Elastic vs Inelastic Collision: How to Tell

Q: What is the difference between an elastic and an inelastic collision?

In an elastic collision, the total kinetic energy of the system is the same before and after. In an inelastic collision, the total kinetic energy decreases, with the difference going to thermal energy, sound and permanent deformation of the objects. Both types conserve total momentum and both conserve total energy, so kinetic energy is the only quantity that separates them. To classify a collision, add up one half m v squared for every object before the impact, add it up again after, and compare the two totals.

Q: How do you know if a collision is elastic or inelastic?

Compute the total kinetic energy before and after and compare. If the two totals match, the collision is elastic; if the total afterwards is smaller, it is inelastic. A faster check in one dimension is to compare the relative speed at which the objects approached with the relative speed at which they separated: those are equal for an elastic collision and smaller after an inelastic one. Do not classify by whether the objects bounced apart, because objects can separate and still have lost kinetic energy.

Q: Is a perfectly inelastic collision the same as an inelastic collision?

A perfectly inelastic collision is a special case of an inelastic collision, not a separate third category. The AP Physics 1 CED defines it at essential knowledge 4.4.A.5 by what the objects do: they stick together and move with the same velocity after the collision. Because they end up sharing one velocity, this outcome loses more kinetic energy than any other outcome available to the same masses at the same approach speeds. It does not lose all of the kinetic energy unless the total momentum was zero to begin with, because the combined object is still moving.

Q: Can a collision be partly elastic?

Yes, and almost every real collision is. Elastic and inelastic mark the two ends of a continuous range rather than a pair of boxes. At one end the kinetic energy total is unchanged; at the other the objects move off together and the loss is the largest the momentum allows. Everything in between loses some kinetic energy and still separates. The usual way to place a collision on that range is the ratio of separation speed to approach speed, which is one for an elastic collision and zero for a perfectly inelastic one, though that ratio is not named in the AP Physics 1 CED.

Q: Are any real collisions perfectly elastic?

Collisions between hard steel spheres, between magnetically repelling carts that never touch, and between gas molecules come close enough to be modeled as elastic. Everyday impacts do not: some kinetic energy always goes into heating the materials, into sound, and into permanent deformation. Treat elastic as a model you verify by comparing kinetic energy totals rather than a default you assume. If a problem does not say a collision is elastic and the numbers do not come out equal, it is not elastic.

Q: Is momentum conserved in an inelastic collision?

Yes. Momentum conservation does not depend on the collision type at all; it depends on the system being isolated, meaning the net external force on it is zero or negligible during the brief impact. This is why the collision type can never be worked out from a momentum calculation, and why writing that momentum is not conserved because a collision is inelastic is a common way to lose a justification mark. The AP Physics 1 CED states the principle at essential knowledge 4.3.B.1: momentum is conserved in all interactions.

Q: Why does an elastic collision give you two equations?

Because elastic is defined by a second conservation statement holding. Every collision gives you conservation of momentum. Being told a collision is elastic tells you the system's total kinetic energy is also unchanged, which is a second independent equation relating the same unknowns. Two objects colliding in one dimension have two unknown final velocities, so two equations close the problem. An inelastic collision gives you momentum only, and you need the problem to supply a second fact, such as one of the final velocities or the statement that the objects stick together.

An elastic collision keeps the system's total kinetic energy the same. An inelastic collision loses some of it to thermal energy, sound and permanent deformation. Both conserve momentum, so kinetic energy is the only quantity that separates them: total it before and after, and compare.

AP Physics: Unit 4 (topics 4.4 Elastic and Inelastic Collisions). AP Physics 1 Unit 4, Linear Momentum, weighted at 10 to 15 percent of the multiple-choice section and estimated at about 10 to 15 class periods. Topic 4.4 has one learning objective, 4.4.A, describe whether an interaction between objects is elastic or inelastic, supported by essential knowledge 4.4.A.1 through 4.4.A.5. The CED lists no boundary statement under Topic 4.4; the constraint on how the topic is examined comes from the Topic 4.3 boundary statement, which excludes questions requiring the solution of simultaneous equations while still expecting students to set them up and reason about them. Suggested skills for Topic 4.4 are 1.B, 2.A, 2.C, 3.A and 3.B.

