Ten lessons that follow the eight course units and the exam skills: kinematics, forces, energy, momentum, rotation, oscillations, fluids and written justification.
A reasoning guide to AP Physics 1: Algebra-Based, not a full course or lab programme. Unit order follows the College Board course page; every example is an original teaching example.
Algebra, basic trigonometry and vector components.
Course outline
Describing motion with graphs and equations
Read position, velocity and acceleration graphs and choose the right kinematic equation.
Forces, free-body diagrams and Newton's laws
Draw free-body diagrams and apply Newton's second law along each axis.
Work, energy and power
Use the work-energy theorem and conservation of energy, and distinguish energy from power.
Momentum, impulse and collisions
Apply impulse-momentum and conserve momentum in collisions.
Torque and rotational dynamics
Compute torque, find equilibrium conditions and apply rotational Newton's second law.
Rotational energy and angular momentum
Use rotational kinetic energy and conservation of angular momentum.
Oscillations and simple harmonic motion
Describe SHM with a spring and a pendulum, and relate period to physical properties.
Fluids: pressure, buoyancy and flow
Use pressure, Archimedes' principle and continuity with Bernoulli's equation.
Choosing between force, energy and momentum
Pick the fastest model for a problem by reading what is asked and what is given.
Exam day and experimental design
Use the exam format and the question types to earn free-response points, including lab design.
Sources and curriculum note
Reviewed October 5, 2026. The exam page says multiple-choice count and timing change from May 2027; confirm on the College Board site for your year.
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. Describing motion with graphs and equations
Learning goal: Read position, velocity and acceleration graphs and choose the right kinematic equation.
Unit 1 is about describing motion before explaining it. Position says where an object is, velocity is the rate of change of position, and acceleration is the rate of change of velocity. Velocity and acceleration are vectors, so sign matters. An object moving in the negative direction and slowing down has negative velocity and positive acceleration, because the velocity change points opposite to the motion. Speeding up means velocity and acceleration have the same sign.
Graphs carry the same information in another form. On a position-time graph, slope is velocity. On a velocity-time graph, slope is acceleration and the area between the curve and the axis is displacement. Area below the axis counts as negative displacement. A horizontal line on a velocity graph means constant velocity and zero acceleration; it does not mean the object is at rest unless that line sits at zero.
For constant acceleration, four equations connect displacement, initial velocity, final velocity, acceleration and time: v = v₀ + at, Δx = v₀t + ½at², v² = v₀² + 2aΔx, and Δx = ½(v₀ + v)t. Each equation leaves out one variable, so list what you know, name what you want and choose the equation that omits the variable you do not have. These equations apply only when acceleration is constant, which is why a graph check comes first.
Free fall near Earth's surface is the standard constant-acceleration case. Ignoring air resistance, the acceleration is about 9.8 m/s² downward at every point, including the top of the path where the velocity is momentarily zero. Projectile motion splits into two independent parts: horizontal motion at constant velocity and vertical motion at constant downward acceleration. Time connects the two parts, so solve the vertical motion for time and use it in the horizontal direction.
Worked example
A ball is thrown straight up at 14 m/s. Use g = 9.8 m/s². How long until it returns to its start?
Take up as positive, so a = −9.8 m/s² and Δx = 0 at return.
Use Δx = v₀t + ½at²: 0 = 14t − 4.9t².
Factor: t(14 − 4.9t) = 0, so t = 0 or t = 14/4.9.
t ≈ 2.9 s is the return time; t = 0 is the launch.
Practice problem and solution
A car goes from rest to 20 m/s in 5.0 s at constant acceleration. What is the acceleration in m/s²?
a = Δv/Δt = 20/5.0 = 4.0 m/s².
Mental model: Slope and area decode graphs. Pick the equation that omits the unknown you do not need.
Common trap: Using constant-acceleration equations when acceleration is changing.
2. Forces, free-body diagrams and Newton's laws
Learning goal: Draw free-body diagrams and apply Newton's second law along each axis.
Unit 2 carries 18% to 23% of the exam, so it repays careful work. A free-body diagram shows one object and only the forces acting on it, drawn as arrows from the object. Typical forces are weight (gravity) downward, normal force perpendicular to a surface, tension along a rope, friction parallel to a surface and applied pushes. Do not draw velocity or acceleration on the diagram, and never draw a force the object exerts on something else.
