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Collisions overview

Topic 3 of 3

Elastic and inelastic outcomes

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.

Optional check 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?
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?
Optional check 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?
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?