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

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
$mv_{1}=2mv_{2}$
$\Rightarrow \, \, \, v_{1}=2v_{2}$
Using conservation of energy
$\frac{1}{2}kx^{2}=\frac{1}{2}mv_{1}^{2}+\frac{1}{2}2mv_{2}^{2}$
$\frac{1}{2}kx^{2}=\frac{1}{2}m\left(2 v_{2}\right)^{2}+mv_{2}^{2}$
$\frac{1}{2}kx^{2}=2mv_{2}^{2}+mv_{2}^{2}$
$3mv_{2}^{2}=\frac{k x^{2}}{2}$
$v_{2}^{2}=\frac{k x^{2}}{6 m}$
$v_{2} \, \sqrt{\frac{k}{6 m}}.x$
Relative velocity $=3v_{2}$
$=3\sqrt{\frac{k}{6 m}}.x$
$= \, \sqrt{\frac{3 k}{2 m}}.x$