A light spring of constant `k` is kept compressed between two blocks of masses `m` and `M` on a smooth horizontal surface. When released, the block 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.
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Consider the two blocks plus the spring to the the system, no external force acts on this suystem in horizontal direction. Hence the linear momentum will remain constant. Suppose, the block of mass `M` moves with a speed `V` and the other block with a speed `v` after losing contact with spring. From conservation of linear momentum in horizontal direction, we have `MV-mv=0` or `V=(mv)/M`........i Initially, the mechanical energy of the system `=1/2MV^(2)` finally the mechanical energy of the system `=1/2mv^(2)+1/2MV^(2)` As there is no friction, mechaical energy will remain conserved, therefore, `1/2m^(2)+1/2MV^(2)=1/2k^(2)`.........ii Solving Eqn i and ii, we get `v=[(kM)/(m(M+m))]^(1/2)x` and `V=[(kM)/(M(M+m))]^(1/2)x`
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