Laws of Motion

1. !
An ice cart of mass 60 kg rests on a horizontal snow patch with coefficient of static friction ". Assuming
that there is no vertical acceleration, find the magnitude of the maximum horizontal force required to
move the ice cart (𝑔 = 9.8 ms#$ )
a) 100 N b) 110 N c) 209 N d) 196 N
2. An 80 kg person is parachuting and is experiencing a downward acceleration of 2.8ms#$ . The mass of the
parachute is 5 kg. The upward force on the open parachute is (Take 𝑔 = 9.8ms#$ )
a) 595 N b) 675 N c) 456 N d) 925 N
3. #!
The rate of the mass of the gas emitted from rear of a rocket is initially 0.1kgs . If the speed of the gas
relative to the rocket is 50 ms#! and mass of the rocket is 2 kg, then the acceleration of the rocket (in
ms#$ ) is
a) 5 b) 5.2 c) 2.5 d) 25
4. A 100 kg block is suspended with the help of three string 𝐴, 𝐵 and 𝐶. The tension in the string 𝐶 is

a) 50 gN b) 100 gN c) 20 gN d) 20 gN
5. If the resultant of all the external forces acting on a system of particles is zero, then form an inertial
frame, one can surely say that
a) Linear momentum of the system does not change in time
b) Kinetic energy of the system does not change in time
c) Angular momentum of the system does not change in time
d) Potential energy of the system does not change in time
6. Two bodies of mass 4 kg and 6 kg are attached to the ends of a string passing over a pulley. The 4 kg mass
is attached to the table by another string. The tension in this string 𝑇! is

a) 19.6 N b) 25 N c) 10.6 N d) 10 N
7. Consider the following statement. When jumping from some height, you should bend your knees as you
come to rest instead of keeping your legs stiff. Which of the following relations can be useful in explaining
the statement?
a) ∆𝐩! = −∆𝐩$ b) ∆𝐸 = −∆(PE + KE) = 0
∆𝐱 ∝ ∆𝐅
c) 𝐅 ∆𝑡 = 𝑚∆𝐯 d)
Where symbols have their usual meaning
8. At a certain instant of time the mass of rocket going up vertically is 100 kg. If it is ejecting 5 kg of gas per
second at a speed of 400 m/s, the acceleration of the rocket would be (Taking 𝑔 = 10 m/s$ )
a) 20 m/s$ b) 10 m/s$ c) 2 m/s$ d) 1 m/s$
9. A bullet of mass 5 g is shot from a gun of mass 5 kg. The muzzle velocity of the bullet is 5000 ms#! . The
recoil velocity of the gun is
a) 0.5 ms#! b) 0.25 ms#! c) 1 ms#! d) Data is insufficient
10. The upper half of an inclined plane of inclination 𝜃 is perfectly smooth while the lower half is rough. A
body starting from the rest at top comes back to rest at the bottom if the coefficient of friction for the
lower half is given by
a) 𝜇 = sin 𝜃 b) 𝜇 = cot 𝜃 c) 𝜇 = 2 cos 𝜃 d) 𝜇 = 2 tan 𝜃

11. A gramophone record is revolving with an angular velocity 𝜔. A coin is placed at a distance 𝑟 from the
centre of the record. The static coefficient of friction is 𝜇. The coin will revolve with the record if
𝜇𝑔 𝜔$ 𝜇𝑔
a) 𝑟 ≥ $ b) 𝑟 = 𝜇𝑔𝜔$ c) 𝑟 < d) 𝑟 ≤ $
𝜔 𝜇𝑔 𝜔
12. A large force is acting on a body for a short time. The impulse imparted is equal to the change in
a) Acceleration b) Momentum c) Energy d) Velocity
13. A ball of mass 1 kg hangs in equilibrium from two strings 𝑂𝐴 and 𝑂𝐵 as shown in figure. What are the
tensions in strings 𝑂𝐴 and 𝑂𝐵? (Take g = 10 ms#$ )

