Two blocks $\mathrm{A}$ and B of masses $m$ and $2m$ respectively, attached at opposite ends of a spring of constant $K$, placed on a smooth horizontal surface. Spring is initially at its natural length $l$. A is given a velocity $2{\mathrm{V}}_0$ and B given velocity ${\mathrm{V}}_0$ as shown.



Maximum separation between $m$ and centre of mass of the system will be:

• A

  $\frac{\mathrm{}l}{3}+\sqrt{\frac{8m{\mathrm{V}}_0^2}{3K}}$
• B

  $\frac{\mathrm{}l}{3}+\sqrt{\frac{2m{\mathrm{V}}_0^2}{3K}}$
• C

  $\frac{2l}{3}+\sqrt{\frac{2m{\mathrm{V}}_0^2}{3K}}$
• Correct answer
  D

  $\frac{2l}{3}+\sqrt{\frac{8m{\mathrm{V}}_0^2}{3K}}$

Solution

$l_1=\frac{2ml}{2m+m}=\frac{2l}{3}$

$K_1{\mathit{\text{l}}}_1=Kl\Rightarrow K_1\left(\frac{2l}{3}\right)=Kl\Rightarrow K_1=\frac{3K}{2}$

$\left|x_{1\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{x}}\right|=\sqrt{\frac{m{\left(2{\mathrm{V}}_0\right)}^2}{\frac{3K}{2}}}=\sqrt{\frac{8m{\mathrm{V}}_0^2}{3K}}$

$\left|d_{1\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{x}}\right|=l_1+x_{1\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{2l}{3}+\sqrt{\frac{8m{\mathrm{V}}_0^2}{3K}}$

Hence, the correct answer is option: 'D'.