AP Physics C: Mechanics · Topic 4.3

Topic 4.3: Conservation of Linear Momentum

Q: When is momentum conserved in AP Physics C Mechanics?

When the net external force on the system you selected is zero, per essential knowledge 4.3.B.2. In the calculus-based course that follows in one step from the derivative form of the second law: the net external force is the time derivative of the system's momentum, so a zero net external force means a zero derivative and therefore a constant momentum at every instant, not just before and after. If the net external force is nonzero, essential knowledge 4.3.B.3 says momentum is transferred between the system and the environment, and 4.3.A.3.iii gives the amount as the impulse exerted on the system.

Q: Does AP Physics C allow two-dimensional collision problems?

Yes, quantitatively. The boundary statement under Topic 4.3 reads that AP Physics C: Mechanics only expects students to quantitatively analyze collisions and interactions in one or two dimensions, and that three-dimensional collisions may be analyzed qualitatively. That is a different boundary from AP Physics 1's, which allows a quantitative treatment only in one dimension, calls the two-dimensional treatment semiquantitative, and excludes exam questions involving solution of simultaneous equations by name. So a two-dimensional collision with two unknown final speeds, solved from the pair of component equations, is a Physics C question and not a Physics 1 one.

Q: Can you solve simultaneous equations on the AP Physics C exam?

Nothing in the AP Physics C: Mechanics course and exam description excludes it, and the phrase does not appear in the document. The exclusion is an AP Physics 1 boundary statement, which says that exam questions involving solution of simultaneous equations are not included in AP Physics 1 although the exam may still ask students to set the equations up. What the Physics C boundary statement in the same position restricts instead is dimensionality: quantitative analysis is expected in one or two dimensions, and three-dimensional collisions may be analyzed qualitatively. Since a two-dimensional collision with two unknowns cannot be analysed quantitatively without solving the pair of component equations, the expectation follows from that.

Q: Why do internal forces not change a system's momentum?

Because their impulses cancel exactly. Essential knowledge 4.3.A.3.i states that the impulse exerted by one object on a second object is equal and opposite to the impulse exerted by the second object on the first, and attributes it to Newton's third law. In the calculus-based course the argument is stronger than a statement about forces: the two contact forces are equal and opposite at every instant, so their integrals over the same interval are equal and opposite whatever shape the force curves have. The forces need not be constant, and the collision need not be brief. Only forces crossing the system boundary can change the total.

Q: How do you check a two-dimensional momentum answer?

Recompute the velocity of the centre of mass from your final answers and compare it with the value before the collision. Essential knowledge 4.3.A.1.i prints the centre-of-mass velocity as the total momentum divided by the total mass, and 4.3.A.1.ii says it is constant in the absence of a net external force, so the two values must agree. This is faster than re-adding all the momenta because the total mass is a common factor. A mismatch means one of the final velocities or one of the signs is wrong, and it usually shows which component is at fault.

Q: Is momentum conserved in an explosion?

Yes. Essential knowledge 4.1.A.3.iii describes an explosion as an interaction in which forces internal to the system move objects within that system apart, and internal forces cannot change the total momentum. So the fragments' momenta sum to whatever the object's momentum was before, which is zero if it started at rest, and the centre of mass carries on exactly as it was. Kinetic energy behaves differently: an explosion increases it, drawing on stored energy inside the system, so an energy audit and a momentum audit give opposite-looking answers about the same event.

Q: How do you know whether to ignore gravity during a collision?

Compare impulses, not forces. Essential knowledge 4.1.A.3.i defines a collision as an interaction where the forces exerted between the involved objects are much larger than the net external force exerted on those objects during the interaction. In AP Physics C: Mechanics you can turn that into a number: work out the external force component along the direction of interest, multiply by the contact time to get the external impulse, and compare it against the momentum change of one of the colliding objects. A few per cent justifies treating the total momentum as constant, and tens of per cent does not. The same collision can pass the test over a two millisecond contact and fail it over a quarter of a second.

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

A system's total momentum is constant when the net external force on it is zero. In AP Physics C: Mechanics that is a theorem, not an axiom: set the net force in the derivative form of the second law to zero and the momentum has zero time derivative. The unit's one boundary statement lives here.

