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FRQ: Conditions for Constant Velocity — The Nested Coffee-Filter Drop

Assessments aligned to 2026 AP Physics 1 standards

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

▤ Scenario

A group of physics students investigates how the steady 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. Figure 1 shows the arrangement.

Figure 1 — The drop column, with the motion detector on the ceiling above the release point.

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. Any other force exerted on a stack by the air is negligible.

Each stack reaches a steady speed well before it reaches the floor, and the detector records that value as the terminal speed v_t. Figure 2 shows the recorded values.

Figure 2 — The terminal speed recorded for each stack.

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

Part A

i.    Consider the three-filter stack while it falls at its terminal speed.

On the following dot that represents the stack, draw and label the forces (not components) that are exerted on the stack. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot. Draw the length of each arrow to represent the magnitude of each force, consistent with the scale used in Figure 3.

Figure 3 — The dot representing the three-filter stack, on a grid of 0.010 N squares.

ii.   Derive an expression for (v_t)², the square of the terminal speed of a stack of mass m. Express your final answer in terms of m, b, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

iii.  Identify the feature of the nesting procedure that allows b to be the same for every stack. Then, using the data in Figure 2, calculate (v_t)² for each of the five stacks, and state, with numerical support drawn from those values, whether the data are consistent with the expression you derived in part A (ii).

Part B

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.”

Consider a stack of ten nested filters, which has twice the mass of the five-filter stack. Let v_10 be the terminal speed of the ten-filter stack, and let v_5 be the terminal speed of the five-filter stack.

Indicate whether v_10 is greater than, less than, or equal to 2v_5 by writing one of the following.

•   v_10 > 2v_5

•   v_10 < 2v_5

•   v_10 = 2v_5

Justify your response.

Then determine what, if anything, the resistive force would have to be proportional to for the student’s claim to be correct.

❖ 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

Earth and the air are the only objects that interact with the falling stack. Earth exerts the gravitational force F_g on the stack, directed downward. The air exerts the resistive force F_R on it, directed upward, opposite to the stack’s velocity, and the Scenario takes this force as the whole of the air’s effect. The diagram therefore carries exactly these two forces.

Figure 4 — The two forces on the three-filter stack at terminal speed, each 0.030 N.

The three-filter stack has mass m = 3.0 × 10⁻³ kg, so F_g = mg = (3.0 × 10⁻³ kg)(10 m/s²) = 0.030 N. At terminal speed the stack is still moving downward, but its velocity is constant, so the net force on it is zero. The resistive force therefore equals the gravitational force in magnitude, F_R = 0.030 N, and on the 0.010 N squares of Figure 3 each arrow is three squares long.

Scoring (2 points):

▸ A1 — 1 point: For drawing the gravitational force directed downward and the resistive force directed upward, as distinct labelled arrows starting on the dot, with no other force.

Scoring Note: Any label that names the force is accepted — F_g, mg, W, weight or gravitational force; F_R, F_air, D, drag, air resistance or resistive force. The response is not required to name the object exerting each force.

▸ A2 — 1 point: For the lengths, on the scale of Figure 3, of the two forces exerted on the stack: each 0.030 N, three squares.

Scoring Note: Each length within half a square of three squares is consistent with the scale, with the two arrows within half a square of each other.

⚠︎ Common error (partial credit): Draws the two forces correctly but with lengths that differ by more than half a square — the gravitational arrow longer, because the stack is still falling, or the resistive arrow longer, because the stack has stopped speeding up — earns 1 of 2 points, A1. A stack moving at constant velocity has zero net force, so the two forces are equal.

✗ Common error (no credit): Adds a third, downward arrow, a square or more long, labelled “force of motion”, “force of falling” or “momentum”, and draws the upward resistive arrow long enough to balance both downward arrows. The extra arrow means A1 is not earned; and the resistive arrow is then longer than the gravitational arrow by more than the half-square tolerance, so A2 is not earned either. No object exerts a “force of motion”, and a stack moving at constant velocity needs no net force at all.

Part A (ii) — Model Answer

At terminal speed the stack moves with constant velocity, so its acceleration is zero. With downward positive, Newton’s second law gives

ΣF = F_g − F_R = m·a = 0
F_R = F_g
b(v_t)² = mg

Solving for the square of the terminal speed:

(v_t)² = mg / b = (g / b)·m

For stacks that share the same b, (v_t)² is directly proportional to m.

