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FRQ: Balanced Forces at Terminal Speed — The Nested Coffee-Filter Drop

Assessments aligned to 2026 AP Physics 1 standards

Question Type: Mathematical Routines (MR)  |  MID-LEVEL  |  10 points

▤ Scenario

A physics class investigates how the terminal speed of a falling object depends on its mass. The group uses identical paper coffee filters, each of mass 1.0 × 10⁻³ kg. The filters nest — one sits snugly inside another — so a stack of n filters has n times the mass of a single filter. Each stack is released from rest just below a motion detector fixed to the ceiling, and the detector records the stack's speed as it falls to the floor.

Air resistance is not negligible here. As a stack falls, the air exerts an upward resistive force on it of magnitude F_R = bv², where v is the stack's speed through the still air and b is a constant that depends only on the size and shape of the falling object and on the properties of the air. The group does not know its value.

Apparatus: the drop column with its ceiling-mounted detector, and how the filters nest.

Each stack reaches a steady speed well before it reaches the floor, and the detector records that value as the terminal speed v_t:

All motion is one-dimensional and vertical. Take the ground as the reference frame and the downward direction as positive. Use g = 10 m/s².

✎ Free Response Questions

(a) Draw a free-body diagram for a stack of filters falling at its terminal speed. Label each force by name and by the object exerting it, and draw the arrows so that their relative lengths show the relative magnitudes. State the principle that fixes those lengths.

(b) Starting from the condition that describes the stack's motion at terminal speed, derive an expression for (v_t)² in terms of the stack's mass m, the gravitational field strength g, and the constant b. State the shape of the graph of (v_t)² against m that your expression predicts, and identify the feature of the nesting procedure that makes b the same for every stack.

(c) Using the recorded data, compute (v_t)² for each of the five stacks. Then state, with numerical support drawn from those values, whether the data are consistent with the prediction you made in part (b).

(d) A student in the group claims: “The data show that terminal speed goes up with mass, so doubling the mass of a stack should double its terminal speed.” Evaluate the claim, using the expression from part (b) and specific values from the recorded data. State the condition on the resistive force that your reasoning depends on, and what that force would instead have to be proportional to for the claim to be correct.

❖ Answer Key & Scoring Guide

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

Part (a) — Model Answer

Two objects interact with the falling stack: Earth and the air. The gravitational force F_g exerted by Earth on the stack is directed downward. The resistive force F_R exerted by the air on the stack is directed upward, opposite to the stack's velocity. No third object acts on the stack, so the diagram carries exactly these two forces.

Free-body diagram at terminal speed. The dashed arrow marks the direction of motion, not a force.

At terminal speed the stack's velocity is constant, so its acceleration is zero and the net force on it is zero. The two forces are therefore equal in magnitude, and the two arrows are drawn the same length.

Scoring (2 points):

▸ 1 point: Shows exactly two forces on the stack — the gravitational force directed downward and the resistive force from the air directed upward — each labelled by name and by the object exerting it, with no additional force shown.

▸ 1 point: Draws the two arrows equal in length and states that the magnitudes are equal because the acceleration is zero while the stack moves at constant terminal speed.

⚠︎ Common error (partial credit): draws the resistive force longer than the gravitational force, reasoning that the stack must be slowing down. The two forces are correctly identified and labelled, so the first criterion is met, but the equal-magnitude criterion is not — earns 1 of 2 points.

✗ Common error (no credit): adds a third, downward arrow labelled “force of falling”, “force of motion” or “momentum”. No object exerts such a force, so the set of forces is wrong, and the diagram no longer shows a zero net force — neither criterion is met.

Part (b) — Model Answer

At terminal speed the stack moves with constant velocity, so the net force on it is zero and the two force magnitudes are equal:

F_R = F_g

b(v_t)² = mg

Solving for (v_t)²:

(v_t)² = (g / b) · m

Both g and b are constant in this experiment, so (v_t)² is directly proportional to m. A graph of (v_t)² against m is therefore predicted to be a straight line through the origin, of slope g/b.

The constant b depends only on the size and shape of the falling object and on the air. Nesting adds mass without changing the stack's shape or the cross-sectional area it presents to the air, so every stack in the experiment falls with the same value of b. That is what makes one straight line through the origin the prediction for all five stacks rather than five unrelated points.

Scoring (3 points):

▸ 1 point: Begins from the terminal-speed condition — the net force on the stack is zero, so F_R = F_g — and writes it as b(v_t)² = mg. This point is earned for the correct starting principle even if the algebra that follows is incorrect.

▸ 1 point: Solves symbolically to (v_t)² = (g/b)·m, showing the rearrangement in symbols before any numbers are used.

▸ 1 point: Predicts a straight line through the origin of slope g/b, and identifies that nesting leaves the stack's size and shape — and therefore b — unchanged while m increases.

