Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal masses, $m$, while C has mass $M$. Block A is given an initial speed $v$ towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically. $\frac{5}{6}$th of the initial kinetic energy is lost in the whole process. What is the value of $M/m$?
Answer: (B) $4$
Momentum is conserved in both collisions, and finally all three blocks move together:
$$mv = (2m + M)v_f \Rightarrow v_f = \frac{mv}{2m+M}$$
Ratio of final to initial kinetic energy:
$$\frac{\tfrac12(2m+M)v_f^2}{\tfrac12 mv^2} = \frac{m}{2m+M}$$
Since $\frac56$ of the energy is lost, this ratio is $\frac16$:
$$2m + M = 6m \Rightarrow \frac{M}{m} = 4$$
Solution by Sreeraj P, M.Sc Physics