Q 11-05-063JEE MainJEE Main 2025 (29 Jan, Shift 1)Medium
As shown below, bob $A$ of a pendulum having a 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 centre. Assuming elastic collision, the magnitude of the velocity of bob $A$ after the collision will be (take $g$ as acceleration due to gravity)
Answer: (D) $\dfrac{1}{3}\sqrt{Rg}$
Speed of $A$ at the lowest point, after falling $R(1 - \cos60^\circ) = R/2$:
$$v = \sqrt{2g\cdot\frac{R}{2}} = \sqrt{gR}$$
Elastic collision of $m$ (moving) with $m/2$ (at rest):
$$v_A' = \frac{m - m/2}{m + m/2}\,v = \frac{1}{3}\sqrt{gR}$$
Solution by Sreeraj P, M.Sc Physics