How do you calculate the loss of kinetic energy in a collision of inelastic bodies?

In one line

Kinetic energy before minus kinetic energy after gives the loss, which depends on \( (1 - e^2) \) and is zero only for a perfectly elastic impact.

Explained simply

When two cars crash, metal bends, things heat up and there is a loud bang. That energy came from the cars' kinetic energy. Momentum is conserved, but kinetic energy is partly turned into heat, sound and deformation.

Technically, for direct central impact, \( \Delta KE = \dfrac{m_1 m_2}{2(m_1 + m_2)}(1 - e^2)(u_1 - u_2)^2 \). For e = 1 there is no loss; for e = 0 (plastic) the loss is maximum.

How it works

1. Find the final velocities from momentum and restitution.
2. Initial KE \( = \tfrac12 m_1u_1^2 + \tfrac12 m_2u_2^2 \).
3. Final KE \( = \tfrac12 m_1v_1^2 + \tfrac12 m_2v_2^2 \).
4. Loss = initial - final. Check with the direct formula.
5. Percentage loss \( = \) loss / initial KE \( \times 100 \).

Key formula

\( \Delta KE = \dfrac{m_1m_2}{2(m_1+m_2)}(1 - e^2)(u_1 - u_2)^2 \). Masses in kg, velocities in m/s, energy in J.

Quick example

Same bodies as before: 2 kg at 6 m/s, 4 kg at -3 m/s, e = 0.5, giving \( v_1 = -3 \), \( v_2 = 1.5 \) m/s. Initial KE \( = 36 + 18 = 54 \) J. Final KE \( = 9 + 4.5 = 13.5 \) J. Loss = 40.5 J (75%). Formula: \( \tfrac{8}{12}(0.75)(81) = 40.5 \) J.

Common mistakes

• Squaring velocities without care for magnitude (KE does not depend on sign, but \( u_1 - u_2 \) does).
• Assuming no loss because momentum is conserved.
• Using e instead of \( 1 - e^2 \) in the formula.

How MU asks it

Usually the last 2 to 3 marks of an impact numerical: 'also find the loss of kinetic energy'. Sometimes a 4 to 5-mark derivation of the formula. Show KE before and after with units, then the loss.