AP Physics 2 study package

AP Physics 2 is a one-year, algebra-based introductory college-level physics course. It covers fluid mechanics, thermodynamics, electricity and magnetism, optics, and quantum, atomic, and nuclear physics. The course emphasizes conceptual understanding, mathematical problem-solving, and the application of physics principles to real-world scenarios. Unlike AP Physics C, no calculus is required — all problems use algebra, geometry, and trigonometry.

Exam Format

The AP Physics 2 exam is 3 hours long and consists of two equally weighted sections:

Section 1: Multiple Choice (50% of score)

• 50 questions in 90 minutes
• Includes discrete questions and sets of 2–3 questions sharing a common stimulus (data, diagram, or scenario)
• Covers all seven course units
• No penalty for guessing — answer every question

Section 2: Free Response (50% of score)

• 4 questions in 90 minutes
• Question types:
  • Mathematical routines (1 question): quantitative problem-solving, algebraic manipulation, and numerical calculation
  • Translation between representations (1 question): converting between diagrams, graphs, equations, and verbal descriptions
  • Experimental design and analysis (1 question): designing or analyzing experiments, collecting and interpreting data
  • Qualitative/quantitative translation (1 question): combining conceptual reasoning with mathematical justification

Resources Allowed

• A calculator (scientific or graphing) is permitted on both sections
• An equation sheet is provided (you do NOT need to memorize formulas)

The Seven Course Units and Their Weightings

Unit	Topic	Exam Weighting
1	Fluids	10–12%
2	Thermodynamics	12–18%
3	Electric Force, Field, and Potential	18–22%
4	Electric Circuits	10–14%
5	Magnetism and Electromagnetic Induction	10–14%
6	Geometric and Physical Optics	12–16%
7	Quantum, Atomic, and Nuclear Physics	10–14%

Key observation: Units 2 and 3 carry the heaviest weight. Prioritize these when studying, but do not neglect any unit — every unit appears on the exam.

Science Practices

The exam tests seven science practices that go beyond memorization:

1. Visual Representations: Creating and interpreting diagrams, graphs, field maps, and circuit schematics
2. Question and Method: Formulating scientific questions and selecting appropriate methods to investigate them
3. Representing Data: Using appropriate data representations (tables, graphs, scatter plots)
4. Data Analysis: Analyzing data to identify patterns, determine relationships, and evaluate sources of error
5. Theoretical Relationships: Explaining physical phenomena using theories, models, and mathematical relationships
6. Mathematical Routines: Solving problems using appropriate mathematical techniques
7. Argumentation: Developing and supporting scientific arguments with evidence and reasoning

Unit-by-Unit Roadmap

Unit 1: Fluids (10–12%)

Core ideas: density, pressure, Pascal's principle, Archimedes' principle, buoyancy, fluid dynamics (continuity equation, Bernoulli's equation), viscosity. Fluids connect to real-world applications like hydraulic systems, airplane lift, and blood flow.

Unit 2: Thermodynamics (12–18%)

Core ideas: temperature scales, thermal expansion, ideal gas law, kinetic molecular theory, first and second laws of thermodynamics, PV diagrams, thermodynamic processes (isothermal, isobaric, isochoric, adiabatic), entropy, heat engines, and refrigerators. This unit bridges mechanics and chemistry.

Unit 3: Electric Force, Field, and Potential (18–22%)

Core ideas: electric charge, Coulomb's law, electric fields, field lines, conductors vs. insulators, electric potential energy, electric potential, equipotential lines, and the relationship between field and potential. This is the highest-weighted unit.

Unit 4: Electric Circuits (10–14%)

Core ideas: current, resistance, Ohm's law, power, series and parallel circuits, Kirchhoff's laws, RC circuits (charging and discharging), ammeters, voltmeters, and internal resistance. Build strong circuit analysis skills.

Unit 5: Magnetism and Electromagnetic Induction (10–14%)

Core ideas: magnetic fields and forces, right-hand rules, force on current-carrying wires, torque on loops, magnetic flux, Faraday's law, Lenz's law, transformers, and electromagnetic waves. Magnetism and induction are tightly connected.

Unit 6: Geometric and Physical Optics (12–16%)

Core ideas: reflection, refraction, Snell's law, total internal reflection, mirrors, thin lenses, ray diagrams, interference (double slit, diffraction grating), single-slit diffraction, thin-film interference, and polarization. Optics requires both conceptual and mathematical mastery.

Unit 7: Quantum, Atomic, and Nuclear Physics (10–14%)

Core ideas: photoelectric effect, photon energy, atomic spectra, Bohr model, wave-particle duality, de Broglie wavelength, nuclear physics (isotopes, half-life, alpha/beta/gamma decay), mass-energy equivalence, and nuclear reactions. This unit is conceptually rich and heavily tested through conceptual FRQs.

Suggested 16-Week Study Plan

Weeks 1–2: Unit 1 — Fluids Weeks 3–4: Unit 2 — Thermodynamics Weeks 5–7: Unit 3 — Electric Force, Field, and Potential (extra week due to weight) Weeks 8–9: Unit 4 — Electric Circuits Weeks 10–11: Unit 5 — Magnetism and Electromagnetic Induction Weeks 12–13: Unit 6 — Geometric and Physical Optics Weeks 14–15: Unit 7 — Quantum, Atomic, and Nuclear Physics Week 16: Full practice exam, targeted review, and exam strategy

Key Advice for Success

1. Understand, don't memorize. The equation sheet has every formula. Your job is knowing when and why to use each one.
2. Draw diagrams constantly. Free-body diagrams, field maps, circuit diagrams, ray diagrams, and PV diagrams are essential tools.
3. Practice dimensional analysis. Checking units catches many errors on the MCQ section.
4. Master the FRQ formats. Practice each type — math routines, representation translation, experimental design, and qualitative/quantitative.
5. Show all work on FRQs. Partial credit is generous. Even if the final answer is wrong, correct physics reasoning earns points.
6. Use your calculator wisely. Know how to use scientific notation, store variables, and solve systems of equations.
7. Review consistently. Physics builds on itself. Review earlier units as you progress.

Scoring

AP scores range from 1 to 5. In recent years, approximately 60–65% of students score a 3 or higher. A score of 3 is considered passing, though many colleges require a 4 or 5 for credit. The mean score typically falls around 2.9–3.1.

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This package provides everything you need: detailed unit notes, practice problems, a full-length practice exam, a summary sheet, exam strategies, a presentation outline, and an audio study script. Work through it systematically, and you will be well prepared for exam day.

A fluid is any substance that can flow — both liquids and gases are fluids. Unlike solids, fluids deform continuously under an applied shear stress.

Density (ρ) is the mass per unit volume:

ρ = m / V

• SI unit: kg/m³
• Water: ρ = 1000 kg/m³
• Mercury: ρ = 13,600 kg/m³
• Air (at STP): ρ ≈ 1.29 kg/m³

  Specific gravity is the ratio of a substance's density to the density of water (dimensionless). Mercury's specific gravity is 13.6.

  Pressure (P) is force per unit area:

  P = F / A

• SI unit: pascal (Pa) = N/m²
• 1 atm = 101,325 Pa ≈ 101.3 kPa
• 1 atm = 760 mmHg = 760 torr

  Pressure in a fluid at rest (hydrostatic pressure):

  P = P₀ + ρgh

  Where P₀ is the pressure at the surface, ρ is the fluid density, g is gravitational acceleration (9.8 m/s²), and h is the depth below the surface. Pressure increases linearly with depth and depends only on depth — not on the shape or total volume of the container.

Gauge Pressure vs. Absolute Pressure

• Absolute pressure = gauge pressure + atmospheric pressure
• Gauge pressure = P - P_atm (what a tire gauge reads)

1.2 Pascal's Principle

Pascal's principle states that a change in pressure applied to an enclosed, incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of the container.

Hydraulic press:

F₁ / A₁ = F₂ / A₂

A small force on a small-area piston creates a large force on a large-area piston. This is how hydraulic brakes, lifts, and presses work.

Worked Example 1: A hydraulic lift has a small piston of area 0.01 m² and a large piston of area 0.5 m². If a force of 200 N is applied to the small piston, what force is exerted by the large piston?

Solution: F₁/A₁ = F₂/A₂ 200 N / 0.01 m² = F₂ / 0.5 m² 20,000 Pa = F₂ / 0.5 F₂ = 20,000 × 0.5 = 10,000 N

The mechanical advantage is 50, meaning the output force is 50 times the input force.

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1.3 Buoyancy — Archimedes' Principle

Archimedes' principle: An object partially or fully submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced.

F_buoy = ρ_fluid × V_displaced × g

Sinking, Floating, and Neutral Buoyancy

• Floats (partially submerged): ρ_object < ρ_fluid. The object displaces less fluid than its own volume.
• Neutrally buoyant: ρ_object = ρ_fluid. The object can remain at any depth.
• Sinks: ρ_object > ρ_fluid. The object displaces fluid equal to its own volume but the weight exceeds the buoyant force.

