JEE Main 2019PhysicsCenter of Mass Momentum and CollisionHardMCQ

JEE Main 2019 — Center of Mass Momentum and Collision Question with Solution

JEE Main 2019 (12 Jan Shift 2)

Question

An alpha- particle of mass m suffers 1- dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing 64% of its initial kinetic energy. The mass of the nucleus is

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Show full solutionCorrect option: B
Correct answer
B4m

Step-by-step explanation



From momentum conservation,

mv+0=-mv1+Mv2   ...(1) 

Using kinetic energy conservation,

12mv2=12mv12+12Mv22⇒12mv2=0.36×12mv2+12Mv22⇒0.64mv2=Mv22⇒v2=0.8vmM 

As it retain 36%,⇒0.36×12mv2=12mv12 ⇒0.36v2=v12 Hence, v1=0.6v
From (1), mv=-m0.6v+M0.8vmM

⇒1=-0.6+0.8Mm

On solving, we get M=4m

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About this question

This is a previous-year question from JEE Main 2019, 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.