This post includes the complete NCERT Exercise Questions and Answers of Class 9 Science Chapter 8 – Force and Laws of Motion. All answers are written in simple and easy language so students can understand important concepts like balanced and unbalanced forces, Newton’s laws of motion, inertia, and momentum. These solutions help in building clear concepts and improving exam preparation.
FORCE AND LAWS OF MOTION
EXERCISE QUESTION ANSWER
Ques 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.
Ques 2: - When a carpet is beaten with a stick, dust comes out of it. Explain.
Ques 3: - Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Ques 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.
Ques 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 it’s mass is 7 tonnes (Hint :- 1 tonne = 1000 kg).
Given, distance covered by the truck (s) = 400 meters
Time taken to cover the distance (t) = 20 seconds
The initial velocity of the truck (u) = 0 m/s
(since it starts from a state of rest).
Acceleration(a) :- ?
Mass (m) :- 7 metric tonnes = 7 x 1000 =7000 Kg
By Second equation of motion
s = ut + ½ at²
Hence acceleration(a) = 2 m/s²
By
Newton's second law of motion
F
= m x a
=> F
= 7000 kg x 2 m/s²
\ F
= 14000 N
So
the force acting on the truck is 14000 N.
Ques 6: - A stone of 1 kg is thrown with a velocity of 20 m/s 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?
Given Mass of the stone (m) = 1kg
Initial velocity (u) = 20m/s
Final (Terminal) velocity (v) = 0 m/s
(the stone reaches a position of rest)
Distance travelled by the stone (s) = 50 m
v² - u² = 2as
v² – u²
=> ———— = a
2s
=> ————— = a
2 x 50
a = - ——
100
By Newton's second law of motion
\ F = - 4N
Negative sign indicates the force is acting opposite to the direction of motion.
F = m x a
\ F = 14000 N (N = kg m/s² )
Ques 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 5000 N, then calculate :
Mass of 1 wagon 2000 kg
Mass of 5 wagons = 5 x 2000
=10,000 kg
Force of friction = 5000 N
(a) the net accelerating force
Net Force = Force of engine - Force of friction
Net Force = 40000 - 5000
Net Force = 35000 N
(b) the acceleration of the train.
We know that
Force = mass x acceleration
We have found Force in the last part Force = 35000 N
Mass = Total Mass
Mass = Mass of Engine + Mass of 5 wagons
Mass = 8000 + 10000
Mass = 18000 kg
Now,
F = m x a => a= F /m
=> a = 35000/18000 = 1.94
\ acceleration (a) = 1.94 m/s²
(c) Mass of all the wagons except wagon 1 is 4 x 2000 = 8000 kg
Thus, force exerted on all the wagons except wagon 1 =
=> F = 8000 x 3.5
\ F = 28000 N
Hence the force exerted by wagon 1 on wagon 2 is 28000 N.
Ques 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/s²?
Acceleration (a) = -1.7 m/s²
By Newton's second law of motion
F = ma
F =1500kg x (-1.7 m/s² )
Hence the force between the automobile and the road is -2550 N, in the opposite direction of the automobile's motion.
Ques 9: - What is the momentum of an object of mass m, moving with a velocity v
?
(c) ½ mv²(d) mv
Ques 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?
Ans :- Since we move the cabinet with constant velocity, its acceleration will be 0.
Since acceleration is 0, no net force will be applied to it.
So, Total Force on body = 0
Now, Total Force on body will be the force applied and Friction
Thus,
Force on body + Friction = 0
200 N + Friction =0
Friction = 0 - 200
Friction = -200 N
Ques 11: - 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.
The reason of not moving the truck is as follows :-
The truck is massive so it has very large inertia Moreover it is parked on the ground there is a frictional force between the truck and ground. The force exerted by us on the truck is insufficient to overcome the force of friction so these are balance force on the truck ( force exerted by us plus force of friction in opposite direction) that is net force and hence the truck doesn't move.
Ques 12: - A hockey ball of mass 200 g travelling at 10 m/s is struck by a hockey stick so as to return it along its original path with a velocity at 5 m/s. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Ans :- Given Mass of the hockey ball = m = 200 g
200
=> ——— kg
1000
Mass of the hockey ball (m) = 0.2
kg
Initial velocity of the ball = u = 10 m/s
Final velocity of the ball = v = -5m/s
Final velocity is negative because it is in the opposite direction.