  Fraction submerged for a floating object:

  fraction submerged = ρ_object / ρ_fluid

Apparent Weight

When an object is submerged, a scale reads its apparent weight:

F_apparent = F_gravity - F_buoy = mg - ρ_fluid × V × g

Worked Example 2: A 2.0 kg aluminum block (density 2700 kg/m³) is submerged in water. What is the buoyant force and the apparent weight?

Solution: V_block = m/ρ = 2.0 / 2700 = 7.41 × 10⁻⁴ m³ F_buoy = ρ_water × V × g = 1000 × 7.41 × 10⁻⁴ × 9.8 = 7.26 N F_apparent = mg - F_buoy = 2.0 × 9.8 - 7.26 = 19.6 - 7.26 = 12.34 N

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1.4 Fluid Dynamics

Fluid dynamics studies fluids in motion. We assume ideal fluids: incompressible, nonviscous, with laminar (non-turbulent) flow.

Continuity Equation (Conservation of Mass)

A₁v₁ = A₂v₂

Where A is the cross-sectional area and v is the fluid velocity. This means fluid flows faster through narrower pipes and slower through wider ones.

Bernoulli's Equation (Conservation of Energy)

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

Each term represents a form of energy per unit volume:

• P = pressure energy
• ½ρv² = kinetic energy density
• ρgh = gravitational potential energy density

  Key consequence: Where velocity is higher, pressure is lower. This explains:

• Airplane lift: Air moves faster over the curved top of a wing, creating lower pressure above.
• Venturi effect: Fluid speeds up in a constriction, pressure drops.
• Spray bottles: Fast-moving air over a tube reduces pressure, drawing liquid up.

  Worked Example 3: Water flows through a horizontal pipe that narrows from a diameter of 4.0 cm to 2.0 cm. If the water speed in the wide section is 1.0 m/s, what is the speed in the narrow section? If the pressure in the wide section is 200 kPa, what is the pressure in the narrow section?

  Solution: A₁ = π(0.02)² = 1.257 × 10⁻³ m² A₂ = π(0.01)² = 3.14 × 10⁻⁴ m² A₁v₁ = A₂v₂ → v₂ = A₁v₁/A₂ = (1.257 × 10⁻³)(1.0)/(3.14 × 10⁻⁴) = 4.0 m/s

  Using Bernoulli's (horizontal pipe, h₁ = h₂): P₁ + ½ρv₁² = P₂ + ½ρv₂² 200,000 + ½(1000)(1.0)² = P₂ + ½(1000)(4.0)² 200,000 + 500 = P₂ + 8000 P₂ = 192,500 Pa = 192.5 kPa

1.5 Viscosity (Conceptual)

Viscosity is a fluid's internal resistance to flow. Honey has high viscosity; water has low viscosity. Viscosity causes energy dissipation in real fluids, which is why Bernoulli's equation (which assumes nonviscous flow) is an idealization.

Laminar vs. turbulent flow:

• Laminar flow: Smooth, orderly layers. Occurs at low velocities and in narrow pipes.
• Turbulent flow: Chaotic, irregular. The Reynolds number predicts the transition.

  For AP Physics 2, you should understand viscosity conceptually — it causes drag forces and energy losses — but you do not need to calculate with it quantitatively.

Common Mistakes

1. Using wrong depth in hydrostatic pressure. Pressure depends on depth below the surface, not height above the bottom. Always measure h from the top surface downward.
2. Confusing volume displaced with total volume. When an object floats, it displaces only a fraction of its volume equal to the mass of the object divided by the fluid density. Only fully submerged objects displace their entire volume.
3. Forgetting that buoyant force depends on fluid density, not object density. F_buoy = ρ_fluid × V_displaced × g. A dense object in water experiences a buoyant force based on water's density, not its own.
4. Applying Bernoulli's equation incorrectly to vertical pipes without the ρgh terms. If the pipe is not horizontal, you must include the gravitational potential energy terms.
5. Assuming pressure is a vector. Pressure is a scalar quantity. It acts equally in all directions at a point in a fluid. Force is a vector; pressure is not.
6. Neglecting atmospheric pressure in gauge vs. absolute problems. Read carefully: if a problem gives "gauge pressure," add atmospheric pressure to get absolute pressure before using it in an equation that requires absolute pressure.

Self-Check Questions

1. A cube of wood (density 600 kg/m³) floats in water. What fraction of the cube is submerged?
2. A submarine at a depth of 50 m in seawater (ρ = 1025 kg/m³) has a circular hatch of area 0.8 m². What force must the hatch withstand from the water pressure?
3. Water flows through a pipe at 3.0 m/s. The pipe narrows to half its original radius. What is the new water speed?
4. A hydraulic system has pistons with radii of 2.0 cm and 20 cm. What force on the small piston is needed to lift a 15,000 N car on the large piston?
5. An object has a mass of 0.5 kg and displaces 3.0 × 10⁻⁴ m³ of water when fully submerged. Does it sink or float? What is the apparent weight?
6. Explain in terms of Bernoulli's principle why a spinning soccer ball curves in the air.

   Answers to self-check questions are embedded in the practice file for this unit.

Temperature measures the average kinetic energy of particles in a substance. Three common scales:

Scale	Freezing of Water	Boiling of Water
Celsius (°C)	0°	100°
Fahrenheit (°F)	32°	212°
Kelvin (K)	273.15	373.15

Conversion formulas:

T_K = T_C + 273.15
T_F = (9/5)T_C + 32

Kelvin is the SI temperature scale. It is an absolute scale — 0 K is absolute zero, the theoretical minimum temperature where all molecular motion ceases.

Thermal expansion occurs because particles vibrate more vigorously at higher temperatures:

• Linear expansion: ΔL = αL₀ΔT (α = coefficient of linear expansion)
• Volume expansion: ΔV = βV₀ΔT (β ≈ 3α for solids)

  These are most relevant for solids (bimetallic strips, expansion joints in bridges) and liquids in containers.

2.2 Ideal Gas Law

The ideal gas law combines Boyle's, Charles's, and Gay-Lussac's laws:

PV = nRT

Where P = pressure (Pa), V = volume (m³), n = number of moles, R = 8.314 J/(mol·K), T = temperature (K, always in Kelvin).

Alternative form (using Boltzmann's constant k_B = 1.38 × 10⁻²³ J/K):

PV = Nk_BT

Where N = total number of molecules.

Gas Laws Summary

• Boyle's Law (constant T): P₁V₁ = P₂V₂ — pressure and volume are inversely proportional
• Charles's Law (constant P): V₁/T₁ = V₂/T₂ — volume and temperature are directly proportional
• Gay-Lussac's Law (constant V): P₁/T₁ = P₂/T₂ — pressure and temperature are directly proportional
• Avogadro's Law (constant P, T): V/n = constant — volume is proportional to moles

2.3 Kinetic Molecular Theory

The kinetic theory of gases connects the macroscopic properties of gases to the microscopic behavior of molecules:

1. Gases consist of many small particles in constant, random motion
2. Collisions between particles and with walls are perfectly elastic
3. The volume of particles is negligible compared to container volume
4. No intermolecular forces act between particles (except during collisions)
5. The average kinetic energy is proportional to temperature:

   KE_avg = (3/2)k_BT = (3/2)(R/N_A)T

   For n moles: KE_total = (3/2)nRT

   The root-mean-square speed of gas molecules:

   v_rms = √(3k_BT/m) = √(3RT/M)

   Where M is the molar mass in kg/mol. Lighter molecules move faster at the same temperature.

2.4 First Law of Thermodynamics

The first law is a statement of conservation of energy for thermodynamic systems:

ΔU = Q - W

Where:

• ΔU = change in internal energy of the system
• Q = heat added to the system (positive if heat enters)
• W = work done by the system (positive if system expands)

  Sign convention (AP Physics 2): W is positive when the gas expands (does work on surroundings). If the gas is compressed, W is negative (or equivalently, work is done on the gas).

  Work done by an ideal gas:

  W = PΔV (at constant pressure) W = area under the curve on a PV diagram

2.5 PV Diagrams and Thermodynamic Processes

A PV diagram graphs pressure (y-axis) vs. volume (x-axis). The area under the curve equals the work done by the gas.

