Question

On a frictionless surface, a block of mass $M$ moving at speed $v$ collides elastically with another block of same mass $M$ which is initially at rest. After collision the first block moves at an angle $\theta$ to its initial direction and has a speed $\frac{v}{3}$. The second block's speed after the collision is

• A. $\frac{3}{\sqrt 2}v$
• B. $\frac{\sqrt3}{ 2}v$
• C. $\frac{2\sqrt 2}{3}v$
• D. $\frac{3}{4}v$

Answer 1

Answer: (C)

The situation is shown in the figure.

Let $v^{\prime}$ be speed of second block after the collision. As the collision is elastic, so kinetic energy is conserved.
According to conservation of kinetic energy,
$\frac{1}{2} M v^{2}+0=\frac{1}{2} M\left(\frac{v}{3}\right)^{2}+\frac{1}{2} M v^{\prime 2} $
$v^{2}=\frac{v^{2}}{9}+v^{\prime 2} $
or $ v^{\prime 2}=v^{2}-\frac{v^{2}}{9}=\frac{9 v^{2}-v^{2}}{9}=\frac{8}{9} v^{2}$
$v^{\prime}=\sqrt{\frac{8}{9} v^{2}}=\frac{\sqrt{8}}{3} v=\frac{2 \sqrt{2}}{3} v$