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A uniform rod $A B$ of length $l$ and mass $m$ is laying on a smooth table. A small particle of mass $m$ strike the rod with a velocity is at $C$ a distance $x$ from the center $O$ . The particle comes to rest after collision. The value of $x$ , so that point $A$ of the rod remains stationary just after collision is

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$\frac{l}{6}$

$\frac{l}{4}$

$\frac{l}{12}$

$\frac{l}{3}$

answer is B.

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Detailed Solution

$m v_{0} + 0 = m \times 0 + m v v = v_{0}$

Also $m v_{0} x = I ω = \frac{m l^{2}}{12} ω$

or $ω = \frac{12 v_{0} x}{l^{2}}$

For end A to be stationary

$v_{A} = v_{0} - ω \frac{l}{2} \to 0 = v_{0} - \frac{12 v_{0} x}{l^{2}} \times \frac{l}{2} \Rightarrow x = \frac{l}{6}$

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