A pendulum bob of mass m connected at 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 a head - on elastic collision with a stationary block of mass 5m, which is kept on a frictionless surface. Mark the correct statement(s) for the instant just after the impact. Acceleration due to gravity is g.

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

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

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

Tension in the string is 3 mg

answer is A, C, D.

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Detailed Solution

The velocity of bob just before the impact is $\sqrt{2 g l}$ along horizontal direction. From momentum conservation,
$Mv = - {mv}_{1} + {5mv}_{2} .$
From coefficient of restitution equation.
$1 = \frac{v_{1} + v_{2}}{v} \Rightarrow v_{1} + v_{2} = v$
Solving above equation we get, $v_{1} = \frac{2 v}{3} a n d v_{2} = \frac{v}{3},$ where $v = \sqrt{2 g l}$
For tension in string, $T - m g = \frac{m v_{1}^{2}}{l}$
$\Rightarrow T = \frac{17}{9} m g$
Let maximum height attained by bob is h, then
$\frac{m v_{1}^{2}}{2} = m g h \Rightarrow h = \frac{4 l}{9}$