a) 5 N, zero b) Zero, N c) 5 N, 5√3 N d) 5√3 N, 5 N

14. A block of mass 𝑚 is placed on a smooth wedge of inclination 𝜃. The whole system is accelerated
horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block ( 𝑔
is acceleration due to gravity ) will be
a) 𝑚𝑔 cos 𝜃 b) 𝑚𝑔 sin 𝜃 c) 𝑚𝑔 d) 𝑚𝑔/ cos 𝜃
15. A body is imparted motion from rest to move in a straight line. If it is then obstructed by an opposite
force, then
a) The body may necessarily change direction
b) The body is sure to slow down
c) The body will necessarily continue to move in the same direction at the same speed
d) None of these
16. A body of mass 6 kg moves in a straight line according to the equation
𝑥 = 𝑡 " − 75𝑡,
Where 𝑥 denotes the distance in metre and 𝑡 the time in second. The force on the body at 𝑡 = 4 s is
a) 64 N b) 72 N c) 144 N d) 36 N
17. Diwali rockets are ejecting 50 g of gases per second at a velocity of 400 ms#! . The accelerating force on
the rocket will be
a) 22 dyne b) 20 N c) 20 dyne d) 100 N
18.
In the figure shown, 𝑚! = 10 kg, 𝑚$ = 6 kg, 𝑚" = 4 kg. If 𝑇" = 40 N, 𝑇$ =?
a) 13 N b) 32 N c) 25 N d) 35 N
19. #!
A 5000 kg rocket is set for vertical firing. The exhaust speed is 800 ms . To give an initial upward
acceleration of 20 ms#$ , the amount of gas ejected per second to supply the needed thrust will be (𝑔 =
10 ms#$ )
a) 127.5 kgs#! b) 187.5 kgs#! c) 185.5 kgs#! d) 137.5 kgs#!
20. The variation of momentum with time of one of the body in a two body collision is shown in fig. The
instantaneous force is maximum corresponding to point
p
S
R

Q
P
t

a) 𝑃 b) 𝑄 c) 𝑅 d) 𝑆
21. A machine gun mounted on a 2000 kg car on a horizontal frictionless surface fires 10 bullets per second.
If 10 g be the mass of each bullet and 500 ms#! , the velocity of each bullet, then the acceleration of the
car will be
1 1 1 1
a) ms#$ b) ms#$ c) ms#$ d) ms#$
10 20 40 60
22. A rocket of mass 100 kg burns 0.1 kg of fuel per second. If velocity of exhaust gas is 1 kms #! , then it lifts
with an acceleration of
a) 1000 ms#$ b) 100 ms#$ c) 10 ms#$ d) 1 ms#$
23. A lift of mass 1000 𝑘𝑔 is moving with an acceleration of 1 𝑚/𝑠 $ in upward direction. Tension developed
in the string, which is connected to the lift is (𝑔 = 9.8 𝑚/𝑠 $ )
a) 9,800 𝑁 b) 10,000 𝑁 c) 10, 800 𝑁 d) 11, 000 𝑁
24. A block is kept on a frictionless inclined surface with angle of inclination α. The incline is given an
acceleration 𝑎 to keep the block stationary. Then 𝑎 is equal to

a) 𝑔/ tan α b) 𝑔 cosec α c) 𝑔 d) 𝑔 tan α

25. The tension in the spring is
5N 5N

a) Zero b) 2.5 𝑁 c) 5 𝑁 d) 10 𝑁
26. 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
B

C
a) Zero b) 13 𝑁 c) 3.3 𝑁 d) 19.6 𝑁
27. A particle of mass 2 𝑘𝑔 is initially at rest. A force acts on it whose magnitude changes with time. The force
time graph is shown below

