As shown below, bob \(A\) of a pendulum having massless string of length ' \(R\) ' is released from…

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 friction less 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.)

1. \(\frac{4}{3} \sqrt{\operatorname{Rg}}\)
2. \(\frac{2}{3} \sqrt{\mathrm{Rg}}\)
3. \(\sqrt{\mathrm{Rg}}\)
4. \(\frac{1}{3} \sqrt{\operatorname{Rg}}\)

Solution

Velocity of a just before hitting :
$\mathrm{u}=\sqrt{2 \mathrm{~g} \frac{\mathrm{R}}{2}}=\sqrt{\mathrm{gR}}$
Just after collision, let velocity of A and B are $\mathrm{v}_1$ and $v_2$ respectively
$\therefore \text { by COM: }$
$\begin{aligned}
& m u=m v_1+\frac{m}{2} v_2 \\ & 2 v_1+v_2=2 u...(i)\\ & e=1=\frac{v_2-v_1}{u} \\ & \Rightarrow v_2-v_1=u..(ii)
\end{aligned}$
From (i) -(ii)
$\Rightarrow 3 \mathrm{v}_1=\mathrm{u} \Rightarrow \mathrm{v}_1=\frac{\mathrm{u}}{3}=\frac{1}{3} \sqrt{\mathrm{gR}}$

Asked in: JEE Main 2025 (29 Jan Shift 1)

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