The one test that classifies a collision

Add up the kinetic energy of every object before the impact, add it up again after, and compare the two totals. That single comparison is the whole classification. Equal totals mean the collision is elastic. A smaller total afterwards means it is inelastic. There is no third measurement to take.

The AP Physics 1 CED writes the two definitions in those terms and no other. Essential knowledge 4.4.A.1 says an elastic collision between objects is one in which the initial kinetic energy of the system is equal to the final kinetic energy of the system. EK 4.4.A.3 says an inelastic collision between objects is one in which the total kinetic energy of the system decreases. Neither definition mentions bouncing, sticking, deformation or how loud the impact was. Those are symptoms. The energy totals are the definition.

Momentum plays no part in the classification, because it behaves identically on both sides of it. That is not a caveat this page needs to argue: the what is conserved in a collision guide makes the case in full, including why the internal impulses are forced to cancel while the internal works are not. Take the result as given here. Momentum is conserved in every collision type, so it can never tell you which type you are looking at.

The practical routine is three lines.

Compute $K_i = \sum \frac{1}{2}m_i v_i^2$ using the speeds before the impact. Speeds, not velocities: kinetic energy is a scalar and the sign of $v$ disappears when you square it.
Compute $K_f$ the same way from the speeds afterwards.
Compare. $K_f = K_i$ is elastic. $K_f < K_i$ is inelastic. $K_f > K_i$ is not a collision at all; something released stored energy, and EK 4.1.A.3.iii calls that an explosion.

The reason people find the classification slippery is that a problem usually hands you the label instead of the numbers, and the label then does work you did not watch it do. Being told "the collision is elastic" is being handed a second equation. Being told "the carts stick together" is being handed a different second fact. The rest of this page is about which fact you get and what it buys.

The classification grid

Read this table down a column when a problem hands you a label, and across a row when you are trying to work out which label applies.

Question you are asking	Elastic	Inelastic
Total kinetic energy afterwards	Equal to the total before	Less than the total before
Symbolic test	$K_f = K_i$	$K_f < K_i$
Total momentum afterwards	Same as before	Same as before
Total energy afterwards	Same as before	Same as before
Where the missing kinetic energy went	Nowhere, none is missing	Thermal energy, sound, permanent deformation
Objects after the impact	Separate, at new speeds	Separate, or stuck together
Relative speed of separation	Equal to the relative speed of approach	Smaller than the relative speed of approach
Conservation equations you may write	Two	One
Words a problem uses	Elastic, perfectly elastic, no energy is lost	Sticks, embeds, couples, locks, crumples, some energy is lost
Physical examples	Hard steel spheres, magnetically repelling carts, gas molecules	Almost every everyday impact
CED essential knowledge	4.4.A.1, 4.4.A.2	4.4.A.3, 4.4.A.4, 4.4.A.5

Three rows in that grid do more work than the rest.

The momentum row is deliberately identical. It is there so you stop looking at it. A question that only asks for the total momentum after an impact cannot be answered wrongly by misclassifying the collision, which is exactly why students go a long way before the classification ever bites them.

The row about the objects afterwards does not classify anything on its own. Sticking together guarantees inelastic. Separating guarantees nothing: two objects can bounce apart and still have lost kinetic energy. A dropped tennis ball bounces and does not come back to the height it was dropped from, and the shortfall is the kinetic energy that did not survive.

The relative-speed row is the fastest hand check there is, and it gets its own section below because it is the one line of the grid that is derived rather than defined.

Approach speed against separation speed

In one dimension, an elastic collision has a property that follows from combining the two conservation statements: the objects separate at the same relative speed at which they approached.

v_1 - v_2 = v_2' - v_1'

The left side is the relative speed of approach, the right the relative speed of separation, with the primes marking the state after impact. Nothing new has been assumed. Write momentum conservation, write kinetic energy conservation, and eliminate the masses between them, and this is what is left.

It is worth having because it turns a squaring-and-summing check into a subtraction. Two carts approach each other at a combined $6.0$ m/s and separate at a combined $6.0$ m/s: elastic. They separate at $4.0$ m/s: inelastic, and you did not touch a square. They separate at $0$ m/s: perfectly inelastic, since separating at zero relative speed is what moving together means.

