Q 11-05-022JEE MainEAMCET 2012 (Engineering)Hard
A ball A of mass $m$ moving along the positive $x$-direction with kinetic energy $K$ and linear momentum $P$ undergoes an elastic head-on collision with a stationary ball B of mass $M$. After the collision, ball A moves along the negative $x$-direction with kinetic energy $K/9$. The final linear momentum of ball B is
Answer: (C) $\dfrac{4P}{3}$
Concept. KE falling to $\dfrac19$ means speed falls to $\dfrac13$; then momentum conservation.
Step 1. A's momentum. $-\dfrac P3$ (now moving backward).
Step 2. B's momentum. $P = -\dfrac P3 + P_B \Rightarrow P_B = \dfrac{4P}{3}$.
Answer. $\dfrac{4P}{3}$ (Option C).
Solution by Sreeraj P, M.Sc Physics