The velocity of the particle after 10 𝑠 is

a) 20 𝑚𝑠 #! b) 10 𝑚𝑠 #! c) 75 𝑚𝑠 #! d) 50 𝑚𝑠 #!
28. A uniform metal chain is placed on a rough table such that one end of chain hangs down over the edge of
the table. When one-third of its length hangs over the edge, the chain starts sliding. Then, the coefficient
of static friction is
3 1 2 1
a) b) c) d)
4 4 3 2
29. A smooth block is released at rest on a 45° incline and then slides a distance 𝑑. The time taken to slide is
𝑛 times as much to slide on rough incline than on a smooth incline. The coefficient of friction is
1 1 1 1
a) µ% = 1 − b) µ% = ƒ1 − c) µ& = 1 − d) µ& = ƒ1 −
𝑛$ 𝑛$ 𝑛$ 𝑛$
30. A 5000 𝑘𝑔 rocket is set for vertical firing. The exhaust speed is 800 𝑚𝑠 #! . To give an initial upward
acceleration of 20 𝑚𝑠 #$ the amount of gas ejected per second to supply the needed thrust will be (𝑔 =
10 𝑚𝑠 #$ )
a) 127.5 𝑘𝑔 𝑠 #! b) 187.5 𝑘𝑔 𝑠 #! c) 185.5 𝑘𝑔 𝑠 #! d) 137.5 𝑘𝑔 𝑠 #!
31. A plumb line is suspended from a ceiling of a car moving with horizontal acceleration of 𝑎. What will be
the angle of inclination with vertical
a) tan#! (𝑎/𝑔) b) tan#! (𝑔/𝑎) c) cos #! (𝑎/𝑔) d) cos #! (𝑔/𝑎)
32. A block of mass 𝑚 on a rough horizontal surface is acted upon by two forces as shown in figure. For
equilibrium of block, the coefficient of friction between block and surface is

𝐹! + 𝐹$ sin θ 𝐹! sin θ + 𝐹$ 𝐹! + 𝐹$ cos θ 𝐹! sin θ − 𝐹$

a) b) c) d)
𝑚g + 𝐹$ cos θ 𝑚g + 𝐹$ sin θ 𝑚g + 𝐹$ sin θ 𝑚g − 𝐹$ cos θ

33. Two masses of 4 𝑘𝑔 and 5 𝑘𝑔 are connected by a string passing through a frictionless pulley and are kept
on a frictionless table as soon as shown in the figure. The acceleration of 5 𝑘𝑔 mass is

4kg

5kg

a) 49 𝑚/𝑠 $ b) 5.44 𝑚/𝑠 $ c) 19.5 𝑚/𝑠 $ d) 2.72 𝑚/𝑠 $

34. Rocket engines lift a rocket from the earth surface because hot gas with high velocity
a) Push against the earth b) Push against the air
c) React against the rocket and push it up d) Heat up the air which lifts the rocket
35. A mass of 100 g strikes the wall with speed 5 ms#! at an angle as shown in figure and it rebounds with
the same speed. If the contact time is 2 × 10#" s, what is the force applied?

a) 250√3 N to right b) 250 N to right c) 250√3 N to left d) 250 N to left

36. A block of mass 4 kg is kept on a rough horizontal surface. The coefficient of static friction is 0.8. If a force
of 19 N is applied on the block parallel to the floor, then the force of friction between the block and floor
is
a) 32 N b) 18 N c) 19 N d) 9.8 N
37. A ball falls from 20 m height on floor and rebounds to 5 m. Time of contact is 0.02 s. Find acceleration
during impact.
a) 1200 ms#$ b) 1000 ms#$ c) 2000 ms#$ d) 1500 ms#$
38. In the arrangement shown in figure, if a force 2 𝑚g is applied at the free end of the rope, the mass 𝑚 will
ascend with an acceleration of

g g
a) b) c) g d) 2 g
3 2

39. A player caught a cricket ball of mass 150 g moving at the rate of 20 ms#! . If the catching process be
completed in 0.1 s, the force of blow exerted by the ball on the hands of the player is
a) 0.3 N b) 30 N c) 300 N d) 3000 N

