Question

Two blocks $A$ and $B$ of mass $m$ and $2m$ 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

• A. $\left(\sqrt{\frac{3k}{2m}}\right)x$
• B. $\left(\sqrt{\frac{5k}{2m}}\right)x$
• C. $\left(\sqrt{\frac{2kx}{m}}\right)x$
• D. $\left(\sqrt{\frac{2km}{2x}}\right)$

Answer 1

The correct option is A $\left(\sqrt{\frac{3k}{2m}}\right)x$

By momentum conservation,

$\Rightarrow m{v}_1-2m{v}_2=0$

$\Rightarrow v_1=2v_2$

By energy conservation,

$\Rightarrow \frac{1}{2}k{x}^2=\frac{1}{2}m{v_1}^2+\frac{1}{2}\left(2m\right){v_1}^2$

$\Rightarrow \frac{6}{2}k{x}^2=m\left(4{v_2}^2\right)+2m{v_2}^2$

$\Rightarrow {v_2}^2=\frac{k}{6m}{x}^2$

$v_2=\sqrt{\frac{k}{6m}}x$

$v_1=2\sqrt{\frac{k}{6m}}x$

Relative velocity $=v_1+v_2$

$=2\sqrt{\frac{k}{6m}}x+\sqrt{\frac{k}{6m}}x$

$=3\sqrt{\frac{k}{6m}}x$

$=\sqrt{\frac{3k}{2m}}x.$

Hence, the answer is $\sqrt{\frac{3k}{2m}}x.$