Two blocks A and B of masses 3 k g and 6 k g are connected by a massless spring of force constant 1800 N m -…

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

$\mu =\frac{m_1{m}_2}{m_1+m_2}=\frac{3\times 6}{3+6}=\frac{18}{9}=2 \mathrm{kg}$

Let $v_r$ be the relative velocity of the two blocks, then from conservation of mechanical energy we get

$\frac{1}{2}\mu v_r^2=\frac{1}{2}k{x}^2$
$v_r=x\sqrt{\frac{k}{\mu }}=5\times {10}^{-2}\sqrt{\frac{1800}{2}}$

$\Rightarrow v_r=1.5 \mathrm{m} {\mathrm{s}}^{-1}$