Eleven algebra-based lessons across the seven units: thermodynamics, electric force and field, circuits, magnetism, optics, waves and modern physics, with interactive explorers.
A reasoning guide to AP Physics 2: Algebra-Based, not a full course. Unit titles and weightings follow the College Board course page. The exam page lists 42 multiple-choice questions in 85 minutes and 4 free-response questions in 95 minutes for May 2027. All numbers in examples are invented for practice, with standard physical constants.
AP Physics 1 or a comparable course, and precalculus at the same time or before.
Course outline
The ideal gas law and kinetic theory
Connect pressure, volume and temperature to molecular motion and count moles and molecules.
The first law of thermodynamics and PV diagrams
Track heat, work and internal energy through gas processes.
Entropy, heat engines and the second law
Use efficiency and the Carnot limit to judge what an engine can do.
Electric force, field and potential
Apply Coulomb's law and uniform-field relations to charges and parallel plates.
Circuits with resistors and capacitors
Analyze series and parallel combinations of resistors and capacitors.
Magnetic forces and induction
Find forces on charges and wires and the emf of a moving conductor.
Geometric optics: mirrors and lenses
Use the thin-lens equation and magnification to locate images.
Refraction, total internal reflection and the speed of light
Use the index of refraction to relate speed, wavelength and bending.
Waves, sound and physical optics
Use wave relations, standing waves, the Doppler effect and double-slit interference.
Photons, the photoelectric effect and matter waves
Use photon energy, the work function and the de Broglie wavelength.
Atomic energy levels, radioactive decay and mass-energy
Use quantized energy levels, half-life and E = mc^2 together.
Sources and curriculum note
Reviewed October 7, 2026. Confirm format on the College Board site for your exam year.
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. The ideal gas law and kinetic theory
Learning goal: Connect pressure, volume and temperature to molecular motion and count moles and molecules.
For an ideal gas, P V = n R T = N k_B T, with temperature in kelvin, R = 8.31 J / (mol K) and k_B = 1.38 x 10^-23 J/K. The number of molecules N equals n times Avogadro's number. Use kelvin: add 273 to a Celsius temperature. A ratio of temperatures in Celsius is meaningless.
Kinetic theory links the macroscopic and microscopic pictures. Pressure comes from molecules colliding with the walls. The average translational kinetic energy per molecule is (3/2) k_B T, which depends only on temperature, so at the same temperature a light gas molecule moves faster than a heavy one. The root-mean-square speed is sqrt(3 k_B T / m).
For changes at fixed amount of gas use P1 V1 / T1 = P2 V2 / T2. Doubling the absolute temperature at constant volume doubles the pressure. Halving the volume at constant temperature doubles the pressure. Always state what is held constant before you pick a form.
A common error is to use gauge pressure, which is measured relative to the atmosphere, in the gas law. The law needs absolute pressure. Also check units: P in pascals, V in cubic meters. One mole at 273 K and 101 kPa occupies about 22.4 liters.
Worked example
A gas has P = 2.0e5 Pa, V = 0.010 m^3 and T = 300 K. How many moles?
n = P V / (R T).
2.0e5 x 0.010 = 2000 J.
R T = 8.314 x 300 = 2494.
n = 0.80 mol.
Practice problem and solution
A gas has P = 2.0e5 Pa, V = 0.010 m^3 and T = 300 K. Number of moles, to 2 decimals?
n = P V / (R T) = 2000 / 2494.2.
Mental model: PV = nRT = NkT in kelvin and absolute pressure. Average KE = (3/2) k T.
Common trap: Using Celsius or gauge pressure in the gas law.
2. The first law of thermodynamics and PV diagrams
Learning goal: Track heat, work and internal energy through gas processes.
The first law says the change in internal energy equals the heat added to the system plus the work done on it: Delta U = Q + W. In the College Board convention, W is work done on the gas, so compressing the gas gives positive W and expansion gives negative W. Say your sign convention before you start.
