5

LAWS OF MOTION

9. According to Galileo's experiment for a double inclined plane,

FACT / DEFINITION TYPE QUESTIONS
if slope of second plane is zero and planes are smooth, then
1. Inertia is the property of a body linked to tendency of a a ball is released from rest on one of the planes rolls down
body and move on the second plane ...X... distance.
(a) to change its position Here, X is
(b) to change its direction (a) zero
(c) to change the momentum (b) infinite
(d) to resist any change in its state (c) equal to length of first plane
2. Physical independence of force is a consequence of (d) None of these
(a) third law of motion (b) second law of motion
10. When a body is stationary
(c) first law of motion (d) all of these (a) there is no force acting on it
3. Newton's first law of motion describes the (b) the force acting on it is not in contact with it
(a) energy (b) work (c) the combination of forces acting on it balances each
(c) inertia (d) moment of inertia other
4. Force depends on (d) the body is in vacuum
(a) change in momentum 11. No force is required for
(b) how fast the change in momentum is brought about (a) an object moving in straight line with constant
(c) Both (a) & (b) (d) None of these velocity
5. Which motion does not require force to maintain it ? (b) an object moving in circular motion
(a) Uniform circular motion (c) an object moving with constant acceleration
(b) Elliptical motion (d) an object moving in elliptical path.
(c) Uniform straight line motion 12. If a stone is thrown out of an accelerated train, then
(d) Projectile motion acceleration of the stone at any instant depends on
6. A ball is travelling with uniform translatory motion. This (a) force acting on it at that instant
means that (b) acceleration of the train
(a) it is at rest. (c) Both (a) & (b) (d) None of these
(b) the path can be a straight line or circular and the ball 13. Which of the following expression is correct?
travels with uniform speed. m
(c) all parts of the ball have the same velocity (magnitude (a) F = ma (b) F =
a
and direction) and the velocity is constant.
(d) the centre of the ball moves with constant velocity a
(c) F = (d) None of these
and the ball spins about its centre uniformly. m
7. An object will continue moving uniformly when 14. Newton’s second law measures the
(a) the resultant force on it is increasing continuously (a) acceleration (b) force
(b) the resultant force is at right angles to its rotation (c) momentum (d) angular momentum
(c) the resultant force on it is zero 15. A reference frame attached to the earth
(d) the resultant force on it begins to decrease (a) is an inertial frame by definition
(b) cannot be an inertial frame because earth is revolving
8. External agencies like gravitational and magnetic forces ...X...
round the sun
exerts force on a body from a distance.
(c) is an inertial frame because Newton's laws are
Here, X refers to
applicable
(a) can (b) cannot (d) is an inertial frame because the earth is rotating about
(c) never (d) None of these its own axis
 EBD_7208
56 LAWS OF MOTION
16. Impulse equals 27. Swimming is possible on account of
(a) rate of change of momentum (a) first law of motion
(b) change in momentum (b) second law of motion
(c) momentum multiplied by time (c) third law of motion
(d) rate of change of force (d) newton's law of gravitation
17. The direction of impulse is 28. Newton’s second and third laws of motion lead to the
(a) same as that of the net force conservation of
(b) opposite to that of the net force
(a) linear momentum (b) angular momentum
(c) same as that of the final velocity
(d) same as that of the initial velocity (c) potential energy (d) kinetic energy
18. A particle of mass m is moving with velocity v1, it is given an 29. Rocket engines lift a rocket from the earth surface,
impulse such that the velocity becomes v2. Then magnitude because hot gases with high velocity
of impulse is equal to (a) push against the air
(a) m( v2 v1 ) (b) m( v1 v 2 ) (b) push against the earth
(c) react against the rocket and push it up
(c) m (v 2 v1 ) (d) 0.5m(v 2 v1 ) (d) heat up the air which lifts the rocket.
19. Impulse is 30. A cannon after firing recoils due to
(a) a scalar quantity (a) conservation of energy
(b) equal to change in the momentum of a body (b) backward thrust of gases produced
(c) equal to rate of change of momentum of a body (c) Newton’s third law of motion
(d) a force (d) Newton’s first law of motion
20. A large force is acting on a body for a short time. The impulse 31. A man is standing at the centre of frictionless pond of ice.
imparted is equal to the change in How can he get himself to the shore?
(a) acceleration (b) momentum (a) By throwing his shirt in vertically upward direction
(b) By spitting horizontally
(c) energy (d) velocity
(c) He will wait for the ice to melt in pond
21. China wares are wraped in straw of paper before packing.
(d) Unable to get at the shore
This is the application of concept of
32. The acceleration of an astronaut is zero once he steps out of
(a) impulse (b) momentum
his accelerated spaceship in the intersteller space. this
(c) acceleration (d) force
statement is in accordance with
22. Which one of the following is not a force?
(a) Newton’s second law of motion
(a) Impulse (b) Tension (b) Newton's first law of motion
(c) Thrust (d) Air resistance (c) Newton's third law of motion
23. In which of the following cases, net force acting on the (d) All of these
body is zero? 33. Law of conservation of momentum follows from
(a) A car moving with uniform velocity (a) Newton's first law of motion
(b) A book lying on the table (b) Newton's second law of motion
(c) Both (a) & (b) (c) Newton's third law of motion
(d) None of these (d) Both (b) & (c)
24. If the net external force on a body is ...X..., its acceleration is 34. A body whose momentum is constant must have constant
zero. Acceleration can be ...Y... only, if there is a net external (a) velocity (b) force
force on the body. Here, X and Y refer to (c) acceleration (d) All of the above
(a) zero, zero (b) zero, non-zero 35. In an explosion, a body breaks up into two pieces of unequal
(c) non-zero, zero (d) non-zero, non-zero masses. In this
25. The same change in momentum about in ...X... time needs (a) both parts will have numerically equal momentum
...Y... force applied. Here, X and Y refer to (b) lighter part will have more momentum
(a) longer, lesser (b) shorter, greater (c) heavier part will have more momentum
(c) both (a) and (b) (d) longer, greater (d) both parts will have equal kinetic energy
26. We can derive Newton’s 36. A jet engine works on the principle of
(a) second and third laws from the first law (a) conservation of mass
(b) first and second laws from the third law (b) conservation of energy
(c) third and first laws from the second law (c) conservation of linear momentum
(d) All the three laws are independent of each other (d) conservation of angular momentum
LAWS OF MOTION 57
37. Which one of the following motions on a smooth plane box as that of the train, keeping it stationary relative to the
surface does not involve force? train. Here, X refers to
(a) Accelerated motion in a straight line (a) kinetic friction (b) static friction
(b) Retarded motion in a straight line (c) limiting friction (d) None of these
(c) Motion with constant momentum along a straight line 45. If s, k and r are coefficients of static friction, kinetic friction
(d) Motion along a straight line with varying velocity and rolling friction, then
38. Identify the correct statement. (a) s < k < f (b) k < r < s
(a) Static friction depends on the area of contact (c) r < k < s (d) r = k = s
(b) Kinetic friction depends on the area of contact 46. It is difficult to move a cycle with brakes on because
(c) Coefficient of kinetic friction does not depend on the (a) rolling friction opposes motion on road
surfaces in contact (b) sliding friction opposes motion on road
(c) rolling friction is more than sliding friction
(d) Coefficient of kinetic friction is less than the coefficient
(d) sliding friction is more than rolling friction
of static friction
47. Which of the following statements about friction is true?
39. If the resultant of all the external forces acting on a system
(a) Friction can be reduced to zero
of particles is zero, then from an inertial frame, one can
(b) Frictional force cannot accelerate a body
surely say that
(c) Frictional force is proportional to the area of contact
(a) linear momentum of the system does not change in
between the two surfaces
