AP Physics C: Mechanics · Unit 4 of 7

Unit 4: Linear Momentum

10-20% of the multiple-choice section4 topics

Topics in this unit

4.1Linear Momentum
4.2Change in Momentum and Impulse
4.3Conservation of Linear Momentum
4.4Elastic and Inelastic Collisions

Linear Momentum is Unit 4 of AP Physics C: Mechanics, 10 to 20 percent of the multiple-choice section over about 11 to 15 class periods. Four topics, six objectives. The net force is the time derivative of momentum, impulse is an integral, and F = ma is a consequence, not a start.

AP Physics: Unit 4 (topics 4.1 Linear Momentum, 4.2 Change in Momentum and Impulse, 4.3 Conservation of Linear Momentum, 4.4 Elastic and Inelastic Collisions). Unit 4 of the current AP Physics C: Mechanics course and exam description, weighted 10 to 20% of the multiple-choice section at about 11 to 15 class periods, the shortest pacing range in the course. Four topics and six learning objectives (4.1.A; 4.2.A, 4.2.B; 4.3.A, 4.3.B; 4.4.A). Exactly one boundary statement, under Topic 4.3: 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. Topics 4.1, 4.2 and 4.4 print none. The calculus differentiator is essential knowledge 4.2.A.1 (net force as the time derivative of momentum, printed on the sheet), 4.2.A.2 and 4.2.B.2.i (impulse as a definite integral of the net force, printed), 4.2.B.2.ii (Newton's second law of motion is a direct result of the impulse-momentum theorem applied to systems with constant mass, giving the chain from dp/dt through m dv/dt to ma, of which only the first equality is printed), and 4.2.B.2.iii (the theorem also describes a system in which the velocity is constant but the mass changes with respect to time, with the net force equal to dm/dt times v, not printed). The AP Physics 1 sheet prints the average-force form of impulse and the ma form of the second law; this sheet prints neither. Topic 4.4 prints no equations at all, and the sheet prints no elastic-collision formula or coefficient of restitution. Both of the CED's sample multiple-choice questions for this unit, Questions 3 and 15, align to 4.2.B and 4.2.B.2, and Question 15 is a variable-mass problem.

What calculus changes: the second law becomes a consequence

The AP Physics 1 equation sheet prints the second law as $\vec{F}_{\text{net}} = \Delta\vec{p}/\Delta t = m\Delta\vec{v}/\Delta t = m\vec{a}$ and impulse as $\vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta\vec{p}$ . The AP Physics C: Mechanics sheet prints neither. It prints these:

\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt} \qquad \vec{J} = \int_{t_1}^{t_2} \vec{F}_{\text{net}}(t)\, dt = \Delta\vec{p}

Essential knowledge 4.2.A.1 states the first in words: the rate of change of a system's momentum is equal to the net external force exerted on that system. Statement 4.2.A.2 defines impulse as the integral of a force exerted on an object or system over a time interval.

Then 4.2.B.2.ii does something the algebra-based course cannot. It reverses the usual logic:

"Newton's second law of motion is a direct result of the impulse-momentum theorem applied to systems with constant mass."

\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}

Read that chain from left to right. $\vec{F} = m\vec{a}$ is not the fundamental statement here. It is what $d\vec{p}/dt$ becomes when $m$ can be pulled through the derivative, and the condition for pulling it through is constant mass. Note also that the sheet prints only the first equality of that chain. The rest is yours to write.

And once you know the condition, you know what happens when it fails. Statement 4.2.B.2.iii gives the other case outright:

"The impulse-momentum theorem also describes the behavior of a system in which the velocity is constant but the mass changes with respect to time."

\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt} = \frac{dm}{dt}\vec{v}

A sand-loaded conveyor belt, a rocket, a truck leaking its load, a chain piling on a scale: none of these can even be posed in an algebra-based course, because $m$ is not a number in them. This equation is not on the equation sheet either. It is in the required content.

So the two directions of calculus split the unit cleanly. Differentiate momentum and you get force, which is 4.2.A.5's statement that the net external force is the slope of a graph of momentum against time. Integrate force and you get impulse, which is 4.2.A.4's statement that the impulse is the area under a graph of net external force against time. Both statements exist in the algebra-based course as slopes and areas of straight-line graphs. Here the graphs may curve.

