How do you solve a direct central impact between two bodies using momentum and restitution?

In one line

Use conservation of momentum and the coefficient of restitution as two equations to find the two unknown final velocities.

Explained simply

Two trolleys rolling towards each other on a track collide head-on. We have two unknowns (their speeds after), so we need two facts: total momentum is unchanged, and the bounciness e tells how fast they separate.

Technically, for direct central impact along a line: (1) \( m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \); (2) \( v_2 - v_1 = e(u_1 - u_2) \).

How it works

1. Choose a positive direction; give velocities in the opposite direction a negative sign.
2. Write the momentum equation.
3. Write the restitution equation.
4. Solve the two simultaneous equations for \( v_1 \) and \( v_2 \).
5. Interpret signs: negative means motion in the negative direction after impact.
6. Special cases: e = 1 with equal masses: the bodies exchange velocities; e = 0: common velocity \( v = (m_1u_1 + m_2u_2)/(m_1 + m_2) \).

Key formula

\( v_1 = \dfrac{m_1u_1 + m_2u_2 - m_2e(u_1 - u_2)}{m_1 + m_2} \), \( v_2 = \dfrac{m_1u_1 + m_2u_2 + m_1e(u_1 - u_2)}{m_1 + m_2} \). Masses in kg, velocities in m/s.

Quick example

\( m_1 = 2 \) kg at 6 m/s, \( m_2 = 4 \) kg at 3 m/s in the opposite direction, e = 0.5. Momentum: \( 2(6) + 4(-3) = 0 = 2v_1 + 4v_2 \). Restitution: \( v_2 - v_1 = 0.5(6 + 3) = 4.5 \). Solving: \( v_2 = 1.5 \) m/s, \( v_1 = -3 \) m/s. Both bodies bounce back.

Common mistakes

• Taking \( u_2 = +3 \) when it moves the other way.
• Writing \( v_1 - v_2 \) instead of \( v_2 - v_1 \) on the separation side.
• Using KE conservation when e is less than 1.

How MU asks it

5 to 8-mark numerical, often combined with 'find the loss of kinetic energy'. Sketch before/after, state the sign convention, write both equations, and box \( v_1 \), \( v_2 \) with directions.