AP Physics C: Mechanics · Topic 2.5

Topic 2.5: Newton's Second Law

Unit 2: Force and Translational Dynamics20-25% of the multiple-choice section

The acceleration of a system's center of mass is proportional to the net force on the system and points the same way. The AP Physics C sheet gives two forms: acceleration equals net force over system mass, and net force equals the rate of change of momentum. It never prints F equals ma.

AP Physics: Unit 2 (topics 2.5 Newton's Second Law). AP Physics C: Mechanics Unit 2, Topic 2.5. One learning objective, 2.5.A, describe the conditions under which a system's velocity changes. Supported by 2.5.A.1 (unbalanced forces are a configuration of forces such that the net force exerted on a system is not equal to zero), 2.5.A.2 (Newton's second law states that the acceleration of a system's CENTER OF MASS has a magnitude proportional to the magnitude of the net force exerted on the system and is in the same direction as that net force, with relevant equation a_sys = sum F / m_sys = F_net / m_sys) and 2.5.A.3 (the velocity of a system's center of mass will only change if a nonzero net EXTERNAL force is exerted on that system). No boundary statement. These three statements are word for word identical to AP Physics 1 Topic 2.5. What differs is the equation sheet: verified against the Table of Information appendix rendered as an image, the AP Physics C: Mechanics sheet prints a_sys = sum F / m_sys AND F_net = dp/dt, and does NOT print ma in any form, whereas the AP Physics 1 sheet prints F_net = delta p / delta t = m delta v / delta t = ma. The derivative form and its consequences are filed by the CED under Unit 4: 4.2.B.2.ii states Newton's second law is a direct result of the impulse-momentum theorem applied to systems with constant mass and prints F_net = dp/dt = m dv/dt = ma, and 4.2.B.2.iii covers a system whose velocity is constant while mass changes with time, printing F_net = dp/dt = (dm/dt) v. The C sheet also prints delta v_x = integral of a_x(t) dt and delta x = integral of v_x(t) dt, which is what makes time-varying-force problems examinable. Suggested skills are 1.B, 2.B, 2.D, 3.A and 3.C, five in total, the joint largest count in the unit; AP Physics 1 lists 1.A, 2.A, 2.D and 3.B for its version. Sample free-response question 2 aligns to 2.5.A. One of the unit's six optional sample instructional activities is on Topic 2.5, determining an unknown mass with known masses, string, a pulley, a metre stick and a stopwatch.

The primitive form is the derivative one, and the sheet says so

Look at what the AP Physics C: Mechanics equation sheet actually prints for the second law. Two lines, and neither of them is $F = ma$ :

\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}} = \frac{\vec{F}_{\text{net}}}{m_{\text{sys}}} \qquad \vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}

Compare the AP Physics 1 sheet, which prints the first of those and then $\vec{F}_{\text{net}} = \dfrac{\Delta\vec{p}}{\Delta t} = m\dfrac{\Delta\vec{v}}{\Delta t} = m\vec{a}$ , with the whole chain spelled out and $m\vec{a}$ on the end. The AP Physics C: Mechanics sheet replaces the finite differences with a derivative and stops before $m\vec{a}$ . Checking the Table of Information appendix directly confirms it: the string $m\vec{a}$ appears nowhere on the mechanics equation table.

The CED explains the relationship in Unit 4 rather than Unit 2. 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

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

Read the order of that chain. The derivative of momentum is the definition; $m\vec{a}$ is what it becomes when you are allowed to pull a constant $m$ out of the derivative. Statement 4.2.B.2.iii then does the other case: 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, with

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

That is the whole reason variable-mass problems exist in this course and not in the algebra-based one. A conveyor belt being loaded with sand, a rope being lifted off a table link by link, a chain piling up: in each case the mass of the moving system changes with time, so $m\vec{a}$ is not available and $d\vec{p}/dt$ is.

