AP Physics C: Mechanics · Topic 4.1

Topic 4.1: Linear Momentum

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

Linear momentum is mass times velocity, and it is a vector. In AP Physics C: Mechanics it is the primitive quantity: the net force is the time derivative of momentum, and force equals mass times acceleration is what that reduces to when the mass is constant.

AP Physics: Unit 4 (topics 4.1 Linear Momentum). Topic 4.1 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 (both figures differ from AP Physics 1's Unit 4, which is 10 to 15% at about 10 to 15 class periods). One learning objective, 4.1.A, and six essential-knowledge statements counting sub-statements: 4.1.A.1 defines p = mv; 4.1.A.2 makes it a vector along the velocity; 4.1.A.3 applies it to collisions and explosions; 4.1.A.3.i defines a collision by internal forces being much larger than the net external force during the interaction; 4.1.A.3.ii grants the object model because only initial and final states are analyzed; 4.1.A.3.iii defines an explosion. Suggested skills 1.C, 2.B, 2.D and 3.B. No boundary statement: Unit 4 prints exactly one, under Topic 4.3. Verified against both CEDs page by page: all six essential-knowledge statements are word for word identical to AP Physics 1's Topic 4.1, so this page is written around what the rest of the calculus-based unit does with momentum rather than around a difference in this topic's own content. The two framework differences are that AP Physics 1's Topic 4.1 prints a boundary statement, that the general term momentum refers specifically to linear momentum, and this one prints none, and that the suggested skills swap 2.C for 2.D. Sheet detail verified against the printed Table of Information: the C: Mechanics sheet prints p = mv, F_net = dp/dt, the impulse integral, v_cm, r_cm as an integral of r dm over the integral of dm, and lambda as the derivative of m with respect to arc length, and it does NOT print m a as a written term anywhere, while the AP Physics 1 sheet prints F_net = delta p over delta t = m delta v over delta t = m a on one line. Not printed on either sheet: K = p squared over 2m. Neither of the framework's two Unit 4 sample multiple-choice questions, 3 and 15, aligns to 4.1.A; both align to 4.2.B.

What Topic 4.1 requires

Topic 4.1 carries one learning objective and six essential-knowledge statements counting sub-statements.

Objective	What it asks
4.1.A	Describe the linear momentum of an object or system

Statement	What it says
4.1.A.1	Linear momentum is defined by the equation $\vec{p} = m\vec{v}$
4.1.A.2	Momentum is a vector quantity and has the same direction as the velocity
4.1.A.3	Momentum can be used to analyze collisions and explosions
4.1.A.3.i	A collision is a model for an interaction where the forces exerted between the involved objects in the system are much larger than the net external force exerted on those objects during the interaction
4.1.A.3.ii	As only the initial and final states of a collision are analyzed, the object model may be used to analyze collisions
4.1.A.3.iii	An explosion is a model for an interaction in which forces internal to the system move objects within that system apart

Suggested skills: 1.C, 2.B, 2.D and 3.B. Topic 4.1 prints no boundary statement. Unit 4 prints exactly one in total, under Topic 4.3.

Unit 4 as a whole is weighted 10 to 20% of the multiple-choice section over about 11 to 15 class periods, and those two figures are not the AP Physics 1 figures for the identically named unit, which are 10 to 15% and about 10 to 15 class periods. Four topics, six learning objectives, and thirty-three essential-knowledge statements once sub-statements are counted individually: six here, ten under 4.2, twelve under 4.3 and five under 4.4.

Be told this plainly, because it will save you time. All six of those statements are word for word the same as the six under AP Physics 1's Topic 4.1. The only differences in the framework are that AP Physics 1's version of this topic prints a boundary statement and this one does not, and that the suggested skill list swaps 2.C for 2.D. So if you have already learned $\vec{p} = m\vec{v}$ in the algebra-based course, you have learned the content of this topic. What changes is everything the rest of the unit does with it, and that begins on the next line.

In this course momentum is the primitive, not the shortcut

Open the AP Physics C: Mechanics equation sheet and look for $\vec{F} = m\vec{a}$ . It is not there. What the printed Table of Information gives you for translational dynamics is these two lines:

\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}} = \frac{\vec{F}_{\text{net}}}{m_{\text{sys}}}

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

The AP Physics 1 sheet, by contrast, prints $\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t} = m\frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$ on a single line, with $m\vec{a}$ as the last term.

