In the figure ball A is released from rest when the spring is at its natural length. For the block B of mass M to leave contact with ground at some stage, the minimum mass of A must be 

Suppose when maximum displacement of the ball is x, when the mass M loses contact with the  ground.
$\begin{array}{l}\therefore \mathrm{kx}=\mathrm{Mg}........\left(1\right)\\ \mathrm{Again},\frac{1}{2}{\mathrm{kx}}^2-\mathrm{mgx}=0\Rightarrow \mathrm{x}=\frac{2\mathrm{mg}}{\mathrm{k}}.......\left(2\right)\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}k\hspace{0.17em}}\frac{2\mathrm{mg}}{\mathrm{k}}=\mathrm{Mg}\Rightarrow \mathrm{m}=\mathrm{M}/2\end{array}$