Be precise about where the CED files each piece, because an exam question will be tagged to one topic. Topic 2.5 itself contains only the constant-mass statement and only the $\vec{a}_{\text{sys}}$ equation. The derivative form and the variable-mass case are Unit 4, Topic 4.2. Both are printed on the same equation sheet and both are fair game on any question, but if you are revising Topic 2.5 in isolation, the second half of this section is a preview rather than a requirement.

What the CED requires of Topic 2.5

One learning objective, three essential-knowledge statements, no boundary statement. Suggested skills 1.B, 2.B, 2.D, 3.A and 3.C, five of them, one of the largest counts in the unit.

Objective 2.5.A: describe the conditions under which a system's velocity changes.

2.5.A.1: unbalanced forces are a configuration of forces such that the net force exerted on a system is not equal to zero.
2.5.A.2: Newton's second law of motion states that the acceleration of a system's center of mass has a magnitude proportional to the magnitude of the net force exerted on the system and is in the same direction as that net force. Its relevant equation is $\vec{a}_{\text{sys}} = \sum \vec{F}/m_{\text{sys}} = \vec{F}_{\text{net}}/m_{\text{sys}}$ .
2.5.A.3: the velocity of a system's center of mass will only change if a nonzero net external force is exerted on that system.

Those three statements are word for word the same in AP Physics 1. Topic 2.5 is one of the two topics in Unit 2 where the required content is identical between the courses, and neither course prints a boundary statement on it.

Notice how the objective is phrased. Topic 2.4 is "describe the conditions under which a system's velocity remains constant" and Topic 2.5 is "describe the conditions under which a system's velocity changes". They are two halves of one idea, split across two topics, and statement 2.5.A.1 defines unbalanced forces as the negation of 2.4.A.2's translational equilibrium. Revise them together.

Three words in 2.5.A.2 and 2.5.A.3 are doing precise work and are worth isolating: center of mass, system, and external.

Why the equation says center of mass

Statement 2.5.A.2 does not say the acceleration of the object. It says the acceleration of a system's center of mass. That qualifier is what makes the law true for things that are not point particles.

A wrench spinning as it flies across a room has every part of it accelerating differently. Its center of mass, though, follows a clean parabola, exactly as a thrown point mass would. The second law makes a claim about that one point and about no other.

The reason is Topic 2.3. Statement 2.3.A.2 says interactions between objects within a system, that is, internal forces, do not influence the motion of a system's center of mass, because every internal force is half of a third-law pair whose other half is also inside the system, so the pair cancels in the sum. What survives is the external forces, and statement 2.5.A.3 draws the conclusion: the velocity of a system's center of mass will only change if a nonzero net external force is exerted on that system.

Three consequences you can use directly.

You may choose the system. Two blocks connected by a string can be treated as one system whose acceleration follows from the external forces alone, with the string tension internal and therefore absent. Then, if you want the tension, isolate one block and apply the law again with the acceleration you already have.
Internal forces cannot accelerate the whole. No arrangement of pushing on yourself moves your center of mass. Statement 2.2.A.1.ii says the same thing from the force side: an object or system cannot exert a net force on itself.
The system may be deformable. Statement 2.1.B.4 lets you model a system as a singular object located at its center of mass, and 2.5.A.2 then applies to that point even if the system is changing shape.

The systems and center of mass page develops where the center of mass actually is, including the integral form for a continuous body.

One vector law, one scalar equation per axis

The second law is a vector statement, and the way you use it is to project it onto axes. In two dimensions,

\sum F_x = m a_x \qquad \text{and} \qquad \sum F_y = m a_y

These are independent. A projectile has $a_x = 0$ and $a_y = -g$ at the same instant, from the same law. Statement 2.4.A.4 states the general version: forces may be balanced in one dimension but unbalanced in another, and the velocity changes only in the direction of the unbalanced force.

Statement 2.2.B.4 tells you how to pick the axes: one axis parallel to the acceleration. Then the other axis has zero acceleration and its equation is an equilibrium equation, which is usually where the normal force or the tension comes from for free.