What the CED requires across Unit 4

Unit 4 of AP Physics C: Mechanics is Linear Momentum. The course and exam description weights it at 10 to 20% of the multiple-choice section and suggests about 11 to 15 class periods, the shortest pacing range in the course. Unit 2 is heaviest at 20 to 25%, Unit 3 next at 15 to 25%, and Units 1, 5, 6 and 7 sit at 10 to 15% each. So Unit 4's band overlaps both of the two heaviest units at its top end and the four lightest at its bottom, which makes it the hardest unit in the course to plan around.

Four topics and six learning objectives.

Topic	Learning objectives	Suggested skills
4.1 Linear Momentum	4.1.A	1.C, 2.B, 2.D, 3.B
4.2 Change in Momentum and Impulse	4.2.A, 4.2.B	1.C, 2.A, 2.C, 3.C
4.3 Conservation of Linear Momentum	4.3.A, 4.3.B	1.B, 1.A, 2.C, 3.A, 3.B
4.4 Elastic and Inelastic Collisions	4.4.A	1.A, 2.B, 2.D, 3.C

Topic 4.3 is the only topic in the unit with five suggested skills, and the only one listing 3.A, creating experimental procedures. Skill 2.A, deriving a symbolic expression, is listed for Topic 4.2 alone.

The CED's framing is that Unit 4 introduces students to the relationships between force, time, impulse, and linear momentum via calculations, data analysis, designing experiments, and making predictions, and that using the law of conservation of linear momentum to analyze physical situations provides students with a more complete picture of forces and opportunities to revisit misconceptions surrounding Newton's third law.

The unit's "Building the Science Practices" page names four skills, 1.B, 2.B, 2.D and 3.A, and leans heavily on experiment: it says inquiry learning, critical thinking and problem-solving skills are best developed when scientific inquiry experiences are designed and implemented with increasing student involvement, and describes practising data analysis by plotting linearized graphs and using the best fit line to make claims about the physical scenario.

The "Preparing for the AP Exam" note ties Unit 4 to the third free-response question, the Experimental Design and Analysis question. It notes that students often struggle with knowing where to start when designing an experiment and benefit from scaffolded opportunities to determine the data needed to answer a scientific question, and that the exam's version of that question also asks students to linearize and analyze data.

The Unit 4 equations, and what the sheet prints

Unit 4 prints nine equation entries across its four topics, and Topic 4.4 prints none at all.

Equation	Where	Printed on the sheet
$\vec{p} = m\vec{v}$	4.1.A.1	yes
$\vec{F}_{\text{net}} = \dfrac{d\vec{p}}{dt}$	4.2.A.1	yes
$\vec{J} = \int_{t_1}^{t_2} \vec{F}_{\text{net}}(t)\, dt$	4.2.A.2	yes
$\Delta\vec{p} = \vec{p} - \vec{p}_0$	4.2.B.1	no
$\vec{J} = \int_{t_1}^{t_2} \vec{F}_{\text{net}}(t)\, dt = \Delta\vec{p}$	4.2.B.2.i	yes
$\vec{F}_{\text{net}} = \dfrac{d\vec{p}}{dt} = m\dfrac{d\vec{v}}{dt} = m\vec{a}$	4.2.B.2.ii	the first equality only
$\vec{F}_{\text{net}} = \dfrac{d\vec{p}}{dt} = \dfrac{dm}{dt}\vec{v}$	4.2.B.2.iii	the first equality only
$\vec{v}_{\text{cm}} = \dfrac{\sum \vec{p}_i}{\sum m_i} = \dfrac{\sum (m_i\vec{v}_i)}{\sum m_i}$	4.3.A.1.i	yes
$\vec{J} = \Delta\vec{p}$	4.3.A.3.iii	yes, as part of the impulse line

Three things are worth pulling out of that.

Topic 4.4 has no equations. Elastic and Inelastic Collisions carries five essential knowledge statements and not one formula, and the sheet prints nothing for it either. There is no elastic-collision velocity formula, no coefficient of restitution, and no relative-velocity result anywhere in this course. An elastic collision is solved by writing conservation of momentum and conservation of kinetic energy as two equations and solving them, every time. Statement 4.4.A.1 is the definition you start from: an elastic collision is one in which the initial kinetic energy of the system equals the final kinetic energy of the system.

There is no average-force version of impulse. The AP Physics 1 sheet prints $\vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta\vec{p}$ . This one does not. If a question wants an average force you get it by dividing the impulse you integrated by the time interval, which is the definition read backwards rather than a formula you are handed.

