Two identical blocks `A` and `B` , each of mass `m` resting on smooth floor are connected by a light spring of natural length `L` and spring constant `k`, with the spring at its natural length. A third identical block `C` (mass `m`) moving with a speed `v` along the line joining `A` and `B` collides with `A`. The maximum compression in the spring is

Explanation

Initially there will be collision between `C` and `A` which is elastic, therefore, by using conservation of momentum we obtain,
`mv_(0)=mv_(A)+mv_(C)" "," "v_(0)=v_(A)+v_(C)`
Since `e=1,v_(0)=v_(A)-v_(C)`
Solving the above two equation, `v_(A)=v_(0)` and `v_(c)=0`
Now `A` will move and compress the spring which in turn acceleration `B` and retard `A` and finally both `A` and `B` will move with same velocity `v`.
(a) Since net external force is zero, therefore momentum of the system (A and B) is conserved.
Hence `mv_(0)=(m+2m)v`
`rArr" "v=v_(0)//3`
(b) If `x_(0)` is the maximum compression, then using energy conservation
`(1)/(2)"mv"_(0)^(2)=(1)/(2)("m+2m")"v"^(2)+(1)/(2)"kx"_(0)^(2)`
`rArr" "(1)/(2)"mv"_(0)^(2)=(1)/(2)(3"m")("v"_(0)^(2))/(9)+(1)/(2)"kx"_(0)^(2)" "rArr" "x_(0)=v_(0)sqrt((2m)/(3k))`
Hence minimum distance `D=l_(0)-x_(0)=l_(0)-v_(0)sqrt((2m)/(3k))`