Essential knowledge 4.2.B.2.ii states the logic explicitly: Newton's second law of motion is a direct result of the impulse-momentum theorem applied to systems with constant mass. Read that as a claim about which statement is derived from which. The law is $\vec{F}_{\text{net}} = d\vec{p}/dt$ . Pull the constant $m$ out of the derivative and $m\,d\vec{v}/dt$ appears, and $d\vec{v}/dt$ is the acceleration, so $m\vec{a}$ drops out as a special case.

That ordering is the reason Topic 4.1 is worth its own page in this course even though its wording is identical to the algebra-based version. In AP Physics 1, momentum arrives in Unit 4 as a new tool for collisions, after $F = ma$ has been the law since Unit 2. Here, momentum is the state variable the law is written about, and it was already needed in Unit 2: Topic 2.5, Newton's second law is where the two forms first meet.

Three consequences you can use:

A net force is a statement about how fast $\vec{p}$ is changing. If you can write $\vec{p}(t)$ , you can differentiate it and you have the net force with no free-body diagram at all. Worked example 1 does this.
When the mass is not constant, $m\vec{a}$ is simply the wrong equation and the derivative form is the only one that survives. That is Topic 4.2 and statement 4.2.B.2.iii.
Momentum and kinetic energy are the two integrals of the same force. The sheet prints $\vec{J} = \int \vec{F}_{\text{net}}(t)\,dt$ and $W = \int_a^b \vec{F} \cdot d\vec{r}$ . Integrate a force over time and you get a change in momentum; integrate it over displacement and you get a change in kinetic energy. Worked example 3 makes that concrete.

Momentum as a function of time (4.1.A.1 with a calculus reading)

Statement 4.1.A.1 defines $\vec{p} = m\vec{v}$ . In a course where velocity is routinely handed to you as a function, that definition immediately produces a function:

\vec{p}(t) = m\vec{v}(t)

and then, by 4.2.A.1, the net force at any instant is its slope. Statement 4.2.A.5 says the same thing in graph language: the net external force exerted on a system is equal to the slope of a graph of the momentum of the system as a function of time.

This is the reading that suggested skill 1.C, creating qualitative sketches of graphs that represent features of a model, is asking for on this topic. Given a described motion, you should be able to sketch $p$ against $t$ with the right slope sign and the right curvature. A few anchors that come up:

Constant velocity. $p$ is a horizontal line. Zero slope, zero net force.
Constant net force. $p$ is a straight line with nonzero slope. Note that $p$ against $t$ is straight here even for a body whose speed is changing, which is what makes this graph easier to read than $v$ against $t$ when the mass is not constant.
A force that dies away. $p$ rises and flattens toward a horizontal asymptote. This is the shape of a cart slowed by a resistive force, and Topic 2.9, Resistive Forces is where that force law lives. For linear drag, $\vec{F} = -b\vec{v} = -(b/m)\vec{p}$ , so the net force is proportional to the momentum itself and the decay is exponential.
A momentum maximum. Wherever $p$ peaks, the slope is zero, so the net force is zero at that instant. Worked example 1 uses this.

The units of momentum are $\mathrm{kg \cdot m/s}$ . The units of impulse are $\mathrm{N \cdot s}$ . Those are the same combination, which is 4.2.B.2.i in disguise, and checking it is a fast way to catch an algebra slip.

The object model, and the criterion for using it (4.1.A.3.i and 4.1.A.3.ii)

Two of the three sub-statements under 4.1.A.3 are permissions, and both come with conditions worth reading carefully.

4.1.A.3.i defines what counts as a collision and it does so with an inequality, not a duration: a collision is a model for an interaction where the forces exerted between the involved objects in the system are much larger than the net external force exerted on those objects during the interaction.

Note what is being compared. Not force against force in the abstract, and not "the collision is short". Internal forces against the net external force, during the interaction. Gravity acting on two colliding carts is real the whole time; the model works because the contact forces during the bump dwarf it. In Physics C you can turn that into a number, because the external impulse is $\int \vec{F}_{\text{ext}}\,dt$ and you can estimate it. Worked example 3 in Topic 4.3 carries that estimate out for carts colliding on an incline.

