AP Physics C: Mechanics · Topic 4.2

Topic 4.2: Change in Momentum and Impulse

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

Impulse is the integral of the net force over time, and it equals the change in momentum. Because it is an integral rather than an average force times an interval, a force that varies through the collision is examinable, and so are systems whose mass changes with time.

AP Physics: Unit 4 (topics 4.2 Change in Momentum and Impulse). Topic 4.2 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, 4.2.A and 4.2.B, and ten essential-knowledge statements counting sub-statements. Suggested skills 1.C, 2.A, 2.C and 3.C. No boundary statement; Unit 4 prints exactly one, under Topic 4.3. This is the topic where the two courses genuinely diverge, verified statement by statement against both CEDs: 4.2.A.1 gives the net force as dp/dt where AP Physics 1's 4.2.A.1 gives delta p over delta t; 4.2.A.2 defines impulse as the integral of a force over a time interval with the equation J = integral from t1 to t2 of F_net(t) dt, where AP Physics 1's 4.2.A.2 defines it as the product of the average force and the time interval, J = F_avg delta t; 4.2.A.3, 4.2.A.4, 4.2.A.5 and 4.2.B.1 are worded identically in the two courses; 4.2.B.2.i is the theorem; 4.2.B.2.ii states that Newton's second law of motion is a direct result of the impulse-momentum theorem applied to systems with constant mass, numbered 4.2.B.3 in AP Physics 1; and 4.2.B.2.iii states that the theorem also describes the behavior of a system in which the velocity is constant but the mass changes with respect to time, with F_net = (dm/dt) v, which has NO AP Physics 1 counterpart because the AP Physics 1 Topic 4.2 boundary statement reads that AP Physics 1 does not require students to quantitatively analyze systems in which the mass of the system changes with respect to time. Note that 4.2.B.2.iii is specifically the constant-velocity case, the mirror of 4.2.B.2.ii's constant-mass case; the general case where both product-rule terms survive is not written out in the framework but is posed by sample multiple-choice question 15. Sheet detail verified against the printed Table of Information: the C: Mechanics sheet prints exactly four momentum lines, p = mv, F_net = dp/dt, J = integral with limits = delta p, and v_cm; it prints neither m a as a written term nor F_avg delta t, and the Calculus table on its third page prints the power and chain rules and the integrals of x^n, e^ax, dx over (x+a), cos(ax) and sin(ax). Both Unit 4 sample multiple-choice questions align here: question 3, skill 2.B, 4.2.B, 4.2.B.2, answer B, a quadratic net force integrated over three seconds; question 15, skill 2.C, 4.2.B, 4.2.B.2, answer C, a truck losing sand under a constant force. Sample free-response question 3 lists 4.2.B alongside 4.4.A.

What Topic 4.2 requires

Topic 4.2 carries two learning objectives and ten essential-knowledge statements counting sub-statements, which is the second largest count in Unit 4 after Topic 4.3.

Objective	What it asks
4.2.A	Describe the impulse delivered to an object or system
4.2.B	Describe the relationship between the impulse exerted on an object or system and the change in momentum of the object or system

Statement	What it says
4.2.A.1	The rate of change of a system's momentum is equal to the net external force exerted on that system. Relevant equation: $\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$
4.2.A.2	Impulse is defined as the integral of a force exerted on an object or system over a time interval. Relevant equation: $\vec{J} = \int_{t_1}^{t_2} \vec{F}_{\text{net}}(t)\, dt$
4.2.A.3	Impulse is a vector quantity and has the same direction as the net force exerted on the system
4.2.A.4	The impulse delivered to a system by a net external force is equal to the area under the curve of a graph of the net external force exerted on the system as a function of time
4.2.A.5	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
4.2.B.1	Change in momentum is the difference between a system's final momentum and its initial momentum. Relevant equation: $\Delta \vec{p} = \vec{p} - \vec{p}_0$
4.2.B.2	The impulse-momentum theorem relates the impulse delivered to an object and the object's change in momentum
4.2.B.2.i	The impulse exerted on an object is equal to the object's change in momentum. Relevant equation: $\vec{J} = \int_{t_1}^{t_2} \vec{F}_{\text{net}}(t)\, dt = \Delta \vec{p}$
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: $\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}$
4.2.B.2.iii	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: $\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt} = \frac{dm}{dt}\vec{v}$

Suggested skills: 1.C, 2.A, 2.C and 3.C. Topic 4.2 prints no boundary statement, which matters here more than it usually does, because its AP Physics 1 counterpart prints one and that boundary statement is the single sharpest difference between the two courses in this unit. Section four below quotes it whole.