Newton's first law says an object keeps its velocity unless a net force acts. Newton's second law says the net force equals mass times acceleration, ΣF = ma, and it holds separately along each axis. Choose axes that line up with the acceleration, resolve forces into components and write one equation per axis. On an incline, tilt the axes so one lies along the slope; the weight component along the slope is mg sin θ and the component into the surface is mg cos θ.
Newton's third law says forces come in pairs acting on different objects, equal in size and opposite in direction. The pair of forces is never on the same free-body diagram, so third-law pairs do not cancel. The Earth pulls you down and you pull the Earth up with the same size of force. Friction has two forms: static friction adjusts up to a maximum μₛN, while kinetic friction is μₖN once sliding begins.
Apparent weight is the normal force, which can differ from mg. In an elevator accelerating upward the normal force is larger than the weight, so you feel heavier; in free fall it is zero. For connected objects, treat each as its own system or the whole chain as one system with the same acceleration magnitude. Check units and signs at the end: if a computed tension is negative, the rope would have to push, which a rope cannot do.
Worked example
A 2.0 kg block is pulled by a 10 N force on a frictionless horizontal surface. Find its acceleration.
Only the 10 N force acts horizontally.
ΣF = ma gives 10 = 2.0a.
a = 5.0 m/s².
Vertical forces cancel, so the block stays on the surface.
Practice problem and solution
A 4.0 kg object has a net force of 12 N. What is its acceleration in m/s²?
a = F/m = 12/4.0 = 3.0 m/s².
Mental model: Free-body diagram, axes, ΣF = ma per axis.
Common trap: Cancelling a third-law pair as if both forces acted on one object.
3. Work, energy and power
Learning goal: Use the work-energy theorem and conservation of energy, and distinguish energy from power.
Unit 3 also carries 18% to 23% of the exam. Work is the transfer of energy by a force acting through a displacement: W = Fd cos θ. Only the component of force along the displacement does work. A force perpendicular to the motion, such as the normal force on a level surface or the tension on a swinging pendulum bob, does zero work. Work can be negative when the force opposes the motion, as friction does on a sliding object.
Kinetic energy is K = ½mv² and gravitational potential energy near Earth is U = mgh relative to a chosen zero. A spring stores ½kx² of elastic potential energy. The work-energy theorem says the net work done on an object equals its change in kinetic energy. This is often the fastest path because it avoids finding acceleration and time; you add up the work done by each force and set it equal to ΔK.
When only conservative forces act, such as gravity and ideal springs, the total mechanical energy of a system stays constant. When friction or other nonconservative forces act, mechanical energy changes by the work they do, and the lost energy appears as thermal energy. Define the system carefully: if the Earth is inside the system, gravity is internal and potential energy is counted; if the object alone is the system, gravity does external work instead.
Power is the rate of energy transfer, P = W/t, or P = Fv for a force acting along a constant velocity, measured in watts. Two machines can do the same work with different power. Energy bar charts help organise problems: draw the initial and final bars for kinetic, potential and thermal energy and check that the totals match. Units are a quick check: joules for energy, watts for power, newtons for force.
Worked example
A 2.0 kg object is released from rest at height 5.0 m. Ignore friction. Use g = 9.8 m/s². Find its speed at the ground.
Initial energy is mgh = 2.0 × 9.8 × 5.0 = 98 J.
At the ground all of it is kinetic: ½mv² = 98.
v² = 2 × 98 ÷ 2.0 = 98.
v ≈ 9.9 m/s.
Practice problem and solution
A motor does 600 J of work in 30 s. What is its average power in watts?
P = W/t = 600/30 = 20 W.
Mental model: Net work equals change in kinetic energy. Energy is conserved once all forms are counted.
Common trap: Calling a force perpendicular to motion productive work.
4. Momentum, impulse and collisions
Learning goal: Apply impulse-momentum and conserve momentum in collisions.
Unit 4 is about momentum, p = mv, a vector pointing along the velocity. Impulse is force times the time it acts, J = FΔt, and equals the change in momentum. The impulse-momentum theorem explains why a longer stopping time lowers the average force: the same momentum change spread over more time needs a smaller force. Airbags and padded landings work this way. On a force-time graph, impulse is the area under the curve.
When no net external force acts on a system, its total momentum is conserved. Choose the system so the collision forces are internal. The two colliding objects exert equal and opposite forces for the same time on each other, so their momentum changes are equal and opposite. Momentum is conserved in every collision of an isolated system, including those where kinetic energy is lost.