The second law is printed in one form only. The sheet's momentum line is $\vec{F}_{\text{net}} = d\vec{p}/dt$ , full stop. The expansions to $m\,d\vec{v}/dt$ , to $m\vec{a}$ and to $(dm/dt)\vec{v}$ all live in the framework and none of them is printed. Elsewhere the sheet gives $\vec{a}_{\text{sys}} = \sum \vec{F}/m_{\text{sys}}$ under Unit 2, which is the constant-mass case rearranged, so the two printed lines together are the two halves of 4.2.B.2.ii.

One place to look that students miss: the Table of Information is more than the Mechanics table. Its separate Calculus table prints $\int x^n\, dx = \frac{1}{n+1}x^{n+1}$ for $n \neq -1$ along with the derivatives and integrals of $e^{ax}$ , $\ln(ax)$ , $\sin(ax)$ and $\cos(ax)$ , which covers every impulse integral a Unit 4 question is likely to set.

Unit 4's one boundary statement, quoted whole

Unit 4 prints exactly one boundary statement, and it sits under Topic 4.3. Topics 4.1, 4.2 and 4.4 print none.

"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."

Both sentences matter. Calculations are capped at two dimensions. Three-dimensional collisions are not excluded from the course, they are moved to qualitative reasoning, which is exactly the same split Topic 1.5 makes for kinematics: quantitative in two dimensions, qualitative in three.

What is not fenced anywhere in this unit is the variable-mass case. No boundary statement limits it and 4.2.B.2.iii states it as required content. What bounds it in practice is the situation the statement names: velocity constant, mass changing with time. That is narrower than the general rocket equation, and it is the case where $d\vec{p}/dt = (dm/dt)\vec{v}$ has one term rather than two.

The rest of the unit's fencing is in the careful wording of the conservation statements, and it is worth reading them slowly because they are built to be read as a decision procedure rather than a rule.

4.3.B.1 and 4.3.A.1.ii give the physics: momentum is conserved in all interactions, and the velocity of a system's centre of mass is constant in the absence of a net external force.
4.3.A.3.ii gives you the choice: a system may be selected so that the total momentum of that system is constant.
4.3.B.2 and 4.3.B.3 give the test: if the net external force on the selected system is zero, the total momentum of the system is constant; if it is nonzero, momentum is transferred between the system and the environment.
4.3.A.3.iii gives the accounting when it is not: if the total momentum of a system changes, that change will be equivalent to the impulse exerted on the system.

So momentum conservation here is not an assumption about collisions. It is a conclusion you earn by drawing the system boundary in the right place, and 4.1.A.3.i says why that boundary usually works: a collision is a model for an interaction where the forces between the objects in the system are much larger than the net external force on those objects during the interaction.

Traps that span more than one topic

Momentum conservation and kinetic energy conservation are not a package. Statement 4.3.B.1 says momentum is conserved in all interactions. Statement 4.4.A.3 says an inelastic collision is one in which the total kinetic energy of the system decreases. Both are true of the same collision. Momentum survives; kinetic energy need not.

Perfectly inelastic is one case of inelastic, not a synonym. Statement 4.4.A.5 defines a perfectly inelastic collision as one in which the objects stick together and move with the same velocity afterward. A collision can be inelastic without the objects sticking, which is 4.4.A.4's statement that some of the initial kinetic energy is transformed by nonconservative forces into other forms of energy.

The area under a curved force graph is not base times height. Statement 4.2.A.4 says the impulse is the area under the curve of net external force against time. For a force pulse that rises and falls, using the peak force multiplied by the duration overestimates the impulse, and the size of the error depends on the shape.

Internal impulses cancel and external ones do not. Statement 4.3.A.3.i says the impulse exerted by one object on a second is equal and opposite to the impulse exerted by the second on the first, and calls it a direct result of Newton's third law. That is why internal forces drop out of a system's total momentum and why the choice of system boundary decides the whole problem.

Mass changing with time is not mass changing with position. Statement 4.2.B.2.iii names a system whose velocity is constant and whose mass changes with respect to time. Reaching for $(dm/dt)\vec{v}$ in a problem where the speed is also changing drops a term, because then both factors in $\vec{p} = m\vec{v}$ are functions of time.

