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Successive impacts and impacts with a wall

A wall behaves like a sphere of infinite mass, so an impact with one uses the restitution equation alone. In a sequence of collisions, a common question is whether the next impact happens at all.

Year FMEDEXCEL 9FM0 FM1 4.2

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.

COMMON MISCONCEPTION

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

Momentum is not conserved for the ball on its own against a fixed surface, since the floor is attached to the Earth and absorbs whatever it needs to. What survives is restitution: the rebound speed is e times the approach speed. Dropped from height h the ball returns to e2h, so successive heights form a geometric sequence of ratio e2.

Impact with a fixed wall

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. Conserving the ball's momentum here would apply a conservation law to a system that was never isolated.

Drop a ball from height h and it arrives at √(2gh), rebounds at e√(2gh), and so reaches e²h. Repeat, and the heights form a geometric sequence of ratio e². The total distance travelled before the ball comes to rest sums to h(1 + e²)/(1 − e²).

Four amber balls at decreasing heights above a floor, joined to it by dashed lines. Two metres, then nought point seven two, then nought point two six: each rebound reaches thirty-six per cent of the height before it.
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. Answering it needs only a comparison of velocities. The two bodies meet again exactly when the one behind is still gaining on the one in front.

Set the work out stage by stage with a fresh diagram each time, carrying the velocities forward with their signs. A negative velocity coming out of one stage is the correct input to the next. Reasoning about directions in words instead loses track of the signs.

Three stages stacked down the page. In the first, sphere A moves right at four metres per second towards B. In the second, both move right after the collision. In the third, B has bounced off the wall and comes back, so the two are closing on each other again.
FIG. 2Three stages: A strikes B, B rebounds from the wall, and the two close on each other again.

GUIDED PRACTICE

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.

ASSESSMENT FOCUS

  • For a wall, use restitution alone. Momentum is not conserved for the ball by itself.
  • Redraw the diagram for each stage instead of tracking 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

e = 4.2/7 = 0.6. Height = 4.2²/(2 × 9.8) = 17.64/19.6 = 0.9 m.

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.

Question workbook

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Question workbook

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Answer book

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15 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the successive impacts and impacts with a wall questions page.

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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.

Open the full revision checklist to see every objective in the course in one place.