AP Physics: Unit 4 (topics 4.3 Conservation of Linear Momentum). Topic 4.3 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. Two learning objectives, 4.3.A and 4.3.B, and twelve essential-knowledge statements counting sub-statements, the most of any topic in Unit 4: 4.3.A.1, 4.3.A.1.i (the printed v_cm equation), 4.3.A.1.ii, 4.3.A.2, 4.3.A.3, 4.3.A.3.i (equal and opposite impulses from Newton's third law), 4.3.A.3.ii, 4.3.A.3.iii (J = delta p), 4.3.A.4, and 4.3.B.1 to 4.3.B.3. Suggested skills 1.A, 1.B, 2.C, 3.A and 3.B, five, more than any other topic in the unit. Verified against both CEDs statement by statement: all twelve are word for word identical to AP Physics 1's Topic 4.3, so the entire difference between the two courses in this topic is the boundary statement. This topic holds Unit 4's ONLY boundary statement, quoted whole: AP Physics C: Mechanics only expects students to quantitatively analyze collisions and interactions in one or two dimensions, and three-dimensional collisions may be analyzed qualitatively. The AP Physics 1 boundary statement in the same position instead reads that AP Physics 1 includes a quantitative and qualitative treatment in one dimension and a semiquantitative treatment in two dimensions, that exam questions involving solution of simultaneous equations are not included in AP Physics 1 although the exam may assess whether students can set the equations up and reason about how changing a mass, speed or angle would affect other quantities, and that AP Physics 2 includes a full treatment in two dimensions for problems that include one unknown final velocity. A full-text search of the C: Mechanics CED finds no occurrence of the simultaneous-equations exclusion, so the C boundary bounds dimensionality only. Sheet detail verified against the printed Table of Information: v_cm, x_cm, p = mv, F_net = dp/dt and the impulse integral are printed, r_cm as the integral of r dm over the integral of dm is printed on this sheet and not on the AP Physics 1 sheet, and the Vectors table on the third page prints component addition and the dot product. The conservation equation itself, sum of p before equals sum of p after, is printed on no AP equation sheet in any of the four courses. Neither Unit 4 sample multiple-choice question aligns here; both align to 4.2.B. Two of the unit's four sample instructional activities are 4.3 Desktop Experiment Tasks.

What Topic 4.3 requires

Topic 4.3 carries two learning objectives and twelve essential-knowledge statements counting sub-statements, more than any other topic in Unit 4, and it holds the unit's only boundary statement.

Objective	What it asks
4.3.A	Describe the behavior of a system using conservation of linear momentum
4.3.B	Describe how the selection of a system determines whether the momentum of that system changes

Statement	What it says
4.3.A.1	A collection of objects with individual momenta can be described as one system with one center-of-mass velocity
4.3.A.1.i	For a collection of objects, the velocity of a system's center of mass can be calculated using $\vec{v}_{\text{cm}} = \frac{\sum \vec{p}_i}{\sum m_i} = \frac{\sum (m_i\vec{v}_i)}{\sum m_i}$
4.3.A.1.ii	The velocity of a system's center of mass is constant in the absence of a net external force
4.3.A.2	The total momentum of a system is the sum of the momenta of the system's constituent parts
4.3.A.3	In the absence of net external forces, any change to the momentum of an object within a system must be balanced by an equivalent and opposite change of momentum elsewhere within the system. Any change to the momentum of a system is due to a transfer of momentum between the system and its surroundings
4.3.A.3.i	The impulse exerted by one object on a second object is equal and opposite to the impulse exerted by the second object on the first. This is a direct result of Newton's third law
4.3.A.3.ii	A system may be selected so that the total momentum of that system is constant
4.3.A.3.iii	If the total momentum of a system changes, that change will be equivalent to the impulse exerted on the system. Relevant equation: $\vec{J} = \Delta\vec{p}$
4.3.A.4	Correct application of conservation of momentum can be used to determine the velocity of a system immediately before and immediately after collisions or explosions
4.3.B.1	Momentum is conserved in all interactions
4.3.B.2	If the net external force on the selected system is zero, the total momentum of the system is constant
4.3.B.3	If the net external force on the selected system is nonzero, momentum is transferred between the system and the environment

Suggested skills: 1.A, 1.B, 2.C, 3.A and 3.B, which is five, more than any other topic in the unit lists.

Every one of those twelve statements is word for word the same as the corresponding statement in AP Physics 1's Topic 4.3. The difference between the two courses in this topic is entirely in the boundary statement, and it is a large one.

The boundary statement, and the AP Physics 1 boundary statement it replaces

AP Physics C: Mechanics, the only boundary statement in Unit 4, quoted whole: "AP Physics C: Mechanics only expects students to quantitatively analyze collisions and interactions in one or two dimensions. Three-dimensional collisions may be analyzed qualitatively."

AP Physics 1, the boundary statement in the same position, quoted whole: "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."

