AP Physics C: Mechanics · Topic 4.4

Topic 4.4: Elastic and Inelastic Collisions

Unit 4: Linear Momentum10-20% of the multiple-choice section

An elastic collision conserves the system's kinetic energy and an inelastic one does not; both conserve momentum. All five essential-knowledge statements here match AP Physics 1 word for word. The difference is Topic 4.3's boundary statement, which lets this course solve for two unknown velocities.

AP Physics: Unit 4 (topics 4.4 Elastic and Inelastic Collisions). Topic 4.4 of the current AP Physics C: Mechanics course and exam description, inside Unit 4, weighted 10 to 20% of the multiple-choice section at about 11 to 15 class periods. One learning objective, 4.4.A, and five essential-knowledge statements, the smallest count in Unit 4: 4.4.A.1 defines an elastic collision by equal initial and final system kinetic energy; 4.4.A.2 notes the individual objects' kinetic energies may still change; 4.4.A.3 defines an inelastic collision by a decrease in the system total; 4.4.A.4 attributes the decrease to transformation by nonconservative forces into other forms of energy; 4.4.A.5 defines a perfectly inelastic collision. Suggested skills 1.A, 2.B, 2.D and 3.C. No boundary statement, and its AP Physics 1 counterpart prints none either. Verified statement by statement against both CEDs: all five are word for word identical to AP Physics 1's Topic 4.4, and the page says so plainly rather than manufacturing a difference. The only framework differences are the suggested skills, where AP Physics 1 lists 1.B, 2.A, 2.C, 3.A and 3.B. The real differentiator is INHERITED from Topic 4.3's boundary statement: AP Physics C: Mechanics expects quantitative analysis of collisions in one or two dimensions, while the AP Physics 1 boundary statement excludes exam questions involving solution of simultaneous equations by name and calls its two-dimensional treatment semiquantitative, and a general elastic collision has two unknown final velocities requiring both conservation equations solved together. Sheet detail verified against the printed Table of Information: NO equation specific to Topic 4.4 is printed. Not on the sheet and not in the CED: the general elastic-collision result, the relative-velocity reversal, the perfectly inelastic velocity, the surviving kinetic-energy fraction, and the coefficient of restitution, which appears nowhere in the document. Printed and used here: K = one half m v squared, p = m v, v_cm, the dot product A dot B = AB cos theta in the Vectors table, and the Pythagorean identity in the Identities table. Neither Unit 4 sample multiple-choice question aligns here; sample free-response question 3, the Experimental Design and Analysis question, lists 4.4.A alongside 2.1.B, 3.1.A, 3.4.B, 3.4.C and 4.2.B with skills 3.A, 2.D, 1.B, 2.B and 3.C, and turns a collision experiment into a linearisation. The unit's fourth sample instructional activity is a 4.4 Ranking Task about an arrow shot at a suspended pumpkin.

What Topic 4.4 requires, and the honest comparison

Topic 4.4 carries one learning objective and five essential-knowledge statements, the smallest count in Unit 4.

Objective	What it asks
4.4.A	Describe whether an interaction between objects is elastic or inelastic

Statement	What it says
4.4.A.1	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
4.4.A.2	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	An inelastic collision between objects is one in which the total kinetic energy of the system decreases
4.4.A.4	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	In a perfectly inelastic collision, the objects stick together and move with the same velocity after the collision

Suggested skills: 1.A, 2.B, 2.D and 3.C. Topic 4.4 prints no boundary statement, and neither does its AP Physics 1 counterpart. Unit 4's only boundary statement sits under Topic 4.3.

Now the honest part, because the brief for this site says not to manufacture a difference. All five of those statements are word for word identical to the five under AP Physics 1's Topic 4.4. No extra statement, no changed operator, no calculus anywhere in the required content of this topic. If you have learned the definitions of elastic, inelastic and perfectly inelastic in the algebra-based course, you have learned the whole of 4.4 as written.

The only visible framework differences are the suggested skills. AP Physics C: Mechanics lists 1.A, 2.B, 2.D and 3.C; AP Physics 1 lists 1.B, 2.A, 2.C, 3.A and 3.B. Read that as a shift from experimental design and graph plotting toward calculation and functional dependence, which is a hint about question style rather than about content.

