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Question

(A) When a particle falls from height $h$,it hits the ground with velocity $v = \sqrt{2gh}$.After the first collision,it rebounds to a height $h_1 = e

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) When a particle falls from height $h$,it hits the ground with velocity $v = \sqrt{2gh}$.After the first collision,it rebounds to a height $h_1 = e^2h$.After the second collision,it rebounds to a height $h_2 = e^2h_1 = e^4h$.In general,the height after the $n^{th}$ rebound is $h_n = e^{2n}h$.The total distance $D$ travelled by the particle is the sum of the initial downward distance and twice the sum of all subsequent rebound heights (since it goes up and down for each rebound):$D = h + 2h_1 + 2h_2 + 2h_3 + \dots$$D = h + 2(e^2h + e^4h + e^6h + \dots)$$D = h + 2he^2(1 + e^2 + e^4 + \dots)$Using the sum of an infinite geometric series $S = \frac{a}{1-r}$ where $a = 1$ and $r = e^2$:$D = h + 2he^2 \left( \frac{1}{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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