Two bodies A and B of masses m and 2m respectively are placed on a smooth floor. They are connected by a light spring of stiffness k.
A third body C of mass m moves with velocity `v_(0)` along the line joining A and B and collides elastically with A. If `l_(o)` be the natural length of the spring then find the minimum separation between the blocks.

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 equations, `v_(A)=v_(0)` and `v_(c )=0`.
Now A will move and compress the spring which in turn accelerate 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.
`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)(3m)(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))`