This is the topic where the calculus-based and algebra-based courses genuinely part company. Three of these ten statements have no counterpart in AP Physics 1, and two more are written with a different operator.

Statement by statement against AP Physics 1, because the wording is what changed

Both courses have a Topic 4.2 called Change in Momentum and Impulse. Here is the required content of each, side by side, read off the two course and exam descriptions.

	AP Physics C: Mechanics	AP Physics 1
Rate of change of momentum	4.2.A.1: $\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$	4.2.A.1: $\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t}$
Definition of impulse	4.2.A.2: the integral of a force over a time interval	4.2.A.2: the product of the average force and the time interval
Impulse equation	$\vec{J} = \int_{t_1}^{t_2} \vec{F}_{\text{net}}(t)\, dt$	$\vec{J} = \vec{F}_{\text{avg}}\Delta t$
Impulse is a vector	4.2.A.3, same wording	4.2.A.3, same wording
Area under force against time	4.2.A.4, same wording	4.2.A.4, same wording
Slope of momentum against time	4.2.A.5, same wording	4.2.A.5, same wording
Change in momentum	4.2.B.1, same wording	4.2.B.1, same wording
The theorem	4.2.B.2 and 4.2.B.2.i	4.2.B.2
Second law as a consequence	4.2.B.2.ii, with $\frac{d\vec{p}}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}$	4.2.B.3, with $\frac{\Delta \vec{p}}{\Delta t} = m\frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$
Changing mass	4.2.B.2.iii, an essential-knowledge statement	excluded by a boundary statement

Read the middle rows first. Where AP Physics 1 writes $\Delta$ , this course writes $d$ , and where AP Physics 1 writes a product, this course writes an integral. That is not a cosmetic upgrade. $\vec{F}_{\text{avg}}\Delta t$ requires you to be handed an average force. $\int \vec{F}_{\text{net}}(t)\,dt$ requires only that you be handed a force, and the exam is then free to give you one that changes throughout the interaction, which is what real collisions do.

Read the last row second. It is the row where one course has a required statement and the other has a prohibition.

The two equation sheets show the same split. The AP Physics C: Mechanics sheet prints exactly four momentum lines, and they are, in the order they appear:

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

The AP Physics 1 sheet prints four momentum lines too. The first and the last are identical. The middle two are $\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t} = m\frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$ and $\vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta \vec{p}$ . So neither $\vec{F}_{\text{avg}}\Delta t$ nor $m\vec{a}$ is printed anywhere on the calculus-based sheet, and neither derivative nor integral is printed anywhere on the algebra-based one.

The second law is the corollary, not the axiom (4.2.B.2.ii)

Statement 4.2.B.2.ii is worth quoting exactly: Newton's second law of motion is a direct result of the impulse-momentum theorem applied to systems with constant mass. The chain it prints runs

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

Read it from the left, and read each equals sign as a separate claim.

$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$ is the law. It is 4.2.A.1, it is printed on the sheet, and it holds with no conditions attached.
$\frac{d\vec{p}}{dt} = m\frac{d\vec{v}}{dt}$ is a product-rule step with one term thrown away. In general $\frac{d}{dt}(m\vec{v}) = m\frac{d\vec{v}}{dt} + \frac{dm}{dt}\vec{v}$ , and this equality keeps only the first term. It needs $\frac{dm}{dt} = 0$ .
$m\frac{d\vec{v}}{dt} = m\vec{a}$ is the definition of acceleration and needs nothing.

So the whole content of the qualifier "applied to systems with constant mass" sits in the middle equality. That is the sentence to be able to reproduce, because it is the reason $\vec{F} = m\vec{a}$ is a special case and $\vec{F}_{\text{net}} = d\vec{p}/dt$ is not.

AP Physics 1 makes the same claim in its statement 4.2.B.3, with $\Delta$ in place of $d$ . The difference is that the algebra-based course then forbids the case where the qualifier bites, and this one requires it.

