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. $\frac{\sqrt{3}}{2}v$
• 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 $v^{\prime 2}=v^2-\frac{v^2}{9}=\frac{9v^2-v^2}{9}=\frac{8}{9}{v}^2$
$v^{\prime }=\sqrt{\frac{8}{9}{v}^2}=\frac{\sqrt{8}}{3}v=\frac{2\sqrt{2}}{3}v$