CHAPTER
:- NEWTON'S LAWS OF MOTION
(SECTION-A)
1. When
a body is stationary-
(A) There is no force acting on it
(B) The force acting on it not in
contact with it
(C) The combination of forces acting
on it balances each other
(D) The body is in vaccum
2. Newton's
Third law is equivalent to the-
(A) law of conservation of linear
momentum
(B) law of conservation of angular
momentum
(C) law of conservation of energy
(D) law of conservation of energy
and mass
3. In
the case of horse pulling a cart, the force that causes the horse to move
forward is the force that :
(A) the horse exerts on the ground
(B) the horse exerts on the cart
(C) the ground exerts on the horse
(D) the cart exerts on the horse
4. 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 ?
(A) 1N (B) 2N (C) 4N (D) 5N
5. A
force of 6N acts on a body at rest of mass 1 kg. During this time, the body
attains a velocity of 30 m/s. The time
for which the force acts on the body is-
(A) 10 seconds (B) 8 seconds
(C) 7 seconds (D) 5 seconds
6. A
bird weighing 1 kg sitting on the base of a wire mesh cage weighing 1.5 kg. The
bird starts flying in the cage. The weight of the bird cage assembly will be
(A) 1250 g (B) 1500 g
(C) 1750 g (D) None of
these
7. A boy having a mass equal to 40 kilograms is standing in an
elevator. The force felt by the feet of the boy will be greatest when the
elevator (g = 9.8 metre/sec2)-
(A) Stands still
(B) Moves downwards at a constant
velocity of 4 metre/sec.
(C) Accelerates downward with an
acceleration equal to 4 metres/sec2
(D) Accelerates upward with an
acceleration equal to 4 metres/sec2
8. A
man weighs 80 kg. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 m/s2. What would be the reading on the scale
? (g = 10 m/s2)
(A) 800 N (B) 1200
N
(C) Zero (D) 400 N
9. A
uniform thick rope of length 5m is kept on frictionless surface and a force of
5N is applied to one of its end. Find tension in the rope at 1m from this end-
(A) 1N (B) 3N (C) 4N (D) 5N
10. For the arraangement shown in figure, pulleys
and strings are ideal Find out the acceleration of 2 kg block
(A) 
(B) 
(C) 
(D)
11. 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.
(A) u cot q (B) u tan q
(C) u sin q (D) u cos q
12. The linear momentum P of a body varies
with time and is given by the equation P = x + yt2 where x and y are
constants. The net force acting on the body for a one dimensional motion is
proportional to-
(A) t2 (B)
a constant
(C) 
(D) t
13. The
average force necessary to stop a hammer having momentum 25 N-s in 0.05 second
is-
(A) 25 N (B) 50 N
(C) 1.25 N (D) 500 N
14. The
mass of ship is 2 × 107 kg. On applying a
force of 25 × 105 N, it is displaced
through 25m. After the displacement, the speed acquired by the ship will be:
(A) 12.5 m/s (B) 5 m/s
(C) 3.7 m/s (D) 2.5 m/s
15. A
body of mass 0.1 kg attains a velocity of 10 ms–1 in 0.1 s. The force acting on the body
is :
(A) 10 N (B) 0.01
N
(C) 0.1 N (D) 100 N
16. A diwali rocket is ejecting 0.05 kg of
gases per second at a velocity of 400 m/sec. The accelerating force on the
rocket is
(A)
20 dynes (B) 20 N
(C)
22 dynes (D) 1000 N
17. A
player takes 0.1 s in catching a ball of mass 150 g moving with velocity of 20
m/s. The force imparted by the ball on the hands of the player is :
(A) 0.3 N (B) 3 N
(C) 30 N (D) 300
N
18. A
lift of mass 1000 kg is moving upwards with an acceleration of 1 m/s2. The tension developed in the string,
which is connected to lift is : (g = 9.8 m/s2)
(A) 9800 N (B) 10800 N
(C) 11000 N (D) 10000 N
19. A
body of mass 0.1 kg attains a velocity of 10 ms–1. in 0.1 s. The force acting on the body