A practical consequence for free-response work. When a Physics C problem gives you a system whose mass is changing, writing $\vec{F}_{\text{net}} = m\vec{a}$ is not an approximation that costs you accuracy; it is the wrong equation, and it will not earn the point. Start from $d\vec{p}/dt$ and differentiate the product honestly. Suggested skill 2.A, deriving a symbolic expression from known quantities by selecting and following a logical mathematical pathway, is listed for this topic, and this chain is the pathway it has in mind.

Changing mass: the case AP Physics 1 rules out (4.2.B.2.iii)

This is the one genuine capability difference in Unit 4, and it is worth being precise about what each course says.

AP Physics C: Mechanics, essential knowledge 4.2.B.2.iii, quoted whole: "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." The relevant equation printed beneath it:

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

AP Physics 1, the boundary statement under its Topic 4.2, quoted whole: "AP Physics 1 does not require students to quantitatively analyze systems in which the mass of the system changes with respect to time."

One course has a required essential-knowledge statement where the other has an explicit exclusion. Nothing about tone or difficulty is doing the work here; the two documents disagree about what may be asked.

Note carefully which half of the product rule 4.2.B.2.iii keeps. It is the mirror image of 4.2.B.2.ii. Where the constant-mass case drops $\frac{dm}{dt}\vec{v}$ and keeps $m\frac{d\vec{v}}{dt}$ , the constant-velocity case drops $m\frac{d\vec{v}}{dt}$ and keeps $\frac{dm}{dt}\vec{v}$ . The framework gives you the two clean limits, one on each side. Between them sits the general case where both terms survive, and the CED's own sample multiple-choice question 15 poses exactly that, so do not read 4.2.B.2.iii as a promise that the velocity will always be constant.

Where the constant-velocity case shows up:

Material landing on a moving belt or being scooped up. Mass joins the system at a steady rate while the speed is held fixed. Worked example 2 does this one.
A hose or a jet striking a surface. The force is the rate at which momentum arrives, $\frac{dm}{dt}v$ , which is why a fire hose pushes back hard without anything accelerating.
A chain being lifted or paid out at constant speed, where the mass in motion changes while the speed does not.

And where both terms survive, which is worth practising because the framework's own sample question is of this kind: a vehicle losing or gaining mass while accelerating. Worked example 3 works one through, with the modelling assumption stated explicitly, because that assumption is where the marks are.

Impulse as an integral, and what to do when there is no average force

Statement 4.2.A.2 defines impulse as the integral of a force exerted on an object or system over a time interval, and 4.2.B.2.i sets that integral equal to the change in momentum. The routine that follows is short:

Declare a positive direction. Impulse is a vector by 4.2.A.3, and in one dimension that means a signed number.
Write the net force as a function of time, or read it off a graph.
Integrate over the stated limits. The Calculus table printed in the same appendix as the equation sheet gives you the antiderivatives the exam expects: $\int x^n dx$ , $\int e^{ax} dx$ , $\int \frac{dx}{x+a} = \ln|x+a|$ , $\int \cos(ax)\,dx$ and $\int \sin(ax)\,dx$ . Those five are a strong hint about the shape of the $F(t)$ functions you will meet.
Set the result equal to $\Delta \vec{p} = \vec{p} - \vec{p}_0$ , which is 4.2.B.1, and solve for whichever end of the motion is unknown.

The average force is an output here, not an input. If a question wants it, it is defined by

F_{\text{avg}} = \frac{J}{\Delta t} = \frac{1}{t_2 - t_1}\int_{t_1}^{t_2} F_{\text{net}}(t)\, dt

which is the mean value of the force over the interval. That reversal of which quantity is given is worth internalising. In AP Physics 1 the average force is given to you and the impulse follows; here the impulse comes first and the average force is what you report afterwards. Worked example 1 finds an average force that is $2/\pi$ of the peak force, which is roughly $64\%$ , and no algebra-based method produces that number.

Two things that stay exactly as they are in the algebra-based course, and you should not spend revision time on the difference:

4.2.A.4, area under the curve. Identical wording in both CEDs. The area under a graph of net external force against time is the impulse. In Physics C the curve is more likely to be a curve, so the area is more likely to need an integral than a triangle, but the statement is the same.
4.2.A.5, slope of the momentum graph. Identical wording in both. The net external force is the slope of a graph of the momentum of the system against time.

