Question

Two blocks A and B of mass $m$ and $2m$ respectively are connected by a massless spring of force constant $k$. They are placed on a smooth horizontal plane. They are stretched by an amount $x$ and then released. The relative velocity of the blocks when the spring comes to its natural length is

• A. $x\sqrt{\frac{3k}{2m}}$
• B. $x\sqrt{\frac{2k}{3m}}$
• C. $x\sqrt{\frac{2k}{m}}$
• D. $x\sqrt{\frac{3k}{m}}$

Answer 1

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

Let speed of block $m$ be $v_1,$ speed of block $2m$ be $v_2,$
Here $v_1,v_2$ are in opposite directions.

From conservation of momentum,
$m{v}_1=2m{v}_2$
$v_1=2v_2$
From conservation of energy,
$\frac{1}{2}k{x}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}\left(2m\right)v_2^2$
$\frac{1}{2}k{x}^2=\frac{1}{2}m(2v_2{)}^2+\frac{1}{2}(2m)v_2^2$
$\frac{1}{2}k{x}^2=\left(3m\right)v_2^2$
$v_2=x\sqrt{\frac{k}{6m}}$
$v_1=2x\sqrt{\frac{k}{6m}},$
Relative velocity
$v_1+v_2=x\sqrt{\frac{3k}{2m}}$
$\therefore \left(a\right)$