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 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 at lowest point just after collision is (17/9) mg

Tension in the string at lowest point just before collision is 3mg

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

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

answer is A, B, C, D.

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Detailed Solution

The velocity of bob just before the impact is $v = \sqrt{2 g l}$ along the horizontal direction

Before collision After Collision
From momentum conservation $m v = - m v_{1} + 5 m v_{2}$
From coefficient of restitution equation, $1 = \frac{v_{1} + v_{2}}{v} \Rightarrow v_{1} + v_{2} = v$
Solving above equations, we get ; $v_{1} = \frac{2 v}{3}$ , $v_{2} = \frac{v}{3}$
For tension in string ; $T - m g = \frac{m v_{1}^{2}}{l} \Rightarrow T = \frac{17}{9} m g$
$T - m g = \frac{m v^{2}}{l} V_{2} = \frac{\sqrt{2 g l}}{3} (T = 3 m g)$
(i) Let the maximum height attained by the bob be $h_{2}$ , then