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) $\dfrac{4P}{3}$
A's kinetic energy falls to $\dfrac{1}{9}$, so its speed falls to $\dfrac{1}{3}$ and it now moves backward: its momentum is $-\dfrac{P}{3}$.
Momentum conservation:
$$P = -\frac{P}{3} + P_B \;\Rightarrow\; P_B = \frac{4P}{3}$$
Solution by Sreeraj P, M.Sc Physics