The ratio of those two numbers has a name, the coefficient of restitution, usually written $e$ :

e = \frac{v_2' - v_1'}{v_1 - v_2}

That symbol is not in the AP Physics 1 CED and no exam question will ask for it by name. It is worth knowing anyway, because it makes the point that elastic and inelastic are the two ends of a continuous range rather than a pair of boxes. An elastic collision is $e = 1$ . A perfectly inelastic one is $e = 0$ . Every real collision sits somewhere between, and the closer $e$ is to zero, the more kinetic energy the impact removed.

One caution about direction. The relation above is written for motion along a line with a sign convention already chosen, and the signs of the velocities matter. Take rightward as positive, put a cart moving at $+5.0$ m/s behind one moving at $+2.0$ m/s, and the approach speed is $5.0 - 2.0 = 3.0$ m/s, not $7.0$ . Adding the magnitudes is only correct when the two objects move toward each other.

Why the label changes how many equations you get

The classification is not a piece of vocabulary. It decides how much algebra the problem gives you, and that is where it shows up on an exam.

A two-object collision in one dimension has two unknown final velocities. Momentum conservation is one equation. One equation and two unknowns does not close.

Told the collision is elastic, you get kinetic energy conservation as a second equation, and the system closes.
Told the objects stick together, you get $v_1' = v_2'$ as the second fact, and the system closes immediately, because two unknowns collapse into one.
Told nothing, you need the problem to hand you one of the final velocities, and then momentum alone finishes the job.

There is a boundary statement under Topic 4.3 that changes how the first of those plays out on the exam, and it is worth quoting in full because summarising it loses the part that matters:

"AP Physics 1 includes a quantitative and qualitative treatment of conservation of momentum in one dimension and a semiquantitative treatment of conservation of momentum in two dimensions. Exam questions involving solution of simultaneous equations are not included in AP Physics 1, but the AP Physics 1 Exam may include questions that assess whether students can set up the equations properly and reason about how changing a given mass, speed, or angle would affect other quantities. AP Physics 2 includes a full treatment of conservation of momentum in two dimensions for problems that include one unknown final velocity."

So the two-equation elastic case is on the syllabus as a setup and a reasoning exercise, not as a grind. You may be asked to write both equations and say what each one expresses. You will not be asked to solve them simultaneously. A question that gives you an elastic collision and expects a number will always have handed you something else too, usually one of the two final velocities, or a mass ratio that makes the arithmetic collapse.

The perfectly inelastic case is the one that stays fully quantitative, because it never needs two equations at all.

Perfectly inelastic is a subset, not a third category

The phrase "elastic, inelastic or perfectly inelastic" reads like three boxes, and it is not. Every collision is either elastic or inelastic. Perfectly inelastic collisions are a subset of the inelastic ones, picked out by what the objects do rather than by how much energy went missing.

EK 4.4.A.5 defines it that way: in a perfectly inelastic collision, the objects stick together and move with the same velocity after the collision. That is a statement about the final velocities, not about the energy. The energy consequence follows, it is not the definition.

The consequence is worth stating because it fixes the range the classification lives in. Hold the masses and the incoming velocities fixed and ask what final states momentum conservation permits. Every one of them has the same total momentum, and among them:

The largest possible final kinetic energy is the elastic one, equal to the initial total. Nothing can exceed it in a collision, because exceeding it would require energy to appear.
The smallest possible final kinetic energy is the perfectly inelastic one. For a fixed total momentum, the kinetic energy of a set of objects is least when they all move at the same velocity, because any other split leaves them moving relative to each other and that relative motion is extra kinetic energy.
Every real collision with those masses and those incoming velocities lands somewhere between those two numbers.

That is why "perfectly" is the right word. It marks the extreme of the inelastic direction, the same way elastic marks the other end of the range.

A second point that trips people: a perfectly inelastic collision does not lose all of the kinetic energy. The combined object is still moving, so it still has kinetic energy. The only way to finish at zero is for the total momentum to have been zero to begin with, which happens in a head-on collision between equal and opposite momenta. Worked example one below puts real numbers on the floor of that range.

When the classification costs a mark

These are the specific errors that follow from getting the label wrong, rather than general advice about being careful.