On a PV diagram, the area under the curve is the magnitude of the work. Moving to larger V (to the right) means the gas does work and W on the gas is negative. For a constant-pressure process the work done by the gas is P Delta V. For a closed cycle the net work is the enclosed area.
For an ideal monatomic gas, internal energy is U = (3/2) n R T, so Delta U depends only on the change in temperature. In an isothermal process Delta U = 0, so Q = -W. In an adiabatic process Q = 0, so Delta U = W. In a constant-volume process W = 0, so Delta U = Q.
A good check is to trace a cycle and confirm that Delta U is zero over the whole loop. Then net heat in equals net work out, which is the basis for heat engines.
Worked example
A gas expands at constant pressure 1.0e5 Pa from 0.010 to 0.030 m^3. Work done by the gas?
W by gas = P Delta V.
Delta V = 0.020.
1.0e5 x 0.020.
2000 J.
Practice problem and solution
A gas absorbs 500 J of heat and does 200 J of work on its surroundings. Change in internal energy in J?
Delta U = Q + W = 500 - 200.
Mental model: Delta U = Q + W. Area under the curve is work. Isothermal: Delta U = 0.
Common trap: Using the wrong sign for work.
3. Entropy, heat engines and the second law
Learning goal: Use efficiency and the Carnot limit to judge what an engine can do.
The second law says the entropy of an isolated system does not decrease. Heat flows spontaneously from hot to cold, never the other way without work. Entropy measures the number of microscopic arrangements consistent with a macroscopic state, so more disorder means higher entropy.
A heat engine takes heat Q_h from a hot reservoir, does work W and rejects Q_c to a cold reservoir. Energy conservation gives W = Q_h - Q_c. The efficiency is e = W / Q_h. No engine can have e = 1, because some heat must be rejected.
The Carnot efficiency, e = 1 - T_c / T_h with temperatures in kelvin, is the upper limit for any engine working between those two reservoirs. A real engine falls short. If a claimed engine beats the Carnot limit, it violates the second law.
For a refrigerator the same bookkeeping applies in reverse: work moves heat from cold to hot, so Q_h = Q_c + W. When you check an answer, confirm that heat rejected is not negative, that efficiency is below the Carnot value, and that energy is conserved.
Worked example
An engine takes 800 J from a hot reservoir and rejects 600 J. Efficiency?
W = 800 - 600 = 200 J.
e = 200 / 800.
= 0.25.
25 percent.
Practice problem and solution
What is the Carnot efficiency, as a percent, between 600 K and 300 K?
1 - 300/600 = 0.50.
Mental model: e = W / Q_h. Carnot limit = 1 - T_c/T_h. Entropy of an isolated system does not decrease.
Common trap: Using Celsius in the Carnot formula.
4. Electric force, field and potential
Learning goal: Apply Coulomb's law and uniform-field relations to charges and parallel plates.
Coulomb's law gives the force between two point charges, F = k q1 q2 / r^2, with k = 8.99 x 10^9 N m^2 / C^2. The electric field of a point charge is E = k q / r^2. The force on a charge q in a field is F = q E, in the direction of E for positive charges and opposite for negative.
Between large parallel plates the field is nearly uniform, E = V / d, pointing from the positive to the negative plate. A charge in that field has constant acceleration a = q E / m. An electron released at rest from the negative plate accelerates toward the positive plate.
Electric potential difference is the change in potential energy per unit charge. A charge q moved through a potential difference Delta V changes its potential energy by q Delta V. When a charge is released from rest, the electric potential energy lost becomes kinetic energy: q Delta V = (1/2) m v^2.
Field lines show direction and strength by density, and equipotential lines are perpendicular to them. Keep three different things apart: force on a charge, field at a point, and potential energy of a charge there.
Worked example
Parallel plates are 2.0 cm apart with 100 V across them. Field strength in V/m?