The Four Basic Processes

1. Isobaric (constant pressure):

• Horizontal line on PV diagram
• W = PΔV
• Q = nC_pΔT (where C_p is molar heat capacity at constant pressure)

  2. Isothermal (constant temperature):

• Follows a hyperbola (PV = constant)
• ΔU = 0 (since T doesn't change for an ideal gas)
• Therefore Q = W

  3. Isochoric/Isovolumetric (constant volume):

• Vertical line on PV diagram
• W = 0 (no volume change)
• Therefore ΔU = Q (all heat changes internal energy)
• Q = nC_vΔT

  4. Adiabatic (no heat exchange):

• Q = 0
• ΔU = -W
• The gas cools as it expands (does work) and heats up as it's compressed
• Steeper curve than isothermal on PV diagram
• Follows PV^γ = constant (γ = C_p/C_v)

Cyclic Processes

In a cyclic process, the system returns to its initial state, so ΔU = 0 and Q_net = W_net. The net work equals the area enclosed by the cycle on the PV diagram. Clockwise cycles do positive net work (heat engines). Counterclockwise cycles do negative net work (refrigerators).

Worked Example 1: A gas expands isobarically from 2.0 L to 5.0 L at a pressure of 300 kPa, then cools isochorically until the pressure drops to 150 kPa. Find the total work done and the change in internal energy if 600 J of heat is removed during the cooling step.

Solution: Step 1 (isobaric expansion): W₁ = PΔV = 300,000 × (5.0 - 2.0) × 10⁻³ = 300,000 × 0.003 = 900 J

Step 2 (isochoric cooling): W₂ = 0 (no volume change)

Total work: W_total = 900 J

For the cooling step: Q₂ = -600 J (heat removed) ΔU₂ = Q₂ - W₂ = -600 - 0 = -600 J

For the expansion step, we need temperature information to find ΔU₁. Using PV = nRT: T₁ = P₁V₁/(nR), T₂ = P₁V₂/(nR) → T₂/T₁ = V₂/V₁ = 5/2 ΔU₁ = nC_v(T₂ - T₁). For a monatomic ideal gas, C_v = (3/2)R. ΔU₁ = (3/2)nR(T₂ - T₁) = (3/2)(P₁V₂ - P₁V₁) = (3/2)(300,000 × 0.005 - 300,000 × 0.002) = (3/2)(1500 - 600) = (3/2)(900) = 1350 J

ΔU_total = 1350 + (-600) = 750 J

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2.6 Second Law of Thermodynamics and Entropy

The second law states that the total entropy of an isolated system never decreases. Natural processes are irreversible and proceed in the direction of increasing entropy.

Entropy (S) is a measure of disorder or the number of available microstates:

ΔS = Q / T   (for a reversible process at constant temperature)

Entropy increases when:

• Heat flows from hot to cold
• A gas expands into a vacuum
• Ice melts
• A building collapses

  The second law explains why:

• Heat flows spontaneously from hot to cold, never the reverse
• No heat engine can be 100% efficient
• You cannot convert heat entirely into work in a cyclic process

2.7 Heat Engines and Refrigerators

Heat Engines

A heat engine takes in heat Q_H from a hot reservoir, does work W, and exhausts heat Q_C to a cold reservoir:

W = Q_H - Q_C
efficiency = W / Q_H = 1 - Q_C / Q_H

Carnot Efficiency

The Carnot cycle is the most efficient possible heat engine operating between two temperatures:

efficiency_Carnot = 1 - T_C / T_H   (temperatures in Kelvin)

No real engine can exceed Carnot efficiency.

Worked Example 2: A heat engine operates between a hot reservoir at 500 K and a cold reservoir at 300 K. If it absorbs 2000 J of heat per cycle, what is the maximum possible work output?

Solution: efficiency_max = 1 - 300/500 = 1 - 0.6 = 0.40 = 40% W_max = 0.40 × 2000 = 800 J

Refrigerators and Heat Pumps

A refrigerator (or heat pump) uses work W to move heat Q_C from a cold reservoir to a hot reservoir:

Q_H = Q_C + W
Coefficient of Performance (COP) = Q_C / W   (for refrigerator)
COP_heat_pump = Q_H / W

Maximum COP (Carnot): COP_max = T_C / (T_H - T_C)

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Common Mistakes

1. Forgetting to convert to Kelvin. All gas law calculations require temperature in Kelvin. A common error is using Celsius in PV = nRT or Carnot efficiency.
2. Confusing sign conventions for work and heat. In AP Physics 2, W is work done by the gas (positive when expanding) and Q is heat added to the gas (positive when entering). Getting signs wrong flips your entire first-law calculation.
3. Assuming ΔU = 0 for all processes. ΔU = 0 only for isothermal processes (ideal gas) or complete cycles. In isobaric and isochoric processes, ΔU is nonzero.
4. Confusing thermal conductivity with specific heat. Thermal conductivity (k) describes how quickly heat transfers through a material. Specific heat (c) describes how much heat is needed to change temperature. They are different properties.
5. Thinking heat engines can exceed Carnot efficiency. The Carnot efficiency is a theoretical maximum. Real engines always have lower efficiency due to irreversibilities. If you calculate an efficiency above Carnot, you made an error.
6. Misidentifying the direction of entropy change. The entropy of the universe always increases, but the entropy of a system can decrease if enough heat is expelled to the surroundings. Consider the whole system when applying the second law.

Self-Check Questions

1. A gas at 300 K is compressed to half its original volume at constant pressure. What is the new temperature?
2. An ideal gas expands isothermally, doing 400 J of work. How much heat was added to the gas, and what is the change in internal energy?
3. A Carnot engine operates between 600 K and 400 K. It absorbs 1500 J from the hot reservoir per cycle. How much work does it produce per cycle, and how much heat is expelled?
4. On a PV diagram, a gas follows a clockwise rectangular cycle. The work done by the gas during the expansion step is 500 J, and the work done on the gas during compression is 300 J. What is the net work output and the efficiency if Q_H = 800 J?
5. Explain why the second law of thermodynamics prevents a refrigerator from cooling a room by leaving its door open.
6. A monatomic ideal gas undergoes an adiabatic expansion. Does its temperature increase, decrease, or stay the same? Explain using the first law.

   Answers to self-check questions are embedded in the practice file for this unit.

Electric charge is a fundamental property of matter. There are two types: positive (protons) and negative (electrons).

• Elementary charge: e = 1.602 × 10⁻¹⁹ C
• Charge is quantized: Q = ne, where n is an integer
• Charge is conserved: the total charge in an isolated system never changes
• Conduction: charging by direct contact — charges transfer between objects
• Induction: charging without contact — a charged object polarizes a nearby conductor, then grounding allows charge redistribution

Conductors vs. Insulators

• Conductors (metals): Free electrons move easily throughout the material. In electrostatic equilibrium:
  • Electric field inside a conductor is zero
  • Any excess charge resides on the surface
  • The electric field at the surface is perpendicular to the surface
  • The surface is an equipotential
• Insulators (rubber, glass): Charges are bound and cannot move freely. Charge redistributes only locally.

3.2 Coulomb's Law

The electric force between two point charges:

F = k|q₁q₂| / r²

Where k = 8.99 × 10⁹ N·m²/C² (Coulomb's constant). The force is attractive for unlike charges and repulsive for like charges. It obeys Newton's third law.

Superposition principle: The net force on a charge is the vector sum of all individual Coulomb forces from every other charge.

Worked Example 1: Two charges, q₁ = +3.0 μC and q₂ = -5.0 μC, are separated by 0.20 m. Find the magnitude and direction of the force on each charge.

Solution: F = k|q₁q₂|/r² = (8.99 × 10⁹)(3.0 × 10⁻⁶)(5.0 × 10⁻⁶)/(0.20)² F = (8.99 × 10⁹)(15 × 10⁻¹²)/0.04 = (134.85 × 10⁻³)/0.04 = 3.37 N

The force is attractive: q₁ is pulled toward q₂, and q₂ is pulled toward q₁. Both forces are 3.37 N in magnitude (Newton's third law).

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3.3 Electric Field

The electric field (E) is the force per unit charge that a test charge would experience:

E = F/q₀ = kQ/r²

• Units: N/C or V/m
• Electric field points away from positive charges and toward negative charges
• Electric field is a vector — add fields vectorially for multiple charges

Electric Field Lines

• Field lines point in the direction of the electric field (from + to -)
• Line density indicates field strength (closer lines = stronger field)
• Lines never cross
• Lines start on positive charges and end on negative charges
• Near a conducting surface, field lines are perpendicular to the surface

Uniform Electric Field

Between two large, parallel, oppositely charged plates, the field is approximately uniform:

E = σ/ε₀ = V/d

Where σ is the surface charge density, ε₀ = 8.85 × 10⁻¹² C²/(N·m²), V is the potential difference, and d is the plate separation.

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3.4 Electric Potential Energy

The electric potential energy (U) of a system of charges is the work required to assemble the charges from infinite separation:

U = kq₁q₂ / r   (for two point charges)

• Like charges (positive U): energy must be added to bring them together
• Unlike charges (negative U): energy is released as they come together

  For a charge q in a uniform field E over distance d:

  U = qEd (if moving against the field) W = -ΔU = qEΔx (work done by the field)

3.5 Electric Potential

Electric potential (V) is the electric potential energy per unit charge:

V = U/q = kQ/r

• Units: volts (V) = J/C
• Potential is a scalar (not a vector) — add potentials algebraically
• Potential due to a positive charge is positive; due to a negative charge is negative

Potential Difference (Voltage)

ΔV = V_B - V_A = -W_by_field / q = -∫E·dl

For a uniform field: ΔV = -Ed (the negative sign means potential decreases in the direction of the field).