⚠︎ Common error (partial credit): writes F_R = F_g correctly but then treats the resistive force as bv rather than bv², reaching v_t = mg/b. The starting principle is stated correctly, so the first criterion is met; the symbolic result and the predicted graph are both wrong — earns 1 of 3 points.

⚠︎ Common error (partial credit): reaches (v_t)² = (g/b)·m and predicts a straight line through the origin, but says nothing about why b is the same for every stack, leaving the proportionality resting on an assumption never stated — earns 2 of 3 points.

✗ Common error (no credit): starts from a kinematic relationship such as (v_t)² = (v_0)² + 2aΔx. No force condition is written, so the derivation never begins from the principle the part asks for and none of the three criteria is met.

Part (c) — Model Answer

Squaring each recorded terminal speed gives

(v_t)² = 0.79, 1.61, 2.37, 3.28, 3.96 m²/s²

for m = 1.0, 2.0, 3.0, 4.0, 5.0 × 10⁻³ kg

The recorded data plotted as (v_t)² against stack mass, with the best-fit line through the origin.

If (v_t)² is proportional to m, the ratio (v_t)²/m must be the same for every stack. Those ratios are 7.9, 8.1, 7.9, 8.2 and 7.9 × 10² m²·s⁻²·kg⁻¹ — an average of 8.0 × 10², with no value more than about 2.5 % away from it. The points therefore lie on a straight line through the origin of that slope, and the data are consistent with the prediction from part (b). The slope also yields b = g/slope = 1.25 × 10⁻² kg/m, a quantity the group never measured directly.

Scoring (2 points):

▸ 1 point: Correctly squares all five recorded speeds — 0.79, 1.61, 2.37, 3.28 and 3.96 m²/s² — and reports the values with units.

▸ 1 point: Makes an explicit quantitative comparison across the five stacks — the ratio (v_t)²/m is near-constant at about 8.0 × 10² m²·s⁻²·kg⁻¹, or an equivalent slope argument — and concludes on that evidence that the data support (v_t)² ∝ m.

⚠︎ Common error (partial credit): squares all five speeds correctly but concludes only that both quantities increase together, with no ratio, slope or other numerical comparison. The first criterion is met, the evidence-based conclusion is not — earns 1 of 2 points.

✗ Common error (no credit): tabulates v_t itself against m, observes that those points bend away from a straight line, and concludes that the model fails. That curvature is exactly what (v_t)² ∝ m predicts on those axes; neither the set of squared values nor a supported conclusion is produced.

Part (d) — Model Answer

The claim is incorrect. From part (b), (v_t)² = (g/b)·m, so

v_t = √(mg / b)

Terminal speed grows as the square root of the mass, not in proportion to it, so doubling m multiplies v_t by √2 ≈ 1.41 rather than by 2.

The data show this directly. From one filter to two the mass doubles and the terminal speed rises from 0.89 m/s to 1.27 m/s, a factor of 1.43. From two filters to four the mass doubles again and the speed rises from 1.27 m/s to 1.81 m/s, a factor of 1.43 as well. Both sit within about 1 % of √2; doubling would have required 1.78 m/s and 2.54 m/s.

The reasoning depends on the resistive force being proportional to the square of the speed. If the air instead exerted a resistive force proportional to the first power of the speed, F_R = cv for some constant c, then balancing the forces at terminal speed would give cv_t = mg, so v_t = mg/c — directly proportional to m, and the student's claim would be exactly right. More generally it is the power of v in the resistive force that fixes how terminal speed scales with mass, and only the first power makes doubling the mass double the speed. The claim fails not because the recorded speeds happen not to double, but because it applies linear reasoning to a relationship fixed by the v² dependence of this particular resistive force.

Scoring (3 points):

▸ 1 point: Concludes that the claim is incorrect and supports it with a mass-doubling pair from the table together with the factor that pair gives, compared against the factor the model predicts.

▸ 1 point: States the relationship that produces the result — v_t = √(mg/b), so v_t ∝ √m and doubling m multiplies v_t by √2 ≈ 1.41.

▸ 1 point: Names the condition the reasoning rests on — that the resistive force is proportional to v² — and determines that a force proportional to the first power of the speed, F_R = cv, would make the claim correct, since cv_t = mg gives v_t = mg/c ∝ m.

⚠︎ Common error (partial credit): states correctly that v_t ∝ √m and that the factor should be √2, but quotes no value from the table and names no condition — earns 1 of 3 points.

⚠︎ Common error (partial credit): rejects the claim using both the √m relationship and a data pair, but stops there, leaving v_t ∝ √m standing as though it held for any falling object — earns 2 of 3 points.

✗ Common error (no credit): rejects the claim on the ground that the recorded speeds do not double, without connecting the observed factor to √2. Numbers are quoted but no relationship is applied, so neither the supported conclusion nor the physical relationship is established.