Newton's Second vs Third Law: The Difference

The second law is about one object: add every force acting on it and divide by its mass to get its acceleration. The third law is about two objects: it pairs a force on A with an equal and opposite force on B. Those two forces never appear on the same free-body diagram, so they never cancel.

AP Physics: Unit 2 (topics 2.2 Forces and Free-Body Diagrams, 2.3 Newton's Third Law, 2.5 Newton's Second Law). AP Physics 1 places the third law at Topic 2.3, before the first law at 2.4 and the second law at 2.5. LO 2.3.A asks students to describe the interaction of two objects using Newton's third law and a representation of paired forces exerted on each object, and EK 2.3.A.1 gives the pair as the force of A on B equal to the negative of the force of B on A. EK 2.3.A.2 adds that internal forces do not influence the motion of a system's center of mass. LO 2.5.A asks students to describe the conditions under which a system's velocity changes; EK 2.5.A.1 defines unbalanced forces, EK 2.5.A.2 states the second law with the relevant equation for the acceleration of a system, and EK 2.5.A.3 restricts changes in center-of-mass velocity to nonzero net external forces. EK 2.2.A.1.i states that a force on an object is always due to interaction with another object, EK 2.2.A.1.ii that an object or system cannot exert a net force on itself, and EK 2.2.B.2 that a free-body diagram shows the forces exerted on the object by the environment. EK 4.2.B.3 derives the second law from the impulse-momentum theorem for constant mass, and EK 4.3.A.3.i gives the impulse form of the third law. The AP Physics 1 equation sheet prints the second law twice and the third law not at all. Unit 2 carries 18 to 23 percent of the multiple-choice section.

One object, or two

Count the objects. That is the whole distinction, and everything else on this page follows from it.

Newton's second law counts one. EK 2.5.A.2: 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. Every force in that sum is a force on the system, and the answer is what that system does.

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

Newton's third law counts two. EK 2.3.A.1: Newton's third law describes the interaction of two objects in terms of the paired forces that each exerts on the other.

\vec{F}_{\text{A on B}} = -\vec{F}_{\text{B on A}}

Read the subscripts. The second law's forces all end in "on the system". The third law's two forces end in "on B" and "on A", and they are on different things by construction.

So the two laws answer different questions. The second law asks what one object does. The third law asks what the other object gets. They are used together constantly, and the failure mode is specific: the moment two forces on two different objects are added into one sum, the sum means nothing.

The glossary states each law on its own at Newton's second law and Newton's third law. This page is about telling them apart in a question that uses both.

Second law vs third law, side by side

	Newton's second law	Newton's third law
Number of objects it talks about	One system	Two, always
What it relates	All forces on that system, to its acceleration	One force on A, to one force on B
The CED statement	EK 2.5.A.2	EK 2.3.A.1
The equation	$\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}}$	$\vec{F}_{\text{A on B}} = -\vec{F}_{\text{B on A}}$
On the AP Physics 1 equation sheet	Yes, twice: the system form and $\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t} = m\vec{a}$	No line at all
Free-body diagrams involved	One	Two, and each force is on a different one
Are the forces equal in magnitude	Only when the situation makes them so	Always, in every situation, with no exceptions
Can the two forces cancel	Yes, that is what equilibrium is	Never. They act on different objects
Do the two forces have the same type	No, a diagram mixes gravity with contact forces	Yes, always. Both gravitational, or both normal, or both frictional
Depends on mass	Yes, directly	No. The pair is equal however different the masses are
What it is for	Getting a number for the acceleration	Getting a force on the other object, and knowing which arrows exist
What breaks it	Adding a force that acts on something else	Claiming a pair on one object

The row about types of force is the fastest test in practice, and section six turns it into a four-question drill.

The case that separates them: a book on a table has four forces and two pairs

One book, one table, one planet. Four forces exist, and they sort into two third law pairs, but only two of the four go on the book's free-body diagram.

Force	Exerted on	Exerted by	Third law partner	On the book's diagram
Gravitational, $mg$ down	The book	Earth	The book pulls Earth up with $mg$	Yes
Normal, $F_N$ up	The book	The table	The book pushes the table down with $F_N$	Yes
Gravitational, $mg$ up	Earth	The book	The first row	No
Normal, $F_N$ down	The table	The book	The second row	No

Now apply each law to that table.

