A pendulum bob of mass m/2 is released from rest from the position shown, at 60° from the vertical, and collides elastically (e=1) with a stationary bob of mass m. Find the speed of the mass m immediately after collision.

Question

G1 Physics · Accepted Answer

The m/2 bob falls through height l(1-cos60°)=l/2, so its speed before collision is u=√(2g·l/2)=√(gl). For a 1D elastic collision with masses m/2 and m, momentum gives (m/2)u=mv₁+(m/2)v₂ and e=1 gives u=v₁-v₂. Solving gives v₁=(2/3)u=(2/3)√(gl).