is :
(A) 10 N (B) 0.01
N
(C) 0.1 N (D) 100 N
20. A
monkey of mass 20 kg 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/s2)
(A) 25 m/s2 (B) 2.5 m/s2
(C) 5 m/s2 (D) 10 m/s2
21. A
spring toy weighing 1 kg on a spring balance suddenly jumps upward. A boy
standing near the toy notices that the scale of the balance reads 1.05 kg. In
this process the maximum acceleration of the toy is- (g = 10m sec–2)
(A) 0.05 m sec–2 (B)
0.5 m sec–2
(C) 1.05 m sec–2 (D)
1 m sec–2
22. A dynamometer D, is connected with to bodies of mass M = 6 kg and m
= 4kg. If two forces F = 20 N & F = 10 N are applied on masses according to
figure then reading of the dynamometer will be -
(A) 10 N (B) 20 N
(C) 6 N (D) 14
N
23. 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 (D) 10 N
24. A
block of mass m1 = 2 kg
on a smooth inclined plane at angle 30° is connected to a second block
of mass m2 = 3 kg by a cord
passing over a frictionless pulley as shown in fig. The acceleration of each
block is- (assume g = 10 m/sec2)
(A) 2 m/sec2 (B) 4 m/sec2
(C) 6 m/sec2 (D) 8 m/sec2
25. A
body, under the action of a force 
, acquires an acceleration of 1 ms–2. The mass of this body must be
(A) 
(B)
10 kg
(C) 20 kg (D) 
26. The mass of a lift is 500 kg. What will be the tension in its cable
when it is going up with an acceleration of 2.0 m/s2-
(A) 5000 N (B) 5600 N
(C) 5900 N (D) 6200 N
27. 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) w (B)
(C) 
(D)
28. Two
blocks of masses M1 and M2 are connected to each other through a
light spring as shown in figure. If we push mass M1 with force F and cause acceleration a1 in mass M1, what will be the magnitude of
acceleration in M2?
(A) F/M2 (B) F/(M1 + M2)
(C) a1 (D) (F – M1a1)/M2
29. Three
blocks of masses 2 kg, 3 kg and 5 kg are connected to each
other with light string and are then placed on a frictionless surface as shown
in the figure. The system is pulled by a force F = 10 N, then tension T1
=
(A) 1N (B)
5 N
(C) 8 N (D) 10 N
30. The
ratio of the weight of a man in a stationary lift & when it is moving
downward with uniform acceleration 'a' is 3 : 2 . The value of
'a' is : (g = acceleration. due to gravity)
(A)
(3/2) g (B) g
(C)
(2/3) g (D) g/3
31. A
body of mass 2 kg is hung on a spring balance mounted vertically in a lift. If
the lift descends with an acceleration equal to the acceleration due to gravity
'g' the reading on the spring balance will be-
(A) 2 kg (B) 4g
kg
(C) 2g kg (D) zero
32. A block of mass m is placed on a smooth
wedge of inclination q.
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 has a magnitude.
(A) mg (B)
mg/cos q
(C) mgcos q (D) mgtan q
33. Block B is moving towards right with constant velocity v0. Velocity of block A with respect to
block B is-
(Assume all pulleys and strings are
ideal)
(A)
v0/2 left (B) v0/2 right
(C) 3/2v0 right (D)
3/2v0 left
34. A
stone is dropped from a height h. It hits the ground with a certain momentum P.
If the same stone is dropped from a height 100% more than the previous height, the momentum when
it hits the ground will change by :
(A) 68% (B) 41%
(C) 200% (D) 100%
35. A person of mass 60 kg is inside a lift of mass
940 kg and presses the button on control panel. The lift starts moving upwards
with an acceleration 1.0 m/s2. If g = 10 ms–2, the tension in
the supporting cable is :
(A) 8600 N (B) 9680 N
(C) 11000 N (D) 1200 N
(SECTION-B)
36. A balloon with mass 'm' is descending
down with an acceleration 'a' (where a < g). How much mass should be removed
from it so that is starts moving up with an acceleration 'a' ?