Second law, on the book. Two forces are on the book: $mg$ down and $F_N$ up. The book is not accelerating, so they sum to zero and therefore $F_N = mg$ . Notice the reason. They are equal because the acceleration is zero, not because of any law about pairs.

Third law, on the two pairs. Rows one and three are a pair, and rows two and four are a pair. Each pair is equal in magnitude and opposite in direction always, whatever the book is doing.

Which means the famous wrong answer is wrong for a structural reason rather than a numerical one. The normal force is not the reaction to the book's weight. Both of those forces act on the book, and a third law pair never has both members on the same object. The reaction to Earth pulling the book down is the book pulling Earth up, which is a force on the planet and belongs on no diagram you will ever draw in this course. The reaction to the table pushing the book up is the book pushing the table down.

The cleanest disproof needs no elevator and no acceleration at all: put a second book on top. The lower book's weight has not changed by a gram, and the normal force from the table has grown by the whole weight of the upper book, so the two are now unequal while the book still sits perfectly still. Two forces that can be made unequal are not a third law pair, because third law pairs are equal with no exceptions. The numbers are worked below, and weight vs normal force takes the same error from the direction of the two forces rather than the two laws.

Why a third law pair can never cancel

Cancellation is a property of a sum, and a sum is only meaningful for forces on the same object. That is the entire argument, and it is worth being able to say in one line, because free-response questions ask for it in exactly that form.

EK 2.2.B.2 sets the rule for what may enter the sum: the free-body diagram of an object or system shows each of the forces exerted on the object by the environment. Forces on other objects are not on it. So the two members of a third law pair are structurally incapable of appearing together in one $\sum \vec{F}$ , and two things that never appear in the same sum cannot cancel within it.

The question people ask next is the right one. If every force has an equal and opposite partner, how does anything ever accelerate? Two answers, and both are useful.

The partner is somewhere else. When you push a shopping trolley, the trolley pushes you back with exactly the same force. That backward force is on you, not on the trolley, so the trolley's own sum still has an unbalanced forward push in it. Look only at the trolley and the third law partner is nowhere in sight.
When both partners are inside your system, they do cancel, and that is correct. EK 2.3.A.2: interactions between objects within a system, the internal forces, do not influence the motion of a system's center of mass. Enclose you and the trolley together and the push genuinely contributes nothing. What accelerates the pair is the friction from the floor on your shoes, which crosses the boundary. See internal vs external forces.

The second law also has a guard rail against the same confusion. EK 2.2.A.1.ii: an object or system cannot exert a net force on itself. A force you exert on your own system is always half of a pair whose other half is also inside, so the two cancel and the sum is untouched.

One more consequence, in momentum language. EK 4.3.A.3.i says the impulse exerted by one object on a second object is equal and opposite to the impulse exerted by the second object on the first, and calls it a direct result of Newton's third law. The two objects therefore get equal and opposite momentum changes, which is why conservation of momentum works at all. The third law is not a curiosity attached to the second law; it is what makes momentum conservation true.

Equal forces, unequal accelerations, and why that is not a contradiction

A 2400 kg truck hits an 800 kg car. Which one pushes harder?

Neither. The third law makes the two forces equal in magnitude at every instant of the collision, with no allowance for the truck being three times heavier. What differs is what each vehicle does with the force it receives, and that is the second law's department.

	Truck, 2400 kg	Car, 800 kg
Force from the other vehicle	48000 N	48000 N, opposite direction
Which law fixed that	Third	Third
Acceleration	$48000/2400 = 20\ \text{m/s}^2$	$48000/800 = 60\ \text{m/s}^2$
Which law gave that	Second	Second
Velocity change over 0.12 s	2.4 m/s	7.2 m/s
Momentum change	$5760\ \text{kg}\cdot\text{m/s}$	$5760\ \text{kg}\cdot\text{m/s}$ , opposite direction

The car's occupants are in far more trouble than the truck's, and no line of that table says the car was pushed harder. The two laws are doing separate jobs in adjacent rows: the third law equalizes the forces, and the second law divides each one by a different mass.