time
(d) Kinetic friction is always greater than rolling friction
(b) kinetic energy of the system does not change in time
48. A thin cushion of air maintained between solid surfaces in
(c) angular momentum of the system does not change in
...X... is another effective way of ...Y... friction. Here, X and Y
time
refer to
(d) potential energy of the system does not change in
time (a) relative motion, reducing
40. Frictional force that opposes relative motion between (b) motion, increasing
surfaces in contact is called ...X... and denoted by ...Y.... (c) relative motion, increasing
Here, X and Y refer to (d) None of these
(a) static friction, fs (b) kinetic friction, fs 49. What are the effects if force is acting on a moving body in a
(c) kinetic friction, fk (d) static friction, fk direction perpendicular to the direction of motion?
41. The coefficient of static friction between two surfaces (a) The speed changes uniformly
depends upon (b) The acceleration changes uniformly
(a) the normal reaction (c) The direction of motion changes
(b) the shape of the surface in contact (d) All of these
(c) the area of contact 50. When a car moves on a level road, then the centripetal force
(d) None of the these required for circular motion is provided by ________
42. A rectangular block is placed on a rough horizontal surface (a) weight of the car
in two different ways as shown, then (b) normal reaction
(c) component of friction between the road & tyres along
the surface.
F (d) All of these
F
51. On a banked road, which force is essential to provide the
necessary centripetal force to a car to take a turn while driving
(a) (b)
at the optimum speed?
(a) friction will be more in case (a) (a) Component of normal reaction
(b) friction will be more in case (b) (b) Component of frictional force
(c) friction will be equal in both the cases (c) Both (a) & (b) (d) None of these
(d) friction depends on the relations among its dimensions. 52. Which of the following forces does not act on a body moving
43. If the normal force is doubled, then coefficient of friction in uniform circular motion?
is (a) Centripetal force (b) Weight of the body
(a) halved (b) tripled (c) Normal reaction (d) Force of friction
(c) doubled (d) not changed 53. A particle revolves round a circular path. The acceleration
44. When a box is in stationary position with respect to train of the particle is inversely proportional to
moving with acceleration, then relative motion is opposed (a) radius (b) velocity
by the ...X.... Which provides the same acceleration to the (c) mass of particle (d) both (b) and (c)
 EBD_7208
58 LAWS OF MOTION
54. A cyclist taking turn bends inwards while a car passenger 61. Which of the following statements is/are correct about action
taking the same turn is thrown outwards. The reason is and reaction forces?
(a) car is heavier than cycle I. Action and reaction are simultancous forces
(b) car has four wheels while cycle has only two II. There is no cause-effect relation between action and
(c) difference in the speed of the two reaction.
(d) cyclist has to counteract the centrifugal force while in III. Action and reaction always on two different body
the case of car only the passenger is thrown by this (a) I only (b) II only
force (c) III only (d) I, II and III
55. A car takes a circular turn with a uniform speed u. If the 62. Which of the following statements is/are incorrect, when a
reaction at inner and outer wheels be denoted by R1 and R2, person walks on a rough surface?
then I. The frictional force exerted by the surface keeps him moving
(a) R1 = R2 (b) R1 < R2 II. The force which the man exerts on the floor keeps him
(c) R1 > R2 (d) None of these moving
56. A cyclist bends while taking turn in order to III. The reaction of the force which the man exerts on floor
(a) reduce friction keeps him moving
(b) provide required centripetal force (a) I only (b) II only
(c) reduce apparent weight (c) I and III (d) I and II
(d) reduce speed 63. Select the wrong statement(s) from the following.
I. Newton's laws of motion hold good for both inertial
STATEMENT TYPE QUESTIONS and non-inertial frames
II. During explosion, linear momentum is conserved
57. Consider the following statements and select the incorrect III. Force of friction is zero when no driving force is applied
statement(s). (a) I only (b) II only
I. To move a football at rest, some one must kick it. (c) I and II (d) II and III
64. Choose the correct statement(s) from the following.
II. To throw a stone upwards, one has to give it an upward
I. Recoiling of a gun is an application of principle of
push.
conservation of linear momentum.
III. A breeze causes the branches of a tree to become
II. Explosion of a bomb is based on second law of motion
stationary.
(a) I only (b) II only
IV. A strong wind can move even heavy objects. (c) I and II (d) None of these
(a) Only I (b) Only III 65. Select the incorrect statement(s) about static friction.
(c) III and IV (d) I and II I. Static friction exists on its own
58. Which of the following statements is/are correct ? II. In the absence of applied force static friction is maximum
I. Newton’s first law of motion defines force III. Static friction is equal and opposite to the applied force
II. Newton’s first law of motion defines inertia upto a certain limit
III. Newton’s first law of motion is a measure of force (a) I only (b) II and III
(a) I only (b) II and III (c) I and III (d) I and II
(c) I and III (d) I and II 66. Select the incorrect statement(s) from the following.
59. Choose the incorrect statement(s) from the following. I. Limiting friction is always greater than the kinetic friction
I. If a body is not in rest position, then the net external II. Limiting friction is always less than the static friction
force acting on it cannot be zero. III. Coefficient of static fiction is always greater than the
II. If the net force acting on a body be zero then the body coefficient of kinetic fiction
will essentially remain at rest. (a) I only (b) I and III
(a) I only (b) II only (c) II and III (d) I and II
(c) I and II (d) None of these
60. There are different types of inertia called MATCHING TYPE QUESTIONS
I. Inertia of rest.
67. Match the column I and II.
II. Inertia of motion.
Column I Column II
III. Inertia of direction.
(A) Inertia (1) 105 gcms–1
IV. Inertia of shape.
(B) Recoil of gun (2) kg f
Choose the correct option.
(C) 1 kg ms–1 (3) Newton’s third law of
(a) I and II (b) I, II and III
motion
(c) I, II, III and IV (d) None of these
(D) Weight (4) Newton’s first law of motion
LAWS OF MOTION 59
(a)(A) (4); (B) (1); C (2); (D) (3) 71. Column I Column II
(b)(A) (4); (B) (3); C (1); (D) (2) (A) Rocket’s work (1) Momentum
(c)(A) (3); (B) (2); C (4); (D) (1) (B) F = ma (2) Uniform motion
(d)(A) (2); (B) (4); C (1); (D) (3) (C) Quantity of motion (3) Conservation of momentum
68. Column I Column II (D) Constant force (4) Newton’s second law
(A) Unbalanced (1) Acts on two different (a) (A) (4); (B) (1); C (2); (D) (3)
bodies (b) (A) (4); (B) (3); C (1); (D) (2)
(B) Action & Reaction (2) Inability to change the (c) (A) (3); (B) (4); C (1); (D) (2)
state (d) (A) (2); (B) (4); C (1); (D) (3)
(C) Inertia (3) mv
(D) Momentum (4) Variable velocity DIAGRAM TYPE QUESTIONS
(a) (A) (4); (B) (1); C (2); (D) (3)
72. Which equation holds true for the given figure?
(b) (A) (1); (B) (2); C (4); (D) (3)
(c) (A) (3); (B) (2); C (4); (D) (1)
(d) (A) (2); (B) (4); C (1); (D) (3) F1
69. Column I Column II
(A) Accelerated motion (1) Newton’s 1st law
(B) Impulse (2) Mass
(C) Law of inertia (3) Force×time F2
(D) Measure of inertia (4) Change in speed and
direction F3
(a) (A) (4); (B) (1); C (2); (D) (3)
(b) (A) (4); (B) (2); C (1); (D) (3) (a) F1 – F2 = F3 (b) F1 + F2 = F3
(c) (A) (3); (B) (4); C (1); (D) (2) (c) F1 + F2 + F3 = 0 (d) F2 + F3 = F1
(d) (A) (4); (B) (3); C (1); (D) (2)
73. A block of mass 4 kg is suspended
70. A light string ABCDE whose extremity A is fixed, has weights
through two light spring balances A
W1 and W2 attached to it at B and C. It passes round a small A
and B. Then A and B will read
smooth peg at D carrying a weight of 300 N at the free end E
as shown in figure. If in the equilibrium position, BC is respectively :
horizontal and AB and CD make 150° and 120° with CB. (a) 4 kg and zero kg
B
Match the columns : (b) zero kg and 4 kg
(c) 4 kg and 4 kg
4kg
(d) 2 kg and 2 kg
74. Which figure shows the correct force acting on the body
D
A sliding down an inclined plane? (m mass, fs force of
150°
friction)
120°
B C E
300 N N B N B
w1 w2 fs fs

