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A uniform rod of AB of mass m and length $l$ is at rest on a smooth horizontal surface. An impulse J is applied to the end B perpendicular to the rod in horizontal direction. Speed of particle P at a distance L/6 from the centre towards A of the rod after time
$\large t = \frac{{\pi ml}}{{12J}}$
is

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$\frac{J}{m}$

$\sqrt 2 \frac{J}{m}$

$\frac{J}{{\sqrt 2 m}}\,\$

answer is D.

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Detailed Solution

mV_cm = J

Moment of J about COM of the rod is equal to change in angular momentum about the same point.

about C.M. tangential velocity

\begin{array}{l} {{\rm{v}}_{\rm{t}}} = \frac{l}{6}\omega = \frac{l}{6}\left( {\frac{{6{\rm{J}}}}{{ml}}} \right) = \frac{{\rm{J}}}{m}\\\\\\ {{\rm{V}}_{\rm{P}}} = \sqrt {{\rm{V}}_{cm}^2 + {\rm{V}}_t^2} = \sqrt 2 \frac{{\rm{J}}}{m} \end{array}

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