JEE Main 2021PhysicsCenter of Mass Momentum and CollisionHardNumerical

JEE Main 2021 — Center of Mass Momentum and Collision Question with Solution

JEE Main 2021 (27 Aug Shift 2)

Question

A bullet of 10 g, moving with velocity v, collides head-on with the stationary bob of a pendulum and recoils with velocity 100 m s-1. The length of the pendulum is 0.5 m and mass of the bob is 1 kg. The minimum value of v in m s-1, so that the pendulum describes a circle. (Assume the string to be inextensible and g=10 m s-2 )

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Show full solutionCorrect answer: 400
Correct answer
400

Step-by-step explanation

Final velocity of system (bullet + mass) should be =5gr

Now M5gr=m(v+100)

thus v=Mm5gr-100=500-100=400 m s-1

v=400 m s-1

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About this question

This is a previous-year question from JEE Main 2021, 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.