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A ball falls freely from a height $\mathrm{h}$ on a rigid horizontal plane. If the coefficient of restitution is e, then the total distance travelled by the ball before hitting the plane second time is
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Verified Answer
The correct answer is:
$h\left(1+2 e^2\right)$
As ball dropped from height $h$, then it will strike the
plane with velocity, $v=\sqrt{2 g h}$. Now after striking the plane, it will bounce back with velocity, $u=e v$
Now it will travel total $2 \mathrm{~h}_1$ distance before
Here $h_1=\frac{e^2 v^2}{2 g}$
So, total distance $=\mathrm{h}+2 \mathrm{~h}_1$
$$
\begin{aligned}
& =\frac{\mathrm{v}^2}{2 \mathrm{~g}}+\frac{2 \mathrm{e}^2 \mathrm{v}^2}{2 \mathrm{~g}} \\
& =\frac{\mathrm{v}^2}{2 \mathrm{~g}}\left[1+2 \mathrm{e}^2\right] \\
& =\frac{2 \mathrm{gh}}{2 \mathrm{~g}}\left[1+2 \mathrm{e}^2\right] \\
& =\mathrm{h}\left(1+2 \mathrm{e}^2\right)
\end{aligned}
$$
plane with velocity, $v=\sqrt{2 g h}$. Now after striking the plane, it will bounce back with velocity, $u=e v$
Now it will travel total $2 \mathrm{~h}_1$ distance before
Here $h_1=\frac{e^2 v^2}{2 g}$
So, total distance $=\mathrm{h}+2 \mathrm{~h}_1$
$$
\begin{aligned}
& =\frac{\mathrm{v}^2}{2 \mathrm{~g}}+\frac{2 \mathrm{e}^2 \mathrm{v}^2}{2 \mathrm{~g}} \\
& =\frac{\mathrm{v}^2}{2 \mathrm{~g}}\left[1+2 \mathrm{e}^2\right] \\
& =\frac{2 \mathrm{gh}}{2 \mathrm{~g}}\left[1+2 \mathrm{e}^2\right] \\
& =\mathrm{h}\left(1+2 \mathrm{e}^2\right)
\end{aligned}
$$
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