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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2025
  • Hard
  • MCQ

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.)

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Answer & Solution

Correct Answer

(D) \(\frac{1}{3} \sqrt{\operatorname{Rg}}\)

Step-by-step Solution

Detailed explanation


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)
\(\Rightarrow 3 \mathrm{v}_1=\mathrm{u} \Rightarrow \mathrm{v}_1=\frac{\mathrm{u}}{3}=\frac{1}{3} \sqrt{\mathrm{gR}}\)
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