1. Given

   • A moving particle of mass $m$ strikes a stationary particle of mass $5m$.
   • Collision is head-on elastic.
   • We need the percentage of initial kinetic energy transferred to the stationary particle.

2. Use standard result for 1D elastic collision

   If a particle of mass $m_1$ with initial speed $u_1$ collides elastically with a stationary particle of mass $m_2$, then the speed of the second particle after collision is $v_2 = \frac{2m_1}{m_1+m_2}u_1$

   Here, $m_1 = m, \qquad m_2 = 5m$

   So, $v_2 = \frac{2m}{m+5m}u_1 = \frac{2m}{6m}u_1 = \frac{u_1}{3}$

3. Initial kinetic energy of first particle $K_i = \frac{1}{2}mu_1^2$

4. Final kinetic energy of second particle $K_2 = \frac{1}{2}(5m)\left(\frac{u_1}{3}\right)^2$ $K_2 = \frac{1}{2}(5m)\frac{u_1^2}{9} = \frac{5}{18}mu_1^2$

5. Fraction of kinetic energy transferred $\frac{K_2}{K_i} = \frac{\frac{5}{18}mu_1^2}{\frac{1}{2}mu_1^2}$ $\frac{K_2}{K_i} = \frac{5}{18}\cdot 2 = \frac{5}{9}$

6. Convert to percentage $\frac{5}{9}\times 100\% = 55.6\%$

7. Option check

   • A: $50.0\%$ ❌
   • B: $66.6\%$ ❌
   • C: $55.6\%$ ✅
   • D: $33.3\%$ ❌

Therefore, the correct answer is C.