A body of mass m₁ = 2 kg moves at v₁ = 4 m/s and collides head-on with a stationary body of mass m₂ = 6 kg. The collision is perfectly elastic. Find the velocities of both bodies after the collision.
m₁ = 2 kg
v₁ = 4 m/s
m₂ = 6 kg
v₂ = 0
Conservation of momentum gives one equation, but there are two unknowns — a second condition is needed.
For a perfectly elastic head-on collision the approach speed equals the separation speed, but with the opposite sign.
A negative velocity means the first body moves backward — this makes physical sense: a light body hits a heavy stationary one.
1. Write down the conservation of momentum:
$$m_1 v_1 = m_1 v_1' + m_2 v_2'$$2. Since the collision is elastic, kinetic energy is also conserved (for a head-on collision this is equivalent to the conservation of relative speed):
$$v_1 - v_2 = v_2' - v_1'$$3. From the second equation: v₂' = v₁ + v₁'. Substitute into the first:
$$m_1 v_1 = m_1 v_1' + m_2 (v_1 + v_1')$$$$v_1' = \frac{m_1 - m_2}{m_1 + m_2} v_1 = \frac{2 - 6}{2 + 6} \cdot 4 = -2 \text{ m/s}$$$$v_2' = \frac{2 m_1}{m_1 + m_2} v_1 = \frac{2 \cdot 2}{8} \cdot 4 = 2 \text{ m/s}$$4. Check: p(before) = 2 * 4 = 8 kg·m/s, p(after) = 2 * (-2) + 6 * 2 = 8 kg·m/s ✓
v₁' = -2 m/s (the body bounces back), v₂' = 2 m/s