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 $$M/m$$?

Initial kinetic energy of the system: $$K_i = \frac{1}{2}mv^2$$

Conservation of linear momentum for the final state (all blocks stick together): $$mv = (m + m + M)v_f \implies v_f = \frac{mv}{2m + M}$$

Final kinetic energy of the system: $$K_f = \frac{1}{2}(2m + M)v_f^2 = \frac{1}{2}(2m + M)\left(\frac{mv}{2m + M}\right)^2 = \frac{m^2v^2}{2(2m + M)}$$

Given fractional loss of kinetic energy is $$\frac{5}{6}$$, the fraction remaining is:

$$\frac{K_f}{K_i} = 1 - \frac{5}{6} = \frac{1}{6}$$

$$\frac{\frac{m^2v^2}{2(2m + M)}}{\frac{1}{2}mv^2} = \frac{1}{6} \implies \frac{m}{2m + M} = \frac{1}{6}$$

$$6m = 2m + M \implies M = 4m$$

$$\frac{M}{m} = 4$$