NCERT Solutions for Class 11 Physics Chapter 4 Laws of Motion

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Physics
Chapter : 5
Chapter Name : Laws of Motion

Q5.1 Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed,
(b) a cork of mass 10 g oating on water,
(c) a kite skillfully held stationary in the sky,
(d) a car moving with a constant velocity of 30 km/h on a rough road,
(e) a high-speed electron in space far from all material objects, and free of electric and
magnetic elds.

Answer. (a) Zero net force
The raindrop is falling with a constant speed. Hence, it acceleration is zero.
Newton's second law of motion, the net force acting on the rain drop is zero.
(b) Zero net force
As per The weight of the cork is acting downward. It is balanced by the buoyant force exerted
by the water in the upward direction. Hence, no net force is acting on the oating cork.
(c) Zero net force
The kite is stationary in the sky, i.e., it is not moving at all. Hence, as per Newton's rst
law of motion, no net force is acting on the kite.
(d) Zero net force
The car is moving on a rough road with a constant velocity. Hence, its acceleration is
zero. As per Newton's second law Of motion, no net force is acting on car.
(e) Zero net force
The high speed electron is free from the in uence of all elds. Hence, no net force is
acting on the electron.

Page : 109 , Block Name : Exercise

Q5.2 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude
of the net force on the pebble,
(a) during its upward motion,
(b) during its downward motion,
(c) at the highest point where it is momentarily at rest. Do your answers change if the pebble
was thrown at an angle of 45° with the horizontal direction?
Ignore air resistance

Answer. 0.5 N, in vertically downward direction, in all cases
Acceleration due to gravity, irrespective of the direction of motion of an object, always

acts downward. The gravitational force is the only force that acts on the pebble in all
three cases. Its magnitude is given by Newton's second law of motion as:
F=m×a
Where,
F = Net force
m = Mass of the pebble = 0.05kg
a = g = 10m / s 2
∴ F = 0.05 × 10 = 0.5N
The net force on the pebble in all three cases is 0.5 N and this force acts in the
downward direction.
If the pebble is thrown at an of 450 with the horizontal, it will have both the
horizontal and vertical components Of velocity. At the highest point, only the vertical
component of velocity becomes zero. However, the pebble will have the horizontal
component of velocity throughout its motion. This component of velocity produces no
effect on the net force acting on the pebble.

Page : 109 , Block Name : Exercise

Q5.3 Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of 36
km/h,
(c) just after it is dropped from the window of a train accelerating with 1 m s − 2 ,
(d) lying on the oor of a train which is accelerating with 1 m s − 2, the stone being at rest
relative to the train. Neglect air resistance throughout.

Answer. (a) l N; vertically downward
Mass of the stone, m 0.1 kg
Acceleration Of the stone, a = g = 10 m / s 2
As per Newton's second law of motion, the net force acting on the stone,
F ma = mg
=0.1 x 10=1 N
(b) l N; vertically downward
The train is moving with-I a constant velocity. Hence, its acceleration is zero in the
direction of its motion, i.e., in tie horizontal direction. Hence, no force is acting on ene
stone in the horizontal direction.
The net force acting on the stone is because of acceleration due to gravity and it always
acts vertically downward. The magnitude of this force is 1 N.
(c) l N; vertically downward
It is given that the train is accelerating at the rate Of 1 m / s 2
Therefore, the net force acting on the stone, F' = ma — -0.1 x 1-0.1 N
This force is acting in the horizontal direction. Now, when the stone is dropped, the
horizontal force F,' stops acting on the stone. This is because of the fact that the force
acting on a body at an instant depends on the situation at that instant and not on earlier

Situations.
Therefore, the net force acting on the stone is given only by acceleration due to gravity.
F = mg = 1N
This force acts vertically downward.
(d) O.1 N, in the direction of motion of the train
The weight of the stone is balanced by the normal reaction of the oor. The only acceleration
is provided by the horizontal motion of the train.
Acceleration of the train, a = 0.1 m / s 2
The net force acting on the stone will be in the direction of motion of the train. Its
magnitude is given by:
F = ma
-0.1 x 1-0.1 N

Page : 109 , Block Name : Exercise

Q5.4 One end of a string of length l is connected to a particle of mass m and the other to a
small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net
force on the particle (directed towards the centre) is :
(i) T
mw 2
(ii) T − l
mw 2
(iii) T + l
(iv) 0

(i)When a particle connected to a string revolves in a circular path around a centre, the
centripetal force is provided by the tension produced in the string. Hence, in the given case,
the net force on the particle is the tension T, i.e
mv 2
F=T= l
Where F is the net force acting on the particle.

Page : 110 , Block Name : Exercise

Q5.5 A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with
a speed of 15 m s − 1. How long does the body take to stop ?

Answer. Retarding force, F = —50 N
Mass of the body, m = 20 kg
Initial velocity of the body, u = 15 m/s
Final velocity Of the body, V = O
using Newton's second law of motion, the acceleration (a) produced in the body can be
calculated as:

F = ma
− 50 = 20 × a
− 50
∴ a = 20 = − 2.5m / s 2
using the rst equation Of motion, the time (t) taken by the body to come to rest can be
calculated as:
v = u + at
−u − 15
∴ t = a = − 2.5 = 6s

Page : 110 , Block Name : Exercise

Q5.6 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s-1 to 3.5
m s − 1 in 25 s. The direction of the motion of the body remains unchanged. What is the
magnitude and direction of the force ?