Those two are inverse operations on the same pair of graphs, and reading them as a pair is what suggested skill 1.C, creating qualitative sketches of graphs, is asking for. Given a force pulse, sketch the momentum curve: it is flat where the force is zero, steepest where the force peaks, and its total rise is the area of the pulse. The impulse-momentum theorem guide owns the step-by-step routine for the constant-mass, average-force version of these problems, so this page does not repeat it.

The traps, and where the marks actually go

Integrating the magnitude instead of the vector. If the net force reverses during the interval, the integral has a negative region and the impulse is the signed total. A ball that arrives at $+8$ m/s and leaves at $-6$ m/s has $\Delta p$ of magnitude $m(14)$ , not $m(2)$ . This is 4.2.A.3 doing its work, and it turns a correct integral into a wrong answer.

Using the peak force as the average force. For any pulse that rises and falls, the average is well below the peak: half the peak for a triangle, $2/\pi$ of it for a half sine. The framework's decision to define impulse as an integral is precisely so that this distinction is examinable.

Forgetting that the integral has limits. Statement 4.2.A.2 prints them, $t_1$ and $t_2$ , and the sheet prints them too. An impulse without an interval is not a number.

Writing $m\vec{a}$ when the mass moves. Covered above, and it is the difference between a correct and an incorrect starting equation on a variable-mass free-response part.

Confusing impulse with momentum. They have the same units and the same dimensions, and 4.2.B.2.i says one equals the change in the other, but momentum is a property of a system at an instant and impulse is something delivered to it over an interval. Impulse compared with momentum is the two-minute version.

Choosing the wrong system so that "net external" changes meaning. Statements 4.2.A.1, 4.2.A.4 and 4.2.A.5 all say net external force. If you draw the system boundary around both colliding objects, the contact forces are internal and contribute nothing. That is the bridge to Topic 4.3.

What the sheet prints for Topic 4.2, line by line

Checked against the printed Table of Information in the AP Physics C: Mechanics course and exam description, including its third page, which carries the Geometry, Trigonometry, Vectors, Calculus and Identities tables that condensed transcriptions of the sheet usually leave out.

Equation	On the C: Mechanics sheet
$\vec{p} = m\vec{v}$	yes
$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$	yes, as its own line
$\vec{J} = \int_{t_1}^{t_2} \vec{F}_{\text{net}}(t)\, dt = \Delta \vec{p}$	yes, definition and theorem on one line, with limits
$\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}} = \frac{\vec{F}_{\text{net}}}{m_{\text{sys}}}$	yes, in the dynamics group
$\int x^n dx$ , $\int e^{ax} dx$ , $\int \frac{dx}{x+a}$ , $\int \cos(ax) dx$ , $\int \sin(ax) dx$	yes, in the Calculus table
$\frac{d}{dx}(x^n) = nx^{n-1}$ and the chain rule	yes, in the Calculus table
$\vec{F}_{\text{net}} = m\vec{a}$ as a written term	no
$\vec{J} = \vec{F}_{\text{avg}}\Delta t$	no, that is the AP Physics 1 sheet
$\vec{F}_{\text{net}} = \frac{dm}{dt}\vec{v}$	no, write it from 4.2.B.2.iii
$\Delta \vec{p} = \vec{p} - \vec{p}_0$	no, though it is printed in the framework under 4.2.B.1
$F_{\text{avg}} = J/\Delta t$	no

Three readings follow. The two equations that define this topic are both printed, so nothing about the calculus needs memorising, only the ability to do the integral. The three equations you may need to write yourself are the variable-mass form, the definition of average force, and the difference form of $\Delta \vec{p}$ ; all three are one line each and all three are quotable from the framework. The Calculus table is on the sheet, which is worth knowing both because it saves you memorising antiderivatives and because its contents tell you what kinds of $F(t)$ the exam considers fair: polynomials, exponentials, reciprocals and sinusoids.

How Topic 4.2 is tested

Unit 4 is weighted 10 to 20% of the multiple-choice section over about 11 to 15 class periods, and the AP Classroom Progress Check for the unit is about 18 multiple-choice questions and 4 free-response questions. Those figures are Physics C figures; the AP Physics 1 unit with the same name is 10 to 15% over about 10 to 15 class periods.

