A particle falls from a height h on a static | vedclass

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Question

(A) The particle falls from height $h$. After the first collision,it rebounds to height $h_1 = e^2 h$.After the second collision,it rebounds to height

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The particle falls from height $h$. After the first collision,it rebounds to height $h_1 = e^2 h$.After the second collision,it rebounds to height $h_2 = e^2 h_1 = e^4 h$,and so on.The total distance $S$ traveled is the sum of the initial fall and all subsequent upward and downward paths:$S = h + 2h_1 + 2h_2 + 2h_3 + \dots$$S = h + 2(e^2 h) + 2(e^4 h) + 2(e^6 h) + \dots$$S = h + 2e^2 h (1 + e^2 + e^4 + \dots)$This is a geometric series with first term $a = 1$ and common ratio $r = e^2$. The sum is $\frac{1}{1 - e^2}$.$S = h + 2e^2 h \left( \frac{1}{1 - e^2} \right)$$S = h \left( 1 + \frac{2e^2}{1 - e^2} \right) = h \left( \frac{1 - e^2 + 2e^2}{1 - e^2} \right)$$S = h \left( \frac{1 + e^2}{1 - e^2} \right)$

$A$ particle falls from a height $h$ on a static horizontal plane and rebounds. If $e$ is the coefficient of restitution,then the total distance traveled by the particle before coming to rest will be:

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