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Two blocks $A$ and $B$ of mass $m$ and $2\, m$ respectively are connected by a massless spring of force constant $k$. They are placed on a smooth horizontal plane. Spring is stretched by an amount $x$ and then released. The relative velocity of the blocks when the spring comes to its natural length is :-
$x\sqrt{\frac{3k}{2m}}$
$x\sqrt{\frac{2k}{3m}}$
$x\sqrt{\frac{2k}{m}}$
$x\sqrt{\frac{3k}{m}}$
Solution
$\mathrm{mv}_{1}=2 \mathrm{mv}_{2}$
$\frac{1}{2} \mathrm{kx}^{2}=\frac{1}{2} \mathrm{mv}_{1}^{2}+\frac{1}{2}(2 \mathrm{m}) \mathrm{v}_{2}^{2}$
$\Rightarrow \mathrm{v}_{1}=2 \mathrm{x} \sqrt{\frac{\mathrm{k}}{6 \mathrm{m}}}, \mathrm{v}_{2}=\mathrm{x} \sqrt{\frac{\mathrm{k}}{6 \mathrm{m}}}$
$\mathrm{v}_{1}+\mathrm{v}_{2}=\mathrm{x} \sqrt{\frac{3 \mathrm{k}}{2 \mathrm{m}}}$