Free Pearson Edexcel International A-level Mathematics · M2

Momentum and restitution

What you’ll learn

  • Apply momentum conservation and Newton’s impact law.
  • Interpret rebound, energy loss and successive impacts.

Understand the idea

For a short direct collision with negligible external impulse, total momentum along the line of impact is conserved. Use signed velocities in one fixed direction. Momentum conservation supplies one equation but usually cannot determine two outgoing velocities on its own.

The coefficient of restitution relates the speed of separation to the speed of approach and provides a second equation. Values range from zero for no relative separation to one for an elastic impact. For a fixed wall, the normal velocity reverses direction and its magnitude is multiplied by ee. Calculate kinetic energies before and after to find the loss.

Key formulae

m1u1+m2u2=m1v1+m2v2m_1u_1+m_2u_2=m_1v_1+m_2v_2

v2−v1=e(u1−u2)(particle 1 approaching particle 2)v_2-v_1=e(u_1-u_2)\quad\text{(particle 1 approaching particle 2)}

v=−eu(direct impact with a fixed wall)v=-eu\quad\text{(direct impact with a fixed wall)}

ΔEloss=Ek,before−Ek,after\Delta E_{\mathrm{loss}}=E_{k,\mathrm{before}}-E_{k,\mathrm{after}}

Adding a stationary mass lowers the shared collision speed

Conditions. A 2 kg object travelling at +6 m s⁻¹ sticks to an initially stationary mass m₂. External impulse is negligible. For the m₂ = 0 endpoint, use the no-added-mass limiting case.

Initially stationary mass / kg; Shared final velocity / m s⁻¹.

Momentum conservation gives v = 12/(2 + m₂). Adding 2 kg gives 3 m s⁻¹, half the initial speed; adding 4 kg gives 2 m s⁻¹. Total momentum remains 12 kg m s⁻¹, but kinetic energy is reduced when a positive stationary mass sticks.

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Worked example

Equal masses collide directly. Their initial velocities are 66 and 0 m s−10\,\mathrm{m\,s^{-1}} and e=1/2e=1/2. Find their outgoing velocities in the initial direction of motion.

  1. Momentum gives v1+v2=6v_1+v_2=6.
  2. Restitution gives v2−v1=(1/2)(6)=3v_2-v_1=(1/2)(6)=3.
  3. Add the equations: 2v2=92v_2=9, then substitute back.

Answer: v1=1.5 m s−1v_1=1.5\,\mathrm{m\,s^{-1}}, v2=4.5 m s−1v_2=4.5\,\mathrm{m\,s^{-1}}.

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