0.18N i in the direction of motion of the body
Mass of the body, m = 3kg
Initial speed of the body, u = 2m / s
Final speed of the body, v = 3.5m / s
Time, t = 25s
Using the first equation of motion, the acceleration (a) produced in the body can be
calculated as:
v = u + at
v−u
∴a= t
3.5 − 2 1.5
= 25
= 25 = 0.06m / s 2
As per Newton's second law of motion, force is given as:
F = ma
= 3 × 0.06 = 0.18N
Since the application of force does not change the direction of the body, the net force
acting on the body is in the direction of its motion.

Page : 110 , Block Name : Exercise

Q5.7 A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the
magnitude and direction of the acceleration of the body.

2m / s 2, at an angle of 37 ∘ with a force of 8N
Mass of the body, m = 5kg
The given situation can be represented as follows:

The resultant Of two forces is given as:
R= √(8) 2 + ( − 6) 2 = √64 + 36 = 10N
θ is the angle made by R with the force of 8N

∴ θ = tan − 1 ( ) −6
8
= − 36.87 ∘

The negative sign indicates that 9 Is in the clockwise direction with respect to the force
of magnitude 8 N.
As per Newton's second law of motion, the acceleration (a) of the body is given as:
F = ma
F 10
∴ a = m = 5 = 2m / s 2

Page : 110 , Block Name : Exercise

Q5.8 The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the
middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is
the average retarding force on the vehicle ? The mass of the three-wheeler is 400 kg and the
mass of the driver is 65 kg.

Initial speed of the three-wheeler, u = 36km / h
Final speed of the three-wheeler, v = 10m / s
Time, t = 4s
Mass of the three- wheeler, m = 400kg
Mass of the driver, m ′ = 65kg
Total mass of the system, M = 400 + 65 = 465kg
Using the first law of motion, the acceleration (a) of the three-wheeler can be calculated
as:
v = u + at
v−u 0 − 10
∴a= t
= 4
= − 2.5m / s 2
The negative sign indicates that the velocity of the three-wheeler is decreasing with

time.
using Newton's second law of motion, the net force acting on the three-wheeler can be
calculated as:
F = Ma
= 465 × ( − 2.5) = − 1162.5N
The negative sign indicates that the force is acting against the direction of motion of the
Three-wheeler.

Page : 110 , Block Name : Exercise

Q5.9 A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of
5.0 m s − 2. Calculate the initial thrust (force) of the blast.

Mass of the rocket, m = 20, 000kg
Initial acceleration, a = 5m / s 2
Acceleration due to gravity, g = 10m / s 2
Using Newton's second law of motion, the net force (thrust) acting on the rocket is given
by the relation:
F − mg = ma
F = m(g + a)
= 20000 × (10 + 5)
= 20000 × 15 = 3 × 10 5N

Page : 110 , Block Name : Exercise

Q5.10 A body of mass 0.40 kg moving initially with a constant speed of 10 m s − 1 to the north
is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the
force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its
position at t = –5 s, 25 s, 100 s.z

Mass of the body, m = 0.40kg
Initial speed of the body, u = 10m / s due north
Force acting on the body, F = − 8.0N
F − 8.0
Acceleration produced in the body, a = m = 0.40 = − 20m / s 2
(i) at t = − 5s
Acceleration, a ′ = 0 and u = 10m / s
1
s = ut + 2 a ′t 2
= 10 × ( − 5) = − 50m
(ii) at t = 25s

Acceleration, a ′′ = − 20m / s 2 and u = 10m / s
1
S ′ = ut ′ + 2 a ′′t 2
1
= 10 × 25 + 2 × ( − 20) × (25) 2

= 250 + 6250 = − 6000m
(iii) Δtt = 100s
For 0 ≤ t ≤ 30s
a = − 20m / s 2
u = 10m / s
1
s 1 = ut + 2 a ′′t 2
1
= 10 × 30 + 2 × ( − 20) × (30) 2
= 300 − 9000
= − 8700m
For 30 ′ < t ≤ 100s
As per the first equation of motion, for t = 30s , final velocity is given as:
v = u + at
= 10 + ( − 20) × 30 = − 590m / s
velocity of the body after 30s = − 590m / s
For motion between 30s to 100s, i.e., in 70s :
1
s 2 = vt + 2 a ′′t 2
= − 590 × 70 = − 41300m
∴ Total distance, x ′′ = s 1 + s 2 = − 8700 − 41300 = − 50000m

Page : 110 , Block Name : Exercise

Q5.11 A truck starts from rest and accelerates uniformly at 2.0 m s − 2. At t = 10 s, a stone is
dropped by a person standing on the top of the truck (6 m high from the ground). What are the
(a) velocity, and (b) acceleration of the stone at t = 11s ? (Neglect air resistance.)

(a) 22.36m / s 1 at an angle of 26.57 ∘ with the motion of the truck

(b) 10m / s 2
(a) Initial velocity of the truck, u = 0
Acceleration, a = 2m / s 2
Time, t = 10s
As per the first equation of motion, final velocity is given as:
v = u + at
= 0 + 2 × 10 = 20m / s

The final velocity of the truck and hence, of the stone is 20m / s .

( )
At t = 11s, the horizontal component v x of velocity, in the absence of air resistance,

remains unchanged, i.e.,
v x = 20m / s

( )
The vertical component v y of velocity of the stone is given by the first equation of

motion as:
v y = u + a yδt
Where, δt = 11 − 10 = 1s and a y = g = 10m / s 2
∴ v y = 0 + 10 × 1 = 10m / s
The resultant velocity (v) of the stone is given as:

2 2
v=
√v + v
x y

= √20 2 + 10 2 = √400 + 100
= √500 = 22.36m / s
Let θ be the angle made by the resultant velocity with the horizontal component of
velocity, v x

∴ tanθ =
() vy
vx

()
θ = tan − 1 20
10

= tan − 1(0, 5)
= 26.57 ∘
(b)When the stone is dropped from the truck, the horizontal force acting on it becornes
zero. However, the stone continues to move under the in uence of gravity. Hence, the
acceleration of the stone is 10 m/s2 and it acts vertically downward.