Put the three courses in a row and the ladder is explicit.

	Momentum in one dimension	Momentum in two dimensions	Simultaneous equations
AP Physics 1	quantitative and qualitative	semiquantitative	excluded by name
AP Physics 2	quantitative	full treatment, for problems with one unknown final velocity	not stated
AP Physics C: Mechanics	quantitative	quantitative, with no cap stated	no exclusion anywhere in the CED

The phrase "solution of simultaneous equations" appears nowhere in the AP Physics C: Mechanics course and exam description. What its boundary statement restricts is dimensionality, and only at three dimensions, where the treatment drops to qualitative.

So the practical difference between the two Topic 4.3 pages is this. In AP Physics 1 a two-dimensional collision is a setup exercise: write the two component equations, reason about how changing a mass or an angle would move things, and stop. Here, writing the two component equations is the start of the problem and solving them for two unknowns is the rest of it. Worked example 1 does exactly that, and it is the shape of problem the algebra-based exam is forbidden to score.

One caution against over-reading this. The absence of an exclusion is not the same as an instruction, and the C: Mechanics boundary statement does not say "simultaneous equations are included". What it says is that quantitative analysis in two dimensions is expected, and a genuine two-dimensional collision with two unknowns cannot be analysed quantitatively any other way.

In this course conservation is a theorem, and the derivation is two lines

AP Physics 1 states 4.3.B.2 as a fact about systems. Here you can prove it, and the proof is short enough to write in a free-response margin.

Start from 4.2.A.1, which is printed on the equation sheet:

\vec{F}_{\text{net,ext}} = \frac{d\vec{p}_{\text{sys}}}{dt}

If $\vec{F}_{\text{net,ext}} = 0$ then $\frac{d\vec{p}_{\text{sys}}}{dt} = 0$ , and a vector whose derivative is zero throughout an interval is constant throughout that interval. That is 4.3.B.2. If instead the net external force is nonzero, integrate both sides over the interaction and 4.2.B.2.i returns $\Delta\vec{p}_{\text{sys}} = \int \vec{F}_{\text{net,ext}}\,dt = \vec{J}$ , which is 4.3.A.3.iii and 4.3.B.3 together.

Three things this framing buys you that the algebra-based statement does not.

Conservation is instantaneous, not just before-and-after. $d\vec{p}/dt = 0$ says the total momentum is constant at every instant of the collision, including halfway through the squash, not merely that the initial and final totals match. Statement 4.3.A.1.ii says the same thing about the centre of mass: its velocity is constant in the absence of a net external force, so during a head-on collision on a frictionless track the centre of mass sails through at unchanged speed while both objects are deforming.

Conservation is a component statement. $d\vec{p}/dt = 0$ is a vector equation, so it holds separately in $x$ and in $y$ . Momentum can be conserved in one direction and not the other, and that is routine: a ball bouncing off a wall conserves the momentum component parallel to the wall and not the perpendicular one, because the wall's normal force has no parallel component.

Internal forces cancel because their impulses do. Statement 4.3.A.3.i gives the reason as Newton's third law: the impulse exerted by one object on a second is equal and opposite to the impulse exerted by the second on the first. In integral form, the two contact forces are equal and opposite at every instant, so their integrals over the same interval are equal and opposite, whatever shape those force curves have. Notice that this argument does not require the forces to be constant, or the collision to be brief, or the two objects to be in contact for the same length of time as anything else in the problem. Topic 2.3, Newton's third law is where the force-level statement lives.

Selecting the system is the method (4.3.A.3.ii and 4.3.B.1 to 4.3.B.3)

Statement 4.3.B.1 says momentum is conserved in all interactions, and statement 4.3.B.3 says that if the net external force on the selected system is nonzero, momentum is transferred between the system and the environment. Those two sit together comfortably only once you notice that the first is a statement about the universe and the second is a statement about your choice of boundary.

Statement 4.3.A.3.ii is the operative one: a system may be selected so that the total momentum of that system is constant. Read it as permission and as instruction. You are being told that choosing the system is part of solving the problem, not a preliminary to it.

The routine, which is what suggested skill 3.B, applying an appropriate law, definition, theoretical relationship, or model to make a claim, is listed here for:

Draw the boundary. Everything inside is the system; everything outside is the environment.
List the forces that cross the boundary. Those are the external ones.
If they sum to zero, or if their total impulse over the interval is negligible against the momentum changes in play, the total momentum is constant and you write one before-and-after equation per direction.
If they do not, the total momentum changes by their impulse, $\vec{J} = \Delta\vec{p}$ , which is 4.3.A.3.iii.

