Elastic and Inelastic Collisions: Study Guide (Physics 1 4.4)

This guide is for the algebra-based Physics 1 course. It joins the momentum ideas of Topic 4.3 to the kinetic energy you met in Unit 3. No calculus is needed.

Two conservation questions for every collision

When two objects collide and no net external force acts on the pair (or the collision is too short for external forces to matter), Topic 4.3 tells you:

total momentum before = total momentum after

That is true for every collision of such a system. A second, separate question is what happens to the kinetic energy of the system:

K_sys = ½m₁v₁² + ½m₂v₂² + …

Kinetic energy is a scalar. It is never negative, and direction does not matter: square the speed. The answer to the kinetic-energy question sorts collisions into types.

Type	Total kinetic energy of the system	What the objects do
Elastic	Same before and after	Separate; individual kinetic energies may change
Inelastic	Decreases	Usually separate, with some kinetic energy transformed
Perfectly inelastic	Decreases (the largest possible loss for that momentum)	Stick together and move with one common velocity

Momentum is conserved in all three rows.

Elastic collisions

In an elastic collision, the total kinetic energy of the system after the collision equals the total before. While the objects are squashed together, kinetic energy is stored briefly (like energy in a compressed spring) and then fully returned as they spring apart.

The system total is what stays the same. One object can lose kinetic energy while the other gains exactly the same amount. Worked example 2 shows this.

Real collisions are never perfectly elastic, but some come close: hardened steel balls, gliders on an air track with magnetic bumpers, and collisions between gas molecules. Treat a collision as elastic only if the question says so or the data show it.

Inelastic collisions

In an inelastic collision, the total kinetic energy of the system decreases. Some of it is not returned as the objects separate. Nonconservative forces during the contact (forces that deform materials, rub surfaces or crush crumple zones) transform it into:

• internal energy, which shows up as a small temperature rise;
• sound;
• energy of permanent deformation (dents, crumpled metal, squashed clay).

Total energy is still conserved. It has only changed form. The phrase “kinetic energy is lost” means lost from kinetic form.

In a perfectly inelastic collision the objects stick together, so after the collision they share one velocity. For a given total momentum, this leaves the least kinetic energy possible: the pair keeps only the kinetic energy of its centre-of-mass motion.

Representations: a velocity–time graph and an energy bar chart

Worked example 1: carts that stick together

Question. Take +x in cart A’s direction. Cart A (0.30 kg) rolls at 2.0 m/s towards cart B (0.20 kg), which is at rest. Hook-and-loop pads make them stick together. Find (a) their velocity just after the collision, (b) the kinetic energy transformed, and (c) a general expression for the fraction of kinetic energy transformed when a moving object sticks to one at rest.

(a) Momentum before = (0.30)(2.0) + 0 = 0.60 kg·m/s. After, both move at v_f: 0.60 = (0.30 + 0.20)v_f, so v_f = +1.2 m/s.

(b) K before = ½(0.30)(2.0)² = 0.60 J. K after = ½(0.50)(1.2)² = 0.36 J. Kinetic energy transformed = 0.60 − 0.36 = 0.24 J (40% of the original).

(c) Let A (mass m₁, speed v) stick to B (mass m₂, at rest). Momentum gives v_f = m₁v / (m₁ + m₂). Then

K_after / K_before = [½(m₁ + m₂)v_f²] / [½m₁v²] = m₁ / (m₁ + m₂)

so the fraction transformed = m₂ / (m₁ + m₂).

Check. m₂ / (m₁ + m₂) = 0.20 ÷ 0.50 = 0.40, matching (b). If the target is very light (m₂ → 0), almost nothing is lost; if it is very heavy, almost all the kinetic energy is transformed.

Worked example 2: is this collision elastic?

Question. Take +x to the right. Glider A (0.25 kg) moves right at 0.60 m/s towards glider B (0.50 kg) at rest on an air track. After they collide, A moves at −0.20 m/s and B at +0.40 m/s. Show that momentum is conserved, and decide whether the collision is elastic.

1. Momentum before: (0.25)(+0.60) + 0 = +0.15 kg·m/s.
2. Momentum after: (0.25)(−0.20) + (0.50)(+0.40) = −0.05 + 0.20 = +0.15 kg·m/s. Conserved.
3. Kinetic energy before: ½(0.25)(0.60)² = 0.045 J (B has none).
4. Kinetic energy after: A: ½(0.25)(0.20)² = 0.005 J. B: ½(0.50)(0.40)² = 0.040 J. Total 0.045 J.

Total kinetic energy is unchanged, so the collision is elastic.

Interpretation. A’s kinetic energy fell from 0.045 J to 0.005 J, and B’s rose from 0 to 0.040 J. Individual kinetic energies changed; only the system total stayed the same. Notice also that A bounced back: the lighter glider rebounds from the heavier one.

You will not be asked to solve for both unknown final velocities of an elastic collision (that needs simultaneous equations). You may be given one final velocity, as here, and asked to find the other and classify the collision.

