What percentage of kinetic energy of a moving particle is...

After collision,

Applying momentum conservation,

$mu = m{v}_{1} + 5m{v}_{2}$ $⇒ u = {v}_{1} + 5{v}_{2} . . .$\left(1\right)$</p><p>Coefficient of restitution,$e = 1 = \dfrac{{v}_{2} - {v}_{1}}{u}$</p><p>$∴ u = {v}_{2} - {v}_{1} . . . \left(2\right)$</p><p>Adding$2u = 6{v}_{2} ⇒ {v}_{2} = \dfrac{u}{3}$and$u = \dfrac{u}{3} - {v}_{1}$</p><p>$∴ {v}_{1} = \dfrac{u}{3} - u = - \dfrac{2u}{3}$</p><p>Percentage change in kinetic energy$= \dfrac{\dfrac{1}{2}{mv}_{1}^{2} - \dfrac{1}{2}{mu}^{2}}{\dfrac{1}{2}{mu}^{2}} \times 100$</p><p>$= \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 %