Where step 3's "negligible" is doing real work, you can quantify it in this course, and worked example 3 does. Statement 4.1.A.3.i defines a collision by the internal forces being much larger than the net external force during the interaction, and with the impulse integral in hand you can compute the ratio instead of asserting it.

Two system choices worth having ready:

Both colliding objects. The contact forces become internal and drop out. This is the choice that makes a collision solvable.
One object only. Now the contact force is external, and its impulse is exactly that object's change in momentum. This is the choice that gets you the force, and it is how the two halves of 4.3.A.3.i are used in practice: compute one object's $\Delta \vec{p}$ , and you have the other's for free with the sign flipped.

The conservation of momentum guide owns the before-and-after routine at the level of arithmetic, and what is conserved in a collision owns the audit of which quantities survive. This page covers the framing the CED asks for instead of repeating either.

Two dimensions, two equations, and the centre of mass

In two dimensions, conservation of momentum is two scalar equations:

\sum m_i v_{ix} \Big|_{\text{before}} = \sum m_i v_{ix} \Big|_{\text{after}} \qquad \sum m_i v_{iy} \Big|_{\text{before}} = \sum m_i v_{iy} \Big|_{\text{after}}

Two equations determine two unknowns. That is the whole of what the C boundary statement licenses beyond the algebra-based course, and it is enough for the standard cases:

Two final speeds, when both final directions are given.
One final speed and one final direction, when the other object's final velocity is fully known.
A third fragment's velocity in an explosion, when the first two are known, since the two components come out directly.

What two equations cannot do is determine four unknowns, so a general two-dimensional collision with both final velocities entirely unknown is not solvable from momentum alone. If a problem looks like that, either a direction has been given, or the collision has been declared elastic, which adds the energy equation as a third. That third equation and its consequences belong to Topic 4.4.

The centre of mass is the cross-check that catches sign errors. Statement 4.3.A.1.i prints

\vec{v}_{\text{cm}} = \frac{\sum \vec{p}_i}{\sum m_i} = \frac{\sum m_i \vec{v}_i}{\sum m_i}

and 4.3.A.1.ii says it is constant with no net external force. So compute $\vec{v}_{\text{cm}}$ from the "before" data and again from your "after" answers. If they differ, one of your final velocities is wrong, and you have found it before the marker did. This is cheaper than re-adding the momenta because the total mass is a common factor you can carry once.

Two further uses of the centre of mass in this topic. In an explosion of a stationary object, $\vec{v}_{\text{cm}} = 0$ before and therefore after, so the fragments' momenta must sum to zero, which is worked example 2. And in any collision the kinetic energy splits into a part carried by the centre-of-mass motion, $\frac{1}{2}M v_{\text{cm}}^2$ , and a part carried by motion relative to it. Only the second part is available to be lost, which is where Topic 4.4's limit on how much a collision can lose comes from. For extended bodies rather than labelled masses, the C: Mechanics sheet also prints $\vec{r}_{\text{cm}} = \frac{\int \vec{r}\, dm}{\int dm}$ , set up in Topic 2.1.

When momentum is not conserved, and how to tell in advance

Statement 4.3.B.1 promises that momentum is conserved in all interactions, but that promise is about a system large enough to include everything doing the pushing. For the system you actually chose, the question is whether external impulse crosses the boundary during the interval you care about, and the answer is usually a matter of degree.

Cases where the external impulse matters:

A long interval. Gravity acting for a second dwarfs gravity acting for two milliseconds. The same collision analysed over the contact time conserves momentum and analysed over the following half-second does not.
A vertical component you forgot. A ball bouncing off the floor changes its vertical momentum by roughly $2mv$ , all of it delivered by the floor. Momentum is conserved for the ball-plus-Earth system and not for the ball.
Friction or a track that is not level. Two carts colliding on an incline have $Mg\sin\theta$ acting on them throughout. Worked example 3 turns that into a percentage.
A held or clamped object. Whatever holds it can supply unlimited impulse through the mount.

Cases where you may drop it, and the wording matters: 4.1.A.3.i says the internal forces must be much larger than the net external force during the interaction. Not much larger than each individual external force, and not "the collision is fast". Fast collisions are the usual reason the criterion is satisfied, but it is the criterion, not the speed, that licenses the model.

The practical test, and it is a Physics C test because it needs an integral: estimate $\left|\int \vec{F}_{\text{ext}}\,dt\right|$ over the contact time and compare it against the momentum change of one of the colliding objects. If the ratio is a few per cent, treating momentum as constant is safe and you should say so in words on a free-response answer. If it is tens of per cent, it is not, and you should say that instead.