(a) sin (b)

sin
mg mg mg mg mg cos
Column I Column II C mg cos A C A
(A) Tension in portion AB, TAB (1) 150 N
(B) Tension in portion BC, TBC (2) 173 N
N B
(C) Weight, W1 (3) 260 N N sin B
mg
(D) Weight, W2 (4) 87 N
(a) (A) (4); (B) (1); C (2); (D) (3) (c) fs (d) fs
sin mg mg cos
(b) (A) (2); (B) (1); C (4); (D) (3) mg mg cos mg A
C A C
(c) (A) (3); (B) (4); C (1); (D) (3)
(d) (A) (4); (B) (3); C (1); (D) (2)
 EBD_7208
60 LAWS OF MOTION
75. For the given situation as shown in the figure, the value of 78. For the system shown in figure, the correct expression is
to keep the system in equilibrium will be m3
m2 m1
F3 F2 F1

T1

m 3F
(a) F3 = F1 + F2 (b) F3 = F F2 F3
1
T2
m 3F m 3F
(c) F3 = m m2 m3 (d) F3 = m
1 1 m2
W = 60N
79. Which of the following is true about acceleration, a for the
system?
(a) 30° (b) 45°
(c) 0° (d) 90° m2
76. The acceleration of the system shown in the figure is given m1
by the expression (ignore force of friction) T T B
F A

a (a) Acceleration is more in A, when force is applied on A.

m1 T (b) Acceleration is more in B, when force is applied on B.
(c) Acceleration is same and does not depend on whether
the force is applied on m1 or m2
T (d) Acceleration depends on the tension in the string.
a 80. A system consists of three masses m1, m2 and m3 connected
m2 by a string passing over a pulley P. The mass m1 hangs
freely and m2 and m3 are on a rough horizontal table (the
m2 g coefficient of friction = ). The pulley is frictionless and of
negligible mass. The downward acceleration of mass m1 is :
(Assume m1 = m2 = m3 = m)
m2g m1g
(a) a = (b) a = g(1 – g )
(m1 m 2 ) (m1 m 2 ) (a)
g m2 m3
P
m1 m2 2g
(c) a = (c) a =
(m1 m 2 ) g (m1 m 2 ) g (b)
3
77. What is the direction of force on the wall due to the ball in
g(1 – 2 )
two cases shown in the figures? (c) m1
3

u g(1 – 2 )
(d)
2
30° 81. The force ‘F’ acting on a particle of mass ‘m’ is indicated by
u u 30° the force-time graph shown below. The change in momentum
u of the particle over the time interval from zero to 8 s is :

(a) (b) 6
3
(a) In (a) force is normal to the wall and in (b) force is
0
F(N)

inclined at 30° to the normal. 2 4 6 8

(b) In (a) force is normal to the wall and in (b) force is –3
inclined at 60° to the normal.
t(s)
(c) In (a) the force is along the wall and in (b) force is
normal to the wall. (a) 24 Ns (b) 20 Ns
(d) In (a) and (b) both the force is normal to the wall. (c) 12 Ns (d) 6 Ns
LAWS OF MOTION 61
82. The motion of a car on a banked road is shown in the figure. 90. Assertion : A bullet is fired from a rifle. If the rifle recoils
The centripetal force equation will be given by freely, the kinetic energy of rifle is more than that of the
bullet.
N cos Reason : In case of rifle bullet system, the law of
N conservation of momentum violates.
v2 91. Assertion : A rocket works on the principle of conservation
a
R of linear momentum.
f Reason : Whenever there is change in momentum of one
mg body, the same change occurs in the momentum of the
second body of the same system but in the opposite
mv 2 2
(a) Nsin + fcos = (b) f = mv direction.
R R 92. Assertion : The two bodies of masses M and m (M > m)
are allowed to fall from the same height if the air resistance
mv 2 mv 2
(c) N cos + f = (d) N sin + f = for each be the same then both the bodies will reach the
R R
earth simultaneously.
Reason : For same air resistance, acceleration of both the
ASSERTION- REASON TYPE QUESTIONS bodies will be same.
Directions : Each of these questions contain two statements, 93. Assertion : A block placed on a table is at rest, because
Assertion and Reason. Each of these questions also has four action force cancels the reaction force on the block.
alternative choices, only one of which is the correct answer. You Assertion : The net force on the block is zero.
have to select one of the codes (a), (b), (c) and (d) given below.
(a) Assertion is correct, reason is correct; reason is a correct 94. Assertion : On a rainy day, it is difficult to drive a car or bus
explanation for assertion. at high speed.
(b) Assertion is correct, reason is correct; reason is not a Reason : The value of coefficient of friction is lowered due
correct explanation for assertion to wetting of the surface.
(c) Assertion is correct, reason is incorrect 95. Assertion : Frictional forces are conservating forces.
(d) Assertion is incorrect, reason is correct. Reason : Potential energy can be associated with frictional
forces.
83. Assertion : Mass is a measure of inertia of the body in 96. Assertion : A man and a block rest on smooth horizontal
linear motion.
surface. The man holds a rope which is connected to block.
Reason : Greater the mass, greater is the force required to
change its state of rest or of uniform motion. The man cannot move on the horizontal surface.
84. Assertion : An object can move with constant velocity if no
net force acts on it.
Reason : No net force is needed to move an object with
constant velocity.
85. Assertion : If the net external force on the body is zero, Reason : A man standing at rest on smooth horizontal surface
then its acceleration is zero. cannot start walking due to absence of friction (The man is
Reason : Acceleration does not depend on force. only in contact with floor as shown).
86. Assertion : For the motion of electron around nucleus,
Newton's second law is used.
Reason : Newton's second law can be used for motion of any
object.
87. Assertion : Impulse of force and momentum are same physi- 97. Assertion: Friction is a necessary evil
cal quantities. Reason: Though friction dissipates power, but without
Reason : Both quantities have same unit. friction we cannot walk.
88. Assertion: A cricketer moves his hands forward to catch a 98. Assertion: There is a stage when frictional force is not needed
ball so as to catch it easily without hurting. at all to provide the necessary centripetal force on a banked
Reason: He tries to decrease the distance travelled by the road.
ball so that it hurts less. Reason: On a banked road, due to its inclination the vehicle
89. Assertion: Same force applied for the same time causes the tends to remain inwards without any chances of skidding.
same change in momentum for different bodies 99. Assertion : Force is required to move a body uniformly
Reason: The total momentum of an isolated system of
along a circle.
interacting bodies remains conserved.
Reason : When the motion is uniform, acceleration is zero.
 EBD_7208
62 LAWS OF MOTION
100. Assertion : Linear momentum of a body changes even 108. A metre scale is moving with uniform velocity. This implies
when it is moving uniformly in a circle. (a) the force acting on the scale is zero, but a torque
Reason : In uniform circular motion, velocity remains about the centre of mass can act on the scale.
constant. (b) the force acting on the scale is zero and the torque
101. Assertion : A cyclist always bends inwards while negotiating acting about centre of mass of the scale is also zero.
a curve. (c) the total force acting on it need not be zero but the
Reason : By bending, cyclist lowers his centre of gravity. torque on it is zero.
(d) neither the force nor the torque need to be zero.
CRITICALTHINKING TYPE QUESTIONS 109. A body of mass M hits normally a rigid wall with velocity V
and bounces back with the same velocity. The impulse
102. A boy, sitting on the topmost birth in the compartment of experienced by the body is
a train which is just going to stop on the railway station, (a) MV (b) 1.5 MV (c) 2 MV (d) zero
drops an apple aiming at the open hand of his brother 110. If rope of lift breaks suddenly, the tension exerted by the
situated vertically below his own hand at a distance of surface of lift (a = acceleration of lift)
2m. The apple will fall (a) mg (b) m(g + a)
(a) in the hand of his brother (c) m(g – a) (d) 0
(b) slightly away from the hand of his brother in the 111. An explosion breaks a rock into three parts in a horizontal
direction of motion of the train plane. Two of them go off at right angles to each other. The
(c) slightly away from the hand of his brother opposite first part of mass 1 kg moves with a speed of 12 ms–1 and the
to the direction of motion of the train second part of mass 2 kg moves with speed 8 ms–1. If the
(d) None of the above third part flies off with speed 4 ms–1 then its mass is
103. A person sitting in an open car moving at constant (a) 5 kg (b) 7 kg
velocity throws a ball vertically up into air. The ball falls
(c) 17 kg (d) 3 kg
(a) outside the car
112. A stationary body of mass 3 kg explodes into three equal
(b) in the car ahead of the person
pieces. Two of the pieces fly off in two mutually
(c) in the car to the side of the person
(d) exactly in the hand which threw it up perpendicular directions, one with a velocity of 3iˆ ms 1

