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FRQ: Conservation of Linear Momentum and System Selection — The Platform Push-Off

Assessments aligned to 2026 AP Physics 1 standards

Mathematical Routines (MR)  |  MID-LEVEL  |  10 points  |  25 min

▤ Scenario

A student of mass m_s = 55.0 kg stands at rest on a low wheeled platform of mass m_p = 25.0 kg on a level concrete floor. She pushes off the platform, leaves it, and lands on the floor a moment later, while the platform rolls away in the opposite direction.

A motion-capture system records the student’s horizontal speed at the instant her feet lose contact with the platform; in the trial analysed below that speed is 0.400 m/s. A speed gate stands about a metre from the platform’s starting position, in the direction the platform travels, and records the platform’s speed as it passes. Figure 1 shows the arrangement before the push.

Figure 1 — The student at rest on the platform, the speed gate, and the sign convention.

All motion is along one horizontal line. Take the floor as the reference frame and the direction in which the student moves as positive. In each trial the student is the only person who touches the platform, and the platform is not attached to anything. Throughout, v_s and v_p name the student’s and the platform’s velocities immediately after the student’s feet lose contact with the platform.

✎ Free Response Questions

Part A

i.    Draw a separate diagram for each of two system choices: the platform alone, and the student and the platform together. Show on each diagram the two horizontal forces that the student’s feet and the platform exert on each other during the push.

•   Draw both the student and the platform on each diagram.

•   On each diagram, draw a closed boundary enclosing the objects in that system.

•   Draw each force as a distinct arrow starting on, and pointing away from, the object on which it is exerted.

•   Label the force exerted on the platform F_p and the force exerted on the student F_s.

For each of the two system choices, state whether each of F_p and F_s is internal to that system, external to it, or not a force on that system at all. Justify your response.

ii.   Using the system consisting of the student and the platform together, and modelling the external horizontal impulse exerted on that system during the push as negligible, derive an expression for v_p, the velocity of the platform immediately after the student’s feet lose contact with it, and then evaluate that expression for the values given in the Scenario.

Express your final answer in terms of m_s, m_p, v_s, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

iii.  The student repeats the push twice more. In each trial the motion-capture system records her horizontal speed as 0.400 m/s, to the resolution it reports. The speed gate reads to the nearest 0.01 m/s. The three recorded platform speeds are shown in Figure 2.

Figure 2 — Platform speed recorded by the gate in each of the three trials.

Compare these measurements with your prediction from part A (ii), and identify the physical factor most likely responsible for the difference between them. Justify your response.

Part B

A second student writes the following about the same push.

“The total horizontal momentum of the student-and-platform system immediately after the push is the same as it was immediately before, so the student’s change in momentum and the platform’s change in momentum are equal in magnitude. It follows that the force the student’s feet exert on the platform and the force the platform exerts on the student’s feet were equal in magnitude at every instant during the push.”

Assume, as in part A (ii), that the external horizontal impulse exerted on the student-and-platform system during the push is negligible. Both of the second student’s conclusions are correct.

Begin by showing how the first of them follows from the system’s total horizontal momentum being unchanged. Then indicate what establishes the second of them — that the two forces were equal in magnitude at every instant — by checking the appropriate line below.

______ The first conclusion, together with the constancy of the system’s total horizontal momentum

______ Newton’s third law, from which the constancy of the system’s total horizontal momentum in turn follows

______ Newton’s third law, which in turn follows from the constancy of the system’s total horizontal momentum

Justify your response.

❖ 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 (i) — Model Answer

Both diagrams show the same two objects and the same two forces. F_p, exerted on the platform, points in the direction the platform will travel; F_s, exerted on the student, points the opposite way. Only the boundary moves between the two diagrams.

Take the platform alone as the system. F_p acts on the platform, which is inside the boundary, so F_p is external to this system, and during the push it is overwhelmingly what changes the platform’s momentum. Its partner F_s acts on the student, who lies outside the boundary, so F_s is not a force on this system at all — it is neither internal nor external to it.

