AP Physics 1 Practice Test 2027: 5 Full Model Papers + Solutions
Use these five original model papers for whole-exam simulation. Each paper contains 42 MCQs in an 85-minute Section I and 4 FRQs in a 95-minute Section II, matching the May 2027 exam structure. Every question has an answer or full solution.
How many full AP Physics 1 practice tests are included?
Five complete original AP Physics 1 model papers are included. Each paper uses the 2027 structure of 42 MCQs and 4 FRQs with answer keys and worked solutions.
- 5 full model papers.
- 210 model-paper MCQs and 20 model-paper FRQs.
- Current 2027 section counts and timing.
Full model papers and topic practice now have different jobs
Choose a model paper when you want a timed final-exam simulation. Choose the MCQ or FRQ subpages when you want focused practice on one exact syllabus topic.
42 MCQs + 4 FRQs per paper, current timing, weighted unit coverage, answers and full solutions.
Select a paper ↓Focused drillMCQ Practice by Topic42 original multiple-choice questions for each of all 43 official topics with direct explanations.
Open topic MCQs →Focused writingFRQ Practice by TopicFour original FRQs for each of all 43 official topics covering the four current response types.
Open topic FRQs →Choose your final-review paper
All five use the same official section lengths but different questions, contexts and FRQ combinations.
They are blueprint-based forecasts, not leaked or official exam questions. Each 42-question MCQ paper uses a balanced count inside the current College Board unit-weighting ranges: Unit 1 = 5, Unit 2 = 8, Unit 3 = 8, Unit 4 = 5, Unit 5 = 5, Unit 6 = 3, Unit 7 = 3, Unit 8 = 5. The four FRQs use one Mathematical Routines, one Translation Between Representations, one Experimental Design & Analysis, and one Qualitative/Quantitative Translation task.
| Unit | Model-paper MCQs | Share of 42 | Official MCQ range |
|---|---|---|---|
| 1 Kinematics | 5 | 11.9% | 10–15% |
| 2 Force & Translational Dynamics | 8 | 19.0% | 18–23% |
| 3 Work, Energy & Power | 8 | 19.0% | 18–23% |
| 4 Linear Momentum | 5 | 11.9% | 10–15% |
| 5 Torque & Rotational Dynamics | 5 | 11.9% | 10–15% |
| 6 Energy and Momentum of Rotating Systems | 3 | 7.1% | 5–8% |
| 7 Oscillations | 3 | 7.1% | 5–8% |
| 8 Fluids | 5 | 11.9% | 10–15% |
AP Physics 1 2027 Model Paper 1
For the most realistic attempt, use the equation/reference sheet allowed for the exam, keep solutions closed, and complete both sections under the current timing.
42 multiple-choice questions
Questions are intentionally mixed across units. Use the Check Answer button only after committing to a choice if you are simulating exam conditions.
A 3 kg object speeds up from 2 m/s to 5 m/s. What is the net work done on it?
Correct answer: D. The work-energy theorem gives W_net = ΔK = ½m(v_f²−v_i²) = 31.5 J.
A constant +25 N force acts for 0.25 s. What impulse is delivered?
Correct answer: C. Impulse is J = FΔt for a constant force, so J = +6.25 N·s.
A fluid sample has mass 0.7 kg and volume 5.0×10⁻⁴ m³. What is its density?
Correct answer: B. Density is ρ = m/V = 1400 kg/m³.
A train moves +20 m/s relative to the ground. A passenger walks toward the rear at 3 m/s relative to the train. What is the passenger's velocity relative to the ground?
Correct answer: D. Use signed relative velocities: v_P/G = v_P/T + v_T/G = 17 m/s.
For an ideal spring-mass oscillator, at what position is the magnitude of velocity greatest?
Correct answer: A. At equilibrium the spring potential energy is minimum and kinetic energy is maximum, so speed is greatest there.
An object moves in a circle of fixed radius. Its speed changes by a factor of 2 while mass and radius stay constant. By what factor does the required centripetal net force change?
