Class 9 - Physics - Force and Laws of Motion


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.


Yes, even when the net external force is 0, an object may travel with constant velocity.

The only condition is that it should move with same constant velocity and it should be moving in the same direction.

For example: - A rain drop falls with constant velocity.


When a carpet is beaten with a stick, dust comes out of it. Explain.


Inertia is the natural tendency of an object to resist a change in its state of motion or of rest. When a carpet is beaten with the stick,

carpet comes in the state of motion but the dust particles are still in the state of rest.

 In order to continue in the position of rest dust particles comes out of carpet.

This is in accordance with Newton’s First law of motion which states that the object

will continue to be in the state of rest unless any external force is applied on it.

 Question 3.

Why is it advised to tie any luggage kept on the roof of a bus with a rope?


When the bus moves in forward direction, it acquires the state of motion.

But the luggage kept on the roof, because of inertia luggage tends to remain in the state of rest.

Therefore when the bus moves in the forward direction the luggage tends to be in its original position and ultimately fall from the roof of the bus.

In order to avoid this, it is advised to tie any luggage kept on the roof of a bus with a rope.

Question 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.


Correct option is (c) there is a force on the ball opposing the motion.

This is because when the ball moves on the ground, the force of friction opposes its movement and after some time it comes to rest.

 Question 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 metric tonnes (Hint: 1 metric tonne = 1000 kg.)



Initial velocity u = 0

Distance s = 400m

Time taken t = 20s

Mass = 7 metric tonnes = 7000kg

  • From second equation of motion: - s =ut + (1/2) at2

= 0 x 20 + (1/2) a x 20 x20

  • a = (2 x 400)/(20 x 20)
  • a = 2m/s2
  • F = ma

= 7000 x 2

F =14000 N

 Question 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?



Mass m = 1kg

Initial velocity u =20m/s

Final velocity v = 0 (finally the stone comes to rest)

Distance s = 50m

Force of friction = Force acting on the stone then it comes to rest.

From third equation of motion: - v2 = u2 + 2as

  • 0 = 20x 20 + 2a x50
  • a = (-) 4 m/s2

The negative (-) sign tells that the boy is retarding. Velocity of the body is gradually decreasing.

F = ma = 1 x (-) 4 = -4N

The (-) ive sign indicates that the force acts opposite to the direction of motion.

This is the frictional force between the stone and the ice.

 Question 7.

An 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:

(a) the net accelerating force;

(b) the acceleration of the train; and

(c) the force of wagon 1 on wagon 2.


  • Force exerted by the engine, F = 40000N

        Frictional force offered by the track, Ff = 5000N

        Net accelerating force, Fa = F – Ff = 40000 – 5000 =35000N.

  • Acceleration of the train = a

        The engine exerts a force of 40000N on all the five wagons.

          Net accelerating force on the wagons, Fa = 35000N

Mass of the wagons, m = mass of a wagon x number of wagons

Mass of wagon = 2000kg

Number of wagons = 5

Therefore m =2000 x 5 =10000kg.

Mass of the engine m’ = 8000kg

Total mass m + m’ = 18000kg

From Newton’s second law of motion:

Fa = ma

a = (Fa/m) = (350000/18000) = 1.944m/s2

  • Mass of all the wagons except wagon 1 is 4 × 2000 = 8000 kg

       Acceleration of the wagons = 3.5 m/s2

       Thus, force exerted on all the wagons except wagon 1

= 8000 × 3.5 = 28000 N

     Therefore, the force exerted by wagon 1 on the remaining four wagons         is 28000 N.

           Hence, the force exerted by wagon 1 on wagon 2 is 28000 N.


Question 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 ms–2?



Mass m =1500kg

Final velocity v=0

Acceleration a = -1.7 m/s2

From Newton’s second law of motion: -

  • F = ma = 1500 x (-) 1.7
  • = (-) 2550N

The (-) ive sign shows that the direction of force is opposite to the direction in which vehicle is moving.

 Question 9.                     

What is the momentum of an object of mass m, moving with a velocity v?

(a) (m) 2 (b) mv2 (c) (1⁄2) mv2   (d) mv


Correct option is (d) mv.

