Q127 / 160 · TS EAMCET 2023 (13 May Shift 1) · Physics

A body falls freely from a height ' $\mathrm{h}$ ' on a fixed horizontal plane and rebounds. If ' $\mathrm{e}$ ' is the coefficient of restitution, the total distance travelled before it comes to rest is

  • Correct answer
    A
    $\mathrm{h}\left[\frac{1+\mathrm{e}^2}{1-\mathrm{e}^2}\right]$
  • B
    $\mathrm{h}\left[\frac{1-\mathrm{e}^2}{1+\mathrm{e}^2}\right]$
  • C
    $\frac{\mathrm{h}}{2}\left[\frac{1-\mathrm{e}^2}{1+\mathrm{e}^2}\right]$
  • D
    $\frac{\mathrm{h}}{2}\left[\frac{1+\mathrm{e}^2}{1-\mathrm{e}^2}\right]$

Solution

When a ball dropped from a height $h$, then it will strick the floor with velocity $v=\sqrt{2 g h}$ Now, after strick the floor it will bounce back with velocity, $\mathrm{u}=\mathrm{ev}$ Now it will travel $2 \mathrm{~h}_1$ distance before again striking the ground. Here $h_1=\frac{e^2 v^2}{2 g}$ again the ball will bounce back with $e^2 v$ velocity. This process continues for long time till ball comes to the rest. So, total distance travelled $=\mathrm{h}+2 \mathrm{~h}_1+2 \mathrm{~h}_2+\ldots$ $$ \begin{aligned} & =\frac{v^2}{2 g}+\frac{2 e^2 v^2}{2 g}+\frac{2 e^4 v^2}{2 g}+\ldots \\ & =\frac{v^2}{2 g}\left[1+2 e^2+2 e^4+2 e^6+\ldots\right] \\ & =\frac{v^2}{2 g}+\frac{2 e^2 v^2}{2 g}-\left[1+e^2+e^4+\ldots\right] \\ & =\frac{v^2}{2 g}+\frac{e^2 v^2}{g}\left(\frac{1}{1-e^2}\right) \\ & =\frac{v^2}{2 g}\left[\frac{1-e^2+2 e^2}{1-e^2}\right]=\frac{v^2}{2 g}\left(\frac{1+e^2}{1-e^2}\right)=h\left(\frac{1+e^2}{1-e^2}\right) \end{aligned} $$

Hence, the correct answer is option: 'A'.

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