A stick of length L and mass M lies on a fric | vedclass

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(D) Let the stick have mass $M$ and length $L$. The ball has mass $m$ and initial velocity $v$. After the elastic collision,the ball comes to rest,and

Vedclass · Accepted Answer

$m=M/4$

Vedclass · Answer

(D) Let the stick have mass $M$ and length $L$. The ball has mass $m$ and initial velocity $v$. After the elastic collision,the ball comes to rest,and the stick moves with linear velocity $v_2$ and angular velocity $\omega$ about its center of mass.$1$. Conservation of linear momentum:$mv = Mv_2 \implies v_2 = \frac{mv}{M} \quad ...(1)$$2$. Conservation of angular momentum about the center of mass of the stick:$mv(L/2) = I\omega = \left(\frac{ML^2}{12}\right)\omega \implies \omega = \frac{6mv}{ML} \quad ...(2)$$3$. Conservation of kinetic energy (elastic collision):$\frac{1}{2}mv^2 = \frac{1}{2}Mv_2^2 + \frac{1}{2}I\omega^2$Substituting $v_2$ and $\omega$:$\frac{1}{2}mv^2 = \frac{1}{2}M\left(\frac{mv}{M}\right)^2 + \frac{1}{2}\left(\frac{ML^2}{12}\right)\left(\frac{6mv}{ML}\right)^2$$mv^2 = \frac{m^2v^2}{M} + \frac{ML^2}{12} \cdot \frac{36m^2v^2}{M^2L^2}$$mv^2 = \frac{m^2v^2}{M} + \frac{3m^2v^2}{M}$$mv^2 = \frac{4m^2v^2}{M}$$1 = \frac{4m}{M} \implies m = \frac{M}{4}$

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