Mass of the object = m

Velocity = v

Momentum = mass x velocity

Therefore momentum = mv.

 Question 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?



Force = 200N

As given constant velocity this implies acceleration a = 0.

  • Fnet = 0 ( From Newton’s second law)

Therefore frictional force Ff = (-) 200N. It should be equal in magnitude but in opposite direction.

 Question 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 m s-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?



Mass m1 = m2 = 1.5kg

Velocity before collision v1 = v2 = 2.5m/s

Before collision linear momentum p1 = m1 v1 + m2 v2

= 1.5 x 2.5 + 1.5 x (-) 2.5 as (v2 is in the opposite direction)

Total momentum before collision p1 = 0

After collision p2 = (m1+m2) v


By conservation of momentum:

Total momentum before collision (p1) = total momentum after collision (p2)

  • 3v =0
  • v= 0

This shows the object won’t move after collision as it won’t have any velocity.

 Question 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.


The truck has a large mass therefore the static friction between the truck and the road is very high.

To move the car, we have to apply force more than static friction.

Therefore, when someone pushes the truck and the truck does not move,

this means applied force gets cancel by the frictional force which is equal in magnitude but in opposite direction.

 Question 13.

A hockey ball of mass 200 g 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.



Mass m = 200g

Velocity v = 10 m s–1

After being struck by the stick:-

Mass m = 200g

Velocity v’ = (-) 5m/s

Therefore change in momentum = (mv – mv’)

=200 x 10-3(10 – (-5))

=3000 x 10-3

Change in momentum = 3kgm/s

 Question 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.



Mass m =10g

Initial velocity u = 150 m/s

Final velocity v =0

Time t = 0.03sec

v = u + at

  • 0 =150 + a x 0.03
  • a = (-) 5000 m/s2

s = ut+ (1/2) at2

= (150 x 0.03) + (1/2) x (-5000) x (0.03)2

s = 2.25m

F = ma = 10 x 10 -3 x (-5000)

F =50N

 Question 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.



Before collision:-

Mass m =10kg

Velocity v =10m/s

Total momentum p1 = mv = 1 x 10 = 10kgm/s

After collision:-

Total mass m’ = 1kg + 5Kg = 6kg.

Total momentum p2 = m’v’ = 6v’

From conservation of momentum: - p1 = p2

  • 10 = 6v’
  • v’ = (10/6) = 1.67m/s

Total momentum after impact p2 = 6 x 1.67 =10kgm/s

 Question 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.



Mass m = 100kg

Initial velocity u = 5m/s

Final velocity v = 8m/s

Time t = 6sec

Initial momentum pi

 Final momentum pf

pi = mu = 100 x 5 = 500kgm/s

pf = mv = 100 x 8 = 800kgm/s

From first equation of motion v = u + at

  • a =(v-u)/t = (8 – 5)/6
  • a = 0.5 m/s2

Therefore F = ma

=100 x (0.5) = 50N

Force exerted F = 50N

 Question 17.

Akhtar, Kiran and Rahul were riding in a motorcar 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.


According to the law of conservation of momentum:

Total Momentum of the system before collision = Momentum of the system after collision.

In this case insect experiences a greater change in its velocity so greater change in momentum. This shows Kiran’s observation is correct.

Motor car has more mass as compared to insect and large velocity. Motorcar continues to move in the same direction even after the collision,

which shows the change in the momentum of the motorcar is very less. Insect experiences more change in its momentum.

Hence, Akhtar’s observation is also correct.

Rahul’s observation is also correct; because the momentum gained by the insect is equal to the momentum lost by the motorcar.

This also happens in accordance to the law of conservation of momentum.

 Question 18.

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 s–2.



Mass of dumbbell m = 10kg

Distance covered = 80cm = 0.8m

Acceleration a = 10m/s2

Initial velocity u =0

From third equation v2= u2 + 2as

v2= 0 + 2 x 10 x 0.8 = 16

  • v = 4m/s

The velocity when it reaches the floor = 4m/s

Momentum transferred to the floor p = mv = 10 x 4 = 40kgm/s




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