104. If a stone of mass 0.05 kg is thrown out a window of a train

and the other with a velocity of 4jˆ ms 1. If the explosion
moving at a constant speed of 100 km/h then magnitude of
occurs in 10–4 s, the average force acting on the third piece
the net force acting on the stone is
in newton is
(a) 0.5 N (b) zero
4
(c) 50 N (d) 5 N (a) (3iˆ 4j)
ˆ 10 y
105. A closed compartment containing gas is moving with same
4
acceleration in horizontal direction. Neglect effect of gravity. (b) (3iˆ 4j)
ˆ 10

Then the pressure in the compartment is 1 4ˆj

(c) (3iˆ 4ˆj) 10 4 x
(a) same everywhere (b) lower in front side 1 3iˆ
4j)
(c) lower in rear side (d) lower in upper side (d) ˆ 104
(3iˆ 4j) i +
(3
106. When an elevator cabin falls down, the cabin and all the (–
1×
bodies fixed in the cabin are accelerated with respect to
(a) ceiling of elevator (b) floor of elevator 113. A spacecraft of mass 100 kg breaks into two when its velocity
(c) man standing on earth is 104 m s–1. After the break, a mass of 10 kg of the space
(d) man standing in the cabin craft is left stationary. The velocity of the remaining part is
107. A monkey is climbing up a rope, then the tension in the rope (a) 103 m s–1 (b) 11.11 × 103 m s–1
(a) must be equal to the force applied by the monkey on (c) 11.11 × 10 m s 2 –1 (d) 104 m s–1
the rope 114. A ball is thrown up at an angle with the horizontal. Then
(b) must be less than the force applied by the monkey on the total change of momentum by the instant it returns to
the rope. ground is
(c) must be greater than the force applied by the monkey (a) acceleration due to gravity × total time of flight
on the rope. (b) weight of the ball × half the time of flight
(d) may be equal to, less than or greater the force applied (c) weight of the ball × total time of flight
by the monkey on the rope. (d) weight of the ball × horizontal range
LAWS OF MOTION 63
115. A spring balance is attached to the ceiling of a lift. A man 122. The force required to just move a body up the inclined plane
hangs his bag on the spring and the spring reads 49 N, when is double the force required to just prevent the body from
the lift is stationary. If the lift moves downward with an sliding down the plane. The coefficient of friction is . The
acceleration of 5m / s 2 , the reading of the spring balance inclination of the plane is
will be (a) tan–1 (b) tan–1 ( /2)
(c) tan 2–1 (d) tan–1 3
(a) 24 N (b) 74 N
(c) 15 N (d) 49 N 123. A hockey player is moving northward and suddenly turns
westward with the same speed to avoid an opponent. The
116. A block of mass m is placed on a smooth wedge of inclination
force that acts on the player is
. The whole system is accelerated horizontally so that the
(a) frictional force along westward
block does not slip on the wedge. The force exerted by the
(b) muscles force along southward
wedge on the block (g is acceleration due to gravity) will be
(c) frictional force along south-west
(a) mg/cos (b) mg cos
(d) muscle force along south-west
(c) mg sin (d) mg
124. The coefficient of static friction s, between block A of mass
117. A person of mass 60 kg is inside a lift of mass 940 kg and
2 kg and the table as shown in the figure is 0.2. What would
presses the button on control panel. The lift starts moving
be the maximum mass value of block B so that the two blocks
upwards with an acceleration 1.0 m/s2. If g = 10 ms–2, the
do not move? The string and the pulley are assumed to be
tension in the supporting cable is
smooth and massless. (g = 10 m/s2)
(a) 8600 N (b) 9680 N
(c) 11000 N (d) 1200 N 2 kg
A
118. Three blocks with masses m, 2 m and 3 m are connected by
strings as shown in the figure. After an upward force F is
B
applied on block m, the masses move upward at constant
speed v. What is the net force on the block of mass 2m? (a) 0.4 kg (b) 2.0 kg
(g is the acceleration due to gravity) (c) 4.0 kg (d) 0.2 kg
(a) 2 mg 125. A conveyor belt is moving at a constant speed of 2m/s. A box
is gently dropped on it. The coefficient of friction between
(b) 3 mg them is µ = 0.5. The distance that the box will move relative to
(c) 6 mg belt before coming to rest on it taking g = 10 ms–2, is
(a) 1.2 m (b) 0.6 m (c) zero (d) 0.4 m
(d) zero
126. The upper half of an inclined plane of inclination is per-
119. The net force on a rain drop falling down with a constant fectly smooth while lower half is rough. A block starting
speed is ________ from rest at the top of the plane will again come to rest at the
(a) weight of drop W bottom, if the coefficient of friction between the block and
(b) viscous drag of air F lower half of the plane is given by
(c) W + F + force of buoyany 2
(a) = (b) = 2 tan
(d) zero tan
120. If two masses (M & m) are connected on a horizontal plane 1
and a force is applied on the combination, then the tension (c) = tan (d) =
tan
T depends on 127. A marble block of mass 2 kg lying on ice when given a velocity
(a) the force applied on the system of 6 m/s is stopped by friction in 10 s. Then the coefficient of
(b) whether force is applied on M or m friction is (Take g = 10 ms–2)
(c) both (a) and (b) (a) 0.06 (b) 0.03
(d) Can’t be predicted. (c) 0.04 (d) 0.01
121. A body is imparted motion from rest to move in a straight 128. A block rests on a rough inclined plane making an angle of
line. If it is then obstructed by an opposite force, then 30° with the horizontal. The coefficient of static friction
(a) the body may necessarily change direction between the block and the plane is 0.8. If the frictional force
(b) the body is sure to slow down on the block is 10 N, the mass of the block (in kg) is
(c) the body will necessarily continue to move in the (take g 10 m / s 2 )
same direction at the same speed
(a) 1.6 (b) 4.0
(d) None of these
(c) 2.0 (d) 2.5
 EBD_7208
64 LAWS OF MOTION
129. A given object takes n times as much time to slide down a 136. A block of mass m is placed on a surface with a vertical cross
45º rough incline as it takes to slide down a perfectly smooth
45º incline. The coefficient of kinetic friction between the x3
section given by y . If the coefficient of friction is 0.5,
object and incline is given by 6
1 1 the maximum height above the ground at which the block
(a) 1 (b)
n 2
1 n2 can be placed without slipping is:

1 1
1 2
(c) 1 (d) (a) m (b) m
6 3
n2 1 n2
1 1
130. The minimum force required to start pushing a body up rough (c) m (d) m
(frictional coefficient ) inclined plane is F1 while the 3 2
minimum force needed to prevent it from sliding down is F2. 137. Two bodies of masses 1 kg and 2 kg moving with same
If the inclined plane makes an angle from the horizontal velocities are stopped by the same force. Then the ratio of
F their stopping distances is
such that tan 2 then the ratio 1 is
F (a) 1 : 2 (b) 2 : 1
(a) 1 (b) 2 2
(c) 3 (d) 4 (c) 2 :1 (d) 1 : 2
131. A particle is acted upon by a force of constant magnitude
which is always perpendicular to the velocity of the 138. A block A of mass m1 rests on a horizontal table. A light
particle. The motion of the particle takes place in a plane. string connected to it passes over a frictionless pulley at the
It follows that edge of table and from its other end another block B of mass
(a) its velocity is constant m2 is suspended. The coefficient of kinetic friction between
(b) its acceleration is constant
the block and the table is µk. When the block A is sliding on
(c) its kinetic energy is constant
(d) it moves in a straight line. the table, the tension in the string is
132. If n bullets each of mass m are fired with a velocity v per
second from a machine gun, the force required to hold the (m 2 – k m1 ) g m1m 2 (1 k )g
(a) (m1 m 2 ) (b) (m1 m 2 )
gun in position is
mv m1m 2 (1 – k )g (m 2 km1 )g
(a) (n + 1) mv (b) (c) (d)
n2 (m1 m 2 ) (m1 m2 )
mv 139. The retarding acceleration of 7.35 m s–2 due to frictional
(c) (d) mnv
n force stops the car of mass 400 kg travelling on a road. The
133. A car moves at a speed of 20 ms–1 on a banked track and coefficient of friction between the tyre of the car and the
describes an arc of a circle of radius 40 3 m. The angle of road is
banking is (g = 10 ms–2 ) (a) 0.55 (b) 0.75
(a) 25° (b) 60°
(c) 0.70 (d) 0.65
(c) 45° (d) 30°
134. A ball of mass 10 g moving perpendicular to the plane of the 140. A hammer weighing 3 kg strikes the head of a nail with a
wall strikes it and rebounds in the same line with the same speed of 2 ms–1 drives it by l cm into the wall. The impulse
velocity. If the impulse experienced by the wall is 0.54 Ns, imparted to the wall is
the velocity of the ball is
(a) 6Ns (b) 3Ns
(a) 27 ms–1 (b) 3.7 ms–1
(c) 54 ms –1 (d) 37 ms–1 (c) 2Ns (d) l2 Ns
135. Two blocks of masses 2 kg and 4 kg are attached by an 141. A balloon with mass ‘m’ is descending down with an
inextensible light string as shown in the figure. If a force of acceleration ‘a’ (where a < g). How much mass should be
120 N pulls the blocks vertically upward, the tension in the removed from it so that it starts moving up with an
string is (take g = 10 ms–2) acceleration ‘a’?
F = 120 N 2ma 2ma
(a) 20 N
4 kg (a) g a (b) g a
(b) 15 N
(c) 35 N ma ma
(c) (d)
(d) 40 N 2 kg g a g a
LAWS OF MOTION 65
142. The time required to stop a car of mass 800 kg, moving at a (a) have a horizontal inward component
speed of 20 ms–1 over a distance of 25 m is (b) be vertical
(a) 2s (b) 2.5s (c) equilibriate the centripetal force
(c) 4s (d) 4.5s
(d) be decreased
143. A particle rests on the top of a hemisphere of radius R.
Find the smallest horizontal velocity that must be imparted 148. A sphere is suspended by a thread of length . What
to the particle if it is to leave the hemisphere without minimum horizontal velocity has to be imparted to the sphere
sliding down is for it to reach the height of the suspension?
(a) g (b) 2 g
(a) gR (b) 2g R
(c) g (d) 2g
(c) 3g R (d) 5g R
149. A car when passes through a bridge exerts a force on it
144. A train is moving with a speed of 36 km/hour on a curved which is equal to
path of radius 200 m. If the distance between the rails is 1.5
m, the height of the outer rail over the inner rail is Mv 2 Mv 2
(a) 1 m (b) 0.5 m (a) Mg (b)
r r
(c) 0.75 m (d) 0.075 m
145. A car moving on a horizontal road may be thrown out of the Mv 2
road in taking a turn (c) Mg – (d) None of these
r
(a) by the gravitational force
150. A bridge is in the from of a semi-circle of radius 40m. The
(b) due to the lack of proper centripetal force
greatest speed with which a motor cycle can cross the bridge
(c) due to the rolling frictional force between the tyre and
without leaving the ground at the highest point is
road
(d) due to the reaction of the ground (g = 10 m s–2) (frictional force is negligibly small)
146. A car sometimes overturns while taking a turn. When it (a) 40 m s–1 (b) 20 m s–1
overturns, it is (c) 30 m s–1 (d) 15 m s–1
(a) the inner wheel which leaves the ground first 151. A particle tied to a string describes a vertical circular motion
(b) the outer wheel which leaves the ground first
of radius r continually. If it has a velocity 3 gr at the
(c) both the wheel leave the ground simultaneously
(d) either wheel will leave the ground first highest point, then the ratio of the respective tensions in
the string holding it at the highest and lowest points is
147. On a railway curve the outside rail is laid higher than the
(a) 4 : 3 (b) 5 : 4
inside one so that resultant force exerted on the wheels of
(c) 1 : 4 (d) 3 : 2
the rail car by the tops of the rails will
 EBD_7208
66 LAWS OF MOTION

FACT / DEFINITION TYPE QUESTIONS acceleration is determined only by the instantaneous