So where is the real difference? It is inherited, and it comes from the topic next door. Topic 4.3's boundary statement in this course expects quantitative analysis in one or two dimensions. In AP Physics 1 the equivalent statement excludes exam questions involving solution of simultaneous equations by name. A general elastic collision has two unknown final velocities and two equations, one from momentum and one from kinetic energy, and solving that pair is exactly what the algebra-based exam is not allowed to ask. That is the whole of the difference, and the next section spells it out.

The general elastic collision, which needs simultaneous equations

Take two objects moving along one line, masses $m_1$ and $m_2$ with initial velocities $u_1$ and $u_2$ , and let the collision be elastic. Statements 4.3.B.2 and 4.4.A.1 give you two equations:

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

\tfrac{1}{2}m_1u_1^2 + \tfrac{1}{2}m_2u_2^2 = \tfrac{1}{2}m_1v_1^2 + \tfrac{1}{2}m_2v_2^2

Two equations, two unknowns, and the second is quadratic. Solving the pair gives

v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1 + \frac{2m_2}{m_1 + m_2}u_2 \qquad v_2 = \frac{m_2 - m_1}{m_1 + m_2}u_2 + \frac{2m_1}{m_1 + m_2}u_1

Neither of those is printed on any AP equation sheet, and neither appears in the CED. They are results you derive, and deriving them is what suggested skill 2.B, calculating an unknown quantity with units from known quantities by selecting and following a logical computational pathway, is asking for here.

The shortcut worth knowing. Rather than substituting the quadratic, rearrange both conservation equations to put like masses together, then divide the energy equation by the momentum equation. The $\frac{1}{2}$ and the masses cancel and what falls out is

u_1 - u_2 = -(v_1 - v_2)

The relative velocity of approach equals the relative velocity of separation, with the sign reversed. That is one linear equation, and pairing it with conservation of momentum turns the whole problem into two linear equations, which is far less error-prone than handling the quadratic. It is also the source of the coefficient of restitution, $e = -\frac{v_1 - v_2}{u_1 - u_2}$ , which is $1$ for an elastic collision and $0$ for a perfectly inelastic one. Neither $e$ nor the relative-velocity result is printed on the sheet or stated in the CED, so treat them as tools rather than as citable authorities on a free-response answer. The collision lab exposes $e$ as a slider so you can watch the two limits and everything between.

Worked example 1 runs this both ways on the same numbers, and checks the kinetic energy to the digits.

Reading 4.4.A.1 and 4.4.A.2 carefully, because they are easy to conflate

Statement 4.4.A.1 defines an elastic collision by the total: the initial kinetic energy of the system is equal to the final kinetic energy of the system. Statement 4.4.A.2 then says something that catches people out: in an elastic collision, the final kinetic energies of each of the objects within the system may be different from their initial kinetic energies.

Both are true at once because the definition is about the sum. In worked example 1 the moving object arrives with all $5.00$ J of the kinetic energy and leaves with $0.918$ J, having handed $4.082$ J to the target. Nothing about that violates 4.4.A.1; the total is unchanged.

Two consequences to have ready.

"Elastic" is not "nothing happened". Energy moves freely between the objects. What does not happen is a decrease in the total.

A collision cannot be judged elastic from one object's numbers. You need the whole system's kinetic energy on both sides. Statement 4.4.A.1 says system, twice.

The parallel statements on the inelastic side are 4.4.A.3, an inelastic collision is one in which the total kinetic energy of the system decreases, and 4.4.A.4, which names the mechanism: some of the initial kinetic energy is not restored to kinetic energy but is transformed by nonconservative forces into other forms of energy. That word "restored" is doing real work. During any collision, kinetic energy is temporarily stored as elastic potential energy in the deformation of both bodies. An elastic collision gives all of it back. An inelastic one does not, because nonconservative forces inside the material have converted part of it to thermal energy and sound. Conservative compared with nonconservative forces is the short version of that distinction.

Note what the CED does not say: it never says kinetic energy is "lost". It says the total decreases and that the difference is transformed into other forms. Total energy is conserved throughout, and writing otherwise on a free-response answer is the kind of phrasing that costs a point.

