Two particles (A and B) of masses m and 2m are joined by a light rigid rod of length L. The system lies on a smooth horizontal table. The particle (A) of mass m is given a sharp impulse so that it acquires a velocity u perpendicular to the rod. Calculate maximum speed of particle B during subsequent motion.

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$\frac{u}{3}$

$\frac{4 u}{3}$

$\frac{2 u}{3}$

$u$

answer is C.

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Detailed Solution

$V_{cm} = \frac{mu}{3 m} = \frac{u}{3}$
In COM frame, the system rotates with speed of $A = \frac{2 u}{3}$ and speed of $B = \frac{u}{3}$

The two figures show the velocities of A and B in COM frame.
In ground frame B will have maximum velocity when it is in position indicated in fig (b)
$∴ {(V_{B})}_{max} = \frac{2 u}{3}$

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