Scoring (2 points):

▸ A3 — 1 point: For including the condition for constant velocity — the net force on the stack is zero, so the resistive force and the gravitational force are equal in magnitude — whether written from Newton’s first law, from Newton’s second law with the acceleration set to zero, as F_R = F_g, or from the work–energy theorem with no change in kinetic energy at constant speed. This point is earned for the correct starting principle alone, even if the steps that follow are incorrect.

▸ A4 — 1 point: For a correct expression for (v_t)² in terms of m, b and g — mg/b or any algebraic equivalent; v_t = √(mg/b) is also accepted.

Scoring Note: A correct, isolated, final expression for (v_t)², or for v_t, earns points A3 and A4. A3 and A4 do not depend on the response in part A (i).

⚠︎ Common error (partial credit): Writes F_R = F_g but uses bv for the resistive force, reaching v_t = mg/b — earns 1 of 2 points, A3. The Scenario gives the resistive force as bv², not bv.

✗ Common error (no credit): Writes a free-fall relation such as (v_t)² = (v_0)² + 2aΔx with a = g. No condition for constant velocity is written — neither zero net force nor zero change in kinetic energy — so A3 is not earned, and the result is not the terminal-speed relation, so A4 is not earned. A stack whose acceleration stayed at g would never reach a steady speed.

Part A (iii) — Model Answer

Nesting adds filters without changing the stack’s outline: a stack of nested filters has the same size and shape as a single filter, so it meets the air in the same way. Because b depends only on size, shape and the air, every stack falls with the same b. The air is the same for every drop as well, but that is a feature of the room, not of the nesting procedure.

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

With the same b for every stack, the expression from part A (ii) makes (v_t)² directly proportional to m, so the ratio (v_t)²/m should be the same for every stack — equivalently, a graph of (v_t)² against m should be a straight line through the origin, of slope g/b. The five ratios are 7.9, 8.1, 7.9, 8.2 and 7.9 × 10² m²/(s²·kg), an average of 8.0 × 10² m²/(s²·kg), with no value more than about 2.5 % away from it.

Figure 5 — The recorded data as (v_t)² against m, with the best-fit line through the origin.

If (v_t)² is proportional to m, the ratio (v_t)²/m is the same for every stack. Here it is, to within about 2.5 %, so the data are consistent with the expression from part A (ii). The best-fit slope, about 800 m²/(s²·kg), also gives b = g/slope ≈ 1.25 × 10⁻² kg/m, a quantity the group never measured directly.

Scoring (3 points):

▸ A5 — 1 point: For indicating that nesting leaves the stack with the same size and shape — the same outline facing the air — as a single filter. Restating that the filters sit snugly inside one another, without saying the stack’s size or shape is therefore unchanged, does not meet this requirement; nor does stating only that the filters are identical, or that every stack falls through the same air.

▸ A6 — 1 point: For calculating (v_t)² correctly for all five stacks — 0.79, 1.61, 2.37, 3.28 and 3.96 m²/s² — and reporting the values with units.

Scoring Note: Values rounded to two significant figures, or carried to more, are accepted; a unit given once in a column heading applies to the whole column. A5 and A6 do not depend on the response in part A (ii).

▸ A7 — 1 point: For a quantitative test of the response’s part A (ii) expression against the response’s own values of (v_t)² for at least three stacks — for example, (v_t)²/m for each stack, the factor by which (v_t)² changes when m doubles, or the slope and intercept of a line drawn through the plotted values — together with the conclusion that test supports. For the correct expression and values, the ratios are nearly equal, about 8.0 × 10² m²/(s²·kg), so the data are consistent with it. A response with no part A (ii) expression may test (v_t)² ∝ m directly.

Scoring Note: A7 does not depend on the response for A5.

⚠︎ Common error (partial credit): Identifies the nesting feature and calculates all five values of (v_t)² correctly, with units, but concludes only that (v_t)² and m increase together, with no ratio, slope or other numerical comparison — earns 2 of 3 points, A5 and A6. Two quantities can both increase without being proportional: v_t itself increases with m and is not proportional to it.