40. A force-time graph for a linear motion of a body is shown in the figure. The change in linear momentum
between 0 and 7 s is

a) 2 Ns b) 3 Ns c) 4 Ns d) 5 NS
41. Two block of masses 7 𝑘𝑔 and 5 𝑘𝑔 are placed in contact with each other on a smooth surface. If a force of
6 𝑁 is applied on the heavier mass, the force on the lighter mass is

a) 3.5 𝑁 b) 2.5 𝑁 c) 7 𝑁 d) 5 𝑁
42. A particle moves in a circular path with decreasing speed. Choose the correct statement
a) Angular momentum remains constant
b) Acceleration 𝐚 Œ⃗ is towards the centre
c) Particle moves in a spiral path with decreasing radius
d) The direction of angular momentum remains constant
43. A box is placed on an inclined plane and has to be pushed down. The angle of inclination is
a) Equal to angle of friction b) More than angle of friction
c) Equal to angle of repose d) Less than angle of repose
44. A body of mass 𝑀 at rest explodes into three pieces, two of which of mass 𝑀/4 each are thrown off in
perpendicular directions with velocities of 3 𝑚/𝑠 and 4 𝑚/𝑠 respectively. The third piece will be thrown
off with a velocity of
a) 1.5 𝑚/𝑠 b) 2.0 𝑚/𝑠 c) 2.5 𝑚/𝑠 d) 3.0 𝑚/𝑠
45. A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2 ∶ 1. The ratio of
their nuclear sizes will be
a) 2!/" : 1 b) 1: 3!/$ c) 3!/$ : 1 d) 1: 2!/"
46. A body moves along a circular path of radius 10𝑚 and the coefficient of friction is 0.5. What should be its
angular speed in rad/s if it is not to slip from the surface (𝑔 = 9.8 𝑚/𝑠 $ )
a) 5 b) 10 c) 0.1 d) 0.7
47. A block is lying static on the floor. The maximum value of static frictional force on the block is 10𝑁. If a
horizontal force of 8 𝑁 is applied to the block, what will be the frictional force on the block
a) 2 𝑁 b) 18 𝑁 c) 8 𝑁 d) 10 𝑁
48. A shell at rest at the origin explodes into three fragments of masses 1 𝑘𝑔, 2𝑘𝑔 and 𝑚 𝑘𝑔. The 1 𝑘𝑔 and
2 𝑘𝑔 pieces fly off with speeds off 5𝑚𝑠 #! along 𝑥-axis and 6𝑚𝑠 #! along 𝑦-axis respectively. If the 𝑚 𝑘𝑔
piece files off with a speed of 6.5 𝑚𝑠 #! , the total mass of the shell must be
a) 4 𝑘𝑔 b) 5 𝑘𝑔 c) 3.5 𝑘𝑔 d) 4.5 𝑘𝑔
 : HINTS AND SOLUTIONS :
1 (d) 7 (c)
Force applied against friction Change of momentum F ∆𝑡 = 𝑚∆𝐯
1 𝑚∆𝐯
𝑓! = µ! 𝑅 = µ! 𝑚𝑔 = × 60 × 9.8 = 196 N ⟹ 𝐅=
3 𝑡
2 (a) By doing so time of change in momentum
Mass of the person 𝑀 = 80 kg increases and impulsive force on knees decreases.
Mass of the parachute 𝑚 = 5 kg 8 (b)
∴ Total mass of the system = 𝑀 + 𝑚 = 85kg The acceleration of a rocket is given by
Downward acceleration 𝑎 = 2.8 ms"# 𝑣 ∆𝑚
𝑎 = h i−𝑔
Let upward force = 𝐹 𝑚 ∆𝑡
Applying Newton’s II law of motion to this system 400 5
= j k − 10
𝐹 = (𝑚 + 𝑀)𝑔 − (𝑚 + 𝑀)𝑎 100 1
or 𝐹 = (𝑚 + 𝑀)(𝑔 − 𝑎) = 20 − 10 = 10 m/s#
𝐹 = 85(9.8 − 2.8)N 9 (a)
= 85 × 7 N Let 𝑚% , 𝑚# be mass of bullet and gun and 𝑢% , 𝑢# be
𝐹 = 595 N their initial velocities and 𝑣% and 𝑣# be their final
velocities. From law of conservation of
momentum.
𝑚% 𝑢% + 𝑚# 𝑢# = 𝑚% 𝑣% + 𝑚# 𝑣#
Given, 𝑚% = 5 g = 5 × 10"& kg, 𝑚# = 5 kg
𝑢% = 0, 𝑢# = 0 (at rest), 𝑣% = 500 ms"%
∴ 𝑚% × 0 + 𝑚# × 0 = 5 × 10"& × 500 + 5 × 𝑣#
2.5
⟹ 𝑣# = − = −0.5 ms"%
5
3 (c)
10 (d)
∆𝑣 𝑣 ∆𝑚
Acceleration 𝑎 = = ∙ For upper half
∆𝑡 𝑚 ∆𝑡
50 𝑣 # = 𝑢# + 2𝑎𝑙/2 = 2(𝑔 sin 𝜃)𝑙/2 = 𝑔𝑙 sin θ
= × 0.1 = 2.5 ms"# For lower half
2
4 (b)
𝑇$ = 100 g l/2
Smoot
h
l/2 Rough

