JEE Main 2025 — Center of Mass Momentum and Collision Question with Solution
JEE Main 2025 (29 Jan Shift 1)
Question
As shown below, bob of a pendulum having massless string of length ' ' is released from to the vertical. It hits another bob 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.)
Choose an option
Show full solutionCorrect option: D
Correct answer
D
Step-by-step explanation
Velocity of a just before hitting : Just after collision, let velocity of A and B are and respectively $\begin{aligned} & m u=m v_1+\frac{m}{2} v_2 \\ & 2 v_1+v_2=2 u ...(i)\\ & e=1=\frac{v_2-v_1}{u} \\ & \Rightarrow v_2-v_1=u..(ii) \end{aligned}$ From (i) -(ii)
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This is a previous-year question from JEE Main 2025, covering the Center of Mass Momentum and Collision chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.