Collisions are classified by what happens to kinetic energy. In an elastic collision, kinetic energy is also conserved. In an inelastic collision some kinetic energy becomes thermal energy or deformation. In a perfectly inelastic collision the objects stick together and move with a common velocity, which is the maximum loss consistent with momentum conservation. Never assume kinetic energy is conserved unless the problem says elastic.
Momentum is a vector, so handle each direction separately. For a two-dimensional collision, write a conservation equation for each axis. Explosions and recoil are the same idea in reverse: a system at rest has zero total momentum, so pieces moving apart have equal and opposite momenta, and the lighter piece moves faster. Centre of mass of an isolated system moves at constant velocity, whatever happens inside it.
Worked example
A 1.0 kg cart at 4.0 m/s hits a stationary 3.0 kg cart and they stick together. Find the final speed.
Initial momentum: 1.0 × 4.0 = 4.0 kg·m/s.
Final mass is 4.0 kg moving at v.
4.0 = 4.0v.
v = 1.0 m/s.
Practice problem and solution
A 2.0 kg ball changes velocity from 3.0 m/s to 8.0 m/s in a straight line. What is the impulse in N·s?
J = mΔv = 2.0 × (8.0 − 3.0) = 10 N·s.
Mental model: Impulse equals Δp. Conserve momentum in isolated systems.
Common trap: Assuming kinetic energy is conserved in every collision.
5. Torque and rotational dynamics
Learning goal: Compute torque, find equilibrium conditions and apply rotational Newton's second law.
Unit 5 extends Newton's laws to rotation. Torque is the tendency of a force to cause rotation, τ = rF sin θ, where r is the distance from the pivot to the point of application and θ is the angle between r and the force. A force along the line through the pivot gives zero torque. A longer lever arm makes the same force more effective, which is why a door handle sits far from the hinge.
An object is in static equilibrium when the net force is zero and the net torque about any point is zero. Choose the pivot cleverly: placing it where an unknown force acts removes that force from the torque equation. Clockwise and counterclockwise torques are assigned opposite signs. A seesaw balances when torques about the pivot match, so a lighter person must sit farther from it.
Rotational inertia, I, measures resistance to angular acceleration and depends on how mass is distributed relative to the axis. Mass farther from the axis contributes more. For point masses I = Σmr², which is why a hoop has a larger I than a disc of the same mass and radius. The rotational form of Newton's second law is Στ = Iα, where α is angular acceleration in rad/s².
Link linear and angular quantities for rolling and circular motion: v = rω, and tangential acceleration a = rα. Angular velocity ω is the rate of change of angle. For constant α the angular kinematic equations mirror the linear ones. A common error is to treat the object's weight as acting at a corner; for a uniform rod it acts at the centre. Draw the pivot, the lever arms and the direction of each torque before writing any equation.
Worked example
A 20 N force acts perpendicular to a wrench 0.30 m from the bolt. Find the torque.
Perpendicular force means sin θ = 1.
τ = rF = 0.30 × 20.
τ = 6.0 N·m.
The direction (clockwise or counterclockwise) depends on the push.
Practice problem and solution
A 40 N force acts perpendicular to a lever at 0.25 m from the pivot. What is the torque in N·m?
τ = rF = 0.25 × 40 = 10 N·m.
Mental model: Torque is rF sin θ. Equilibrium needs zero net force and zero net torque.
Common trap: Using the full force when only part of it is perpendicular.
6. Rotational energy and angular momentum
Learning goal: Use rotational kinetic energy and conservation of angular momentum.
Unit 6 is smaller on the exam, 5% to 8%, but it reuses ideas you already know. A rotating object has kinetic energy K = ½Iω². A rolling object has both translational energy ½mv² and rotational energy ½Iω², so it needs more total energy to reach a given speed than a sliding object. That is why a solid disc beats a hoop down an incline: less of its energy goes into rotation.
Angular momentum of a rotating object is L = Iω, and for a point mass moving in a circle L = mvr. Torque is the rate of change of angular momentum, so a net external torque of zero means L stays constant. This is conservation of angular momentum. A skater pulling in her arms reduces I, and since L is constant, ω must rise. Kinetic energy rises too, because the skater does work pulling the arms in.
Angular impulse mirrors linear impulse: torque times time equals the change in angular momentum. In a collision of a spinning disc with another disc dropped on it, no external torque acts about the axis, so Iω before equals the combined I times the new ω after. The collision is inelastic, so rotational kinetic energy is lost as thermal energy even though angular momentum is conserved.
