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Early and unofficial 2026 solutions

Early Solutions to the 2026 AP Physics 1: Algebra-Based FRQs | Step by Step

A complete student-friendly walkthrough of all four released FRQs, including projectile graphs, fluid flow, momentum vectors, center-of-mass motion, friction experiments, linearization, rotational work, units, and physical interpretation.

4 complete FRQs19 parts and subparts9 visual placeholdersAll derivations shown
Important: These are independent, early solutions prepared from the uploaded detailed-solution PDF and checked against the released 2026 question paper. They are not official College Board scoring guidelines. Equivalent correct derivations and experimental methods may also earn credit.

How to use these AP Physics 1 FRQ solutions

Begin every derivation with the governing principle, substitute the relevant conditions, keep signs tied to directions, and state the units and physical meaning of the result. In this guide, bold italic symbols represent vectors; ordinary italic symbols represent scalar components or magnitudes. The positive x-direction is right and the positive y-direction is upward unless stated otherwise.

Principle before algebraName constant-acceleration kinematics, momentum conservation, work-energy, or impulse-momentum before manipulating equations.
Graphs carry physicsUse slope for velocity or a model parameter, preserve continuity at an event, and label axes with units.
Explain the mechanismFor qualitative comparisons, describe how area, inertia, equal interaction forces, or energy transfer changes the outcome.
AP Physics 1 past FRQsAP Physics 1 score calculatorWhy AP Physics 1 is challengingAP study hub

Mathematical routines

FRQ 1: Fountain projectile and volume flow rate

Water exits a circular fountain nozzle at speed v0 and angle θ0, reaches height h1, and returns to launch height at time tf.

System definition and variables

v0x = v0 cos θ0v0y = v0 sin θ0a = (0, −g)nozzle radius = r0

Air resistance is neglected. Gravity is the only acceleration after the droplet leaves the nozzle.

2026 AP Physics 1 FRQ 1 fountain projectile diagram showing a water droplet launched from a nozzle at angle theta zero and reaching maximum height h one
The fountain projectile system diagram.

A(i)Horizontal and vertical velocity graphs

Task: Sketch vx(t) and vy(t) from 0 to tf.

Fundamental principleax = 0 and ay = −g.
vx(t) = v0 cos θ0vy(t) = v0 sin θ0 − gt
  1. The horizontal component stays constant and positive because horizontal acceleration is zero.
  2. The vertical component decreases linearly with slope −g.
  3. At maximum height, vy = 0. Symmetry places this at tf/2.
  4. At tf, vy = −v0 sin θ0.
AP-ready answervx is a horizontal line above zero. vy begins at +v0 sin θ0, crosses zero at tf/2, and ends at −v0 sin θ0.

A(ii)Derive the exit speed

Task: Express v0 in terms of θ0, h1, and constants.

Constant-acceleration kinematicsvy2 = v0y2 + 2ayΔy.
  1. From nozzle to maximum height: vy = 0, v0y = v0 sin θ0, ay = −g, and Δy = h1.
  2. Substitute and isolate the positive speed.
0 = (v0 sin θ0)2 − 2gh1v02 sin2θ0 = 2gh1v0 = √(2gh1)sin θ0
gh1 has units m2/s2, so its square root has units m/s.
AP-ready answerv0 = √(2gh1)sin θ0 m/s

A(iii)Derive the volume flow rate

Task: Express Q using r0, θ0, h1, and constants.

Volume flow rateQ = Av, where A is cross-sectional area and v is the speed through the cross section.
A = πr02Q = πr02v0Q = πr02√(2gh1)sin θ0
AP-ready answerQ = πr02√(2gh1)sin θ0, with units m3/s.

BSmaller nozzle at the same flow rate

Task: Compare the new maximum height h2 with h1 and give qualitative reasoning.

  1. A smaller nozzle radius gives a smaller cross-sectional area.
  2. Because Q = Av is unchanged, the water must leave the smaller area at a greater speed.
  3. The exit angle is unchanged, so the upward velocity component is greater.
  4. Gravity needs more time and vertical distance to reduce that component to zero.
h = (v sin θ0)22g
AP-ready answerh2 > h1. The narrower opening requires faster water to maintain the same flow rate, giving each droplet more upward kinetic energy.

Translation between representations

FRQ 2: Collision, momentum, energy, and center of mass

Disk R has mass m0 and initially moves right at v0. Disk S has mass 3m0 and starts at rest. After collision, R moves left at v0/2.

AMomentum vectors after the collision

Task: Determine both final momentum vectors and draw scaled arrows.

