A uniform rod AB of length L and mass M is ly | vedclass

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(B) After the collision,the rod will have both translational and rotational motion about its centre of mass $(C.M.)$.From the conservation of linear m

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$L/6$

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(B) After the collision,the rod will have both translational and rotational motion about its centre of mass $(C.M.)$.From the conservation of linear momentum:$m v_0 = M V_{CM} \implies V_{CM} = \frac{m v_0}{M} \dots(1)$From the conservation of angular momentum about the $C.M.$:$m v_0 x = I \omega = \left(\frac{M L^2}{12}\right) \omega$$\omega = \frac{12 m v_0 x}{M L^2} \dots(2)$For point $A$ to be stationary,its velocity must be zero. The velocity of point $A$ is the vector sum of the translational velocity of the $C.M.$ and the rotational velocity about the $C.M.$:$V_A = V_{CM} - \omega \left(\frac{L}{2}\right) = 0$$V_{CM} = \omega \left(\frac{L}{2}\right) \dots(3)$Substituting equations $(1)$ and $(2)$ into equation $(3)$:$\frac{m v_0}{M} = \left(\frac{12 m v_0 x}{M L^2}\right) \left(\frac{L}{2}\right)$$\frac{m v_0}{M} = \frac{6 m v_0 x}{M L}$$1 = \frac{6x}{L} \implies x = \frac{L}{6}$

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