Problem: Find the ratio of masses of two particles after an elastic collision in which one particle of mass m1 rebounds with half of its original speed after making a head-on collision with another stationary particle of mass m2.

Solution:
Step 1: Apply conservation of momentum
- Before the collision: m1u1 + m2u2 = (m1 + m2)v
- After the collision: m1v1 + m2v2 = (m1 + m2)v

Step 2: Apply conservation of kinetic energy
- Before the collision: (1/2)m1u1^2 + (1/2)m2u2^2 = total kinetic energy
- After the collision: (1/2)m1v1^2 + (1/2)m2v2^2 = total kinetic energy

Step 3: Use the fact that m1 rebounds with half of its original speed
- v1 = u1/2

Step 4: Substitute v1 in the equations from Step 1 and Step 2
- Before the collision: m1u1 + m2u2 = (m1 + m2)v
- (1/2)m1u1^2 + (1/2)m2u2^2 = total kinetic energy
- After the collision: (1/8)m1u1^2 + (1/2)m2v2^2 = total kinetic energy
- m1u1 + m2u2 = (m1 + m2)(u1/2 + v2)
- (1/2)m1u1^2 + (1/2)m2u2^2 = (1/8)m1u1^2 + (1/2)m2v2^2

Step 5: Simplify the equations
- m1u1 + m2u2 = (m1 + m2)(u1/2 + v2)
- (1/2)m1u1^2 + (1/2)m2u2^2 = (1/8)m1u1^2 + (1/2)m2v2^2
- 2m1u1 + 4m2u2 = (m1 + m2)u1 + 2(m1 + m2)v2
- (3/8)m1u1^2 = (1/2)m2v2^2

Step 6: Solve for v2 in terms of u1
- 2m1u1 + 4m2u2 = (m1 + m2)u1 + 2(m1 + m2)v2
- 2m1u1 + 4m2u2 = m1u1 + m2u1 + 2(m1 + m2)v2
- 2m2u2 = (m1/2)u1 + (m2/2)u1 + (m1 + m2)v2
- v2 = (2m2u2 - (m1/2)u1 - (m2/2)u1)/(m1 + m2)

Step 7