MathsFurther Mechanics 1 › Successive impacts and impacts with a wall

Successive impacts and impacts with a wall

A wall is a sphere of infinite mass, so only the restitution equation survives. Chain a few impacts together and the question becomes whether the next one happens at all.

Year FMEDEXCEL 9FM0 FM1

Builds on Direct impact and Newton's law of restitution and Kinematics with constant acceleration.

IN THIS TOPIC

  • Apply restitution to an impact with a fixed wall or floor.
  • Follow a chain of impacts and decide whether a further collision occurs.
  • Handle repeated bounces, including the total distance travelled.

WHAT YOU PROBABLY THINK

When a ball bounces off the floor, momentum is conserved, so its speed is unchanged.

A wall takes momentum away

In an impact with a fixed surface, momentum is not conserved for the ball on its own: the wall is attached to the Earth and absorbs whatever it needs to. What survives is the restitution equation, which for a direct impact says the rebound speed is e times the approach speed. The opening claim applies the wrong conservation law to a system that was never isolated.

For a ball dropped from height h, the speed on arrival is √(2gh), the rebound speed is e√(2gh), and the height reached is therefore e²h. Repeating gives a geometric sequence of heights with ratio e², so the total distance travelled before the ball comes to rest sums to h(1 + e²)/(1 − e²).

A ball dropped from 2 m with e = 0.6: each rebound reaches e² of the height before it2 m0.72 m0.259 m0.0933 meach height is 0.36 of the lasttotal distance travelled = 2(1.36)/0.64 = 4.25 m
FIG. 1Successive bounce heights forming a geometric sequence with ratio e squared.

WORKED EXAMPLE

A bouncing ball

A ball is dropped from 2 m onto a floor with e = 0.6. Find the height of the first rebound and the total distance travelled before it stops bouncing.

The first rebound reaches 0.6² × 2 = 0.72 m.

Successive heights are 2, 0.72, 0.259, ... with common ratio 0.36.

Total distance = 2 + 2(0.72 + 0.259 + …) = 2 + 2(0.72/0.64) = 4.25 m, since every height after the first is travelled twice.

Does it happen again?

In a chain of impacts, the answer to each stage feeds the next, and the question usually ends by asking whether a further collision occurs. That is a comparison of velocities, not a new principle: the two bodies meet again exactly when the one behind is moving faster towards the other than the other is moving away.

Set the work out stage by stage with a fresh diagram each time, carrying the velocities forward with their signs. A negative velocity that comes out of one stage is the correct input to the next, and trying to reason about directions in words instead is where the marks go.

Three stages: A strikes B, B rebounds from the wall, and the two approach again at 3.4 m/swallABA hits B4ABafter impact1.63.6ABB returns1.6−1.8
FIG. 2Three stages: A strikes B, B rebounds from the wall, and the two close on each other again.

YOUR TURN

Two collisions and a wall

A sphere A of mass 3 kg moving at 4 m/s strikes a stationary sphere B of mass 2 kg with e = 0.5. B then strikes a wall, rebounding with coefficient of restitution 0.5. Show that A and B collide again.

Show the working

First impact. Momentum: 3(4) = 3vA + 2vB. Restitution: vB − vA = 0.5(4) = 2.

Substituting: 5vA = 8, so vA = 1.6 m/s and vB = 3.6 m/s.

At the wall B rebounds at 0.5 × 3.6 = 1.8 m/s back towards A.

A is still moving forwards at 1.6 and B is now moving backwards at 1.8, so they approach at 3.4 m/s and must collide again.

THE EXAM BIT

  • For a wall, use restitution alone; momentum is not conserved for the ball by itself.
  • Redraw the diagram for each stage rather than trying to track everything on one.
  • To decide whether a further impact occurs, compare velocities with their signs, not speeds.
  • For repeated bounces, identify the geometric ratio e² and say whether you are summing heights or distances.

CHECK YOURSELF

A ball strikes a floor at 7 m/s and rebounds at 4.2 m/s. Find e, and the height it reaches, taking g = 9.8 m/s².

Show a hint

Restitution first, then the usual constant-acceleration result.

Show the answer

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Against a fixed surface only restitution applies: the rebound speed is e times the approach speed, and momentum is not conserved for the ball alone.

In a chain of impacts, carry signed velocities forward; a further collision happens exactly when the bodies are still approaching.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

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  • Apply restitution to an impact with a fixed wall or floor.
  • Follow a chain of impacts and decide whether a further collision occurs.
  • Handle repeated bounces, including the total distance travelled.

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