4.1.A.3.ii is the permission to forget the internal structure: as only the initial and final states of a collision are analyzed, the object model may be used to analyze collisions. So during a collision you do not need to know how the deformation propagated, where the contact patch was, or how the force varied, provided you only want the before and after. Notice that this permission is exactly what fails the moment a question asks about the middle of the collision, and Physics C questions do ask about the middle. When they do, you are back to $\vec{F}_{\text{net}}(t)$ and $\vec{p}(t)$ , and the object model has been replaced by a force function.

4.1.A.3.iii covers explosions: an explosion is a model for an interaction in which forces internal to the system move objects within that system apart. Momentum handles it with no changes at all, because the internal forces cancel in the total whichever way they point. Kinetic energy does not: an explosion increases it, drawing on stored energy inside the system, which is why the two quantities give different answers and why 4.4 exists.

The momentum of a system, and of a body you cannot treat as a point

Statement 4.3.A.2 says the total momentum of a system is the sum of the momenta of the system's constituent parts, and 4.3.A.1.i gives the printed equation for the velocity of its centre of mass:

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

Rearranged, that says $\sum \vec{p}_i = M_{\text{total}}\,\vec{v}_{\text{cm}}$ . A system of any number of parts has the momentum of a single object of the total mass moving at the centre-of-mass velocity. That equation is identical on the AP Physics 1 sheet.

Where the two courses part company is what you are able to do when the body is continuous rather than a set of labelled masses. The C: Mechanics sheet prints, in the same appendix, two more lines that the AP Physics 1 sheet does not have:

\vec{r}_{\text{cm}} = \frac{\int \vec{r}\, dm}{\int dm} \qquad \lambda = \frac{d}{d\ell}m(\ell)

With those you can find the mass and the centre of mass of a rod whose density varies along its length, and therefore its momentum when it translates. Topic 2.1, Systems and Center of Mass is where those integrals are set up in detail; worked example 2 below applies them to a momentum question so you can see the join.

One caution on notation. $\int dm$ in that printed line is just the total mass $M$ . Students sometimes read it as a quantity to be cancelled against the numerator. It is not: the numerator is a weighted integral and the denominator is a plain one.

Momentum against kinetic energy, in a course that prints both integrals

Both quantities are built from mass and velocity, both are conserved under the right conditions, and choosing the wrong one costs you the whole question. The distinction is sharper in Physics C than in the algebra-based course, because here you can see it in the two printed integrals.

	Linear momentum	Kinetic energy
Definition	$\vec{p} = m\vec{v}$ , statement 4.1.A.1	$K = \frac{1}{2}mv^2$
Kind	vector, so it has a direction and components	scalar, so no direction and no cancellation
Built by integrating the net force over	time: $\vec{J} = \int \vec{F}_{\text{net}}(t)\,dt$	displacement: $W = \int_a^b \vec{F} \cdot d\vec{r}$
Constant when	the net external force is zero, 4.3.B.2	the total work done on the system is zero
In a perfectly inelastic collision	unchanged	decreases, 4.4.A.3
In an explosion	unchanged	increases, 4.1.A.3.iii

The relation between them, worth memorising because it is not printed: $K = \frac{p^2}{2m}$ , so $p = \sqrt{2mK}$ . Both follow in one line from the two definitions.

The consequence of the middle row is the practical one. Deliver the same impulse to two different masses and they end up with the same momentum but different kinetic energies, the smaller mass getting more. Do the same amount of work on them and they end up with the same kinetic energy but different momenta, the larger mass getting more. Worked example 3 runs both directions with numbers. This is also the honest answer to "why does a heavy slow truck do more damage than a light fast bullet with the same momentum", and the answer is that they do not have the same kinetic energy.

Reference frames, and why the momentum of a system is never one number

Velocity is measured relative to a frame, so momentum is too. The printed exam conventions in the same appendix as the equation sheet state that the frame of reference of any problem is assumed to be inertial unless otherwise stated, which tells you the frames you will be asked to work in are the ones where $\vec{F}_{\text{net}} = d\vec{p}/dt$ holds unchanged.

Two habits keep this from becoming a source of sign errors.

Declare a positive direction before you write a single momentum. In one dimension a momentum is a signed number, and the sign is the whole content of 4.1.A.2. Once declared, do not change it partway through, including in the "after" line of a collision.

