Pause here: To build true exam stamina, attempt these questions on your own before checking the model answers.

FRQ: Banked-Curve Circular Motion — Two Curves, One Rainstorm

Assessments aligned to 2026 AP Physics 1 standards

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

▤ Scenario

A vehicle-testing facility has two curves that share the same radius, R = 50 m. Curve A is level. On its dry surface the coefficient of static friction between the car’s tyres and the road is μ_s = 0.80. Curve B is banked at a constant angle θ = 30° to the horizontal. It is built so that a car travelling at the curve’s design speed can round it with no friction force exerted on the car.

Figure 1 — Cross-sections of the two curves, viewed along the direction of travel.

The same car, of mass m = 1500 kg, is driven around each curve at constant speed; on Curve B it travels at the design speed. Air resistance and rolling resistance are negligible.

Model the car as a particle. Use an inertial reference frame fixed to the ground. Both circular paths lie in horizontal planes, and every force exerted on the car lies in the vertical plane that contains the car and the centre of its circular path. For horizontal forces take the direction toward the centre of the curve as positive, and for vertical forces take upward as positive. Use g = 10 m/s².

✎ Free Response Questions

Part A

i.    On the following dots, which represent the car on Curve A and the car on Curve B, draw and label the forces (not components) that are exerted on the car on each curve. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.

Figure 2 — Dots representing the car on each curve, in the view of Figure 1.

ii.   Derive an expression for the design speed v_B of Curve B. Express your final answer in terms of R, θ, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

iii.  Calculate the maximum speed v_A at which the car can round dry Curve A without sliding. Then calculate the design speed v_B of Curve B.

Part B

A sudden downpour leaves both road surfaces effectively frictionless. The car enters Curve B on its 50 m circle, moving horizontally along the circle at 12 m/s.

Indicate whether the car’s distance from the centre of the curve increases, decreases, or remains constant just after the car enters the curve.

______ Increases

______ Decreases

______ Remains constant

Justify your answer. In your justification, include qualitative reasoning beyond mathematical derivations or expressions.

❖ 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 exerts the gravitational force on the car, F_g = mg = (1500 kg)(10 m/s²) = 1.5 × 10⁴ N, downward, on both curves. The road exerts the normal force F_N, perpendicular to its surface, and, where one is needed, a static friction force f_s along its surface.

On level Curve A the normal force is vertical and, with no vertical acceleration, balances the gravitational force: F_N = 1.5 × 10⁴ N, upward. A vertical force has no component toward the centre, so the static friction force, horizontal and directed toward the centre, is the only force that turns the car. Its size, mv²/R, is set by the car’s speed.

On Curve B the car travels at the design speed, so no friction force is exerted on it and only F_g and F_N act. The normal force is perpendicular to the banked road, so it tilts from the vertical toward the centre by the bank angle, 30°. It is larger than the gravitational force, 1.7 × 10⁴ N (part A (ii)), because only its vertical component balances F_g.

Figure 3 — The forces on the car on each curve; F_g and F_N to one scale, f_s set by the speed.

On each curve the net force points toward the centre: on Curve A it is the friction force, on Curve B the horizontal component of the normal force. No separate “centripetal force” is drawn, because that name belongs to the net inward force the real forces already provide.

Scoring (2 points):

▸ A1 — 1 point: For drawing, on the Curve A dot, a downward gravitational force, an upward normal force and a friction force directed horizontally toward the centre of the curve, as distinct labelled arrows starting on the dot, and no other force arrow starting on the dot.

▸ A2 — 1 point: For drawing, on the Curve B dot, a downward gravitational force and a normal force tilted from the vertical toward the centre of the curve, as distinct labelled arrows starting on the dot, and no other force arrow starting on the dot.

Scoring Note: Any label that names the force is accepted — F_g, mg, W, weight or gravitational force; F_N, N or normal force; f_s, f, F_f or friction. The response is not required to name the object exerting each force, and arrow lengths are not scored. The tilt of the normal force on Curve B need not be measured, but the arrow must lean toward the centre and must not be horizontal. An arrow for the velocity, the acceleration or a “centripetal force” drawn from a dot counts as another force arrow. A1 and A2 do not depend on any other part.