𝑙
⇒ 0 = 𝑢# + 2𝑔(sin 𝜃 − 𝜇 cos 𝜃)
5 (a) 2
Since there in no resultant external force, linear ⇒ −𝑔𝑙 sin 𝜃 = 𝑔𝑙(sin 𝜃 − 𝜇 cos 𝜃)
momentum of the system remains constant ⇒ 𝜇 cos 𝜃 = 2 sin 𝜃 ⇒ 𝜇 = 2 tan 𝜃
6 (a) 11 (d)
For body of mass 6 kg Here friction force provides centripetal force so
𝑇 = 6g = 6 × 9.8 = 58.8 N 𝑓 = 𝑚 𝜔# 𝑟 but 𝑓 ≤ 𝜇𝑚𝑔
'(
For body of mass 4 kg So 𝑚𝜔# 𝑟 ≤ 𝜇𝑚𝑔 ⇒ 𝑟 ≤ )!
𝑇 − 𝑇% = 4g = 4 × 9.8 = 39.2 N 12 (b)
𝑇% = 𝑇 − 39.2 If a large force 𝐹 acts for a short time 𝑑𝑡 the
= 58.8 − 39.2 = 19.6 N impulse imparted 𝐼 is
𝑑𝑝
𝐼 = 𝐹. 𝑑𝑡, = . 𝑑𝑡
𝑑𝑡
 𝐼 = 𝑑𝑝 = change in momentum 𝑚% + 𝑚#
𝑇# = h i × 𝑇&
13 (c) 𝑚% + 𝑚# + 𝑚&
Various forces acting on the ball are as shown in 10 + 6
=h i × 40 = 32 N
figure. The three concurrent forces are in 10 + 6 + 4
equilibrium. Using Lami’s theorem 19 (b)
𝑇% 𝑇# Thrust force by rocket
= 𝑑𝑚
sin 150° sin 120° 𝐹+ = 𝑣, h− i (upwards)
10 𝑑𝑡
= Weight of the rocket
sin 90°
𝑇% 𝑇# 10 𝑤 = 𝑚𝑔 (downwards)
= =
sin 30° sin 60° 1 Net force on the rocket
∴ 𝑇% = 10 sin 30 ° 𝐹-./ = 𝑓+ − 𝑤
= 10 × 0.5 = 5N −𝑑𝑚
𝑇# = 10 sin 60° ⟹ 𝑚𝑎 = 𝑣, h i − 𝑚𝑔
𝑑𝑡
and = 10 ×
√&
= 5√3 N −𝑑𝑚 𝑚(𝑔 + 𝑎)
# ⟹ 𝑚𝑎 = 𝑣, h i=
𝑑𝑡 𝑣,
14 (d)
R
∴ Rate of the ejected per second
ma cosq
5000(10 + 20) 5000 × 30
= =
q
ma a 800 800
= 187.5 kgs"%
q mg cosq
mg sin q + 20 (c)
q mg ma sin q 01
𝐹= 0+
, so the force is maximum when slope of
When the whole system is accelerated towards
graph is maximum
left then pseudo force (𝑚𝑎) works on a block
21 (c)
towards right
2345. 3- /7. 584
For the condition of equilibrium Acceleration of the car = 98:: 3; /7. 584
𝑔 sin 𝜃 𝑚𝑛𝑣 0.01 × 10 × 500 "# 5
𝑚𝑔 sin 𝜃 = 𝑚𝑎 cos 𝜃 ⇒ 𝑎 = = = ms = ms"#
cos 𝜃 𝑀 2000 200
