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FRQ: Newton’s Third Law and Internal Forces — The Tractor–Trailer Tug-of-War

Assessments aligned to 2026 AP Physics 1 standards

Qualitative/Quantitative Translation (QQT)  |  MID-LEVEL  |  8 points  |  20 min

▤ Scenario

A small tractor tows a loaded trailer along a flat, straight farm track. The two are joined by a rigid coupling bar whose sensor records the magnitude of the force in the bar, as shown in Figure 1. The bar’s mass, air resistance, and the rolling resistance of both vehicles are negligible.

Figure 1 — The tractor, the trailer and the instrumented coupling bar between them.

A student holds the tractor at one fixed engine setting, so the forward drive force F_drive that the ground exerts on the tractor’s drive wheels stays constant at 3600 N on every run, and the tractor’s own mass stays at m_tractor = 600 kg. The trailer is then loaded to four different masses m_trailer, and for each load the acceleration and the sensor reading are recorded, as shown in Figure 2.

Figure 2 — The four runs: recorded acceleration and coupling-bar force for each trailer load.

Let F_1 be the magnitude of the force exerted on the trailer by the tractor through the coupling bar, and let F_2 be the magnitude of the force exerted on the tractor by the trailer through the bar.

All motion is one-dimensional and every quantity is measured in the ground frame. Take the tractor’s direction of travel as the positive direction.

✎ Free Response Questions

Part A

Consider the run with m_trailer = 1200 kg, at an instant while the tractor and the trailer are speeding up.

Indicate whether F_2 is greater than, less than, or equal to F_1 by writing one of the following.

•   F_2 > F_1

•   F_2 < F_1

•   F_2 = F_1

Justify your answer using qualitative reasoning beyond referencing equations.

Part B

Starting with Newton’s second law, derive an expression for F_1. Express your final answer in terms of F_drive, m_tractor, m_trailer, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

Part C

Justify how your derived equation in part B is or is not consistent with your reasoning in part A.

❖ Answer Key & Scoring Guide

▸ earns credit  ⚠︎ common error, partial credit  ✗ common error, no credit

Points are independent and there are no deductions. ▸ criteria follow College Board’s published scoring standard; ⚠︎ and ✗ lines are TheScienceCube teaching commentary, not College Board rubric.

Part A — Model Answer

F_2 = F_1

The tractor and the trailer pull on each other through the coupling bar. The bar’s mass is negligible, so the net force on it is zero: the tractor pulls the bar forward exactly as hard as the trailer pulls it back. By Newton’s third law the bar pulls back on the tractor, and forward on the trailer, just as hard. So F_1 and F_2 have equal magnitudes, and the two pulls behave as a Newton’s third law pair between the vehicles themselves. The sensor’s single reading fits this: a bar of negligible mass carries the same force along its whole length, so one reading gives the size of both pulls.

The equality holds whatever the masses and whatever the motion. The trailer on this run has twice the tractor’s mass, but that cannot make it pull back harder than it is pulled: a heavier load makes both pulls larger together — the recorded force in Figure 2 rises with the load — never one larger than the other. Nor does speeding up need one pull to beat the other, because the two pulls are exerted on different vehicles. F_1 is the only horizontal force exerted on the trailer — the trailer’s own pull, F_2, is exerted on the tractor — so the trailer has a forward net force and speeds up. The tractor speeds up because the ground’s forward drive force on it is larger than the trailer’s backward pull F_2. The two pulls are exerted on different vehicles, so they can never cancel on either one of them.

Figure 3 — Horizontal forces on each vehicle at the 1200 kg load, one scale: F_1 equals F_2.

The size of the pair is set by the forces from outside and the motion they produce, not the other way round. If the trailer’s brakes were set so that the ground pushed back on the trailer with 3600 N, the net external force on the pair would be zero and neither vehicle would speed up — yet the two pulls would still be equal, and larger than on this run: 3600 N each.

