Elastic and Inelastic Collisions: Study Guide (Physics C: Mechanics 4.4)

Topic 4.3 showed that total momentum is the same just before and just after a collision. That is true for every collision of an isolated system, so momentum cannot tell collisions apart. Kinetic energy can. This topic sorts collisions by what happens to the system’s total kinetic energy, and shows how to use that extra information.

Two checks for every collision

For a system of colliding objects, compare just before with just after (Topic 4.3):

1. Momentum: P_before = P_after (vector, in components). This holds for elastic and inelastic collisions alike.
2. Kinetic energy: compare K_before = Σ ½mᵢuᵢ² with K_after = Σ ½mᵢvᵢ².

Here u stands for velocities before and v for velocities after. The result of check 2 gives the type:

Type	Total kinetic energy	What the objects do
Elastic	K_after = K_before	separate
Inelastic	K_after < K_before	usually separate
Perfectly inelastic	K_after < K_before, by the largest amount allowed	stick together, one common velocity

Kinetic energy is a scalar, so add the kinetic energies of the objects as plain numbers. Never add them as vectors.

Elastic collisions

In an elastic collision the system’s total kinetic energy after equals the total before. That does not mean each object keeps its own kinetic energy. Energy is usually transferred from one object to the other. In Worked example 1, an elastic outcome would leave the first glider with only 2.2 mJ of its 54 mJ; the rest moves to the second glider.

Elastic collisions happen when the contact forces are close to conservative: the objects deform and spring back fully, like stiff steel spheres, or never touch at all, like carts with repelling magnets. During the contact, some kinetic energy is stored as elastic potential energy, so the system’s kinetic energy dips and then recovers. “Elastic” is a statement about before and after only.

Solving a one-dimensional elastic collision

Take objects 1 and 2 on the x-axis, with velocities u₁, u₂ before and v₁, v₂ after. Two equations hold:

• Momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, so m₁(u₁ − v₁) = m₂(v₂ − u₂).
• Kinetic energy: ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂², so m₁(u₁² − v₁²) = m₂(v₂² − u₂²).

Factor the second as m₁(u₁ − v₁)(u₁ + v₁) = m₂(v₂ − u₂)(v₂ + u₂) and divide by the first (allowed when the objects really do collide, so u₁ ≠ v₁). This leaves u₁ + v₁ = v₂ + u₂, which rearranges to

v₂ − v₁ = −(u₂ − u₁)

The relative velocity reverses: the objects separate as fast as they approached. This linear equation replaces the quadratic energy equation. Solve it with the momentum equation. For object 2 initially at rest (u₂ = 0):

v₁ = (m₁ − m₂)u₁ / (m₁ + m₂) and v₂ = 2m₁u₁ / (m₁ + m₂)

Check the limits. Equal masses: v₁ = 0 and v₂ = u₁, so the objects swap velocities. A very heavy target (m₂ ≫ m₁): v₁ ≈ −u₁, so the light object bounces straight back. A very light target (m₂ ≪ m₁): v₂ ≈ 2u₁.

Inelastic and perfectly inelastic collisions

In an inelastic collision the total kinetic energy decreases. During contact, nonconservative forces act between the objects: forces that permanently bend metal, crush foam, rub surfaces or make them vibrate. These forces transform some of the kinetic energy into other forms: internal (thermal) energy of the objects, sound, and the energy stored in permanent deformation. That energy does not come back as kinetic energy when the objects separate.

When the objects stick together, the collision is perfectly inelastic: they leave with one shared velocity. Momentum conservation fixes it:

v = (m₁u₁ + m₂u₂) / (m₁ + m₂) = v_cm

The joined object moves at the center-of-mass velocity, which no collision can change.

Why sticking loses the most

Split the system’s kinetic energy into two parts: K = ½Mv_cm² + K_rel. The first part belongs to the motion of the center of mass. It cannot change, because v_cm is constant. The second part, K_rel, belongs to the motion of the objects relative to the center of mass. Only K_rel is available to be transformed. When the objects stick, they have no motion relative to the center of mass, so all of K_rel is gone. For two objects:

K_lost (perfectly inelastic) = ½ · m₁m₂/(m₁ + m₂) · |u₁ − u₂|²

This is a useful extension rather than a required formula. You can always get the same answer by finding K_before and K_after directly.

A quick test in one dimension: approach and separation speeds

For a head-on collision, compare how fast the objects separate after with how fast they approached before:

• Separation speed equal to approach speed: elastic (the relative velocity reversed).
• Separation speed smaller, but not zero: inelastic.
• Separation speed zero: perfectly inelastic (they move together).