The routine, in the order that works:

Choose the system and draw its free-body diagram, following the Topic 2.2 boundary statement: forces only, straight arrows from the dot, same-direction forces side by side.
Choose axes with one along the acceleration, and declare the positive direction in writing.
Write one equation per axis. Write both even if you think you need one.
Solve symbolically. On the AP Physics C: Mechanics exam this is where the marks are, because skill 2.A carries 25 to 30 percent of the multiple-choice section.
Substitute numbers last, and check the limits of your expression before you trust them.

Step 5 is worth expanding, because it is what skill 2.D is testing. Skill 2.D is predict new values or factors of change of physical quantities using functional dependence between variables, and it is one of Topic 2.5's suggested skills. A question that gives you no numbers at all and asks what happens to the acceleration if the net force triples and the mass doubles is a pure 2.D question, and the answer comes straight from the structure of $\vec{a} = \vec{F}_{\text{net}}/m$ : the acceleration becomes $3/2$ of what it was.

Traps this topic sets

Using $m\vec{a}$ when the mass is changing. This is the trap the C course is equipped to set. If sand is falling onto a moving cart, or a rocket is burning fuel, or a chain is being lifted off a pile, the system's mass is a function of time and $\vec{F}_{\text{net}} = m\vec{a}$ is simply the wrong equation. Go back to $\vec{F}_{\text{net}} = d\vec{p}/dt$ and differentiate the product properly.

Confusing the net force with an individual force. The left side of the law is the vector sum of every force on the system. A question that gives you the applied force and asks for the acceleration is not giving you the net force unless it says friction and everything else is absent.

Using the wrong mass. $m_{\text{sys}}$ is the mass of whatever you drew the diagram for. If you drew the two-block system, use the total; if you drew one block, use that block's mass and remember to include the force the other block exerts on it.

Reading acceleration as velocity. The law relates force to acceleration, not to velocity. An object can be moving left while accelerating right, which is what happens whenever something is slowing down. Zero velocity does not mean zero acceleration: a ball at the top of its flight has $v = 0$ and $a = g$ downward.

Treating the equation as scalar. The direction claim in 2.5.A.2 is part of the law: the acceleration is in the same direction as the net force. Not the same direction as the velocity, and not the same direction as the largest single force.

Forgetting that the net force is instantaneous. If the forces change with time, so does the acceleration. That is when you stop solving algebraically and start integrating, using $\Delta v_x = \int a_x(t)\, dt$ and $\Delta x = \int v_x(t)\, dt$ , both printed on the equation sheet. The second worked example does exactly this.

If you want the algebra-based version of this topic

The three essential-knowledge statements are identical, so the physics you need for Topic 2.5 itself is the same in both courses. If you are taking AP Physics 1, read AP Physics 1 Topic 2.5: Newton's Second Law. If you are taking AP Physics C: Mechanics, stay here, because the sheet you will be handed is different and so are the problems that difference makes possible.

	AP Physics 1 Topic 2.5	AP Physics C Topic 2.5
Objectives	2.5.A	2.5.A
Statements	2.5.A.1 to 2.5.A.3, same text	2.5.A.1 to 2.5.A.3, same text
Boundary statement	none	none
Sheet: system form	$\vec{a}_{\text{sys}} = \sum\vec{F}/m_{\text{sys}}$	same
Sheet: momentum form	$\vec{F}_{\text{net}} = \Delta\vec{p}/\Delta t = m\Delta\vec{v}/\Delta t = m\vec{a}$	$\vec{F}_{\text{net}} = d\vec{p}/dt$
Is $m\vec{a}$ printed?	yes	no
Variable-mass systems	not in the course	4.2.B.2.iii
Time-varying force	not integrated	$\Delta v_x = \int a_x(t)\, dt$ , printed
Suggested skills	1.A, 2.A, 2.D, 3.B	1.B, 2.B, 2.D, 3.A, 3.C

The suggested-skill lists differ more here than in most shared topics. AP Physics C adds 3.A, create experimental procedures, and 1.B, create quantitative graphs including plotting data, where AP Physics 1 lists 1.A and 3.B. That points at the second law arriving as a laboratory question: vary the force, measure the acceleration, plot one against the other, and interpret the slope as $1/m$ .