∴ Force exerted by the wedge on the block 1
𝑅 = 𝑚𝑔 cos 𝜃 + 𝑚𝑎 sin 𝜃 = ms"#
40
𝑔 sin 𝜃 22 (d)
𝑅 = 𝑚𝑔 cos 𝜃 + 𝑚 h i sin 𝜃
cos 𝜃 𝑑𝑚
Given that, = 0.1 kgs"% ;
𝑚𝑔(cos # 𝜃 + sin# 𝜃) 𝑑𝑡
=
cos 𝜃 mass of the rocket=100 kg
𝑚𝑔 and 𝑣 = 1 kms "% = 1000 ms"%
𝑅=
cos 𝜃 𝑑𝑚
15 (b) Thrust on the rocket, 𝐹 = 𝑣 = 1000 × 0.1
𝑑𝑡
Opposite force causes retardation Now, 𝐹 = 𝑀𝑎
16 (c) 1000 × 0.1
𝑣 = 3𝑡 # − 75; 𝑎 = 6𝑡 ∴ 𝑎= = 1 ms"#
100
𝐹 = 𝑚𝑎 = 36𝑡. At 𝑡 = 4 s, 𝐹 = 144 N 23 (c)
17 (b) 𝑇 = 𝑚(𝑔 + 𝑎) = 1000(9.8 + 1) = 10800 𝑁
The accelerating force of the rocket 24 (d)
∆𝑚 In the frame of wedge, the force diagram of block
= upward thrust = ∙𝑣
∆𝑡 is shown in figure. From free body diagram of
∆𝑚 wedge,
Given, = 50 × 10"& kgs"% , 𝑣 = 400 ms"%
∆𝑡
So, accelerating force = 50 × 10& × 400 = 20 N
18 (b)
Let 𝑎 be the acceleration of each block. Then,
𝑇& = (𝑚% + 𝑚# + 𝑚& )𝑎 … . (i)
and 𝑇# = (𝑚% + 𝑚# )𝑎 … . (ii) For block remain stationary,
from Eq. (i) and (ii), we get
 𝑚𝑎 cos α = 𝑚𝑔 sin α 1
or 1 − µ! =
𝑎 = 𝑔 tan α 𝑛#
25 (c) 1
or µ! = 1 − #
5𝑁 force will not produce any tension in spring 𝑛
30 (b)
without support of other 5𝑁 force. So here the
𝑢𝑑𝑚
tension in the spring will be 5 𝑁 only 𝐹= = 𝑚(𝑔 + 𝑎)
𝑑𝑡
26 (b) 𝑑𝑚 𝑚(𝑔 + 𝑎) 5000 × (10 + 20)
Tension between 𝑚# and 𝑚& is given by ⇒ = =
𝑑𝑡 𝑢 800
= 187.5 𝑘𝑔/𝑠
31 (a)
𝑇 cos 𝜃 = 𝑚𝑔
m1 𝑇 sin 𝜃 = 𝑚𝑎
m2
𝑎
T tan 𝜃 =
𝑔
m3
2𝑚% 𝑚&
𝑇= ×𝑔
𝑚% + 𝑚# + 𝑚&
2×2×2
= × 9.8 = 13 𝑁
2+2+2
27 (d)
% ∴ 𝜃 = tan"% (𝑎/𝑔)
Velocity after 10 sec. = < 32 (a)
Area enclosed between 𝐹 − 𝑡 graph and time axis 𝑅 = 𝑚g + 𝐹# cos θ, 𝑓 = 𝜇𝑅
% % %
Area = # ‹# × 2 × 10 + 2 × 10 + # × 30 × 2 + 𝑓 = 𝜇(𝑚g + 𝐹# cos θ)
%
× 4 × 20Œ
#
= 50 𝑚/𝑠
28 (d)