Scoring (3 points):

▸ A1 — 1 point: For indicating F_2 = F_1.

▸ A2 — 1 point: For a justification that indicates that the two pulls are equal in magnitude because they are the forces the tractor and the trailer exert on each other through the bar (Newton’s third law), or because the bar, whose mass is negligible, pulls on the two vehicles equally hard.

▸ A3 — 1 point: For a justification that indicates that the equality does not depend on the vehicles’ masses or on their motion — for example, that the heavier trailer pulls back no harder than it is pulled, or that the pair can speed up with equal pulls because the pulls are exerted on different vehicles and never act together on either one.

Scoring Note: Addressing either the masses or the motion is sufficient for A3. Stating that the pulls act on different vehicles, so they cannot cancel on either one, meets it by the motion route; a bare “always” or “at every instant” does not. A response in which the two pulls cancel on the trailer or on the tractor does not meet A3 by the motion route.

Scoring Note: A2 and A3 are content criteria, not route criteria; any wording with that content is sufficient, and the words “third law” are not required.

⚠︎ Common error (partial credit): Indicating F_2 = F_1 and citing Newton’s third law, but never addressing the trailer’s greater mass or the speeding up — earns 2 of 3 points, for A1 and A2.

⚠︎ Common error (partial credit): Indicating F_2 = F_1 as a third-law pair, but arguing that the equal pulls cancel on the trailer, so that the drive force carries the trailer forward, and never addressing the masses — earns 2 of 3 points, for A1 and A2. The drive force is exerted on the tractor, and the two pulls are exerted on different vehicles; they cancel only in the net force on the two vehicles taken together.

⚠︎ Common error (partial credit): Indicating F_2 = F_1 only because the two vehicles share one acceleration, because the bar is rigid, or because the sensor gives a single reading, and never addressing the masses — earns 1 of 3 points, for A1. Rigidity gives the vehicles one motion, and equal accelerations with unequal masses need unequal net forces; the pulls are equal because the bar’s mass is negligible — a bar with mass, speeding up, would pull back on the tractor harder than it pulls the trailer forward.

✗ Common error (no credit): Indicating F_2 < F_1 because the tractor must out-pull the trailer or the trailer could not speed up — including a response that calls the pulls an action–reaction pair and has the action win — earns 0 points. The trailer’s pull is exerted on the tractor, so it cannot hold the trailer back; F_1 alone accelerates the trailer.

✗ Common error (no credit): Indicating F_2 > F_1, either because the heavier trailer pulls harder than it is pulled or from the data — for example F_2 = F_drive − m_tractor·a = 3600 N − 600 kg × 1.98 m/s² = 2412 N against the recorded 2370 N — earns 0 points. A heavier trailer makes both pulls larger together, never one larger than the other; and a 42 N gap built from two recorded columns is measurement scatter, which reverses at 900 kg (2142 N against 2190 N).

Part B — Model Answer

Treat the tractor, the bar and the trailer as one system, of mass m_tractor + m_trailer. The bar’s mass is negligible and it lies inside this system, so the pulls it exerts on the two vehicles are internal to it; being equal and opposite, they cancel in the net force on the two vehicles taken together. The track is flat, the vertical forces balance, and air resistance and rolling resistance are negligible, so the only horizontal force exerted on the system from outside it is the ground’s drive force on the tractor’s wheels. Newton’s second law for the system gives

F_net = m·a
F_drive = (m_tractor + m_trailer)·a
a = F_drive / (m_tractor + m_trailer)

Figure 4 — The pair as one system at the 1200 kg load: only the drive force acts from outside.

The rigid bar keeps the two vehicles together, so both share this acceleration. Now isolate the trailer. F_1 is the only horizontal force exerted on it, so

F_1 = m_trailer·a = F_drive·m_trailer / (m_tractor + m_trailer)

Equivalently, the second law for each vehicle alone gives F_1 = m_trailer·a for the trailer and F_drive − F_2 = m_tractor·a for the tractor; setting F_2 = F_1 and eliminating a gives the same result. F_1 is the only horizontal force on the trailer, so it equals the trailer’s mass times the acceleration the drive force gives the whole pair.