If you want the algebra-based treatment of this unit

AP Physics 1 has a Unit 4 with the same title and the same four topic titles. If that is your course, the page you want is AP Physics 1 Unit 4: Linear Momentum. If you are in AP Physics C: Mechanics, this is the page.

	AP Physics 1 Unit 4	AP Physics C: Mechanics Unit 4
Multiple-choice weighting	10 to 15%	10 to 20%
Topic titles 4.1 to 4.4	identical	identical
Second law on the sheet	$\vec{F}_{\text{net}} = \Delta\vec{p}/\Delta t = m\vec{a}$	$\vec{F}_{\text{net}} = d\vec{p}/dt$
Impulse on the sheet	$\vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta\vec{p}$	$\vec{J} = \int_{t_1}^{t_2} \vec{F}_{\text{net}}(t)\, dt = \Delta\vec{p}$
Variable-mass systems	not in the course	essential knowledge 4.2.B.2.iii
Force from a curved $F$ against $t$ graph	area of straight-line segments	a definite integral

All four topic titles match, so the two pages look identical from their headings and are not. The clearest way to tell which course a momentum question belongs to is to look at the force: if it is a number, either course can do it, and if it is a function of time, only this one can.

If you are in AP Physics 1, everything on this page about integrals and derivatives is beyond your exam, and the Physics 1 unit page is written for you. The conservation half of the unit is genuinely shared, so the site's conservation of momentum guide, its impulse-momentum theorem guide, its guide to what is conserved in a collision and the momentum and collision calculator serve both courses. All of them assume a constant mass and, where impulse appears, an average force, which is exactly the regime in which the two courses agree.

How Unit 4 is assessed

The AP Physics C: Mechanics exam is 3 hours long. Section I is 42 multiple-choice questions in 85 minutes for 50% of the score. Section II is 4 free-response questions in 95 minutes for the other 50%, one of each type in a fixed order: Mathematical Routines, Translation Between Representations, Experimental Design and Analysis, and Qualitative/Quantitative Translation. A four-function, scientific, or graphing calculator is allowed on both sections. Unit 4's Progress Check runs about 18 multiple-choice questions and 4 free-response questions, one of each type.

The CED's own sample questions are unusually informative about this unit, because both of its sample multiple-choice questions land on the same essential knowledge statement. Question 3 and Question 15 both align to learning objective 4.2.B and essential knowledge 4.2.B.2, with skills 2.B and 2.C respectively. In other words, both of the sample multiple-choice questions for Unit 4 are about the impulse-momentum theorem, and neither is about collisions.

Question 3 gives the net force on an object as $F(t) = At^2 + B$ with numerical constants and asks for the change in momentum over a stated interval, which is $\int F\,dt$ and nothing else. Question 15 puts a truck initially at rest with a load of sand, pushes it with a constant force while the sand leaves at a constant rate, and asks which graph best shows the truck's velocity against time. That second question is 4.2.B.2.iii in graph form: while the mass is falling the same force produces a growing acceleration, so the velocity curve bends upward, and once the sand is gone the mass is constant and the curve becomes a straight line. No algebra-based course can ask it.

Unit 4 also appears in one of the four sample free-response questions, the Experimental Design and Analysis question, which aligns to 4.2.B and 4.4.A alongside objectives from Units 2 and 3. Its scenario is two carts on a horizontal track, one given an initial speed and one at rest with a spring attached, and it asks what quantities to measure, how to reduce experimental uncertainty, and which axes will linearise the data.

Four optional sample instructional activities are listed: one on Topic 4.2, two on Topic 4.3, one on Topic 4.4, none on 4.1. The Topic 4.2 one presents a scenario where an object's motion changes and has students find the applied force three separate ways, using Newton's laws, the work-energy theorem, and the impulse-momentum theorem, then draw a free-body diagram. Three routes to the same number is the point.

Impulse from a force that is a function of time

A 0.50 kg ball at rest is struck, and the net force on it during contact is $F(t) = c\,t(T - t)$ for $0 \leq t \leq T$ , with $c = 2.4 \times 10^5 \ \mathrm{N/s^2}$ and $T = 0.050$ s. Find (a) the impulse delivered, (b) the ball's speed just after contact, (c) the peak force and the average force, and (d) the momentum halfway through the contact.