Perfectly inelastic collisions, and the floor on how much can go (4.4.A.5)

Statement 4.4.A.5 defines the case: in a perfectly inelastic collision, the objects stick together and move with the same velocity after the collision. That single common velocity is what makes these the easiest collisions to solve, because momentum conservation alone determines it. For a mass $m_1$ at $u_1$ striking a stationary $m_2$ :

v = \frac{m_1 u_1}{m_1 + m_2}

The kinetic-energy fraction that survives then comes out clean:

\frac{K_f}{K_i} = \frac{\frac{1}{2}(m_1 + m_2)v^2}{\frac{1}{2}m_1u_1^2} = \frac{m_1}{m_1 + m_2}

The initial speed cancels completely. That is a functional-dependence result of the kind suggested skill 2.D asks for: for a target at rest, what fraction of the kinetic energy survives depends only on the mass ratio, not on how fast the projectile was going. A light projectile into a heavy target keeps almost none of it; two equal masses keep exactly half.

A perfectly inelastic collision loses the most kinetic energy that momentum permits, and it does not lose all of it. Both halves of that sentence matter. The reason is the centre of mass. The system's kinetic energy always splits into a part carried by the centre-of-mass motion, $\frac{1}{2}Mv_{\text{cm}}^2$ , and a part carried by motion relative to it. Conservation of momentum fixes $v_{\text{cm}}$ , so the first part cannot be touched by any collision at all. Only the second part is available to be transformed, and sticking together is what reduces it to zero. So the kinetic energy that survives is exactly $\frac{1}{2}Mv_{\text{cm}}^2$ , and it is zero only when the total momentum was zero to begin with, as in a head-on collision between equal and opposite momenta.

That derivation is a Physics C-style route to a result the algebra-based course reaches by arithmetic. The what is conserved in a collision guide owns the arithmetic route and the energy audit; this page gives the centre-of-mass argument for why the answer has the form it does.

The three cases side by side, all with a target at rest:

	Momentum	Total kinetic energy	Final velocities
Elastic, 4.4.A.1	constant	constant	two unknowns, need both equations
Inelastic, 4.4.A.3	constant	decreases	need one final velocity given
Perfectly inelastic, 4.4.A.5	constant	decreases by the most momentum allows	one unknown, momentum alone suffices
Explosion, 4.1.A.3.iii	constant	increases	depends on the fragment count

Elastic collisions in two dimensions, and the equal-mass result

Topic 4.3's boundary statement expects quantitative analysis of collisions in one or two dimensions, so a two-dimensional elastic collision is fair game. The equation count is what to watch: two momentum components plus one energy equation is three equations, against four unknowns for two unrestricted final velocity vectors. One more piece of information is always supplied, usually a final direction.

The case worth memorising, because it comes up repeatedly and because its derivation is three lines of vector algebra, is equal masses with one target at rest. Conservation of momentum gives $\vec{u}_1 = \vec{v}_1 + \vec{v}_2$ after the common mass cancels, and conservation of kinetic energy gives $u_1^2 = v_1^2 + v_2^2$ . Square the first as a dot product:

u_1^2 = \vec{v}_1 \cdot \vec{v}_1 + 2\,\vec{v}_1 \cdot \vec{v}_2 + \vec{v}_2 \cdot \vec{v}_2 = v_1^2 + v_2^2 + 2\,\vec{v}_1 \cdot \vec{v}_2

Compare with the energy equation and $\vec{v}_1 \cdot \vec{v}_2 = 0$ . The two objects separate at exactly $90^\circ$ to each other, whatever the impact geometry, provided neither final velocity is zero. The dot-product definition $\vec{A}\cdot\vec{B} = AB\cos\theta$ that makes the last step readable is printed in the Vectors table on the third page of the equation-sheet appendix, one of the pages usually dropped from condensed copies of the sheet.

That result collapses the four unknowns to two the moment one final direction is given, which is worked example 3. It also explains a familiar observation: a cue ball striking a stationary ball of the same mass off centre leaves at right angles to it, and a head-on hit stops the cue ball dead, which is the $90^\circ$ result in the limiting case where one final speed is zero.

For unequal masses in two dimensions there is no comparably tidy rule, and the route is the one from Topic 4.3: write the two component equations, add the energy equation if the collision is elastic, and solve. That is more algebra than the algebra-based courses are asked for, and it is the direct consequence of the different boundary statement.

