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Question

(A) The particle falls from height $h$. The first impact velocity is $v_0 = \sqrt{2gh}$.After the first collision,the rebound velocity is $v_1 = ev_0$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The particle falls from height $h$. The first impact velocity is $v_0 = \sqrt{2gh}$.After the first collision,the rebound velocity is $v_1 = ev_0$. The height reached is $h_1 = \frac{v_1^2}{2g} = e^2h$.The particle travels $h$ downwards,then $h_1$ upwards and $h_1$ downwards.After the second collision,it reaches $h_2 = e^2h_1 = e^4h$,traveling $h_2$ up and $h_2$ down.The total distance $D$ is given by:$D = h + 2h_1 + 2h_2 + 2h_3 + ...$$D = h + 2(e^2h + e^4h + e^6h + ...)$$D = h + 2h(e^2 + e^4 + e^6 + ...)$Using the sum of an infinite geometric series $S = \frac{a}{1-r}$ where $a = e^2$ and $r = e^2$:$D = h + 2h \left( \frac{e^2}{1-e^2} \right)$$D = h \left( 1 + \frac{2e^2}{1-e^2} \right) = h \left( \frac{1-e^2+2e^2}{1-e^2} \right) = h \left( \frac{1+e^2}{1-e^2} \right)$.

$A$ 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

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