Two blocks `m_(1) and m_(2) ` are connected by a spring of force constant K and are placed on a frictionless horizontal surface. Initially the spring is given extension `x_(0)` when the sysetem is released from rest, find the distance moved by two blocks before they again comes to rest.

Explanation

Letm`x_(1) and x_(2)` be the distances travelled by the blocks. Both the blocks are initially at rest so initial velocity of the centre of mass is zero. Net external force on the system is zero in the horizontal direction hence acceleration of centre of mass remains zero and thus centre of mass remains at rest. So we can conclude that both the blocks will come to rest simultaneously, At this point we can use conservation of mechanical energy to understand that final compression of the spring will also be xo. Hence total distance travelled by both the blocks must be 2x, and we can write the following equation:
`x_(1)+x_(2)=2x_(0) " "...(1)`
We know that centre of mass is at rest hence displacement of the centre of mass is zero. Both the blocks me and my are displaced by X, and X respectively in opposite directions, hence for zero displacement of centre of mass we can write the following equation:
`m_(1)x_(1)=m_(2)x_(2)" "....(2)`
Solving the above two equations we get
`x_(1)=(2m_(2)x_(0))/(m_(1)+m_(2)) and , x_(2)=(2m_(1)x_(0))/(m_(1)+m_(2))`