A pendulum bob of mass m connected to the end of an ideal string of length l is released from rest from horizontal position as shown in the figure. At the lowest point, the bob makes an elastic collision with a stationary block of mass 5m, which is kept on a frictionless surface. Mark out the correct statement(s) for the instant just after the impact.

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Tension in the string is $\frac{17}{9} m g$

Tension in the string is $3 m g$

The velocity of the block is $\frac{\sqrt{2 g l}}{3}$

The maximum height attained by the pendulum bob after impact is (measured from the lowest position) $\frac{4 l}{9}$

answer is A, C, D.

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Detailed Solution

Just after collision,

$v_{m} = (\frac{m - 5 m}{m + 5 m}) \sqrt{2 g l} = - \frac{2}{3} \sqrt{2 g l} v_{5 m} = (\frac{2 \times m}{m + 5 m}) \sqrt{2 g l} = \frac{\sqrt{2 g l}}{3} T - m g = \frac{m v_{m}^{2}}{l} = \frac{m}{l} (\frac{8 g l}{9}) ∴ T = \frac{17}{9} m g h_{m} = \frac{v_{m}^{2}}{2 g} = \frac{4 l}{9}$