Relationship Between E and V

E = -ΔV/Δx   (in one dimension)

The electric field points in the direction of decreasing potential. This is a critically important relationship tested frequently on the AP exam.

Equipotential Lines

• Lines (or surfaces) of constant electric potential
• Electric field lines are always perpendicular to equipotential lines
• No work is required to move a charge along an equipotential
• For a point charge, equipotentials are concentric circles
• For a uniform field, equipotentials are parallel planes (or lines in 2D)

  Worked Example 2: A proton (q = +1.6 × 10⁻¹⁹ C) moves from a point where V = 100 V to a point where V = 50 V. What is the change in the proton's electric potential energy?

  Solution: ΔU = qΔV = (1.6 × 10⁻¹⁹ C)(50 - 100) = (1.6 × 10⁻¹⁹)(-50) = -8.0 × 10⁻¹⁸ J

  The potential energy decreases by 8.0 × 10⁻¹⁸ J. By conservation of energy, the kinetic energy increases by this amount (assuming no other forces).

3.6 Key Connections

1. Force and field: F = qE. The field tells you the force that would act on any charge placed there.
2. Field and potential: E = -ΔV/Δx. The field is the negative spatial rate of change of potential.
3. Potential and potential energy: U = qV. The potential energy depends on both the potential at a point and the charge placed there.
4. Work and potential energy: W = -ΔU. The work done by the electric field equals the negative change in potential energy.

   These four relationships form the backbone of Unit 3. Master them, and most problems become applications of one or more of these connections.

Common Mistakes

1. Confusing electric field and electric force. The electric field exists at a point regardless of whether a test charge is there. The force depends on both the field and the charge: F = qE. A common error is using E = kQ/r² to find force without multiplying by the test charge.
2. Adding electric potentials as vectors. Electric potential is a scalar quantity. You add potentials algebraically, not vectorially. Electric fields are vectors and require vector addition.
3. Forgetting the negative sign in E = -ΔV/Δx. The electric field points from high potential to low potential. Forgetting the negative sign gives the wrong direction.
4. Confusing potential energy and potential. U = qV. A point at high potential can still have low potential energy if the charge is negative. Always distinguish between V (volts) and U (joules).
5. Incorrectly applying Coulomb's law to non-point charges. Coulomb's law applies to point charges or spherically symmetric charge distributions (where r is measured from the center). For extended objects, you need integration (not required on AP Physics 2, but the concept is tested).
6. Drawing field lines from negative to positive. Electric field lines point from positive to negative charges (they show the direction a positive test charge would move). Students sometimes reverse this.

Self-Check Questions

1. Two identical positive charges are placed 0.10 m apart. At the midpoint between them, what is the electric field and what is the electric potential?
2. An electron moves from a point at 20 V to a point at 80 V. Does its kinetic energy increase or decrease? By how much?
3. A -2.0 μC charge is placed in a uniform electric field of 500 N/C pointing to the right. What is the force on the charge, and in which direction does it accelerate?
4. Draw the electric field lines and equipotential lines for a positive point charge.
5. A charge q₁ = +4.0 μC is at the origin, and q₂ = -3.0 μC is at x = 0.30 m. At what point on the x-axis is the electric field zero?
6. Explain why the electric field inside a conductor in electrostatic equilibrium is zero, and why all excess charge resides on the surface.

   Answers to self-check questions are embedded in the practice file for this unit.

Current (I) is the rate of flow of electric charge:

I = ΔQ/Δt

• SI unit: ampere (A) = C/s
• Conventional current flows from positive to negative terminal (direction positive charges would move)
• Electron flow is opposite to conventional current
• Current is the same everywhere in a series circuit
• Current splits at junctions in a parallel circuit

  For current to flow, you need a potential difference (voltage source) and a closed path (complete circuit).

4.2 Resistance and Ohm's Law

Resistance (R) opposes the flow of current:

V = IR   (Ohm's law)

• SI unit: ohm (Ω) = V/A
• Ohm's law applies to ohmic materials (linear V-I relationship)
• Non-ohmic materials (diodes, LEDs) do not follow V = IR

  Factors affecting resistance:

  R = ρL/A

  Where ρ = resistivity (material property), L = length, A = cross-sectional area. Longer wires have more resistance; thicker wires have less resistance.

Resistivity and Temperature

For most conductors, resistance increases with temperature:

ρ = ρ₀[1 + α(T - T₀)]

Where α is the temperature coefficient of resistivity. Semiconductors (like thermistors) behave oppositely — their resistance decreases with temperature.

4.3 Power and Energy

Electrical power:

P = IV = I²R = V²/R

• SI unit: watt (W) = J/s
• Energy: E = Pt = IVt (measured in joules or kilowatt-hours)
• 1 kWh = 3.6 × 10⁶ J

  Worked Example 1: A 60 W light bulb is connected to a 120 V source. What is the current through the bulb, and what is its resistance?

  Solution: P = IV → I = P/V = 60/120 = 0.50 A V = IR → R = V/I = 120/0.50 = 240 Ω

  (Or directly: P = V²/R → R = V²/P = 14400/60 = 240 Ω)

4.4 Series Circuits

In a series circuit, all components are connected in a single path:

• Current is the same through all components: I_total = I₁ = I₂ = I₃ = ...
• Voltage splits across components: V_total = V₁ + V₂ + V₃ + ...
• Equivalent resistance: R_eq = R₁ + R₂ + R₃ + ...

  Series resistors always produce a larger equivalent resistance than any individual resistor.

  Voltage divider: In a series circuit, the largest resistor gets the largest voltage drop (since V = IR and I is the same).

4.5 Parallel Circuits

In a parallel circuit, components are connected across the same two points:

• Voltage is the same across all branches: V_total = V₁ = V₂ = V₃ = ...
• Current splits at junctions: I_total = I₁ + I₂ + I₃ + ...
• Equivalent resistance: 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ + ...

  Parallel resistors always produce a smaller equivalent resistance than any individual resistor.

  Current divider: In a parallel circuit, the branch with the smallest resistance carries the largest current.

4.6 Kirchhoff's Laws

Kirchhoff's Junction Rule (KCL): The sum of currents entering a junction equals the sum of currents leaving it. This is conservation of charge.

ΣI_in = ΣI_out

Kirchhoff's Loop Rule (KVL): The sum of all voltage changes around any closed loop is zero. This is conservation of energy.

ΣΔV = 0 around any closed loop

Procedure for solving complex circuits:

1. Assign current directions to each branch (if you guess wrong, the current will come out negative)
2. Apply the junction rule at each junction
3. Apply the loop rule around each independent loop
4. Solve the system of equations

   Worked Example 2: A 12 V battery is connected to R₁ = 4 Ω and R₂ = 6 Ω in series. Find the current, voltage across each resistor, and power dissipated by each.

   Solution: R_eq = 4 + 6 = 10 Ω I = V/R_eq = 12/10 = 1.2 A V₁ = IR₁ = 1.2 × 4 = 4.8 V V₂ = IR₂ = 1.2 × 6 = 7.2 V (Check: 4.8 + 7.2 = 12.0 V ✓) P₁ = I²R₁ = (1.2)²(4) = 5.76 W P₂ = I²R₂ = (1.2)²(6) = 8.64 W P_total = 5.76 + 8.64 = 14.4 W (Check: P = VI = 12 × 1.2 = 14.4 W ✓)

4.7 RC Circuits

An RC circuit contains a resistor and capacitor in series with a voltage source.

Capacitor Charging

When a switch closes, charge builds up on the capacitor:

Q(t) = Cε(1 - e^(-t/RC))
I(t) = (ε/R)e^(-t/RC)

Capacitor Discharging

When the voltage source is removed, the capacitor discharges through the resistor:

Q(t) = Q₀e^(-t/RC)
I(t) = (Q₀/RC)e^(-t/RC)

Time Constant

τ = RC

• τ is the time constant — the time for the charge to reach ~63.2% of its final value (charging) or decay to ~36.8% of its initial value (discharging)
• After 5τ, a capacitor is considered fully charged (~99.3%) or fully discharged (~0.7%)
• At t = τ: Q = 0.632Q_max (charging) or Q = 0.368Q₀ (discharging)

4.8 Meters and Internal Resistance

Ammeters

• Measure current
• Connected in series with the component
• Have very low (ideally zero) internal resistance so they don't affect the circuit

Voltmeters

• Measure voltage (potential difference)
• Connected in parallel across the component
• Have very high (ideally infinite) internal resistance so minimal current flows through them

Internal Resistance of Batteries

Real batteries have internal resistance (r). The terminal voltage is:

V_terminal = ε - Ir

Where ε is the emf (ideal voltage), I is the current, and r is the internal resistance. As the current increases, the terminal voltage decreases.