E = V / d.
d = 0.020 m.
100 / 0.020.
5000 V/m.
Practice problem and solution
An electron is accelerated through 100 V from rest. Its kinetic energy in eV?
KE = q Delta V = 100 eV.
Mental model: F = qE, E = V/d, energy change = q Delta V.
Common trap: Confusing field direction with force direction on a negative charge.
5. Circuits with resistors and capacitors
Learning goal: Analyze series and parallel combinations of resistors and capacitors.
For resistors, series adds: R = R1 + R2, and parallel adds reciprocals: 1 / R = 1 / R1 + 1 / R2. For capacitors it is the opposite way round: parallel adds, C = C1 + C2, and series adds reciprocals. Make a table of both so you do not swap them.
In series the current is the same through each element. In parallel the voltage is the same across each branch. Ohm's law V = I R applies to each resistor. For capacitors, Q = C V. In a series combination of capacitors each carries the same charge.
Power dissipated by a resistor is P = I V = I^2 R = V^2 / R. A capacitor stores energy U = (1/2) C V^2 and does not dissipate power in steady state. When a capacitor is first connected through a resistor, it behaves like a short circuit, and after a long time like an open circuit.
Reduce the circuit in steps and then expand back out. At each step check that Kirchhoff's rules hold: charge conservation at junctions and energy conservation around loops.
Worked example
A 10 uF and a 20 uF capacitor are in parallel. Equivalent capacitance?
Parallel adds.
10 + 20.
= 30 uF.
Check units.
Practice problem and solution
A 6 uF and a 3 uF capacitor are in series. Equivalent capacitance in uF?
1/C = 1/6 + 1/3 = 1/2, so C = 2 uF.
Mental model: Resistors: series add. Capacitors: parallel add. Power P = I V.
Common trap: Swapping the series and parallel rules.
6. Magnetic forces and induction
Learning goal: Find forces on charges and wires and the emf of a moving conductor.
A charge q moving with speed v perpendicular to a magnetic field B feels a force F = q v B, perpendicular to both v and B. The right-hand rule gives the direction for a positive charge. The force does no work, so the speed does not change, but the direction does, and the path is circular.
A wire of length L carrying current I in a field B perpendicular to it feels F = B I L. It is the sum of the forces on the moving charges inside it. Reverse the current or the field and the force reverses. Two parallel wires with currents in the same direction attract.
A conducting rod of length L moving at speed v perpendicular to a field B develops an emf of B L v between its ends. If the rod is part of a closed circuit with resistance R, the current is B L v / R. Faraday's law states that an emf is induced when the magnetic flux through a loop changes.
Lenz's law gives the direction: the induced current opposes the change in flux. Always say whether the flux is increasing or decreasing and in which direction. The induced field opposes that change.
Worked example
A 0.30 m wire carries 5.0 A in a 0.40 T field perpendicular to it. Force?
F = B I L.
0.40 x 5.0 x 0.30.
= 0.60 N.
Perpendicular to wire and field.
Practice problem and solution
A 0.50 m rod moves at 3.0 m/s perpendicular to a 0.20 T field. Motional emf in volts, to 2 decimals?
emf = B L v = 0.20 x 0.50 x 3.0.
Mental model: F = qvB, F = BIL, emf = BLv. Induced current opposes the flux change.
Common trap: Applying the right-hand rule to a negative charge without reversing.
7. Geometric optics: mirrors and lenses
Learning goal: Use the thin-lens equation and magnification to locate images.
For a thin lens or a spherical mirror, 1 / f = 1 / d_o + 1 / d_i, where d_o is the object distance and d_i is the image distance. The magnification is m = -d_i / d_o. A negative magnification means the image is inverted. A positive d_i means a real image on the opposite side of a lens, or the same side as the object for a mirror.