A momentum of zero is not a special physical state. For any system there is a frame in which the total momentum is zero, namely the one moving with the centre of mass, and 4.3.A.1.ii says the centre of mass moves at constant velocity in the absence of a net external force, so that frame is inertial and legitimate. In it, the two carts of a head-on collision arrive with equal and opposite momenta whatever their masses. The kinetic energy in that frame is smaller than in any other, by exactly $\frac{1}{2}M v_{\text{cm}}^2$ , and that leftover is the part no collision can remove. That is where the "how much kinetic energy must a perfectly inelastic collision lose" result in Topic 4.4 comes from.

What does not change with the frame: whether momentum is conserved, and whether a collision is elastic. Both are statements about a difference between two states, so the frame cancels.

How Topic 4.1 is tested, and where the algebra-based page is

Topic 4.1 is not the topic that gets its own question. Of the fifteen sample multiple-choice questions in the course framework, the two that sit in Unit 4 are questions 3 and 15, and the published alignment table puts both under learning objective 4.2.B, not 4.1.A. Sample free-response question 3, the Experimental Design and Analysis question, aligns to a list of objectives that includes 4.2.B and 4.4.A, again not 4.1.A.

Read that as a hint about how 4.1 appears: as the first line of a longer problem. You will be asked for a momentum in order to be asked for something else. The skills listed for it fit that role. Skill 2.B, calculating an unknown quantity from known quantities, and skill 2.D, predicting new values or factors of change using functional dependence between variables, are both about getting a number out and then reasoning about how it scales. Skill 2.D replaces the 2.C that AP Physics 1 lists here, which is a small but real signal: this course wants you to say how $p$ and $K$ scale with mass and speed, not just to compare two situations.

The unit's essential questions in the framework are worth reading as a set, because all three are momentum questions in disguise: why does water moving one way push a ship the other, why are cannon barrels so much longer and heavier than cannonballs, and why might a person land in the water instead of on the dock when trying to exit a canoe.

If you are in the algebra-based course, read the [AP Physics 1 Topic 4.1 page](/ap-physics-1/unit-4-linear-momentum/4-1-linear-momentum) instead. It covers the same six statements at the level your exam asks for, with the reference-frame and vector-sign work in algebra, and it is the page written for your course. This page is for AP Physics C: Mechanics students, who need momentum as a function of time, the derivative form of the second law, and the centre-of-mass integrals.

Where to go next. Topic 4.2 is the calculus core of the unit and the reason the sheet looks the way it does. Topic 4.3 turns $d\vec{p}/dt$ into a conservation law. The conservation of momentum guide owns the before-and-after routine, the momentum collision calculator checks arithmetic, and the collision lab lets you set the two masses, the two initial velocities and a coefficient of restitution and watch the momentum total stay put while the kinetic energy does not. The Unit 4 hub lists all four topics.

Momentum as a function of time, and reading the net force off its slope

A 2.0 kg cart moves along a straight track with velocity $v(t) = \left(3.0\ \mathrm{m/s^2}\right)t - \left(0.50\ \mathrm{m/s^3}\right)t^2$ , valid from $t = 0$ to $t = 4.0$ s. Take the direction of motion at $t = 0$ as positive. Find (a) $p(t)$ , (b) the net force at $t = 1.0$ s and at $t = 4.0$ s, (c) the instant at which the momentum is largest and its value there, and (d) the change in momentum from $t = 0$ to $t = 4.0$ s, checked two ways.