Deciding elastic or inelastic from data, and the linearisation the exam uses

Suggested skill 3.C, justifying or supporting a claim using evidence from experimental data, physical representations, or physical principles or laws, is listed for this topic. In practice the claim is almost always "this collision was or was not elastic", and the evidence is a kinetic-energy comparison.

The procedure:

Compute the total kinetic energy of the system before, summing over every object. Momentum is not involved in this step and should not be checked here.
Compute the total after, the same way.
Compare. Equal within experimental uncertainty means elastic by 4.4.A.1. A decrease means inelastic by 4.4.A.3. An increase means either a measurement error or an interaction that was not a collision at all, and 4.1.A.3.iii names that second case an explosion.

Two things that do not decide it. Whether the objects bounced apart: an inelastic collision can still leave the objects separated, just with less kinetic energy than before. And whether momentum was conserved: momentum is conserved in all three cases, so it carries no information about elasticity at all. That second point is worth rereading, because momentum and elasticity are routinely conflated.

The linearisation the framework actually uses. Sample free-response question 3, the Experimental Design and Analysis question, aligns to learning objective 4.4.A along with 4.2.B, and it turns a collision experiment into a straight-line graph twice over. In its first part, students investigate whether the fraction of kinetic energy remaining after a collision, $K_{\text{total},f}/K_{\text{total},i}$ , depends on the initial speed of the incoming cart, and are asked which quantities to measure and what to plot. In its later part, an impulse device delivers the same impulse to carts of different total mass and students measure the maximum height reached on a ramp. Since a fixed impulse gives $v = J/M$ , the height satisfies $h = \frac{J^2}{2gM^2}$ , so a plot of $h$ against $1/M^2$ is a straight line whose slope is $J^2/2g$ , and the impulse comes out of the slope.

Practise both moves. Asking whether a ratio depends on a variable means plotting the ratio against that variable and looking for a nonzero slope. Extracting a constant from a nonlinear relationship means choosing axes that make it linear, then reading the constant out of the gradient.

What the sheet prints for Topic 4.4

Checked against the printed Table of Information in the AP Physics C: Mechanics course and exam description, including the third page carrying the Geometry, Trigonometry, Vectors, Calculus and Identities tables.

Equation	On the C: Mechanics sheet
$K = \frac{1}{2}mv^2$	yes
$\vec{p} = m\vec{v}$	yes
$\vec{v}_{\text{cm}} = \frac{\sum \vec{p}_i}{\sum m_i} = \frac{\sum m_i \vec{v}_i}{\sum m_i}$	yes
$\vec{A}\cdot\vec{B} = AB\cos\theta$	yes, in the Vectors table
$\Delta K = \sum W_i = \sum F_{\parallel,i}d_i$	yes
$v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1 + \dots$ , the elastic-collision result	no
$u_1 - u_2 = -(v_1 - v_2)$ , relative velocity reversal	no
Coefficient of restitution $e$	no, and it appears nowhere in the CED
$v = \frac{m_1u_1}{m_1 + m_2}$ , the perfectly inelastic result	no
$\frac{K_f}{K_i} = \frac{m_1}{m_1 + m_2}$	no
Any equation at all specific to Topic 4.4	no

Topic 4.4 is the only topic in Unit 4 whose framework text prints no relevant equation, and the sheet prints nothing specific to it either. Everything on this page is derived from the two conservation statements plus the definition of kinetic energy. That is the reason the topic rewards deriving fluency rather than recall, and the reason suggested skill 2.B sits on it.

One consequence for exam technique: if you want the general elastic result on the exam, you either derive it in the moment from the two conservation equations, which takes a few lines, or you carry the relative-velocity shortcut in your head and derive it from that in one line. The second is faster, and the derivation of the shortcut itself is a division of one conservation equation by the other.

How Topic 4.4 is tested, and where the algebra-based page is

Unit 4 is weighted 10 to 20% of the multiple-choice section over about 11 to 15 class periods, against 10 to 15% and about 10 to 15 class periods for the identically named AP Physics 1 unit. The unit's Progress Check is about 18 multiple-choice questions and 4 free-response questions.

Neither of the framework's two Unit 4 sample multiple-choice questions aligns to 4.4; both align to learning objective 4.2.B. Sample free-response question 3 does list 4.4.A in its alignment, alongside 2.1.B, 3.1.A, 3.4.B, 3.4.C and 4.2.B, with skills 3.A, 2.D, 1.B, 2.B and 3.C. So in the framework's own sample, Topic 4.4 shows up on the free-response section, inside an experimental-design question, and not as a standalone multiple-choice item.

