Page No 129: - Chapter 9 Force & Laws Of Motion class 9 ncert solutions Science - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

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.

Answer:

The truck has a large mass. Therefore, the static friction between the truck and the road is also very high. To move the car, one has to apply a force more than the static friction. Therefore, when someone pushes the truck and the truck does not move, then it can be said that the applied force in one direction is cancelled out by the frictioal force of equal amount acting in the opposite direction.

Therefore, the student is right in justifying that the two opposite and equal cancel each other.

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.

Answer:

Mass of the hockey ball, m = 200 g = 0.2 kg

Hockey ball travels with velocity, v₁ = 10 m/s

Initial momentum = mv₁

Hockey ball travels in the opposite direction with velocity, v₂ = −5 m/s

Final momentum = mv₂

Change in momentum = mv₁ − mv₂ = 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_.

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.

Answer:

Now, it is given that the bullet is travelling with a velocity of 150 m/s.

Thus, when the bullet enters the block, its velocity = Initial velocity, u = 150 m/s

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)

(Negative sign indicates that the velocity of the bullet is decreasing.)

According to the third equation of motion:

v² = u² + 2as

0 = (150)² + 2 (−5000) s

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/s²

F = ma = 0.01 × 5000 = 50 N

Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.

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.

Answer:

Mass of the object, m₁ = 1 kg

Velocity of the object before collision, v₁ = 10 m/s

Mass of the stationary wooden block, m₂ = 5 kg

Velocity of the wooden block before collision, v₂ = 0 m/s

∴ Total momentum before collision = m₁ v₁ + m₂ v₂

= 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 v₂ = (m₁ + m₂) v

1 (10) + 5 (0) = (1 + 5) v

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 = 

Hence, velocity of the combined object after collision =

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.

Answer:

Initial velocity of 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 =

Initial momentum of the object is 500 kg m s⁻¹.

Final momentum of the object is 800 kg m s⁻¹.

Force exerted on the object is 50 N.

Question 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.

Answer:

According to the law of conservation of momentum:

Momentum of the car and insect system before collision = Momentum of the car and insect system after collision

Hence, the change in momentum of the car and insect system is zero.

The insect gets stuck on the windscreen. This means that the direction of the insect is reversed. As a result, the velocity of the insect changes to a great amount. On the other hand, the car continues moving with a constant velocity. Hence, Kiran’s suggestion that the insect suffers a greater change in momentum as compared to the car is correct. The momentum of the insect after collision becomes very high because the car is moving at a high speed. Therefore, the momentum gained by the insect is equal to the momentum lost by the car.

Akhtar made a correct conclusion because the mass of the car is very large as compared to the mass of the insect.

Rahul gave a correct explanation as both the car and the insect experienced equal forces caused by the Newton’s action-reaction law. But, he made an incorrect statement as the system suffers a change in momentum because the momentum before the collision is equal to the momentum after the collision.

Question 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/s²

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:

v² = u² + 2as

v² = 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