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)

First, we note that a particle of mass $$m$$ moving with velocity $$u$$ strikes a stationary particle of mass $$5m$$ in a head-on elastic collision, and we wish to find the percentage of kinetic energy transferred.

Next, we apply conservation of momentum, which gives $$mu = mv_1 + 5mv_2.$$ Dividing by $$m$$ yields $$u = v_1 + 5v_2 \quad \cdots(1).$$

Since the collision is elastic, conservation of kinetic energy implies $$\tfrac{1}{2}mu^2 = \tfrac{1}{2}mv_1^2 + \tfrac{1}{2}(5m)v_2^2.$$ Dividing by $$\tfrac{1}{2}m$$ leads to $$u^2 = v_1^2 + 5v_2^2 \quad \cdots(2).$$

Now, using the coefficient of restitution condition for a perfectly elastic collision, the relative velocity of separation equals the relative velocity of approach: $$v_2 - v_1 = u \quad \cdots(3).$$

Substituting from equation (3) into equation (1) allows us to solve for the velocities. Since $$v_1 = v_2 - u$$, equation (1) becomes $$u = (v_2 - u) + 5v_2 = 6v_2 - u,$$ so $$2u = 6v_2$$ and therefore $$v_2 = \frac{u}{3}.$$ Then equation (3) gives $$v_1 = \frac{u}{3} - u = \frac{-2u}{3}.$$

Next, we calculate the kinetic energy transferred to the initially stationary particle of mass $$5m$$. The initial kinetic energy of the system is $$KE_i = \frac{1}{2}mu^2,$$ and the final kinetic energy of the $$5m$$ particle is $$KE_2 = \frac{1}{2}(5m)\left(\frac{u}{3}\right)^2 = \frac{1}{2}(5m)\left(\frac{u^2}{9}\right) = \frac{5mu^2}{18}.$$

Therefore, the percentage of kinetic energy transferred is $$\text{Percentage} = \frac{KE_2}{KE_i} \times 100 = \frac{\frac{5mu^2}{18}}{\frac{mu^2}{2}} \times 100 = \frac{5}{18} \times 2 \times 100 = \frac{10}{18} \times 100 = \frac{5}{9} \times 100 \approx 55.6\%.$$ Thus, the correct answer is Option C.