---

Common Mistakes

1. Confusing series and parallel rules. In series, current is the same and voltage splits. In parallel, voltage is the same and current splits. Students frequently mix these up.
2. Calculating equivalent resistance incorrectly for parallel circuits. The formula is 1/R_eq = 1/R₁ + 1/R₂ + ..., not R_eq = 1/R₁ + 1/R₂. Always remember to take the reciprocal at the end.
3. Forgetting that current flows out of the positive terminal of a battery. Conventional current flows from the positive terminal, through the external circuit, and back into the negative terminal.
4. Misapplying Ohm's law to the entire circuit. Ohm's law (V = IR) applies to individual resistors and the equivalent resistance of a circuit. It does NOT apply directly to capacitors.
5. Confusing the time constant with the total time. τ = RC is NOT the time to fully charge or discharge. It's the time for ~63% charging or ~37% remaining. Full charge/discharge takes about 5τ.
6. Incorrectly placing ammeters and voltmeters. Ammeters in series, voltmeters in parallel. Putting an ammeter in parallel creates a short circuit; putting a voltmeter in series means it reads zero.

Self-Check Questions

1. Three 6 Ω resistors are connected: two in parallel, and the combination in series with the third. What is the equivalent resistance?
2. A 9 V battery with internal resistance 0.5 Ω is connected to a 4 Ω external resistor. What is the terminal voltage of the battery?
3. An RC circuit has R = 100 kΩ and C = 10 μF. What is the time constant? How long until the capacitor is approximately fully charged?
4. A 100 W light bulb and a 60 W light bulb are connected in series to a 120 V source. Which bulb is brighter? Explain.
5. In a parallel circuit with a 10 Ω and a 20 Ω resistor connected to a 30 V battery, what is the total current and the current through each resistor?
6. Explain why an ideal ammeter has zero resistance and an ideal voltmeter has infinite resistance.

   Answers to self-check questions are embedded in the practice file for this unit.

Magnetic fields (B) are created by moving charges (currents) and permanent magnets. The SI unit is the tesla (T). A smaller unit is the gauss (1 T = 10⁴ G). Earth's magnetic field is approximately 0.5 G = 5 × 10⁻⁵ T.

• Magnetic field lines point from north to south outside a magnet
• Lines form closed loops (they always return to the south pole)
• Line density indicates field strength
• Lines never cross

5.2 Force on a Moving Charge

A charge moving through a magnetic field experiences a force:

F = qvB sin θ

Where q is the charge, v is the velocity, B is the magnetic field, and θ is the angle between v and B.

Key properties:

• The force is maximum when v is perpendicular to B (θ = 90°)
• The force is zero when v is parallel to B (θ = 0° or 180°)
• The force is always perpendicular to both v and B
• The magnetic force does no work (since F ⊥ v, W = F·d·cos 90° = 0)

  Right-hand rule for positive charges: Point fingers in the direction of v, curl them toward B; your thumb points in the direction of F. For negative charges (electrons), the force is in the opposite direction (use the left hand, or reverse the right-hand result).

Circular Motion in a Magnetic Field

Since the magnetic force is always perpendicular to velocity, a charged particle in a uniform magnetic field moves in a circle (or helix if there is a velocity component along B).

Setting magnetic force equal to centripetal force:

qvB = mv²/r
r = mv/(qB)

The radius is proportional to momentum and inversely proportional to the field strength and charge. Faster, heavier particles have larger radii.

Worked Example 1: A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) moves at 3.0 × 10⁶ m/s perpendicular to a 0.50 T magnetic field. Find the radius of its circular path and its period.

Solution: r = mv/(qB) = (1.67 × 10⁻²⁷)(3.0 × 10⁶)/((1.6 × 10⁻¹⁹)(0.50)) r = (5.01 × 10⁻²¹)/(8.0 × 10⁻²⁰) = 0.0626 m = 6.26 cm

T = 2πr/v = 2π(0.0626)/(3.0 × 10⁶) = 1.31 × 10⁻⁷ s

Note that the period depends only on m, q, and B (not on v or r): T = 2πm/(qB) = 2π(1.67 × 10⁻²⁷)/((1.6 × 10⁻¹⁹)(0.50)) = 1.31 × 10⁻⁷ s ✓

5.3 Force on a Current-Carrying Wire

A wire carrying current in a magnetic field experiences a force:

F = BIL sin θ

Where B is the field, I is the current, L is the length of wire in the field, and θ is the angle between the current direction and B.

Right-hand rule for wires: Point fingers in the direction of current (I), curl toward B; thumb gives force direction.

5.4 Torque on a Current Loop

A rectangular current loop in a uniform magnetic field experiences a torque:

τ = NIAB sin θ

Where N is the number of turns, I is the current, A is the area of the loop, and θ is the angle between the normal to the loop and the magnetic field.

• Maximum torque when the plane of the loop is parallel to B (θ = 90°)
• Zero torque when the plane of the loop is perpendicular to B (θ = 0°) — this is the equilibrium position

  This principle is the basis for electric motors.

5.5 Magnetic Flux

Magnetic flux measures the amount of magnetic field passing through an area:

Φ = BA cos θ

Where θ is the angle between B and the normal to the area.

• SI unit: weber (Wb) = T·m²
• Maximum flux when B is perpendicular to the surface (θ = 0°)
• Zero flux when B is parallel to the surface (θ = 90°)

5.6 Faraday's Law

Faraday's law of electromagnetic induction: A changing magnetic flux induces an emf:

ε = -NΔΦ/Δt

Where N is the number of turns in the coil. The negative sign is Lenz's law (see below).

An emf can be induced by:

• Changing the magnetic field strength (B)
• Changing the area (A)
• Changing the angle between B and the area (θ)
• Moving a conductor through a field (motional emf: ε = BLv)

5.7 Lenz's Law

Lenz's law: The induced current flows in a direction that opposes the change in magnetic flux that produced it.

• If flux is increasing, the induced current creates a field that opposes the increase
• If flux is decreasing, the induced current creates a field that supports the remaining flux
• Lenz's law is a consequence of conservation of energy

  Worked Example 2: A coil of 200 turns with area 0.05 m² is in a 0.4 T magnetic field. The field is reduced to zero in 0.02 s. Find the induced emf.

  Solution: ΔΦ = Φ_final - Φ_initial = 0 - (0.4)(0.05)cos 0° = -0.020 Wb ΔΦ/Δt = -0.020/0.02 = -1.0 Wb/s ε = -NΔΦ/Δt = -(200)(-1.0) = 200 V

  The positive emf means the induced current opposes the decrease in flux (Lenz's law).

5.8 Transformers

A transformer changes AC voltage levels using two coils sharing a magnetic core:

V_s/V_p = N_s/N_p

Where V and N are the voltage and number of turns in the secondary (s) and primary (p) coils.

• Step-up transformer: N_s > N_p → V_s > V_p
• Step-down transformer: N_s < N_p → V_s < V_p
• Power conservation (ideal): V_pI_p = V_sI_s
• Transformers only work with AC (changing flux); they do not work with DC

5.9 Electromagnetic Waves (Maxwell's Equations)

Maxwell's equations predict that accelerating charges produce electromagnetic waves — oscillating electric and magnetic fields that propagate through space at:

c = 1/√(μ₀ε₀) = 3.0 × 10⁸ m/s

Properties of EM waves:

• E and B fields are perpendicular to each other and to the direction of propagation (transverse wave)
• All EM waves travel at c in a vacuum
• They carry energy and momentum
• The EM spectrum includes (in order of increasing frequency): radio, microwave, infrared, visible, ultraviolet, X-ray, gamma

Common Mistakes

1. Using the wrong hand for negative charges. The right-hand rule gives the force direction for positive charges. For electrons, the force is opposite — either use your left hand or reverse the right-hand result. This is one of the most common mistakes on the exam.
2. Confusing the angle in F = qvB sin θ. The angle θ is between the velocity vector and the magnetic field vector, NOT between the force and the field. If v and B are perpendicular, θ = 90° and sin θ = 1 (maximum force).
3. Forgetting that magnetic force does no work. Since F is perpendicular to v, W = Fd cos 90° = 0. A magnetic field can change a particle's direction but never its speed or kinetic energy.
4. Confusing flux and field. Magnetic flux (Φ = BA cos θ) depends on the angle. A field can be present (B ≠ 0) but the flux through an area can be zero if the field is parallel to the surface.
5. Misapplying Lenz's law. Lenz's law opposes the change in flux, not the flux itself. If flux is increasing, oppose the increase. If flux is decreasing, oppose the decrease. Students sometimes try to oppose the flux itself.
6. Assuming transformers work with DC. Transformers require changing magnetic flux, which means they only work with AC. A DC voltage applied to a transformer produces zero output voltage after the initial transient.

