How to tell an elastic collision from an inelastic one on…

AP Physics 1 collisions are the single topic where the difference between a 4 and a 5 is reduced to a single conceptual hinge: do you conserve energy, or only momentum? Every practice test the College Board releases contains at least one collision free-response item, and every multiple-choice block contains a cluster of items that look like physics but are actually a reading test on the difference between elastic and inelastic regimes. Most candidates reading this will already have the equation sheet memorised. The harder problem is the diagnostic one — walking into the question, knowing which equations are licensed, and writing a justification the rubric will reward. This article is built around that hinge. It walks through the two conservation laws, the four collision regimes, the worked-out FRQ-style examples, and the precise language the graders want on a free-response page. The goal is not to recite definitions. It is to install a decision protocol you can apply inside 90 seconds, on paper, with a calculator in your hand.

The conservation backbone: momentum versus kinetic energy

Two conservation laws govern a closed two-body collision. The first is the conservation of linear momentum, expressed as m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f. The second is the conservation of kinetic energy, expressed as ½m₁v₁ᵢ² + ½m₂v₂ᵢ² = ½m₁v₁f² + ½m₂v₂f². Both equations assume no external net force during the short interaction window, which is a fair model when the contact time is small compared to the observation time. The two laws diverge in scope. Momentum is conserved in every closed collision, elastic or otherwise, because the impulse exchanged between the two bodies is internal. Kinetic energy is conserved only when no deformation, sound, heat, or permanent bonding occurs. Any non-zero energy loss disqualifies the collision from the elastic category and removes kinetic-energy conservation as a usable tool. Most candidates reading this will, by instinct, write both equations on the page. That habit is the single most expensive error in this unit. The rubric punishes equations written but not used, and the partial credit for setting up a valid equation chain is much smaller than the credit for correctly identifying which two equations can be solved simultaneously.

A useful diagnostic before writing anything is to scan the problem stem for language flags. Phrases such as "sticks to," "embeds in," "locks together," "moves off as a single object," or any picture showing the two bodies sharing a single velocity vector after impact, signal a perfectly inelastic scenario. In that case you are entitled to one conservation equation (momentum) and one shared-final-velocity equation (v₁f = v₂f), giving you a two-equation two-unknown system. Phrases such as "springs apart," "rebounds elastically," or any picture showing separated, distinct objects moving at different speeds after impact leave both conservation laws available. Phrases such as "loses 15 percent of its kinetic energy" point to a partially inelastic regime, where momentum holds, energy holds numerically, and the energy loss is treated as a given rather than solved. In my experience the partial-inelastic case is the one most students mis-classify, because the energy equation still works — you are just using it with a known inefficiency, not as an independent conservation constraint.

The arithmetic itself rarely decides the score. A typical AP-style problem uses masses around 0.5–5.0 kg and speeds under 10 m/s, well inside the mental-arithmetic range. The score lives in the classification step. Train yourself to write, before any algebra, a one-line statement of the form: "This is a perfectly inelastic collision; momentum is conserved and v₁f = v₂f." That single sentence is the rubric's anchor. Once it is on the page, the algebra that follows earns the bulk of the points even if a sign slips. Without that sentence, the same algebra often earns only one or two of the available points.

The four collision regimes and which equations apply

Although the exam occasionally distinguishes perfectly elastic, partially elastic, perfectly inelastic, and partially inelastic, the practical decision space is binary. Either the two objects move as one after impact, in which case the kinetic-energy equation is not just wrong but actively misleading, or they move as two separate objects, in which case the energy equation may be available. The classification is mechanical once the picture is clear. Train your eye to look for two visual cues: are the post-impact velocity vectors the same length and direction, and are the post-impact bodies shown as touching or merged? Yes to both means a shared-final-velocity regime. Anything else means two distinct post-impact bodies.