Writing kinetic energy conservation in an inelastic problem. This is the expensive one, because it produces a clean-looking number that is wrong. If the objects stick together, or the problem says energy was lost, $\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$ before does not equal the total after and any equation built on it is void. Use momentum.
Writing that momentum is not conserved because the collision is inelastic. Momentum conservation depends on the system being isolated, not on the collision type. This costs a justification mark rather than a numerical one, and it is the most common wrong sentence in a free-response answer on this topic.
Calling a collision elastic because the objects bounced apart. Separating is necessary for elastic, not sufficient. Check the energy totals.
Calling a collision inelastic because one object slowed down. EK 4.4.A.2 says outright that in an elastic collision the final kinetic energies of each of the objects within the system may be different from their initial kinetic energies. A cue ball that stops dead after hitting an identical ball has lost all of its own kinetic energy, and the collision can still be elastic, because the other ball gained exactly that much.
Writing that the lost kinetic energy was destroyed. Say it was transformed into thermal energy, sound and permanent deformation. A boundary statement under Topic 3.4 sets that expectation: AP Physics 1 expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces.
Assuming perfectly inelastic means zero final kinetic energy. Only true when the total momentum was zero.
Using signed velocities inside the kinetic energy sum and then losing the square. $\frac{1}{2}(2.0)(-3.0)^2$ is $9.0$ J, not $-9.0$ J. Kinetic energy is never negative.
Comparing the kinetic energy of one object instead of the system total. The definitions in 4.4.A.1 and 4.4.A.3 are about the system.

What the two share, and why that hides the difference

Elastic and inelastic collisions agree on more than they disagree on, and the agreement is the reason the distinction goes unnoticed for so long.

They agree that total momentum is unchanged. They agree that total energy is unchanged. They agree that the individual objects' kinetic energies can change freely. They agree on the velocity of the system's centre of mass, which EK 4.3.A.1.ii says is constant in the absence of a net external force and which therefore does not notice the collision at all. Exactly one quantity, the system's total kinetic energy, behaves differently, and it is the last thing most problems ask about.

Two consequences follow.

A momentum-only question is blind to the classification. "Two carts collide on a frictionless track. What is the total momentum of the system afterwards?" has the same answer whatever happened during the impact. So does "which way is the combined system moving" and "what is the velocity of the centre of mass". A student can answer a page of those correctly with no idea what elastic means.

The distinction appears the moment a question asks for a speed with two unknowns, or asks for an energy. That is the point at which the two labels give different numbers, and it arrives without warning.

There is one more coincidence worth naming. When a very heavy object strikes a very light one, or a very light one strikes a very heavy one, the fraction of kinetic energy that survives an inelastic impact is set by the mass ratio alone, and at the extremes it approaches the elastic answer or falls to almost nothing. The guide on what is conserved in a collision derives that fraction. Between the extremes, the two labels part company by a wide margin, and worked example two below shows a case where they differ by a factor of three.

What the CED asks, and how the exam frames it

This is Topic 4.4 of AP Physics 1 Unit 4, Linear Momentum. Unit 4 carries 10 to 15 percent of the multiple-choice section and an estimated 10 to 15 class periods.

Topic 4.4 has a single learning objective, 4.4.A: describe whether an interaction between objects is elastic or inelastic. Note the verb. It asks you to classify, not to solve. Five pieces of essential knowledge sit under it:

4.4.A.1 defines an elastic collision as one in which the initial kinetic energy of the system is equal to the final kinetic energy of the system.
4.4.A.2 notes that in an elastic collision the final kinetic energies of each of the objects within the system may be different from their initial kinetic energies.
4.4.A.3 defines an inelastic collision as one in which the total kinetic energy of the system decreases.
4.4.A.4 says that in an inelastic collision some of the initial kinetic energy is not restored to kinetic energy but is transformed by nonconservative forces into other forms of energy.
4.4.A.5 defines a perfectly inelastic collision as one in which the objects stick together and move with the same velocity after the collision.

The CED lists no boundary statement under Topic 4.4. The constraint that shapes the questions comes from Topic 4.3, quoted in full above. The suggested skills for Topic 4.4 are 1.B, 2.A, 2.C, 3.A and 3.B, and the presence of 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim, is a fair description of what this topic tests: you apply a definition to a set of numbers and state which label fits.

Nothing on the AP Physics 1 equation sheet solves an elastic collision. For this whole area the sheet prints $\vec{p} = m\vec{v}$ , $\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t} = m\frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$ , $\vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta \vec{p}$ , $\vec{v}_{\text{cm}} = \frac{\sum \vec{p}_i}{\sum m_i}$ and $K = \frac{1}{2}mv^2$ . There is no elastic-collision formula and no restitution coefficient. The classification has to come from you comparing two totals.

