As shown below, bob A of a pendulum having massless string of length 'R' is released from 60° to the vertical. It hits another bob B of half the mass that is at rest on a friction less table in the centre. Assuming elastic collision, the magnitude of the velocity of bob A after the collision will be (take g as acceleration due to gravity)

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$\sqrt{Rg}$

$\frac{1}{3} \sqrt{Rg}$

$\frac{2}{3} \sqrt{Rg}$

$\frac{4}{3} \sqrt{Rg}$

answer is A.

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Detailed Solution

Velocity of a just before hitting :
$u = \sqrt{2 g \frac{R}{2}} = \sqrt{gR}$
Just after collision, let velocity of A and B are v₁ and v₂ respectively
$∴$ by COM:
$\begin{array}{l} mu = {mv}_{1} + \frac{m}{2} v_{2} \\ 2 v_{1} + v_{2} = 2 u . . . . . . . . (i) \\ e = 1 = \frac{v_{2} - v_{1}}{u} \\ \Rightarrow v_{2} - v_{1} = u . . . . . . . . (ii) \end{array}$
From (i) –(ii)
$\Rightarrow 3 v_{1} = u \Rightarrow v_{1} = \frac{u}{3} = \frac{1}{3} \sqrt{gR}$

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