Page : 110 , Block Name : Exercise

Q5.12 A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into
oscillation. The speed of the bob at its mean position is 1 m s − 1 . What is the trajectory of the
bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean

position

Answer. (a) Vertically downward
(b) Parabolic path
(a) At the extreme position, the velocity of the bob becomes zero. If the string is cut at
this moment, then the bob will fall vertically on the ground.
(b)At the mean position, the velocity of the bob is 1 m/s. The direction of this velocity is
tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position,
then it will trace a projectile path having the horizontal component of velocity only.
Hence, it will follow a parabolic path.

Page : 110 , Block Name : Exercise

Q5.13 A man of mass 70 kg stands on a weighing scale in a lift which is moving
(a) upwards with a uniform speed of 10 m s − 1 ,
(b) downwards with a uniform acceleration of 5 m s − 2 ,
(c) upwards with a uniform acceleration of 5 m s − 2 . What would be the readings on the scale
in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under
gravity ?

(a) Mass of the man, m = 70kg
Acceleration, a = 0
Using Newton's second law of motion, we can write the equation of motion as:
R − mg = ma
Where, ma is the net force acting on the man.
As the lift is moving at a uniform speed, acceleration a = 0
∴ R = mg
= 70 × 10 = 700N
700 700
¿Reading on the weighing scale = g = 10 = 70kg
(b) Mass of the man, m = 70kg
Acceleration, a = 5m / s 2 downward
Using Newton's second law of motion, we can write the equation of motion as:
R + mg = ma
R = m(g − a)
= 70(10 − 5) = 70 × 5
= 350N
350 350
∴ Reading on the weighing scale = g = 10 = 35kg
(c) Mass of the man, m = 70kg
Acceleration, a = 5m / s 2 upward
Using Newton's second law of motion, we can write the equation of motion as:

R − mg = ma
R = m(g + a)
= 70(10 + 5) = 70 × 15
= 1050N
1050 1050
∴ Reading on the weighing scale = g
= 10 = 105kg
(d) When the lift moves freely under gravity, acceleration a = g
Using Newton's second law of motion, we can write the equation of motion as:
R + mg = ma
R = m(g − a)
= m(g − g) = 0
0
∴ Reading on the weighing scale = g = 0kg

The man will be in a state of weightlessness.

Page : 110 , Block Name : Exercise

Q5.14 Figure 5.16 shows the position-time graph of a particle of mass 4 kg. What is the (a)
force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s ? (Consider one-
dimensional motion only)

Answer. (a) For t < 0
It can be observed from the given graph that the position of the particle is coincident with the
time axis. It indicates that the displacement of the particle in this time interval is zero. Hence,
the force acting on the particle is zero.
For t > 4 s
It can be observed from the given graph that the position of the particle is parallel to the time
axis. It indicates that the particle is at rest at a distance of 3 m from the origin. Hence, no force
is acting on the particle.
For 0 < t < 4
It can be observed that the given position-time graph has a constant slope. Hence, the
acceleration produced in the particle is zero. Therefore, the force acting on the particle is zero.
(b) At t = 0
Impulse = Change in momentum
= mv – mu
Mass of the particle, m = 4 kg
Initial velocity of the particle, u = 0

3
Final velocity of the particle, v = 4 m / s

( ) 3
∴Impulse = 4 4 − 0 = 3kgm / s

At t = 4 s
3
Initial velocity of the particle, u = 4 m / s
Final velocity of the particle, v = 0

( )
∴ Impulse = 4 0 − 4
3
= − 3kgm / s

Page : 110 , Block Name : Exercise

Q5.15 Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface
are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along
the direction of string. What is the tension in the string in each case?

Horizontal force, F = 600N
Mass of body A, m 1 = 10kg
Mass of body B, m 2 = 20kg
Total mass of the system, m = m 1 + m 2 = 30kg
Using Newton's second law of motion, the acceleration (a) produced in the system can
be calculated as:
F = ma
F 600
∴ a = m = 30 = 20m / s 2
V-Vhen force F is applied on body A:

The equation of motion can be written as:
F − T = m 1a
∴ T = F − m 1a
= 600 − 10 × 20 = 400N… (i)
When force F is applied on body B:

The equation of motion can be written as:
F − T = m 2a
T = F − m 2a
∴ T = 600 − 20 × 20 = 200N… (ii)

Page : 110 , Block Name : Exercise

Q5.16 Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string
that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the
string when the masses are released.

Answer. The given system of two masses and a pulley can be represented as shown in the
following gure:

Smaller mass, m 1 = 8kg
Larger mass, m 2 = 12kg
Tension in the string = T
Mass m 2′ owing to its weight, moves downward with acceleration a , and mass m 1 moves
upward.
Applying Newton's second law of motion to the system of each mass:
For mass m 1 .
The equation of motion can be written as:
T − m 19 = ma…(i)
Eor mass m 2 .
The equation of motion can be written as:
m 2g − T = m 2a…(ii)
Adding equations (i) and (ii), we get:

(m2 − m1 )g = (m1 + m2 )a

∴a=
( )m2 − m1
m1 + m2
g

= ( ) 12 − 8
12 + 8
4
× 10 = 20 × 10 = 2m / s 2

Therefore, the acceleration of the masses is 2m / s 2.
Substituting the value of a in equation (ii), we get:

m 2g − T = m 2 m + m g
1 2 ( )
m2 − m1

(
T = m2 − m + m
1 2
m 22 − m 1m 2

) g

=
( ) 2m 1m 2
m1 + m2
g

= ( 2 × 12 × 8
12 + 8 ) × 10

2 × 12 × 8
= 20
× 10 = 96N
Therefore, the tension in the string is 96 N.