For the solving routines rather than the classification, the conservation of momentum guide has the before-and-after method, and the momentum and collision calculator will run the energy audit for a pair of masses and velocities. Topic 4.4 carries the CED framing, and the momentum and collisions practice set has problems to classify.

One before-state, three endings, three labels

A $3.0$ kg cart moving at $4.0$ m/s to the right strikes a $1.0$ kg cart at rest on a frictionless track. Classify each of these three outcomes and find the kinetic energy lost in each. Take rightward as positive throughout. (a) The carts end up moving at $2.0$ m/s and $6.0$ m/s. (b) The carts stick together. (c) The carts end up moving at $2.5$ m/s and $4.5$ m/s.

Set the before-state once, because all three endings share it. Total momentum $p_i = (3.0)(4.0) + (1.0)(0) = 12\ \text{kg}\cdot\text{m/s}$ . Total kinetic energy $K_i = \frac{1}{2}(3.0)(4.0)^2 + 0 = 24\ \text{J}$ .
Ending (a). Momentum check first: $(3.0)(2.0) + (1.0)(6.0) = 6.0 + 6.0 = 12\ \text{kg}\cdot\text{m/s}$ , unchanged, so the outcome is physically allowed. Now the classification test: $K_f = \frac{1}{2}(3.0)(2.0)^2 + \frac{1}{2}(1.0)(6.0)^2 = 6.0 + 18 = 24\ \text{J}$ . Equal to $K_i$ to every digit, so this collision is elastic. Cross-check with the relative-speed rule: approach $4.0 - 0 = 4.0$ m/s, separation $6.0 - 2.0 = 4.0$ m/s. Equal, as an elastic collision requires.
Ending (b). Sticking together means one common velocity, so momentum alone finishes it: $v_f = \frac{12}{3.0 + 1.0} = 3.0\ \text{m/s}$ . Then $K_f = \frac{1}{2}(4.0)(3.0)^2 = 18\ \text{J}$ . Less than $24$ J, so this is inelastic, and because the objects moved off together it is the perfectly inelastic case. Kinetic energy lost: $24 - 18 = 6.0\ \text{J}$ , which is $25$ percent of the original. Separation speed is $0$ , as it must be for objects moving together.
Ending (c). Momentum check: $(3.0)(2.5) + (1.0)(4.5) = 7.5 + 4.5 = 12\ \text{kg}\cdot\text{m/s}$ , unchanged again. Energy test: $K_f = \frac{1}{2}(3.0)(2.5)^2 + \frac{1}{2}(1.0)(4.5)^2 = 9.375 + 10.125 = 19.5\ \text{J}$ . Less than $24$ J, so inelastic, but not perfectly so, because the carts separated. Lost: $24 - 19.5 = 4.5\ \text{J}$ , which is $18.75$ percent. Separation speed $4.5 - 2.5 = 2.0$ m/s against an approach of $4.0$ m/s, so $e = 0.50$ , halfway between the two extremes.
Line the three up. Momentum after impact: $12$ , $12$ , $12\ \text{kg}\cdot\text{m/s}$ , identical, which is why the momentum row of the grid classifies nothing. Kinetic energy after impact: $24$ , $18$ , $19.5\ \text{J}$ . The elastic ending sits at the ceiling of the range and the perfectly inelastic one at the floor, with the partially inelastic ending in between. No outcome with these masses and this approach speed can finish above $24$ J or below $18$ J.

(a) Elastic: $K_f = 24$ J, nothing lost. (b) Perfectly inelastic: $v_f = 3.0$ m/s, $K_f = 18$ J, $6.0$ J lost. (c) Inelastic: $K_f = 19.5$ J, $4.5$ J lost. All three conserve momentum at $12\ \text{kg}\cdot\text{m/s}$ , and $18$ J to $24$ J is the full range of final kinetic energies momentum allows.

The label decides how much algebra you get

A $0.50$ kg ball moving at $8.0$ m/s strikes a stationary $1.5$ kg ball head on. Set up the problem twice: once told the collision is elastic, once told the balls stick together. Show which version an AP Physics 1 exam can ask for a number. Rightward is positive.

