Question 93 of 130

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)

A$50.0$B$66.6$C$55.6$D$33.3$

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Answer: (C) $55.6$

After collision,

Applying momentum conservation,
$mu=m{v}_{1}+5m{v}_{2}$ $\Rightarrow u={v}_{1}+5{v}_{2}...\left(1\right)$
Coefficient of restitution, $e=1=\dfrac{{v}_{2}-{v}_{1}}{u}$
$∴u={v}_{2}-{v}_{1}...\left(2\right)$
Adding $2u=6{v}_{2}$ $\Rightarrow {v}_{2}=\dfrac{u}{3}$ and $u=\dfrac{u}{3}-{v}_{1}$
$∴{v}_{1}=\dfrac{u}{3}-u=-\dfrac{2u}{3}$
Percentage change in kinetic energy $=\dfrac{\dfrac{1}{2}{mv}_{1}^{2}-\dfrac{1}{2}{mu}^{2}}{\dfrac{1}{2}{mu}^{2}}\times 100$
$=\dfrac{{v}_{1}^{2}-{u}^{2}}{{u}^{2}}\times 100=\dfrac{\dfrac{4{u}^{2}}{9}-{u}^{2}}{{u}^{2}}\times 100$ $=\dfrac{-5}{9}\times 100$ $=55.6$