(A) 
(B)
(C) 
(D) 
37. 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)
(A) 200 N (B) 100 N
(C) 140 N (D) 70 N
38. 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:
(A) 
(B) 
(C) 
(D) 
39. Assertion : On a rainy day, it is difficult
to drive a car or bus at high speed.
Reason : The value of coefficient of
friction is lowered due to wetting of the surface.
(A)
If both assertion and reason are true and the reason is the correct explanation
of the assertion.
(B)
If both assertion and reason are true but reason is not the correct explanation
of the assertion.
(C)
If assertion is true but reason is false.
(D)
If the assertion and reason both are false.
40. Assertion : Linear momentum of
a body changes even when it is moving uniformly in a circle.
Reason : Force required to move a body
uniformly along a straight line is zero.
(A) If
both assertion and reason are true and the reason is the correct explanation of
the assertion.
(B)
If both assertion and reason are true but reason is not the correct explanation
of the assertion.
(C)
If assertion is true but reason is false.
(D)
If the assertion and reason both are false.
41. A uniform rope of length l lies
on a table. If the coefficient of friction is 
, then the maximum length 
of the part of
this rope which can overhang from the edge of the table without sliding down is
(A)
(B) 
(C)
(D) 
42. A heavy uniform chain lies on a horizontal table-top. If the coefficient
of friction between the chain and table surface is 0.25, then the maximum
fraction of length of the chain, that can hang over one edge of the table is
(A)
20% (B)
25%
(C)
35% (D)
15%
43. Which one of the following statements is correct
(A)
Rolling friction is greater than sliding friction
(B)
Rolling friction is less than sliding friction
(C)
Rolling friction is equal to sliding friction
(D)
Rolling friction and sliding friction are same
44. The maximum speed that can be achieved without skidding by a car on a
circular unbanked road of radius R
and coefficient of static friction , is
(A)
(B) 
(C)
(D) 
45. Consider a car
moving along a straight horizontal road with a speed of 72 km/h. If the coefficient
of kinetic friction between the tyres and the road is 0.5, the shortest
distance in which the car can be stopped is 
(A)
30 m (B) 40 m
(C)
72 m (D) 20 m
46. On the horizontal surface of a truck (m = 0.6), a block of mass 1 kg is placed. If the truck is accelerating at the rate of 5m/sec2 then frictional force
on the block will be
(A) 5 N (B) 6 N
(C) 5.88 N (D) 8 N
47. When a body is moving on a surface, the force of friction is called
(A) Static
friction
(B) Dynamic
friction
(C) Limiting
friction
(D) Rolling
friction
48. A block of mass 
is resting on a
rough horizontal surface for which the coefficient of friction is 0.2. When a
force 
is applied, the
acceleration of the block will be 
(A) 
(B) 
(C) 
(D) 
49. When a body is
placed on a rough plane inclined at an angle 
to the
horizontal, its acceleration is
(A)
(B)
(C)
(D)
50. Match column I with
column II and select the correct option from the given codes.