Both of the framework's Unit 4 sample multiple-choice questions align to this topic, and neither aligns to 4.1, 4.3 or 4.4.

Sample question 3 gives the net force on an object moving along a straight line as a quadratic function of time and asks for the change in momentum over a three second interval. Skill 2.B, learning objective 4.2.B, essential knowledge 4.2.B.2, published answer B. There is no average force anywhere in the stem, and there is no way to reach the answer except by integrating. This is the question that the algebra-based course cannot pose.

Sample question 15 is a variable-mass problem. A truck initially at rest with a load of sand is pushed by a constant force while the sand leaves the truck, with the mass that has left given as a linear function of time, and the question asks which graph shows the truck's velocity against time. Skill 2.C, learning objective 4.2.B, essential knowledge 4.2.B.2, published answer C. The reasoning is that the remaining mass falls, so the acceleration rises, so the velocity curve is concave up until the sand runs out and straight afterwards. Worked example 3 below works the same shape of problem numerically.

Sample free-response question 3, the Experimental Design and Analysis question, is also partly a 4.2 question: its published alignment lists learning objective 4.2.B alongside 4.4.A. It gives students an impulse device that delivers the same impulse in every trial to carts of different total mass, has them measure the maximum height reached on a ramp, and asks them to linearise the data and extract the impulse from the slope. The physics is that a fixed impulse gives $v = J/M$ , so the height satisfies $h = \frac{J^2}{2gM^2}$ , and $h$ plotted against $1/M^2$ is a straight line of slope $J^2/2g$ . Suggested skills 1.B and 3.A appear in that question's skill list, and they are the experimental-design skills the unit's Developing Understanding page says Unit 4 is the natural place to build.

If you are in the algebra-based course, read the [AP Physics 1 Topic 4.2 page](/ap-physics-1/unit-4-linear-momentum/4-2-change-in-momentum-and-impulse) instead. It is written for an exam where impulse is an average force multiplied by a time, where graph areas are triangles and rectangles, and where changing-mass systems are excluded by name. Nothing on this page about integrals or the product rule is on your exam. This page is for AP Physics C: Mechanics students.

Where to go next. Topic 4.1 is the definition this topic differentiates, Topic 4.3 is what happens when the net external force is zero, and Topic 6.3 is the whole of this topic rewritten with torque in place of force. The momentum collision calculator checks arithmetic and the collision lab lets you vary the two masses, the two initial velocities and a coefficient of restitution. The Unit 4 hub lists all four topics.

A force pulse that is a curve: impulse, average force, and the momentum function

A $0.50$ kg cart at rest on a frictionless track is struck by a plunger that exerts a force along the track of $F(t) = F_0 \sin\!\left(\frac{\pi t}{T}\right)$ for $0 \le t \le T$ , with $F_0 = 45.0$ N and $T = 0.120$ s, and no force before or after. Take the direction of the push as positive. Find (a) the impulse delivered, (b) the cart's final speed, (c) the average force, compared with the peak force, and (d) the momentum as a function of time during the pulse, and the speed at the halfway instant $t = T/2$ .