Related on this site: the Newton's second law glossary entry, the net force guide and the net force calculator, plus Topic 2.9 on resistive forces, where the second law becomes a differential equation.

How Topic 2.5 is tested

Unit 2 carries 20 to 25 percent of the multiple-choice section across about 15 to 25 class periods, the highest minimum weighting of any unit in AP Physics C: Mechanics (only Unit 3, at 15 to 25 percent, reaches the same 25 percent ceiling). Topic 2.5 has one of the unit's nineteen learning objectives, and its five suggested skills are the joint largest count in the unit, tied with Topics 2.7 and 2.9.

Those skills are 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.B, calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

Skill 2.B carries 20 to 25 percent of the multiple-choice section and 2.D carries 10 to 15 percent, so between them Topic 2.5's two Science Practice 2 skills account for 30 to 40 percent of that section. Skill 3.A is not assessed on the multiple-choice section at all; it appears only in free response, and specifically in question 3, Experimental Design and Analysis.

The CED's sample free-response question 2, Translation Between Representations, worth 12 points, aligns to 2.5.A among its seven learning objectives.

The unit's own "Building the Science Practices" page names the second law as its example of what the unit is for: it says the skill of making claims can be developed throughout the unit by providing students with opportunities such as having them make predictions about the acceleration of a system based on the forces exerted on that system, and then justifying those predictions with appropriate physics principles.

One of the six optional sample instructional activities the CED lists for Unit 2 is on Topic 2.5: students are given an object of unknown mass, along with known masses, string, a pulley, a metre stick and a stopwatch, and asked to determine the unknown mass. That is a second-law experiment with skill 3.A written all over it, and it is worth doing rather than reading.

For practice on the surrounding material, the forces and Newton's laws practice set is the closest fit on this site.

Sand falling onto a moving belt: the case where ma fails

A conveyor belt moves horizontally at a constant $v = 1.8$ m/s. Sand falls vertically onto it at a steady rate $dm/dt = 4.5$ kg/s. Find the extra horizontal force the motor must supply to keep the belt at constant speed, and compare the mechanical power delivered with the rate at which the sand gains kinetic energy.

Identify why this is not an $m\vec{a}$ problem. The belt's speed is constant, so if you wrote $F = ma$ you would get $F = 0$ and conclude the motor does nothing. That is wrong, and the reason is that the moving system's mass is increasing with time.
Declare the convention: positive in the direction of belt motion, horizontally.
Use the primitive form printed on the sheet, $\vec{F}_{\text{net}} = d\vec{p}/dt$ , and differentiate the product $p = mv$ properly: $\dfrac{dp}{dt} = m\dfrac{dv}{dt} + v\dfrac{dm}{dt}$ .
Here $dv/dt = 0$ , so the first term vanishes and only the second survives: $F = v\dfrac{dm}{dt}$ . This is exactly the case essential knowledge 4.2.B.2.iii names, the system in which the velocity is constant but the mass changes with respect to time.
Substitute: $F = (1.8)(4.5) = 8.1$ N. The sand arrives with zero horizontal velocity and has to be brought up to 1.8 m/s, and doing that at 4.5 kilograms per second takes a steady 8.1 newtons.
Now the energy comparison, which is the part that surprises people. The mechanical power the motor delivers against this force is $P = Fv = (8.1)(1.8) = 14.58$ W.
The rate at which the sand gains kinetic energy is $\dfrac{d}{dt}\left(\tfrac{1}{2}mv^2\right) = \tfrac{1}{2}v^2\dfrac{dm}{dt} = \tfrac{1}{2}(1.8)^2(4.5) = \tfrac{1}{2}(3.24)(4.5) = 7.29$ W.
Exactly half the motor's work becomes kinetic energy of the sand. The other 7.29 W is dissipated as the sand slips on the belt before it comes up to speed. The factor of two is not an accident of these numbers: it comes out of the algebra for any $v$ and any $dm/dt$ .
Check the units at every stage. $v\,dm/dt$ has units of $\mathrm{m/s} \times \mathrm{kg/s} = \mathrm{kg\,m/s^2}$ , which is newtons. Power is $\mathrm{N} \times \mathrm{m/s} = \mathrm{W}$ .
The general lesson: whenever a problem says mass is being added to or removed from the moving system, go back to $d\vec{p}/dt$ and differentiate the product. AP Physics 1 cannot pose this question, because its sheet stops at $m\vec{a}$ and its framework never separates the two terms.