Now take the student and the platform together as the system. Both members of the pair act on objects inside the boundary, so both are internal. Internal forces cannot change the total momentum of the system that contains both of them, which is why part A (ii) can set the system’s total momentum after the push equal to its value before. What decides the classification is where the boundary falls, not which object touches which.

Figure 3 — The same interaction pair under two system choices; only the boundary moves.

Scoring (2 points):

▸ A1 — 1 point: For two diagrams that each show both the student and the platform, with F_p and F_s drawn as distinct arrows, each starting on the object it is exerted on and pointing away from it and each labelled as specified, and with a closed boundary on each diagram enclosing the correct objects — the platform alone in one, the student and the platform together in the other.

▸ A2 — 1 point: For the classification that follows from those boundaries: for the platform alone, F_p is external because it acts on an object inside the boundary while F_s does not act on that system at all; for the student and the platform together, both are internal because both act on objects inside the boundary.

Scoring Note: A2 may be earned independently of the response in A1. A classification given with no reason, or with a reason that does not turn on where the boundary falls, does not earn A2. A1 through A4 are scored on the response’s own work and carry no consistency credit, there being no upstream result to carry from.

⚠︎ Common error (partial credit): Draws both diagrams, both arrows and both boundaries correctly, but calls the forces internal in both cases, on the grounds that the student is in contact with the platform throughout — earns 1 of 2 points, A1. Contact does not decide the classification; the boundary does, and F_s lands outside the first boundary however firmly her feet are planted.

⚠︎ Common error (partial credit): Draws both boundaries and classifies both systems correctly on the boundary, but draws only F_p, omitting the force exerted on the student — earns 1 of 2 points, A2. A1 requires both members of the pair on both diagrams.

✗ Common error (no credit): Draws a single free-body diagram of the student standing on the platform, showing the gravitational force and the normal force, with no system boundary and neither member of the horizontal pair, and offers no classification — earns 0 of 2 points. Only one diagram is drawn, with no boundary and neither member of the pair, so A1 is not earned, and no classification is given, so A2 is not earned either.

Part A (ii) — Model Answer

Take the system to be the student and the platform together. Before the push both are at rest, so the total horizontal momentum of the system is zero. With the external horizontal impulse on the system during the push modelled as negligible, the system’s total horizontal momentum immediately after the push equals its value immediately before:

p_i = p_f
0 = m_s·v_s + m_p·v_p
v_p = −(m_s / m_p)·v_s

Evaluating for the values given in the Scenario,

v_p = −(55.0 kg / 25.0 kg)(+0.400 m/s) = −0.880 m/s

The platform moves at 0.880 m/s in the direction opposite to the student’s motion. The mass ratio sets the split, so the lighter platform leaves with the greater speed.

Scoring (3 points):

▸ A3 — 1 point: For starting from conservation of linear momentum applied to the student-and-platform system. A3 is awarded for the correct starting principle alone, before any algebra is performed. The expected statement is that the system’s total horizontal momentum immediately after the push equals its value immediately before, which is zero.

▸ A4 — 1 point: For reaching a symbolic expression for v_p in terms of m_s, m_p and v_s before substituting numbers. The expected expression is v_p = −(m_s / m_p)·v_s.

Scoring Note: A correct, isolated, final expression for v_p earns points A3 and A4.

▸ A5 — 1 point: For a numerical value obtained by correct substitution into the response’s own symbolic expression, with the direction stated or carried by the sign. The expected value is −0.880 m/s, or 0.880 m/s in the direction opposite to the student’s motion.

⚠︎ Common error (partial credit): Writes 0 = m_s·v_s + m_p·v_p, rearranges it correctly to v_p = −(m_s / m_p)·v_s, then reports the answer as a bare speed of 0.880 m/s with no sign and no direction — earns 2 of 3 points, A3 and A4. Momentum is a vector and the direction is part of the answer.