Worked example 3: zero total momentum, but not zero kinetic energy

Question. Take +x to the right. A 0.40 kg ball moving at +3.0 m/s meets a 0.60 kg ball moving at −2.0 m/s head-on. Afterwards the 0.40 kg ball moves at −1.5 m/s. Find the other ball’s velocity and classify the collision.

1. Momentum before: (0.40)(3.0) + (0.60)(−2.0) = 1.2 − 1.2 = 0.
2. After: (0.40)(−1.5) + (0.60)v₂ = 0, so −0.60 + 0.60v₂ = 0 and v₂ = +1.0 m/s.
3. K before: ½(0.40)(3.0)² + ½(0.60)(2.0)² = 1.8 + 1.2 = 3.0 J.
4. K after: ½(0.40)(1.5)² + ½(0.60)(1.0)² = 0.45 + 0.30 = 0.75 J.

The kinetic energy fell by 2.25 J (75%), so the collision is inelastic but not perfectly inelastic: the balls separate.

Interpretation. Zero total momentum does not mean zero kinetic energy. If these balls had stuck together, momentum would require them to stop dead, and all 3.0 J would have been transformed. That is the extreme case of a perfectly inelastic collision.

Comparing scenarios with the expressions

Symbolic results let you predict a change without new numbers. For a moving object (m₁, speed v) that sticks to one at rest (m₂):

• Double the launch speed. v_f doubles, since v_f = m₁v / (m₁ + m₂). Both kinetic energies are proportional to v², so the kinetic energy transformed is 4 times as large, but the fraction transformed, m₂ / (m₁ + m₂), does not change.
• Make the target heavier. v_f gets smaller and the fraction transformed gets larger. With m₂ = m₁ the fraction is ½; with m₂ = 3m₁ it is ¾.
• Swap which object moves (same speed). The momentum and so v_f change, and so does the fraction: a heavy object hitting a light one keeps most of its kinetic energy.

Always say which quantities you are holding constant when you make a comparison like this.

Explosions: the reverse of sticking together

In an explosion (Topic 4.3), internal forces push parts apart, so the system’s kinetic energy increases. The extra comes from stored energy, such as a compressed spring or chemical energy. In Topic 4.3’s Worked example 2, the 3.0 kg cart at 12 m/s had 216 J of kinetic energy; after the spring split it, the parts had 162 J + 81 J = 243 J. The spring supplied 27 J. Momentum was conserved throughout.

Testing whether a collision is elastic

An experiment needs the masses and the velocities just before and just after.

1. Measure each mass with a balance.
2. Use motion sensors (or photogates) to measure each velocity shortly before and shortly after the collision, on a level, low-friction track.
3. Calculate the total momentum and total kinetic energy before and after.
4. Repeat for a range of launch speeds. Plot K_after against K_before. For an elastic collision the slope should be 1. For two equal carts that stick, Worked example 1’s result predicts a slope of m₁/(m₁ + m₂) = 0.5.

Check momentum as well: if momentum does not agree within experimental uncertainty, external forces or measurement errors are too large to trust the energy result.

Limiting cases worth knowing

• Equal masses, elastic, target at rest. The moving object stops and the target moves off with the original velocity. Check: momentum mv = 0 + mv, and kinetic energy ½mv² = 0 + ½mv².
• Elastic bounce off a very heavy wall. The object rebounds with nearly the same speed. Its momentum change is about 2mv, but its kinetic energy is unchanged.
• Stick together with zero total momentum. The pair ends at rest, and all the kinetic energy is transformed (Worked example 3).
• Light object hits a heavy one and sticks. Most of the kinetic energy is transformed (Worked example 1(c)).

Common misconceptions

• “Momentum is lost in an inelastic collision.” No. Kinetic energy decreases; momentum is conserved in every collision of an isolated system.
• “Energy is destroyed in an inelastic collision.” Kinetic energy is transformed into internal energy, sound and deformation. Total energy is conserved.
• “In an elastic collision each object keeps its kinetic energy.” Only the system total is the same (Worked example 2).
• “If the objects bounce apart, the collision is elastic.” Bouncing apart is not enough; compare the totals (Worked example 3).
• “Use energy conservation to find the speed just after a sticky collision.” You cannot: kinetic energy is not conserved. Use momentum for the collision, then energy for any later motion, such as rolling up a ramp.
• Adding kinetic energies with signs. Kinetic energy is never negative. Square the speed; direction does not matter.
• Rounding too early. Small differences in kinetic energy can vanish if you round velocities before squaring.

Where this leads

This topic completes Unit 4. Unit 5 begins rotational motion, where similar ideas return for spinning objects: continue with Topic 5.1, Rotational Kinematics. Angular momentum and its conservation follow in Unit 6. Before moving on, try the practice questions, then use the revision notes and the checklist. You can also review Topic 4.3, Conservation of Linear Momentum, or return to the course roadmap.