Length of the chain hanging from the table
𝜇= =
Length of the chain lying on the table Also, 𝑓 = 𝐹% + 𝐹# sin θ
𝑙/3 𝑙/3 1 Equating, 𝜇(𝑚g + 𝐹# cos θ)
= = =
𝑙 − 𝑙/3 2𝑙/3 2 = 𝐹% + 𝐹# sin θ
> ?> :@- A
" !
or 𝜇 = <B?> 53: A
!
29 (a)
33 (b)
When friction absent
𝑚# 5 49
𝑎% = 𝑔 sin θ 𝑎= ×𝑔 = × 9.8 =
𝑚% + 𝑚# 4+5 9
1
∴ 𝑠% = 𝑎# 𝑡## … . (i) = 5.44 𝑚/𝑠 #
2
When friction in present 34 (c)
𝑎# = 𝑔 sin θ − µ! 𝑔 cos θ It works on the principle of conservation of
1 momentum
∴ 𝑠# = 𝑎# 𝑡## … . (ii) 35 (a)
2
From Eqs. (i) and (ii), we have
1 1
𝑎% 𝑡%# = 𝑎# 𝑡##
2 2
#
or 𝑎% 𝑡% = 𝑎# (𝑛𝑡% )# (∵ 𝑡# = 𝑛𝑡# )
or 𝑎% = 𝑛# 𝑎#
𝑎# 𝑔 sin θ − µ! 𝑔 cos θ 1
or = = #
𝑎% 𝑔 sin θ 𝑛 Change in the velocity = 𝑣 sin θ − (−𝑣 sin θ) =
𝑔 sin 45° − µ! 𝑔 cos 45° 1 2 sin θ
or = #
𝑔 sin 45° 𝑛 Change in the momentum
∆𝑝 = 2 𝑚 𝑣 sin θ
 ∆𝑝 1 1
∴ Force applied 𝐹 = = − + 4 − 2 + 1 − = 2 Ns
∆𝑡 2 2
2 × 100 × 10"& × 5 sin θ 60° 41 (b)
=
2 × 10"& Newton second law
√3 1𝑚
= 100 × 5 × 𝐹 = 𝑚𝑎 ⇒ 6 = (7 + 5)𝑎 ; 𝑎 = ; 𝐹′ → 5 𝑘𝑔
2 2 𝑠#
%
= 250√3 N (To the right) Now, 𝐹 C = 5 × # = 2.5 𝑁
36 (a) 42 (d)
𝑓 = µ𝑚g = 0.8 × 4 × 10 = 32 N Angular momentum is an axial vector, so its
Applied force 𝐹 < 𝑓 therefore, answer will be (a) direction is along the axis, perpendicular to the
37 (d) plane of motion which is not changing because of
According to Newton’s second law of motion, change of speed. Therefore, the direction of
∆𝑝 angular momentum remains the same and its
𝐹=
∆𝑡 magnitude may vary
Also, 𝐹 = 𝑚𝑎 43 (d)
𝑝# − 𝑝%
∴ 𝑚𝑎 = Because if the angle of inclination is equal to or
∆𝑡
𝑚# − (−𝑚𝑣% ) more than angle of repose then box will
or 𝑎 = automatically slide down the plane
𝑚∆𝑡
𝑣# + 𝑣% 44 (c)
or 𝑎 = D
∆𝑡 Momentum of one piece = ×3