$F = v\,dm/dt = 8.1$ N. The motor delivers $Fv = 14.6$ W, while the sand gains kinetic energy at $\frac{1}{2}v^2\,dm/dt = 7.29$ W, exactly half. The rest is dissipated as the sand slips up to speed.

A force that changes with time, integrated twice

A $2.0$ kg block on a frictionless horizontal surface starts at rest at the origin. From $t = 0$ to $t = 4.0$ s a horizontal force $F(t) = (6.0\ \mathrm{N/s})t - (1.5\ \mathrm{N/s^2})t^2$ acts on it. Find the acceleration as a function of time, the time of maximum acceleration, and the velocity and position at $t = 4.0$ s.

Declare the convention: positive in the direction of $F$ at small $t$ , where the force is positive.
Apply 2.5.A.2 at each instant. The mass is constant here, so $a(t) = \dfrac{F(t)}{m} = \dfrac{6.0t - 1.5t^2}{2.0} = 3.0t - 0.75t^2$ , in $\mathrm{m/s^2}$ with $t$ in seconds.
Maximum acceleration: differentiate and set to zero. $\dfrac{da}{dt} = 3.0 - 1.5t = 0$ gives $t = 2.0$ s, and $a(2.0) = 3.0(2.0) - 0.75(4.0) = 6.0 - 3.0 = 3.0\ \mathrm{m/s^2}$ .
Note also where the acceleration returns to zero: $3.0t = 0.75t^2$ gives $t = 4.0$ s. So the force vanishes exactly at the end of the interval, and the speed is greatest there.
Velocity, using the printed relation $\Delta v_x = \int a_x(t)\, dt$ and the initial condition $v = 0$ at $t = 0$ : $v(t) = \displaystyle\int_0^t (3.0t' - 0.75t'^2)\, dt' = 1.5t^2 - 0.25t^3$ .
At $t = 4.0$ s: $v = 1.5(16) - 0.25(64) = 24 - 16 = 8.0$ m/s.
Position, using $\Delta x = \int v_x(t)\, dt$ with $x = 0$ at $t = 0$ : $x(t) = \displaystyle\int_0^t (1.5t'^2 - 0.25t'^3)\, dt' = 0.5t^3 - 0.0625t^4$ .
At $t = 4.0$ s: $x = 0.5(64) - 0.0625(256) = 32 - 16 = 16$ m.
Cross-check the velocity by impulse, which is the same integral seen from the momentum side. $\displaystyle\int_0^{4} F\, dt = \int_0^4 (6.0t - 1.5t^2)\, dt = 3.0(16) - 0.5(64) = 48 - 32 = 16$ N s. Dividing by $m = 2.0$ kg gives 8.0 m/s, which matches.
Cross-check the average. Constant-acceleration kinematics does not apply here, so the average velocity is not $(v_0 + v)/2 = 4.0$ m/s. The actual average is $16/4.0 = 4.0$ m/s, which happens to agree for this particular cubic; that coincidence is worth noticing precisely so you do not rely on it.
Why the kinematic equations are out. The three printed equations $v_x = v_{x0} + a_x t$ and its partners all assume $a_x$ is constant. Here it is a function of time, so the only route is the two integral relations, which is exactly why the sheet prints them.