⚠︎ Common error (partial credit): Draws Curve B’s normal force vertical and adds a horizontal arrow toward the centre, labelled friction or centripetal force, to turn the car, with Curve A’s diagram correct — earns 1 of 2 points, A1. The road pushes perpendicular to its own surface, so on the bank the normal force itself tilts toward the centre.

⚠︎ Common error (partial credit): Draws Curve B’s normal force as two arrows, one vertical and one horizontal, with Curve A’s diagram correct — earns 1 of 2 points, A1. A free-body diagram shows each force as a single arrow, not as its components.

⚠︎ Common error (partial credit): Draws Curve A’s friction force pointing away from the centre, as if it had to balance an inward pull, with Curve B’s diagram correct — earns 1 of 2 points, A2. Left to itself the car would go straight on, which carries it outward across the road, so static friction points toward the centre.

✗ Common error (no credit): Adds an outward “centrifugal force” to each diagram, balancing the inward force. Each diagram then carries an arrow that no object exerts, so neither A1 nor A2 is earned. A car rounding a curve accelerates toward the centre, so the forces on it cannot balance.

Part A (ii) — Model Answer

Apply Newton’s second law to the car on Curve B, taking components horizontally and vertically. The car moves on a horizontal circle at constant speed, so its acceleration is horizontal, directed toward the centre, with magnitude v²/R, and its vertical acceleration is zero. The gravitational force is vertical, so the horizontal component of the normal force is the whole net force:

Vertical:  F_N cos θ − mg = 0
Horizontal:  F_N sin θ = mv²/R

Figure 4 — The normal force on Curve B and its components: the vertical one balances F_g.

Dividing the horizontal equation by the vertical one eliminates both F_N and m:

tan θ = v²/(gR)
v_B = √(gR tan θ)

The mass cancels, so the design speed is the same for every vehicle, from a motorcycle to a loaded truck. The normal force on this car is F_N = mg/cos θ = (1.5 × 10⁴ N)/cos 30° = 1.7 × 10⁴ N, larger than its weight.

Scoring (2 points):

▸ A3 — 1 point: For including Newton’s second law applied to the forces on the car on Curve B with an acceleration of magnitude v²/R — for example, F_N sin θ = mv²/R, or any equation setting a force or force component on the car equal to mv²/R, or to its nonzero component in the chosen direction, such as mg sin θ = m(v²/R) cos θ. This point is earned for the starting principle alone, even if the steps that follow are incorrect.

▸ A4 — 1 point: For a correct relation between the speed and the bank angle — for example, F_N cos θ = mg together with F_N sin θ = mv²/R, mg sin θ = m(v²/R) cos θ along the slope, or tan θ = v²/(gR). This point is not earned by a relation that does not follow from the response’s own equations.

Scoring Note: A correct, isolated, final expression for v_B earns points A3 and A4; a correct final expression preceded only by an unapplied equation from the reference information, by a relation equivalent to it such as tan θ = v²/(gR), or only by words, is treated as isolated. Work, in equations or in words, that balances the real forces against an outward “centrifugal force” is not Newton’s second law: it does not meet A3, and a final expression that follows it is not treated as isolated. A3 and A4 do not depend on the response in part A (i).

⚠︎ Common error (partial credit): Writes F_N sin θ = mv²/R correctly but takes F_N = mg cos θ, as for a block at rest on an incline — earns 1 of 2 points, A3. The car accelerates horizontally, so part of its acceleration is perpendicular to the road, and the forces perpendicular to the road do not balance.

⚠︎ Common error (partial credit): Sets the whole normal force equal to mv²/R, alongside F_N cos θ = mg — earns 1 of 2 points, A3. Only the horizontal component of the normal force points toward the centre.

⚠︎ Common error (partial credit): Balances the horizontal component of the normal force against an outward “centrifugal force” of size mv²/R, with F_N cos θ = mg, and reaches tan θ = v²/(gR) — earns 1 of 2 points, A4. In the ground’s frame nothing pushes the car outward; the horizontal component of the normal force is the net force, and it produces the acceleration toward the centre.

