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 $
