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 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.)

Solution

Physics sir · 4 steps

First let gravity give A its speed at the bottom, then apply the elastic-collision formula for unequal masses.

1. Height dropped. Released at 60°, bob A starts at a height R - R cos 60° = R/2 above its lowest point, where it meets B.
2. Speed before impact. Energy conservation gives 1/2mu² = mgR/2, so u = √gR. The collision happens with the string vertical, so u is horizontal along the line joining the bobs.

   v_A = (m_A - m_B)/(m_A + m_B) u = (m - m/2)/(m + m/2) u = 1/3u

3. Check with B. B moves off with v_B = 2m_A/(m_A + m_B)u = 4/3u. Then momentum is m·u/3 + m/2·4u/3 = mu, and kinetic energy is also conserved, as an elastic collision needs.
4. So. Bob A continues forward with speed 1/3√Rg, which is option (b).

The idea to keep. In a 1D elastic collision with the target at rest, the striker keeps (m₁ - m₂)/(m₁ + m₂) of its speed and the target gets 2m₁/(m₁ + m₂) of it. Get the speed at impact first.

Answer from the NTA final answer key. Solution by PrepWiser.