Page : 111 , Block Name : Exercise

Q5.17 A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates
into two smaller nuclei the products must move in opposite directions.

Let m, m 1t and m 2 be the respective masses of the parent nucleus and the two daughter
nuclei. The parent nucleus is at rest.
Initial momentum of the system (parent nucleus) =0
Let v 1 and v 2 be the respective velocities of the daughter nuclei having masses m 1 and
m2 .
Total linear momentum of the system after disintegration = m 1v 1 + m 2v 2
According to the law of conservation of momentum:
Total initial momentum = Total final momentum
0 = m 1v 1 + m 2 + v 2
− m 2v 2
v1 = m1

Here, the negative sign indicates that the fragments of the parent nucleus move in
directions opposite to each other.

Page : 111 , Block Name : Exercise

Q5.18 Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m
s − 1 collide and rebound with the same speed. What is the impulse imparted to each ball due to
the other ?

Mass of each ball = 0.05kg
Initial velocity of each ball = 6m / s
Magnitude of the initial momentum of each ball, p i = 0.3kgm / s
After collision, the balls change their directions of motion without changing the
magnitudes of their velocity.
Final momentum of each ball, p f = − 0.3kgm / s
Impulse imparted to each ball = Change in the momentum of the system
= pf − pi
= − 0.3 − 0.3 = − 0.6kgm / s
The negative sign indicates that the impulses imparted to the balls are opposite in
direction.

Page : 111 , Block Name : Exercise

Q5.19 A shell of mass 0.020 kg is red by a gun of mass 100 kg. If the muzzle speed of the shell
is 80 m s − 1, what is the recoil speed of the gun ?

Mass of the gun, M = 100kg
Mass of the shell, m = 0.020kg
Muzzle speed of the shell, v = 80m / s
Recoil speed of the gun = V
Both the gun and the shell are at rest initially.
Initial momentum of the system = mv − MV
Here, the negative sign appears because the directions of the shell and the gun are
opposite to each other.
According to the law of conservation of momentum:
Final momentum = Initial momentum
mv
∴V=
M
0.020 × 80
= = 0.016m / s
100 × 1000

Page : 111 , Block Name : Exercise

Q5.20 A batsman de ects a ball by an angle of 45° without changing its initial speed which is
equal to 54 km/h. What is the impulse imparted to the ball ? (Mass of the ball is 0.15 kg.)

Answer. The given situation can be represented as shown in the followIng gure.

Where,
AO = Incident path of the ball
OB = Path followed by the ball after deflection
∠AOB = Angle between the incident and deflected paths of the ball = 45
∠AOP = ∠BOP = 22.5 ∘ = θ
Initial and final velocities of the ball = v
Horizontal component of the initial velocity = vcosθ along RO
Vertical component of the initial velocity = vsinθ along PO
Horizontal component of the final velocity = vcosθ along os
vertical component of the final velocity = vsinθ along op
The horizontal components of velocities suffer no change. The vertical components of
velocities are in the opposite directions.
∴ Impulse imparted to the ball = Change in the linear momentum of the ball
= mvcosθ − ( − mvcosθ)
= 2mvcosθ
Mass of the ball, m = 0.15kg
Velocity of the ball, v = 54km / h = 15m / s
∴ Impulse = 2 × 0.15 × 15cos22.5 ∘ = 4.16kgm / s

Page : 111 , Block Name : Exercise

Q5.21 A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius
1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string ?
What is the maximum speed with which the stone can be whirled around if the string can

withstand a maximum tension of 200 N ?

Mass of the stone, m = 0.25kg
Radius of the circle, r = 1.5m
40 2
Number of revolution per second, n = 60 = 3 rps
Angular velocity, ω = 2πn
The centripetal force for the stone is provided by the tension T, in the string, i.e.,
T = F Cemiripetal
mv 2
= = mrω 2 = mr(2πn) 2
r

(
= 0.25 × 1.5 × 2 × 3.14 × 3
2
) 2

= 6.57N
Maximum tension in string,T max = 200N
2
mv max
T max = r
T max × r
∴ v max =
√ m

√
200 × 1.5
=
0.25

= √1200 = 34.64m / s
Therefore, the maximum speed of the stone is 34.64 m/s.

Page : 111 , Block Name : Exercise

Q5.22 If, in Exercise 5.21, the speed of the stone is increased beyond the maximum
permissible value, and the string breaks suddenly, which of the following correctly describes
the trajectory of the stone after the string breaks : (a) the stone moves radially outwards, (b)
the stone ies off tangentially from the instant the string breaks, (c) the stone ies off at an
angle with the tangent whose magnitude depends on the speed of the particle ?

Answer. (b)When the string breaks, the stone will move in the direction of the velocity at that
instant. According to the rst law of motion, the direction of velocity vector is tangential
to the path Of the stone at that instant. Hence, the stone will y Off tangentially from the
instant the string breaks.

Page : 111 , Block Name : Exercise

Q5.23 Explain why
(a) a horse cannot pull a cart and run in empty space,

(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
(c) it is easier to pull a lawn mower than to push it,
(d) a cricketer moves his hands backwards while holding a catch.

