JEE Main 2022PhysicsCenter of Mass Momentum and CollisionMediumMCQ

JEE Main 2022 — Center of Mass Momentum and Collision Question with Solution

JEE Main 2022 (27 Jun Shift 1)

Question

What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass?
(Assume the collision to be head-on elastic collision)

Choose an option

Show full solutionCorrect option: C
Correct answer
C55.6%

Step-by-step explanation

After collision,

Applying momentum conservation,

mu=mv1+5mv2  ⇒u=v1+5v2   ...1 

Coefficient of restitution, e=1=v2-v1u

∴ u=v2-v1   ...2

Adding  2u=6v2  ⇒v2=u3 and  u=u3-v1

∴ v1=u3-u=-2u3

Percentage change in kinetic energy =12mv12-12mu212mu2×100

=v12-u2u2×100=4u29-u2u2×100  =-59×100  =55.6%

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About this question

This is a previous-year question from JEE Main 2022, 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.