Self-Check Questions

1. An electron moves at 5.0 × 10⁶ m/s to the right through a 0.3 T magnetic field pointing into the page. What is the magnitude and direction of the magnetic force on the electron?
2. A wire carrying 8.0 A of current is placed at a 30° angle to a 0.6 T magnetic field. If the wire is 0.5 m long, what force does it experience?
3. A circular loop of radius 0.10 m with 50 turns is in a 0.2 T magnetic field. The field increases to 0.8 T in 0.05 s. What is the induced emf?
4. Explain using Lenz's law the direction of induced current when a bar magnet's north pole is pushed into a coil of wire.
5. A step-up transformer has 100 turns on the primary and 500 turns on the secondary. If the primary voltage is 120 V and the primary current is 10 A, what are the secondary voltage and current (ideal case)?
6. Why can't a magnetic field change the kinetic energy of a charged particle? Use the definition of work in your explanation.

   Answers to self-check questions are embedded in the practice file for this unit.

Light exhibits wave-particle duality. In this unit, we treat light primarily as a wave to explain reflection, refraction, interference, and diffraction.

• Speed of light in vacuum: c = 3.0 × 10⁸ m/s
• Speed in a medium: v = c/n, where n is the index of refraction
• Wavelength in a medium: λ = λ₀/n (wavelength decreases in a medium; frequency stays the same)

6.2 Reflection

Law of reflection: The angle of incidence equals the angle of reflection, both measured from the normal:

θ_i = θ_r

• Specular reflection: smooth surface, clear image (mirror)
• Diffuse reflection: rough surface, scattered light (paper, wall)

6.3 Refraction and Snell's Law

When light passes from one medium to another, it changes speed and direction (bends). Snell's law governs this bending:

n₁ sin θ₁ = n₂ sin θ₂

Where n is the index of refraction and θ is measured from the normal.

• Light bends toward the normal when entering a denser medium (n₂ > n₁, so θ₂ < θ₁)
• Light bends away from the normal when entering a less dense medium (n₂ < n₁, so θ₂ > θ₁)

Total Internal Reflection (TIR)

When light travels from a denser to a less dense medium (n₁ > n₂), there exists a critical angle θ_c above which all light is reflected:

θ_c = sin⁻¹(n₂/n₁)

Conditions for TIR:

1. Light must travel from a medium of higher n to a medium of lower n
2. The angle of incidence must exceed the critical angle

   Applications: fiber optics, prisms, diamond brilliance (n_diamond ≈ 2.42, θ_c ≈ 24.4°)

   Worked Example 1: Light travels from water (n = 1.33) to air (n = 1.00). What is the critical angle for total internal reflection?

   Solution: θ_c = sin⁻¹(n₂/n₁) = sin⁻¹(1.00/1.33) = sin⁻¹(0.752) = 48.8°

   Any light hitting the water-air boundary at an angle greater than 48.8° (measured from the normal) will be totally internally reflected.

6.4 Mirrors

Plane Mirrors

• Image is virtual, upright, same size, and located behind the mirror at the same distance as the object is in front
• Left-right reversal (parity)

Spherical Mirrors

Concave mirrors (converging):

• Focal length f = R/2 (positive)
• Can form real or virtual images depending on object position

  Convex mirrors (diverging):

• Focal length f = R/2 (negative for convex)
• Always form virtual, upright, reduced images

Mirror Equation

1/f = 1/d_o + 1/d_i
magnification: m = -d_i/d_o = h_i/h_o

Sign conventions (AP convention):

• f is positive for concave, negative for convex
• d_o is always positive (real object)
• d_i is positive for real images (in front of mirror), negative for virtual (behind)
• m is positive for upright, negative for inverted

Ray Diagrams for Concave Mirrors

Three principal rays:

1. Parallel ray → reflects through focal point
2. Ray through focal point → reflects parallel to axis
3. Ray through center of curvature → reflects back on itself

6.5 Thin Lenses

Converging (convex) lenses: thicker in the middle, positive focal length, can form real or virtual images

Diverging (concave) lenses: thinner in the middle, negative focal length, always form virtual, upright, reduced images

Thin Lens Equation (same as mirror equation)

1/f = 1/d_o + 1/d_i
m = -d_i/d_o

Ray Diagrams for Converging Lenses

1. Parallel ray → refracts through focal point on the far side
2. Ray through the center → passes straight through
3. Ray through focal point on the near side → refracts parallel to the axis

6.6 Interference

Double-Slit Interference (Young's Experiment)

When light passes through two narrow slits, it produces an interference pattern of bright and dark fringes:

Bright fringes (constructive): d sin θ = mλ    (m = 0, ±1, ±2, ...)
Dark fringes (destructive): d sin θ = (m + ½)λ  (m = 0, ±1, ±2, ...)

Where d is the slit separation and λ is the wavelength. For small angles (sin θ ≈ tan θ ≈ y/L):

y_bright = mλL/d
y_dark = (m + ½)λL/d

Diffraction Gratings

A diffraction grating has many slits (thousands per cm), producing sharper, more widely separated maxima:

d sin θ = mλ   (same equation as double slit)

6.7 Single-Slit Diffraction

Even a single slit produces a diffraction pattern. The central maximum is the brightest and widest. Minima occur at:

a sin θ = mλ    (m = ±1, ±2, ±3, ...)

Where a is the slit width. Note: m = 0 is NOT a minimum for single-slit diffraction — the central maximum spans from m = -1 to m = +1.

6.8 Thin-Film Interference

Light reflecting off the top and bottom surfaces of a thin film can interfere. The key question is whether there is a phase shift upon reflection:

• Reflection off a medium of higher n → phase shift of ½λ (equivalent to π radians)
• Reflection off a medium of lower n → no phase shift

  Worked Example 2: A soap bubble (n = 1.33) in air has a thin film of thickness 200 nm. What wavelength of light is most strongly reflected (constructive interference)?

  Solution: Light reflects off the top surface (air→soap, n increases → ½λ phase shift) and bottom surface (soap→air, n decreases → no phase shift). Total phase shift from reflections = ½λ. For constructive interference, the path difference (2nt) must provide another ½λ to make a total of λ (full wavelength):

  2nt = (m + ½)λ for m = 0, 1, 2, ... For m = 0: λ = 2(1.33)(200 × 10⁻⁹)/(0.5) = 1064 nm (infrared, not visible) For m = 1: λ = 2(1.33)(200 × 10⁻⁹)/(1.5) = 355 nm (ultraviolet)

  So this thickness does not strongly reflect visible light. A thinner film would be needed.

6.9 Polarization

Polarization is the restriction of light wave oscillations to a single plane.

• Unpolarized light oscillates in all directions perpendicular to propagation
• Polarizing filters (like Polaroid) transmit only one polarization direction
• Malus's law: I = I₀ cos²θ (where θ is the angle between the polarizer axis and the light's polarization direction)
• Two perpendicular polarizers block all light
• Light reflected at Brewster's angle is partially polarized

Common Mistakes

1. Measuring angles from the surface instead of the normal. All angles in reflection and refraction (θ_i, θ_r, θ_c) are measured from the normal (perpendicular) to the surface, not from the surface itself. This is the single most common mistake in optics.
2. Forgetting that wavelength changes in a medium. When light enters a medium with index n, its wavelength decreases (λ = λ₀/n) but its frequency stays the same. Using the vacuum wavelength inside a medium gives wrong answers for interference calculations.
3. Confusing the conditions for constructive and destructive interference in thin films. The rules depend on the number of phase shifts (0, 1, or 2). You must analyze each surface reflection individually before applying the interference condition.
4. Misidentifying real and virtual images. Real images form where light rays actually converge (d_i > 0, image can be projected). Virtual images form where light rays only appear to converge (d_i < 0, cannot be projected).
5. Using the wrong sign convention for mirrors and lenses. Concave mirrors and converging lenses have positive f. Convex mirrors and diverging lenses have negative f. Mixing these up inverts your entire solution.
6. Confusing single-slit and double-slit conditions. For double slit, d sin θ = mλ gives bright fringes. For single slit, a sin θ = mλ gives dark fringes. These look similar but mean opposite things.

Self-Check Questions

1. Light travels from glass (n = 1.52) to water (n = 1.33). Does it bend toward or away from the normal? What is the critical angle if light travels from water to glass?
2. An object is placed 15 cm in front of a concave mirror with focal length 10 cm. Find the image distance, magnification, and describe the image.
3. A double-slit experiment uses light of wavelength 500 nm with slit separation 0.040 mm. The screen is 2.0 m away. How far apart are adjacent bright fringes?
4. Why is there no total internal reflection when light travels from air into glass?
5. Unpolarized light of intensity 100 W/m² passes through two polarizing filters. The first has its axis vertical; the second has its axis at 30° from vertical. What is the transmitted intensity?
6. A single slit of width 0.10 mm is illuminated with 600 nm light. What is the angular width of the central maximum?

