A particle of mass m and momentum P moves on a smooth horizontal table and collides directly and elastically with a similar particle (of mass m) having momentum –2p. The loss (–) or gain (+) in the kinetic energy of the first particle in the collision is

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$- \frac{p^{2}}{4 m}$

$+ \frac{p^{2}}{2 m}$

$zero$

$+ \frac{3 p^{2}}{2 m}$

answer is C.

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Detailed Solution

$Let | P | = mv = p$

in elastic head on collision with equal masses, velocities are interchanged.

Gain in kinetic energy of first $\frac{1}{2} m [(2 v)^{2} - v^{2}]$

$= \frac{3}{2} m v^{2} = (\frac{3}{2} m) {(\frac{p}{m})}^{2} = \frac{3 p^{2}}{2 m}$

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