$a(t) = 3.0t - 0.75t^2$ , maximum $3.0\ \mathrm{m/s^2}$ at $t = 2.0$ s. At $t = 4.0$ s, where the force has fallen back to zero, $v = 8.0$ m/s and $x = 16$ m.

Frequently asked questions

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

No. The AP Physics C: Mechanics Table of Information prints two forms of the second law and neither is F equals ma. It prints the acceleration of a system as the sum of the forces divided by the system mass, and separately the net force as the derivative of momentum with respect to time. The AP Physics 1 sheet is the one that spells out the full chain ending in ma. Essential knowledge 4.2.B.2.ii of the C framework does print that chain, describing Newton's second law as a direct result of the impulse-momentum theorem applied to systems with constant mass, but the exam booklet stops at the derivative.

Why is the net force written as dp/dt rather than ma in AP Physics C?

Because the derivative of momentum is the general statement and ma is a special case of it. Essential knowledge 4.2.B.2.ii of the AP Physics C: Mechanics framework says Newton's second law is a direct result of the impulse-momentum theorem applied to systems with constant mass, and prints the chain from the derivative of p, through m times the derivative of v, to ma. Pulling the mass out of the derivative is only legal when the mass is constant. Essential knowledge 4.2.B.2.iii covers 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.

How do you solve a variable-mass problem in AP Physics C?

Start from the net force equals the derivative of momentum, then differentiate the product mv using the product rule, which gives m times dv/dt plus v times dm/dt. Each term matters in a different situation. If the mass is constant, the second term vanishes and you recover ma. If the velocity is constant while mass is being added, as with sand falling onto a moving conveyor belt, the first term vanishes and the force is the velocity multiplied by the rate of mass change. That second case is exactly what essential knowledge 4.2.B.2.iii of the AP Physics C: Mechanics framework describes, and AP Physics 1 has no counterpart to it.

Why does Newton's second law refer to the center of mass?

Because that is the one point whose motion the law can predict for an extended or deformable system. Essential knowledge 2.5.A.2 states that the acceleration of a system's center of mass has a magnitude proportional to the magnitude of the net force exerted on the system and is in the same direction as that net force. Individual parts of a spinning or flexing object accelerate in all sorts of ways, but the center of mass does not care, because internal forces cancel in pairs by Newton's third law. Essential knowledge 2.3.A.2 states that cancellation directly, and 2.5.A.3 draws the conclusion that only a nonzero net external force can change the velocity of a system's center of mass.

Is Newton's second law the same in AP Physics 1 and AP Physics C?

The topic content is identical. Both frameworks have one learning objective, 2.5.A, with the same three essential knowledge statements in the same words, and neither prints a boundary statement. What differs is the equation sheet and therefore the problems that can be set. The AP Physics C sheet prints the net force as the derivative of momentum instead of the finite-difference chain ending in ma, and it also prints that the change in velocity is the integral of acceleration with respect to time. Those two lines together make time-varying forces and variable-mass systems examinable in the C course and not in the algebra-based one.

What do you do when the force on an object changes with time?

Integrate. The kinematic equations printed on the sheet all assume constant acceleration, so they do not apply. Instead divide the force function by the mass to get the acceleration as a function of time, then use the printed relations that the change in velocity is the integral of acceleration with respect to time and the displacement is the integral of velocity with respect to time. Apply the initial conditions at each integration to fix the constant. Equivalently, the change in momentum is the integral of the net force over time, which the sheet also prints as the definition of impulse.

Does zero velocity mean zero acceleration?

No. Newton's second law relates the net force to the acceleration, not to the velocity, so the two are independent at any instant. A ball at the very top of its flight has zero velocity and an acceleration of g downward, because gravity is still the only force on it. A mass at the extreme of a spring oscillation has zero velocity and its maximum acceleration. Essential knowledge 2.5.A.2 makes the direction claim explicit: the acceleration is in the same direction as the net force, not the same direction as the velocity.