# TEXTBOOK ANSWERS AND SOLUTIONS OF CBSE CLASS IX SCIENCE Chapter 9 Force And Laws Of Motion

1) An object experiences a net zero
external unbalanced force. Is it possible for the object to be travelling with
a non-zero velocity? If yes, state the conditions that must be placed on the
magnitude and direction of the velocity. If no, provide a reason.
ANSWER:-Yes, an object may travel with a
non-zero velocity even when the net external force on it is zero. A rain drop
falls down with a constant velocity. The weight of the drop is balanced by the
up thrust and the
velocity of air.
2) When a carpet is beaten with a
stick, dust comes out of it. Explain.
ANSWER:-When the carpet is beaten, it is
suddenly set into motion. The dust particles tend to
remain at rest due to
inertia of rest, therefore the dust comes
out of it.
3) Why is it advised to tie any luggage
kept on the roof of a bus with a rope?
ANSWER:-When a bust starts suddenly, the
lower part of the luggage kept on the roof being in
contact with the bus
begins to move forward with the speed of bus, but the upper part tends to
remain at rest due to inertia of rest. Therefore, the upper part is left behind
and hence luggage falls backward. So, it is advised to tie any luggage kept on
the roof of a bus with a rope.
4) A batsman hits a cricket ball which
then rolls on a level ground. After covering a short distance, the ball comes
to rest. The ball slows to a stop because
(a) The batsman did not hit the ball
hard enough.
(b) Velocity is proportional to the
force exerted on the ball.
(c) There is a force on the ball
opposing the motion.
(d) There is no unbalanced force on
the ball, so the ball would want to come to rest.
ANSWER:-The ball slows down and comes to
rest due to opposing forces of air resistance

and frictional force on the ball opposing its motion. Therefore the choice (c) there is a force on the ball opposing the
motion is correct.
5) A truck starts from rest and rolls
down a hill with a constant acceleration. It travels a distance of 400 m in 20
s. Find its acceleration. Find the force acting on it if its mass is 7 metric
tonnes (Hint: 1 metric tonne = 1000kg).
ANSWER:-Initial velocity, u = 0 Distance travelled, s =
400 m Time taken, t = 20 s
We know, s = ut + ½ at2
Or, 400 = 0 + ½ a (20) 2
Or, a = 2 ms –2
Now, m = 7 MT = 7000 kg, a = 2 ms–2
Or, F = ma = 7000 × 2 = 14000N
6) A stone of 1 kg is thrown with a
velocity of 20 m s −1 across the frozen surface of a lake and comes to rest
after travelling a distance of 50 m. What is the force of friction between the
stone and the ice?
u= 20 m/s
Final velocity of the stone, v=0
Distance covered by the stone,
s = 50 m
Since, v 2 – u 2
2 = 2as,
Or, 0 – 202 = 2a × 50,
Or, a = – 4 ms-2
Force of friction, F = ma = – 4N
7) A 8000 kg engine pulls a train of 5
wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force
of 40000 N and the track offers a friction force of 5000N, then calculate:
(a) The net accelerating force;
(b) The acceleration of the train; and
(C) the force of wagon 1 on wagon 2.
ANSWER 🙁 a) Force exerted by the engine, F = 40000 N
Frictional force offered by the track, Ff = 5000 N
Net accelerating force, Fa = F − Ff = 40000 − 5000 =
35000 N
Hence, the net accelerating force is 35000 N.
(b) Acceleration of the train =a
The engine exerts a force of 40000 N on all the five
wagons.
Net accelerating force on the wagons, Fa = 35000 N
Mass of the wagons, m = Mass of a wagon x Number of
wagons
Mass of a wagon = 2000 kg
Number of wagons = 5
m = 2000 × 5 = 10000 kg
Total mass, M = m = 10000 kg
From Newton’s second law of motion:
F a = Ma
a=Fam
= 35000 10000    = 3.
5 ms-2
Hence, the acceleration of the wagons and the train is
3.5 m/ s 2.
