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 inita speed v towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically `(5)/(6)th` of the initial kinetic energy is lost in whole process. What is balue of M//M?

Options

A. 2
B. 4
C. 5
D. 3

Answer

B

Explanation

Collision between A and B , ( inelastic collision )
`mv+0 = (m+m) v. rArr v. = v/2 " "…(i)`
velocity of combined mass of A and B (2 m ) is v.
Loss of kinetic energy
`Delta K.E_(1) = 1/2 (2m) (v/2)^(2) - (1/2mv^(2)+0)=-1/4 mv^(2)`
Thus this mass again collides with C inelastically .
`2mv. = (2m+M) v"`
`v" = (2mv.)/(2m+M) = (m/(2m+M))v ` (From (i))
Loss of kinetic energy in this collision
`DeltaK.E_(2) =1/2(2m+M)(v") ^(2)-1/2(2m)(v.)^(2)`
`=1/2 (2m+M) (m^(2)v^(2))/((2m+M)^(2) -1/2 2m(v/2)^(2)`
`=-1/2((2mM)/(2m+M)) (v/2)^(2)`
Given , `DeltaK.E_(1)+DeltaK.E_(2)=5/6 (DeltaK.E_(i))`
` = 1/2mv^(2)+1/2((2mM)/(2m+M)) (v^(2))/4 = 5/6 (1/2mv^(2)) rArr M/m = 4`