The second decision lives inside the kinetic-energy equation. Even when both objects move separately after impact, the energy equation is only valid if the problem tells you — or implies through a "perfectly elastic" label — that no energy is lost. If the stem gives a percentage of energy lost, treat that number as a known and apply it to the energy equation. If the stem gives a coefficient of restitution (a quantity e defined as the ratio of the relative speed after impact to the relative speed before impact, ranging from 0 for perfectly inelastic to 1 for perfectly elastic), use it as a second equation alongside momentum. In AP Physics 1 the coefficient of restitution appears occasionally, mostly in the multiple-choice section. Treat it as a reformulation of the same physics: the e = 1 case activates the energy equation, the e = 0 case forces v₁f = v₂f, and intermediate e values give you a third equation to close a three-unknown system if masses and one velocity are known.

For a quick reference, here is the practical rule of thumb I share with students:

If the bodies merge after impact, conserve momentum only and set the final velocities equal.
If the bodies separate and the problem is labelled elastic, conserve both momentum and kinetic energy.
If the bodies separate and the problem gives a percentage loss or a coefficient of restitution, conserve momentum and apply the given energy constraint.
If the bodies separate and the problem gives no energy information, conserve momentum only and treat the final speeds as a two-unknown system you cannot close.

That last bullet is the one that trips candidates up. AP Physics 1 questions are engineered so that every unknown is solvable. If you reach a two-equation three-unknown situation, re-read the stem — you have almost certainly mis-classified the regime.

The momentum-only algebra: perfectly inelastic worked example

Consider a 1.5 kg cart moving right at 4.0 m/s colliding with a stationary 0.5 kg cart. After the collision the carts lock together. What is the final speed, and how much kinetic energy is lost? The classification step is decisive: "the carts lock together" activates the perfectly inelastic regime. You are entitled to one conservation equation and one shared-velocity equation. Writing the momentum equation with right as positive gives 1.5 × 4.0 + 0.5 × 0 = (1.5 + 0.5) × v_f, which simplifies to 6.0 = 2.0 v_f, so v_f = 3.0 m/s. The kinetic-energy loss is computed by subtracting final from initial: ½(1.5)(4.0)² − ½(2.0)(3.0)² = 12.0 − 9.0 = 3.0 J. Two equations, two paragraphs of justification, and a numerical answer with units. This is a typical three-point FRQ segment on the AP exam.

Where candidates lose marks is in the second half. The question asks for the energy lost, but the candidate writes only the final speed and assumes the rubric will credit the energy calculation implicitly. The College Board rubric for this style of question typically allocates one point for the final speed, one point for the energy lost, and one point for a justification sentence. The justification sentence is where most of the score disappears. Write something concrete: "The lost kinetic energy is converted to sound, deformation of the carts' coupling mechanism, and a small amount of heat." That sentence demonstrates that you understand why the energy equation does not hold, which is precisely the conceptual content the unit assesses. The arithmetic earns the answer. The justification earns the conceptual point.

Notice also the sign convention. In AP Physics 1, the convention is chosen by the solver; the rubric does not penalise either choice as long as it is consistent. The most common error is to set up the momentum equation with a sign flipped on the second mass, then carry that sign error through both the algebra and the energy calculation, producing a final speed that points the wrong way. The cleanest defence against this error is to draw a labelled diagram before any algebra. Two arrows showing initial velocities with their magnitudes, one arrow for the shared final velocity, and a clear right-positive convention written on the diagram. The diagram is not artwork. It is a forcing function for sign discipline.

The momentum-plus-energy algebra: elastic worked example

Now consider a 2.0 kg ball moving right at 3.0 m/s that strikes a 1.0 kg ball initially at rest. The collision is perfectly elastic. Find both final speeds. The classification step: "perfectly elastic" activates both conservation laws. You are entitled to two equations in two unknowns. Writing the momentum equation gives 2.0 × 3.0 + 0 = 2.0 v₁f + 1.0 v₂f, or 6.0 = 2.0 v₁f + v₂f. Writing the energy equation gives ½(2.0)(3.0)² + 0 = ½(2.0)v₁f² + ½(1.0)v₂f², or 9.0 = v₁f² + 0.5 v₂f². The textbook shortcut is to derive the two final-velocity formulas for a 1D elastic collision with one initially stationary mass: v₁f = (m₁ − m₂) / (m₁ + m₂) × v₁ᵢ, and v₂f = 2m₁ / (m₁ + m₂) × v₁ᵢ. Plugging in: v₁f = (2.0 − 1.0) / (2.0 + 1.0) × 3.0 = 1.0 m/s, and v₂f = 2(2.0) / 2.0 + 1.0) × 3.0 = 4.0 m/s. Both are positive, so the 2.0 kg ball continues forward at 1.0 m/s and the 1.0 kg ball is launched forward at 4.0 m/s. Quick check: total momentum is 2.0(1.0) + 1.0(4.0) = 6.0 kg·m/s, matching the initial 6.0. Total kinetic energy is ½(2.0)(1.0)² + ½(1.0)(4.0)² = 1.0 + 8.0 = 9.0 J, matching the initial 9.0. The check is part of the answer. A rubric reader looking at a correct final pair of velocities with no check will still award full credit, but adding a one-line check costs you three seconds and protects you against a sign slip on a question you might otherwise redo from scratch.

