question_answer

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]

A) \[\frac{{{v}_{o}}}{2}\]

B) \[\frac{{{v}_{o}}}{\sqrt{2}}\]

C) \[\frac{{{v}_{o}}}{4}\]

D) \[\sqrt{2}{{v}_{o}}\]

Correct Answer: D

Solution :

[d]

From C.O.L.M.

\[m{{v}_{0}}=m{{v}_{1}}+m{{v}_{2}}\]       (1)

And \[K.{{E}_{f}}=\frac{3}{2}K.E{{.}_{i}}\]

\[\frac{1}{2}mv_{1}^{2}+\frac{1}{2}mv_{2}^{2}=\frac{3}{2}\left( \frac{1}{2}mv_{0}^{2} \right)\]

\[v_{1}^{2}+v_{2}^{2}=\frac{3}{2}V_{0}^{2}\]                  (2)

Solving equation (1) and (2) we get

\[{{v}_{rel}}={{v}_{2}}-{{v}_{1}}=\sqrt{2}{{v}_{0}}\]