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 2
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 4 kg object speeds up from 2 m/s to 4 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) = 24 J.
An object moves along the x-axis from x = -4 m to x = +11 m. What is its displacement?
Correct answer: A. Displacement is final position minus initial position: Δx = x_f − x_i = 15 m.
A rotating object has I = 1.6 kg·m² and ω = 5 rad/s. What is its rotational kinetic energy?
Correct answer: D. Rotational kinetic energy is K_rot = ½Iω² = 20 J.
For energy accounting of a roller-coaster car descending a hill, 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.
A point on a rotating disk is 0.34 m from the axis and the disk rotates at 6 rad/s. What is the point's tangential speed?
Correct answer: C. Tangential speed and angular speed are related by v = rω.
Water flows steadily from a pipe of cross-sectional area 4.0 cm² at 2.4 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 fluid sample has mass 0.8 kg and volume 5.0×10⁻⁴ m³. What is its density?
Correct answer: B. Density is ρ = m/V = 1600 kg/m³.
In a level water pipe that narrows, 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.
For two skaters pushing apart on smooth ice, 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.
A force of 14 N is applied perpendicular to a wrench 0.24 m from the pivot. What is the torque magnitude?
Correct answer: C. For a perpendicular force, τ = rF = 3.36 N·m.
In water (ρ = 1000 kg/m³), what is the gauge-pressure increase 1.4 m below the surface? Take g = 10 m/s².
Correct answer: C. Hydrostatic pressure change is ΔP = ρgΔh = 14000 Pa.
An ideal spring oscillator is instantaneously at x = +A/2, 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 4 kg object is lifted vertically by 5 m near Earth’s surface. Take g = 10 m/s². What is the increase in gravitational potential energy?
Correct answer: D. ΔU_g = mgΔy = 200 J.
A 1.4 kg cart moving at +6 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 = +2.47 m/s.
A rigid object has rotational inertia 2.4 kg·m² and experiences net torque 8 N·m. What is its angular acceleration?
Correct answer: D. Rotational dynamics gives τ_net = Iα, so α = τ_net/I = 3.33 rad/s².
A swimmer pushes water backward with her hands. Which force is the Newton’s-third-law partner?
Correct answer: C. Newton’s-third-law forces are the same interaction on two different objects and have equal magnitude with opposite direction.
A 4 kg cart moves at +5 m/s. What is its linear momentum?
Correct answer: B. Linear momentum is p = mv, including the velocity direction, so p = +20 kg·m/s.
A constant horizontal force of 14 N acts in the direction of motion while an object moves 5 m. How much work does the force do?
Correct answer: B. For a force parallel to displacement, W = Fd = 70 J.
For an ideal horizontal mass-spring system, 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.
A wheel of radius 0.26 m rolls without slipping with angular speed 6 rad/s. What is the center-of-mass speed?
Correct answer: B. For rolling without slipping, v = ωR = 1.56 m/s.
A 55 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) = 660 N.
A spring with k = 160 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| = 8 N.
A frictionless block slides down an incline at 37°. Use sin(37°) ≈ 0.6 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 = 6 m/s² down the incline.
A 5 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 = 20 N.
A constant +30 N force acts for 0.3 s. What impulse is delivered?
Correct answer: C. Impulse is J = FΔt for a constant force, so J = +9 N·s.
A 4 kg cart experiences a constant net horizontal force of 12 N. What is the cart's acceleration?
Correct answer: B. Newton’s second law gives a = F_net/m = 3 m/s².
A figure skater spinning with arms extended pulls both arms close to the body. 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 horizontal beam is in rotational equilibrium. A downward 16 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.
An object displaces 0.0024 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 = 24 N.
A spring with k = 240 N/m is compressed 0.10 m. How much elastic potential energy is stored?
Correct answer: A. Spring potential energy is U_s = ½kx² = 1.2 J.
A cart starts from rest and descends a frictionless track through a vertical height of 2.2 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) = 6.63 m/s.
A motor transfers 800 J of energy in 7 s. What is its average power?