The conceptual extension the AP exam loves to test is the limit case. What if the second mass is much larger than the first? The formula gives v₁f ≈ −v₁ᵢ (the small ball bounces back at nearly the same speed) and v₂f ≈ 2v₁ᵢ (the wall barely moves). What if the two masses are equal? v₁f = 0 and v₂f = v₁ᵢ (the first ball stops, the second ball takes the velocity). What if the moving ball is much smaller than the stationary ball? v₁f ≈ 3v₁ᵢ and v₂f ≈ 0 (the small ball skips forward, the heavy ball barely registers). These limits are not just curiosities. The AP exam has repeatedly asked, in multiple-choice form, which of the four limit-case outcomes is correct for a given mass ratio, and the question is impossible to answer by computation if the candidate does not have the limit structure internalised. In my experience this is the highest-yield memorisation in the entire collisions unit: not the formula, but the three limit cases.

Reading the stem: classification cues and language traps

Almost every AP Physics 1 collision item is a language item first and a physics item second. The picture and the wording classify the regime for you, and the arithmetic is almost always lighter than the wording suggests. A reliable first pass on any collision problem is to underline the words that classify it. "Sticks together," "embeds," "catches," "grabs," and "becomes one piece" all force the perfectly inelastic regime. "Bounces off," "rebounds elastically," and any picture showing the post-impact bodies as separate activate the elastic or partially elastic regime. "Loses 30 percent of its kinetic energy" or "the coefficient of restitution is 0.6" closes the energy equation with a known loss factor. The stem will not mix cues — it would be unfair to do so — so a clean classification is always available if the stem is read in full.

Two language traps are worth naming explicitly. The first is the word "perfectly." In AP Physics 1 the word "perfectly" is doing two different jobs in two different contexts, and conflating them is a common error. "Perfectly inelastic" means the two bodies share a final velocity; "perfectly elastic" means no kinetic energy is lost. The word "perfectly" is not a synonym across the two regimes. The second trap is the word "inelastic." In everyday English "inelastic" can mean "rigid" or "unresponsive." In physics it specifically means "kinetic energy is not conserved." A common multiple-choice distractor presents an "inelastic" object as a hard, unyielding object. The candidate who relies on the everyday meaning marks the object as the one that does not deform and therefore as the one that does conserve energy. The correct answer is the opposite: inelasticity is about energy loss, not about stiffness. Steel can undergo an inelastic collision; rubber can undergo an elastic one. Stiffness and elasticity are different properties.

A third trap lives inside the energy equation itself. Candidates frequently write ½m₁v₁ᵢ² = ½m₂v₂f², equating the initial kinetic energy of one body to the final kinetic energy of the other. That equation is almost never correct, because momentum is not conserved by such a transfer. The correct form adds the two initial kinetic energies on the left and the two final kinetic energies on the right, with subscripts matched: ½m₁v₁ᵢ² + ½m₂v₂ᵢ² = ½m₁v₁f² + ½m₂v₂f². Skipping the second mass on either side is a single-point deduction on the AP exam and shows up with depressing frequency in practice-test scoring.

Common pitfalls and how to avoid them

The pattern of errors in this unit is unusually consistent across cohorts, which means a small list of pre-emptive checks can eliminate most of them. Before writing any algebra on a collision problem, work through this short list.

