Question

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

Answer 1

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".

⇒ $KE_1=0.7\times \frac{mv^2}{2}$    

⇒ $KE_1=0.7\times \frac{0.10\times (20)^2}{2}$

⇒ $KE_1=0.7\times \frac{0.10\times 400}{2}$

⇒ $KE_1=14$ Joules (J).

Rebound velocity "v1".

⇒ $KE_1=\frac{m(v_1)^2}{2}$

⇒ $v_1 = \sqrt{\frac{2KE_1}{m} }$

⇒ $v_1 = \sqrt{\frac{2\times 14}{0.10} }$

⇒ $v_1=16.73$

⇒ $v_1=-16.73$ m/s ...as it rebounds.

Change in momentum Δp.

⇒ $\triangle p= m\triangle v$

⇒ $\triangle p= 0.10\times (20-(-16.73)$

⇒ $\triangle p= 0.10\times (20+16.73)$

⇒ $\triangle p= 0.10\times (36.73)$

⇒ $\triangle p = 3.673$ Kg.m/s

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