A light spring of spring constant k is kept compressed between two blocks of masses m and M on a smooth horizontal surface. When released, the blocks acquire velocities in opposite directions. The spring loses contact with the blocks when it acquires natural length. If the spring was initially compressed through a distance x find the final speeds of the two blocks.

Explanation

Consider the two blocks plus the spring to be the system. No exterN/Al force acts on this system in horizontal direction. Hence, the linear momentum will remain constant. As the spring is light, it has no linear momentum. Suppose the block of mass M moves with a speed V and the other block with a speed v after losing contact with the spring. As the blocks are released from rest the initial momemtum is zero. The fiN/Al momentum is `MV-mv` towards right. Thus,
`MV-mv=0 or V=m/Mv`......... i
initially the energy of the system `=1/2 kx^2`.
FiN/Ally, the energy of the system `=1/2mv^2+1/2MV^2`
As there is no friction
`1/2mv^2+1/2MV^2=1/2kx^2`....ii
Using i and ii
`mv^2(1+m/M)=kx^2`
or, `v=sqrt((kM)/(m(M+m))x)`
and `V=sqrt((km)/(M(M+m)_)x`