Answer. (a) In order to pull a cart, a horse pu±ies the ground backward with some force. The
ground in turn exerts an equal and opposite reaction force upon the feet of the horse.
This reaction force causes the horse to move forward.
An empty *'ace is devoid Of any such reaction force. Therefore, a horse cannot pull a
cart and run in empty space.
(b) When a speeding bus stops suddenly, the lower portion Of a passenger's body, which
is in contact with the seat, suddenly comes to rest. However, the upper portion tends to
remain in motion (as per the rst law of motion). As a result, the passenger's upper
body is thrown forward in the direction in which the bus was moving.
(c) While pulling a lawn mower, a force at an angle 9 is applied on it, as shown in the
following gure.

The vertical cornponent of this applied force acts upward. This reduces the effective
weight of the mower.
On the other hand, while pushing a lawn mower, a force at an angle 9 is applied on it, as
shown in the following gure.

In this case, the vertical component of the applied force acts in the direction of the
weight Of the mower. This increases the effective weight of the mower.
Since the effective weight of the lawn mower is lesser in the rst case, pulling the lawn
mower is easier than pushing it.

(d) According to Newton's second law of motion, we have the equation of motion:
Δv
F = ma = m Δt

Where,
F = Stopping force experienced by the cricketer as he catches the ball
m = Mass of the ball
Δt = Time of impact of the ball with the hand
It can be inferred from equation (i) that the impact force is inversely proportional to the
impact time, i.e.,
1
F ∝ Δt
Equation (ii) shows that the force experienced by the cricketer decreases if the time of
impact increases and vice versa.
While taking a catch, a cricketer moves his hand backward so as to increase the time of
impact (Δt). This is turn results in the decrease in the stopping force, thereby preventing
the hands of the cricketer from getting hurt.

Page : 111 , Block Name : Exercise

Q5.24 Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable
physical context for this motion. What is the time between two consecutive impulses received
by the body ? What is the magnitude of each impulse ?

Answer. A ball rebounding between two walls located between at x O and x — 2 cm; after every

2 s, the ball receives an impulse of magnitude 0.08 x 10 − 2 kg m/s from the walls
The given graph shows that a body changes its direction of motion after every 2 s.
Physically, this situation can be visualized as a ball rebounding to and fro between two
stationary walls situated between positions x = O and x = 2 cm. Since the slope Of the x-
t graph reverses after every 2 s, the ball collides with a wall after every 2 s. Therefore,
ball receives an impulse after every 2 s.
Mass of the ball, m = 0.04 kg
( 2 − 0 ) × 10 − 2
u= (2−0)
= 10 − 2m / s
Velocity of the ball before collision, u = 10 − 2 m/s
Velocity of the ball after collision, v = − 10 − 2 m/s
(Here, the negative sign arises as the ball reverses its direction of motion.)
Magnitude Of impulse = Change in momentum

= | mv − mu |
= | 0.04(v − u) |

| (
= 0.04 − 10 − 2 − 10 − 2 )|
= 0.08 × 10 − 2kgm / s
= | mv − mu |
= | 0.04(v − u) |

| (
= 0.04 − 10 − 2 − 10 − 2 )|
= 0.08 × 10 − 2kgm / s

Page : 111 , Block Name : Additional Exercises

Q5.25 Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt
that is accelerating with 1 m s − 2. What is the net force on the man? If the coef cient of static
friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can
the man continue to be stationary relative to the belt ? (Mass of the man = 65 kg.)

Mass of the man, m = 65kg
Acceleration of the belt, a = 1m / s 2
Coefficient of static friction, μ = 0.2
The net force F t acting on the man is given by Newton's second law of motion as:
F ret = ma = 65 × 1 = 65N
The man will continue to be stationary with respect to the conveyor belt until the net
force on the man is less than or equal to the frictional force f st exerted by the belt, i.e.,
F ′net = f s

ma ′ = μmg
∴ a ′ = 0.2 × 10 = 2m / s 2
Therefore, the maximum acceleration of the belt up to which the man can stand
stationary is 2m / s 2 .

Page : 111 , Block Name : Additional Exercises

Q5.26 A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The
net forces at the lowest and highest points of the circle directed vertically downwards are :
[Choose the correct alternative]

T 1 and v 1 denote the tension and speed at the lowest point. T 2 and v 2 denote corresponding
values at the highest point.

Answer.(a)The free body diagram of the stone at the lowest point is shown in the following
gure.

According to Newton’s second law of motion, the net force acting on the stone at this point is
equal to the centripetal force, i.e.,
2
mv 1
F ret = T − mg = R . . . (i)
Where, v 1 = Velocity at the lowest point
The free body diagram of the stone at the highest point is shown in the following gure.

Using Newton’s second law of motion, we have:

2
mv 2
T + mg = R …(ii)
Where, v 2 = Velocity at the highest point
It is clear from equations (i) and (ii) that the net force acting at the lowest and the highest
points are respectively (T – mg) and (T + mg)

Page : 112 , Block Name : Additional Exercises

Q5.27 A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s − 2. The crew and
the passengers weigh 300 kg. Give the magnitude and direction of the
(a) force on the oor by the crew and passengers,
(b) action of the rotor of the helicopter on the surrounding air,
(c) force on the helicopter due to the surrounding air

(a) Mass of the helicopter, m h = 1000kg
Mass of the crew and passengers, m p = 300kg
Total mass of the system, m = 1300kg
Acceleration of the helicopter, a = 15m / s 2
Using Newton's second law of motion, the reaction force R, on the system by the floor
can be calculated as:
R − m p9 = ma
= m p(g + a)
= 300(10 + 15) = 300 × 25
= 7500N
since the helicopter is accelerating vertically upward, the force on the floor
by the crew and passengers is 7500N, directed downward.
(b) Using Newton's second law of motion, the reaction force R ′ , experienced by the
helicopter can be calculated as:
R ′ − mg = mq
= m(g + a)
= 1300(10 + 15) = 1300 × 25
= 32500N
The reaction force experienced by the helicopter from the surrounding air is acting
upward. Hence, as per Newton's third law of motion, the action of the rotor on the
surrounding air will be 32500N, directed downward.
(c) The force on the helicopter due to the surrounding air is 32500 N, directed upward.

