Question

A ball is dropped from height $H$ on to a horizontal surface. If the coefficient of restitution is $e$, then the total time after which it comes to rest is

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

Answer 1

The correct option is B $\sqrt{\frac{2H}{g}}\left(\frac{1+e}{1-e}\right)$

If a body drops from height $h_1$ and bounces back to $h_2$ then

$e=\sqrt{\frac{h_2}{h_1}}$

$h_2=e^2{h}_1$

time elapsed between falling from height $h_1$ and going back to $h_2$

$\frac{2\times \sqrt{2g{h}_2}}{g}=t_1$

$t_2=\frac{2\times \sqrt{2g{h}_3}}{g}$

${\sum }_{n=1}^{n=\infty }{t}_i=\frac{2}{g}\times \sqrt{2g}(\sqrt{h_n}+\sqrt{h_{n-1}}........\sqrt{h_2})+\frac{\sqrt{2gH}}{g}$

$=\frac{2}{g}\times \sqrt{2g}(e^{n-1}...........e^2\sqrt{h_1}+e\sqrt{h_1})+\frac{\sqrt{2gH}}{g}$

$=\frac{2}{g}\times \sqrt{2g}\times \sqrt{h_1}\times \frac{1}{1-e}+\frac{\sqrt{2gH}}{g}$

now $\sqrt{h_1}=e\sqrt{H}$

$t\Rightarrow t=\frac{2}{g}\sqrt{2g}\times e\sqrt{H}\times \frac{1}{1-e}+\frac{\sqrt{2gH}}{g}=\frac{\sqrt{2gH}}{g}\left(\frac{1+e}{1-e}\right)$