E
√2 × 10 × 20 + √2 × 10 × 5 D
∴ 𝑎= Momentum of the other piece = ×4
0.02 E
20 + 10 FD ! HD
or 𝑎 = = 1500 ms"# ∴ Resultant momentum = ˜ + 𝑀# =
0.02 %G E
38 (d) The third piece should also have the same
Force 2𝑚g applied at the free end of the string momentum
acts on mass 𝑚. Therefore, its acceleration Let its velocity be 𝑣, then
Force 5𝑀 𝑀 5
𝑎= = × 𝑣 ⇒ 𝑣 = = 2.5 𝑚/𝑠𝑒𝑐
mass 4 2 2
2𝑚g 45 (d)
= = 2g
𝑚 law of conservation of momentum gives
39 (b) 𝑚% 𝑣% = 𝑚# 𝑣#
Force, 𝑚% 𝑣#
⟹ =
𝑚(𝑢 − 𝑣) 150 × 10"& (20 − 0 𝑚# 𝑣%
𝐹= = 4
𝑡 0.1
But, 𝑚 = 𝜋 𝑟 & ρ
= 30 N 3
40 (a) or 𝑚 ∝ 𝑟 &
∆𝑝 𝑚% 𝑟%& 𝑣#
Since, 𝐹 = ∴ = =
∆𝑡 𝑚# 𝑟#& 𝑣%
or ∆𝑝 = 𝐹 ∆𝑡 𝑟% 1 %/&
We can say that momentum between 0 to 7 s is ⟹ =h i
𝑟# 2
equal to the vector area enclosed by the force-
∴ 𝑟% : 𝑟# = 1: 2%/&
time graph from 0 to 7 s. So, Change in linear
46 (d)
momentum
For moving on circular path without slipping,
= vector area of triangle OAB + vector area of
centripetal force must equal frictional force
square BCDE+ vector area of triangle EFG +
That is,
vector area of square GHIJ + vector area of
𝑚𝑣 #
triangle JKL =𝜇𝑚𝑔
1 1 𝑟
= j × 1 × (−1)k + [2 × 2] + j × 2 × (−2)k ⇒ 𝑚𝑟𝜔# = 𝜇 𝑚 𝑔 [∴ 𝑣 = 𝑟 𝜔]
2 2
1 ⇒ 𝑟 𝜔# = 𝜇 𝑔
+ [1 × 1] + j × 1 × (−1)k
2
 48 (b)
𝜇𝑔 0.5 × 9.8
∴ 𝜔=˜ =Ÿ = 0.7 𝑟𝑎𝑑/𝑠 Resolve momentum 6.5 𝑚 along 𝑥 and 𝑦 axes and
𝑟 10
equate
47 (c)

The limiting force of friction is

𝑓= = 10 𝑁 ∴ 6.5 𝑚 cos 𝜃 = 5 × 1
As 𝐹 = 8𝑁 < 𝑓= , therefore, block does not move. and 6.5 𝑚 sin 𝜃 = 6 × 2
As static friction is a self adjusting force, therefore ⇒ (6.5 𝑚)# = (5)# + (12)#
the frictional force on the block is 8 N ⇒ 6.5𝑚 = 13 ⇒ 𝑚 = 2 𝑘𝑔
∴ Total mass = 1 + 2 + 2 = 5𝑘𝑔