9478 / 2027
Collisions overview

Lesson 3 of 3

Elastic and inelastic outcomes

  • I can apply momentum conservation to inelastic and perfectly elastic interactions between two bodies in one dimension.Syllabus

    Syllabus 9478, 6(c). Apply momentum conservation to inelastic and perfectly elastic interactions between two bodies in one dimension.

  • I can use equality of relative approach and separation speeds for a perfectly elastic two-body collision.Syllabus

    Syllabus 9478, 6(d). Use equality of relative approach and separation speeds for a perfectly elastic two-body collision.

  • I can distinguish conserved total momentum in an isolated interaction from the kinetic energy, which usually changes.Syllabus

    Syllabus 9478, 6(e). Distinguish conserved total momentum in an isolated interaction from the kinetic energy, which usually changes.

Make a guess. You are not marked.

A trolley moving at 4.0 m/s hits an identical trolley at rest, and they stick together. What fraction of the kinetic energy remains?

  1. One quarter, because the speed has halved and kinetic energy goes with v2.
  2. One half: they move off together at 2.0 m/s with twice the mass.
  3. None of it is lost, because momentum is conserved in the collision.
Show the answer

One half: they move off together at 2.0 m/s with twice the mass.

Momentum is conserved: m x 4.0 = 2m x v, so v = 2.0 m/s. Kinetic energy goes from 1/2 m (16) = 8m to 1/2 (2m)(4) = 4m, which is one half.

Momentum conservation is the starting condition, not a classification of the collision. A perfectly elastic collision also conserves total kinetic energy; an inelastic collision does not.

Use the same two-cart model with right positive: A is left of B, with mass 0.60 kg and initial velocity +3.0 m/s; B has mass 0.40 kg and initial velocity -2.0 m/s. Assume a one-dimensional collision with negligible external impulse.

Initial total momentum = 0.60(3.0) + 0.40(-2.0) = +1.0 kg m/s
Initial total kinetic energy = ½(0.60)(3.0)2 + ½(0.40)(2.0)2
= 3.5 J

Momentum gives 0.60vA + 0.40vB = 1.0. With two unknown final velocities, another condition is needed to predict an outcome. Do not assume kinetic energy is conserved merely to obtain another equation.

Three possible outcomes from the same initial account

Initially: A is left, 0.60 kg at +3.0 m/s; B is right, 0.40 kg at -2.0 m/s. Total momentum is +1.00 kg m/s and kinetic energy is 3.50 J. The approach speed is 3.0 - (-2.0) = 5.0 m/s.

Right is positive. Each alternative below assumes negligible net external horizontal impulse for the A + B system, so every outcome retains +1.00 kg m/s total momentum. Their kinetic energies differ.

1. Perfectly elastic

1. Perfectly elasticAfter the same approaching-cart initial state, cart A moves left at negative one metre per second and cart B moves right at positive four metres per second. The blue arrow lengths are 22 and 88 drawing units, on the same velocity scale as the other cart views. The two carts remain inside the named A-plus-B system boundary. Total momentum is positive one kilogram metre per second and kinetic energy is 3.50 joules, equal to its initial value. Relative separation speed is four minus negative one, or five metres per second, equal to the original approach speed.System: cart A + cart BA0.60 kg-1.0 m/sB0.40 kg+4.0 m/sPositive direction

Total kinetic energy: 3.50 J. Kinetic energy is conserved. Separation speed is 4.0 - (-1.0) = 5.0 m/s, equal to the initial approach speed.

2. Inelastic, with the carts separating

2. Inelastic, with the carts separatingAfter the same initial state, cart A has zero velocity and therefore no velocity arrow. Cart B moves right at positive 2.5 metres per second, shown by a 55-unit blue arrow. The carts separate; they have not stuck together. Total momentum is still positive one kilogram metre per second. Kinetic energy is 1.25 joules, so 2.25 joules has been transferred from kinetic energy to other stores and surroundings as appropriate. Relative separation speed is 2.5 metres per second, less than the five-metres-per-second approach speed.System: cart A + cart BA0.60 kgv = 0B0.40 kg+2.5 m/sPositive direction

Total kinetic energy: 1.25 J. Kinetic energy decreases by 2.25 J even though the carts separate. Their separation speed is 2.5 m/s; the elastic relative-speed equality does not apply.

3. Sticking together

3. Sticking togetherAfter the same initial state, the two carts are joined and share positive one metre per second velocity, represented by one 22-unit blue arrow. The same named A-plus-B system boundary encloses both. Total momentum is positive one kilogram metre per second and kinetic energy is 0.50 joules, a decrease of three joules. Their relative separation speed is zero because they remain joined. This is one type of inelastic collision, not the definition of every inelastic collision.System: cart A + cart BA0.60 kgB0.40 kgBoth: +1.0 m/sPositive direction

Total kinetic energy: 0.50 J. Kinetic energy decreases by 3.0 J. The carts have one common velocity, so their relative separation speed is zero.

All blue cart arrows share the same velocity scale, including the common arrow for joined carts. A zero velocity has no direction arrow. Cart positions and sizes are schematic; these are state comparisons, not force diagrams.

Inelastic does not always mean sticking. Only the perfectly elastic outcome preserves the initial kinetic energy and the equality of approach and separation speeds.

All three final states use the same incoming carts and positive direction. Total momentum agrees in every case, while final kinetic energy and relative separation speed differ.

Perfectly elastic: use the relative speeds

A perfectly elastic collision conserves total kinetic energy as well as momentum. In a one-dimensional two-body collision, the relative speed of approach equals the relative speed of separation.