That is also the resolution to the oldest objection in mechanics. A horse pulls a cart; the cart pulls the horse back just as hard; so how does the pair ever move? Because those two forces are on two different objects and never meet in a sum. On the cart, the horse's pull competes with rolling resistance. On the horse, the cart's backward pull competes with the friction the ground exerts forward on its hooves. Both objects accelerate because both of those separate competitions come out positive, and the third law pair between them is not part of either competition. The full arithmetic is the third worked example below.

Four questions that identify a third law partner

Given a force, and asked for its Newton's third law partner, run these in order. Any one of them settles most exam cases on its own.

Rewrite the force as "X on Y". The partner is "Y on X". Nothing else about it needs to be decided. The gravitational force of Earth on the book has the partner the gravitational force of the book on Earth. This single reversal answers the question every time it is asked directly.
Check the two objects are different. If your candidate partner acts on the same object as the original, it is not the partner. This is the test that kills the normal-force-and-weight answer in one step.
Check the type matches. A third law pair is always the same interaction seen from both ends, so it is always the same kind of force. Two gravitational, two normal, two frictional, two tension. If one candidate is gravitational and the other is a contact force, they are not a pair, because they are not even the same interaction.
Look for a counterexample to equality. Third law pairs are equal in magnitude in every situation with no exceptions. If you can invent any arrangement, any acceleration, any extra object, that makes your two forces unequal, they were never a pair. Stack a second book on the first and the normal force and the weight part company, which settles it.

And the reverse drill, for identifying which law a question wants:

The question says	The law you need
Find the acceleration	Second
Find the net force	Second
Name the reaction force	Third
Compare the forces the two objects exert on each other	Third
Compare the accelerations of the two objects	Both: third for the forces, second for the accelerations
Explain why the object does not move	Second, with the net force zero, which is Newton's first law as a special case
Explain how something can push off something else	Third, then second on each object separately

The habit that prevents most of these errors is drawing two diagrams instead of one. Put the pair on two separate dots and it becomes visually impossible to add them together. How to draw a free-body diagram covers the mechanics.

Where it costs a mark

Calling the normal force the reaction to the weight. Both act on the same object. Asked for the partner of the normal force on a block, the answer is the downward push the block exerts on the surface, and answering the weight scores nothing.
Claiming a third law pair cancels. They act on two different objects, so they never enter the same sum. A justification that says the forces cancel so nothing moves has used the wrong law.
Saying the heavier object exerts more force. It does not. The third law equalizes the magnitudes regardless of mass, and the mass difference shows up in the accelerations instead.
Putting a force on the wrong diagram. EK 2.2.B.2 restricts the diagram to forces exerted on the object by the environment. A force the object exerts on something else belongs on that other thing's diagram.
Drawing force components as extra arrows. The Topic 2.2 boundary statement says AP Physics 1 only expects students to depict the forces exerted on objects, not the force components, and that individual forces must be drawn as individual straight arrows originating on the dot and pointing in the direction of the force, with same-direction forces drawn side by side rather than overlapping.
Using the second law with the mass of the wrong object. The sum and the mass must belong to the same system. Mixing the net force on the pair with the mass of one part is a fast way to a wrong acceleration.
Forgetting the second law is a vector equation. It is one equation per axis, and EK 2.2.B.4 recommends choosing a coordinate system with one axis parallel to the direction of acceleration, which is why an incline gets tilted axes.
Explaining an acceleration with an internal force. A car's engine, a person's muscles and a rocket's own frame are all inside. EK 2.2.A.1.ii forbids a system exerting a net force on itself, so name the external force from the ground, the road or the exhaust.
Treating the third law as approximate during a collision. It is exact at every instant, including while the forces are changing violently. EK 4.3.A.3.i carries it into impulses, which is why the two momentum changes match to the digit.

What the CED requires, and what the sheet prints

The two laws sit in adjacent topics of AP Physics 1 Unit 2, and the ordering is deliberate: the third law comes first, at Topic 2.3, before the first law at 2.4 and the second law at 2.5. The framework builds the idea that a force is an interaction before it builds the idea that forces add.

Topic 2.3, Newton's Third Law. LO 2.3.A is to describe the interaction of two objects using Newton's third law and a representation of paired forces exerted on each object. Note the phrase paired forces exerted on each object: the learning objective itself asks for two representations. EK 2.3.A.1 gives $\vec{F}_{\text{A on B}} = -\vec{F}_{\text{B on A}}$ , EK 2.3.A.2 draws the internal-force consequence, and EK 2.3.A.3 applies the law to tension, defining it as the macroscopic net result of forces that segments of a string, cable, chain, or similar system exert on each other in response to an external force. The suggested skills for the topic are 1.A, 2.D, 3.B and 3.C.

