JEE Main On 08 April 2018 Question 23

Question: In a collinear collision, a particle with an initial speed $ v _{o} $ 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: [JEE Main Online 08-04-2018]

Options:

A) $ \frac{v _{o}}{2} $

B) $ \frac{v _{o}}{\sqrt{2}} $

C) $ \frac{v _{o}}{4} $

D) $ \sqrt{2}v _{o} $

Show Answer

Answer:

Correct Answer: D

Solution:

  • From C.O.L.M. $ mv _0=mv _1+mv _2 $ ?(1) And $ K.E _{f}=\frac{3}{2}K.E{. _{i}} $ $ \frac{1}{2}mv_1^{2}+\frac{1}{2}mv_2^{2}=\frac{3}{2}( \frac{1}{2}mv_0^{2} ) $ $ v_1^{2}+v_2^{2}=\frac{3}{2}V_0^{2} $ ?(2) Solving equation (1) and (2) we get $ v _{rel}=v _2-v _1=\sqrt{2}v _0 $