A spring of force constant k is kept in the compressesd condition between two blocks masses m and M on the smooth surface of a table as shown in figure .When the spring is released , both the blocks move in opposite directions .When the spring attains the orginal (normal ) position . both blocks lose the constacts with the spring . If x is the intial compression of the spring find the speeds of block while getting detached from the spring .

Explanation

Suppose V and v be the velocity of the block of mass M and m respectively when they lose the contact with spring . Here the initial momnetum is zero and final momentum is MV - mv According to the law of conservation of momentum
` MV - mv = 0 `
` :. V = (mv)/M `
` :. ` The potential energy stored in spring will be the intial energy of the system ` = 1/2 kx^(2)`
Final energy of the system ` = 1/2 mv^(2) +1/2 Mv^(2)`
(Neglecting frictional forces )
` :. 1/2 mv^(2)+1/2 Mv^(2)=1/2 kx^(2)`
` :. mv^(2) [1+m/M] =kx^(2)`
` :. mv^(2) [(M+m)/M] = kx^(2)`
` :. v^(2) = (kM)/(m(M+m)).x^(2)`
` :. v = x sqrt((kM)/(m(M+m)))`