Page : 112 , Block Name : Additional Exercises

Q5.28 A stream of water owing horizontally with a speed of 15 m s − 1 gushes out of a tube of

cross-sectional area 10 − 2 m 2, and hits a vertical wall nearby. What is the force exerted on the
wall by the impact of water, assuming it does not rebound ?

Speed of the water stream, v = 15m / s
Cross-sectional area of the tube, A = 10 − 2m 2
Volume of water coming out from the pipe per second,
V = AV = 15 × 10 − 2m 3 / s
Density of water, ρ = 10 3kg / m 3
Mass of water flowing out through the pipe per second = ρ × V = 150kg / s
The water strikes the wall and does not rebound. Therefore, the force exerted by the
water on the wall is given by Newton's second law of motion as:
ΔP
F = Rate of change of momentum = Δt
mv
= t
= 150 × 15 = 2250N

Page : 112 , Block Name : Additional Exercises

Q5.29 Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m.
Give the magnitude and direction of
(a) the force on the 7 th coin (counted from the bottom) due to all the coins on its top,
(b) the force on the7 th coin by the eighth coin,
(c) the reaction of the 6 th coin on the 7 th coin

(a) Force on the seventh coin is exerted by the weight of the three coins on its top.
Weight of one coin = mg
Weight of three coins = 3mg
Hence, the force exerted on the 7 th coin by the three coins on its top is 3mg. This force
acts vertically downward.
(b) Force on the seventh coin by the elghth coin is because of the weight of the eighth
coin and the other two coins (ninth and tenth) on its top.
Weight of the eighth coin = mg
Weight of the ninth coin = mg
Weight of the tenth coin = mg
Total weight of these three coins = 3mg
Hence, the force exerted on the 7 th coin by the eighth coin is 3mg . This force acts
vertically downward.
(c) The 6th coin experiences a downward force because of the weight of the four coins
(7th, e, 90', and 100') on its top.
Therefore, the total downward force experienced by the 6th coin is 4mg.

As per Newton's third law of motion, the 6+ coin will produce an equal reaction force on
the 7th coin, but in the opposite direction. Hence, the reaction force of the 6th coin on the
7th coin is of magnitude 4mg. This force acts in the upward direction.

Page : 112 , Block Name : Additional Exercises

Q5.30 An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at
15°. What is the radius of the loop ?

5
Speed of the aircraft, v = 720km / h = 720 × 18 = 200m / s

Acceleration due to gravity, g = 10m / s 2
Angle of banking, θ = 15 ∘
For radius r, of the loop, we have the relation:
v2
tanθ = rg
v2
r = gtan θ
200 × 200 4000
= 10 × tan 15 = 0.268
= 14925.37m
= 14.92km

Page : 112 , Block Name : Additional Exercises

Q5.31 A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The
mass of the train is 106 kg. What provides the centripetal force required for this purpose —
The engine or the rails ? What is the angle of banking required to prevent wearing out of the
rail ?

Radius of the circular track, r = 30m
Speed of the train, v = 54km / h = 15m / s
Mass of the train, m = 10 6kg
The centripetal force is provided by the lateral thrust of the rail on the wheel. As per
Newton's third law of motion, the wheel exerts an equal and opposite force on the rail.
This reaction force is responsible for the wear and rear of the rail
The angle of banking θ, is related to the radius (r) and speed (v) by the relation:
v2
tanθ = rg

( 15 ) 2 225
= 30 × 10 = 300

θ = tan − 1(0.75) = 36.87 ∘
Therefore, the angle of banking is about 36.87 ∘ .

Page : 112 , Block Name : Additional Exercises

Q5.32 A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig.
5.19. What is the action on the oor by the man in the two cases ? If the oor yields to a
normal force of 700 N, which mode should the man adopt to lift the block without the oor
yielding ?

750N and 250N in the respective cases; Method (b)
Mass of the block, m = 25kg
Mass of the man, M = 50kg
Acceleration due to gravity, g = 10m / s 2
Force applied on the block, F = 25 × 10 = 250N
Weight of the man, W = 50 × 10 = 500N
Case (a): When the man lifts the block directly
In this case, the man applies a force in the upward direction. This increases his apparent
weight.
Action on the floor by the man = 250 + 500 = 750N
Case (b): When the man lifts the block using a pulley
In this case, the man applies a force in the downward direction. This decreases his
apparent welght.
∴ Action on the floor by the man = 500 − 250 = 250N
If the floor can yield to a normal force of 700N, then the man should adopt the second
method to easily lift the block by applying lesser force.

Page : 112 , Block Name : Additional Exercises

Q5.33 A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum
tension of 600 N. In which of the following cases will the rope break: the monkey
(a) climbs up with an acceleration of 6 m s − 2
(b) climbs down with an acceleration of 4 m s − 2
(c) climbs up with a uniform speed of 5 m s − 1
(d) falls down the rope nearly freely under gravity? (Ignore the mass of the rope).