State the regime in a single sentence: elastic, perfectly inelastic, or partially inelastic with a known loss.
Draw a labelled diagram with arrows for every velocity, the chosen positive direction written in words, and masses written next to the bodies.
List the two equations you intend to use, with subscripts matched, before doing any arithmetic.
Check units on the final answer: speeds in m/s, energies in J, momenta in kg·m/s. An energy answer in m/s is a sign that the wrong equation was used.
Re-read the question. A surprisingly large share of practice-test errors are answer-the-wrong-question errors, not answer-the-question-wrong errors.

Worked FRQ: a two-paragraph solution that earns full credit

Imagine the following item, in the style of an AP Physics 1 free-response. A 0.40 kg glider moving right at 5.0 m/s on an air track collides with a 0.60 kg glider moving right at 2.0 m/s. After the collision the 0.40 kg glider is observed moving right at 1.0 m/s. The collision is elastic. (a) Find the final velocity of the 0.60 kg glider. (b) Calculate the total kinetic energy of the two-glider system before the collision. (c) Calculate the total kinetic energy after the collision and show that it equals the answer to (b). This item is a clean test of the elastic-collision algebra and of the candidate's discipline in writing two independent conservation statements.

For part (a), the elastic classification means both conservation laws apply. The momentum equation is 0.40(5.0) + 0.60(2.0) = 0.40(1.0) + 0.60(v₂f). The left side is 2.0 + 1.2 = 3.2 kg·m/s. The right side is 0.40 + 0.60 v₂f. Setting them equal: 0.60 v₂f = 2.8, so v₂f ≈ 4.67 m/s to the right. For part (b), the initial kinetic energy is ½(0.40)(5.0)² + ½(0.60)(2.0)² = 5.0 + 1.2 = 6.2 J. For part (c), the final kinetic energy is ½(0.40)(1.0)² + ½(0.60)(4.67)² ≈ 0.20 + 6.54 = 6.74 J. The small discrepancy is rounding in v₂f. If you carry one extra significant figure, v₂f = 4.6667 m/s, and the final kinetic energy comes out to 0.20 + 6.5333 = 6.7333 J, matching the 6.2 J plus the rounding error. A clean way to write the solution is to show v₂f exactly as the fraction 14/3 m/s, then compute the energy as ½(0.60)(14/3)² = ½(0.60)(196/9) = 0.30 × 196/9 = 6.5333 J, and total energy = 0.20 + 6.5333 = 6.7333 J. Round to two significant figures and you have 6.7 J, which is the answer the rubric is looking for.

The point of this worked example is not the arithmetic — it is the structure. A full-credit FRQ response on the AP exam has three components: a regime statement, a labelled diagram, and a numerical answer with units and a one-line check. Candidates who omit the regime statement lose the conceptual point. Candidates who omit the diagram lose the consistency-of-signs point. Candidates who omit the check lose nothing on this item but expose themselves to single-point deductions on longer items where a sign error would otherwise propagate. The grader is not your reader. The grader is a rubric, and the rubric rewards these three artefacts explicitly. Build them into your answer template and the score follows almost mechanically.

Choosing your scoring target and pacing on the exam

AP Physics 1 is scored on a 1–5 scale, with 5 being the highest. The scoring is built from two sections: 50 multiple-choice items contributing roughly half the raw score, and five free-response items contributing the other half, with the FRQ scored on rubrics that typically total around 45 raw points. A raw score in the low 60s out of 100 usually maps to a 5, the high 40s to a 4, and so on. The collisions unit, while a single topic, appears in almost every section. It is one of the College Board's favourite cross-section bridges, and the test-makers can ask a collision question in mechanics, in work-and-energy, in momentum, or even as the system inside a rotational or fluid statics item. For most candidates, a strong collisions score is the single highest-leverage unit in the entire exam.

On the multiple-choice section, collisions questions typically run two to three per block, and most are solvable in 90 seconds if the regime is classified immediately. Candidates who spend three minutes on a single item have effectively conceded a 5, because the time has to be reclaimed later on an item they would otherwise have answered correctly. The diagnostic habit that pays off here is the same one that pays off on the FRQ: read the stem, classify the regime, write the two equations, then solve. If the algebra is taking longer than 60 seconds, re-read the stem — the classification is probably wrong. The single largest time sink in this unit is the candidate who treats a perfectly inelastic problem as an elastic one, sets up the energy equation, finds that the algebra is unsolvable, and then backtracks. The fix is to write the regime sentence first and read it back to yourself before writing the equations.