Suggested skill 3.A, creating experimental procedures that are appropriate for a given scientific question, is listed on this topic and nowhere else in Unit 4 except through the free-response question. It is listed here because designing a momentum experiment is mostly designing away the external impulse: level the track, use an air track or low-friction wheels, and measure velocities close enough in time to the collision that drag has not yet mattered.

What the sheet prints for Topic 4.3

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

Equation	On the C: Mechanics sheet
$\vec{v}_{\text{cm}} = \frac{\sum \vec{p}_i}{\sum m_i} = \frac{\sum m_i \vec{v}_i}{\sum m_i}$	yes, and identically on the AP Physics 1 sheet
$\vec{x}_{\text{cm}} = \frac{\sum m_i \vec{x}_i}{\sum m_i}$	yes
$\vec{r}_{\text{cm}} = \frac{\int \vec{r}\, dm}{\int dm}$	yes, and not on the AP Physics 1 sheet
$\vec{p} = m\vec{v}$	yes
$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$	yes
$\vec{J} = \int_{t_1}^{t_2}\vec{F}_{\text{net}}(t)\,dt = \Delta\vec{p}$	yes
$\vec{C} = \vec{A} + \vec{B}$ and $\vec{C} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j}$	yes, in the Vectors table
$\vec{A}\cdot\vec{B} = AB\cos\theta$	yes, in the Vectors table
$\sum \vec{p}_{\text{before}} = \sum \vec{p}_{\text{after}}$ as its own line	no, and it is on no AP sheet
$\vec{J} = \Delta\vec{p}$ as a standalone line	no, only as the right-hand end of the impulse line
$\vec{p}_{\text{sys}} = M\vec{v}_{\text{cm}}$	no, though it is the centre-of-mass line rearranged

Two readings. The conservation equation itself is not printed on any AP equation sheet, in any of the four courses, so you write it from 4.3.B.2 every time; what is printed is the machinery it is built from. The Vectors table is on this sheet, which matters for a two-dimensional problem: component addition and the unit-vector form are printed, so the vector bookkeeping is supported even though the conservation statement is not. That table is one of the ones commonly dropped from condensed copies of the sheet, so check a printed CED appendix rather than a transcription before concluding an equation is missing.

How Topic 4.3 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 AP Physics 1 unit of the same name. 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 4.3; both align to 4.2.B. That is not a sign the topic is lightly tested, since the sample set is fifteen questions for the whole course. It does say something about the shape of 4.3 questions: conservation of momentum is more often the middle step of a longer problem than a question in itself.

The framework's sample instructional activities point at where 4.3 does get tested directly, and two of the four in this unit are 4.3 activities, both labelled Desktop Experiment Tasks. In the first, students use conservation of momentum to determine the launch speed of a projectile that is too fast to measure directly, by firing it into a stationary freely movable object. In the second, they launch a spring-loaded cart, find the kinetic energy it gained, then make it collide elastically with a second cart and predict where the second cart lands off the end of the track. Both are the same move: use momentum to get a velocity you cannot measure, then hand that velocity to another part of physics.

Suggested skills 1.A and 1.B, creating diagrams and creating quantitative graphs with appropriate scales and units including plotting data, are listed here, and 1.B appears in the skill list of the framework's sample free-response question 3, the Experimental Design and Analysis question. Practising the plot-and-linearise routine on momentum data is directly examinable.

If you are in the algebra-based course, read the [AP Physics 1 Topic 4.3 page](/ap-physics-1/unit-4-linear-momentum/4-3-conservation-of-linear-momentum) instead. Its exam poses one-dimensional collisions quantitatively, two-dimensional ones only semiquantitatively, and never asks you to solve simultaneous equations. The two-dimensional worked examples on this page are outside its boundary statement. This page is for AP Physics C: Mechanics students.

Where to go next. Topic 4.2 is where the derivative form this page starts from comes from, Topic 4.4 adds the energy equation, and Topic 6.4 is this argument again with torque and angular momentum. The momentum collision calculator checks arithmetic, and the collision lab lets you set both masses, both initial velocities and a coefficient of restitution and watch the momentum total hold while the kinetic energy does not. The Unit 4 hub lists all four topics.

A two-dimensional collision with two unknown final speeds

On a frictionless horizontal surface, puck A of mass $0.20$ kg slides at $8.0$ m/s along the $+x$ axis and strikes puck B of mass $0.30$ kg, which is at rest. After the collision, puck A moves at $30.0^\circ$ above the $+x$ axis and puck B moves at $45.0^\circ$ below it. Find both final speeds, then decide whether the collision was elastic.