Conservation of linear momentumΣpx,i = Σpx,f because external horizontal impulse is zero.
pi = m0v0pR,f = m0(−v0/2) = −12m0v0m0v0 = −12m0v0 + pS,fpS,f = +32m0v0
2026 AP Physics 1 FRQ 2 momentum-vector diagram with Disk R pointing left at half the reference momentum and Disk S pointing right at one and one-half times the reference momentum
Replaceable placeholder for the scaled final momentum vectors.
AP-ready answerpR,f = −(1/2)m0v0 x̂;   pS,f = +(3/2)m0v0 x̂. R points left; S points right.

BKinetic energy of Disk S

Task: Begin with momentum conservation and derive KS.

m0v0 = m0(−v0/2) + 3m0vS,fvS,f = v0/2KS = 12(3m0)(v0/2)2 = 38m0v02
Check: KR = (1/8)m0v02 and KS = (3/8)m0v02, summing to the initial (1/2)m0v02.
AP-ready answerKS = (3/8)m0v02

CExtend the position-versus-time graph

Task: Draw and label R, S, and center-of-mass lines from t1 to 2t1.

Let the collision position be xc = v0t1. All three curves pass through the collision point (t1, xc).

xR(t) = xc − (1/2)v0(t − t1)xS(t) = xc + (1/2)v0(t − t1)vcm = m0v04m0 = v0/4xcm(t) = xc + (1/4)v0(t − t1)
2026 AP Physics 1 FRQ 2 position versus time graph after collision showing Disk R sloping downward, Disk S upward, and the center of mass continuing with unchanged positive slope
Replaceable placeholder for the completed position-time graph.
AP-ready answerAt 2t1: xR = (1/2)xc, xS = (3/2)xc, and xcm = (5/4)xc. The center-of-mass slope does not change.

DCompare momentum-change magnitudes

Task: Compare |ΔpR| and |ΔpS|.

Newton's third law and impulse-momentumThe interaction forces are equal in magnitude and opposite in direction at every instant, and they act for the same time.
JR = −JSΔpR = −ΔpS
AP-ready answer|ΔpR| = |ΔpS|. The vectors are opposite, but their magnitudes are equal.

Experimental design and analysis

FRQ 3: Experimental determination of kinetic friction

Design a meterstick-only experiment, select graph axes, linearize rough-incline data, draw a best-fit line, and extract μk.

Experiment 1: Frictionless curved ramp and rough horizontal surface

Define h as vertical release height and d as horizontal distance from the bottom of the ramp to where the block stops. The same block starts from rest; only h changes.

2026 AP Physics 1 FRQ 3 experimental setup with a block released from height h on a frictionless curved ramp and stopping after distance d on a rough horizontal surface
Replaceable placeholder for the curved-ramp and rough-surface setup.

A(i)Quantities to measure

Task: Identify meterstick measurements that can determine μk from a linear graph.

  1. Measure the vertical release height h above the horizontal surface.
  2. Release the same block from rest.
  3. Measure the horizontal stopping distance d from the bottom of the ramp to the block's final position.
  4. Repeat for several values of h.
AP-ready answerMeasure vertical release height h and horizontal stopping distance d, varying only h.

A(ii)Reduce experimental uncertainty

Task: Briefly describe a method.

Repeat several trials at each release height and average the measured stopping distances. Use the same reference point on the block for every distance and a broad range of heights so measured distances are large compared with the meterstick's smallest division.

AP-ready answerPerform repeated trials at each h, average d, measure from a consistent reference point, and use a wide measurement range.

B(i)Choose graph axes

Task: Select horizontal and vertical axes for a linear graph.

Work-energy with frictionKi + Ui + Wfr = Kf + Uf.
fk = μkmgWfr = −μkmgdmgh − μkmgd = 0h = μkd

This matches y = mx + b with an ideal intercept of zero.

AP-ready answerPlot d on the horizontal axis and h on the vertical axis.

B(ii)Relate the graph feature to μk

Task: State how to obtain the coefficient from the graph.

slope = ΔhΔd = μk
2026 AP Physics 1 FRQ 3 linear graph of vertical release height h versus horizontal stopping distance d with slope equal to the coefficient of kinetic friction
Replaceable placeholder for the h-versus-d graph.
AP-ready answerThe slope of the best-fit line equals μk.

Experiment 2: Rough incline and photogate

A block starts from rest a distance d up a rough ramp inclined at θ = 30°. A photogate measures speed v near the bottom. The supplied model is

v2 = [2g(sin θ − μk cos θ)]d.