(a) Statement 4.1.A.1 gives $p = mv$ , and $m$ is constant, so multiply through: $p(t) = (2.0)\left(3.0t - 0.50t^2\right) = \left(6.0t - 1.0t^2\right)\ \mathrm{kg \cdot m/s}$ with $t$ in seconds.
(b) Statement 4.2.A.1 gives $F_{\text{net}} = dp/dt$ , and the sheet's Calculus table prints the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ . Differentiating: $F_{\text{net}}(t) = 6.0 - 2.0t$ , in newtons.
At $t = 1.0$ s, $F_{\text{net}} = 6.0 - 2.0 = 4.0$ N, positive, so the momentum is still growing.
At $t = 4.0$ s, $F_{\text{net}} = 6.0 - 8.0 = -2.0$ N. The net force has reversed. Nothing in the problem mentions a force at all, and there is no free-body diagram: the force came out of the momentum function.
(c) The momentum is largest where its slope is zero, which by 4.2.A.5 is where the net force is zero: $6.0 - 2.0t = 0$ gives $t = 3.0$ s.
$p(3.0) = 6.0(3.0) - 1.0(3.0)^2 = 18.0 - 9.0 = 9.0\ \mathrm{kg \cdot m/s}$ . Check the endpoints so you know it is a maximum and not an inflection: $p(0) = 0$ and $p(4.0) = 24.0 - 16.0 = 8.0\ \mathrm{kg \cdot m/s}$ , both smaller.
(d) Direct subtraction: $\Delta p = p(4.0) - p(0) = 8.0 - 0 = 8.0\ \mathrm{kg \cdot m/s}$ , which is 4.2.B.1.
Now as an impulse, which is 4.2.B.2.i: $\Delta p = \int_0^{4.0} (6.0 - 2.0t)\,dt = \left[6.0t - t^2\right]_0^{4.0} = 24.0 - 16.0 = 8.0\ \mathrm{kg \cdot m/s}$ . The two agree, as they must.
Sketch check for skill 1.C. The $p$ against $t$ curve is a downward parabola through the origin, peaking at $(3.0,\ 9.0)$ and falling to $8.0$ at $t = 4.0$ . Its slope starts at $+6.0$ N, passes through zero at $t = 3.0$ s, and is $-2.0$ N at the end, so the force graph is a straight line crossing zero at $t = 3.0$ s. The area under that straight line from 0 to 4.0 s is the $8.0\ \mathrm{kg \cdot m/s}$ from part (d), with the positive region from 0 to 3.0 s outweighing the negative region after it.

(a) $p(t) = \left(6.0t - 1.0t^2\right)\ \mathrm{kg \cdot m/s}$ . (b) $F_{\text{net}} = 4.0$ N at $t = 1.0$ s and $-2.0$ N at $t = 4.0$ s. (c) Largest at $t = 3.0$ s, where $p = 9.0\ \mathrm{kg \cdot m/s}$ and the net force is zero. (d) $\Delta p = 8.0\ \mathrm{kg \cdot m/s}$ , whether taken as $p(4.0) - p(0)$ or as the integral of the net force over the interval.

The momentum of a rod whose density is not uniform

A thin rod of length $L = 1.5$ m has linear mass density $\lambda(x) = \lambda_0\left(1 + \frac{x}{L}\right)$ with $\lambda_0 = 0.40$ kg/m, where $x$ is measured from the light end. The rod slides along a frictionless surface, translating without rotating, at $2.4$ m/s parallel to its own length. Find (a) its total mass, (b) the position of its centre of mass, and (c) its momentum.

(a) The sheet prints $\lambda = \frac{d}{d\ell}m(\ell)$ , so $dm = \lambda\,dx$ and the total mass is $M = \int_0^L \lambda(x)\,dx$ .
$M = \int_0^L \lambda_0\left(1 + \frac{x}{L}\right) dx = \lambda_0\left[x + \frac{x^2}{2L}\right]_0^L = \lambda_0\left(L + \frac{L}{2}\right) = \frac{3}{2}\lambda_0 L$ .
$M = 1.5(0.40)(1.5) = 0.90$ kg. Sanity check: the density runs from $0.40$ kg/m at one end to $0.80$ kg/m at the other, averaging $0.60$ kg/m over $1.5$ m, which is $0.90$ kg.
(b) The sheet prints $\vec{r}_{\text{cm}} = \frac{\int \vec{r}\, dm}{\int dm}$ , and the denominator is the $M$ from part (a). In one dimension, $x_{\text{cm}} = \frac{1}{M}\int_0^L x\,\lambda(x)\,dx$ .
$\int_0^L x\lambda_0\left(1 + \frac{x}{L}\right) dx = \lambda_0\left[\frac{x^2}{2} + \frac{x^3}{3L}\right]_0^L = \lambda_0 L^2\left(\frac{1}{2} + \frac{1}{3}\right) = \frac{5}{6}\lambda_0 L^2$ .
$x_{\text{cm}} = \frac{\frac{5}{6}\lambda_0 L^2}{\frac{3}{2}\lambda_0 L} = \frac{5L}{9}$ . Both $\lambda_0$ and one power of $L$ cancel, so the answer is a pure fraction of the length, which is the kind of result skill 2.A rewards.
$x_{\text{cm}} = \frac{5(1.5)}{9} = 0.833$ m from the light end. Check it against the midpoint: $0.75$ m would be a uniform rod, and the heavy end pulls the centre of mass past it, as it should.
(c) Because the rod translates without rotating, every part has the same velocity, so 4.3.A.1.i gives $v_{\text{cm}} = 2.4$ m/s and the total momentum is $p = M v_{\text{cm}}$ .
$p = (0.90)(2.4) = 2.2\ \mathrm{kg \cdot m/s}$ , to two significant figures, in the direction of motion. Keeping the extra digit, $2.16\ \mathrm{kg \cdot m/s}$ .
Notice which part of the answer needed the centre of mass and which did not. For the momentum of a translating body only the total mass matters, so part (b) was not needed for part (c). Part (b) is what you need the moment the rod is struck off centre, because then the parts no longer share a velocity and the centre of mass is the point whose motion still obeys $\vec{F}_{\text{net}} = d\vec{p}/dt$ with the total mass.