   Answers to self-check questions are embedded in the practice file for this unit.

The photoelectric effect demonstrated that light behaves as particles (photons), not just waves. When light shines on a metal surface, electrons can be ejected.

Key observations (cannot be explained by classical wave theory):

1. Electrons are ejected only if the light frequency exceeds a threshold frequency (f₀)
2. The maximum kinetic energy of ejected electrons depends on frequency, not intensity
3. Increasing intensity increases the number of ejected electrons, not their energy
4. Electrons are ejected essentially instantaneously, even at low intensity

   Einstein's explanation: Light consists of photons, each with energy:

   E = hf = hc/λ

   Where h = 6.626 × 10⁻³⁴ J·s (Planck's constant), f is frequency, c = 3.0 × 10⁸ m/s, and λ is wavelength.

   Photoelectric equation:

   KE_max = hf - φ

   Where φ (the work function) is the minimum energy needed to eject an electron from the metal surface. φ = hf₀.

5. If hf < φ: no electrons are ejected regardless of intensity
6. If hf = φ: electrons are just barely ejected with KE_max = 0
7. If hf > φ: KE_max = hf - φ

   Worked Example 1: Light with wavelength 400 nm strikes a sodium surface (work function φ = 2.28 eV). What is the maximum kinetic energy of ejected photoelectrons?

   Solution: E_photon = hc/λ = (6.626 × 10⁻³⁴)(3.0 × 10⁸)/(400 × 10⁻⁹) = 4.97 × 10⁻¹⁹ J Convert to eV: E_photon = 4.97 × 10⁻¹⁹ / 1.602 × 10⁻¹⁹ = 3.10 eV KE_max = 3.10 - 2.28 = 0.82 eV

7.2 Atomic Spectra and the Bohr Model

When atoms absorb or emit light, they produce line spectra — discrete wavelengths rather than a continuous spectrum. Each element has a unique spectral fingerprint.

Bohr Model of the Hydrogen Atom

• Electrons orbit the nucleus in allowed energy levels (stationary states)
• Electrons do not radiate energy while in a stationary state
• Photon emission: electron drops from higher to lower level: ΔE = E_high - E_low = hf
• Photon absorption: electron jumps from lower to higher level: hf = E_high - E_low

  Energy levels of hydrogen:

  E_n = -13.6 eV / n² (n = 1, 2, 3, ...)

• Ground state (n = 1): E₁ = -13.6 eV
• First excited state (n = 2): E₂ = -3.4 eV
• Ionization energy: 13.6 eV (energy needed to remove electron from ground state)

  Spectral series:

• Lyman series (to n = 1): ultraviolet
• Balmer series (to n = 2): visible
• Paschen series (to n = 3): infrared

  Worked Example 2: What is the energy, frequency, and wavelength of the photon emitted when a hydrogen atom transitions from n = 3 to n = 2?

  Solution: E₃ = -13.6/9 = -1.51 eV E₂ = -13.6/4 = -3.40 eV ΔE = E₃ - E₂ = -1.51 - (-3.40) = 1.89 eV f = ΔE/h = (1.89 × 1.602 × 10⁻¹⁹)/(6.626 × 10⁻³⁴) = 4.57 × 10¹⁴ Hz λ = c/f = (3.0 × 10⁸)/(4.57 × 10¹⁴) = 6.56 × 10⁻⁷ m = 656 nm (red light — Balmer Hα line)

7.3 Wave-Particle Duality

The de Broglie hypothesis states that all matter has wave properties:

λ = h/p = h/(mv)

Where h is Planck's constant, p is momentum, m is mass, and v is velocity.

• Massive, slow-moving objects have negligible wavelengths
• Electrons (small mass) have significant wavelengths, comparable to atomic spacing
• This explains why electron beams produce diffraction patterns

  Worked Example 3: What is the de Broglie wavelength of an electron (m = 9.11 × 10⁻³¹ kg) moving at 2.0 × 10⁶ m/s?

  Solution: λ = h/(mv) = (6.626 × 10⁻³⁴)/((9.11 × 10⁻³¹)(2.0 × 10⁶)) = 3.64 × 10⁻¹⁰ m = 0.364 nm

  This is comparable to the spacing between atoms in a crystal, which is why electrons can be diffracted by crystals.

7.4 Nuclear Physics

Structure of the Nucleus

• The nucleus contains protons (positive charge, +e) and neutrons (no charge)
• Atomic number (Z): number of protons — determines the element
• Mass number (A): number of protons + neutrons
• Isotopes: same Z, different A (same element, different number of neutrons)
• Nuclear notation: ᴬ_ZX (e.g., ²³⁵₉₂U)

Radioactive Decay

Alpha decay (α): Emits ⁴₂He (alpha particle = 2 protons + 2 neutrons)

• Mass number decreases by 4, atomic number decreases by 2
• Example: ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He

  Beta-minus decay (β⁻): Neutron converts to proton, emitting electron and antineutrino

• Mass number unchanged, atomic number increases by 1
• Example: ¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ν̄ₑ

  Beta-plus decay (β⁺): Proton converts to neutron, emitting positron and neutrino

• Mass number unchanged, atomic number decreases by 1

  Gamma decay (γ): Emits a high-energy photon; no change in Z or A

• Nucleus drops from excited state to lower energy state

Half-Life

The half-life (t₁/₂) is the time for half of a radioactive sample to decay:

N = N₀(½)^(t/t₁/₂)

After n half-lives: N = N₀/(2ⁿ)

Half-lives elapsed	Fraction remaining
1	1/2
2	1/4
3	1/8
4	1/16
5	1/32

Mass-Energy Equivalence

Einstein's famous equation:

E = mc²

• A small amount of mass converts to a large amount of energy
• Mass defect: the mass of a nucleus is less than the sum of its individual nucleon masses
• Binding energy: the energy equivalent of the mass defect; energy required to separate a nucleus into individual nucleons
• 1 atomic mass unit (u) = 931.5 MeV/c²

Nuclear Reactions

Fission: A heavy nucleus splits into lighter nuclei, releasing energy

• Example: ²³⁵U + n → ²³⁶U* → ¹⁴¹Ba + ⁹²Kr + 3n + energy
• Requires a critical mass of fissile material
• Powers nuclear reactors and weapons

  Fusion: Light nuclei combine to form a heavier nucleus, releasing energy

• Example: ²¹H + ³₁H → ⁴₂He + n + 17.6 MeV
• Powers the Sun and stars
• Requires extremely high temperatures (millions of K) to overcome Coulomb repulsion

Common Mistakes

1. Confusing intensity and frequency in the photoelectric effect. Increasing intensity (more photons) increases the number of ejected electrons but NOT their maximum kinetic energy. Only increasing frequency (shorter wavelength) increases KE_max.
2. Forgetting to convert eV to joules (or vice versa). Many quantum problems mix eV and joules. Know the conversion: 1 eV = 1.602 × 10⁻¹⁹ J. Always check units before calculating.
3. Using the wrong transition for energy level calculations. Emitted photon energy is E_higher - E_lower (always positive). Absorbed photon energy is also E_higher - E_lower. Make sure you subtract the more negative (lower) energy from the less negative (higher) energy.
4. Confusing alpha and beta decay products. Alpha decay reduces mass number by 4 and atomic number by 2. Beta-minus decay increases atomic number by 1 but does not change mass number. Gamma decay changes neither.
5. Misinterpreting half-life. After 2 half-lives, 1/4 of the original sample remains (not 0). The sample never fully decays — it approaches zero asymptotically.
6. Confusing binding energy per nucleon with total binding energy. The most stable nuclei (like iron-56) have the highest binding energy per nucleon, not the highest total binding energy. Fission and fusion both move toward this peak.

Self-Check Questions

1. Light with wavelength 250 nm strikes a metal surface with work function 3.8 eV. Are photoelectrons emitted? If so, what is their maximum kinetic energy?
2. A hydrogen atom is in the n = 4 state. What is the shortest wavelength photon it can emit?
3. An electron and a proton are each moving at 1.0 × 10⁶ m/s. Which has the longer de Broglie wavelength?
4. ²³⁸U undergoes two alpha decays and one beta-minus decay. What is the final nucleus?
5. A sample of ¹³¹I has a half-life of 8.0 days. How much of a 100 g sample remains after 32 days?
6. Explain why nuclear fusion requires extremely high temperatures but releases enormous energy once it begins.

   Answers to self-check questions are embedded in the practice file for this unit.

Question 1. A rectangular tank contains water to a depth of 2.0 m. The absolute pressure at the bottom of the tank is most nearly:

(A) 2.0 × 10⁴ Pa
(B) 1.2 × 10⁵ Pa
(C) 2.0 × 10⁵ Pa
(D) 3.2 × 10⁵ Pa

Question 2. A cube of aluminum (density 2700 kg/m³) is suspended from a string and fully submerged in water. The tension in the string is 15 N. What is the mass of the cube?