Topic 2.5, Newton's Second Law. LO 2.5.A is to describe the conditions under which a system's velocity changes. EK 2.5.A.1 defines unbalanced forces as a configuration of forces such that the net force exerted on a system is not equal to zero, EK 2.5.A.2 states the law itself, and EK 2.5.A.3 restricts changes in the velocity of a system's center of mass to nonzero net external forces. Its suggested skills are 1.A, 2.A, 2.D and 3.B.

On the equation sheet the asymmetry is total. The AP Physics 1 sheet prints the second law twice, once as $\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}} = \frac{\vec{F}_{\text{net}}}{m_{\text{sys}}}$ and once, in the momentum group, as $\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t} = m\frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$ . It prints nothing at all for the third law, on any of the four course sheets, because the third law tells you which forces exist rather than what to substitute. Its work is done when you build the diagram. EK 4.2.B.3 connects the two printed lines, saying Newton's second law of motion is a direct result of the impulse-momentum theorem applied to systems with constant mass, and the Topic 4.2 boundary statement adds 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 full sheet is here.

Unit 2 carries 18 to 23 percent of the AP Physics 1 multiple-choice section. Topic 2.3 Newton's Third Law and Topic 2.5 Newton's Second Law have the full framing, and how to find net force owns the second law's arithmetic.

Truck meets car: the third law sets the force, the second law sets the damage

A 2400 kg truck collides head on with an 800 kg car. During the 0.12 s of contact the truck exerts an average force of 48000 N on the car. Find the force the car exerts on the truck, each vehicle's acceleration, each vehicle's change in velocity and each vehicle's change in momentum. Say which law produced each answer.

Declare the convention: positive is the truck's original direction of travel. Both vehicles are treated as objects.
Third law first, because it needs no masses. EK 2.3.A.1 gives $\vec{F}_{\text{car on truck}} = -\vec{F}_{\text{truck on car}}$ , so the car exerts 48000 N on the truck, backward. The mass difference does not enter this line at all.
Second law on the car, whose only significant horizontal force during the impact is that 48000 N: $a_{\text{car}} = (48000\ \text{N})/(800\ \text{kg}) = 60\ \text{m/s}^2$ , directed forward relative to the truck's motion.
Second law on the truck: $a_{\text{truck}} = (48000\ \text{N})/(2400\ \text{kg}) = 20\ \text{m/s}^2$ , directed backward. Three times the mass, one third of the acceleration.
Velocity changes over the 0.12 s of contact: $\Delta v_{\text{car}} = (60\ \text{m/s}^2)(0.12\ \text{s}) = 7.2\ \text{m/s}$ and $\Delta v_{\text{truck}} = (20\ \text{m/s}^2)(0.12\ \text{s}) = 2.4\ \text{m/s}$ .
Momentum changes: $\Delta p_{\text{car}} = (800\ \text{kg})(7.2\ \text{m/s}) = 5760\ \text{kg}\cdot\text{m/s}$ and $\Delta p_{\text{truck}} = (2400\ \text{kg})(2.4\ \text{m/s}) = 5760\ \text{kg}\cdot\text{m/s}$ , in opposite directions.
Check both against the impulse: $J = F\Delta t = (48000\ \text{N})(0.12\ \text{s}) = 5760\ \text{N}\cdot\text{s}$ , which matches each momentum change exactly. This is EK 4.3.A.3.i, the impulse form of the third law.
Check the system. The two momentum changes sum to zero, so the two-vehicle system's total momentum is unchanged, exactly as it must be when the only forces on the pair are internal.

The car pushes back with 48000 N, the same number, because the third law admits no exceptions for mass. Accelerations are 60 m/s squared for the car and 20 m/s squared for the truck, from the second law. The momentum changes are equal at 5760 kg m/s and opposite. Every asymmetry in this problem came from the second law dividing by different masses, and none of it came from the forces, which were identical.

Two stacked books: the normal force and the weight come apart with nothing moving

A 1.4 kg book lies on a table with a 0.90 kg book resting on top of it. Nothing moves. Find every force on each book, find the normal force from the table on the lower book, and use the result to show that the normal force and the weight of the lower book are not a Newton's third law pair.

