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Question

(D) The change in momentum during each collision is the impulse imparted to the floor.For the first impact, the ball arrives with momentum $p$ (downwa

Vedclass · Accepted Answer

$p(\frac{1+e}{1-e})$

Vedclass · Answer

(D) The change in momentum during each collision is the impulse imparted to the floor.For the first impact, the ball arrives with momentum $p$ (downward) and rebounds with $ep$ (upward). The change in momentum is $\Delta p_1 = p - (-ep) = p(1+e)$.For the second impact, the ball arrives with momentum $ep$ (downward) and rebounds with $e^2p$ (upward). The change in momentum is $\Delta p_2 = ep - (-e^2p) = ep(1+e)$.For the third impact, the change is $\Delta p_3 = e^2p(1+e)$, and so on.The total momentum imparted is the sum of all these impulses:$\Delta p_{total} = p(1+e) + ep(1+e) + e^2p(1+e) + \dots$$\Delta p_{total} = p(1+e) [1 + e + e^2 + \dots]$Using the sum of an infinite geometric series $S = \frac{1}{1-e}$ for $|e| $\Delta p_{total} = p(1+e) \cdot \frac{1}{1-e} = p(\frac{1+e}{1-e})$.

$A$ ball falls vertically onto a floor with momentum $p$ and then bounces repeatedly. If the coefficient of restitution is $e$, then the total momentum imparted by the ball on the floor is:

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