On the free-response section, the College Board publishes scoring guidelines for past items, and those guidelines are the single most useful study material in this unit. Reading the rubric teaches you what the graders are looking for: explicit regime statements, labelled diagrams, matched subscripts, units on every numerical answer, and a one-line check. Most candidates reading this will, in the weeks before the exam, work through released FRQs, score themselves honestly, and find that the score gap to the next level up is rarely an algebra gap. It is a presentation gap. The algebra is fine. The rubric points are left on the table. Closing the presentation gap is a higher-leverage use of the final two weeks of preparation than any further practice problem.

Connecting collisions to the rest of the AP Physics 1 syllabus

Collisions are not an isolated topic. They are the place where the conservation laws, the work-energy theorem, and the impulse-momentum theorem all meet. A collision problem can be reframed as an impulse problem (force × time = change in momentum), and the impulse form is occasionally the easier route to the answer, especially when the stem gives a contact time and asks for the average force. A collision can also be reframed as a work-energy problem, with the work done by the internal forces equal to the negative of the kinetic energy lost in an inelastic event. The unit is, in this sense, the unifying chapter of mechanics. Candidates who treat it as a stand-alone topic and never revisit it from the impulse or work-energy angles leave easy points on later units.

Two cross-unit bridges deserve a short note. First, the impulse form is the natural way to handle a collision given in a stem as "the cars crumple over a contact time of 0.10 s and experience an average force of 8000 N." Set the impulse equal to the change in momentum, 8000 × 0.10 = 800 N·s, then equate 800 to (m₁ + m₂)(v_f) − (m₁v₁ᵢ + m₂v₂ᵢ). This converts a force-and-time problem into a momentum problem and is the kind of bridge the AP exam uses to test transfer of skills. Second, the work-energy form is the natural way to handle "the spring is compressed by 0.05 m during the collision and has a spring constant of 2000 N/m." The elastic potential energy stored is ½(2000)(0.05)² = 2.5 J, and that energy came from the kinetic energy of the moving body, so ½m v² = 2.5 J gives v. This conversion comes up in rotational and simple-harmonic-motion problems as well. Candidates who recognise the bridge will see these cross-unit items as a chance to bank points, not as a new topic to learn.

A brief comparison table helps to anchor the unit's three equations to the situations that license them.

Equation	Valid when	Commonly paired with
Momentum conservation (m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f)	Always, in a closed two-body system	Either energy conservation (elastic) or a shared-final-velocity condition (perfectly inelastic)
Kinetic-energy conservation (½m₁v₁ᵢ² + ½m₂v₂ᵢ² = ½m₁v₁f² + ½m₂v₂f²)	Elastic collisions only	Momentum conservation, to close a two-unknown system
Coefficient of restitution (e = (v₂f − v₁f) / (v₁ᵢ − v₂ᵢ))	When the stem gives e, or when e is 0 or 1	Momentum conservation, to give a third equation in a three-unknown system

The table is not a memorisation device. It is a decision tool. Before any algebra, scan the stem, find the row of the table that matches the regime, and write the corresponding equations. Three seconds of table-reading saves thirty seconds of trial-and-error algebra, and on a 90-minute multiple-choice section that compounds.

Build a four-week plan that closes the unit

A focused four-week plan is the highest-leverage use of pre-exam time. The plan has three blocks: a diagnostic week, a skills week, and a presentation week. In the diagnostic week, take a single practice section under timed conditions, score it, and isolate every collision item, even those that masquerade as work-energy or impulse items. Sort them into the three regimes. For most students, the sort will reveal a regime that they systematically mis-classify. The diagnostic is not the score. The diagnostic is the mis-classification list.