Relative speed describes how quickly the distance between the bodies decreases or increases. It is non-negative. The individual velocities in the calculation still retain their signs.

Before contact, A moves right and B moves left, so the gap closes at:

Approach speed = uA - uB
= 3.0 - (-2.0) = 5.0 m/s

After a separating collision, with B still to A's right, the separation speed is vB - vA. For this perfectly elastic outcome:

vB - vA = 5.0
0.60vA + 0.40vB = 1.0

Substitute vB = vA + 5.0 into the momentum equation:

0.60vA + 0.40(vA + 5.0) = 1.0
vA = -1.0 m/s,   vB = +4.0 m/s

A reverses to the left, while B reverses to the right. Check both conserved quantities:

pfinal,total = 0.60(-1.0) + 0.40(4.0) = +1.0 kg m/s
Ek,final = ½(0.60)(1.0)2 + ½(0.40)(4.0)2
= 0.30 + 3.20 = 3.50 J

The final separation speed is 4.0 - (-1.0) = 5.0 m/s, matching the approach speed. Check the bodies' left/right arrangement when forming a relative speed rather than adding or subtracting velocity symbols without reference to the motion.

Inelastic does not require sticking

Consider a supplied alternative outcome: vA = 0 and vB = +2.5 m/s. The data provide both final velocities; they are not uniquely predicted by momentum conservation alone.

pfinal,total = 0 + 0.40(2.5) = +1.0 kg m/s
Ek,final = 0 + ½(0.40)(2.5)2 = 1.25 J

The total momentum is unchanged, while kinetic energy decreases by 3.5 - 1.25 = 2.25 J. This collision is inelastic. B moves away from stationary A at a relative speed of 2.5 m/s, so the carts separate.

If instead they stick, both velocities are +1.0 m/s, as in the joined-cart calculation. Final kinetic energy is 0.50 J and the kinetic decrease is 3.0 J. Their relative separation speed is zero. Sticking is a perfectly inelastic outcome, not the definition of every inelastic collision.

In these supplied inelastic cases, energy transfers from kinetic energy to other stores or to the surroundings. Total energy is still conserved. The equality of approach and separation speeds belongs to the perfectly elastic case and must not be imposed on either inelastic alternative.

Try it step by step

Work it out in steps: Total kinetic energy before in J, then Total kinetic energy after in J, then Elastic or inelastic?

  1. Problem 1. A 2.0 kg trolley at 3.0 m/s hits a stationary 1.0 kg trolley and they move off together at 2.0 m/s. Is the collision elastic?
  2. Problem 2. A 1500 kg car at 20 m/s locks onto a stationary 1000 kg car and both move at 12 m/s. Is the collision elastic?
  3. Problem 3. A 1.0 kg ball moving at 4.0 m/s hits an identical stationary ball head on. After the collision the first ball is at rest and the second moves at 4.0 m/s. Is the collision elastic?
Show the answer

Problem 1.

Total kinetic energy before in J: KE before = 1/2 x 2.0 x 3.02 = 9.0 J

Total kinetic energy after in J: KE after = 1/2 x 3.0 x 2.02 = 6.0 J

Elastic or inelastic?: Kinetic energy has fallen by 3.0 J, so the collision is inelastic. Momentum is still conserved.

Problem 2.

Total kinetic energy before in J: KE before = 1/2 x 1500 x 202 = 300000 J

Total kinetic energy after in J: KE after = 1/2 x 2500 x 122 = 180000 J

Elastic or inelastic?: Kinetic energy has fallen by 120 kJ, so the collision is inelastic.

Problem 3.

Total kinetic energy before in J: KE before = 1/2 x 1.0 x 4.02 = 8.0 J

Total kinetic energy after in J: KE after = 0 + 1/2 x 1.0 x 4.02 = 8.0 J

Elastic or inelastic?: Total kinetic energy is unchanged, so the collision is perfectly elastic. The relative speed of separation, 4.0 m/s, equals the relative speed of approach.

Check your understanding

Cart A (0.60 kg, initially +3.0 m/s) collides with B (0.40 kg, initially -2.0 m/s) with negligible external impulse. Afterwards A is at rest and B moves right at +2.5 m/s. Which description is correct?

  1. Inelastic and separating: momentum is conserved while kinetic energy decreases.
  2. Perfectly elastic, because the total momentum remains +1.0 kg m/s.
  3. Impossible, because an inelastic collision must leave both carts moving together.
Show the answer

Inelastic and separating: momentum is conserved while kinetic energy decreases.

Final momentum is 0 + 0.40 x 2.5 = +1.0 kg m/s, matching the initial total. Kinetic energy decreases by 2.25 J, and B moves away from stationary A. Inelastic does not require sticking.

Check your understanding

Before contact, A is to the left of B with u_A = +3.0 m/s and u_B = -2.0 m/s. Which relative-speed statement is valid for the perfectly elastic separating outcome?

  1. v_B - v_A = 5.0 m/s here, because this particular outcome is perfectly elastic.
  2. v_B - v_A = 5.0 m/s for every collision, including the sticking outcome.
  3. v_A + v_B = 5.0 m/s, because the approach speeds add.
Show the answer

v_B - v_A = 5.0 m/s here, because this particular outcome is perfectly elastic.

The initial approach speed is 3 - (-2) = 5.0 m/s. For a perfectly elastic collision it equals the final separation speed. Momentum conservation then supplies the other equation needed to find both final velocities.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

A trolley of mass 2.0 kg moving at 3.0 m s-1 collides with a stationary trolley of mass 1.0 kg. They stick together. Calculate their common velocity and the kinetic energy lost, and state the type of collision. [4]

Saved on this browser only. Anyone using this browser can see it.