⚠︎ Common error (partial credit): Inverts the mass ratio, writing v_p = −(m_p / m_s)·v_s and obtaining −0.182 m/s — earns 2 of 3 points, A3 and A5. The starting principle is stated correctly, and A5 is awarded on the response’s own expression; only A4 is lost.

✗ Common error (no credit): Argues that the two forces of the push are equal and opposite, so the two objects must move apart at equal speeds, and reports 0.400 m/s for the platform — earns 0 of 3 points. The forces are indeed equal and opposite, but equal forces on unequal masses produce unequal speeds; the response starts from Newton’s third law rather than from conservation of momentum applied to the combined system, so A3 is not earned, no symbolic expression is written, so A4 is not earned, and there is no expression to substitute into, so A5 is not either.

Part A (iii) — Model Answer

The three readings average

(0.83 m/s + 0.81 m/s + 0.82 m/s) / 3 = 0.820 m/s

against the predicted speed of 0.880 m/s from part A (ii) — a shortfall of 0.060 m/s, about 7 % of the prediction.

That shortfall is not scatter. The gate reads to the nearest 0.01 m/s; the three readings span only two least counts and none lies more than one least count from their mean, so their tight agreement on its own says little. What does say something is that all three sit below the prediction, by between five and seven least counts, with their mean six least counts below it. A difference that size, repeated in the same direction in every trial, points to a systematic effect rather than to random uncertainty.

The most likely factor is a small resistive external horizontal force on the system — rolling resistance at the wheels and their bearings. It acts during the push and again over the metre the platform rolls before reaching the gate. A resistive force opposes the motion of the object it acts on, so it points opposite to the platform’s velocity throughout, and the impulse it delivers can only reduce the magnitude of the platform’s momentum. It therefore accounts for readings that fall below the prediction and could never produce one above it. Momentum is conserved in every interaction, so what a small external force changes is whether this system’s total momentum stays constant.

Scoring (2 points):

▸ A6 — 1 point: For a quantitative comparison of the three readings, treated together rather than singly, with the response’s own part A (ii) prediction, stating the size of the difference. Award the point wherever that comparison is correctly carried out, whatever difference it leads to. The expected figures are a mean of 0.820 m/s against a predicted speed of 0.880 m/s — the magnitude of the part A (ii) value — a shortfall of about 0.06 m/s.

Scoring Note: A6 is a content criterion, not a route criterion. Any treatment of the three readings as a set satisfies it — their mean, their range, or the statement that all three lie below the prediction — provided the size of the difference is also stated. Quantifying the spread is not required for A6.

▸ A7 — 1 point: For naming a small external horizontal force on the system, such as rolling resistance at the wheels, and reasoning from the direction in which that force acts to the sign of the difference between the readings and the response’s own part A (ii) prediction. The expected answer is rolling resistance, which opposes the platform’s motion and so can only place the readings below the prediction. A7 may be earned independently of the response in part A (ii).

Scoring Note: A7 requires the force to be named and its direction used. The word “external” alone does not earn A7 — part A (ii)’s prompt already supplies it.

⚠︎ Common error (partial credit): Compares the values correctly but attributes the difference to random measurement uncertainty — earns 1 of 2 points, A6. Random uncertainty is as likely to place a reading above the prediction as below it, and would not hold all three readings five to seven least counts low.

⚠︎ Common error (partial credit): Identifies the resistive force and reasons correctly from its direction, but compares the prediction with the single closest reading, 0.83 m/s, without treating the three together or stating the size of the difference — earns 1 of 2 points, A7. A6 requires the readings as a set.

⚠︎ Common error (partial credit): Compares the values correctly and names the resistive force, but gives no directional reason and instead attributes the shortfall to the interaction failing to conserve momentum — earns 1 of 2 points, A6. A7 is lost because the direction of the force is never used. Momentum is conserved in every interaction; what an external force changes is whether the chosen system’s momentum stays constant, so the accurate statement is that an external impulse acted on the system.