C(i)Vertical-axis label for linearization

Task: The horizontal axis is d. Choose a vertical quantity and units.

The model has the form v2 = (constant)d, so plotting v2 makes the data linear.

AP-ready answerVertical axis: v2 (m2/s2).

C(ii)Calculate and plot transformed data

Task: Square each measured speed and plot v2 versus d.

Measured and transformed rough-ramp data
d (m)v (m/s)v2 (m2/s2)
0.200.290.0841
0.300.380.1444
0.400.410.1681
0.500.490.2401
0.600.520.2704
AP-ready answerPlot the five coordinate pairs (d, v2) listed in the table with a consistent numerical scale.

C(iii)Draw a best-fit line

Task: Draw a line representing the trend, not a point-to-point connection.

v2 = (0.4683 m/s2)d − 0.0059 m2/s2

The small intercept is close to the zero intercept predicted by the physical model. A hand-drawn best fit may differ slightly.

AP-ready answerA representative best-fit slope is s ≈ 0.468 m/s2.

DExperimental coefficient of kinetic friction

Task: Use the best-fit slope and θ = 30° to calculate μk.

s = 2g(sin θ − μk cos θ)μk = sin θ − s/(2g)cos θμk = sin 30° − 0.4683/[2(9.8)]cos 30° = 0.5000 − 0.023890.8660 ≈ 0.5498
The coefficient of friction is dimensionless. Slopes around 0.45-0.48 m/s2 still give μk close to 0.55.
AP-ready answerμk ≈ 0.550

Qualitative/quantitative translation

FRQ 4: Rotational work and angular speed

Toys X and Y start from rest. Equal forces F0 pull equal string lengths ℓ0 from equal axle radii r0. Their inertias satisfy IY = (1/2)IX.

AQualitatively compare ωY and ωX

Task: Compare the final angular speeds without relying only on equations.

  1. The same force acts through the same pulled distance, so the same work and energy are transferred to each toy.
  2. Toy Y has smaller rotational inertia, so it is less resistant to a change in rotational motion.
  3. With equal energy input, the toy with smaller inertia must rotate faster. Equivalently, equal torque produces greater angular acceleration for Y.
AP-ready answerωY > ωX

BDerive ωX

Task: Begin with rotational work-energy and express the result using F0, ℓ0, and IX.

Rotational work-energy theoremWext = ΔKrot.
  1. No slipping gives ℓ0 = r0Δφ, so Δφ = ℓ0/r0.
  2. The torque magnitude is τ = F0r0.
  3. Work is τΔφ = (F0r0)(ℓ0/r0) = F0ℓ0; the axle radius cancels.
  4. Set that work equal to the final rotational kinetic energy.
F0ℓ0 = 12IXωX2ωX2 = 2F0ℓ0IXωX = √2F0ℓ0IX
AP-ready answerωX = √[2F0ℓ0/IX], with units rad/s.

CCheck consistency with Part A

Task: Apply the derived relation to IY = (1/2)IX.

ωY = √2F0ℓ0IYωY = √2F0ℓ0(1/2)IX = √2 ωX
2026 AP Physics 1 FRQ 4 comparison of rotating Toys X and Y receiving equal work, with Toy Y having half the rotational inertia and square root of two times the angular speed
Replaceable placeholder for the rotating-toy comparison diagram.
AP-ready answerωY = √2 ωX > ωX, fully consistent with the qualitative reasoning.

Final AP Physics 1 response checklist

  • State the fundamental principle before the derivation.
  • Keep vector directions and algebraic signs consistent.
  • Define every variable before using it.
  • Show substitutions and include units on numerical results.
  • Label graph axes, numerical scales, and plotted lines clearly.
  • Explain what each graph slope represents physically.
  • Use continuity at collisions or transitions where required.
  • Give qualitative mechanisms beyond equations when requested.
  • Distinguish a vector from its scalar magnitude.
  • Explain the physical meaning of each final result.

Continue your AP Physics 1 review

Rework each released FRQ without the solution, then compare your principles, signs, graph features, units, and justifications with this guide.

More AP Physics 1 FRQsEstimate your AP Physics 1 scoreAP Physics 1 study strategyBrowse the AP study hub

Content basis: the user-supplied AP Physics 1: Algebra-Based 2026 FRQ Detailed Solutions PDF. Prompt wording, values, figures, and subparts were cross-checked against the released 2026 question paper. This is an independent educational guide, not an official scoring document.