A rod of length L and mass M is free to move | vedclass

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(A) Let $V$ be the velocity of the center of mass of the rod after the collision. By conservation of linear momentum:$mv = MV \implies V = \frac{mv}{M

Vedclass · Accepted Answer

$\frac{ML^2}{(L^2 + 12d^2)}$

Vedclass · Answer

(A) Let $V$ be the velocity of the center of mass of the rod after the collision. By conservation of linear momentum:$mv = MV \implies V = \frac{mv}{M}$By conservation of angular momentum about the center of mass of the rod:$mvd = I\omega \implies \omega = \frac{mvd}{I}$,where $I = \frac{ML^2}{12}$Assuming the collision is elastic,by conservation of kinetic energy:$\frac{1}{2}mv^2 = \frac{1}{2}MV^2 + \frac{1}{2}I\omega^2$Substituting $V$ and $\omega$:$mv^2 = M(\frac{mv}{M})^2 + I(\frac{mvd}{I})^2$$mv^2 = \frac{m^2v^2}{M} + \frac{m^2v^2d^2}{I}$Dividing by $mv^2$:$1 = \frac{m}{M} + \frac{md^2}{I} = \frac{m}{M} + \frac{md^2}{ML^2/12} = \frac{m}{M}(1 + \frac{12d^2}{L^2})$$1 = \frac{m}{M}(\frac{L^2 + 12d^2}{L^2})$$m = \frac{ML^2}{L^2 + 12d^2}$

$A$ rod of length $L$ and mass $M$ is free to move on a frictionless surface. $A$ ball of mass $m$ moving with speed $v$ strikes the rod as shown in the figure. What should be the mass of the ball so that it remains at rest after the collision?

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