Case (a)
Mass of the monkey, m = 40kg
Acceleration due to gravity, g = 10m / s
Maximum tension that the rope can bear, T max = 600N

Acceleration of the monkey, a = 6m / s 2 upward
Using Newton's second law of motion, we can write the equation of motion as:
T − mg = ma
∴ T = m(g + a)
= 40(10 + 6)
= 640N
since T > T maxt the rope will break in this case.

Case(b)
Acceleration of the monkey, a = 4m / s 2 downward
Using Newton's second law of motion, we can write the equation of motion as:
mg − T = ma
∴ T = m(g − a)
= 40(10 − 4)
= 240N
since T < T maxt the rope will not break in this case.

Case (C)
The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is
zero, i.e., a = O,
using Newton's second law of motion, we can write the equation of motion as:
T − mg = ma
T − mg = 0
∴ T = mg
= 40 × 10
= 400N
since T < T maxt the rope will not break in this case.

Case (d)
When the monkey falls freely under gravity, its will acceleration become equal to the
acceleration due to gravity, i.e., a = g
Using Newton's second law of motion, we can write the equation of motion as:
∴ T = m(g − g) = 0
since T < T maxt the rope will not break in this case.

Page : 113 , Block Name : Additional Exercises

Q5.34 Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table
against a rigid wall (Fig. 5.21). The coef cient of friction between the bodies and the table is
0.15. A force of 200 N is applied horizontally to A. What are (a) the reaction of the partition (b)
the action-reaction forces between A and B ? What happens when the wall is removed? Does
the answer to (b) change, when the bodies are in motion? Ignore the difference between µs
and µk

(a) Mass of body A, m A = 5kg
Mass of body B, m 8 = 10kg
Applied force, F = 200N
Coefficient of friction, μ s = 0.15
The force of friction is given by the relation:

(
fs = μ mA + mB g )
= 0.15(5 + 10) × 10
= 1.5 × 15 = 22.5N leftward
Net force acting on the partition = 200 − 22.5 = 177.5N rightward
As per Newton's third law of motion, the reaction force of the partition will be in the
direction opposite to the net applied force.
Hence, the reaction of the partition will be 177.5N, in the leftward direction.
(b) Force of friction on mass A:
f A = μm Ag
= 0.15 × 5 × 10 = 7.5N leftward
Net force exerted by mass A on mass B = 200 − 7.5 = 192.5N rightward
As per Newton's third law of motion, an equal amount of reaction force will be exerted by
mass B on mass A, l.e., 192.5N acting leftward.
When the wall is removed, the two bodies will move in the direction of the applied force
Net force acting on the moving system = 177.5N
The equation of motion for the system of acceleration a , can be written as:

(
Net force = m A + m B a )
Net force
∴ a = m +m
A B
177.5 177.5
= 5 + 10 = 15 = 11.83m / s 2
Net force causing mass A to move:
F A = m Aa
= 5 × 11.83 = 59.15N
This force will act in the direction of motion. As per Newton's third law of motion, an
equal amount of force will be exerted by mass B on mass A r i.e., 133.3N, acting
opposite to the direction of motion.

Page : 113 , Block Name : Additional Exercises

Q5.35 A block of mass 15 kg is placed on a long trolley. The coef cient of static friction
between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 m s − 2 for
20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a
stationary observer on the ground, (b) an observer moving with the trolley

(a) Mass of the block, m = 15kg
coefficient of static friction, μ = 0.18
Coefficient of static friction, μ = 0.18
Acceleration of the trolley, a = 0.5m / s 2
As per Newton's second law of motion, the force (F) on the block caused by the motion
of the trolley is given by the relation:
F = ma = 15 × 0.5 = 7.5N
This force is acted in the direction of motion of the trolley.
Force of static friction between the block and the trolley:
f = μmg
= 0.18 × 15 × 10 = 27N
The force of static friction between the block and the trolley is greater than the applied
external force. Hence, for an observer on the ground, the block will appear to be at rest.
When the trolley moves with uniform velocity there will be no applied external force.
Only the force of friction will act on the block in this situation.
(b) An observer, moving with the trolley, has some acceleration. This is the case of non-
inertial frame Of reference. The frictional force, acting on the trolley backward, is
opposed by a pseudo force of the sarne magnitude. However, this force acts in the
opposite direction. Thus, the trolley will appear to be at rest for the observer moving
with the trolley.

Page : 113 , Block Name : Additional Exercises

Q5.36 The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the
open end as shown in Fig. 5.22. The coef cient of friction between the box and the surface
below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s − 2. At
what distance from the starting point does the box fall off the truck? (Ignore the size of the
box).

Mass of the box, m = 40kg
Coefficient of friction, μ = 0.15
Initial velocity, u = 0
Acceleration, a = 2m / s 2

Distance of the box from the end of the truck, s ′ = 5m
As per Newton's second law of motion, the force on the box caused by the accelerated
F motion of the truck is given by:
F = ma
= 40 × 2 = 80N
As per Newton's third law of motion, a reaction force of 80N is acting on the box in the
backward direction. The backward motion of the box is opposed by the force of friction f
acting between the box and the floor of the truck. This force is given by:
f = μmg
= 0.15 × 40 × 10 = 60N
= 0.15 × 40 × 10 = 60N
∴ Net force acting on the block:
F net = 80 − 60 = 20N backward
The backward acceleration produced in the box is given by
F mt 20
= = m = 40 = 0.5m / s 2
a back
Using the second equation of motion, time t can be calculated as:
1
s ′ = ut + 2 a bad t 2
1
5 = 0 + 2 × 0.5 × t 2

∴ t = √20s
Hence, the box will fall from the truck after √20s from start.

