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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 is (17/9) mg

The maximum height attained by the pendulum bob after impact is (measured from the lowest position) 4l/9.

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

Tension in the string is 3 mg.

answer is A, D.

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Detailed Solution

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

From momentum conservation,

$mv = - {mv}_{1} + 5 {mv}_{2}$

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, $T - mg = \frac{{mv}_{1}^{2}}{l}$

$\Rightarrow T = \frac{17}{9} mg$

Let the maximum height attained by the bob be h, then $\frac{{mv}_{1}^{2}}{2} = mgh$

$\Rightarrow h = \frac{4 l}{9}$

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