A converging lens has positive focal length. An object beyond the focal point gives a real inverted image. An object inside the focal length gives a virtual, upright, magnified image on the same side as the object, which is how a magnifying glass works. A diverging lens always gives a smaller virtual upright image.
For a concave mirror the focal length is half the radius of curvature and the rules mirror those of a converging lens. Ray diagrams back up the equation: draw one ray parallel to the axis through the focal point and one ray through the center of the lens. The crossing point is the image.
Check limits. As the object moves far away, the image forms at the focal point. When d_o = f, no image forms because the rays emerge parallel. Always keep the sign convention stated.
Worked example
A converging lens has f = 10 cm and the object is at 30 cm. Image distance?
1/d_i = 1/10 - 1/30.
= 3/30 - 1/30 = 2/30.
d_i = 15 cm.
Real and inverted, m = -0.5.
Practice problem and solution
A converging lens has f = 10 cm and the object is at 30 cm. Image distance in cm?
1/d_i = 1/10 - 1/30 = 1/15.
Mental model: 1/f = 1/do + 1/di, m = -di/do. Inside f the image is virtual.
Common trap: Dropping the sign convention.
8. Refraction, total internal reflection and the speed of light
Learning goal: Use the index of refraction to relate speed, wavelength and bending.
The index of refraction of a medium is n = c / v, where c is the speed of light in vacuum, 3.00 x 10^8 m/s, and v is the speed in the medium. Light slows in glass or water, so n is greater than 1. Frequency stays the same across a boundary, so the wavelength shortens: lambda in the medium equals lambda in vacuum divided by n.
Snell's law says n1 sin(theta1) = n2 sin(theta2), with angles measured from the normal. Light entering a higher index bends toward the normal and entering a lower index bends away from it. The ray never changes if it enters along the normal.
When light goes from a higher to a lower index at a large enough angle, there is no refracted ray and all of the light reflects. The critical angle satisfies sin(theta_c) = n2 / n1, and total internal reflection happens beyond it. This is how optical fibers trap light.
A habit that helps: first say whether the ray enters a slower or faster medium, then predict the bending, then calculate. If sin(theta2) comes out above 1, there is no refracted ray.
Worked example
Glass has n = 1.50. Speed of light in glass?
v = c / n.
3.00e8 / 1.50.
= 2.00e8 m/s.
Slower than in vacuum.
Practice problem and solution
What is the speed of light in water (n = 1.33), in units of 10^8 m/s, to 2 decimals?
v = 3.00 / 1.33.
Mental model: n = c/v. n1 sin1 = n2 sin2. Frequency does not change across a boundary.
Common trap: Measuring angles from the surface instead of the normal.
9. Waves, sound and physical optics
Learning goal: Use wave relations, standing waves, the Doppler effect and double-slit interference.
Every wave obeys v = f lambda. Sound in air at room temperature moves at about 343 m/s. Waves superpose: where crests meet crests the amplitude adds (constructive interference), and where a crest meets a trough it cancels (destructive interference).
A string fixed at both ends supports standing waves with wavelengths lambda = 2 L / n, so frequencies f_n = n v / (2 L) for n = 1, 2, 3. A tube open at both ends has the same pattern, and a tube closed at one end supports only odd harmonics with f = n v / (4 L) for n = 1, 3, 5.
The Doppler effect shifts the observed frequency when the source and observer move relative to each other. A source approaching an observer gives a higher frequency. For a moving source and stationary observer, f_obs = f_s v / (v - v_s) approaching and f_s v / (v + v_s) receding.
In a double-slit experiment, bright fringes occur where the path difference is a whole number of wavelengths: d sin(theta) = m lambda. For small angles on a screen at distance L, the fringe spacing is Delta y = lambda L / d. A single slit produces a wide central maximum, and narrower slits spread it more.
Worked example
A string of 0.80 m has wave speed 200 m/s. Fundamental frequency?
f1 = v / (2L).
200 / 1.60.
= 125 Hz.
