A mass m moves with a velocity v and collides elastically with another identical mass. After collision the first mass moves with velocity v /3 direction perpendicular to the initial direction of motion. Find the speed of the second mass after collision

In x-directionmv1+0=0+mvx or vx=v1=v In y-direction 0+0=mv3+mvy or vy=v3 ∴ Velocity of second mass after collision v′=v32+v2=43v2=23v

A mass m moves with a velocity v and collides elastically with another identical mass. After collision the first mass moves with velocity v /3 direction perpendicular to the initial direction of motion. Find the speed of the second mass after collision

In x-directionmv1+0=0+mvx or vx=v1=v In y-direction 0+0=mv3+mvy or vy=v3 ∴ Velocity of second mass after collision v′=v32+v2=43v2=23v

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A mass m moves with a velocity v and collides elastically with another identical mass. After collision the first mass moves with velocity v / $\sqrt{3}$ direction perpendicular to the initial direction of motion. Find the speed of the second mass after collision

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$\frac{V}{\sqrt{3}}$

$\sqrt{3 v}$

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

answer is C.

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Detailed Solution

In x-direction

${mv}_{1} + 0 = 0 + m v_{x} o r v_{x} = v_{1} = v I n y - d i r e c t i o n 0 + 0 = m (\frac{v}{\sqrt{3}}) + m v_{y} o r v_{y} = \frac{v}{\sqrt{3}} ∴ V e l o c i t y of second mass afte r collision v^{'} = \sqrt{{(\frac{v}{\sqrt{3}})}^{2} + v^{2}]} = \sqrt{(\frac{4}{3} v^{2})} = \frac{2}{\sqrt{3}} v$