The shared before-state: $p_i = (0.50)(8.0) = 4.0\ \text{kg}\cdot\text{m/s}$ and $K_i = \frac{1}{2}(0.50)(8.0)^2 = 16\ \text{J}$ .
Elastic version, the setup. Momentum: $0.50v_1' + 1.5v_2' = 4.0$ . Kinetic energy: $\frac{1}{2}(0.50)v_1'^2 + \frac{1}{2}(1.5)v_2'^2 = 16$ , or $0.25v_1'^2 + 0.75v_2'^2 = 16$ . Two equations, two unknowns. The Topic 4.3 boundary statement puts solving that pair simultaneously outside AP Physics 1, so an exam question here would ask you to write both equations and say what each expresses, or to reason about which ball ends up faster and why.
Elastic version, the answer, for checking. The pair is satisfied by $v_1' = -4.0$ m/s and $v_2' = +4.0$ m/s. Verify both equations rather than trusting the algebra. Momentum: $(0.50)(-4.0) + (1.5)(4.0) = -2.0 + 6.0 = 4.0\ \text{kg}\cdot\text{m/s}$ , matches. Kinetic energy: $\frac{1}{2}(0.50)(4.0)^2 + \frac{1}{2}(1.5)(4.0)^2 = 4.0 + 12 = 16\ \text{J}$ , matches. Relative speeds: approach $8.0$ m/s, separation $4.0 - (-4.0) = 8.0$ m/s, equal. The light ball rebounds, which is what happens whenever a lighter object strikes a heavier one elastically.
Sticking version. Now the second fact is $v_1' = v_2' = v_f$ , and momentum alone closes it in one line: $v_f = \frac{4.0}{0.50 + 1.5} = 2.0\ \text{m/s}$ . No second equation was needed and no simultaneous solving happened, which is why this version is fully fair game for a numerical exam question.
Now audit the energy for the sticking version. $K_f = \frac{1}{2}(2.0)(2.0)^2 = 4.0\ \text{J}$ , so $16 - 4.0 = 12\ \text{J}$ of kinetic energy went to thermal energy, sound and deformation. That is $75$ percent of the original, against $0$ percent for the elastic ending. Same masses, same approach speed, same conserved momentum of $4.0\ \text{kg}\cdot\text{m/s}$ , and the final kinetic energies differ by a factor of four.

Elastic: $v_1' = -4.0$ m/s, $v_2' = +4.0$ m/s, $K_f = 16$ J, nothing lost, and the two-equation setup is beyond what AP Physics 1 asks you to solve. Perfectly inelastic: $v_f = 2.0$ m/s from momentum alone, $K_f = 4.0$ J, $12$ J lost.

Classifying a bounce from two heights

A $0.058$ kg tennis ball is dropped from rest at $1.25$ m onto a hard floor and rebounds to $0.80$ m. Classify the collision between ball and floor, find the kinetic energy lost, and find the ratio of separation speed to approach speed. Use $g = 9.8\ \text{m/s}^2$ and take down as positive.

Find the speed just before impact from the drop. Starting from rest, $v^2 = 2g h$ , so $v_{\text{down}} = \sqrt{2(9.8)(1.25)} = \sqrt{24.5} = 4.9\ \text{m/s}$ to two significant figures.
Find the speed just after impact from the rebound height, running the same relation backwards: $v_{\text{up}} = \sqrt{2(9.8)(0.80)} = \sqrt{15.68} = 4.0\ \text{m/s}$ to two significant figures. Both speeds are worth carrying unrounded, $4.9497$ and $3.9598$ m/s, because the next step squares them.
Apply the test. $K_i = \frac{1}{2}(0.058)(24.5) = 0.711\ \text{J}$ and $K_f = \frac{1}{2}(0.058)(15.68) = 0.455\ \text{J}$ . Because the $\frac{1}{2}mv^2$ and $mgh$ routes agree, you can shortcut both: $K_i = mgh_i = (0.058)(9.8)(1.25) = 0.711\ \text{J}$ and $K_f = mgh_f = (0.058)(9.8)(0.80) = 0.455\ \text{J}$ . The total went down, so the collision is inelastic.
Quantify it. Kinetic energy lost $= 0.711 - 0.455 = 0.256\ \text{J}$ . As a fraction, $\frac{K_f}{K_i} = \frac{h_f}{h_i} = \frac{0.80}{1.25} = 0.64$ , so $64$ percent survived and $36$ percent became thermal energy in the ball and the floor, plus the sound you heard. Notice that the mass cancelled out of that fraction entirely: the ratio of rebound height to drop height is the energy verdict, whatever the ball weighs.
The speed ratio is the square root of the energy ratio, because kinetic energy goes as $v^2$ : $\frac{v_{\text{up}}}{v_{\text{down}}} = \sqrt{0.64} = 0.80$ . The ball separates from the floor at $80$ percent of the speed at which it arrived, so $e = 0.80$ , closer to the elastic end than to the sticking end but not at it.
One thing this example does that the cart examples do not: it classifies the collision without ever resolving what the system is. Momentum is awkward here, because the ball alone is not isolated and the floor is bolted to the Earth. The energy test needed none of that. Two heights were enough.