This test needs only velocities, so it is quick to apply to sensor data. It agrees with the kinetic energy test because, for a fixed total momentum, only the relative motion carries the kinetic energy that can be transformed. When data have uncertainties, decide whether a small difference in speeds, or a small fall in kinetic energy, is larger than the measurements can explain before you call a collision inelastic.

Background: explosions run the other way

In an explosion, stored energy (a compressed spring, a chemical reaction) becomes kinetic energy, so the system’s kinetic energy increases. Momentum is still conserved. This topic’s three labels describe collisions, where kinetic energy can only stay the same or fall.

Worked example 1: classifying a glider collision

Question. Take +x along a level air track. Glider A (0.30 kg) moves at +0.60 m/s towards glider B (0.20 kg), which is at rest. Just after the collision, B moves at +0.60 m/s. Find (a) A’s velocity after, (b) whether the collision is elastic, and (c) how the result compares with the elastic and perfectly inelastic outcomes.

1. System: both gliders. The track is level and frictionless, so P_x is conserved.
2. P_before = 0.30 × 0.60 = 0.18 kg·m/s.
3. (a) 0.18 = 0.30v_A + 0.20 × 0.60, so 0.30v_A = 0.06 and v_A = +0.20 m/s. A keeps moving forward, more slowly.
4. (b) K_before = ½ × 0.30 × 0.60² = 0.054 J (54 mJ). K_after = ½ × 0.30 × 0.20² + ½ × 0.20 × 0.60² = 0.006 + 0.036 = 0.042 J (42 mJ). K fell, so the collision is inelastic. 0.012 J became internal energy and sound.
5. (c) Elastic limit: v_A = (0.30 − 0.20)(0.60) ÷ 0.50 = +0.12 m/s and v_B = 2(0.30)(0.60) ÷ 0.50 = +0.72 m/s. K = 2.16 + 51.84 = 54 mJ, as required.
6. Perfectly inelastic limit: v = 0.18 ÷ 0.50 = +0.36 m/s, K = ½ × 0.50 × 0.36² = 0.0324 J (32.4 mJ).

Interpretation. The measured outcome lies between the two limits (Figure 1). Another sign: the gliders separate at 0.60 − 0.20 = 0.40 m/s, slower than their approach speed of 0.60 m/s. In an elastic collision the two speeds would be equal.

Worked example 2: two pucks that stick, in two dimensions

Question. On a level air table, take +x east and +y north. Puck 1 (0.20 kg) slides east at 3.0 m/s. Puck 2 (0.30 kg) slides north at 2.0 m/s. Both have sticky edges, and they lock together when they meet. Find (a) their velocity just after and (b) the fraction of kinetic energy transformed.

1. System: both pucks; no horizontal external force, so P_x and P_y are conserved.
2. P_x = 0.20 × 3.0 = 0.60 kg·m/s; P_y = 0.30 × 2.0 = 0.60 kg·m/s.
3. (a) Combined mass 0.50 kg: v_x = 0.60 ÷ 0.50 = 1.2 m/s and v_y = 1.2 m/s. So v = √(1.2² + 1.2²) ≈ 1.7 m/s at 45° north of east.
4. (b) K_before = ½(0.20)(3.0)² + ½(0.30)(2.0)² = 0.90 + 0.60 = 1.5 J. K_after = ½(0.50)(1.70)² ≈ 0.72 J. Transformed: 0.78 J, which is 52% of the original.

Check with the extension formula. The relative velocity has size √(3.0² + 2.0²), so |u₁ − u₂|² = 13 m²/s². ½ × (0.20 × 0.30 ÷ 0.50) × 13 = ½ × 0.12 × 13 = 0.78 J. The two methods agree.

Common misconceptions

• “Inelastic means momentum is not conserved.” Momentum is conserved in both types. The labels describe kinetic energy only.
• “In an elastic collision each object keeps its kinetic energy.” Only the total is kept. Energy moves between objects.
• “Kinetic energy is conserved during an elastic collision.” It dips during contact while energy is stored elastically; it is restored afterwards.
• “Perfectly inelastic means all the kinetic energy is lost.” Only the energy of motion relative to the center of mass is lost. If v_cm ≠ 0, the joined object still moves.
• Adding kinetic energies as vectors. Kinetic energy is a scalar. Add ½mv² values for all objects.
• Using the elastic formulas without checking the type. v₂ − v₁ = −(u₂ − u₁) holds only if kinetic energy is conserved.
• “The lost energy is destroyed.” It is transformed into internal energy, sound and deformation; total energy is conserved.

Where this leads

Collisions often start or end other motion. A common pattern is “collision, then energy”: use momentum across the short collision, then energy conservation (Unit 3) for the motion afterwards, never energy across an inelastic collision. Unit 5 begins with Topic 5.1, Rotational Kinematics. Try the practice questions now, then use the revision notes and the checklist. To review conservation of momentum itself, go back to Topic 4.3.