Answer:
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".
⇒
⇒
⇒
⇒ Joules (J).
Rebound velocity "v1".
⇒
⇒
⇒
⇒
⇒ m/s ...as it rebounds.
Change in momentum Δp.
⇒
⇒
⇒
⇒
⇒ Kg.m/s
The magnitude of the change in momentum of the stone is 3.673 kg.m/s.