Correct answer: C. Average power is energy transferred per time: P = ΔE/Δt = 114.29 W.
Two packages leave the same level platform simultaneously with different horizontal speeds and zero initial vertical velocity. Ignoring air resistance, how do their landing times compare?
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 force-time graph is a triangle with base 0.6 s and peak height 40 N. What is the impulse magnitude?
Correct answer: B. Impulse equals the area under F(t). The triangle area is ½(base)(height) = 12 N·s.
A cart undergoes constant acceleration. Its velocity changes from +8 m/s to +2 m/s in 6 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 = 30 m.
For an ideal mass-spring oscillator, the mass changes by a factor of 9 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.
Two masses attract gravitationally. Their separation changes by a factor of 3 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 3 changes F by 1/(3)² = 0.111.
A train moves +22 m/s relative to the ground. A passenger walks toward the front 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 = 25 m/s.
A moving object’s speed changes by a factor of 3 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 = 9.
A cart has velocity +6 m/s and acceleration -2 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.
Two rotating objects have the same total mass and outer radius. Object B moves equal mass from the center toward the edge. 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.
An object moves in a circle of fixed radius. Its speed changes by a factor of 3 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 (3)² = 9.
Open quick MCQ answer key for Model Paper 2
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.
Circular motion on a flat curve
A 900 kg car travels around a flat circular curve of radius 45 m. The coefficient of static friction is 0.50. Use g = 10 m/s².
- Determine the maximum static-friction force.
- Determine the maximum speed before slipping.
- Explain why the car’s velocity changes even when its speed is constant.
Open full solution and scoring checkpoints
- N = mg = 9000 N for g=10, so f_s,max = 0.50(9000)=4500 N.
- Set mv²/r = f_s,max: v = √(f_s,max r/m)=√225≈15 m/s.
- Velocity is a vector; its direction continuously changes around the curve.
Practice scoring checkpoints only; these are not official College Board point allocations.
Force-time pulse
A force on a 0.40 kg cart rises linearly from 0 to 20 N during the first 0.20 s, then falls linearly to 0 by 0.40 s.
- Sketch the force-time graph.
- Determine the impulse.
- If the cart starts at −2.0 m/s, determine its final velocity.
- Sketch a qualitative momentum-time graph.
Open full solution and scoring checkpoints
- The force-time graph is a triangle with base 0.40 s and height 20 N.
- Impulse is area = ½(0.40)(20)=4.0 N·s.
- p_i = 0.40(−2)=−0.8 kg·m/s; p_f=3.2, so v_f=3.2/0.40=8.0 m/s.
- Momentum increases smoothly with slope equal to force, reaching its steepest slope at the pulse peak.
Practice scoring checkpoints only; these are not official College Board point allocations.
Spring oscillation method
Use a spring, several known masses, a meterstick, and a timer to determine the spring constant from oscillations.
- State what to measure for each mass.
- Choose a graph that should be linear.
- Explain how k is obtained from the graph.
- Give one way to reduce timing uncertainty.
Open full solution and scoring checkpoints
- Measure mass m and period T, preferably timing many cycles and dividing by the number of cycles.
- From T² = (4π²/k)m, plot T² versus m.
- The slope is 4π²/k, so k = 4π²/slope.
- Time 10–20 oscillations per trial and repeat trials to reduce reaction-time fractional error.
Practice scoring checkpoints only; these are not official College Board point allocations.
Fluid flow through a narrowing pipe
Water flows horizontally from a wide pipe into a section with half the cross-sectional area.
- If the wide-section speed is 2.0 m/s, determine the narrow-section speed.
- Compare static pressures in the two sections.
- Explain the pressure comparison using energy per unit volume.
- Predict what happens to the speed if the narrow area is reduced again.
Open full solution and scoring checkpoints
- Continuity: A_1v_1=A_2v_2, so halving area doubles speed to 4.0 m/s.
- The narrow section has lower static pressure.
- At equal height, an increase in kinetic-energy density ½ρv² is balanced by a decrease in pressure term P.
- Reducing area further increases flow speed for steady incompressible flow.
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.