A block possessing kinetic energy $K$ collides head-on  with a stationary spring-block system and rebounds in opposite direction with kinetic energy $K\text{'}$ . The masses of all the blocks are equal. Calculate the energy of oscillations of the spring block system.

If $u$ is the speed of the block, then

           $\frac{1}{2}m{u}^2 = K$ or $u=\sqrt{\frac{2k}{m}}$ .

The speed of block with which it rebounds, $u\text{'}=\sqrt{\frac{2k\text{'}}{m}}$

The maximum energy that can be stored in the spring will be the energy of oscillations. At the instant of maximum compression, let speed of each connected block is $v$ . By conservation of momentum, we have

          $mu=m(-u\text{'})+2m{v}_c$    where $v_c is the velocity of centre of mass of the spring mass system.$

or      $m\sqrt{\frac{2k}{m}} = -m\sqrt{\frac{2k\text{'}}{m}}+2m{v}_c$     …(i)

For elastic collision,

                 $K = K\text{'}+\frac{1}{2}\times 2m\times {v_c}^2+E_{oscialltion}$    …(ii)

After solving above equations, we get

                $E_{oscialltion} = \left[\frac{k-3k\text{'}-2\sqrt{kk\text{'}}}{2}\right]$