The unit's fourth sample instructional activity is a 4.4 Ranking Task, and it is a good self-test. An arrow is shot at a pumpkin hanging from the ceiling on a long string, in three scenarios: the arrow bounces back off the pumpkin, the arrow sticks in the pumpkin and travels with it, and the arrow passes straight through. Students rank the final speeds of the pumpkin and justify the ranking with evidence. The answer follows from momentum alone: the bounce transfers the most momentum to the pumpkin because the arrow's own momentum change is largest when it reverses, sticking transfers less, and passing through transfers least of the three. Notice that the ranking is settled entirely by 4.3, and that the elastic-or-inelastic labels do not decide it. That is the point of the activity.

Two habits that pay on 4.4 questions:

Answer symbolically first, then substitute. The suggested skills here lean on calculation and functional dependence, and a symbolic answer makes the dependence visible. The kinetic-energy fraction being independent of the initial speed is invisible until you have the algebra.
Say which quantity you are conserving and why. Momentum, because the net external force is negligible over the contact time by 4.1.A.3.i. Kinetic energy, only if the problem says elastic or the data show it. Writing that sentence is often worth a point on its own.

If you are in the algebra-based course, read the [AP Physics 1 Topic 4.4 page](/ap-physics-1/unit-4-linear-momentum/4-4-elastic-and-inelastic-collisions) instead. Its required content is genuinely the same five statements, so that page is not a lesser version of this one; it is written for an exam that will not ask you to solve the pair of simultaneous equations, and it spends its space on classifying collisions from data, which is where its own suggested skills point. This page is for AP Physics C: Mechanics students who need the general solution, the relative-velocity result and the two-dimensional case.

Where to go next. Topic 4.3 holds the boundary statement that governs this topic, Topic 4.2 is where the impulse in the free-response question comes from, and Topic 6.5, Rolling is where collisions meet rotation. The conservation of momentum guide owns the before-and-after routine, elastic compared with inelastic collisions is the two-minute version, and the momentum collision calculator checks arithmetic. The Unit 4 hub lists all four topics.

A general one-dimensional elastic collision, solved two ways

On a frictionless track a $0.40$ kg glider moving at $5.0$ m/s strikes a stationary $1.00$ kg glider, and the collision is elastic. Take the initial direction of motion as positive. Find both final velocities, first from the two conservation equations directly and then using the relative-velocity result, and verify that the kinetic energy is conserved to the digits you print.

Label $m_1 = 0.40$ kg, $u_1 = 5.0$ m/s, $m_2 = 1.00$ kg, $u_2 = 0$ . Two unknowns, $v_1$ and $v_2$ .
Route one, the two conservation equations. Momentum, from 4.3.B.2: $(0.40)(5.0) = 0.40v_1 + 1.00v_2$ , so $2.0 = 0.40v_1 + 1.00v_2$ .
Kinetic energy, from 4.4.A.1: $\frac{1}{2}(0.40)(5.0)^2 = \frac{1}{2}(0.40)v_1^2 + \frac{1}{2}(1.00)v_2^2$ , so $5.00 = 0.20v_1^2 + 0.50v_2^2$ .
Solving the pair gives the standard result $v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1$ and $v_2 = \frac{2m_1}{m_1 + m_2}u_1$ for a target at rest.
$v_1 = \frac{0.40 - 1.00}{1.40}(5.0) = \frac{-0.60}{1.40}(5.0) = -\frac{15}{7} = -2.1429$ m/s, so $-2.14$ m/s: the light glider rebounds, as it must against a heavier target.
$v_2 = \frac{2(0.40)}{1.40}(5.0) = \frac{0.80}{1.40}(5.0) = \frac{20}{7} = 2.8571$ m/s, so $2.86$ m/s forward.
Route two, the relative-velocity shortcut. Divide the energy equation by the momentum equation after grouping like masses and the result is $u_1 - u_2 = -(v_1 - v_2)$ . Here $5.0 - 0 = v_2 - v_1$ , so $v_2 = v_1 + 5.0$ .
Substitute into momentum: $2.0 = 0.40v_1 + 1.00(v_1 + 5.0) = 1.40v_1 + 5.0$ , so $1.40v_1 = -3.0$ and $v_1 = -2.1429$ m/s, matching. Two linear equations rather than one quadratic, and no risk of picking the wrong root.
Check the momentum. $0.40(-2.1429) + 1.00(2.8571) = -0.8571 + 2.8571 = 2.000\ \mathrm{kg \cdot m/s}$ , equal to the initial $2.0\ \mathrm{kg \cdot m/s}$ .
Check the kinetic energy, which is the check that matters for 4.4.A.1. Before: $\frac{1}{2}(0.40)(5.0)^2 = 5.000$ J. After: $\frac{1}{2}(0.40)(2.1429)^2 + \frac{1}{2}(1.00)(2.8571)^2 = 0.20(4.5918) + 0.50(8.1633) = 0.9184 + 4.0816 = 5.0000$ J. Equal to four decimal places, as an elastic collision requires.
Note 4.4.A.2 in the numbers: the first glider went from $5.000$ J to $0.918$ J and the second from $0$ to $4.082$ J. Each object's kinetic energy changed a great deal; the system total did not change at all.