✗ Common error (no credit): Treats the car as being in equilibrium because its speed is constant, sets the net force on it to zero, and concludes that the bank can hold the car at any speed. A3 is not earned, because no force or force component is set equal to mv²/R or to a nonzero component of it, and A4 is not earned, because no relation between the speed and the bank angle is found. Moving on a circle at constant speed is still accelerating.

Part A (iii) — Model Answer

Curve A. On the level road the normal force balances the gravitational force, F_N = mg, and static friction is the only force toward the centre. The car can follow the curve as long as the friction it needs, mv²/R, does not exceed the maximum available, μ_s F_N = μ_s mg = (0.80)(1.5 × 10⁴ N) = 1.2 × 10⁴ N. The maximum speed is the one at which the two are equal:

μ_s mg = m(v_A)²/R
v_A = √(μ_s gR) = √((0.80)(10 m/s²)(50 m)) = 20 m/s

Curve B. Evaluating the expression from part A (ii), with tan 30° = 0.577:

v_B = √(gR tan θ) = √((10 m/s²)(50 m)(0.577)) = 17 m/s

Side by side, v_A = √(μ_s gR) and v_B = √(gR tan θ) differ only in μ_s and tan θ, so in the speed formula tan θ takes the place of μ_s. The roles differ, though: μ_s sets the highest speed at which friction can hold the car on the level curve, and any lower speed works too, while tan θ fixes the one speed at which the bank holds the car with no friction at all.

Scoring (3 points):

▸ A5 — 1 point: For applying Newton’s second law to the car on dry Curve A with the maximum static friction force as the net force toward the centre — μ_s mg = mv²/R, or an equivalent equation. This point is earned for the correct equation alone.

▸ A6 — 1 point: For a maximum speed on dry Curve A of 20 m/s. This point is not earned by a value that follows from an incorrect equation.

▸ A7 — 1 point: For a design speed for Curve B of 17 m/s, or a value consistent with the response’s expression in part A (ii).

Scoring Note: A6 and A7 accept values that round to 20 m/s and 17 m/s, or 19.8 m/s and 16.8 m/s with g = 9.8 m/s². A5 requires the equation to be shown, symbolically or with numbers substituted; a correct value of v_A alone does not meet it. A5 and A6 do not depend on the responses in parts A (i) and A (ii).

⚠︎ Common error (partial credit): Uses the level-road relationship for Curve B as well, v = √(μ_s gR), and reports 20 m/s for both curves instead of evaluating the part A (ii) expression, with Curve A correct — earns 2 of 3 points, A5 and A6. At its design speed the bank exerts no friction, so μ_s plays no part in v_B.

⚠︎ Common error (partial credit): Adds friction to the bank, v_B = √(gR(tan θ + μ_s)) = 26 m/s, instead of evaluating the part A (ii) expression, with Curve A correct — earns 2 of 3 points, A5 and A6. The design speed is the speed at which the bank needs no friction at all.

✗ Common error (no credit): Sets the net force toward the centre equal to the car’s weight, mv²/R = mg, and reports √(gR) = 22 m/s for both curves instead of evaluating the part A (ii) expression. A5 is not earned, because the net force on Curve A is the friction force, not the weight, and 22 m/s is neither speed, so neither A6 nor A7 is earned.

Part B — Model Answer

The car’s distance from the centre decreases: it slides down the bank, toward the inside of the curve.

The rain removes friction from both curves, but not the normal force, which is a contact force of a different kind. On level Curve A the two remaining forces are vertical, so nothing turns the car and it goes straight on, off the outside of the curve, whatever its speed. On Curve B the normal force is still perpendicular to the road and tilted toward the centre, so the car does still turn.

