An elastic collision conserves momentum and kinetic energy
You will be able to: Use both conservation laws to find one-dimensional outgoing velocities.
What extra equation makes an elastic collision solvable?
On an ideal low-friction track, a moving cart can strike an identical resting cart and transfer its velocity: the first stops and the second moves away. Total momentum and total kinetic energy both remain unchanged.
A useful starting point: When objects stick, share momentum—not speed →
Words and symbols before equations
- Elastic collision
- A collision with the same total kinetic energy before and after.
- Initial velocities u₁, u₂
- Signed velocities just before the interaction.
- Final velocities v₁, v₂
- Signed velocities just after the interaction.
- Relative velocity
- One velocity minus the other, measured along the same axis.
What this picture assumes
Isolated one-dimensional perfectly elastic collision of approaching point objects. Total momentum and total kinetic energy are conserved. A common velocity-arrow scale is used before and after; arrows are not force vectors.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Final velocities: left -1 m/s, right 2 m/s. Total P = 3 kg·m/s before and after; K_i = 4.5 J, K_f = 4.5 J; converted 0 J. Elastic: total K is restored.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
For an isolated pair, momentum gives m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. Elasticity adds ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂². Each individual object can gain or lose K while the pair’s total remains fixed.
Factor differences of squares in the energy equation and use the momentum exchange. For a nontrivial one-dimensional collision this yields u₁ − u₂ = −(v₁ − v₂): approach and separation relative speeds match. This relation plus momentum provides the two equations for two unknown final velocities.
Solving gives v₁ = [(m₁−m₂)u₁ + 2m₂u₂]/(m₁+m₂) and v₂ = [2m₁u₁ + (m₂−m₁)u₂]/(m₁+m₂). Equal masses exchange velocities. If a light object strikes a much heavier stationary one, it nearly reverses while the heavy object gains a small forward speed.
A worked example, step by step
A 1 kg cart moving at +3 m/s hits a stationary 2 kg cart elastically on a line. Find both outgoing velocities and check the two conserved totals.
- Momentum equation: 3 = v₁ + 2v₂ in kg·m/s.
- Relative-speed equation: v₂ − v₁ = 3 m/s, so v₂ = v₁ + 3.
- Substitute to get 3 = 3v₁ + 6, giving v₁ = −1 m/s and v₂ = +2 m/s.
- Final momentum is −1 + 4 = 3 kg·m/s. Final K = ½(1)(1) + ½(2)(4) = 4.5 J, matching initial K.
Elasticity preserves total K, not the kinetic energy or velocity of each object separately.
Does a bounce automatically mean an elastic collision?
Compare with an explanation
No. Objects can separate with less total K than before. You must compare the energies.
Predict. Change one thing. Explain.
Keep cart 2 initially at rest and vary its mass. Compare equal masses with a heavier target. Check both the momentum and kinetic-energy totals as the first cart’s final velocity changes sign.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Final velocities: left -1 m/s, right 2 m/s. Total P = 3 kg·m/s before and after; K_i = 4.5 J, K_f = 4.5 J; converted 0 J. Elastic: total K is restored.
Isolated one-dimensional perfectly elastic collision of approaching point objects. Total momentum and total kinetic energy are conserved. A common velocity-arrow scale is used before and after; arrows are not force vectors.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant impulse, system boundary, momentum or calculus relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA 3 kg cart at +2 m/s strikes a stationary 1 kg cart elastically. Derive the final velocities from momentum and relative speed, then verify K.
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Compare with the answer and four-point rubric
- 1 point: Momentum gives 6 = 3v₁ + v₂.
- 1 point: Elastic relative speed gives v₂ − v₁ = 2.
- 1 point: Solving yields v₁ = +1 m/s and v₂ = +3 m/s.
- 1 point: K_f = ½(3)(1²) + ½(1)(3²) = 6 J, equal to initial ½(3)(2²).
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Which two totals are preserved for an isolated elastic collision?
Momentum and kinetic energy.
RECALL 2Can one object gain K in an elastic collision?
Yes, while the other loses the same amount.
RECALL 3What additional condition makes the one-dimensional solution unique?
Elasticity supplies the energy or equivalent relative-speed equation.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
An elastic collision conserves momentum and kinetic energy
- P_i = P_f and K_i = K_f for an isolated elastic pair.
- In a one-dimensional nontrivial elastic collision: v₂ − v₁ = u₁ − u₂.
- Equal masses exchange their initial velocities.
Remember: Elasticity preserves total K, not the kinetic energy or velocity of each object separately.
Conditions: Isolated one-dimensional perfectly elastic collision of approaching point objects. Total momentum and total kinetic energy are conserved. A common velocity-arrow scale is used before and after; arrows are not force vectors.
Refresh Kid · AP Physics C: Mechanics Unit 4 (official Unit 4) · Objectives 4.4.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 4.4, objectives 4.4.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026 alongside the Fall 2026 clarifications. This is Mechanics Unit 4: Linear Momentum. The unit covers Topics 4.1–4.4. Calculus connects force to momentum derivatives and impulse integrals. Collision calculations use one or two dimensions; the optional spatial fragment diagram is qualitative. Changing-mass examples explicitly account for momentum carried across a boundary, so dp/dt is not used blindly for an open system. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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