✗ Common error (no credit): Concludes that the readings confirm the prediction, on the grounds that 0.880 m/s is the largest recoil speed conservation of momentum allows and every reading lies below it — earns 0 of 2 points. No difference is quantified, so A6 is not earned, and no cause is offered, so A7 is not earned. 0.880 m/s is the prediction of an idealised model rather than a ceiling: a reading above it would mean the model was wrong in the other direction, not that a law had been broken.

Part B — Model Answer

The first conclusion does follow from the unchanged total momentum. If the system’s total horizontal momentum immediately after the push equals its value immediately before, then

Δp_student + Δp_platform = 0
Δp_student = −Δp_platform

so the two changes are equal in magnitude and opposite in direction.

The second conclusion is established by Newton’s third law, from which the constancy of the system’s total horizontal momentum in turn follows — the second of the three lines.

What the step joining the two conclusions leaves out is the difference between a quantity accumulated over the whole push and a quantity at a single instant. Equal and opposite momentum changes mean equal and opposite impulses — equal products of average force and contact time, or equivalently equal areas under the two graphs of force against time. Two forces acting over the same interval can deliver equal impulses while differing at any particular moment, so comparing the endpoints of the push places no constraint at all on what happens within it. Equal impulses are an accumulated result, so they fix nothing about the forces at any single instant.

Newton’s third law does fix them: the force the student’s feet exert on the platform and the force the platform exerts on the student’s feet are equal in magnitude and opposite in direction at every instant, not merely on average. The two statements are not independent, and the second student has them in the wrong order — because the two forces are equal and opposite at every instant, the impulses they deliver over the push are equal and opposite, and because those impulses are equal and opposite, a system containing both objects keeps its total momentum.

Figure 4 — Force on each object during the push; the two shaded areas are equal, 22.0 N·s.

Scoring (3 points):

▸ B1 — 1 point: For obtaining the first conclusion from the system’s total horizontal momentum being unchanged, written as a relation between the two momentum changes — Δp_student + Δp_platform = 0, so Δp_student = −Δp_platform, or the same relation in words. B1 requires the two changes to be related to each other and to the unchanged total; merely restating that they are equal in magnitude, which the second student has already asserted, does not earn B1.

▸ B2 — 1 point: For identifying what the step joining the two conclusions leaves out — that equal and opposite momentum changes fix only the two impulses, the products of average force and contact time, or equivalently the areas under the two graphs of force against time, and place no constraint on the two forces at any single instant during the push.

▸ B3 — 1 point: For checking the second line.

Scoring Note: B3 is the selection alone; the reasoning that supports it is scored by B1 and B2. A checked line that the response’s own justification contradicts does not earn B3. B1, B2 and B3 may each be earned independently of the response in part A.

⚠︎ Common error (partial credit): Obtains the first conclusion correctly and identifies that equal momentum changes fix only the impulses, but checks the third line, making Newton’s third law a consequence of the unchanged system momentum — earns 2 of 3 points, B1 and B2. The dependence runs the other way: the equal and opposite impulses, and with them the constancy of an isolated system’s momentum, follow from the third law.

⚠︎ Common error (partial credit): Obtains the first conclusion correctly and checks the second line, but does not distinguish the impulses from the forces at an instant, arguing instead that the second conclusion is true simply because Newton’s third law is always true — earns 2 of 3 points, B1 and B3.

⚠︎ Common error (partial credit): Identifies correctly that the unchanged system momentum constrains only the impulses, but concludes that the second conclusion therefore cannot be justified at all, checks no line, and writes no relation between the two momentum changes — earns 1 of 3 points, B2. The response identifies the gap correctly but neither affirms the second conclusion nor checks a line.

✗ Common error (no credit): Checks the first line, endorsing the argument as it stands and asserting that equal forces follow directly from the system’s momentum being conserved, without ever writing the relation between the two momentum changes — earns 0 of 3 points. This is the confusion the question is built on: a quantity accumulated over the whole push is treated as though it constrained every instant within it.

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