Question

A ball is let to fall from a height $h_s$. It makes ${}^{\prime }{n}^{\prime }$ collisions with the horizontal ground. If after ${}^{\prime }{n}^{\prime }$ collisions it rebounds with a velocity $v_a$ and the ball rises to a height $h_a$, then the coefficient of restitution for the collision is :

Answer 1

Height = $h_0$

collision is = $v_n$

height = $h_n$

coefficient of restitution = e

The velocity of the earth before and after collision may be assumed zero.

Hence $e^n=\frac{v_1}{v_0}\times \frac{v_2}{v_1}\times \frac{v_3}{v_2}\times .........\frac{v_n}{v_{n-1}}$

$v_n$ = velocity after $n^{th}$ rebound

$v_0$ = velocity with which the ball strikes the earth first time

Hence $e^n$ = $\frac{v_n}{v_0}$

= $\frac{\sqrt{2g{h}_n}}{\sqrt{2g{h}_0}}$

$e^n=\sqrt{\frac{h_n}{h_0}}$