Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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
PhysicsCenter of Mass Momentum and CollisionJEE MainJEE Main 2018 (08 Apr)
Options:
  • A v02
  • B v04
  • C  2 v0
  • D v02
Solution:
1047 Upvotes Verified Answer
The correct answer is:  2 v0

KE2=32mv02. Conservation of momentum,  v 1 + v 2 = v 0 . Conservation of energy, 12mv12+12mv22=12 × 32mv02
v12+v22=32v02v12+v22+2v1v2=v0232v02+2v1v2=v02
2v1v2=12v02(v1v2)2=32v02+12v02 =2 v 0 2
v1v2=2v0.

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.