(a) Statement 4.2.A.2 gives the impulse as $J = \int_{t_1}^{t_2} F_{\text{net}}(t)\,dt$ . The plunger force is the only horizontal force, so it is the net force.
The Calculus table on the sheet prints $\int \sin(ax)\,dx = -\frac{1}{a}\cos(ax)$ , here with $a = \pi/T$ .
$J = \int_0^{T} F_0 \sin\!\left(\frac{\pi t}{T}\right) dt = F_0\left[-\frac{T}{\pi}\cos\!\left(\frac{\pi t}{T}\right)\right]_0^{T} = -\frac{F_0 T}{\pi}\left(\cos\pi - \cos 0\right) = \frac{2F_0 T}{\pi}$ .
$J = \frac{2(45.0)(0.120)}{\pi} = \frac{10.8}{\pi} = 3.4377\ \mathrm{N \cdot s}$ , so $3.44\ \mathrm{N \cdot s}$ to three significant figures.
(b) By 4.2.B.2.i the impulse is the change in momentum, and the cart started at rest, so $p = 3.4377\ \mathrm{kg \cdot m/s}$ and $v = p/m = 3.4377/0.50 = 6.8755$ m/s, so $6.88$ m/s.
(c) The average force is the impulse divided by the interval: $F_{\text{avg}} = J/T = 3.4377/0.120 = 28.648$ N, so $28.6$ N.
Symbolically that is $F_{\text{avg}} = \frac{2F_0T/\pi}{T} = \frac{2F_0}{\pi}$ , so the average is $2/\pi = 0.6366$ of the peak, about $64\%$ , whatever the numbers are. Using the peak force of $45.0$ N in place of the average would overstate the impulse by $57\%$ .
(d) Integrating from $0$ to a general $t$ instead: $p(t) = \frac{F_0 T}{\pi}\left(1 - \cos\frac{\pi t}{T}\right)$ . Check the endpoint: at $t = T$ the cosine is $-1$ , giving $\frac{2F_0T}{\pi}$ , which is part (a).
At $t = T/2 = 0.0600$ s the cosine is zero, so $p = \frac{F_0T}{\pi} = \frac{5.4}{\pi} = 1.7189\ \mathrm{kg \cdot m/s}$ , exactly half the total, and $v = 1.7189/0.50 = 3.44$ m/s.
Check 4.2.A.5 by differentiating: $\frac{dp}{dt} = \frac{F_0T}{\pi}\cdot\frac{\pi}{T}\sin\frac{\pi t}{T} = F_0\sin\frac{\pi t}{T}$ , which is the applied force. So the slope of the momentum curve returns the force, as 4.2.A.5 says it must.
Sketch check for skill 1.C: the momentum curve starts flat at the origin, is steepest at $t = T/2$ where the force peaks, and flattens again at $t = T$ where the force returns to zero. It is an S shape, not a straight line, and half the momentum has been delivered by the halfway instant only because this pulse is symmetric.

(a) $J = \frac{2F_0T}{\pi} = 3.44\ \mathrm{N \cdot s}$ . (b) $6.88$ m/s. (c) $F_{\text{avg}} = \frac{2F_0}{\pi} = 28.6$ N, which is $64\%$ of the $45.0$ N peak. (d) $p(t) = \frac{F_0T}{\pi}\left(1 - \cos\frac{\pi t}{T}\right)$ , giving $1.72\ \mathrm{kg \cdot m/s}$ and $3.44$ m/s at $t = T/2$ , and differentiating it returns the applied force.

Gravel onto a moving belt: the constant-velocity variable-mass case

Gravel falls vertically from a hopper onto a horizontal conveyor belt at a steady $25$ kg/s. The belt runs at a constant $1.6$ m/s. Ignore friction in the belt's own bearings. Find (a) the extra horizontal force the motor must supply to keep the belt at constant speed, (b) the power that force delivers, (c) the rate at which the gravel gains kinetic energy, and (d) account for the difference.

Take the belt's direction of travel as positive. The gravel arrives with zero horizontal velocity and leaves the loading point moving with the belt, so the system whose momentum is changing is the gravel already on the belt.
(a) The belt's speed is constant and its load's mass is growing, which is exactly the situation essential knowledge 4.2.B.2.iii describes: a system in which the velocity is constant but the mass changes with respect to time.
$F_{\text{net}} = \frac{dp}{dt} = \frac{d}{dt}(mv) = \frac{dm}{dt}v$ , because $v$ is constant and the other product-rule term vanishes.
$F = (25)(1.6) = 40$ N. Note that this is a force with no acceleration anywhere in the problem. Writing $F = ma$ here gives zero, which is wrong, and that is the concrete reason 4.2.B.2.ii carries its qualifier.
(b) Power delivered by that force: $P = Fv = (40)(1.6) = 64$ W.
(c) The gravel's kinetic energy grows as mass arrives and is brought up to belt speed: $\frac{dK}{dt} = \frac{d}{dt}\left(\tfrac{1}{2}mv^2\right) = \tfrac{1}{2}\frac{dm}{dt}v^2 = \tfrac{1}{2}(25)(1.6)^2 = \tfrac{1}{2}(25)(2.56) = 32$ W.
(d) The motor supplies $64$ W and the gravel gains $32$ W of kinetic energy, so $32$ W is unaccounted for, exactly half.
It is dissipated by the sliding friction between each newly landed lump of gravel and the belt surface while the lump is being accelerated up to belt speed. The belt surface moves further than the gravel does during that skid, and the product of friction force and the difference in displacements is the missing energy.
The factor of exactly one half is general, not a feature of these numbers: with $F = \frac{dm}{dt}v$ the input power is $\frac{dm}{dt}v^2$ and the kinetic energy rate is $\frac{1}{2}\frac{dm}{dt}v^2$ , whatever $\frac{dm}{dt}$ and $v$ are.
Momentum, by contrast, is fully accounted for with no loss term at all. That asymmetry between the momentum audit and the energy audit is the same one that separates elastic from inelastic collisions in Topic 4.4.

