JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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.)

- A \(\frac{4}{3} \sqrt{\operatorname{Rg}}\)
- B \(\frac{2}{3} \sqrt{\mathrm{Rg}}\)
- C \(\sqrt{\mathrm{Rg}}\)
- D \(\frac{1}{3} \sqrt{\operatorname{Rg}}\)
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 \(\therefore \text { by COM: }\)…
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