Two blocks A and B of equal masses (m = 10kg) are connected by a light spring of spring constant k= 150 N/m. The system is in equilibrium. The minimum value of initial downward velocity $V_0$ of the block B for which the block A bounce up is $\frac{20}{\sqrt{3n}}=m/s$ . Find the value of n
 

At equilibrium the compression of spring to balance the weight of B is 

150 x =100

x=2/3

to lift the weight of A another compression of 2/3 m is required as both have the same weight.

so total compression becomes 4/3 m

then $\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$

$\frac{1}{2}10{\left(\frac{20}{\sqrt{3n}}\right)}^2=\frac{1}{2}150{\left(\frac{4}{3}\right)}^2$

solving for n we get n=5