As shown below, bob A of a pendulum having massless string of length \( R \) is released from \( 60^\circ \) to the vertical. It hits another bob B of half the mass that is at rest on a frictionless table in the center. 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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In elastic collisions, both kinetic energy and momentum are conserved. For pendulum collisions, use energy conservation for pre-collision velocities and momentum conservation for post-collision velocities.
Based on the principle of conservation of mechanical energy, the initial velocity of bob A before impact is determined as:
\[
v_A = \sqrt{2gR(1 - \cos 60^\circ)} = \sqrt{gR}.
\]
Given that the collision is elastic and bob B's mass is half that of bob A, conservation of momentum and energy can be applied to find the final velocity of bob A post-collision.
The final velocity of bob A following the elastic collision, calculated using relative velocities and momentum conservation, is:
\[
v'_A = \frac{2}{3} \sqrt{Rg}.
\]
Therefore, the correct solution is \( \frac{2}{3} \sqrt{Rg} \).