Checking against the data (beyond what part B asks). At 600 kg the derived results give a = 3600 N / 1200 kg = 3.00 m/s² and F_1 = 3600 N × 600 kg / 1200 kg = 1800 N, against the recorded 2.94 m/s² and 1780 N, and every run in Figure 2 lies within about 2 percent of its prediction, some above and some below. Counting the bar’s forward pull on the trailer as an outside force on the pair would instead predict (3600 N + 1780 N) / 1200 kg = 4.48 m/s² at 600 kg, far above the 2.94 m/s² recorded.

Scoring (3 points):

▸ B1 — 1 point: For a multistep derivation that includes Newton’s second law for the tractor, the trailer, or the two together. A final expression for F_1 on its own is not a multistep derivation.

▸ B2 — 1 point: For a Newton’s second law relation for one vehicle in which that vehicle’s own mass multiplies its acceleration and its net horizontal force is correct — F_1 = m_trailer·a for the trailer, or F_drive − F_2 = m_tractor·a (or F_drive − F_1 = m_tractor·a) for the tractor. The acceleration may appear as the response’s own expression for it. A correct final expression for F_1, in any equivalent form, also shows this relation.

Scoring Note: B1 and B2 carry no consistency credit. B2 is not affected by an error in an acceleration the response derived, but an isolated incorrect final expression does not meet B2.

▸ B3 — 1 point: For a correct expression for F_1 in terms of F_drive, m_tractor and m_trailer, such as F_1 = F_drive·m_trailer / (m_tractor + m_trailer) or F_1 = F_drive / (1 + m_tractor/m_trailer). Where B2 was not earned, an expression that follows correctly from the response’s own incorrect single-vehicle relation, where F_1 or F_2 appears in that relation, also earns this point; any other error prevents it.

Scoring Note: A correct, isolated, final expression for F_1 earns points B2 and B3.

⚠︎ Common error (partial credit): Finding the common acceleration wrongly — from one vehicle’s mass, a = F_drive/m_tractor or a = F_drive/m_trailer, or by counting the bar’s pull as a second external force on the pair, F_drive + F_1 = (m_tractor + m_trailer)·a — and then writing F_1 = m_trailer·a correctly — earns 2 of 3 points, for B1 and B2. The drive force is the only horizontal external force on the pair; the two pulls are internal to it.

⚠︎ Common error (partial credit): Isolating the tractor and writing F_1 = m_tractor·a, which gives F_1 = F_drive·m_tractor / (m_tractor + m_trailer) — earns 2 of 3 points, for B1 and B3. The tractor has two horizontal forces exerted on it, so its net force is F_drive − F_2.

⚠︎ Common error (partial credit): Writing the trailer’s second law with the drive force as its net force, F_drive = m_trailer·a, and taking F_1 = F_drive without writing F_1 = m_trailer·a, or writing the pair’s second law correctly and then stating F_1 = F_drive — earns 1 of 3 points, for B1. The drive force is exerted on the tractor; the only horizontal force on the trailer is F_1.

✗ Common error (no credit): Stating F_1 = F_drive because the bar passes the whole drive force to the trailer, with no second-law relation written for either vehicle or for the pair — earns 0 points. The drive force is exerted on the tractor; F_1 alone accelerates the trailer.

Part C — Model Answer

The derived equation is consistent with the reasoning in part A.