(c) Mass of all the wagons except wagon 1 is 4 ×
2000 =8000 kg
Acceleration of the wagons = 3.5 m/s 2
Thus, force exerted on all the wagons except wagon 1
= 8000 × 3.5 = 28000 N
8) An automobile vehicle has a mass of
1500 kg. What must be the force between the vehicle and road if the vehicle is
to be stopped with a negative acceleration of 1.7 m s −2?
automobile vehicle, m= 1500 kg
Final velocity, v= 0 (finally the automobile stops)
Acceleration of the automobile, a = −1.7 ms−2
From Newton’s second law of motion:
Force = Mass x Acceleration = 1500 x (−1.7) = −2550 N
Hence, the force between the automobile and the road
is −2550 N, in the direction opposite to the motion of the automobile.
9) What
is the momentum of an object of mass m, moving with a velocity v ?
(a) (mv)
2 (b) 1/2mv2 (c)mv
of the object = m
Velocity =
v
Momentum =
Mass x Velocity
Momentum =
mv
10) Using a horizontal force of 200 N,
we intend to move a wooden cabinet across a floor at a constant velocity. What
is the friction force that will be exerted on the cabinet?
move with constant velocity only when the net
force on it is zero. Therefore, force of
friction on the cabinet = 200 N, in a
direction opposite to the direction of motion of the cabinet.
11) Two objects, each of mass 1.5 kg are
moving in the same straight line but in opposite directions. The velocity of
each object is 2.5 ms−1 before the collision during which they stick together.
What will be the velocity of the combined object after collision?
ANSWER:-Mass of one of the objects, m1=
1.5 kg
Mass of the other object, m2 =1.5 kg
Velocity of m1 before collision,
u1 = 2.5 m/s
Velocity of m2, moving in opposite direction before
collision, u 2 = −2.5 m/s
Let v be the velocity of the combined object after
collision.
By the law of conservation of momentum,
Total momentum after collision = Total momentum before
collision,
Or, (m1 + m2) v = m1u1 +m2u2
Or, (1.5 + 1.5) v = 1.5 × 2.5 +1.5 × (–2.5) [negative
sign as moving in opposite direction]
Or, v = 0 ms–1
12) According to the third law of
motion when we push on an object, the object pushes back on us with an equal
and opposite force. If the object is a massive truck parked along the roadside,
it will probably not move. A student justifies this by answering that the two
opposite and equal forces cancel each other. Comment on this logic and explain
why the truck does not move.
ANSWER:-When we push a massive truck,
the force of friction between its tyres and the road is very large and so the
truck does not move.
13) A hockey ball of mass 200g
travelling at 10 m s −1 is struck by a hockey stick so as to return it along
its original path with a velocity at 5 m s -1. Calculate the change of momentum
occurred in the motion of the hockey ball by the force applied by the hockey
stick.
ANSWER:-Mass of the hockey ball, m =200
g = 0.2 kg
Hockey ball travels with velocity, v1 =
10 m/s
Initial momentum = mv 1
Hockey ball travels in the opposite
direction with
velocity, v2 = −5 m/s
Final momentum = mv2
Change in momentum = mv 1 − mv 2 = 0.2 [10 − (−5)] =
0.2 (15) = 3 kg m s −1
Hence, the change in momentum of the hockey
ball is 3 kg m s −1.
14) A bullet of mass 10 g travelling
horizontally with a velocity of 150 m s −1 strikes a stationary wooden block
and comes to rest in 0.03 s. Calculate the distance of penetration of the
bullet into the block. Also calculate the magnitude of the force exerted by the
wooden block on the bullet.
Final velocity, v= 0 (since the bullet finally comes to
rest)
Time taken to come to rest, t=0.03 s
According to the first equation of motion, v = u + at
Acceleration of the bullet, a 0 = 150 + (a × 0.03 s) a
= -150 / 0.03 = -5000 m/s2
bullet is decreasing.)