force and not by any history of the motion of the
1. (d) Inertia is defined as the ability of a body to oppose any particle. Therefore, the moment the stone is thrown out
change in its state of rest or of uniform motion. of an accelerated train, it has no horizontal force and
2. (c) Newton's first law of motion is related to the physical acceleration, if air resistance is neglected.
independence of force. 13. (a)
3. (c) Newton's first law of motion defines the inertia of
dp
body. It states that every body has a tendency to 14. (b) F
remain in its state (either rest or motion) due to its dt
inertia. 15. (b) The frame of reference which are at rest or in uniform
4. (c) According to Newton’s 2nd law of motion motion are called inertial frames while frames which
are accelerated with respect to each other are non–
change in momentum inertial frames. Spinning or rotating frames are
F=
time accelerated frame, hence these are non-inertial frames.
Thus force depends directly on the rate of change of 16. (b) Impulse = Force × time duration. …(1)
momentum. According to Newton’s second law
5. (c) According to Newton's second law of motion,
Change in momentum
F = ma Force = …(2)
time duration
When body is moving uniformly along a straight line
and there is no force of friction, acceleration / Force × time = change in momentum
retardation of the body a = 0, i.e., Impulse = change is momentum.
F = ma = 0 17. (a)
i.e, no external force is required.
For accelerated motion, force is necessary. In uniform 18. (a) Impulse = change in momentum = m v 2 m v1
circular motion, elliptical motion and projectile motion P
direction of velocity changes due to which force is 19. (b) F and impulse = F. T
t
imposed.
6. (c) 20. (b) If a large force F acts for a short time dt the impulse
imparted I is
7. (c) The body will continue accelerating until the resultant
force acting on the body becomes zero. dp
I F.dt .dt
8. (a) External agencies like gravitational and magnetic forces dt
can exert force on body from distance. When ball is I = dp = change in momentum
21. (a)
released from some height. Earth exerts gravitational
22. (a) Impulse is not a force.
force from distance, ball moves faster with time. Impulse = Force × Time duration
9. (b) 23. (c) A book lying on the table is acted by its weight
downwards and a reaction upwards.
A car moving on a road, has an applied forward force
and force of friction acts backwards. Thus it moves
with constant velocity.
Force, F = ma, if a = 0, then Fnet = 0
On second plane ball will move with constant velocity 24. (b) If the net external force on a body is zero, its acceleration
because no external force is there to provide acceleration is zero.
or retardation. Acceleration can be non-zero only if there is net external
10. (c) From Newton’s second law if Fi 0 then the body is force on the body. This is concluded from Newton's
first law of motion.
in translational equilibrium.
11. (a) No force is required for an object moving in straight P P
line with constant velocity or for non acceleration 25. (c) As fext , if constant
t t
motion.
i.e, fext t = constant. If force is small time taken is
12. (a) Since force at a point at any instant is related to the
more and if force is large time taken is less.
acceleration at that point, at that instant and
LAWS OF MOTION 67
26. (c) 48. (a) A thin cushion of air maintained between solid surfaces
27. (c) Swimming is a result of pushing water in the opposite in relative motion is another effective way or reducing
direction of the motion. friction
28. (a) Newton’s second and third laws of motion leads to the
conservation of linear momentum. Inflated balloon
29. (c) Hot gases with high velocity react against the rocket Plastic disc with hole
and push it up.
30. (c) The gun applied a force F12 on the bullet in forward
Air cushion
direction & according to Newton’s third law bullet surface
applies a reaction force on gun F21 in backward
Because of air cushion between plastic disc and surface,
direction. But the recoil speed of gun is very low in
there is very less friction between plastic disc and
comparison to bullet due to large mass.
surface. So plastic disc can be moved on surface with
31. (b)
very less frictional dissipation of energy. This is
32. (b) Since no nearby stars are there to exert gravitational because friction between solid and air is very small.
force on the astronaut, so the net force acting on him is
49. (c) When force is applied on a moving body in a direction
zero when he moves out of the spaceship. Thus in perpendicular to the direction of motion, then it takes a
accordance with first law of motion the acceleration of circular path. Thus the direction of motion changes
astronaut will be zero. without changes in the speed.
33. (d) 50. (c) Normal reaction N = weight mg thus the centripetal
34. (a) It works on the principle of conservation of linear force required by the car for circular motion is provided
momentum. by the component of the force of friction b/w the road
35. (a) If m1, m2 are masses and u1, u2 are velocity then by and the car tyres.
conservation of momentum m 1u1 + m 2u2 = 0 or
N
| m1u1 | | m2 u 2 |
36. (c)
37. (c) Motion with constant momentum along a straight line. f
According to Newton's second law rate of change of
momentum is directly proportional to force applied. mg
38. (d) k < s coefficient of static friction is always greater 51. (a) Optimum speed is given by V0 = (Rg tan )1/2 on a
than kinetic friction. banked road, the normal reaction’s component is
39. (a) Since there in no resultant external force, linear enough to provide the necessary centripetal force to a
momentum of the system remains constant. car driven at optimum speed.
40. (c) Frictional force that opposes relative motion between 52. (a) Material forces like friction, gravitational force etc. act
surfaces in contact is called kinetic friction and denoted on the body and provide the centripetal force. The
by fk. centripetal force cannot be regarded as any kind of
force acting externally. It is simple name given to the
force of friction force that provides inward radial acceleration to a body
41. (a) Coefficient of static friction =
normal reaction in circular motion.
Therefore, coefficient of static friction depends upon v2 1
the normal reaction. 53. (a) Acceleration (centripetal) a i.e., a
r r
42. (c) Friction does not depend on area of surfaces in 54. (d)
contact. 55. (b) Due to centrifugal force, the inner wheel will be left up
43. (d) Coefficient of friction is independent of normal force. when car is taking a circular turn. Due to this, the reaction
44. (b) When a box is in stationary position with respect to on outer wheel is more than that on inner wheel.
train moving with acceleration, then relative motion is 56. (b) The cyclist bends while taking turn in order to provide
opposed by the static friction. necessary centripetal force.
45. (c)
46. (d) When brakes are on, the wheels of the cycle will slide STATEMENT TYPE QUESTIONS
on the road instead of rolling there. It means the sliding
friction will come into play instead of rolling friction. 57. (b) A breeze causes branches of tree to swing. In general
The value of sliding friction is more than that of rolling force is required to put a stationary object in motion.
friction. 58. (d) Newton’s 2nd law of motion gives F = ma. Thus it is a
47. (d) measure of force. Newton’s first law of motion simply
gives a qualitative definition of force.
 EBD_7208
68 LAWS OF MOTION
59. (d) If a body is moving with a constant velocity then the Applying Lami's equation, we have
net force on the body is zero. Also if net force is zero,
TBC W2 300
the body may be moving uniformly along a straight = =
sin150 sin120 sin 90
line. Thus both the given statements are false.
60. (b) There are three types of inertia. TAB W1 TBC
Inertia of rest : The resistance of a body to change its and
sin 90
=
sin150
=
sin120
state of rest is called inertia of rest.
After simplifying, we get
Inertia of motion : The resistance of a body to change
its state of motion is called inertia of motion. TAB = 173 N, TBC = 150 N, W1 = 87 N, W2 = 260 N.
Inertia of direction : The resistance of a body to change 71. (c) (A) (3); (B) (4); C (1); (D) (2)
its direction of motion is called inertia of direction.
61. (d) Action & reaction forces act simultaneously: There is DIAGRAM TYPE QUESTIONS
no cause effect relation between action and reaction as
any of the two mutual forces can be called action and 72. (c) Equilibrium under three concurrent forces F1, F2and F3
the other reaction since action & reaction act on requires that vector sum of the three forces is zero.
different bodies, so they cannot be cancelled out. F1 + F2 + F3 = 0.
62. (c) When the men push the rough surface on walking, R F1 F2 F1
then surface (from Newton’ third Law) applies reaction R
force in forward direction. It occurs because there is F3 R (In eqbm)
friction between men & surface. If surface is frictionless
(such as ice), then it is very difficult to move on it. F3 (F1 F2 )
F2
63. (a) Newton's laws of motion are applicable only for inertial
F1 F2 F3 = 0 F3
frames. All refrence frames present on surface of earth
are supposed to be inertial frame of refrence.
64. (a) According to third law of motion bullet experiences a
force F then, give experiences an equal and opposite 73. (c) In series each spring will have same force.
force F. According to second law, F t is change in Here it is 4 kg-wt.
momentum of the bullet, then – F t is change in
momentum of the gun. Since initially both are at rest, 74. (b) If a body slides down, then the force of friction acts
the final momentum = 0. Pb + Pg = 0. Thus the total upwards along the plane weight(mg) act vertically
momentum of (bullet + gun) is conserved. downwards.
65. (d) The static friction comes into play, the moment there is 75. (b)
an applied force. As the applied force increases, static
friction also increases, remaining equal and opposite
to the applied force upto a certain limit. But if the applied T1 cos
T1
force increases so much, it overcomes the static friction
and the body starts moving.
66. (c) Limiting friction is the maximum static friction beyond T1 sin T2
which the object starts moving. It decreases a little bit 2m
before the object comes into motion. Thus limiting T2
friction is less than the kinetic friction.
W = 60N
MATCHING TYPE QUESTIONS
In eqbm T1cos = T2 = 60N. …(1)
67. (b) (A) (4); (B) (3); C (1); (D) (2) T1sin = 60 N …(2)
68. (a) (A) (4); (B) (1); C (2); (D) (3) tan = 1
69. (d) (A) (4); (B) (3); C (1); (D) (2) = 45°.
70. (b) (A) (2) ; (B) (1) ; (C) (4) ; (D) (3) 76. (a) Equations of motion of m1 & m2 are as:
T = m1a …(1)
300N
m2g – T = m2a …(2)
A Adding eqn. (i) and (ii)
150° 120°
C
m2g = (m1 + m2) a
B
TBC TBC
m2g
a= m m
2
w1 w2
LAWS OF MOTION 69
77. (d) Case (a) ASSERTION- REASON TYPE QUESTIONS
(Px)i = mu Py(initial) = 0
(Px)f = f = –mu Py(final) = 0
F
Impulse = P = –2mu (along x –axis) 83. (a) According to Newton's second law of motion a =
Impulse = 0 along y–axis m
parallaly in case (b) i.e. magnitude of the acceleration produced by a
(Px)i = mu cos30° (Py)i = –mu sin30° given force is inversely proportional to the mass of
(Px)f = f = –mu cos30° (Py)f = –mu sin30°
the body. Higher is the mass of the body, lesser will
Impulse = –2mu cos 30°(along x-axis)
Impulse = 0 (along y–axis) be the acceleration produced i.e. mass of the body is
Force and impulse are in the same direction the force a measure of the opposition offered by the body to
on wall due to the ball is normal to the wall along change a state, when the force is applied i.e. mass of
positive x–direction in both (a) & (b) case. a body is the measure of its inertia.
78. (c) Common acceleration of system is 84. (a)
a=
F 85. (c) According to Newton's second law
m1 m2 m3
force
Acceleration = i.e. if net external force on the
m 3F mass
Force on m3 is F3 = m3 × a = m m2 m3 body is zero then acceleration will be zero
1
86. (c) Newton's second law can not be used for any object.
F 87. (d) Impulse and momentum are different quantities, but
79. (c) a = m m2
1 both hae same unit (N–s).
So the acceleration is same whether the force is applied 88. (d) A cricket player moves his hands backward to increase
on m1 or m2. the time interval for reducing the momentum of the ball
80. (c) Acceleration to zero. Thus the ball does not hit him hard as force is
Net force in the direction of motion directly proportional to change of momentum.
=
Total mass of system 89. (b) According to 2nd law of motion;

