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)

$\frac{2\sqrt 2}{3}v$

Answer 2

The situation is shown in the figure.

Using energy conservation, $\frac{1}{2} m _{1} v _{1}^{2}+\frac{1}{2} m _{2} v _{2}^{2}=\frac{1}{2} m _{1} v _{1}^{\prime 2}+\frac{1}{2} m _{2} v _{2}^{\prime 2}$ here, $\frac{1}{2} Mv ^{2}+0=\frac{1}{2} M ( v / 3)^{2}+\frac{1}{2} Mv _{2}^{\prime 2}$
or $v _{2}^{\prime 2}=\frac{8}{9} v ^{2}$
or $v_{2}^{\prime}=\frac{2 \sqrt{2}}{3} v$