Initial Momentum
Initial momentum = Mass x Initial velocity
= 0.2 x 10
\ Initial momentum = 2 kg m/s
Final Momentum
Final momentum = Mass x Final velocity
= 0.2 x -5
\ Final Momentum = -1 kg m/s
Change in momentum = Final momentum - Initial momentum
Change in momentum = ( -1 -2 )kg m/s
Change in momentum = -3 kg m/s
Thus, the magnitude of change in momentum is 3 kg m/s
(Magnitude is only value, without the sign)
Ques 13: - A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s 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.
Ans :- Given, mass of the bullet (m) = 10g =10/1000= 0.01 kg
Initial velocity of the bullet (u) = 150 m/s
Terminal(Final) velocity of the bullet (v) = 0 m/s
(as the bullet comes to rest)
Time period (t) = 0.03 s
By first equation
of motion
v = u + at
v – u
=> a = ———
t
(0-150) m
=> a = ——————
0.03
- 15000 m
=> a = ———————
3 s²
=> a = - 5000 m/s²
By Second
equation of motion
* S = ut + ½ at²
=> S = 150 m/s x 0.03 s + 1/2 (-5000 m/s²) x (0.03)²
=> S = 4.50 m - 2500 (m/s²) x 0.0009 s²
=> S = 4.50 m - 2.25 m
\ S = 2.25 m
Distance of penetration of bullet is 2.25 m
By second law of motion,
F = m x a => F =
0.01 kg x (-5000 m/s²)
F = -50 N
Magnitude of force is 50 N.
Ques 14: - An object of mass 1 kg travelling in a straight line with a velocity of 10 m/s 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.
Ans :- Given, mass of the object (m1) = 1kg
Initial velocity of the object (u1) = 10 m/s
Mass
of stationary wooden block = 5kg
Mass
of the resulting object(m2) = mass of the object + Mass of
stationary wooden block
Mass of the resulting object(m2) = 1 kg + 5 kg = 6kg
Let velocity of the combined object = v
Total momentum Just before impact = m1 x u =1 kg x 10 m/s = 10 kg m/s
As per the law of conservation of momentum,
total momentum just before the collision = total momentum just after the collision.
total momentum just after the collision = 10 kg m/s
Therefore
total momentum just before the collision = total momentum just after the
collision = 10 kg m/s
total momentum just after the collision = total momentum of the combined object = 10 kg m/s
total momentum of the combined object = Mass of the resulting object(m₂) x velocity of the combined object
10
kg m/s = 6 kg x v
=> v = 10/6 =10 kg m/s
v = 1.66 m/s
The
resulting object moves with a velocity of 1.66 meters per second.
Initial
velocity of the block (u2) = 0
Velocity
of the resulting object (v) =?
= (1kg) × (10m/s) + 0 =10
kg.m/s
Therefore,
the total momentum post the collision is also 10 kg.m/s
Now,
(m₁ + m₂) × v = 10 kg.m/s
10 kg m/s
=> v = ——————
6 kg
\ v =1.66
m/s
Ques 15: - An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s to 8 m/s in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
ANS
:- Mass of the object m =
100 kg
Initial velocity u = 5 m/s
Final velocity v = 8 m/s
Time t = 6s
Initial momentum p1 = m x u
Initial momentum p1 = 100 x 5
Initial momentum p1 = 500 kg m/ s
Final momentum P2 = m x v
Final momentum P2 = 100 x 8 = 800kg m/s
Force exerted on the object F = ma
m(v-u)
=> F = —————
t
100(8-5)
=> F = —————
6
100
x 3
=> F = —————
6
100
=> F = —————
2
F = 50 N
Ques 16: - How much momentum will a dumb-bell 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/s².
Ans
:- Given, mass of the dumb-bell (m) = 10kg
Distance
covered (s) = 80cm =80/100 = 0.8m
Initial
velocity (u) = 0 (it is dropped from a position of rest)
Acceleration (a) = 10 m/s²
Terminal
velocity (v) =?
Momentum
of the dumb-bell when it hits the ground = mv
By
third equation of motion
v² – u² = 2as
v² – 0 = 2 (10 m/s²) (0.8m) = 16 m² / s²
v
= 4 m/s
The
momentum transferred by the dumb-bell to the floor = m x v
=
(10kg) × (4 m/s)
=
40 kg. m/s
By practising these NCERT solutions,
students can build a strong foundation for higher classes and score better in
exams. Always revise formulas like F = ma and understand examples rather
than memorizing them
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