Two blocks `A` and `B` of masses in and `2m`, respectively, are connected with the help of a spring having spring constant, `k` as shown in Fig. Initially, both the blocks arc moving with same velocity `v` on a smooth horizontal plane with the spring in its natural length. During their course of motion, block `B` makes an inelastic collision with block `C` of mass `m` which is initially at rest. The coefficient of restitution for the collision is `1//2`. The maximum compression in the spring is

Options

A. `sqrt((2m)/k)`
B. will never be attained
C. `sqrt(m/(12k))v`
D. `sqrt(m/(6k))v`

Answer

D

Explanation

For collision of `B` and `C`

`2mv=2mv_(1)+mv_(2)`
`1/2=(v_(2)-v_(1))/vimplies2v_(2)-2v_(1)=v`
Solving above equation `v_(1)=v/2` and `v_(2)=v`
Now for blocks `A` and `B` plus spring system, using reduced mass concept and applying work energy theorem, let maximum compresion in spring be `x_(0)` and that the time of maximum compression relative velocity of blocks be zero. reduced mass is given by
`"mu"=(mxx2m)/(3m)=(2m)/3`
`0-(muxx(v-v//2)^(2))/2=-(kx_(0)^(2))/2impliesx_(0)=(sqrt(m/(6k)))v`