Question

A particle falls from a height h on a fixed horizontal plate and rebounds. If e is the coefficient of restitution, the total distance travelled by the particle before it stops rebounding is

• A. $\frac{h(1+e^2)}{(1-e^2)}$
• B. $\frac{h(1-e^2)}{(1+e^2)}$
• C. $\frac{h(1-e^2)}{2(1+e^2)}$
• D. $\frac{h(1+e^2)}{2(1-e^2)}$

Answer 1

Answer: (A)

. $\frac{h(1+e^2)}{(1-e^2)}$.

As ball dropped fro height '$h$', then it will strike the floor with velocity,

$v=\sqrt{2gh}$

Now after striking the floor it will bounce back with velocity $\left(u\right)=ev$

now it will travel total $2h_1$ distance before again striking the ground, where

$h_1=\frac{u^2}{2g}=\frac{e^2{v}^2}{2g}$

again the ball will bounce back with $e^{2}v$ velocity and now it will travel $2h_2$ distance before striking the ground, where

$h_2=\frac{e^4{v}^2}{2g}$

For long time till ball comes to the rest.

So total distance travelled = $h+2h_1+2h_2+.......$

$=\frac{v^2}{2g}+2\frac{e^2{v}^2}{2g}+2\frac{e^4{v}^2}{2g}+.......$

$=\frac{v^2}{2g}+2\frac{e^2{v}^2}{2g}(1+e^2+e^4+e^3+........)$

$=\frac{v^2}{2g}+\frac{e^2{v}^2}{g}\left(\frac{1}{1-e^2}\right)$

$=\left(\frac{1-e^2+2e^2}{1-e^2}\right)\frac{v^2}{2g}$

$=\left(\frac{1+e^2}{1-e^2}\right)\frac{v^2}{2g}$

$=\left(\frac{1+e^2}{1-e^2}\right)h$.