Question

A particle falls from a height 'h' upon a fixed horizontal plane and rebounds. If 'e' is the coefficient of restitution the total distance travelled before rebounding has stopped is :-

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

Answer 1

The correct option is A $h\left(\frac{1+e^2}{1-e^2}\right)$

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 (u)=$ev$

now it will travel total $2h_1$ distance before again striking the ground. where $h_1$=$\frac{e^2{v}^2}{2g}$

again the ball will bounce back with $e^{2}v$ velocity and now it will travel $2h_2$ distance before striking the ground,where $h_1$=$\frac{e^4{v}^2}{2g}$ and this process continues for long time till ball comes to the rest.

So total distance travelled=$h+2h_1+{2h}_2+.....$

=$\frac{v^2}{2g}+2\frac{e^2{v}^2}{2g}+2\frac{e^4{v}^2}{2g}+2\frac{e^6{v}^2}{2g}+........$

=$\frac{v^2}{2g}+2\frac{e^2{v}^2}{2g}(1+e^2+e^4+e^6+....)$

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

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

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

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