All subjects · Mathematics · WME02 Mechanics 2

Impacts with a fixed surface and bouncing

Edexcel IAL Mathematics · WME02 Mechanics 2

A particle hitting a fixed smooth wall or floor at right angles rebounds with a smaller speed. Spec 4.3 says impacts with a plane surface will not be oblique, so the motion is always perpendicular to the surface.

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Rebound speed

For a particle hitting a fixed surface at right angles with speed u:

rebound speed = eu.

The impulse from the surface has magnitude m(u + eu) = mu(1 + e), because the velocity reverses. The wall does not move, so momentum is not conserved for the particle alone.

Bouncing balls

A ball dropped from height h hits the ground at speed sqrt(2gh) and rebounds at e sqrt(2gh). It then rises to height e2 h. After the second bounce it rises to e4 h, and so on. The time between bounces is 2 × rebound speed / g.

Kinetic energy lost

Loss at each bounce = (1/2)m u2 - (1/2)m(eu)2 = (1/2)m u2(1 - e2).

Worked example

A ball is dropped from rest from a height of 2.5 m onto a horizontal floor. e = 0.6. Find the height it rebounds to.

  1. Speed hitting the floor: sqrt(2 × 9.8 × 2.5) = 7.00 m/s.
  2. Rebound speed: 0.6 × 7 = 4.20 m/s.
  3. Height: 4.202/(2 × 9.8) = 0.90 m (or e2 h = 0.36 × 2.5).

Answer: 0.90 m

Key idea: Rebound speed = e × impact speed. Impulse = mu(1 + e). Dropped from h: rebounds to e2 h.

Check yourself

  1. A sphere moving at 5 m/s hits a fixed smooth wall at right angles. e = 0.4. Find its rebound speed.
    Show answer

    2 m/s. 0.4 × 5 = 2 m/s.

  2. For the same sphere (mass 0.4 kg, 5 m/s, e = 0.4), find the magnitude of the impulse from the wall.
    Show answer

    2.8 N s. I = 0.4(2 - (-5)) = 2.8 N s.

  3. A ball is dropped from 10 m onto a horizontal floor. e = 0.5. Find the height of its first bounce.
    Show answer

    2.5 m. e2 h = 0.25 × 10 = 2.5 m.

  4. A ball dropped from 3.6 m rebounds to 0.9 m. Find e.
    Show answer

    0.5. e2 = 0.9/3.6 = 0.25, so e = 0.5.

  5. A ball is dropped from 4.9 m onto a floor. e = 0.5. Find the time between its first and second bounces.
    Show answer

    1 s. Impact speed sqrt(96.04) = 9.8 m/s, rebound 4.9 m/s. Time up and down = 2 × 4.9/9.8 = 1 s.

  6. A 0.4 kg sphere hits a wall at 5 m/s and rebounds at 2 m/s. Find the kinetic energy lost.
    Show answer

    4.2 J. (1/2)(0.4)(25 - 4) = 4.2 J.

  7. The ball dropped from 2.5 m with e = 0.6 bounces twice. Find the height it reaches after the second bounce, to 3 significant figures.
    Show answer

    0.324 m. e4 h = 0.1296 × 2.5 = 0.324 m.

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