Declare the convention: positive is upward. Take the two books as separate objects, so each gets its own free-body diagram.
Weights: upper book $F_{g,\text{up}} = (0.90\ \text{kg})(9.8\ \text{m/s}^2) = 8.82\ \text{N}$ down; lower book $F_{g,\text{low}} = (1.4\ \text{kg})(9.8\ \text{m/s}^2) = 13.72\ \text{N}$ down.
Upper book, second law with zero acceleration. Two forces: its weight down and the normal force from the lower book up. So $N_{\text{low on up}} = 8.82\ \text{N}$ upward.
Third law on that contact: the upper book pushes down on the lower book with 8.82 N. Same magnitude, opposite direction, and the two forces sit on two different diagrams.
Lower book, second law with zero acceleration. Three forces: its own weight 13.72 N down, the upper book's push 8.82 N down, and the table's normal force $N_{\text{table}}$ up. So $N_{\text{table}} = 13.72 + 8.82 = 22.54\ \text{N}$ upward.
Now compare the two forces people call a pair. On the lower book the upward normal force is 22.54 N and the downward weight is 13.72 N. They differ by 8.82 N while the book sits motionless on a table with zero acceleration.
Conclude. A third law pair is equal in magnitude in every situation, so two forces that can be made unequal by placing a book on top were never a pair. What is actually true here is that the three forces on the lower book sum to zero, which is the second law with $a = 0$ .
Name the real partners for completeness. The partner of the lower book's 13.72 N weight is a 13.72 N upward gravitational pull the book exerts on Earth. The partner of the table's 22.54 N push is a 22.54 N downward push the lower book exerts on the table.
Check the whole stack as one system: total weight $(0.90 + 1.4)(9.8) = (2.3)(9.8) = 22.54\ \text{N}$ , balanced by the table's 22.54 N. The internal 8.82 N pair has cancelled, as EK 2.3.A.2 requires.

The table pushes up on the lower book with 22.54 N while that book's weight is 13.72 N, a gap of 8.82 N with nothing accelerating anywhere. That single fact disproves the claim that the normal force is the reaction to the weight, without needing an elevator or any motion at all. The two forces were only ever equal in the simpler arrangement because the second law made them equal there, and a coincidence produced by $a = 0$ is not a law about pairs.

The horse and the cart: why an equal and opposite pair still lets both move

A 600 kg horse pulls a 250 kg cart along level ground. Rolling resistance on the cart totals 300 N. The pair accelerates forward at 0.40 m/s squared. Find the tension in the harness, find the friction force the ground exerts forward on the horse, and explain why the cart pulling back on the horse does not prevent the acceleration.

Declare the convention: positive is forward, the direction of the acceleration. Treat the horse and the cart as two objects, and take the harness as an ideal connector so the tension pulling the cart forward and the tension pulling the horse backward have the same magnitude $T$ .
Second law on the cart. Two horizontal forces: the harness pulls it forward with $T$ , and rolling resistance pushes it back with 300 N. So $T - 300 = m_{\text{cart}} a = (250\ \text{kg})(0.40\ \text{m/s}^2) = 100\ \text{N}$ , giving $T = 400\ \text{N}$ .
Third law on the harness. The cart therefore pulls backward on the horse with exactly 400 N. Same number, opposite direction, different object.
Second law on the horse. Two horizontal forces: the ground pushes its hooves forward with a friction force $F_{\text{ground}}$ , and the cart pulls back with 400 N. So $F_{\text{ground}} - 400 = m_{\text{horse}} a = (600\ \text{kg})(0.40\ \text{m/s}^2) = 240\ \text{N}$ , giving $F_{\text{ground}} = 640\ \text{N}$ .
Answer the paradox. The 400 N backward pull on the horse is real and it is exactly matched by the 400 N forward pull on the cart. But those two forces are on two different objects, so neither one is cancelled by the other. On the horse, 640 N forward beats 400 N backward. On the cart, 400 N forward beats 300 N backward. Both competitions come out positive, and both objects accelerate.
Check by taking horse and cart as one system, total mass 850 kg. The harness tension is now internal and disappears. The external horizontal forces are the 640 N ground friction forward and the 300 N rolling resistance backward: $640 - 300 = 340\ \text{N}$ , and $340/850 = 0.40\ \text{m/s}^2$ , matching the given acceleration exactly.
Note what the third law did and did not do. It supplied the 400 N on the horse, once the cart's equation had produced it. It never told anyone whether the system accelerates, because that is not a question about a pair.

