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 collision,the ball comes to rest,and the sti

Vedclass · Accepted Answer

$m = \frac{M}{4}$

Vedclass · Answer

(D) Let the stick have mass $M$ and length $l$. The ball has mass $m$ and initial velocity $v$. After the collision,the ball comes to rest,and the stick moves with linear velocity $V$ and angular velocity $\omega$ about its center of mass.$1$. Conservation of linear momentum:$mv = MV \implies V = \frac{mv}{M} \quad \dots(1)$$2$. Conservation of angular momentum about the center of mass of the stick:The ball hits at a distance $l/2$ from the center.$mv(l/2) = I\omega = \left(\frac{Ml^2}{12}\right)\omega \implies \omega = \frac{6mv}{Ml} \quad \dots(2)$$3$. Conservation of kinetic energy (elastic collision):$\frac{1}{2}mv^2 = \frac{1}{2}MV^2 + \frac{1}{2}I\omega^2$Substitute $V$ and $\omega$:$mv^2 = M\left(\frac{mv}{M}\right)^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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