Harmonics 250, 375 and so on.
Practice problem and solution
A closed-end tube has L = 0.50 m and sound speed 340 m/s. Fundamental frequency in Hz?
f1 = v / (4L) = 340 / 2.0.
Mental model: v = f lambda. Double slit spacing = lambda L / d. Doppler: approaching source raises pitch.
Common trap: Using the wrong harmonic series for a closed tube.
10. Photons, the photoelectric effect and matter waves
Learning goal: Use photon energy, the work function and the de Broglie wavelength.
A photon has energy E = h f = h c / lambda. Using h c = 1240 eV nm gives the energy in electron volts when the wavelength is in nanometers. A 400 nm photon has E = 1240 / 400 = 3.1 eV.
In the photoelectric effect, light ejects electrons from a metal only if the photon energy exceeds the work function phi. The maximum kinetic energy is K_max = h f - phi. Below the threshold frequency there is no emission however bright the light is. A brighter beam increases the number of electrons, not their maximum energy.
The stopping potential V_s is the reverse voltage that just stops the fastest electrons, so e V_s = K_max. A plot of K_max against frequency is a line with slope h, and its intercept on the frequency axis gives the threshold frequency. This was strong evidence that light has particle-like behavior.
Matter also has wave behavior. The de Broglie wavelength is lambda = h / p, with p the momentum. An electron accelerated through 100 V has lambda of about 0.12 nm, comparable to atomic spacing, which is why electrons diffract from crystals.
Worked example
Light of 400 nm hits a metal with phi = 2.3 eV. Maximum kinetic energy?
E = 1240 / 400 = 3.1 eV.
K = E - phi.
3.1 - 2.3.
0.8 eV.
Practice problem and solution
Light of 400 nm hits a metal with phi = 2.3 eV. Stopping potential in volts, to 1 decimal?
K_max = 0.8 eV, so V_s = 0.8 V.
Mental model: E = 1240/lambda in eV nm. K_max = hf - phi. lambda = h/p.
Common trap: Thinking brighter light increases the maximum kinetic energy.
11. Atomic energy levels, radioactive decay and mass-energy
Learning goal: Use quantized energy levels, half-life and E = mc^2 together.
In the Bohr model of hydrogen the energy levels are E_n = -13.6 eV / n^2. A photon is emitted when the atom drops from level n_i to n_f, with energy equal to the difference. The wavelength follows from lambda = 1240 eV nm / E. A drop from n = 3 to n = 2 releases 1.89 eV, giving 656 nm, the red hydrogen line.
Because the levels are discrete, atoms emit and absorb only certain wavelengths. That is why each element has a distinct line spectrum. An electron with exactly enough energy to jump a gap can be absorbed, and if the photon energy does not match a gap, it is not absorbed by that transition.
Radioactive decay is random but follows a predictable pattern for large numbers. The number of undecayed nuclei after time t is N = N0 (1/2)^(t / T_half). After one half-life half remain, after two a quarter, and after three an eighth. Nuclear reactions conserve charge and nucleon number.
Mass and energy are related by E = m c^2. In a nuclear reaction, the mass lost converts to energy. With 1 atomic mass unit equal to 931.5 MeV / c^2, a mass defect of 0.0300 u gives 27.9 MeV. Binding energy per nucleon explains why fission of heavy nuclei and fusion of light nuclei both release energy.
Worked example
A hydrogen atom drops from n = 3 to n = 2. Photon energy in eV?
E3 = -1.51 eV.
E2 = -3.40 eV.
Difference.
1.89 eV.
Practice problem and solution
A sample has a half-life of 8 days. What fraction remains after 24 days, as a decimal?
Three half-lives: (1/2)^3 = 0.125.
Mental model: E_n = -13.6/n^2. N = N0 (1/2)^(t/T). E = mc^2 with 931.5 MeV per u.
Common trap: Subtracting the levels in the wrong order.