Set up the axes and note the count: two unknowns, $v_A$ and $v_B$ , and two component equations. Under the AP Physics C: Mechanics boundary statement this is a quantitative two-dimensional problem, so solving the pair is the expected route.
Conservation in $y$ , where the total was zero before: $0 = m_A v_A \sin 30.0^\circ - m_B v_B \sin 45.0^\circ$ .
$0.20 v_A (0.50000) = 0.30 v_B (0.70711)$ , so $0.10 v_A = 0.21213 v_B$ and $v_A = 2.1213\, v_B$ .
Conservation in $x$ : $m_A v_{A0} = m_A v_A \cos 30.0^\circ + m_B v_B \cos 45.0^\circ$ , so $(0.20)(8.0) = 1.60 = 0.20 v_A (0.86603) + 0.30 v_B (0.70711)$ .
$1.60 = 0.17321 v_A + 0.21213 v_B$ . Substitute $v_A = 2.1213 v_B$ : $1.60 = 0.17321(2.1213)v_B + 0.21213 v_B = 0.36742 v_B + 0.21213 v_B = 0.57956 v_B$ .
$v_B = 1.60/0.57956 = 2.7607$ m/s, so $2.76$ m/s at $45.0^\circ$ below the $+x$ axis. Then $v_A = 2.1213(2.7607) = 5.8564$ m/s, so $5.86$ m/s at $30.0^\circ$ above it.
Check the $x$ equation with the answers: $0.20(5.8564)(0.86603) + 0.30(2.7607)(0.70711) = 1.0144 + 0.5856 = 1.6000$ , matching the initial $1.60\ \mathrm{kg \cdot m/s}$ .
Check with the centre of mass, which by 4.3.A.1.ii must be unchanged. Before: $\vec{v}_{\text{cm}} = \frac{1.60}{0.50} = 3.20$ m/s along $+x$ . After: the same $1.600\ \mathrm{kg \cdot m/s}$ over the same $0.50$ kg, and the $y$ momenta cancel, so $3.20$ m/s along $+x$ again.
Now the energy audit, which momentum conservation alone cannot supply. $K_i = \frac{1}{2}(0.20)(8.0)^2 = 6.40$ J.
$K_f = \frac{1}{2}(0.20)(5.8564)^2 + \frac{1}{2}(0.30)(2.7607)^2 = 0.10(34.297) + 0.15(7.6215) = 3.4297 + 1.1432 = 4.573$ J.
$K_f < K_i$ , so by 4.4.A.3 the collision is inelastic. It lost $6.40 - 4.573 = 1.83$ J, which is $28.5\%$ of the initial kinetic energy. Momentum was conserved throughout, which is 4.3.B.2 doing its job and is entirely independent of the energy result.
Worth noticing: the given angles were the extra information that made two unknowns solvable with two equations. Had the problem given neither final direction, four unknowns would face two equations and momentum alone would not settle it.

$v_A = 5.86$ m/s at $30.0^\circ$ above the $+x$ axis and $v_B = 2.76$ m/s at $45.0^\circ$ below it. The centre-of-mass velocity is $3.20$ m/s along $+x$ before and after. Kinetic energy falls from $6.40$ J to $4.57$ J, a loss of $1.83$ J or $28.5\%$ , so the collision is inelastic.

An explosion into three fragments, and where the energy came from

A $10.0$ kg object sits at rest on frictionless ice and explodes into exactly three fragments that all move horizontally. Fragment 1, of mass $2.0$ kg, moves at $6.0$ m/s along $+x$ . Fragment 2, of mass $3.0$ kg, moves at $4.0$ m/s along $+y$ . Find the velocity of fragment 3, the energy released, and the subsequent motion of the centre of mass.