For rolling without slipping, the point of contact is momentarily at rest and v = rω links the quantities. Use energy for incline problems: initial potential energy equals the sum of translational and rotational kinetic energy at the bottom. Check the direction of angular momentum with the right-hand rule when a problem has several rotating pieces, and keep track of signs consistently for clockwise and counterclockwise.
Worked example
A disc with I = 2.0 kg·m² spins at 6.0 rad/s. A second disc with I = 1.0 kg·m² is dropped on it and they spin together. Find the final ω.
Initial L = 2.0 × 6.0 = 12 kg·m²/s.
Final I = 3.0 kg·m².
12 = 3.0ω.
ω = 4.0 rad/s.
Practice problem and solution
A disc has I = 0.50 kg·m² and ω = 4.0 rad/s. What is its rotational kinetic energy in joules?
K = ½Iω² = 0.5 × 0.50 × 16 = 4.0 J.
Mental model: K = ½Iω². Zero external torque keeps L constant.
Common trap: Believing kinetic energy is conserved when angular momentum is.
7. Oscillations and simple harmonic motion
Learning goal: Describe SHM with a spring and a pendulum, and relate period to physical properties.
Unit 7 covers repeating motion. Simple harmonic motion happens when the restoring force is proportional to displacement and points back to equilibrium, F = −kx for a spring. The motion repeats with period T, the time for one cycle, and frequency f = 1/T. The amplitude is the maximum displacement. Acceleration is largest at the ends of the motion where the restoring force is largest, and speed is largest at equilibrium.
For a mass on a spring, T = 2π√(m/k). A heavier mass makes the period longer; a stiffer spring makes it shorter. The period does not depend on amplitude for ideal SHM, which is why a clock pendulum can keep time as it slows. For a simple pendulum swinging through small angles, T = 2π√(L/g), depending on length and gravity but not on mass.
Energy in SHM trades between kinetic and potential. At maximum displacement the speed is zero, so all energy is potential: ½kA². At equilibrium the potential energy is zero and kinetic energy is at its maximum. Total energy stays constant for an ideal system and grows with the square of the amplitude. A position-time graph is sinusoidal; the velocity graph is shifted by a quarter period, and acceleration is opposite in sign to position.
A vertical spring has the same period as a horizontal one with the same mass and spring constant; gravity only shifts the equilibrium position. To test the period law in the lab, measure the time for many cycles and divide, which reduces timing error. A graph of T² against mass for a spring gives a straight line with slope 4π²/k. Damping removes energy and makes amplitude decay but changes the period only slightly for light damping.
Worked example
A 0.50 kg mass on a spring with k = 50 N/m oscillates. Find its period.
Use T = 2π√(m/k).
m/k = 0.50/50 = 0.010.
√0.010 = 0.10.
T = 2π × 0.10 ≈ 0.63 s.
Practice problem and solution
An oscillator completes 20 cycles in 10 seconds. What is its period in seconds?
T = time per cycle = 10/20 = 0.50 s.
Mental model: T = 2π√(m/k) for springs and 2π√(L/g) for pendulums.
Common trap: Thinking a larger amplitude lengthens the period in ideal SHM.
8. Fluids: pressure, buoyancy and flow
Learning goal: Use pressure, Archimedes' principle and continuity with Bernoulli's equation.
Unit 8 covers 10% to 15% of the exam. Pressure is force per unit area, P = F/A, measured in pascals. In a static fluid, pressure increases with depth: P = P₀ + ρgh, where ρ is the fluid density and h is the depth. Pressure at the same depth is the same in all directions and does not depend on the container's shape. This is why water towers use height to create pressure.
The buoyant force on an object in a fluid equals the weight of the fluid it displaces: F_B = ρ_fluid V_displaced g. An object floats when the buoyant force can balance its weight, which happens when its average density is less than the fluid's. A floating object displaces fluid equal to its own weight, but a fully submerged object displaces its own volume. Apparent weight in a fluid is weight minus buoyant force.
For flowing fluids, the continuity equation states that for an incompressible fluid, A₁v₁ = A₂v₂, so the fluid speeds up in a narrower pipe. Volume flow rate Q = Av is the same at every point of a single pipe. This conserves mass: whatever volume enters per second must leave per second.
Bernoulli's equation relates pressure, speed and height along a streamline: P + ½ρv² + ρgh is constant. Where the fluid moves faster, its pressure is lower at the same height. This explains why a narrow section of horizontal pipe has lower pressure than a wide one, not higher. Apply Bernoulli carefully: it assumes steady, incompressible, nonviscous flow, and it is an energy-conservation statement per unit volume.
