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

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

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

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

$\frac{3}{4} v$

answer is C.

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Detailed Solution

Since the collision is elastic, the total kinetic energy is conserved.
If v’ is the final velocity of the second particle,

$\Rightarrow \frac{1}{2} {mv}^{2} + \frac{1}{2} m (0)^{2} = \frac{1}{2} m {(\frac{v}{3})}^{2} + \frac{1}{2} m {(v^{'})}^{2}$

$\Rightarrow {(v^{'})}^{2} = \frac{8}{9} v^{2} \Rightarrow v^{'} = \frac{2 \sqrt{2}}{3} v$

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