Two blocks A and B of masses 3kg and 6kg are connected by a...

Let $\mu$ is reduced mass of the two blocks, then

$\mu = \dfrac{{m}_{1}{m}_{2}}{{m}_{1} + {m}_{2}} = \dfrac{3 \times 6}{3 + 6} = \dfrac{18}{9} = 2kg$

Let ${v}_{r}$ be the relative velocity of the two blocks, then from conservation of mechanical energy we get

$\dfrac{1}{2}\mu{v}_{r}^{2} = \dfrac{1}{2}k{x}^{2}$
${v}_{r} = x\sqrt{\dfrac{k}{\mu}} = 5 \times {10}^{- 2}\sqrt{\dfrac{1800}{2}}$

$⇒ {v}_{r} = 1 . 5m{s}^{- 1}$