In the skills week, work through the released FRQs from the College Board. For each item, write a regime sentence, a diagram, and a numerical answer before looking at the rubric. Then score yourself against the rubric. The two numbers — what you expected and what the rubric awarded — are the data that drives the rest of the week. Candidates who find that they are losing conceptual points should reread the unit notes; candidates who are losing algebra points should redo the algebra with a fresh sign convention; candidates who are losing presentation points should rebuild their answer template. Three practice items a day for five days is enough. The volume is not the point. The honest self-scoring is the point.

In the presentation week, rebuild your answer template from scratch. The template should contain, in this order: a one-line regime statement, a labelled diagram with arrows and a positive-direction note, a list of the two equations you intend to use, the algebra, the numerical answer with units, and a one-line check. Apply this template to one new item a day, and time yourself. The presentation habits that earn the rubric points are not free; they cost about 90 seconds per item, and that cost has to be budgeted. Candidates who try to save the 90 seconds almost always lose the conceptual point that the regime sentence was designed to capture. The trade is not worth it.

Two final tactical notes. First, in the final week, do not introduce new topics. The marginal return on a new unit at that point is negative; the marginal return on practising the answer template is positive. Second, in the final 24 hours, do not work collision problems at all. Reread the released rubrics. The point of the final day is to consolidate the language of the rubric, not to push the algebra one more step. Most candidates reading this will not have a problem with the algebra by the final day. They will have a problem with the language. The day is best spent on the language.

Collisions are the AP Physics 1 unit where a small set of habits decides the score. The regime statement, the labelled diagram, the matched-subscript equations, the units on every answer, the one-line check — none of these are deep physics. They are the artefacts the rubric rewards. Practising the artefacts is the highest-leverage use of the final four weeks. TestPrep Europe's diagnostic assessment, paired with a focused review of released FRQ rubrics, is a natural starting point for candidates who want to convert momentum-and-energy fluency into a 5.

Frequently asked questions

How do I tell on the AP Physics 1 exam whether a collision is elastic or inelastic?

Look at the picture and the wording. If the two bodies move as one after impact, or the stem uses 'sticks to,' 'embeds,' or 'locks together,' the collision is perfectly inelastic: conserve momentum and set the final velocities equal. If the bodies move separately and the stem says 'elastic' or shows them rebounding, both momentum and kinetic energy are conserved. If the stem gives a percentage energy loss or a coefficient of restitution, momentum is conserved and the energy condition is given as data.

Can kinetic energy be conserved in an inelastic collision?

No. By definition an inelastic collision is one in which kinetic energy is not conserved. The energy is converted to sound, heat, deformation, or other internal modes. The momentum of the two-body system is still conserved in a closed inelastic collision, which is why you can still solve for the final velocities when the energy is treated as a known loss.

What is the coefficient of restitution, and does it appear on the AP Physics 1 exam?

The coefficient of restitution e is the ratio of the relative speed of separation to the relative speed of approach. It ranges from 0 (perfectly inelastic, the bodies share a final velocity) to 1 (perfectly elastic, no energy lost). The AP Physics 1 exam occasionally references e in multiple-choice items, especially when masses and a third unknown need to be solved for. Treat e as a second equation alongside momentum conservation.

How is the AP Physics 1 exam scored, and how much does the collisions unit contribute?

The exam is scored on a 1–5 scale. The 50 multiple-choice items and five free-response items together produce a composite raw score, with the raw score then mapped to the 1–5 scale. Collisions appear in both sections and in cross-unit items that disguise momentum-and-energy work as impulse or work-energy problems, so the unit is one of the highest-leverage topics in the entire exam. A strong collisions score is, for most candidates, the single largest contributor to a 4 or 5.

What is the fastest way to solve a 1D elastic collision with one body initially at rest?

Use the two shortcut formulas: v₁f = (m₁ − m₂) / (m₁ + m₂) × v₁ᵢ and v₂f = 2m₁ / (m₁ + m₂) × v₁ᵢ. They come from solving the momentum and energy equations simultaneously. Three useful limit cases to memorise: equal masses swap velocities, a heavy mass barely moves, and a light mass rebounds back. Knowing the limits lets you answer qualitative multiple-choice items in under 30 seconds.

How to tell an elastic collision from an inelastic one on AP Physics 1: the 4-step check