⚠︎ Common error (partial credit): Says only that the filters are identical, then calculates the five values of (v_t)² correctly, with units, tests the correct part A (ii) expression with the ratio (v_t)²/m, and concludes that the data are consistent with it — earns 2 of 3 points, A6 and A7. Identical filters could be stacked into different shapes; it is nesting that keeps the stack’s outline, and with it b, unchanged.

✗ Common error (no credit): Says that b grows with the number of filters, because a heavier stack pushes harder on the air; then, with the correct expression from part A (ii), tests v_t rather than (v_t)² against m — finding v_t/m falling from 890 to 398 (m/s)/kg — and concludes that the model fails. The nesting feature is not identified, so A5 is not earned; no squared values are calculated, so A6 is not earned; and the proportionality tested is not the one the expression predicts, so A7 is not earned. The Scenario makes b depend only on size, shape and the air, and a falling v_t/m is exactly what (v_t)² ∝ m implies.

Part B — Model Answer

v_10 < 2v_5

From part A (ii), (v_t)² = (g/b)·m, so

v_t = √(mg / b)

Nesting keeps b the same for the ten-filter stack, so terminal speed is proportional to the square root of the mass: doubling m multiplies v_t by √2 ≈ 1.41, not by 2. The ten-filter stack’s terminal speed is therefore about √2 × 1.99 m/s ≈ 2.8 m/s, well below 2v_5 = 3.98 m/s.

The same conclusion follows from the forces alone. At 2v_5 the resistive force on the ten-filter stack would be b(2v_5)² = 4bv_5², four times the five-filter stack’s weight and so twice the ten-filter stack’s own weight. At that speed the net force on it would point upward, so it could not fall steadily at 2v_5, and its terminal speed must be less.

The recorded data show the same pattern. 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 it rises from 1.27 m/s to 1.81 m/s, again a factor of 1.43. Both sit within about 1 % of √2; doubling would have required 1.78 m/s and 2.54 m/s. The model predicts the same factor for the five-to-ten doubling.

Figure 6 — Terminal speed against mass: about 2.8 m/s at ten filters, not the claimed 3.98 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, 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. The claim fails because it applies linear reasoning to a relationship fixed by the v² dependence of this resistive force. It is the power of v in the resistive force that fixes how terminal speed scales with mass. Only the first power makes doubling the mass double the speed.

Scoring (3 points):

▸ B1 — 1 point: For indicating v_10 < 2v_5.

▸ B2 — 1 point: For a justification showing that doubling the mass from five to ten filters raises the terminal speed by less than a factor of 2. Examples of acceptable responses may include the following: from the part A (ii) relationship, v_t is proportional to √m, so doubling m multiplies v_t by √2 ≈ 1.41; doubling the weight doubles the resistive force needed at terminal speed, and because that force is proportional to v², v rises by a factor of √2; at 2v_5 the resistive force would be four times the five-filter stack’s weight — twice the ten-filter stack’s — so at that speed the net force on it would be upward and it could not fall steadily, making its terminal speed less than 2v_5; or a mass-doubling pair from Figure 2 with the factor it gives, about 1.43 rather than 2, together with the statement that the five-to-ten doubling scales by the same factor.

▸ B3 — 1 point: For determining that a resistive force proportional to the first power of the speed, F_R = cv (c a constant), would make the student’s claim correct.

Scoring Note: B1, B2 and B3 are scored independently of the responses in Part A and of one another.

⚠︎ Common error (partial credit): Indicates v_10 < 2v_5 and justifies it correctly, but gives no resistive force that would make the claim correct, or states that none could, leaving v_t ∝ √m standing as though it held for any falling object — earns 2 of 3 points, B1 and B2.

⚠︎ Common error (partial credit): Indicates v_10 < 2v_5 with no justification, or with the observation only that heavier stacks fall faster, and gives no resistive force that would make the claim correct — earns 1 of 3 points, B1. That terminal speed rises with mass is exactly what the claim assumes; it does not show the rise is less than proportional.

✗ Common error (no credit): Indicates v_10 = 2v_5, reasoning that terminal speed goes up with mass so doubling the mass doubles it — the student’s claim — and gives no resistive force that would make the claim correct. None of the three points is earned. A rising trend is not a proportionality: the data’s own doubling pairs give 1.43, not 2.

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