Harness tension 400 N, ground friction on the horse 640 N forward. The cart really does pull back on the horse with the full 400 N, and the horse accelerates anyway, because the ground supplies 640 N in the other direction. The third law pair never appeared in the same equation, and the whole-system check at 0.40 m/s squared confirms the two-object treatment. This is why the two laws are used together: the third law hands you a force on the second object, and the second law is what you do with it once you have it.

Frequently asked questions

What is the difference between Newton's second law and Newton's third law?

The second law is about one object and the third law is about two. The second law says the acceleration of a system's center of mass is proportional to the net force on that system and points in the same direction, so it adds up every force acting on one thing. The third law says that when two objects interact, each exerts a force on the other that is equal in magnitude and opposite in direction, so it links a force on object A to a force on object B. Because the third law's two forces act on different objects, they never appear in the same free-body diagram and never cancel each other. The second law's forces all act on the same object and can cancel.

Is the normal force the Newton's third law reaction to weight?

No, and it is the error the third law produces most often. Both the weight and the normal force act on the same object, while a third law pair always acts on two different objects. The reaction to Earth pulling a book down is the book pulling Earth up with the same force, which is a force on the planet. The reaction to the table pushing the book up is the book pushing the table down. A quick proof needs no motion at all: stack a second book on the first and the lower book's weight is unchanged while the normal force from the table grows by the upper book's weight, so the two are now unequal. Third law pairs are equal in every situation.

Why do Newton's third law forces not cancel out?

Because cancellation only happens inside a sum, and a sum only makes sense for forces acting on the same object. AP Physics 1 EK 2.2.B.2 says a free-body diagram shows the forces exerted on the object by the environment, so a force on some other object is not on that diagram and cannot be added there. When you push a trolley, the trolley pushes you back just as hard, but that backward force is on you, so the trolley's own sum still has an unbalanced forward force in it and the trolley accelerates. If you enclose both you and the trolley in one system, the pair does cancel, and EK 2.3.A.2 says so: internal forces do not influence the motion of a system's center of mass.

If the forces are equal, why does the smaller object accelerate more?

Because two different laws are doing two different jobs. The third law fixes the two forces at the same magnitude, and nothing about mass enters that statement. The second law then divides each force by a different mass, so a 48000 N force gives an 800 kg car 60 m/s squared and gives a 2400 kg truck only 20 m/s squared. Every asymmetry in a collision comes from the second law, and none of it comes from the forces. The momentum changes still match exactly, at 5760 kg m/s each in this example, because the same force acted for the same length of time on both.

How do I find the Newton's third law partner of a force?

Write the force in the form force of X on Y, then swap the names: the partner is the force of Y on X. That single reversal answers the question every time it is asked directly, and it also builds in the checks. The partner acts on a different object, always. It is the same type of force, always, because it is the same interaction viewed from the other end, so a gravitational force pairs with a gravitational force and a normal force pairs with a normal force. And it is equal in magnitude in every situation, so if you can construct any arrangement in which your two candidate forces differ, they were never partners.

Which law do I use for a problem with two connected objects?

Usually both, in a fixed order. Use the second law on the whole system first, treating the two objects as one, because the connecting force is internal and drops out, which gives the acceleration in one line. Then, if the question wants the connecting force itself, use the second law again on one object alone, where that force is now external. The third law is what lets you use the same symbol for the pull on each object, since the force one exerts on the other is equal and opposite. Choosing the boundary is covered at object vs system, and the tension case has its own routine in the how to find tension guide.

Does Newton's third law appear on the AP Physics equation sheet?

No, on none of the four course sheets. The second law appears twice on the AP Physics 1 sheet, as the system form giving the acceleration as the net force divided by the system mass, and again in the momentum group as the net force equal to the rate of change of momentum and to mass times acceleration. The third law is printed nowhere, and that is not an oversight. It is a statement about which forces exist rather than a formula to substitute into, so its work is done at the diagram stage. The CED does state it symbolically in EK 2.3.A.1 as the force of A on B being the negative of the force of B on A.