Thre block A, B and C are lying on a smooth horizontal surface, as shwon in the figure. A and B have equal masses m while C has mass M. Block A is given an initial seed v towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically and `5/6`th of the initial kinetic energy is lost whole process. If the value of M/m is `4/x`. FInd the value of x.

Explanation

Kinetic energy of block A
`k_(1)=1/2mv_(0)^(2)`
`:.` From principle of linear momentum conservation
`mv_(0)=(2m+M)v_(f)Phi v_(f)=(mv_(0))/(2m+M)`
According to question of `5/6` th the initial kinetic energy is lost in whole process.
`:.(k_(j))/(k_(f))=6implies(1/2mv_(0)^(2))/(1/2(2m+M)((mv_(0))/(2m+M))^(2))=6`
`Phi(2m+M)/m=6:.M/m=4`