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FRQ: Newton’s Second Law for Connected Systems — The Cargo Deck Crates

Assessments aligned to 2026 AP Physics 1 standards

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

▤ Scenario

A dock worker moves two crates along a level, frictionless cargo deck. Crate A has mass m_A = 4.0 kg and crate B has mass m_B = 6.0 kg. A horizontal rope attached to the front of crate A pulls it with tension P. A light, inextensible cable runs backward from crate A to crate B, so crate B trails directly behind crate A and the two crates move together, as shown in Figure 1. Use g ≈ 10 N/kg.

Figure 1 — The two crates on the level, frictionless deck, with the rope attached to crate A.

All motion is one-dimensional along the deck. Take the direction of the pull as positive, and treat the deck as an inertial reference frame.

✎ Free Response Questions

Part A

The crates are pulled on Earth with P = 30 N. The crates, the cable and the deck are then taken to the Moon, where g_Moon ≈ 1.6 N/kg, and the crates are pulled again with P = 30 N on the same level, frictionless deck. Let T_E be the tension in the connecting cable on Earth, and let T_M be the tension in the connecting cable on the Moon.

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

•   T_M > T_E

•   T_M < T_E

•   T_M = T_E

Justify your answer using qualitative reasoning beyond referencing equations.

Part B

Starting with Newton’s second law, derive an expression for the tension T in the connecting cable. Express your final answer in terms of P, m_A, m_B, 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

T_M = T_E

What changes on the Moon is the gravitational field, and with it the gravitational force on each crate: each crate’s weight falls to about a sixth of its Earth value. But the gravitational force is vertical. On the level deck each crate’s weight is balanced by the upward normal force the deck exerts, which is smaller on the Moon by the same factor, and because the deck is frictionless the smaller normal force changes no horizontal force either. Neither vertical force has a horizontal component, so neither enters the net horizontal force on either crate.

The horizontal motion is set by the pull and by the crates’ masses. The pull is 30 N in both places. Mass — the property that sets how strongly each crate resists a change in its motion — is the same on the Moon as on Earth. The same pull therefore gives the two crates the same acceleration, and the cable exerts the only horizontal force on crate B, so its pull must give crate B’s unchanged mass that same acceleration: it pulls exactly as hard as it did on Earth. Mass, not weight, sets the acceleration that a given net force produces.

A prediction that the tension will be smaller “because everything weighs less there” has the fact right and the physics wrong. The crates do weigh less, but their weights are vertical, and the cable’s pull has only to give crate B’s mass its acceleration — crate B’s weight plays no part in that.

Figure 2 — Crate B on Earth and on the Moon, one scale: the vertical pair shrinks, T does not.

Scoring (3 points):

▸ A1 — 1 point: For indicating T_M = T_E.

▸ A2 — 1 point: For a justification that indicates that the crates’ masses — their inertia — are the same on the Moon as on Earth.

▸ A3 — 1 point: For a justification that indicates that the gravitational force is vertical, so it contributes nothing to the net horizontal force on either crate.

Scoring Note: A2 and A3 are content criteria, not route criteria; any wording with that content is sufficient, and the normal force need not be mentioned.

Scoring Note: A2 and A3 are judged on the reasoning a response gives. A response that also writes an equation is not penalised for it; an equation or a numerical substitution offered with no accompanying reasoning — including the bare observation that g does not appear in an expression for T — satisfies neither criterion.

Scoring Note: A2 and A3 may each be earned independently of the option indicated in A1.

⚠︎ Common error (partial credit): Indicating T_M = T_E and justifying it only by the unchanged 30 N pull, or by the general rule that tension does not depend on gravity — earns 1 of 3 points, for A1. The rule is false in general: a single vertical cord holding a hanging crate at rest exerts a tension equal to the crate’s weight, which is smaller on the Moon. Here gravity drops out because the pull is fixed at 30 N rather than depending on g, and because the crates’ weights are vertical, so on this frictionless deck they enter no horizontal force.

⚠︎ Common error (partial credit): Indicating T_M = T_E and stating that mass does not change on the Moon, but never saying why the smaller weight does not matter — earns 2 of 3 points, for A1 and A2.

⚠︎ Common error (partial credit): Indicating T_M = T_E and establishing that the gravitational force is vertical and contributes nothing horizontally, but never addressing the crates’ masses — earns 2 of 3 points, for A1 and A3.

✗ Common error (no credit): Indicating T_M < T_E because everything weighs less on the Moon — for example by scaling the Earth tension by g_Moon/g, so that 18 N becomes about 2.9 N — earns 0 points. The crates do weigh less, but their weights are vertical; scaling the horizontal cable tension as though it were a weight confuses weight with mass, and it is mass that sets how much acceleration the pull produces.

Part B — Model Answer

Treat the two crates and the cable between them as one system of mass m_A + m_B. Because the cable is light, it pulls forward on crate B and backward on crate A with equal magnitudes; both pulls are internal to the system, so neither enters its net force. The deck is level and frictionless and the vertical forces balance, so the only horizontal external force on the system is the rope’s pull P. Newton’s second law for the system gives

F_net = m·a
P = (m_A + m_B)·a
a = P / (m_A + m_B)

The cable is inextensible, so both crates share this acceleration. Now isolate crate B. The rope pulls on crate A, not on crate B, so the cable exerts the only horizontal force on crate B:

T = m_B·a = m_B·P / (m_A + m_B)

Figure 3 — Horizontal forces on the two-crate system and on crate B alone, to one scale.

Equivalently, applying the second law to each crate on its own gives P − T = m_A·a and T = m_B·a, and eliminating a gives the same result. The rope pulls crate A forward with P and the cable pulls it back with T, so the net force on crate A is P − T = m_A·a — nothing is “lost” in the cable. The cable’s pull is the only horizontal force on crate B, so T is the fixed fraction m_B/(m_A + m_B) of P.

