JEE Main | 2018Momentum and CollisionEasy

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 collision, is :

Options

A.

2v0

B.

v02

C.

v02

D.

v04

Solution

initial 

12mv12+12mv22=3212mv02
⇒  V12+V22=32V02    ...(1)
from momentum conservation
mv0 = m(v1 + v2)      ...(2)
(v1 + v2)2 = v02 
⇒ v12+v22+2v1v2 = v02
2v1v2 = -v022
(v1 - v2)2 = v12+v22-2v1v2=32v02+v022
v1 - v2 = 2v0

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