A 0.100-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wall and rebounds with 70.0% of its initial kinetic energy. What is the magnitude of the change in momentum of the stone

Change in momentum of the stone is 3.673 kg.m/s.

Explanation:

Given:

Mass of the ball on the horizontal the surface, m = 0.10 kg

Velocity of the ball with which it hits the stone, v = 20 m/s

According to the question it rebounds with 70% of the initial kinetic energy.

We have to find the change in momentum i.e Δp

Before that:

We have to calculate the rebound velocity with which the object rebounds.

Lets say that the rebound velocity be "v1" and KE remaining after the object rebounds be "KE1".

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⇒

⇒

⇒ Joules (J).

Rebound velocity "v1".

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⇒ m/s ...as it rebounds.

Change in momentum Δp.

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⇒

⇒

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⇒ Kg.m/s

The magnitude of the change in momentum of the stone is 3.673 kg.m/s.

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what is the speed of a softball that travels 20m in 2secs

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