2 T cos = 150 T=
cos
When string become straight becomes 90º
36. Sol. (2)
The length of string AB is constant.
speed A and B along the string are same u sin = V
V
u sin = V u=
sin

37. Sol. (2)

From constrained relation 10sin = acos a = 10 tan

20 − 10 10 g
38. Sol (1) (a)system = = =
3 3 3

39. Sol.(2) Since rod is rigid, its length can’t increase.

velocity of approach of A and B point of rod is zero.
u sin – v cos = 0
v = u tan
40. Sol.(4)
net driving force 5g 5g
a= = =
total mass in motion 5+4 9

41. Sol.(2) Key Idea : Force applied on the object is rate of change of momentum.
According to Newton's 2nd law, force applied on an object is equal to rate of change of momentum.

dp
That is F=
dt
dv
or F=m ......(i)
dt
Given , m = 3 kg, t = 3s, F = (6t 2 ˆi + 4t ˆj) N
Substituting these values in Eq. (i), we get

dv
F = (6t 2 ˆi + 4t ˆj) = 3
dt
1 2 ˆ
or dv = (6t i + 4t ˆj) dt
3
Now, taking integration of both sides, we get
t 1
dv = (6t 2 ˆi + 4t ˆj) dt
0 3

1 3
v= (6t 2 ˆi + 4t ˆj) dt
3 0
but t = 3s
1 3
v= (6t 2 ˆi + 4t ˆj) dt
3 0
3
1 6t 2 ˆi + 4t
2
ˆj
or v=
3 3 2
0
1
or v = [2(3)2 ˆi + 2(3)2 ˆj]
3
1
or v = [54iˆ + 15ˆj]
3
or v = 18iˆ + 6ˆj
42. Sol.(3)
30 – T2 = 3 a [Newton’s II law for 3 kg block]
T2 – T1 = 6 a [Newton’s II law for 6 kg block]
T1 – 10 = 1 a [Newton’s II law for 1 kg block]
By adding three equations
30 – 10 = 10 a a = 2 m/s2.

43. Sol.(3)
w – f = ma w – ma = f
m m a
w 1– a = f w 1– a =f w 1– =f
w mg g

44. Sol. (3) Impulse = Change in momentum = m(v2 – v1) …(i)

Again impulse = Area between the graph and time axis
1 1
= 2 4 + 2 4 + (4 + 2.5) 0.5 + 2 2.5
2 2
= 4 + 8 + 1.625 + 5 ...(ii)
From (i) and (ii), m(v 2 − v1 ) = 18.625
18.625 18.625
v2 = + v1 = + 5 = 14.25 m / s
m 2

45. Sol.(1) The vertical component of acceleration of mass 1 and mass 2 are
a1 = g sin260 , a2 = g sin230
since vertical displacement for both masses is 1m, the block with larger acceleration will reach the
base of wedge first. Hence block of mass m1 shall reach base of wedge first.

46. Sol.

47. Sol. (c)Initially particle was at rest. By the application of force its momentum increases.
Final momentum of the particle = Area of F - t graph
mu = Area of semi circle
r2 r1r2 (F0 ) (T / 2) F0 T
mu = = = u=
2 2 2 4m
48. Sol. (b)In elastic one dimensional collision particle rebounds with same speed in opposite direction
i.e. change in momentum = 2mu
But Impulse = F T = Change in momentum
2mu
F0 T = 2mu F0 =
T

49. Sol. (c)Impulse = Area between force and time graph and it is maximum for graph (III) and (IV)

50. Sol. Answer (3)

51. Sol. Answer (3)

52. Sol. Answer (4)

53. Sol. Answer (3)

54. Sol. Answer (3)

55. Sol. Answer (3)

56. Sol. Answer (2)

57. Sol. Answer (2)

58. Sol. Answer (2)

59. Sol. Answer (4)

60. Sol. Answer (1)

61. Sol. Answer (4)

(1) Action and reaction act on the different bodies.
(2) Example : Gravitational force, coulomb force
(3) 3rd law is irrespective of the state of motion
62. Sol. Answer (2)

63. Sol. Answer (4)

64. Sol. Answer (2)

dx 2
65. Sol. (2) V1 = = =1
dt 1 2
dx
V2 = = −1
dt 2
Impulse = | P| = |m(V2 – V1)| = |0.4 (–1 –1)| = 0.8 Ns

66. Sol.(4) Vertical component of acceleration of A

a1 = (g sin ). sin
3
= g sin 60º . sin 60º = g .
4
That for B
1
a2 = g sin 30º . sin 30º = g
4
3g g g
(aAB)⊥ = – = = 4.9 m/s2
4 4 2
67. Ans. (3)
Sol. F = ma = F0 e–bt
dv F0 −bt
= e
dt m
v t
F0
dv = e −bt dt
m
0 0

t
F0 e−bt
v=
m −b
0

v=
F0
mb
(
1 − e−bt )
68. Ans. (2)
Sol.
T
45°

10kg F

100N
T
= 100
2
T
=F
2
F = 100 N

69. Ans. (3)

dp dp
Sol. F= Kt =
dt dt
3P t
Kt 2 P
dP = Ktdt 3P – P = t= 2
2 K
P 0

70. Ans. (3)

Sol.

T 45°

F
100N
100N
10kg
 T
= 100
2
T
=F
2
F = 100N

71. Ans. (4)

Sol. Net force on the particle is zero
a =0
v = remains constant

72. Sol.(b) If m1 and m2 are masses of blocks then tension T in the string as well as spring are
2 m1 m2
T= g
m1 + m2
T1 = 2g T2 = 2.4 g T3 = 1.33 g
T2 > T1 > T3 or x2 > x1 > x3

73. Sol. (1)

After string is cut, FBD of m

mg
a= = g
m
FBD of 2m (when string is cut tension in the spring takes finite time to become zero. How ever
tension in the string immediately become zero.)

v 2 25
74. Sol. (3) a== = 2.5 m/s2
2s 10
For 6 kg : – F – 2T = 6a

For 2 kg : – T – 2g = 2 (2 a)
From (1) & (2) F = 75 N
F
= 5.
15

3mg

2m

2mg
3mg − 2mg g
a= =
2m 2
75. Sol. (3)
Reading of spring balance is same as tension in the balance.

T = 10 g = 98 N
T=2a [Newton’s II law for 2 kg block] [2 kg]