Fragment 3's mass is what is left: $10.0 - 2.0 - 3.0 = 5.0$ kg.
Statement 4.1.A.3.iii calls an explosion an interaction in which forces internal to the system move objects within that system apart. Internal forces only, so by 4.3.B.2 the total momentum is constant, and it was zero.
Momenta of the two known fragments: $\vec{p}_1 = (2.0)(6.0)\hat{i} = 12.0\hat{i}$ and $\vec{p}_2 = (3.0)(4.0)\hat{j} = 12.0\hat{j}$ , both in $\mathrm{kg \cdot m/s}$ .
Total must remain zero, so $\vec{p}_3 = -\left(12.0\hat{i} + 12.0\hat{j}\right)\ \mathrm{kg \cdot m/s}$ . Both components come out directly; no simultaneous solving is needed here because the unknown is a whole vector and the two equations hand over its two components.
$|\vec{p}_3| = \sqrt{12.0^2 + 12.0^2} = 12\sqrt{2} = 16.971\ \mathrm{kg \cdot m/s}$ , and $v_3 = 16.971/5.0 = 3.394$ m/s, so $3.39$ m/s.
Direction: the components are equal and both negative, so $45.0^\circ$ below the $-x$ axis, which is $225^\circ$ measured counterclockwise from $+x$ .
Energy released, which is the kinetic energy that did not exist before: $K_1 = \frac{1}{2}(2.0)(6.0)^2 = 36.0$ J, $K_2 = \frac{1}{2}(3.0)(4.0)^2 = 24.0$ J, and $K_3 = \frac{1}{2}(5.0)(3.394)^2 = \frac{1}{2}(5.0)(11.52) = 28.8$ J.
Total $K_f = 36.0 + 24.0 + 28.8 = 88.8$ J, against $K_i = 0$ . So $88.8$ J came out of stored energy inside the object. Momentum is unchanged and kinetic energy has increased, which is the signature of an explosion and the reason 4.3.B.1 and the energy statements have to be read separately.
The centre of mass: $\vec{v}_{\text{cm}} = \frac{\sum \vec{p}_i}{\sum m_i} = \frac{\vec{0}}{10.0} = \vec{0}$ , before and after. Statement 4.3.A.1.ii says it stays constant with no net external force, so the centre of mass of the three fragments remains at the point where the object was, permanently, while all three fragments fly away from it.
Check the arithmetic on that claim: $12.0 + 0 - 12.0 = 0$ in $x$ and $0 + 12.0 - 12.0 = 0$ in $y$ . Both zero, as required.

Fragment 3 has mass $5.0$ kg and moves at $3.39$ m/s at $45.0^\circ$ below the $-x$ axis, carrying momentum $-12.0\hat{i} - 12.0\hat{j}$ in $\mathrm{kg \cdot m/s}$ . The explosion released $88.8$ J of kinetic energy from internal stores. The centre of mass has zero velocity before and after and never moves.

Is momentum conserved on this incline? Put a number on it

Two carts with a combined mass of $1.3$ kg collide on a track tilted at $20.0^\circ$ to the horizontal. During the collision the $0.50$ kg cart's velocity along the track changes by $1.8$ m/s. Contact lasts $2.5$ ms. Use $g = 9.8\ \mathrm{m/s^2}$ . Decide whether treating momentum as constant is justified, and say what would change if the interaction lasted $0.25$ s instead.

Take the positive direction as down the slope, and work along the track, since that is the direction in which the collision happens and in which gravity has an unbalanced component. The normal force is perpendicular to the track and contributes nothing along it.
Statement 4.1.A.3.i sets the criterion: the internal forces must be much larger than the net external force during the interaction. Along the track the net external force on the two-cart system is the gravitational component $Mg\sin\theta$ ; the track is assumed low-friction.
External force: $Mg\sin\theta = (1.3)(9.8)\sin 20.0^\circ = (1.3)(9.8)(0.34202) = 4.357$ N.
External impulse over the contact time, which is the quantity that actually matters: $J_{\text{ext}} = (4.357)(0.0025) = 0.01089\ \mathrm{N \cdot s}$ .
Internal impulse, taken from the momentum change of one cart, since 4.3.A.3.i says the other is equal and opposite: $J_{\text{int}} = m\,\Delta v = (0.50)(1.8) = 0.90\ \mathrm{N \cdot s}$ .
Ratio: $0.01089/0.90 = 0.0121$ , so the external impulse is about $1.2\%$ of the internal one. Treating the total momentum as constant across the collision introduces an error of roughly one per cent, and the collision model of 4.1.A.3.i is comfortably justified. Say so in words on a free-response answer rather than leaving it implied.
Now stretch the interaction to $0.25$ s, a hundred times longer, with everything else the same. $J_{\text{ext}} = (4.357)(0.25) = 1.089\ \mathrm{N \cdot s}$ , which is $121\%$ of the internal impulse.
At that point the momentum of the two-cart system is not approximately constant; it is dominated by the external impulse, and 4.3.B.3 is the statement that applies: momentum is transferred between the system and the environment, by $\Delta\vec{p} = \vec{J}$ .
The lesson is that "momentum is conserved in a collision" is a statement about a ratio, not about collisions as a category. Nothing about the carts changed between the two cases. Only the interval did.
Two ways to make a real experiment satisfy the criterion, which is suggested skill 3.A: level the track so that $\sin\theta = 0$ and the external component vanishes, or measure the velocities within a few milliseconds either side of contact so the interval stays short. The first is better, because it removes the external force rather than limiting its exposure.

