Question

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): 
 

• A. \( \frac{1}{3} \sqrt{Rg} \)
• B. \( \sqrt{Rg} \)
• C. \( \frac{2}{3} \sqrt{Rg} \)
• D. \( \frac{4}{3} \sqrt{Rg} \)

Hint:

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.

Answer 1

The Correct Option is C

Using the principle of conservation of mechanical energy, the velocity of bob A just before the collision is given by: \[ v_A = \sqrt{2gR(1 - \cos 60^\circ)} = \sqrt{gR}. \] Since the collision is elastic and the mass of bob B is half that of bob A, we can use the conservation of momentum and energy to solve for the final velocity of bob A after the collision. The final velocity of bob A, after the elastic collision, can be derived using the relative velocities and momentum conservation. The final velocity is given by: \[ v'_A = \frac{2}{3} \sqrt{Rg}. \] Thus, the correct answer is \( \frac{2}{3} \sqrt{Rg} \).