$v_1 = -\frac{15}{7} = -2.14$ m/s, so the light glider rebounds, and $v_2 = \frac{20}{7} = 2.86$ m/s forward. Momentum is $2.000\ \mathrm{kg \cdot m/s}$ before and after, and the kinetic energy is $5.0000$ J before and after, with the split changing from $5.000$ J and $0$ to $0.918$ J and $4.082$ J.

The same two gliders, stuck together: how much must go

Repeat the previous collision with the gliders sticking together instead of bouncing: a $0.40$ kg glider at $5.0$ m/s into a stationary $1.00$ kg glider, perfectly inelastic. Find the common final velocity, the fraction of kinetic energy that survives, and show that the surviving amount is exactly the kinetic energy of the centre-of-mass motion.

Statement 4.4.A.5 says the objects stick together and move with the same velocity, so there is one unknown and momentum alone determines it.
$v = \frac{m_1u_1}{m_1 + m_2} = \frac{(0.40)(5.0)}{1.40} = \frac{2.0}{1.40} = 1.4286$ m/s, so $1.43$ m/s forward.
Kinetic energy after: $K_f = \frac{1}{2}(1.40)(1.4286)^2 = 0.70(2.0408) = 1.4286$ J, so $1.43$ J.
Kinetic energy before is the same $5.000$ J as in the elastic case, since the initial state is identical.
Fraction surviving: $\frac{1.4286}{5.000} = 0.2857$ , so $28.6\%$ survives and $71.4\%$ , that is $3.571$ J, is transformed by nonconservative forces into other forms, which is 4.4.A.4.
Check that fraction against the symbolic result: $\frac{K_f}{K_i} = \frac{m_1}{m_1 + m_2} = \frac{0.40}{1.40} = 0.2857$ . It matches, and notice the $5.0$ m/s does not appear. Doubling the approach speed would quadruple both kinetic energies and leave the fraction untouched, which is the functional dependence skill 2.D is after.
Now the centre-of-mass check. By 4.3.A.1.i, $v_{\text{cm}} = \frac{\sum p_i}{\sum m_i} = \frac{2.0}{1.40} = 1.4286$ m/s, and 4.3.A.1.ii says it is unchanged by the collision.
Kinetic energy of the centre-of-mass motion: $\frac{1}{2}Mv_{\text{cm}}^2 = \frac{1}{2}(1.40)(1.4286)^2 = 1.4286$ J, exactly the $K_f$ found above.
That is not a coincidence. Sticking together means both objects end up moving at the centre-of-mass velocity, so the motion relative to the centre of mass is gone entirely and only the centre-of-mass part of the kinetic energy remains. Since momentum conservation fixes $v_{\text{cm}}$ , no collision can remove that part, so no collision can lose more than this one did.
Compare the three outcomes of the identical initial state: elastic keeps $5.000$ J, this perfectly inelastic case keeps $1.429$ J, and any partially inelastic outcome falls between. The lower bound is not zero, because the system was moving as a whole and had to keep moving as a whole.

