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In the figure shown the cart of mass 6 m is initially at rest on a smooth horizontal surface. A particle of mass m is attached to the end of the light rigid rod which can rotate freely about end A in a vertical plane. If the rod is released from rest from the horizontal position as shown, find the velocity $(v_{rel})$ of the particle with respect to the cart when the rod is vertical.

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$\sqrt{\frac{14}{3} g l}$

$\sqrt{\frac{8}{3} g l}$

$\sqrt{\frac{7}{3} g l}$

$\sqrt{\frac{7}{6} g l}$

answer is A.

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Detailed Solution

Using conservation of momentum :

$m (v_{r} - v) = 6 mv$

$\Rightarrow v_{r} = 7 v \dots .... (1)$

Now, using conservation of energy :

$m g l = \frac{1}{2} m (6 v)^{2} + \frac{1}{2} 6 m v^{2}$

$\Rightarrow v = \sqrt{\frac{g l}{21}}$

$∴ v_{r} = 7 v = \sqrt{\frac{7 g l}{3}}$

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