Question

A body from height h is dropped. If the coefficient of restitution is e, then calculate the height achieved after one bounce.

• A. $h_{1}=e^{2} h$
• B. $h_{1}=e^{4} h$
• C. $ h_{1}=eh $
• D. $h_{1}=\frac{h}{e}$

Answer 1

Answer: (A)

When a body falls from a height $h$,
it strikes the ground with a velocity $u=\sqrt{2 g h}$.
Let it rebouneds with a velocity $v$ and rise to a height hi.
Therefore, $u=\sqrt{2 g h_{1}}$
$\therefore e=\frac{v}{u}=\sqrt{\left(\frac{h_{1}}{h}\right)}$
If the successive heights to which the body rebounds again and again are
$h_{2}, h_{3}, \ldots$ then $e=\sqrt{\frac{h_{1}}{h}}=\sqrt{\frac{h_{2}}{h_{1}}}=\sqrt{\frac{h_{3}}{h_{2}}}=\ldots$
Clearly, $h_{1}=e^{2} h, h_{2}=e^{4} h . . .$ and so on.
Similarly, after nth rebound $h_{n}=\left(e^{2 n}\right) h$
Note: The degree of elasticity of a collision is determined by a quantity, called coefficient of restitution or coefficient of resislience of the collision. It is defined as the ratio of relative velocity of separation after collision to the relative velocity of approach before collision. It is represented by e.