(a) $F = \frac{dm}{dt}v = 40$ N, with no acceleration involved. (b) $P = Fv = 64$ W. (c) $\frac{dK}{dt} = \frac{1}{2}\frac{dm}{dt}v^2 = 32$ W. (d) The missing $32$ W, exactly half of the input in every case of this kind, is dissipated by friction as each arriving lump skids up to belt speed.

A cart that loses mass while it accelerates

A $4.0$ kg cart carries $6.0$ kg of sand on a level frictionless track. At $t = 0$ a constant horizontal force of $20$ N is applied and the sand simultaneously begins to drain straight down through a hole at a steady $1.5$ kg/s, leaving with no horizontal velocity relative to the cart. Find (a) the instant the sand runs out, (b) the acceleration at $t = 0$ and just before that instant, (c) the cart's speed when the sand runs out, and (d) compare with what a constant-mass treatment would give.

State the model first, because this is where the reasoning marks are. The sand leaves moving horizontally at the cart's own speed, so it carries its horizontal momentum away with it and exerts no horizontal thrust. The system to track is the cart plus the sand still aboard, and for that system $F_{\text{net}} = m(t)\frac{dv}{dt}$ with the applied force unchanged.
Take the direction of the applied force as positive. $m(t) = 10.0 - 1.5t$ kilograms, with $t$ in seconds.
(a) The sand is gone when $1.5t = 6.0$ , so $t_1 = 4.0$ s, leaving $m = 4.0$ kg.
(b) $a(0) = \frac{20}{10.0} = 2.0\ \mathrm{m/s^2}$ , and just before $t_1$ , $a = \frac{20}{4.0} = 5.0\ \mathrm{m/s^2}$ . The acceleration rises by a factor of $2.5$ during the drain, which is the ratio of the masses.
(c) $\frac{dv}{dt} = \frac{20}{10.0 - 1.5t}$ , so $v(t_1) = \int_0^{4.0}\frac{20\,dt}{10.0 - 1.5t}$ .
The Calculus table on the sheet prints $\int \frac{dx}{x + a} = \ln|x + a|$ , which is what this needs after a sign adjustment for the $-1.5t$ : $\int_0^{4.0}\frac{dt}{10.0 - 1.5t} = -\frac{1}{1.5}\Big[\ln(10.0 - 1.5t)\Big]_0^{4.0} = \frac{1}{1.5}\ln\frac{10.0}{4.0}$ .
$\ln 2.5 = 0.91629$ , so the integral is $0.91629/1.5 = 0.61086$ s per kilogram of the appropriate units, and $v = 20(0.61086) = 12.217$ m/s, so $12.2$ m/s.
Symbolically, $v(t_1) = \frac{F_0}{C}\ln\!\left(\frac{m_0}{m_1}\right)$ with $C$ the drain rate. The speed depends on the ratio of the masses through a logarithm, which is a functional dependence skill 2.D could ask you to reason about, and which no constant-mass formula produces.
(d) A constant-mass treatment at the initial $10.0$ kg gives $v = \frac{20}{10.0}(4.0) = 8.0$ m/s, and at the final $4.0$ kg it gives $20$ m/s. The true answer, $12.2$ m/s, sits between them and is closer to the light-end value than a simple average would suggest, because the cart spends its later, lighter seconds accelerating hardest.
Sketch check for skill 1.C: because the acceleration grows throughout the drain, $v$ against $t$ is concave up from $0$ to $4.0$ s, and after $4.0$ s the mass is fixed at $4.0$ kg so the acceleration is a constant $5.0\ \mathrm{m/s^2}$ and the graph becomes a straight line. That concave-up-then-straight shape is exactly the answer the framework marks correct on its sample multiple-choice question 15.