According to the third equation of motion:
v 2= u 2+ 2 as
0 = (150) 2+ 2 (-5000)
= 22500 / 10000
= 2.25 m
Hence, the distance of penetration of
the bullet into the block is 2.25 m.
From Newton’s second law of motion:
Force, F = Mass × Acceleration
Mass of the bullet, m = 10 g =0.01 kg
Acceleration of the bullet, a = 5000 m/s 2
F = ma = 0.01 × 5000 = 50 N
15) An object of mass 1 kg travelling
in a straight line with a velocity of 10 m s −1 collides with, and sticks to, a
stationary wooden block of mass 5 kg. Then they both move off together in the
same straight line. Calculate the total momentum just before the impact and
just after the impact. Also, calculate the velocity of the combined object.
ANSWER:-Mass of the object, m1 = 1 kg
Velocity of the object before collision, v 1 = 10 m/s
Mass of the stationary wooden block, m2 = 5 kg
Velocity of the wooden block before collision, v 2 = 0
m/s
Total momentum before collision = m1 v1 + m2 v
2
= 1 (10) + 5 (0) = 10 kg m s −1
It is given that after collision, the object and the wooden
block stick together.
Total mass of the combined system = m1 + m2
Velocity of the combined object = v
According to the law of conservation of
momentum:
Total momentum before collision = Total momentum
after collision
m1 v1 + m2 v 2 = (m1 + m2) v
1 (10) + 5 (0) = (1 + 5) v
v = 10 / 6
= 5 / 3
The total momentum after collision is
also 10 kg m/s.
Total momentum just before the impact =
10 kg m s −1
Total momentum just after the impact = (m1 + m2) v = 6
× 5 / 3 = 10 kg ms-1
16) An object of mass 100 kg is
accelerated uniformly from a velocity of 5 m s −1 to 8 m s −1 in 6 s. Calculate
the initial and final momentum of the object. Also, find the magnitude of the
force exerted on the object.
u= 5 m/s
Final velocity of the object, v = 8 m/s
Mass of the object, m = 100 kg
Time take by the object to accelerate, t = 6 s
Initial momentum = mu = 100 × 5 = 500 kg m s −1
Final momentum = mv = 100 × 8 = 800 kg m s −1
Force exerted on the object, F
= mv – mu / t
= m (v-u) / t
= 800 – 500
= 300 / 6
= 50 N
17)Akhtar, Kiran and Rahul were riding
in a motorocar that was moving with a high velocity on an expressway when an
insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran
started pondering over the situation. Kiran suggested that the insect suffered
a greater change in momentum as compared to the change in momentum of the motorcar
(because the change in the velocity of the insect was much more than that of
the motorcar). Akhtar said that since the motorcar was moving with a larger
velocity, it exerted a larger force on the insect. And as a result the insect
died. Rahul while putting an entirely new explanation said that both the motorcar
and the insect experienced the same force and a change in their momentum.
Comment on these suggestions
.
that the insect suffered a
greater change in momentum as compared to the change in
momentum of the motor car is wrong.
The suggestion made by Akhtar that the
motor car exerted a larger force on the insect because of large velocity of
motor car is also wrong. The explanation put

forward by Rahul is correct. On collision of insect
with motor car, both experience the same force as action and reaction are
always equal and opposite. Further, changes in their momenta are also the same.
Only the signs of changes in momenta are
opposite.
18) How much momentum will a dumbbell
of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take
its downward acceleration to be 10 m s −2.
ANSWER:-Mass of the dumbbell, m = 10 kg
Distance covered by the dumbbell, s = 80 cm = 0.8
m
Acceleration in the downward direction, a = 10 m/s 2
Initial velocity of the dumbbell, u = 0
Final velocity of the dumbbell (when it was about to hit
the floor) = v
According to the third equation of motion:
v 2 = u2 + 2as
v 2 = 0 + 2 (10) 0.8
v = 4 m/s
Hence, the momentum with which the dumbbell hits
the floor is
= mv = 10 × 4 = 40 kg m s −1