Solution of Force and Law of Motion (Page No - 76 & 77) - Physics by Lakhmir Singh, Class 9 | Extra Documents & Tests for Class 9 PDF Download

Question 39:
(a) State the law of conservation of momentum.
(b) Discuss the conservation of momentum in each of the following cases :
(i) a rocket taking off from ground.
(ii) flying of a jet aeroplane.
Solution :
a) According to the law of conservation of momentum: When two (or more) bodies act upon one another, their total momentum remains constant (or conserved) provided no external forces are acting. It means that when one body gains momentum, then some other body loses an equal amount of momentum i.e. momentum is neither created nor destroyed.
b)
a. Rocket taking off from the ground The chemicals inside the rocket burn and produce very high velocity blast of hot gases. These gases pass out through the tail nozzle of the rocket in the downward direction with tremendous speed and the rocket moves up to balance the momentum of the gases. The gases have a very high velocity ang hence a very large momentum. An equal momentum is imparted to the rocket in the opposite direction, so that it goes up with a high velocity.
b. Flying of jet aeroplane
In jet aeroplanes, a large volume of gases produced by the combustion of fuel is allowed to escape through a jet in backward direction. Due to high velocity, the backward rushing gases have a large momentum. They impart an equal and opposite momentum to the jet aeroplane due to which it moves forward with a great speed.

Question 40:
(a) If a balloon filled with air and its mouth untied/ is released with its mouth in the downward direction, it
moves upwards. Why ?
(b) An unloaded truck weighing 2000 kg has a maximum acceleration of 0.5 m/s². What is the maximum acceleration when it is carrying a load of 2000 kg ?
Solution :
a) If of a balloon filled with compressed air and its mouth untied is released with its mouth in the downward direction, the balloon moves in the upward direction because the air present in the balloon rushes out in the downward direction. The equal and opposite reaction of downward going air pushes the balloon upwards.
b) Mass of the unloaded truck, m₁= 2000 kg
Acceleration a₁ = 0.5 m/s²
Mass of loaded truck, m₂ = 2000+ 2000 = 4000 kg
Acceleration a₂
m₁ x a₁ = m₂ x a₂

Lakhmir Singh Physics Class 9 Solutions Page No:77 

Question 51:
Why are car seat-belts designed to stretch some what in a collision ?
Solution :
Car seat-belts are somewhat stretchable so as to increase the time taken by the passengers to fall forward. Due to this, the rate of change of momentum of passengers is reduced and hence less stopping force acts on them. So the passengers do not get hurt.

Question 52:
The troops (soldiers) equipped to be dropped by parachutes from an aircraft are called paratroopers. Why do paratroopers roll on landing ?

Solution :
The paratroopers roll on landing to increase the time taken to reduce the momentum of their body. Thus, the rate of change of momentum is reduced and hence less force is exerted on their legs and they do not get hurt.

Question 53:
Why would an aircraft be unable to fly on the moon ?
Solution :
An aircraft needs air because air moving under the wings of aircraft is strong enough to hold it up and air is also required to burn the fuel in aircraft engines. Since there is no air on moon, an aircraft cannot fly on moon.

Question 54:
Explain why it is possible for a small animal to fall from a considerable height without any injury being caused when it reaches the ground.
Solution :
It is possible for a small animal to fall from a considerable height without being injured because a small animal has small mass, so the momentum produced is less. When the small animal falls to the ground with less momentum, less opposing force of ground acts on it and hence no injury is caused to it

Question 55:
A boy of mass 50 kg running at 5 m/s jumps on to a 20 kg trolley travelling in the same direction at 1.5 m/s. What is their common velocity ?
Solution :
Mass of the boy, m₁= 50 kg
Speed of boy, u₁ = 5 m/s
Mass of trolley m₂ = 20 kg
Speed of trolley u₂ =1.5 m/s
Combined mass of boy and trolley, m = 20+ 50 = 70 kg
Combined velocity v
Acc. to the law of conservation of momentum
m₁u₁ + m₂u₂ = mv
50 x 5 + 20 x 1.5 = 70 x v

Question 56:
A girl of mass 50 kg jumps out of a rowing boat of mass 300 kg on to the bank, with a horizontal velocity of 3 m/s. With what velocity does the boat begin to move backwards ?
Solution :
Mass of the boat m_b = 300 kg
Velocity of boat v_b
Mass of girl m_g = 50 kg
Velocity of girl v_g = 3 m/s
Acc. to the law of conservation of momentum
m_bv_b=m_gv_g
300 x v_b = 50 x 3 

Question 57:
A truck of mass 500 kg moving at 4 m/s collides with another truck of mass 1500 kg moving in the same direction at 2 m/s. What is their common velocity just after the collision if they move off together ?
Solution :
Mass of first truck, m₁= 500 kg
Speed of first truck, u₁ = 4 m/s
Mass of second truck, m₂ = 1500 kg
Speed of second truck, u₂ =2 m/s
Combined mass of both trucks, m = 1500 + 500 = 2000 kg
Combined velocity v
Acc. to the law of conservation of momentum
m₁u₁ + m₂u₂ = mv
500 x 4 + 1500 x 2 = 2000 x v 

Question 58:
A ball X of mass 1 kg travelling at 2 m/s has a head-on collision with an identical ball Y at rest. X stops and Y moves off. Calculate the velocity of Y after the collision.
Solution :
Mass of the ball x, m₁= 1 kg
Speed of ball x, u₁ = 2 m/s
Mass of ball y, m₂ = 1 kg
Speed of ball y, u₂ =0 m/s (at rest)
Velocity of ball x after collision, v₁ = 0 m/s
Velocity of ball y after collision, v₂
Acc. to the law of conservation of momentum
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
1 x 2 + 1 x 0 = 1 x 0 + 1 x v₂ 

Question 59:
A heavy car A of mass 2000 kg travelling at 10 m/s has a head-on collision with a sports car B of mass 500 kg. If both cars stop dead on colliding, what was the velocity of car B ?
Solution :
Mass of car A, m₁= 2000 kg
Speed of car A, v₁ = 10 m/s
Mass of car B, m₂ = 500 kg
Speed of car B, v₂
Acc to law of conservation of momentum
m₁v₁ = m₂v₂
2000 x 10 = 500 x v₂ 

Question 60:
A man wearing a bullet-proof vest stands still on roller skates. The total mass is 80 kg. A bullet of mass 20 grams is fired at 400 m/s. It is stopped by the vest and falls to the ground. What is then the velocity of the man ?
Solution :
Mass of the man, m₁= 80 kg
Speed of man, v₁
Mass of bullet m₂ = 20 g = 0.02 kg
Speed of bullet v₂ = 400 m/s
Acc to law of conservation of momentum
m₁v₁ = m₂v₂
80 x v₁ = 0.02 x 400