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(C) Let the mass of the moving particle be $m_1 = m$ and its initial velocity be $u_1 = u_0$.Let the mass of the stationary particle be $m_2 = 5m$ and

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$55.5$

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(C) Let the mass of the moving particle be $m_1 = m$ and its initial velocity be $u_1 = u_0$.Let the mass of the stationary particle be $m_2 = 5m$ and its initial velocity be $u_2 = 0$.For a head-on elastic collision,the final velocity $V_2$ of the second particle is given by:$V_2 = \frac{2m_1 u_1}{m_1 + m_2} = \frac{2m u_0}{m + 5m} = \frac{2m u_0}{6m} = \frac{u_0}{3}$.The kinetic energy transferred to the stationary particle is its final kinetic energy $K_2 = \frac{1}{2} m_2 V_2^2$.$K_2 = \frac{1}{2} (5m) \left(\frac{u_0}{3}ight)^2 = \frac{5}{2} m \frac{u_0^2}{9} = \frac{5}{18} m u_0^2$.The initial kinetic energy of the moving particle is $K_1 = \frac{1}{2} m u_0^2$.The percentage of kinetic energy transferred is $\frac{K_2}{K_1} 	imes 100$.$\frac{\frac{5}{18} m u_0^2}{\frac{1}{2} m u_0^2} 	imes 100 = \frac{5}{18} 	imes 2 	imes 100 = \frac{5}{9} 	imes 100 = 55.55\% \approx 55.5\%$.

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 (in $\%$)? (Assume the collision to be head-on elastic collision)

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