question_answer

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)  \[\frac{2\sqrt{2}}{3}v\]          

B)  \[\frac{3}{4}v\]               

C)  \[\frac{3}{\sqrt{2}}v\]             

D)  \[\frac{\sqrt{3}}{2}v\]

Correct Answer: A

Solution :

According to law of conservation of kinetic energy, we have \[\frac{1}{2}M{{v}^{2}}+0=\frac{1}{2}M{{\left( \frac{v}{3} \right)}^{2}}+\frac{1}{2}Mv_{2}^{2}\] Þ       \[{{v}^{2}}=\frac{{{v}^{2}}}{9}+v_{2}^{2}\] Þ        \[{{v}^{2}}-\frac{{{v}^{2}}}{9}=v_{2}^{2}\Rightarrow \frac{8{{v}^{2}}}{9}\] Velocity of second block after collision \[{{v}_{2}}=\frac{2\sqrt{2}}{3}v\]