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Particles 1 and 2 of masses m and 2m respectively are moving along x and y axes respectively with speed V each as shown in figure. They collide at the origin and coalesce. Then the energy lost in the process is

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$\frac{1}{4} m V^{2}$

$\frac{3}{4} m V^{2}$

$\frac{2}{3} m V^{2}$

$\frac{1}{6} m V^{2}$

answer is C.

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Detailed Solution

Let $V^{1}$ be the velocity of the combined mass. Also let the angle between $V^{1}$ and x axis be $θ$ .

$∴ 3 m . V^{1} \cos θ = m V and 3 m V^{1} \sin θ = 2 m V$

$∴ {(V^{1})}^{2} = {(\frac{V}{3})}^{2} + {(\frac{2 V}{3})}^{2} \Rightarrow {(V^{1})}^{2} = \frac{5}{9} V^{2}$

$∴$ Energy lost $= (\frac{1}{2} m V^{2} + \frac{1}{2} 2 m V^{2}) - \frac{1}{2} 3 m {(V^{1})}^{2}$

$= \frac{3}{2} m V^{2} - \frac{5}{6} m V^{2} = \frac{2}{3} m V^{2}$

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