The distance s, travelled by the truck in √20s is given by the relation:
1
s = ut + 2 at 2
1
= 0 + 2 × 2 × (√20) 2
= 20m

Page : 113 , Block Name : Additional Exercises

1
Q5.37 A disc revolves with a speed of 33 3 rev/min, and has a radius of 15 cm. Two coins are
placed at 4 cm and 14 cm away from the centre of the record. If the coef cient of friction
between the coins and the record is 0.15, which of the coins will revolve with the record ?

Coin placed at 4cm from the centre
Mass of each coin = m
Radius of the disc, r = 15cm = 0.15m

1 100 5
Frequency of revolution, v = 33 3 rev/min = 3 × 60 = 9 rev/s
coefficient of friction, μ = 0.15
In the given situation, the coin having a force of friction greater than or equal to the
centripetal force provided by the rotation of the disc will revolve with the disc. If this is not the
case, then the coin will slip from the disc.
Coin placed at 4 cm:
Radius of revolution, r‘ = 4 cm = 0.04 m
Angular frequency, ω = 2πν
22 5
= 2 × 7 × 9 = 3.49s − 1
Frictional force, f = μmg = 0.15 × m × 10 = 1.5m N
Centripetal force on the coin:
F cent.
= mr ′ω 2
= m × 0.04 × (3.49) 2
r''= 0.49m N
Since f > Fcent, the coin will revolve along with the record.
Coin placed at 14 cm:
Radius,
= 14 cm = 0.14 m
Angular frequency, ω = 3.49 s–1
Frictional force, f‘ = 1.5m N
Centripetal force is given as:
F cent.
= mr ′′ω 2
= m × 0.14 × (3.49) 2
= 1.7mN
Since f < F cent., the coin will slip from the surface of the record.

Page : 113 , Block Name : Additional Exercises

Q5.38 You may have seen in a circus a motorcyclist driving in vertical loops inside a
‘deathwell’ (a hollow spherical chamber with holes, so the spectators can watch from outside).
Explain clearly why the motorcyclist does not drop down when he is at the uppermost point,
with no support from below. What is the minimum speed required at the uppermost position
to perform a vertical loop if the radius of the chamber is 25 m ?

Answer. In a death-well, a motorcyclist does not fall at the top point Of a vertical loop because
both the force of normal reaction and the weight of the motorcyclist act downward and are
balanced by the centripetal force. This situation is shown in the following gure.

The net force acting on the motorcyclist is the sum of the normal force and the force
due to gravity (Fg mg).
The equation of motion for the centripetal acceleration a, , can be written as:

( )
The net force acting on the motorcyclist is the sum of the normal force F n and the force

(
due to gravity F 9 = mg )
The equation of motion for the centripetal acceleration a cr can be written as:
F net = ma c
F N + F g = ma c
mv 2
F N + mg = r
Normal reaction is provided by the speed of the motorcyclist. At the minimum speed

(v min ), Fn = 0
(v min ), Fn = 0
2
mv min
mg = r

∴ v min = √rg

= √25 × 10 = 15.8m / s

Page : 113 , Block Name : Additional Exercises

Q5.39 A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of
radius 3 m rotating about its vertical axis with 200 rev/min. The coef cient of friction between
the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to
enable the man to remain stuck to the wall (without falling) when the oor is suddenly
removed ?

Mass of the man, m = 70kg
Radius of the drum, r = 3m
coefficient of friction, μ = 0.15

200 10
Frequency of rotation, v = 60 = 3 rev/s
The necessary centripetal force required for the rotation of the man is provided by the

normal force F n . ( )
When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the
man (mg) acting downward is balanced by the frictional force

(f = μFn ) acting upward.
Hence, the man will not fall until:
mg < f
mg < μF 11 = μmrω 2

g < μrω 2
g
ω>
√ μr

The minimum angular speed is given as:
g
ω min =
√ μr

10
=
√ 0.15 × 3
= 4.71 rad s − 1

Page : 113 , Block Name : Additional Exercises

Q5.40 A thin circular loop of radius R rotates about its vertical diameter with an angular
frequency ω. Show that a small bead on the wire loop remains at its lowermost point for ω ≤ g /
R . What is the angle made by the radius vector joining the centre to the bead with the vertical
downward direction for ω = 2g / R ? Neglect friction.

Let the radius vector joining the bead with the centre make an angle θ, with the vertical
downward direction.

OP = R = Radius of the circle
N = Normal reaction
The respective vertical and horizontal equations of forces can be written as:
mg = Ncosθ…(i)

|
m ω 2 = Nsinθ…(ii)
In △OPQ, we have:
I
sinθ = R
Substituting equation (iii) in equation (ii), we get:
m(Rsinθ)ω 2 = Nsinθ
mRω 2 = N…(iv)
Substituting equation (iv) in equation (i), we get:
Substituting equation (iv) in equation (i), we get:
mg = mRω 2cosθ
g
cosθ = …(v)
Re 2
g
since cosθ ≤ 1, the bead will remain at its lowermost point for ≤1
Rω 2
g
ω≤
√ R

2g 2g
For ω=
√ R
ω2 = R

On equating equations (v) and (vi), we get:
On equating equations (v) and (vi), we get
2g g
R
= Rcos θ
1
cosθ = 2

∴ θ = cos − 1(0.5) = 60 ∘

Page : 113 , Block Name : Additional Exercises