Correct answer: D. Because F_c = mv²/r, the force scales as the square of speed. The factor is (2)² = 4.
An object moves along the x-axis from x = -3 m to x = +9 m. What is its displacement?
Correct answer: A. Displacement is final position minus initial position: Δx = x_f − x_i = 12 m.
In water (ρ = 1000 kg/m³), what is the gauge-pressure increase 1.2 m below the surface? Take g = 10 m/s².
Correct answer: C. Hydrostatic pressure change is ΔP = ρgΔh = 12000 Pa.
A 1.2 kg cart moving at +5 m/s sticks to a 2.0 kg cart initially at rest. Neglect external impulse. What is their common velocity?
Correct answer: D. Conservation of momentum gives m₁v₁ = (m₁+m₂)v_f, so v_f = +1.88 m/s.
A constant horizontal force of 12 N acts in the direction of motion while an object moves 4 m. How much work does the force do?
Correct answer: B. For a force parallel to displacement, W = Fd = 48 J.
An ideal spring oscillator is instantaneously at x = +A, where x=0 is equilibrium. In which direction is its acceleration?
Correct answer: D. For SHM, a ∝ −x, so acceleration always points from the displaced position back toward equilibrium.
A spring with k = 220 N/m is compressed 0.10 m. How much elastic potential energy is stored?
Correct answer: A. Spring potential energy is U_s = ½kx² = 1.1 J.
An object displaces 0.0022 m³ of water. Take ρ_water = 1000 kg/m³ and g = 10 m/s². What is the buoyant-force magnitude?
Correct answer: D. Archimedes’ principle gives F_B = ρ_fluid g V_displaced = 22 N.
A force of 12 N is applied perpendicular to a wrench 0.22 m from the pivot. What is the torque magnitude?
Correct answer: C. For a perpendicular force, τ = rF = 2.64 N·m.
A rotating object has I = 1.4 kg·m² and ω = 4 rad/s. What is its rotational kinetic energy?
Correct answer: D. Rotational kinetic energy is K_rot = ½Iω² = 11.2 J.
Two masses attract gravitationally. Their separation changes by a factor of 2 while both masses remain constant. By what factor does the force change?
Correct answer: A. Gravitational force follows F ∝ 1/r², so changing r by factor 2 changes F by 1/(2)² = 0.25.
For an ideal mass-spring oscillator, the mass changes by a factor of 4 while k stays constant. By what factor does the period change?
Correct answer: C. Because T = 2π√(m/k), the period follows the square root of m and the inverse square root of k.
A cart undergoes constant acceleration. Its velocity changes from +7 m/s to +1 m/s in 5 s. What is its displacement during the interval?
Correct answer: C. For constant acceleration, average velocity is (v_i+v_f)/2. Multiplying by time gives Δx = 20 m.
A motor transfers 700 J of energy in 6 s. What is its average power?
Correct answer: C. Average power is energy transferred per time: P = ΔE/Δt = 116.67 W.
For two carts on a nearly frictionless track, why can total system momentum be approximately conserved during the short interaction?
Correct answer: A. Internal forces transfer momentum between system parts; negligible external impulse keeps the total momentum nearly constant.
Two rotating objects have the same total mass and outer radius. Object B has more of its mass concentrated near the rim. Which has the greater rotational inertia?
Correct answer: A. Rotational inertia weights each mass element by r², so placing more mass farther from the axis increases I.
A 50 kg rider stands on a scale in an elevator accelerating upward at 2 m/s². Take g = 10 m/s². What does the scale read?
Correct answer: B. For upward acceleration, N − mg = ma, so N = m(g+a) = 600 N.
A book rests on a table. Which force is the Newton’s-third-law partner to the table’s upward normal force on the book?
Correct answer: C. Newton’s-third-law forces are the same interaction on two different objects and have equal magnitude with opposite direction.
A spinning student on a low-friction stool pulls two masses inward. With negligible external torque, what happens?
