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A rod AB of mass m and length l lies on a smooth horizontal surface, end A is pivoted at a fixed point so that the rod can rotate on the horizontal surface about a vertical surface through A. Now particle of mass m moving with velocity v strikes the rod normally at end B and sticks to it. Find the angular velocity of the rod.

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$\frac{2 v}{3 l}$

$\frac{3 v}{4 l}$

$\frac{4 v}{3 l}$

$\frac{3 v}{2 l}$

answer is B.

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Detailed Solution

Conserving angular momentum,

$(\frac{1}{3} {ml}^{2} + {ml}^{2}) ω = mvl \Rightarrow ω = \frac{3}{4} \frac{v}{l}$

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