(a) $t_1 = 4.0$ s. (b) $2.0\ \mathrm{m/s^2}$ at the start and $5.0\ \mathrm{m/s^2}$ just before the sand runs out. (c) $v = \frac{F_0}{C}\ln\frac{m_0}{m_1} = \frac{20}{1.5}\ln 2.5 = 12.2$ m/s. (d) A constant-mass treatment gives $8.0$ m/s using the initial mass and $20$ m/s using the final mass, so it brackets rather than answers the question.

Frequently asked questions

What is impulse in AP Physics C Mechanics?

Essential knowledge 4.2.A.2 defines impulse as the integral of a force exerted on an object or system over a time interval, and the equation sheet prints that integral with its limits. Statement 4.2.B.2.i then sets it equal to the change in momentum of the system, which is the impulse-momentum theorem. Statement 4.2.A.3 adds that impulse is a vector with the same direction as the net force. Its units are newton seconds, the same combination as kilogram metres per second. The definition differs from the algebra-based course, which defines impulse as the product of an average force and a time interval.

Why is impulse an integral in AP Physics C but a product in AP Physics 1?

Because the two courses can ask different questions. AP Physics 1 defines impulse as the average force multiplied by the time interval, which requires the average force to be supplied in the problem. AP Physics C: Mechanics defines it as the integral of the net force over time, which requires only that the force be given as a function of time or as a graph. That lets the calculus-based exam specify a force that varies throughout a collision and still ask for an exact answer. The average force then becomes an output, found by dividing the impulse by the interval, rather than an input.

Does AP Physics C test changing-mass systems like rockets?

Yes, and this is a real difference between the two courses. Essential knowledge 4.2.B.2.iii states that 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, and it prints the net force as the rate of change of mass multiplied by the velocity. The AP Physics 1 boundary statement for its Topic 4.2 says the opposite, that AP Physics 1 does not require students to quantitatively analyze systems in which the mass of the system changes with respect to time. The framework's own sample multiple-choice question 15 for AP Physics C: Mechanics is a variable-mass problem, a truck losing sand while being pushed by a constant force.

Is Newton's second law the same as the impulse-momentum theorem?

In AP Physics C: Mechanics the second law is derived from it. Essential knowledge 4.2.B.2.ii states that Newton's second law of motion is a direct result of the impulse-momentum theorem applied to systems with constant mass, and prints the chain from net force to the time derivative of momentum to mass times acceleration. The first equality holds always. The step from the derivative of momentum to mass times the derivative of velocity discards the term in which the mass changes, so it requires constant mass. When the mass is not constant, force equals mass times acceleration is not an approximation but the wrong equation, and you start from the derivative form instead.

How do you find average force from a force versus time graph?

Find the area under the curve first, which by essential knowledge 4.2.A.4 is the impulse delivered, then divide that impulse by the length of the time interval. That quotient is the average force, and it is the mean value of the force over the interval rather than the midpoint of the highest and lowest values. For a pulse that rises to a peak and falls back, the average is well below the peak: exactly half the peak for a triangular pulse, and two divided by pi, about 64 percent of the peak, for a half sine pulse.

What does the area under a force versus time graph represent?

The impulse delivered to the system by that force, per essential knowledge 4.2.A.4, which is worded identically in the AP Physics C: Mechanics and AP Physics 1 course descriptions. If the graph shows the net external force, the area is also the change in the system's momentum, by 4.2.B.2.i. Area below the time axis counts as negative, so a force that reverses during the interval delivers a signed total, not a sum of magnitudes. The inverse reading is essential knowledge 4.2.A.5: the slope of a graph of momentum against time is the net external force.

Is F = ma or J = F average times delta t on the AP Physics C equation sheet?

Neither. The printed Table of Information for AP Physics C: Mechanics gives four momentum lines: momentum as mass times velocity, the net force as the time derivative of momentum, the impulse as a definite integral of the net force that equals the change in momentum, and the centre-of-mass velocity. Mass times acceleration appears nowhere on it as a written term, and neither does an average force multiplied by a time interval. Both of those forms are printed on the AP Physics 1 sheet instead. The calculus-based sheet does print a Calculus table with the derivative and integral rules the exam assumes.