What the rain removes on Curve B is the only force that could act up the slope. The normal force changes size but always acts perpendicular to the road, so along the slope the only force is the component of the gravitational force, which points down the slope and has the same size at every speed. At the design speed, 17 m/s, the car needs a net force toward the centre of mv²/R = mg tan θ = 8.7 × 10³ N, and the part of it that lies along the slope is exactly this pull. At 12 m/s the car needs only mv²/R = (1500 kg)(12 m/s)²/(50 m) = 4.3 × 10³ N, about half as much, so the circle also requires only about half as much along the slope. The pull is unchanged, so it is more than the circle requires. On a dry bank, static friction acting up the slope would make up the difference; on the wet bank nothing can, and the car’s path bends more sharply than the 50 m circle.

Figure 5 — Plan view of the car’s path just after it enters each wet curve at 12 m/s.

Below the design speed, the pull down a frictionless bank is more than the circle requires, so the car slides down the bank, toward the centre. The common shortcut of taking the inward force to be mg tan θ assumes the vertical forces still balance. They do not, even at the instant of entry: the car’s velocity is still horizontal, but its acceleration already has a downward component. The inward force is nevertheless more than the 4.3 × 10³ N the circle needs.

A frictionless bank holds a car on a given circle at one speed only. To hold a 12 m/s car on the 50 m circle, Curve B would need a bank angle with tan θ = v²/(gR) = 0.288, which is 16°; at 30°, 12 m/s is the design speed of a circle of radius v²/(g tan θ) = 25 m. That is where 12 m/s would be the design speed, not where the car goes: it speeds up as it slides down, so it never circles at 12 m/s on that line.

Scoring (3 points):

▸ B1 — 1 point: For indicating “Decreases”.

▸ B2 — 1 point: For indicating one of the following: the car’s speed, 12 m/s, is less than the design speed of Curve B; at 12 m/s the net force toward the centre that the car needs to follow the 50 m circle, mv²/R, is smaller than it is at the design speed; or a 12 m/s car would need a bank shallower than 30° (about 16°) to hold the 50 m circle with no friction, or the 30° bank holds a 12 m/s car with no friction only on a smaller circle (radius about 25 m).

▸ B3 — 1 point: For indicating one of the following: the component of the gravitational force down the slope is more than the 50 m circle requires along the slope at 12 m/s; or the net force on the car toward the centre, or its acceleration toward the centre, is greater than the 50 m circle requires at 12 m/s.

Scoring Note: B1, B2 and B3 are scored independently of the responses in part A and of one another. B2 may be stated in words or in symbols, for example v < v_B, and its values are not required. B3 requires a comparison, between a force or acceleration acting toward the centre or down the slope and what the 50 m circle requires in the same direction at 12 m/s, that names in words what is compared — for example, “the inward component of the normal force is more than the net force the circle needs”; a comparison written only in symbols, such as F_N sin θ > mv²/R, does not meet it, and neither does stating only that the gravitational force pulls the car down the slope and that no friction holds it. The resulting motion need not be stated, and the size of the force is not required: a comparison stated in words that takes the inward force to be mg tan θ meets the requirement. A single comparison of mg tan θ with mv²/R meets B3 when mg tan θ is described as the inward force on the car, or B2 when it is described as the net force needed at the design speed, but not both.

⚠︎ Common error (partial credit): Indicates “Decreases” and states that 12 m/s is below the design speed, but makes no comparison of a force or acceleration with what the circle requires, or makes it only in symbols, such as F_N sin θ > mv²/R — earns 2 of 3 points, B1 and B2.

⚠︎ Common error (partial credit): Indicates “Decreases”, reasoning only that the gravitational force pulls the car down the slope and no friction holds it — earns 1 of 3 points, B1. Both facts hold at the design speed as well, where the car does follow the circle.

⚠︎ Common error (partial credit): Indicates “Remains constant”, reasoning that 12 m/s is below the 17 m/s design speed, so the curve can hold the car — earns 1 of 3 points, B2. The design speed is not a speed limit: a frictionless bank holds the 50 m circle at 17 m/s only.

✗ Common error (no credit): Indicates “Increases”, reasoning that with the friction gone nothing can turn the car, so it goes straight on and off the outside of the curve. That happens on wet Curve A, where both remaining forces are vertical; on the bank the normal force survives the rain, tilted toward the centre. No comparison of the kinds B2 and B3 list is made, so neither B2 nor B3 is earned, and B1 is not earned.