Question

A light spring of force constant $K$ is held between two blocks of masses $m$ and $2m$. The two blocks and the spring system rests on a smooth horizontal floor. Now the blocks are moved towards each other compressing the spring by $x$ and then they are suddenly released. Then find the relative velocity between the blocks when the spring attains its natural length.

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

Answer 1

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

Potential energy of the spring is being converted into kinetic energy of system.
Potential energy of the spring before released $=\frac{1}{2}K{x}^2$
K.E. of the system in Center of mass frame $=\frac{1}{2}\mu v_{\text{relative}}^2$, where $\mu =\frac{m\times 2m}{m+2m}=\frac{2m}{3}$ is reduced mass of the system.
Now, $\frac{1}{2}\mu v_{\text{relative}}^2=\frac{1}{2}K{x}^2$
$v_{\text{relative}}^2=\frac{3}{2m}K{x}^2$
$v_{\text{relative}}=\left(\sqrt{\frac{3K}{2m}}\right)x$