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(D) Let the stick have mass $M$ and length $L$. The ball of mass $m$ hits the end of the stick with velocity $v$ and comes to rest.$1$. Conservation o

Vedclass · Accepted Answer

$m = M/4$

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(D) Let the stick have mass $M$ and length $L$. The ball of mass $m$ hits the end of the stick with velocity $v$ and comes to rest.$1$. Conservation of linear momentum: $mv = MV$,where $V$ is the velocity of the center of mass of the stick. So,$V = \frac{mv}{M}$.$2$. Conservation of angular momentum about the center of mass of the stick: $mv(L/2) = I\omega$,where $I = \frac{ML^2}{12}$ is the moment of inertia of the stick about its center. Thus,$mv(L/2) = \frac{ML^2}{12} \omega$,which gives $\omega = \frac{6mv}{ML}$.$3$. Conservation of kinetic energy (elastic collision): $\frac{1}{2}mv^2 = \frac{1}{2}MV^2 + \frac{1}{2}I\omega^2$.Substituting $V$ and $\omega$: $\frac{1}{2}mv^2 = \frac{1}{2}M(\frac{mv}{M})^2 + \frac{1}{2}(\frac{ML^2}{12})(\frac{6mv}{ML})^2$.$\frac{1}{2}mv^2 = \frac{1}{2}\frac{m^2v^2}{M} + \frac{1}{2}(\frac{ML^2}{12})(\frac{36m^2v^2}{M^2L^2})$.$mv^2 = \frac{m^2v^2}{M} + \frac{3m^2v^2}{M}$.$1 = \frac{m}{M} + \frac{3m}{M} = \frac{4m}{M}$.Therefore,$m = M/4$.

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