Class 9 Science

Force and Laws of Motion

In Text Question Answer

Question:- 1 - Which of the following has more inertia:

(a) a rubber ball and a stone of the same size?
(b) a bicycle and a train?
(c) a five rupees coin and a one-rupee coin?

Answer: (a) A stone, (b) A train, (c) A five rupees coin.

Explanation: Inertia is associated with mass. Objects having more mass have more inertia.

Question 2: In the following example, try to identify the number of times the velocity of the ball changes:

“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”. Also identify the agent supplying the force in each case.

Answer: The velocity of football changes four times. First, when a football player kicks to another player, second when that player kicks the football to the goalkeeper. Third when the goalkeeper stops the football. Fourth when the goalkeeper kicks the football towards a player of his own team.

Agent supplying the force

First case – First player
Second case – Second player
Third case – Goalkeeper
Fourth case – Goalkeeper

Question 3: Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

Answer: The answer of this cause lies behind the Newton’s First Law of Motion. Initially, leaves and tree both are in rest. But when the tree is shaken vigorously, tree comes in motion while leaves have tendency to be in rest. Thus, because of remaining in the position of rest some of the leaves may get detached from a tree if we vigorously shake its branch.

Question 4: Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?

Answer: In a moving bus, passengers are in motion along with bus. When brakes are applied to stop a moving bus, bus comes in the position of rest. But because of tendency to be in the motion a person falls in forward direction.

Similarly, when a bus is accelerated from rest, the tendency to be in rest, a person in the bus falls backwards.

Question 5: If action is always equal to the reaction, explain how a horse can pull a cart.

Answer: Horse pushes the ground in backward direction. In reaction to this, the horse moves forward and cart moves along with horse in forward direction.

In this case when horse tries to pull the cart in forward direction, cart pulls the horse in backward direction, but because of the unbalanced force applied by the horse, it pulls the cart in forward direction.

Question 6: Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

Answer: When large amount of water is ejected from a hose at a high velocity, according to Newton’s Third Law of Motion, water pushes the hose in backward direction with the same force. Therefore, it is difficult for a fireman to hold a hose in which ejects large amount of water at a high velocity.

Question 7: From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m/s. Calculate the initial recoil velocity of the rifle.

Answer: Given,

Mass of rifle (m1) = 4 kg
Initial velocity of rifle (u1) = 0
Mass of bullet (m2) = 50 g = 50/1000 kg = 0.05 kg
Initial velocity of bullet (u2) = 0
Final velocity (v2) of bullet = 35 m/s
Final velocity [Recoil velocity] of rifle (v1)=?

We know that,

`=>4\ kg xx 0+0.5\ kgxx0` `=4\ kgxxv_1+0.05\ kgxx35ms^(-1)`
`=>0=4\ kgxxv_1+1.75\ kg m//s`
`=>-4\ kgxxv_1=1.75\ kg m//s`
`=>v_1=(1.75\ kg\ m//s)/(-4\ kg)`
`=-0.4375\ m//s` `~~-0.44m//s`

Here negative sign of velocity of rifle shows that rifle moves in the opposite direction of the movement of bullet. Therefore, recoil velocity of rifle is equal to 0.44 m/s.

Question 8: Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m/s and 1 m/s, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m/s. Determine the velocity of the second object.


Mass of first object (m1) = 100 g = 100/1000 kg = 0.1 kg
Initial velocity of first object (u1) = 2 m/s
Final velocity of first object after collision (v1) = 1.67 m/s
Mass of second object (m2) = 200 g = 200/1000 kg = 0.2 kg
Initial velocity of second object (u2) = 1 m/s
Final velocity of second object after collision (v2) =?

We know that,

`=>0.1\ kgxx2m//s+0.2\ kgxx1m//s` `=0.1\ kg xx1.67m//s+0.2\ kg xxv_2`
`=>0.2\ kg\ m//s + 0.2\ kg\ m//s` `=0.167\ kg\ m//s +0.2\ kg xx v_2`
`=>0.4\ kg\ m//s - 0.167\ kg\ m//s` `=0.2\ kgxxv_2`
`=>v_2=(0.233\ kg\ m//s)/(0.2\ kg)`

Thus, velocity of the second object after collision `=1.165 ms^(-1)`