|
|
Column I |
|
Column II |
|
P. |
Definition of
force |
(i) |
Newton's third
law |
|
Q. |
Measure of force |
(ii) |
Impulse |
|
R. |
Effect of force |
(iii) |
Newton's second
law |
|
S. |
Recoiling of gun |
(iv) |
Newton's first
law |
(A)
P – (ii), Q – (i), R – (iii), S – (iv)
(B)
P – (i), Q – (ii), R – (iii), S – (iv)
(C)
P – (iv), Q – (iii), R – (ii), S – (i)
(D)
P – (iv), Q – (ii), R – (iii), S – (i)
CHAPTER :- NEWTON'S LAWS OF MOTION
ANSWER KEY
1. (C) 2. (A) 3. (C) 4. (C) 5. (D) 6. (B) 7. (D)
8. (B) 9. (C) 10. (A) 11. (B) 12. (D) 13. (D) 14. (D)
15. (A) 16. (B) 17. (C) 18. (B) 19. (A) 20. (B) 21. (B)
22. (D) 23. (B) 24. (B) 25. (D) 26. (C) 27. (C) 28. (D)
29. (C) 30. (D) 31. (D) 32. (B) 33. (B) 34. (B) 35. (C)
36. (A) 37. (B) 38. (C) 39. (A) 40. (B) 41. (C) 42. (A)
43. (B) 44. (D) 45. (B) 46. (A) 47. (B) 48. (A) 49. (C)
50. (C)
SOLUTIONS
SECTION-A
1. (C)
2. (A)
3. (C)
Sol. While the horse pulling a cart, the horse exerts a force on the
ground, therefore from the third law of newton, the ground will also exerts a
force on the horse that causes the horse to move forward.
4. (C)
Sol.
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]
Þ N = 4 Newton
5. (D)
Sol. v
= u + at
Þ 30 = 0 + × t
Þ 30 = × t
Þ t = 5 sec.
6. (B)
Sol. When
bird starts flying in the cage, the weight of the bird is not measured.
Therefore, weight of the bird cage assembly is now 1.5 kg or 1500 g.
7. (D)
Sol. when acclerated upward N – mg = ma
Þ N = m (g + a)
8. (B)
Sol. Key Idea : When lift is moving upwards, it weighs more
than actual weight of man by a factor of ma.
Mass of man M = 80 kg
acceleration of lift, a = 5 m/s2
When lift is moving upwards, the
reading of weighing scale will be equal to R. The equation of motion gives
R – Mg = Ma
or R
= Mg + Ma = M (g + a)
\ R = 80 (10 + 5) = 80 × 15 = 1200 N
9. (C)
Sol.
Equation of motion
F – T = × a ....(1)
T = × a ....(2)
Solving (1) and (2)
T = 4 N
10. (A)
Sol
11. (B)
Sol. Since rod is rigid, its length can’t increase.
\ velocity of approach of A and B point of rod
is zero.
Þ u
sin q – v cos q = 0
Þ v
= u tan q
12. (D)
sol. F = = 0 + 2yt
So, F is proportional to t
13. (D)
Sol. PL = 25, Pf = 0, Dt = 0.05
F = = = = 500 N
14. (D)
Sol. Here : Mass of ship m = 2 × 107 kg,
Force F - 25 × 105 N
Displacement s = 25 m
According to the Newton’s second law
of motion
F = ma
or a
= = 
The relation for final velocity is
v2 = u2 + 2as
or v2
= 0 + 2 × (12.5 × 10–2) × 25
or v2
= 
= 2.5 m/s
15. (A)
Sol. F = ma = m × 
= 0.1 × 
= 10 N
16. (B)
Sol. 
17. (C)
Sol. Key Idea : The force imparted (or impulse) by
the ball to the hands of the player equal to the rate of change of linear
momentum.
Force imparted = Rate of change of
momentum
or F
= 
or F
=
or F
= 
Here m = 0.150 kg, v1 = 20 m/s v2 = 0
Dt = 0.1 s
\ F
= 
= 30 N
18. (B)
Sol. Key Idea : The tension in the string during
upward motion increases from weight of lift due to its upward acceleration.
when lift moves upward with same
acceleration then
T – mg = ma
or T
= m (g + a)
Given m = 1000 kg, a = 1 m/s2, g = 9.8 m/s2
Thus T = 1000 (9.8 + 1)
= 1000 × 10.8
= 10800 N
19. (A)
Sol. F = ma = m ×
= 0.1 × = 10 N
20. (B)
Sol. Maximum bearable tension in the rope
T = 25 × 10 = 250 N
From the figure,
T – mg = ma
or a
= 
Given
m = 20 kg, g = 10 m/s2,
T = 250 N
Hence a = 
= 
= 2.5 m/s2
21. (B)
Sol. 