Correct answer: A. With negligible external torque, angular momentum L = Iω is conserved, so I and ω change inversely.
A point on a rotating disk is 0.32 m from the axis and the disk rotates at 5 rad/s. What is the point's tangential speed?
Correct answer: C. Tangential speed and angular speed are related by v = rω.
A 4 kg block rests on a horizontal surface with μ_s = 0.40. Take g = 10 m/s². What is the maximum static-friction magnitude?
Correct answer: D. Maximum static friction is f_s,max = μ_s N = μ_s mg = 16 N.
A 3 kg object is lifted vertically by 4 m near Earth’s surface. Take g = 10 m/s². What is the increase in gravitational potential energy?
Correct answer: D. ΔU_g = mgΔy = 120 J.
A spring with k = 140 N/m is stretched 0.050 m beyond its relaxed length. What is the magnitude of the spring force?
Correct answer: C. Hooke’s law gives |F_s| = k|x| = 7 N.
In a horizontal Venturi tube, the fluid speed is greater in the narrower region. How does the static pressure there compare with the wider region?
Correct answer: B. For steady ideal flow at equal height, Bernoulli’s relation pairs greater speed with lower static pressure.
A 3 kg cart moves at +4 m/s. What is its linear momentum?
Correct answer: B. Linear momentum is p = mv, including the velocity direction, so p = +12 kg·m/s.
A moving object’s speed changes by a factor of 2 while mass stays constant. By what factor does its kinetic energy change?
Correct answer: C. K = ½mv², so kinetic energy scales with the square of speed: factor = 4.
A frictionless block slides down an incline at 30°. Use sin(30°) ≈ 0.5 and g = 10 m/s². What is its acceleration along the incline?
Correct answer: A. Along the incline, ma = mg sinθ, so a = g sinθ = 5 m/s² down the incline.
Water flows steadily from a pipe of cross-sectional area 4.0 cm² at 2.2 m/s into a 2.0 cm² section. What is the speed in the narrower section?
Correct answer: A. For steady incompressible flow, A₁v₁ = A₂v₂. Halving the area doubles the speed.
A force-time graph is a triangle with base 0.5 s and peak height 35 N. What is the impulse magnitude?
Correct answer: B. Impulse equals the area under F(t). The triangle area is ½(base)(height) = 8.75 N·s.
Two balls roll horizontally off the same table at the same instant. Ball B has twice the horizontal launch speed of Ball A. Ignoring air resistance, which statement is correct?
Correct answer: A. Horizontal and vertical motions share the same time. With identical vertical initial conditions and the same gravitational acceleration, the flight time is independent of horizontal speed.
A horizontal beam is in rotational equilibrium. A downward 14 N force acts 0.25 m left of the pivot. What downward force acting 0.50 m right of the pivot balances it?
Correct answer: B. For rotational equilibrium, clockwise and counterclockwise torques have equal magnitude: F_L r_L = F_R r_R.
A 3 kg cart experiences a constant net horizontal force of 9 N. What is the cart's acceleration?
Correct answer: B. Newton’s second law gives a = F_net/m = 3 m/s².
A wheel of radius 0.23 m rolls without slipping with angular speed 5 rad/s. What is the center-of-mass speed?
Correct answer: B. For rolling without slipping, v = ωR = 1.15 m/s.
A cart starts from rest and descends a frictionless track through a vertical height of 2 m. Take g = 10 m/s². What is its speed at the bottom?
Correct answer: B. Conservation of mechanical energy gives mgh = ½mv², so v = √(2gh) = 6.32 m/s.
A rigid object has rotational inertia 2.2 kg·m² and experiences net torque 7 N·m. What is its angular acceleration?
Correct answer: D. Rotational dynamics gives τ_net = Iα, so α = τ_net/I = 3.18 rad/s².
A cart has velocity +5 m/s and acceleration -1.5 m/s² at an instant. What is happening to its speed?
Correct answer: B. Velocity and acceleration have opposite signs, so the acceleration opposes the motion and the speed decreases.