(a) $M = \frac{3}{2}\lambda_0 L = 0.90$ kg. (b) $x_{\text{cm}} = \frac{5L}{9} = 0.833$ m from the light end, past the geometric midpoint. (c) $p = M v_{\text{cm}} = 2.2\ \mathrm{kg \cdot m/s}$ along the direction of travel.

Same impulse or same work: which quantity is fixed and which is not

Two carts, one of mass $0.40$ kg and one of mass $1.20$ kg, start from rest on a frictionless track. First give each cart the same impulse of magnitude $2.4\ \mathrm{N \cdot s}$ . Then, in a separate trial, do the same amount of work, $7.2$ J, on each cart instead. Find each cart's final speed, momentum and kinetic energy in both trials, and state the general scaling.

Take the direction of the push as positive. Both carts start at rest, so in each trial $\Delta p = p$ and $\Delta K = K$ .
Same impulse. By 4.2.B.2.i the impulse is the change in momentum, so both carts end with $p = 2.4\ \mathrm{kg \cdot m/s}$ regardless of mass.
Speeds: $v = p/m$ gives $2.4/0.40 = 6.0$ m/s for the light cart and $2.4/1.20 = 2.0$ m/s for the heavy one.
Kinetic energies from $K = p^2/2m$ : $\frac{(2.4)^2}{2(0.40)} = \frac{5.76}{0.80} = 7.2$ J and $\frac{5.76}{2.40} = 2.4$ J. Check against $\frac{1}{2}mv^2$ : $0.5(0.40)(36) = 7.2$ J and $0.5(1.20)(4.0) = 2.4$ J. They agree.
So with the momentum pinned, $K \propto 1/m$ , and the light cart carries three times the kinetic energy of the heavy one because it is three times lighter.
Same work. By the work-energy theorem both carts end with $K = 7.2$ J regardless of mass.
Momenta from $p = \sqrt{2mK}$ : $\sqrt{2(0.40)(7.2)} = \sqrt{5.76} = 2.40\ \mathrm{kg \cdot m/s}$ and $\sqrt{2(1.20)(7.2)} = \sqrt{17.28} = 4.157\ \mathrm{kg \cdot m/s}$ .
Speeds: $6.0$ m/s for the light cart, unchanged from the first trial because $2.4\ \mathrm{N \cdot s}$ and $7.2$ J happen to describe the same push for that mass, and $4.157/1.20 = 3.46$ m/s for the heavy one.
With the kinetic energy pinned, $p \propto \sqrt{m}$ , and the ratio of momenta is $\sqrt{1.20/0.40} = \sqrt{3} = 1.73$ , matching $4.157/2.40 = 1.73$ .
The general statement, which is what skill 2.D asks you to produce: a fixed force-time product fixes the momentum and hands the extra kinetic energy to the lighter object; a fixed force-distance product fixes the kinetic energy and hands the extra momentum to the heavier one. The sheet prints both integrals, $\vec{J} = \int \vec{F}_{\text{net}}(t)\,dt$ and $W = \int_a^b \vec{F} \cdot d\vec{r}$ , and this is the difference between them.