1.05 g – 1 × g = 1 × a Þ a = 0.5 m/s2
22. (D)
Sol. 1 ms–2
F – T = m.a
20 – T = 6 (1)
T = 14 N
23. (B)
Sol Since downward force along the inclined
plane = 
= 25 N
24. (B)
Sol. a = 
a = 
= 4 m/s2
25. (D)
Sol. Key Idea : According to
Newton's second law of motion force = mass × acceleration.
Here
| F | = 
= 
N a = 1
ms–2
\ m = 
= kg
26. (C)
Sol. T – mg = ma Þ T = m (g + a) = 500 (9.8 + 2) = 5900 N
27. (C)
Sol.
w – f = ma w – ma = f
w = f w = f
w = f
28. (D)
Sol. 
F – k x = M1 a1 [Newton’s II law for M1]
kx = M2a2 [Newton’s II law for M2]
By adding both equations.
F = M1a1 + M2a2
Þ a2 =
29. (C)
Sol.
so T1 = (3 + 5) (1)system = 8 × 1 = 8 N
30. (D)
Sol. Weight
of man in stationary lift is mg.
mg – N = ma [Newton’s II law for man]
Þ N =m (g – a)
Weight of man in moving lift is
equal to N.
Þ Þ a =
31. (D)
Sol.
Reading of spring balance is tension
2g – T = 2 a
2g – 2a = T (a = g)
0 = T
32. (B)
Sol. ma cosq = mg
sinq
a
= g tanq
N
= mg cosq + ma sinq
=
mg cosq
+ 
=
33. (B)
Sol. 
l'1
+ l'2 + l'3 = 0
(–v + v0) + (–v + v0) + (0 + v0) = 0
3 v0 = 2 v Þ v = 
Þ VAB = VA – VB = V – V0
= 
34. (B)
Sol. When
stone hits the ground momentum 
when some stone dropped from 2h
(100% of initial) then momentum 
=
Which is is changed by 41% of
initial.
35. (C)
Sol. 
T – 1000g = 1000 × 1
T = 1000 × 11
SECTION-B
36. (A)
Sol. mg
– F = ma .....(1)
F – (m – m')g = (m – m')a
from (1)
F – mg + m¢g = ma – m¢a
mg – ma – mg + m¢g = ma – m¢a
m¢(g + a) = 2ma
m¢ = 
37. (B)
Sol. 
F = 100 N
38. (C)
Sol. F = 
Þ Kt =
Þ 3P – P = 
Þ
t = 
39. (A)
Sol. The correct option
is A Both Assertion and Reason are correct and Reasion is the correct
explanation for Assertion
The value of coefficient of friction is lowered due to wetting of the surface.
Hence the frictional force becomes less and vehicle takes longer to stop after
sliding for some distance.
40. (B)
Sol. In
uniform circular motion of a body the speed remains constant but velocity
changes as direction of motion changes.
As linear momentum = mass × velocity, therefore linear
momentum of a body changes in a circle.
On the other hand, if the body is moving uniformly along
a straight line then its velocity remains constant and hence acceleration is
equal to zero. So force is equal to zero.
41. (C)
Sol. For given condition we can apply
direct formula 
42. (A)
Sol. 
of l.
43. (B)
44. (D)
Sol. In the given condition the required centripetal force is provided by
frictional force between the road and tyre.
45. (B)
Sol. 
46. (A)
Sol. 
Pseudo
force on the block = 
Pseudo
is less then limiting friction hence static force of friction = 5 N.
47. (B)
48. (A)
Sol. 
Kinetic
friction =
Acceleration
of the block 
49. (C)
50. (C)
Sol. Newton's first law
defines force. Newton's second law given us a measure of force. Impulse given us
the effect of force. Recoiling of gun is accounted for by Newton's 3rd law.
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