m1g (m 2 m3 )g g P1 P2
= = (1 2 ) F1 = F2 =
m1m 2 m3 3 t1 t2
( m1 = m2 = m3 = m given) F1 × t1 = F2 × t2
81. (c) Change in momentum,
P1 = P2
p = Fdt Thus the same force for the same time causes the same
= Area of F-t graph change in momentum for different bodies.
= ar of – ar of + ar of
90. (d) Law of conservation of linear momentum is correct
1 when no external force acts. When bullet is fired form
= 2 6 3 2 4 3
2 a rifle then both should possess equal momentum
= 12 N-s
82. (a) p2
N cos but different kinetic energy. E = Kinetic
2m
N energy of the rifle is less than that of bullet because
f cos E 1/m
N sin 91. (a)
f 92. (a) The force acting on the body of mass M are its weight
mg
Mg acting vertically downward and air resistance F
f sin
acting vertically upward.
Clearly form the figure, N sin and f cos contribute to
F
the centripetal force. Acceration of the body , a g
M
2
N sin + f cos = mv Now M > m, therefore, the body with larger mass will
R have great acceleration and it will reach the ground first.
 EBD_7208
70 LAWS OF MOTION
93. (d) The net force on the block is zero, but action cannot 106. (c) When an elevator cabin falls down, it is accelerated down
cancel the reaction because these two act on different with respect to earth i.e. man standing on earth.
bodies. 107. (a)
94. (a) On a rainy day, the roads are wet. Wetting of roads
108. (b)
lowers the coefficient of friction between the types
109. (c) Impulse experienced by the body
and the road. Therefore, grip on a road of car reduces
and thus chances of skidding increases. = change in momentum
95. (d) = MV – (–MV)
96. (d) The man can exert force on block by pulling the rope. = 2MV.
The tension in rope will make the man move. Hence 110. (d) If rope of lift breaks suddenly, then acceleration
statement-1 is false. becomes equal to g so that tension T = m(g – g) = 0
97. (a) Friction causes wear & tear and loss of energy, so it is
an evil but without friction walking. Stopping a vehicle
etc. would not be possible. So it is necessary for us. 111. (a) y
98. (c) The assertion is true for a reason that when the car is 2 kg m2
driven at optimum speed. Then the normal reaction
8 m/sec Presultant
component is enough to provide the centripetal force.
99. (b) When a body is moving in a circle, its speed remains
12 m/sec
same but velocity changes due to change in the m1
c x
direction of motion of body. According to first law of /se 1 kg
motion, force is required to change the state of a 4m
3
body. As in circular motion the direction of velocity m
of body is changing so the acceleration cannot be
zero. But for a uniform motion acceleration is zero (for
rectilinear motion).
100. (c) In uniform circular motion, the direction of motion Presultant = 122 162
changes, therefore velocity changes.
= 144 256 = 20
As P = mv therefore momentum of a body also
changes uniform circular motion. m3v3 = 20 (momentum of third part)
101. (c) The purpose of bending is to acquire centripetal
force for circular motion. By doing so component of 20
or, m3 = = 5 kg
normal reaction will counter balance the centrifugal 4
force. 112. (d) According to law of conservation of momentum the
third piece has momentum
CRITICALTHINKING TYPE QUESTIONS
1 –(3iˆ 4j)
ˆ kg ms–1
102. (b) The apple will fall slightly away from the hand of his
brother in the direction of motion of the train due to Impulse = Average force × time
inertia of motion. When train is just going to stop, Impulse
the boy and his brother slows down with train but Average force
time
the apple which is in free fall continue to move with y
the same speed and therefore, falls slightly away
from the hand in the direction of motion of the train. Change in momentum
103. (d) Horizontal velocity of ball and person are same so time
1 4ˆj
both will cover equal horizontal distance in a given x
1 3iˆ
interval of time and after following the parabolic path –(3iˆ 4ˆj)kg ms –1 4j)
the ball falls exactly in the hand which threw it up. i +
10 –4 s (3
104. (a) After the stone is thrown out of the moving train, the (–
only force acting on it is the force of gravity i.e. its 1×
weight. 113. (b) From law of conservation of momentum
F = mg = 0.05 × 10 = 0.5 N.
MV = m1v1 + m2v2
105. (b) The pressure on rear side would be more due to fictious
force on the rear face. Consequently the pressure in Here, M = 100 kg, v = 104 m s–1
the front side would be lowered. m1 = 10 kg, v1 = 0
m2 = 90 kg, v2 = ?
LAWS OF MOTION 71
120. (b) When force is applied on m1
4
100 10 then T = m2a and when force is applied on m2, then
100 × 104 = 10 × 0 + 90 × v2 v2
90 T = m1a. Thus value of T is different for each case. And
v2 = 11.11 × 103 m s–1. it depends on whether the force is applied on m1, or m2.
114. (c) Change in momentum of the ball
= mv sin – (– mv sin ) 121. (b) Opposite force causes retardation.
= 2 mv sin 122. (d) In case (a) In case (b)
2vsin
= mg
g N mN N
F2
= weight of the ball × total time of flight
F1
115. (a) 49 mg mg
Mass 5 kg
9.8 mg Cos mN Sin mg Cos S in