Declare the convention: positive is the direction of the blow, and the ball starts with $p_0 = 0$ .
(a) Statement 4.2.A.2 defines the impulse as the integral of the net force over the interval: $J = \int_0^T c\,t(T-t)\, dt = c\left[\dfrac{Tt^2}{2} - \dfrac{t^3}{3}\right]_0^T = cT^3\left(\dfrac{1}{2} - \dfrac{1}{3}\right) = \dfrac{cT^3}{6}$ .
Do the symbolic answer first, then the number. $T^3 = (0.050)^3 = 1.25 \times 10^{-4} \ \mathrm{s^3}$ , so $J = \dfrac{(2.4 \times 10^5)(1.25 \times 10^{-4})}{6} = \dfrac{30}{6} = 5.0$ N s.
(b) Statement 4.2.B.2.i gives $\vec{J} = \Delta\vec{p}$ , and $p_0 = 0$ , so $p = 5.0$ kg m/s and $v = \dfrac{p}{m} = \dfrac{5.0}{0.50} = 10$ m/s.
(c) The force peaks where $dF/dt = 0$ , which is $c(T - 2t) = 0$ , so $t = T/2 = 0.025$ s. There $F = c\left(\dfrac{T}{2}\right)\left(\dfrac{T}{2}\right) = \dfrac{cT^2}{4} = \dfrac{(2.4 \times 10^5)(2.5 \times 10^{-3})}{4} = 150$ N.
The average force is the impulse divided by the interval: $F_{\text{avg}} = \dfrac{J}{T} = \dfrac{5.0}{0.050} = 100$ N. Note that this course's sheet prints no $\vec{J} = \vec{F}_{\text{avg}}\Delta t$ line, so the average force is something you extract from the integral rather than something you start with.
Check the ratio symbolically: $\dfrac{F_{\text{peak}}}{F_{\text{avg}}} = \dfrac{cT^2/4}{cT^2/6} = \dfrac{6}{4} = 1.5$ . A parabolic pulse always has a peak exactly 1.5 times its mean, independent of $c$ and $T$ , and 150/100 confirms the arithmetic.
That ratio is also the warning: multiplying the peak force by the contact time would have given $150 \times 0.050 = 7.5$ N s, half again too large, and a speed of 15 m/s instead of 10 m/s.
(d) Integrate to the halfway point rather than assuming half the impulse arrives in half the time. $p(t) = c\left(\dfrac{Tt^2}{2} - \dfrac{t^3}{3}\right)$ , so at $t = T/2$ , $p = c\left(\dfrac{T^3}{8} - \dfrac{T^3}{24}\right) = \dfrac{cT^3}{12} = 2.5$ kg m/s.
It is exactly half, and that is a consequence of the pulse being symmetric about $t = T/2$ , not a general rule. For an asymmetric pulse it would not be.

(a) $J = cT^3/6 = 5.0$ N s. (b) $v = 10$ m/s. (c) Peak force 150 N at $t = 0.025$ s, average force 100 N, a ratio of exactly 1.5 for any parabolic pulse. (d) $p = 2.5$ kg m/s at $t = T/2$ , half the total, because this pulse is symmetric.

A variable-mass system, which the algebra-based course cannot pose

Sand falls from a stationary hopper onto a horizontal conveyor belt at a constant rate $dm/dt = 15$ kg/s. The belt moves at a constant $v = 2.0$ m/s. Find (a) the horizontal force needed to keep the belt moving at constant speed, (b) the power the motor delivers, (c) the rate at which the sand's kinetic energy increases, and (d) account for the difference.

Declare the convention: positive is the belt's direction of travel. The sand lands with zero horizontal velocity and leaves the loading point moving at 2.0 m/s with the belt.
(a) This is the case named by 4.2.B.2.iii, a system whose velocity is constant while its mass changes with respect to time. The belt's speed does not change, so $d\vec{v}/dt = 0$ and the whole of $d\vec{p}/dt$ comes from the mass term: $F_{\text{net}} = \dfrac{dp}{dt} = \dfrac{dm}{dt}v = (15)(2.0) = 30$ N.
Notice what would have gone wrong with $F = ma$ . The belt's acceleration is zero, so $ma = 0$ , and the answer would have been that no force is needed. The reason $F = ma$ fails is stated at 4.2.B.2.ii: it is the impulse-momentum theorem applied to systems with constant mass, and this system does not have one.
(b) Power delivered by that force at the belt's speed: $P = Fv = (30)(2.0) = 60$ W.
(c) The sand's kinetic energy grows as new sand is brought up to speed: $\dfrac{dK}{dt} = \dfrac{1}{2}\dfrac{dm}{dt}v^2 = \dfrac{1}{2}(15)(2.0)^2 = 30$ W.
(d) So the motor supplies 60 W and the sand gains kinetic energy at 30 W. The missing 30 W, exactly half, is dissipated while each grain slips on the belt before it matches the belt's speed.
That half is not a coincidence of these numbers. Symbolically, $Fv = (dm/dt)v^2$ and $dK/dt = \frac{1}{2}(dm/dt)v^2$ , so the ratio is always 2 for this situation regardless of the rate or the speed.
One more figure for scale: after 4.0 s of loading onto an initially empty belt there are $(15)(4.0) = 60$ kg of sand on it, carrying $p = mv = (60)(2.0) = 120$ kg m/s, which is the same as $\int F\, dt = (30)(4.0) = 120$ N s. The impulse and the momentum agree, which is 4.2.B.2.i as a check on the whole calculation.