Inelastic. $K_i = 0.711$ J, $K_f = 0.455$ J, so $0.256$ J of kinetic energy was lost, $36$ percent of the original. The separation speed is $0.80$ of the approach speed, and the surviving energy fraction equals the height ratio, $0.64$ .

Frequently asked questions

What is the difference between an elastic and an inelastic collision?

In an elastic collision, the total kinetic energy of the system is the same before and after. In an inelastic collision, the total kinetic energy decreases, with the difference going to thermal energy, sound and permanent deformation of the objects. Both types conserve total momentum and both conserve total energy, so kinetic energy is the only quantity that separates them. To classify a collision, add up one half m v squared for every object before the impact, add it up again after, and compare the two totals.

How do you know if a collision is elastic or inelastic?

Compute the total kinetic energy before and after and compare. If the two totals match, the collision is elastic; if the total afterwards is smaller, it is inelastic. A faster check in one dimension is to compare the relative speed at which the objects approached with the relative speed at which they separated: those are equal for an elastic collision and smaller after an inelastic one. Do not classify by whether the objects bounced apart, because objects can separate and still have lost kinetic energy.

Is a perfectly inelastic collision the same as an inelastic collision?

A perfectly inelastic collision is a special case of an inelastic collision, not a separate third category. The AP Physics 1 CED defines it at essential knowledge 4.4.A.5 by what the objects do: they stick together and move with the same velocity after the collision. Because they end up sharing one velocity, this outcome loses more kinetic energy than any other outcome available to the same masses at the same approach speeds. It does not lose all of the kinetic energy unless the total momentum was zero to begin with, because the combined object is still moving.

Can a collision be partly elastic?

Yes, and almost every real collision is. Elastic and inelastic mark the two ends of a continuous range rather than a pair of boxes. At one end the kinetic energy total is unchanged; at the other the objects move off together and the loss is the largest the momentum allows. Everything in between loses some kinetic energy and still separates. The usual way to place a collision on that range is the ratio of separation speed to approach speed, which is one for an elastic collision and zero for a perfectly inelastic one, though that ratio is not named in the AP Physics 1 CED.

Are any real collisions perfectly elastic?

Collisions between hard steel spheres, between magnetically repelling carts that never touch, and between gas molecules come close enough to be modeled as elastic. Everyday impacts do not: some kinetic energy always goes into heating the materials, into sound, and into permanent deformation. Treat elastic as a model you verify by comparing kinetic energy totals rather than a default you assume. If a problem does not say a collision is elastic and the numbers do not come out equal, it is not elastic.

Is momentum conserved in an inelastic collision?

Yes. Momentum conservation does not depend on the collision type at all; it depends on the system being isolated, meaning the net external force on it is zero or negligible during the brief impact. This is why the collision type can never be worked out from a momentum calculation, and why writing that momentum is not conserved because a collision is inelastic is a common way to lose a justification mark. The AP Physics 1 CED states the principle at essential knowledge 4.3.B.1: momentum is conserved in all interactions.

Why does an elastic collision give you two equations?

Because elastic is defined by a second conservation statement holding. Every collision gives you conservation of momentum. Being told a collision is elastic tells you the system's total kinetic energy is also unchanged, which is a second independent equation relating the same unknowns. Two objects colliding in one dimension have two unknown final velocities, so two equations close the problem. An inelastic collision gives you momentum only, and you need the problem to supply a second fact, such as one of the final velocities or the statement that the objects stick together.