Question

Two blocks $A$ and $B$ of masses $m$ and $2m$ are connected by a massless spring of force constant $k$ and are placed on a smooth horizontal plane. The 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. $x\sqrt{\frac{3k}{2m}}$
• B. $x\sqrt{\frac{2k}{3m}}$
• C. $x\sqrt{\frac{k}{3m}}$
• D. $x\sqrt{\frac{2k}{m}}$

Answer 1

Answer: (A)

Let us assume the velocity of block $A$ and $B$ are $v_1$ and $v_2$ respectively.
Relative velocity will be $v_1+v_2$
Using conservation of linear momentum
$m{v}_1=2m{v}_2$
$\Rightarrow v_1=2v_2$
Using conservation of energy
$\frac{1}{2}k{x}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}2m{v}_2^2$
$\frac{1}{2}k{x}^2=\frac{1}{2}m{\left(2v_2\right)}^2+m{v}_2^2$
$\frac{1}{2}k{x}^2=2m{v}_2^2+m{v}_2^2$
$3m{v}_2^2=\frac{k{x}^2}{2}$
$v_2^2=\frac{k{x}^2}{6m}$
$v_2\sqrt{\frac{k}{6m}}.x$
Relative velocity $=3v_2$
$=3\sqrt{\frac{k}{6m}}.x$
$=\sqrt{\frac{3k}{2m}}.x$