Three blocks A, B, and C are lying on a smooth horizontal surface. A and B have equal masses, m, while C has mass M. Block A is given a speed v towards B, due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically. 5/6 of the initial kinetic energy is lost in the whole process. What is the value of M/m?

Let the initial kinetic energy be KE_initial = (1/2)mv².
After the first collision (A and B), the combined mass is 2m, and the velocity is v' such that momentum is conserved:
mv = 2mv'
=> v' = v/2
The kinetic energy after the first collision is KE_1 = (1/2)(2m)(v/2)² = (1/4)mv²
After the second collision (2m and M), the combined mass is 2m + M, and the velocity is v'' such that momentum is conserved:
2mv' = (2m + M)v''
2m(v/2) = (2m + M)v''
=> mv = (2m + M)v''
=> v'' = mv/(2m + M)
The kinetic energy after the second collision is KE_2 = (1/2)(2m + M)(v'')² = (1/2)(2m + M)(mv/(2m + M))² = (1/2)m²v²/(2m + M)
Given that 5/6 of the initial kinetic energy is lost, the final kinetic energy is (1/6) of the initial kinetic energy:
KE_2 = (1/6)KE_initial
(1/2)m²v²/(2m + M) = (1/6)(1/2)mv²
Dividing both sides by (1/2)mv²:
m/(2m + M) = 1/6
6m = 2m + M
4m = M
M/m = 4