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(A) The velocity of each bead just before hitting the pan is $v = \sqrt{2gh}$.Given $g = 10 \ m/s^2$ and $h = 5 \ m$, we have $v = \sqrt{2 	imes 10 \

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$3$

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(A) The velocity of each bead just before hitting the pan is $v = \sqrt{2gh}$.Given $g = 10 \ m/s^2$ and $h = 5 \ m$, we have $v = \sqrt{2 	imes 10 	imes 5} = 10 \ m/s$.Since the beads bounce back to their original height, the collision is perfectly elastic, and the velocity just after the collision is also $v = 10 \ m/s$ upwards.The change in momentum for one bead is $\Delta p = m(v - (-v)) = 2mv$.Given $m = 15 \ g = 0.015 \ kg$, $\Delta p = 2 	imes 0.015 	imes 10 = 0.3 \ kg \ m/s$.The rate of change of momentum (force) exerted on the pan by $n = 100 \ 	ext{beads per second}$ is $F = n 	imes \Delta p = 100 	imes 0.3 = 30 \ N$.To keep the pointer at zero, the weight of the mass $M$ placed in the other pan must balance this force: $Mg = F$.$M 	imes 10 = 30 \implies M = 3 \ kg$.

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