Same $2.4\ \mathrm{N \cdot s}$ impulse: both reach $p = 2.4\ \mathrm{kg \cdot m/s}$ , at $6.0$ m/s and $7.2$ J for the $0.40$ kg cart and $2.0$ m/s and $2.4$ J for the $1.20$ kg cart, so $K \propto 1/m$ . Same $7.2$ J of work: both reach $K = 7.2$ J, at $2.40\ \mathrm{kg \cdot m/s}$ and $4.16\ \mathrm{kg \cdot m/s}$ , so $p \propto \sqrt{m}$ , a ratio of $\sqrt{3}$ .

Frequently asked questions

What is linear momentum in AP Physics C Mechanics?

Essential knowledge 4.1.A.1 defines linear momentum as mass times velocity, printed on the equation sheet as a vector equation. Its units are kilogram metres per second. Statement 4.1.A.2 adds that it is a vector with the same direction as the velocity. The definition is identical to the one in the algebra-based course. What differs is its role: statement 4.2.A.1 makes the net force the time derivative of momentum, so momentum is the quantity the law of motion is written about, and the familiar force equals mass times acceleration is what that reduces to when the mass is constant.

Is F = ma on the AP Physics C Mechanics equation sheet?

Not in that form. The printed Table of Information for AP Physics C: Mechanics gives the system acceleration as the net force divided by the system mass, and separately gives the net force as the time derivative of momentum. The product form with mass times acceleration as a written term appears on the AP Physics 1 sheet, on the same line as the net force and the change in momentum over the change in time, and not on the calculus-based one. Essential knowledge 4.2.B.2.ii explains why: it treats force equals mass times acceleration as a direct result of the impulse-momentum theorem applied to systems with constant mass, so it is a consequence rather than a starting point.

How is Topic 4.1 different in AP Physics C than in AP Physics 1?

Its required content is not different. All six essential-knowledge statements under learning objective 4.1.A are word for word the same in both courses. The framework differences are small: AP Physics 1 attaches a boundary statement to this topic saying that the general term momentum refers specifically to linear momentum, and AP Physics C: Mechanics prints none, and the suggested skill list swaps functional dependence for comparison between scenarios. The real difference is what the rest of the unit does with momentum, starting with Topic 4.2, where the net force becomes a derivative and impulse becomes an integral.

What does the slope of a momentum versus time graph represent?

The net external force exerted on the system, per essential knowledge 4.2.A.5. In the calculus-based course that statement is the graphical form of the definition of force itself, since 4.2.A.1 gives the net force as the time derivative of momentum. A horizontal stretch means zero net force and constant momentum. A peak in the momentum curve means the net force passes through zero at that instant. A straight sloped line means a constant net force, and that stays true even when the mass of the system is changing, which is one reason a momentum graph is easier to read than a velocity graph for a system that loses or gains mass.

Can you find momentum from a velocity function?

Yes, and it is one of the standard moves in AP Physics C: Mechanics. If the velocity is given as a function of time and the mass is constant, multiply through to get the momentum as a function of time. Differentiating that function then gives the net force at any instant with no free-body diagram, by essential knowledge 4.2.A.1, and integrating the net force over an interval gives the change in momentum, by 4.2.B.2.i. The momentum is largest at the instant its derivative vanishes, which is the instant the net force is zero.

What is the difference between momentum and kinetic energy?

Momentum is a vector, mass times velocity, and its components can cancel; kinetic energy is a scalar, one half mass times speed squared, and it cannot. In AP Physics C: Mechanics the cleanest way to keep them apart is by which integral of the net force produces each: integrating force over time gives impulse and therefore a change in momentum, while integrating force over displacement gives work and therefore a change in kinetic energy. Both integrals are printed on the equation sheet. The two are linked by kinetic energy equals momentum squared divided by twice the mass, so at equal momentum the lighter object has more kinetic energy, and at equal kinetic energy the heavier object has more momentum.

What counts as a collision in the AP Physics C framework?

Essential knowledge 4.1.A.3.i states that a collision is a model for an interaction where the forces exerted between the involved objects in the system are much larger than the net external force exerted on those objects during the interaction. The criterion is a comparison between internal and net external forces, not a statement about how short the interaction is. Statement 4.1.A.3.ii then grants the object model, because only the initial and final states of a collision are analyzed. In the calculus-based course you can check the criterion numerically by comparing the external impulse over the contact time against the momentum change of one of the objects.