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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
$\frac{3}{\sqrt2}v$
B
$\frac{2\sqrt2}{3}v$
C
$\frac{3}{4}v$
D
$\frac{\sqrt3}{2}v$
Detailed Solution
In an elastic collision kinetic energy is conserved:
$\frac{1}{2}Mv^2 = \frac{1}{2}M\left(\frac{v}{3}\right)^2 + \frac{1}{2}Mv'^2$
$v' = \sqrt{v^2 - \frac{v^2}{9}} = \frac{2\sqrt2}{3}v$
$\frac{1}{2}Mv^2 = \frac{1}{2}M\left(\frac{v}{3}\right)^2 + \frac{1}{2}Mv'^2$
$v' = \sqrt{v^2 - \frac{v^2}{9}} = \frac{2\sqrt2}{3}v$