For energy accounting of a falling ball, which system choice allows gravitational potential energy to be treated as internal energy of the system?
Correct answer: A. Gravitational potential energy is associated with the interaction between Earth and the object, so the useful system contains both.
Open quick MCQ answer key for Model Paper 1
Full reasoning remains available under each MCQ. The quick key is for scoring after a timed attempt.
4 free-response questions
One prompt is provided for each current AP Physics 1 FRQ task type. Write all parts before opening the solution.
Cart, ramp, and spring
A 2.0 kg cart starts from rest 1.8 m vertically above the bottom of a frictionless track and compresses a horizontal spring of k = 320 N/m. Use g = 10 m/s².
- Determine the cart speed at the bottom before it contacts the spring.
- Determine the maximum spring compression.
- If the cart mass were doubled with the same release height, state whether maximum compression would increase, decrease, or stay the same and justify.
Open full solution and scoring checkpoints
- Use mgh = ½mv²: v = √(2gh) = √36 = 6.0 m/s for g = 10 m/s².
- At maximum compression the cart is momentarily at rest, so mgh = ½kx². x = √(2mgh/k) = √(72/320) ≈ 0.474 m.
- Compression increases by √2 because x ∝ √m when h and k are fixed.
Practice scoring checkpoints only; these are not official College Board point allocations.
Velocity graph to motion story
A cart has velocity +6 m/s at t = 0. Its velocity decreases linearly to −2 m/s at t = 8 s.
- Sketch the velocity-time graph and the corresponding constant acceleration-time graph.
- Determine the acceleration.
- Determine when the cart changes direction.
- Use signed area to determine displacement from 0 to 8 s.
Open full solution and scoring checkpoints
- v-t is a straight line from (0,6) to (8,−2); a-t is a horizontal line below zero.
- a = (−2−6)/8 = −1.0 m/s².
- Set v = 6−t = 0, giving t = 6 s.
- Signed area = average velocity × time = [(6 + (−2))/2]×8 = 16 m.
Practice scoring checkpoints only; these are not official College Board point allocations.
Measure kinetic friction
Design an experiment using a block, added masses, a force sensor, and a horizontal surface to determine the coefficient of kinetic friction.
- Identify the independent and dependent variables.
- Describe how to collect at least five useful trials.
- State what graph to make and how μ_k is obtained.
- Identify one important control or uncertainty issue.
Open full solution and scoring checkpoints
- Independent variable: normal force, changed by adding mass. Dependent variable: kinetic-friction force while moving at constant speed.
- For each mass, pull at approximately constant speed and record the steady force; repeat or use several mass values.
- Plot F_k versus N. For F_k = μ_kN, the best-fit slope is μ_k.
- Keep the same contact surfaces and pull horizontally; sensor zeroing and maintaining constant speed are major uncertainty/control issues.
Practice scoring checkpoints only; these are not official College Board point allocations.
Momentum and impulse collision
A 0.50 kg cart moving +4.0 m/s collides and sticks to a 1.0 kg cart initially at rest.
- Calculate the common velocity immediately after the collision.
- Calculate the initial and final kinetic energies.
- Explain why momentum can be conserved while kinetic energy is not.
- Describe the sign of the impulse on each cart during the collision.
Open full solution and scoring checkpoints
- Momentum conservation: v_f = (0.50×4.0)/(1.50) = +1.33 m/s.
- K_i = ½(0.50)(4.0²)=4.0 J; K_f = ½(1.50)(1.33²)≈1.33 J.
- Internal nonconservative deformation/thermal processes convert some mechanical energy, while negligible external impulse preserves system momentum.
- The moving cart receives a negative impulse and the initially stationary cart receives an equal-magnitude positive impulse.
Practice scoring checkpoints only; these are not official College Board point allocations.
Do not repeat another full paper immediately if one topic caused most of the errors. Move to the focused banks first, repair the weakness, then return to a different model paper for transfer practice.