Worked example
Water flows at 2.0 m/s in a pipe of area 0.010 m² and enters a section of area 0.0050 m². Find the speed in the narrow section.
Continuity: A₁v₁ = A₂v₂.
0.010 × 2.0 = 0.0050 × v₂.
v₂ = 0.020/0.0050.
v₂ = 4.0 m/s.
Practice problem and solution
A pipe has area 0.020 m² and water moves at 3.0 m/s. What is the volume flow rate in m³/s?
Common trap: Expecting higher pressure where the fluid moves faster.
9. Choosing between force, energy and momentum
Learning goal: Pick the fastest model for a problem by reading what is asked and what is given.
Many AP Physics 1 problems can be solved in more than one way, but one way is usually much shorter. Force methods, built on ΣF = ma, are best when you need acceleration, tension, normal force or friction at one moment. They need the forces at that instant and work well for constant acceleration over time.
Energy methods are best when the question links speeds and positions, such as the speed at the bottom of a track or the compression of a spring. They skip time and the path shape, as long as you account for nonconservative work like friction. Write the initial and final energy, include thermal energy if friction acts, and solve for the unknown.
Momentum methods are best for short interactions: collisions, explosions and recoil. During the brief interaction, internal forces are huge and complicated, but they cancel for the system, so total momentum before equals total momentum after. After you find the velocities just after the collision, switch to energy or forces for the motion that follows, such as a pendulum swinging up or a block sliding to rest.
Real problems chain methods. A bullet embeds in a block hanging from a string: momentum conservation for the collision, then energy conservation for the swing. A block slides down a ramp and hits another: energy for the slide, momentum for the collision, then forces or energy for the stopping distance. Label each stage, state which quantity is conserved in it, and carry only the speed from one stage to the next.
Worked example
A 0.020 kg bullet at 200 m/s embeds in a 1.98 kg block at rest on a smooth surface. Find the speed just after.
The collision is brief, so use momentum.
Initial momentum: 0.020 × 200 = 4.0 kg·m/s.
Final mass: 2.00 kg.
v = 4.0/2.00 = 2.0 m/s.
Practice problem and solution
A 2.0 kg cart at 3.0 m/s sticks to a 1.0 kg cart at rest. What is the final speed in m/s?
6.0 = 3.0v, so v = 2.0 m/s.
Mental model: Force for acceleration, energy for speed versus position, momentum for collisions. Chain stages.
Common trap: Using energy conservation straight through an inelastic collision.
10. Exam day and experimental design
Learning goal: Use the exam format and the question types to earn free-response points, including lab design.
The AP Physics 1 exam has two sections. The College Board exam page says the number of multiple-choice questions and the section timing change starting with the May 2027 exam, so read the current page for your year: it lists Section I as 42 questions in 85 minutes for 50% of the score, and Section II as 4 free-response questions in 95 minutes for 50%. Calculators are permitted and the free response is handwritten in a booklet.
The course framework groups skills into creating representations, mathematical routines and scientific questioning and argumentation. In the free-response section, the framework gives roughly 20% to 35% of points to representations, 30% to 40% to mathematical routines and 35% to 45% to scientific questioning and argumentation. That means diagrams and written justification together carry most of the free-response points, so practise explaining, not only calculating.
The free-response section includes an experimental design question. State the question, name the quantity you will measure and the equipment, describe what you vary and what you hold constant, and say how you will analyse the data. A graph that is linear in the variables of the physical law lets you read a constant from the slope. For example, plot T² against m for a spring and find k from the slope.
Strong written justifications name the principle, apply it to the situation and reach the claim. Say 'the net force is zero, so by Newton's first law the velocity is constant', not 'it moves at constant speed because the forces are balanced'. Show symbolic work before substituting numbers, include units, and check that the answer has a sensible size and sign. For a qualitative question, comparing before and after states in a sentence is often enough.
Worked example
Design a short test of how a spring's period depends on mass.
Hang a mass on a spring and time ten oscillations; divide by ten for T.
Repeat with several masses, keeping the same spring.
Plot T² against m.
A straight line through the origin supports T² = (4π²/k)m, and the slope gives k.
Practice problem and solution
In the spring experiment, which graph is best for finding k: T versus m, or T² versus m? Type T2 or T.
T² versus m, because T² = (4π²/k)m is linear.
Mental model: Know the format. Justify with principles. Design experiments with linear graphs.
Common trap: Writing a conclusion without naming the physical law behind it.