JEE Main 2018PhysicsCenter of Mass Momentum and CollisionHardMCQ

JEE Main 2018 — Center of Mass Momentum and Collision Question with Solution

JEE Main 2018 (08 Apr)

Question

In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after the collision, is

Choose an option

Show full solutionCorrect option: C
Correct answer
C 2 v0

Step-by-step explanation

KE2=32mv02. Conservation of momentum,  v 1 + v 2 = v 0 . Conservation of energy, 12mv12+12mv22=12 × 32mv02
⇒v12+v22=32v02⇒v12+v22+2v1v2=v02⇒32v02+2v1v2=v02
⇒2v1v2=−12v02⇒(v1−v2)2=32v02+12v02 =2 v 0 2
⇒v1−v2=2v0.

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Center of Mass Momentum and Collision chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2018, 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.