Both gliders move off together at $\frac{10}{7} = 1.43$ m/s. Kinetic energy falls from $5.000$ J to $1.429$ J, so $28.6\%$ survives and $71.4\%$ is transformed, and the surviving fraction $\frac{m_1}{m_1+m_2}$ is independent of the approach speed. The $1.429$ J that remains is exactly $\frac{1}{2}Mv_{\text{cm}}^2$ , which is the part momentum conservation forbids any collision from removing.

Equal masses in two dimensions: the ninety degree result

On a frictionless surface a $0.25$ kg puck moving at $4.0$ m/s strikes an identical stationary $0.25$ kg puck in a glancing elastic collision. The incoming puck is deflected $35.0^\circ$ from its original direction. Find the direction of the struck puck and both final speeds, then verify momentum and kinetic energy.

Set the original direction of motion as $+x$ and take the incoming puck's deflection as being above that axis.
Direction first, from the equal-mass result. With $m_1 = m_2$ , momentum gives $\vec{u}_1 = \vec{v}_1 + \vec{v}_2$ and kinetic energy gives $u_1^2 = v_1^2 + v_2^2$ . Squaring the first as a dot product gives $u_1^2 = v_1^2 + v_2^2 + 2\vec{v}_1\cdot\vec{v}_2$ , so $\vec{v}_1\cdot\vec{v}_2 = 0$ .
A zero dot product with two nonzero speeds means the angle between the final velocities is $90.0^\circ$ , using the printed $\vec{A}\cdot\vec{B} = AB\cos\theta$ . So the struck puck leaves at $90.0 - 35.0 = 55.0^\circ$ below the $+x$ axis. The result holds for any deflection angle, not just this one.
Speeds next. Momentum in $y$ , which was zero before: $v_1\sin 35.0^\circ = v_2 \sin 55.0^\circ$ , so $0.57358 v_1 = 0.81915 v_2$ and $v_2 = 0.70021 v_1$ .
Momentum in $x$ : $4.0 = v_1\cos 35.0^\circ + v_2\cos 55.0^\circ = 0.81915 v_1 + 0.57358(0.70021)v_1 = 0.81915v_1 + 0.40163v_1 = 1.22078 v_1$ .
$v_1 = 4.0/1.22078 = 3.2766$ m/s, so $3.28$ m/s at $35.0^\circ$ above the axis, and $v_2 = 0.70021(3.2766) = 2.2943$ m/s, so $2.29$ m/s at $55.0^\circ$ below it.
Because the separation angle is $90.0^\circ$ , the two speeds are just $u_1\cos 35.0^\circ = 4.0(0.81915) = 3.2766$ m/s and $u_1\sin 35.0^\circ = 4.0(0.57358) = 2.2943$ m/s, which is the fastest route once you know the result.
Check momentum in $x$ : $0.25\left[3.2766(0.81915) + 2.2943(0.57358)\right] = 0.25\left[2.6840 + 1.3160\right] = 0.25(4.0000) = 1.000\ \mathrm{kg \cdot m/s}$ , against $0.25(4.0) = 1.00\ \mathrm{kg \cdot m/s}$ before.
Check momentum in $y$ : $0.25\left[3.2766(0.57358) - 2.2943(0.81915)\right] = 0.25\left[1.8794 - 1.8794\right] = 0$ exactly, since both products equal $4.0\sin 35.0^\circ\cos 35.0^\circ$ , against zero before.
Check kinetic energy: before, $\frac{1}{2}(0.25)(4.0)^2 = 2.000$ J. After, $\frac{1}{2}(0.25)\left[3.2766^2 + 2.2943^2\right] = 0.125\left[10.736 + 5.264\right] = 0.125(16.000) = 2.000$ J. Equal, as 4.4.A.1 requires, and the exactness comes from $\cos^2\theta + \sin^2\theta = 1$ , an identity printed in the same appendix.
One boundary check. This problem had four unknowns at the outset, reduced to three by the given deflection angle, and settled by three equations: two momentum components and one energy equation. That is a quantitative two-dimensional collision, which Topic 4.3's boundary statement in this course expects and the AP Physics 1 boundary statement does not permit.

The struck puck leaves at $55.0^\circ$ on the other side of the original line, so the two pucks separate at exactly $90.0^\circ$ , a result that holds for equal masses in any elastic glancing collision with the target at rest. The speeds are $3.28$ m/s and $2.29$ m/s. Momentum checks to $1.000\ \mathrm{kg \cdot m/s}$ in $x$ and zero in $y$ , and the kinetic energy is $2.000$ J before and after.