When lift is moving downward mg mg

Apparent weight = 5(9.8 – 5) = 5 × 4.8 = 24 N
116. (a) N = m a sin + mg cos ......(1)
mg sin = F1 – N
also m g sin = m a cos ......(2)
N = mg cos
from (2) a = g tan
mg sin + mg cos = F1
sin 2 In second case (b)
N mg mg cos ,
cos
N + F2 = mg sin
mg mg cos – F2 = mg sin
or N
cos or F2 = mg sin – mg cos
m but F1 = 2F2
ac
os therefore mg sin + mg cos
N = 2(mg sin – mg cos )
ma
mg sin = 3 mg cos
cos
mg m or tan = 3 or = tan–1 (3 )
cos g
ma mg sin
123. (c) Frictional force is always opposite to the direction of
motion
117. (c) a=1
N

W E
m = 1000 kg
S

124. (a) mBg = s mAg { mAg = s mAg}

mB = s mA
Total mass = (60 + 940) kg = 1000 kg
or mB = 0.2 × 2 = 0.4 kg
Let T be the tension in the supporting cable, then
125. (d) Frictional force on the box f = mg
T – 1000g = 1000 × 1
Acceleration in the box
T = 1000 × 11 = 11000 N
118. (d) v = constant a = g = 5 ms–2

so, a = 0, Hence, Fnet = ma = 0 v2 = u2 + 2as

119. (d) When a rain drop falls down with the constant speed, 0 = 22 + 2 × (5) s
its weight is balanced by the upward viscous drag of 2
s=– w.r.t. belt
air and the force of buoyancy. Thus the net force acting 5
on it is zero. distance = 0.4 m
 EBD_7208
72 LAWS OF MOTION
126. (b) For the upward motion of the body
mg sin f1 F1

S/2 or, F1 = mg sin + mg cos

h
oot S/2 sin For the downward motion of the body,
Sm
mg sin – f 2 F2
S/2
ugh S/2 sin or F2 = mg sin – mg cos
Ro
F1 sin cos
For upper half of inclined plane =
F2 sin cos
v2 = u2 + 2a S/2 = 2 (g sin ) S/2 = gS sin
For lower half of inclined plane tan 2 3
3
0 = u2 + 2 g (sin – cos ) S/2 tan 2
– gS sin = gS ( sin – cos ) 131. (c) It is a case of uniform circular motion in which
2 sin = cos velocity and acceleration vectors change due to
change in direction. As the magnitude of velocity
2 sin
= = 2 tan remains constant, the kinetic energy is constant.
cos
132. (d) By Newton’s second law of motion
6 F = n(mv) = nmv
127. (a) a g [using v = u + at]
10
v2 20 2
6 6 133. (d) Angle of banking is tan =
= 0.06 rg 40 3 10
10 g 10 10
1
tan =
128. (c) fs 3
N = 30º
134. (a) As the ball, m = 10 g = 0.01 kg rebounds after striking
mg the wall
Change in momentum = mv – (–mv) = 2 mv
Inpulse = Change in momentum = 2mv

mg sin fs ( for body to be at rest) Impulse 0.54 N s 1

= 27 m s
2m 2 × 0.01 kg
m 10 sin 30 10
m 2.0 kg 135. (d) Acceleration of the system
T
2s 2s F 4g 2g 120 40 20
n a=
129. (a) We have 4 2 6
g(sin cos ) gsin
2 kg a
2s 2s n 2 10 ms 2
g (sin cos ) g sin From figure
T – 2g = 2a 2g
1 2 FBD of block
here = 45º n 2 or (1 1 / n ) T = 2 (a + g) = 2 (10 + 10)
1
= 40 N
130. (c) F1 N2 136. (a) At limiting equilibrium,
N1
2
F

= tan
f2 m
dy x 2 y
tan = =
mg sin
dx 2
mg sin
f1 mg cos mg cos (from question)
mg mg
Coefficient of friction = 0.5
LAWS OF MOTION 73

x2 2ma
0.5 Therefore m =
2 g a

x=+1 u v
3
142. (b) As we know, S = t
x 1 2
Now, y m
6 6
0 20
137. (a) Energy of both bodies is given by 2
t = 25 t = 2.5s
KE1 = F.S1
143. (a) The velocity should be such that the centripetal
KE2 = F.S2
acceleration is equal to the acceleration due to gravity
As force is equal
v2
S1 m1v12 m1 1 g or v gR
v1 v2 R
S2 m 2 v 22 m2 2
144. (d) tan = v2 / rg, tan = H / 1.5, r = 200 m, b = 1.5 m
138. (b) For the motion of both the blocks v = 36 km/hour = 36 × (5/18) = 10 m/s.
m1a = T – Putting these values, we get H = 0.075 m.
km1g
145. (b) It means that car which is moving on a horizontal road
m2g – T = m2a
& the necessary centripetal force, which is provided
a
T by friction (between car & road) is not sufficient.
k m1g m1
If is friction between car and road, then max speed of
k safely turn on horizontal road is determined from figure.

m2
a
N
m2g car of
f mass m
m 2g – k m1g
a= m1 m 2 mg

m 2 g – k m1g
m2g – T = (m2) m1 m 2
solving we get tension in the string N = mg ...(i)
m1mg (1 k )g mv 2
T= f ...(ii)
m1 m 2 r
F Where f is frictional force between road & car, N is the
139. (b) As we know, coefficient of friction =
N normal reaction exerted by road on the car. We know
ma a that
= (a = 7.35 m s–2 given)
mg g f sN s mg ......(iii)
7.35 where s is static friction
= 0.75 so from eq (ii) & (iii) we have
9.8
140. (a) As we know, |impulse| = |change in momentum| mv 2
s mg v2 s rg or v s rg
= |p2 – p1| r
= |0 – mv1| = |0 – 3 × 2| = 6 Ns & v max s rg

141. (a) Let upthrust of air be Fa then If the speed of car is greater than vmax at that road,
For downward motion of balloon then it will be thrown out from road i.e., skidding.
Fa = mg – ma 146. (a) The car over turn, when reaction on inner wheel of car
mg – Fa = ma is zero, i.e., first the inner wheel of car leaves the ground
For upward motion (where G is C.G of car, h is height of C.G from the ground,
Fa – (m – m)g = (m – m)a f1 & f2 are frictional force exerted by ground on inner &
 EBD_7208
74 LAWS OF MOTION
outer wheel respectively). The max. speed for no over Where is angle of banking of rail track, N is normal
turning is reaction exerted by rail track on rail.
G It is clear from the equation (i) & (ii) that N cos balance
N1 N2 the weight of the train & N sin provide the necessary
h centripetal force to turn.
f1 B
A If width of track is (OB) & h (AB) be height of outside
inner wheel f2 mg outer wheel
of car of track from the inside, then
of car
2a h v2 v2
tan or h .................(iv)
rg rg
gra So it is clear from the above analysis that if we increase
v max
h the height of track from inside by h metre then resultant
where r is radius of the path followed by car for turn & 2a force on rail is provided by railway track & whose
direction is inwards.
is distance between two wheels of car (i.e., AB)
147. (a) If the outside rail is h units higher than inside of rail track 1
148. (d) mv 2 m g or v (2 g )
as shown in figure then 2
N cos = mg....................(i) 149. (c) Force exerted by a car when passes through a bridge

Mv 2
F Mg .
N Ncos Train
r
1
sin A 150. (b) v gr 10 40 20 m s
N
O
h 151. (c) Tension at the highest point
mg B
inside mv2
Ttop – mg 2mg ( vtop = 3gr )
r
trrack
Rail

outside Tension at the lowest point

Tbottom = 2mg + 6mg = 8mg
mv 2 Ttop
N sin ...................(ii) 2mg 1
r .
Tbottom 8mg 4
v2
& tan ....................(iii)
rg