The external impulse along the track is $Mg\sin\theta\,\Delta t = 0.0109\ \mathrm{N \cdot s}$ against an internal impulse of $0.90\ \mathrm{N \cdot s}$ , a ratio of $1.2\%$ , so conservation of momentum is justified to about one per cent. Stretch the interaction to $0.25$ s and the external impulse becomes $1.09\ \mathrm{N \cdot s}$ , about $121\%$ of the internal one, and momentum is no longer even approximately constant.

Frequently asked questions

When is momentum conserved in AP Physics C Mechanics?

When the net external force on the system you selected is zero, per essential knowledge 4.3.B.2. In the calculus-based course that follows in one step from the derivative form of the second law: the net external force is the time derivative of the system's momentum, so a zero net external force means a zero derivative and therefore a constant momentum at every instant, not just before and after. If the net external force is nonzero, essential knowledge 4.3.B.3 says momentum is transferred between the system and the environment, and 4.3.A.3.iii gives the amount as the impulse exerted on the system.

Does AP Physics C allow two-dimensional collision problems?

Yes, quantitatively. The boundary statement under Topic 4.3 reads that AP Physics C: Mechanics only expects students to quantitatively analyze collisions and interactions in one or two dimensions, and that three-dimensional collisions may be analyzed qualitatively. That is a different boundary from AP Physics 1's, which allows a quantitative treatment only in one dimension, calls the two-dimensional treatment semiquantitative, and excludes exam questions involving solution of simultaneous equations by name. So a two-dimensional collision with two unknown final speeds, solved from the pair of component equations, is a Physics C question and not a Physics 1 one.

Can you solve simultaneous equations on the AP Physics C exam?

Nothing in the AP Physics C: Mechanics course and exam description excludes it, and the phrase does not appear in the document. The exclusion is an AP Physics 1 boundary statement, which says that exam questions involving solution of simultaneous equations are not included in AP Physics 1 although the exam may still ask students to set the equations up. What the Physics C boundary statement in the same position restricts instead is dimensionality: quantitative analysis is expected in one or two dimensions, and three-dimensional collisions may be analyzed qualitatively. Since a two-dimensional collision with two unknowns cannot be analysed quantitatively without solving the pair of component equations, the expectation follows from that.

Why do internal forces not change a system's momentum?

Because their impulses cancel exactly. Essential knowledge 4.3.A.3.i states that the impulse exerted by one object on a second object is equal and opposite to the impulse exerted by the second object on the first, and attributes it to Newton's third law. In the calculus-based course the argument is stronger than a statement about forces: the two contact forces are equal and opposite at every instant, so their integrals over the same interval are equal and opposite whatever shape the force curves have. The forces need not be constant, and the collision need not be brief. Only forces crossing the system boundary can change the total.

How do you check a two-dimensional momentum answer?

Recompute the velocity of the centre of mass from your final answers and compare it with the value before the collision. Essential knowledge 4.3.A.1.i prints the centre-of-mass velocity as the total momentum divided by the total mass, and 4.3.A.1.ii says it is constant in the absence of a net external force, so the two values must agree. This is faster than re-adding all the momenta because the total mass is a common factor. A mismatch means one of the final velocities or one of the signs is wrong, and it usually shows which component is at fault.

Is momentum conserved in an explosion?

Yes. Essential knowledge 4.1.A.3.iii describes an explosion as an interaction in which forces internal to the system move objects within that system apart, and internal forces cannot change the total momentum. So the fragments' momenta sum to whatever the object's momentum was before, which is zero if it started at rest, and the centre of mass carries on exactly as it was. Kinetic energy behaves differently: an explosion increases it, drawing on stored energy inside the system, so an energy audit and a momentum audit give opposite-looking answers about the same event.

How do you know whether to ignore gravity during a collision?

Compare impulses, not forces. Essential knowledge 4.1.A.3.i defines a collision as an interaction where the forces exerted between the involved objects are much larger than the net external force exerted on those objects during the interaction. In AP Physics C: Mechanics you can turn that into a number: work out the external force component along the direction of interest, multiply by the contact time to get the external impulse, and compare it against the momentum change of one of the colliding objects. A few per cent justifies treating the total momentum as constant, and tens of per cent does not. The same collision can pass the test over a two millisecond contact and fail it over a quarter of a second.