What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of $5$ times its mass?
(Assume the collision to be head-on elastic collision)

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TestHub · Accepted Answer

After collision,

Applying momentum conservation,

$m u = m v_{1} + 5 m v_{2}$ $\Rightarrow u = v_{1} + 5 v_{2} . . . (1)$

Coefficient of restitution, $e = 1 = \frac{v_{2} - v_{1}}{u}$

$∴ u = v_{2} - v_{1} . . . (2)$

Adding $2 u = 6 v_{2}$ $\Rightarrow v_{2} = \frac{u}{3}$ and $u = \frac{u}{3} - v_{1}$

$∴ v_{1} = \frac{u}{3} - u = - \frac{2 u}{3}$

Percentage change in kinetic energy $= \frac{\frac{1}{2} {mv}_{1}^{2} - \frac{1}{2} {mu}^{2}}{\frac{1}{2} {mu}^{2}} \times 100$

$= \frac{v_{1}^{2} - u^{2}}{u^{2}} \times 100 = \frac{\frac{4 u^{2}}{9} - u^{2}}{u^{2}} \times 100$ $= \frac{- 5}{9} \times 100$ $= 55.6 %$

Physics - Center of Mass Question with Solution | TestHub

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