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Loss of kinetic energy in collisions

Edexcel IAL Mathematics · WME02 Mechanics 2

Momentum is conserved in a collision, but kinetic energy is usually not: some is converted into heat, sound and deformation. M2 asks you to find this loss, and sometimes to use it to find e.

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Finding the loss

Loss of KE = total KE before - total KE after = Σ(1/2)m u2 - Σ(1/2)m v2.

The loss is zero only when e = 1. It is greatest when e = 0. Subtract the right way round: the KE after is smaller.

Impulse in a collision

The impulse on B equals its change in momentum, m_B(v_B - u_B). The impulse on A is equal in size and opposite in direction (Newton's third law). Pearson's examiners note that if your velocity for one particle is wrong, using the other particle can still earn the marks.

Working with letters

With velocities in terms of u, expand carefully: (x + y)2 is x2 + 2xy + y2, not x2 + y2 (an error noted in the June 2025 examiners' report). Factorise at the end to give the answer in the requested form.

Worked example

Using the collision A (2 kg, 6 m/s) with B (3 kg, at rest), e = 0.5, where afterwards v_A = 0.6 m/s and v_B = 3.6 m/s, find the loss of kinetic energy.

  1. KE before = (1/2)(2)(36) = 36 J.
  2. KE after = (1/2)(2)(0.36) + (1/2)(3)(12.96) = 0.36 + 19.44 = 19.8 J.
  3. Loss = 36 - 19.8 = 16.2 J.

Answer: 16.2 J

Key idea: Loss of KE = KE before - KE after. Impulse on one particle = its change in momentum. Zero loss only when e = 1.

Check yourself

  1. A (2 kg, 5 m/s) and B (1 kg, 4 m/s) collide head on with e = 0.6, and afterwards move at 0.2 m/s and 5.6 m/s in A's original direction. Find the loss of kinetic energy, to 3 significant figures.
    Show answer

    17.3 J. Before: 25 + 8 = 33 J. After: 0.04 + 15.68 = 15.72 J. Loss = 17.3 J.

  2. A 2 kg particle moving at 6 m/s coalesces with a 4 kg particle at rest. Find the loss of kinetic energy.
    Show answer

    24 J. Common speed 2 m/s. Before 36 J, after (1/2)(6)(4) = 12 J. Loss 24 J.

  3. In the worked example, find the magnitude of the impulse on B.
    Show answer

    10.8 N s. I = 3 × (3.6 - 0) = 10.8 N s.

  4. A (1 kg, 8 m/s) catches B (3 kg, 2 m/s) with e = 0.5; afterwards v_A = 1.25 m/s and v_B = 4.25 m/s. Find the loss of kinetic energy, to 3 significant figures.
    Show answer

    10.1 J. Before: 32 + 6 = 38 J. After: 0.78125 + 27.09375 = 27.875 J. Loss = 10.1 J.

  5. A (2 kg) moving at 4 m/s hits B (2 kg) at rest. The loss of kinetic energy is 4 J. Find e, to 3 significant figures.
    Show answer

    0.707. v_A = 2(1 - e), v_B = 2(1 + e). KE after = 4(1 - e)2 + 4(1 + e)2 = 8 + 8e2. 16 - (8 + 8e2) = 4, so e2 = 0.5 and e = 0.707.

  6. For which value of e is no kinetic energy lost in a collision?
    • e = 0
    • e = 1
    • e = 0.5
    • Kinetic energy is never lost
    Show answer

    e = 1. Perfectly elastic collisions conserve kinetic energy.

  7. Where does the kinetic energy lost in a collision mainly go?
    • Extra momentum
    • Gravitational potential energy
    • It is destroyed
    • Heat, sound and permanent deformation
    Show answer

    Heat, sound and permanent deformation. Energy is conserved overall; it changes into other forms during the impact.

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Last updated 8 November 2026. Spotted a mistake? Tell us.