Checking against measurement (beyond what part B asks). For these crates m_B/(m_A + m_B) = 6.0 kg / 10.0 kg = 0.60. With P = 30 N, a = 3.0 m/s² and T = 0.60 × 30 N = 18 N. In a separate set of trials on this deck, a student varied P and measured T each time: 6.1, 11.9, 17.9 and 24.1 N at P = 10, 20, 30 and 40 N. Every trial lies close to T = 0.60P, the prediction for these crates — 17.9 N against 18 N at P = 30 N — and well below the line T = P.

Figure 4 — The student’s measurements lie near T = 0.60P for these crates, below T = P.

Scoring (3 points):

▸ B1 — 1 point: For beginning a derivation with Newton’s second law for crate A, crate B or the two-crate system; the starting relation alone is sufficient, but a final expression for T on its own is not a derivation.

▸ B2 — 1 point: For a Newton’s second law relation for one crate in which that crate’s own mass multiplies its acceleration and its net horizontal force is correct in terms of T — for example T = m_B·a for crate B or P − T = m_A·a for crate A. The acceleration may appear as an expression, as in T = m_B·P / (m_A + m_B).

Scoring Note: B1 and B2 are judged on their own and carry no consistency credit; B2 is not affected by an error in the acceleration.

▸ B3 — 1 point: For a correct expression for T in terms of P, m_A and m_B; a form with the masses’ values substituted, such as T = 0.60P, does not qualify. Where B2 was not earned, an expression that follows correctly from the response’s own incorrect single-crate relation also earns this point; any other error prevents it. The expected result is T = m_B·P / (m_A + m_B), or an equivalent form such as T = P − m_A·P / (m_A + m_B).

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

⚠︎ Common error (partial credit): Finding the common acceleration wrongly — with only one crate’s mass, a = P/m_A or a = P/m_B (7.5 or 5.0 m/s² for P = 30 N), or by treating the cable tension as an external force on the two-crate system, P − T = (m_A + m_B)·a — and then writing T = m_B·a correctly — earns 2 of 3 points, for B1 and B2. P is the only horizontal external force on the two-crate system, and the cable’s pulls on the two crates are internal to it.

⚠︎ Common error (partial credit): Isolating crate A and writing T = m_A·a, which gives T = m_A·P / (m_A + m_B), or 12 N for P = 30 N — earns 2 of 3 points, for B1 and B3. Crate A has two horizontal forces exerted on it — the rope’s pull P forward and the cable’s pull T backward — so its net force is P − T.

✗ Common error (no credit): Stating T = P because the cable “passes the pull along” to crate B, with no second-law relation written for any crate or for the system — earns 0 points. The rope pulls on crate A only; the cable exerts the only horizontal force on crate B, and that force accelerates crate B alone.

Part C — Model Answer

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

Part A’s reasoning was that the Moon’s weaker gravity changes only vertical forces and leaves the crates’ masses unchanged. The part B equation says the same thing in symbols:

T = m_B·P / (m_A + m_B)

contains no g. The tension depends only on the pull and on the two masses. On the Moon the pull is still 30 N and the masses are unchanged, so the equation gives the same 18 N there as on Earth: T_M = T_E.

g is absent because the pull is set directly rather than by gravity, and because on a level, frictionless deck no crate’s weight enters any horizontal force. Change the first of those and g enters at once. If the rope were replaced by a cord running over an ideal pulley at the end of the deck to a hanging block of mass m_h = 3.0 kg, then, taking crate A, crate B and the hanging block together along the cord’s path, the only external force driving all three would be the gravitational force on the hanging block, m_h·g, and the cord itself would pull crate A with less than that. The cable tension would become

T = m_B·m_h·g / (m_A + m_B + m_h)

which falls from 13.8 N on Earth to 2.2 N on the Moon, in the ratio g_Moon/g. With that change, a prediction that the tension would be smaller on the Moon “because everything weighs less there” would be right — but only because the gravitational force on the hanging block now drives the motion; the crates’ smaller weights still play no part. g can reach T only through an external force along the deck that depends on gravity, and on this level, frictionless deck the only external force along it is a pull fixed at 30 N.

Figure 5 — A cord to a hanging 3.0 kg block replaces the rope, and T now scales with g.

Scoring (2 points):

▸ C1 — 1 point: For attempting to address how T, in the response’s own part B equation (whether or not that equation is correct), depends on g or on the change from Earth to the Moon.

Scoring Note: It is not necessary to use the dependence correctly to earn this point. The response only needs to relate T, as it appears in the part B expression, to g or to a quantity that does or does not change on the Moon, using functional-dependence language such as “does not appear in”, “is independent of”, “is proportional to”, or “depends only on”.

▸ C2 — 1 point: For correctly using that dependence to decide whether the part B expression 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 from what actually changes on the Moon.

Scoring Note: Stating that g does not appear in the expression is enough for the dependence; the response need not also state that P and the masses are unchanged.

⚠︎ Common error (partial credit): Noting that g does not appear in the part B expression, or that T is therefore the same on the Moon, but giving no verdict on whether the equation agrees with part A — earns 1 of 2 points, for C1.

⚠︎ Common error (partial credit): Arguing that T in the part B expression changes on the Moon because the masses in it are smaller there, whichever verdict on part A follows — earns 1 of 2 points, for C1. Mass is the same on the Moon; only the gravitational and normal forces are smaller.

✗ Common error (no credit): Asserting that the equation agrees, or does not agree, with part A with no statement relating T to g, to P or to the masses — earns 0 points. Part C asks what the derived equation says.

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