(a) 30 N, from $F = (dm/dt)v$ . (b) 60 W. (c) 30 W. (d) The other 30 W is dissipated as the sand slips before matching the belt's speed, and the factor of 2 between the two rates holds for any loading rate and belt speed. $F = ma$ gives zero here, because 4.2.B.2.ii holds only for constant mass.

Frequently asked questions

How much of the AP Physics C Mechanics exam is Unit 4?

Unit 4, Linear Momentum, is weighted at 10 to 20% of the multiple-choice section of the AP Physics C: Mechanics exam, and the course and exam description suggests about 11 to 15 class periods for it, the shortest pacing range in the course. Unit 2 is heaviest at 20 to 25% and Unit 3 next at 15 to 25%, while Units 1, 5, 6 and 7 each sit at 10 to 15%. Unit 4's band is the widest-overlapping in the course: at its top end it rivals Unit 3 and at its bottom end it matches the four lightest units. Its Progress Check runs about 18 multiple-choice questions and 4 free-response questions.

Why is F = dp/dt rather than F = ma in AP Physics C?

Because the course treats the momentum form as the more general statement and derives the familiar one from it. Essential knowledge 4.2.A.1 says the rate of change of a system's momentum equals the net external force on that system, and the equation sheet prints exactly that. Statement 4.2.B.2.ii then says Newton's second law of motion is a direct result of the impulse-momentum theorem applied to systems with constant mass, and shows the chain from the derivative of momentum through mass times the derivative of velocity to mass times acceleration. The condition is constant mass. Statement 4.2.B.2.iii gives the other case, a system whose velocity is constant while its mass changes with time, where the net force equals the rate of change of mass multiplied by the velocity.

What is the difference between AP Physics C Unit 4 and AP Physics 1 Unit 4?

All four topic titles are identical, so the difference is the mathematics and one extra idea. AP Physics 1 is given impulse as an average force multiplied by a time interval and the second law as the change in momentum over the change in time, both printed on its equation sheet. AP Physics C: Mechanics is given impulse as a definite integral of the net force over time and the second law as the time derivative of momentum, and prints neither of the algebra-based forms. Physics C also adds variable-mass systems at essential knowledge 4.2.B.2.iii, which cannot be posed in an algebra-based course at all. Weightings are 10 to 15% for Physics 1 and 10 to 20% for Physics C.

Which Unit 4 equations are on the AP Physics C Mechanics equation sheet?

The sheet prints momentum as mass times velocity, the net force as the time derivative of momentum, impulse as the definite integral of the net force over time set equal to the change in momentum, and the velocity of the centre of mass as the total momentum divided by the total mass. It does not print the definition of the change in momentum as final minus initial, any average-force version of impulse, or the expanded forms of the second law, including the constant-mass chain that ends in mass times acceleration and the variable-mass version. Topic 4.4, Elastic and Inelastic Collisions, has no equations at all in the framework and none on the sheet, so elastic collisions are solved by writing conservation of momentum and conservation of kinetic energy as two simultaneous equations.

Does AP Physics C Mechanics cover three-dimensional collisions?

Only qualitatively. The single boundary statement in Unit 4, 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. So calculations stop at two dimensions while reasoning does not. That is the same split the course makes for kinematics, where the Topic 1.5 boundary statement caps quantitative analysis of motion at two dimensions. Topics 4.1, 4.2 and 4.4 print no boundary statement, which means nothing in the framework fences off the variable-mass case in Topic 4.2.