A pendulum bob of mass $m$ connected to the end of an ideal string to 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.

Tension in the string is

\frac{17}{9} m g

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Tension in the string is 3 mg.

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The velocity of the block is

\frac{\sqrt{2 g l}}{3}

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The maximum height attained by the pendulum bob after impact is (measured from the lowest position)

\frac{4 l}{9}

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Solution

The correct options are
B Tension in the string is $\frac{17}{9} m g$
C The velocity of the block is $\frac{\sqrt{2 g l}}{3}$
D The maximum height attained by the pendulum bob after impact is (measured from the lowest position) $\frac{4 l}{9}$
Just after collision:
$v_{m} = (\frac{m - 5 m}{m + 5 m}) \sqrt{2 g l} = \frac{- 2}{3} \sqrt{2 g l} v_{5 m} = (\frac{2 m}{m + 5 m}) \sqrt{2 g l} = \frac{1}{3} \sqrt{2 g l} T = \frac{m v^{2}}{l} + m g = \frac{17 m g}{9}$
Now, using energy conservation:
$m g h = \frac{1}{2} m v^{2} h = \frac{v^{2}}{2 g} = \frac{4 l}{9}$

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