Frequently asked questions

What is the difference between elastic and inelastic collisions in AP Physics C?

Essential knowledge 4.4.A.1 defines an elastic collision as one in which the initial kinetic energy of the system equals the final kinetic energy of the system, and 4.4.A.3 defines an inelastic collision as one in which the total kinetic energy of the system decreases. Statement 4.4.A.4 names the mechanism for the decrease: some of the initial kinetic energy is not restored to kinetic energy but is transformed by nonconservative forces into other forms of energy. Momentum is conserved in both kinds whenever the net external force is negligible, so momentum carries no information at all about which kind a collision was. These definitions are worded identically in AP Physics C: Mechanics and AP Physics 1.

Is Topic 4.4 different in AP Physics C than in AP Physics 1?

The required content is not. All five essential-knowledge statements under learning objective 4.4.A are word for word the same in both courses, and neither course prints a boundary statement on this topic. The difference is inherited from Topic 4.3. The AP Physics C: Mechanics boundary statement there expects quantitative analysis of collisions in one or two dimensions, while the AP Physics 1 one excludes exam questions involving solution of simultaneous equations by name. A general elastic collision has two unknown final velocities and needs both the momentum and the kinetic-energy equation solved together, so the calculus-based exam can ask for it and the algebra-based exam cannot.

How do you solve an elastic collision with two unknown final velocities?

Write conservation of momentum and conservation of kinetic energy, then avoid the quadratic by dividing one by the other. Grouping like masses and dividing the energy equation by the momentum equation gives the relative-velocity result: the relative velocity of approach equals the relative velocity of separation with the sign reversed. Pair that linear equation with conservation of momentum and you have two linear equations in two unknowns, with no risk of selecting the wrong root. For a target at rest the answers reduce to the incoming velocity multiplied by the mass difference over the mass sum for the projectile, and by twice the projectile mass over the mass sum for the target.

Is the elastic collision formula on the AP Physics C equation sheet?

No. The printed Table of Information for AP Physics C: Mechanics carries no equation specific to Topic 4.4 at all: no general elastic-collision result, no relative-velocity reversal, no coefficient of restitution and no perfectly inelastic velocity formula. The coefficient of restitution does not appear anywhere in the course and exam description either. What is printed is the machinery you build them from, namely momentum as mass times velocity, kinetic energy as one half mass times speed squared, the centre-of-mass velocity, and the dot product in the Vectors table. Topic 4.4 rewards deriving fluency rather than recall for exactly that reason.

How much kinetic energy does a perfectly inelastic collision lose?

As much as conservation of momentum permits, which is not all of it. For a projectile of mass m1 striking a stationary target of mass m2 the surviving fraction is m1 divided by the sum of the masses, and the initial speed cancels out entirely, so the fraction depends only on the mass ratio. The reason it cannot reach zero is the centre of mass: the system's kinetic energy splits into a part carried by centre-of-mass motion and a part carried by motion relative to it, momentum conservation fixes the first part, and sticking together removes only the second. The surviving energy is exactly one half the total mass times the square of the centre-of-mass speed.

Why do equal masses separate at 90 degrees in an elastic collision?

Because the dot product of the two final velocities must vanish. With equal masses and one object initially at rest, conservation of momentum says the initial velocity vector equals the sum of the two final velocity vectors, and conservation of kinetic energy says the square of the initial speed equals the sum of the squares of the final speeds. Squaring the vector equation produces an extra cross term of twice the dot product of the two final velocities, so comparing the two results forces that dot product to zero. Two nonzero vectors with a zero dot product are perpendicular, so the separation angle is ninety degrees whatever the impact geometry.

Can you tell a collision is elastic from the momentum?

No, and conflating the two is a routine error. Momentum is conserved in elastic collisions, in inelastic collisions, in perfectly inelastic collisions and in explosions alike, whenever the net external force on the system is negligible over the interaction. The only test for elasticity is a kinetic-energy comparison: total the kinetic energy of every object in the system before, total it after, and see whether the two agree within experimental uncertainty. Whether the objects bounced apart or stayed together is also not the test, since an inelastic collision can leave the objects separated with less kinetic energy than they started with.