Part A found the two pulls equal and argued that equal pulls still let both vehicles speed up, because each pull is exerted on a different vehicle. In the part B equation the fraction m_trailer/(m_tractor + m_trailer) is less than 1 for any tractor with mass, so F_1 is always less than F_drive. The trailer pulls back on the tractor with that same F_1, so the ground’s forward push on the tractor always exceeds the trailer’s backward pull, and the tractor keeps a forward net force, F_drive − F_1 = m_tractor·a. Had part B given F_1 equal to or greater than F_drive, equal pulls would have left the tractor no forward net force, and the equation would not have been consistent with part A.

The total-mass denominator agrees. The bracket in F_1 = m_trailer × [F_drive / (m_tractor + m_trailer)] is the acceleration that the drive force alone gives the whole pair, which is right only because the two pulls, being equal and opposite, cancel in the net force on the two vehicles taken together. An equation that rests on that cancellation cannot be consistent with unequal pulls.

Loading the trailer more heavily raises the fraction while the drive force stays at 3600 N, so F_1 climbs toward F_drive — the recorded force rises from 1210 N at 300 kg to 2370 N at 1200 kg — and both pulls rise together. It never reaches 3600 N, because the tractor has mass of its own: to speed up with the trailer it needs a forward net force, so the ground’s push must always exceed the trailer’s pull back by m_tractor·a. The tractor speeds up because the ground pushes it forward harder than the trailer pulls it back, never because it out-pulls the trailer.

Figure 5 — F_1 rises toward F_drive = 3600 N with the load; the gap is the tractor’s net force.

Scoring (2 points):

▸ C1 — 1 point: For attempting to address how F_1 depends on F_drive or on the masses in the response’s own part B expression, whether or not it is correct — for example by comparing F_1 with F_drive, by describing how F_1 changes with the masses, or by interpreting the expression’s numerator or denominator.

Scoring Note: It is not necessary to use the expression correctly to earn this point. The response only needs to use functional-dependence language — such as “fraction of”, “increases with”, “numerator” or “denominator” — to relate F_1 to F_drive or to the masses.

▸ C2 — 1 point: For correctly using the part B expression to decide whether it agrees with part A — its answer or its reasoning — judged against the response’s own part B expression and its own part A response. Award the point for either verdict, consistent or not consistent, wherever it follows correctly.

Scoring Note: Correct uses include that F_1 is less than F_drive for any tractor with mass, so an equal backward pull still leaves the tractor a forward net force (a part B expression giving F_1 ≥ F_drive is not consistent with equal pulls on a pair that is speeding up); that the total-mass denominator treats the drive force as the only force accelerating the pair, which holds only if the two pulls cancel in the net force on the two vehicles together (so a correct part B expression is not consistent with a part A that found F_2 ≠ F_1); and that F_1 rises with m_trailer while F_drive stays fixed, so a heavier load raises both pulls together, never one above the other.

⚠︎ Common error (partial credit): Showing that F_1 is less than F_drive, or that it rises toward F_drive with the load, but giving no verdict on whether the equation agrees with part A — earns 1 of 2 points, for C1.

⚠︎ Common error (partial credit): Keeping part A’s F_2 < F_1 and calling the part B equation consistent because F_2 < F_1 < F_drive still leaves the tractor a forward net force — earns 1 of 2 points, for C1. A correct part B equation holds only if the two pulls cancel in the net force on the two vehicles together, which unequal pulls would not.

⚠︎ Common error (partial credit): Reading F_1 < F_drive as the tractor out-pulling the trailer, or the rise of F_1 with m_trailer as the heavier trailer pulling harder than it is pulled, or as the drive force growing with the load — earns 1 of 2 points, for C1. F_drive is the ground’s push on the tractor, not a pull on the trailer; a heavier load needs a larger F_1, the trailer pulls back with that same F_1, and the drive force stays at 3600 N.

✗ Common error (no credit): Asserting that the equation agrees, or does not agree, with part A with no statement about what the part B expression shows, or stating that it cannot bear on part A because F_2 does not appear in it — earns 0 points. Its total-mass denominator holds only if the two pulls cancel, and F_1 < F_drive decides whether equal pulls still let the tractor speed up.