(A) 1.53 kg
(B) 2.06 kg
(C) 2.59 kg
(D) 3.12 kg

Question 3. Water flows through a horizontal pipe. At point A, the pipe has a cross-sectional area of 0.010 m² and the water flows at 2.0 m/s. At point B, the pipe narrows to 0.004 m². If the pressure at point A is 150 kPa, what is the pressure at point B?

(A) 120 kPa
(B) 128 kPa
(C) 134 kPa
(D) 142 kPa

Question 4. A wooden block (density 600 kg/m³) floats in water. If the block has a volume of 0.005 m³, what is the buoyant force on the block?

(A) 10 N
(B) 29.4 N
(C) 49 N
(D) 50 N

Question 5. In a hydraulic press, a force of 50 N is applied to a piston of radius 2.0 cm. The output piston has a radius of 20 cm. What is the maximum force the press can exert?

(A) 100 N
(B) 500 N
(C) 2500 N
(D) 5000 N

Question 6. A small hole is punched in the side of a water tank at a depth h below the surface. The water exits horizontally. As the depth h increases, the horizontal range of the water stream:

(A) increases only
(B) decreases only
(C) increases then decreases
(D) decreases then increases

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Free Response Question

A large tank is filled with water (density 1000 kg/m³) to a height of 3.0 m above a small hole near the bottom of the tank, as shown in the diagram. The hole has a cross-sectional area of 2.0 cm². Atmospheric pressure is 1.01 × 10⁵ Pa.

(a) Calculate the speed at which water exits the hole. Use Bernoulli's equation, treating the water surface as point 1 and the hole as point 2. Assume the tank is large enough that the water surface velocity is approximately zero.

(b) Calculate the volume flow rate of water leaving the tank in m³/s.

(c) A solid sphere of density 2500 kg/m³ and volume 400 cm³ is dropped into the tank. Calculate the buoyant force acting on the sphere when it is fully submerged.

(d) Will the sphere float or sink? Calculate the normal force exerted by the bottom of the tank on the sphere when it comes to rest at the bottom.

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Answers to Self-Check Questions from Unit 1 Notes

SCQ 1: Fraction submerged = ρ_object/ρ_fluid = 600/1000 = 0.60, or 60% of the cube is submerged.

SCQ 2: P_gauge = ρgh = (1025)(9.8)(50) = 502,250 Pa. P_abs = P_atm + P_gauge = 101,325 + 502,250 = 603,575 Pa. F = P_abs × A = 603,575 × 0.8 = 482,860 N ≈ 4.83 × 10⁵ N.

SCQ 3: If the radius is halved, the area is quartered (A = πr²). By continuity: A₁v₁ = A₂v₂, so v₂ = 4v₁ = 12 m/s.

SCQ 4: F₁/A₁ = F₂/A₂ → F₁/(π(0.02)²) = 15000/(π(0.20)²) → F₁ = 15000 × (0.02)²/(0.20)² = 15000 × 0.01 = 150 N.

SCQ 5: ρ_object = m/V = 0.5/(3.0 × 10⁻⁴) = 1667 kg/m³. Since 1667 > 1000, the object sinks. F_buoy = ρ_water × V × g = 1000 × 3.0 × 10⁻⁴ × 9.8 = 2.94 N. F_apparent = mg - F_buoy = 0.5(9.8) - 2.94 = 4.9 - 2.94 = 1.96 N.

SCQ 6: A spinning ball drags air with it. On one side, the ball's spin adds to the airflow speed; on the other side, it subtracts. By Bernoulli's principle, where air speed is higher, pressure is lower. The pressure difference creates a net force perpendicular to the ball's velocity, causing it to curve. This is the Magnus effect.

Question 1. A gas in a sealed container is heated from 300 K to 600 K at constant volume. What happens to the pressure of the gas?

(A) It doubles
(B) It triples
(C) It is halved
(D) It remains the same

Question 2. An ideal gas undergoes an isothermal expansion. Which of the following is true about the gas during this process?

(A) Internal energy increases
(B) Internal energy decreases
(C) Internal energy remains constant
(D) Heat is not transferred

Question 3. A heat engine operates between a hot reservoir at 500 K and a cold reservoir at 300 K. What is the maximum possible efficiency?

(A) 20%
(B) 33%
(C) 40%
(D) 60%

Question 4. In a PV diagram, a gas undergoes a process represented by a straight horizontal line from left to right. This process is:

(A) Isothermal
(B) Isobaric
(C) Isochoric
(D) Adiabatic

Question 5. The average kinetic energy of the molecules in a sample of an ideal gas at 300 K is doubled. The new temperature of the gas is:

(A) 150 K
(B) 300 K
(C) 600 K
(D) 900 K

Question 6. During an adiabatic compression of an ideal gas, which of the following is necessarily true?

(A) The temperature of the gas decreases
(B) The temperature of the gas increases
(C) Heat flows into the gas
(D) The pressure of the gas decreases

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Free Response Question

A monatomic ideal gas is taken through the cyclic process ABCA shown on a PV diagram:

• Process A→B: Isobaric expansion from V = 2.0 L to V = 5.0 L at P = 4.0 atm
• Process B→C: Isochoric (constant volume) pressure decrease to P = 2.0 atm
• Process C→A: Isothermal compression back to the initial state

  Use 1 atm = 1.01 × 10⁵ Pa and 1 L = 10⁻³ m³.

  (a) Calculate the work done by the gas during process A→B.

  (b) Calculate the change in internal energy during process A→B.

  (c) Calculate the heat added to the gas during process A→B.

  (d) Calculate the total work done by the gas during the complete cycle ABCA.

  (e) Is this cycle a heat engine or a refrigerator? Justify your answer using the net work.

Answers to Self-Check Questions from Unit 2 Notes

SCQ 1: By Gay-Lussac's law (constant V): P₁/T₁ = P₂/T₂ → P₂ = P₁(T₂/T₁) = P₁(600/300) = 2P₁. The pressure doubles.

SCQ 2: Isothermal means constant temperature. For an ideal gas, internal energy depends only on temperature, so ΔU = 0. By the first law: ΔU = Q - W → 0 = Q - W → Q = W. Since the gas expands (positive work), heat must be added: Q = W = 400 J. The change in internal energy is zero.

SCQ 3: efficiency_Carnot = 1 - T_C/T_H = 1 - 300/500 = 1 - 0.6 = 0.40 = 40%. W = efficiency × Q_H = 0.40 × 1500 = 600 J. Q_C = Q_H - W = 1500 - 600 = 900 J.

SCQ 4: Net work = W_expansion - W_compression = 500 - 300 = 200 J (clockwise → positive net work). Efficiency = W_net/Q_H = 200/800 = 0.25 = 25%.

SCQ 5: A refrigerator moves heat from inside the fridge to outside. With the door open, it removes heat from the room and dumps it back into the same room. The net effect is zero cooling (actually heating, since the motor adds waste heat). The second law prevents creating a temperature difference without doing work, and no useful temperature difference is maintained.

SCQ 6: Temperature decreases. In an adiabatic expansion, Q = 0, so ΔU = -W. The gas does positive work (expands), so ΔU is negative, meaning temperature decreases. For compression, the opposite occurs: work is done on the gas, so ΔU is positive and temperature increases.

Question 1. Two identical conducting spheres, one with charge +Q and the other uncharged, are brought into contact and then separated. What is the charge on each sphere?

(A) +Q and 0
(B) +Q/2 and +Q/2
(C) +Q and -Q
(D) +2Q and -Q

Question 2. A positive charge +q is placed at the midpoint between two identical negative charges -Q. What is the net force on the +q charge?

(A) Zero
(B) Directed toward one of the -Q charges
(C) Directed away from both -Q charges
(D) Perpendicular to the line connecting the -Q charges

Question 3. The electric potential at a point P due to a positive point charge is V. If the distance from the charge to P is doubled, the electric potential at P becomes:

(A) 2V
(B) V/2
(C) 4V
(D) V/4

Question 4. A proton is released from rest in a uniform electric field pointing to the right. Which of the following describes the proton's motion?

(A) It accelerates to the left
(B) It accelerates to the right
(C) It moves at constant velocity to the right
(D) It remains at rest

Question 5. Which of the following statements about a conductor in electrostatic equilibrium is NOT true?

(A) The electric field inside the conductor is zero
(B) Any excess charge resides on the surface
(C) The electric field at the surface is parallel to the surface
(D) The entire conductor is at the same electric potential

Question 6. A charge of -3.0 μC moves from a point at potential +50 V to a point at potential -20 V. What is the change in the charge's electric potential energy?

(A) +90 μJ
(B) -90 μJ
(C) +210 μJ
(D) -210 μJ

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Free Response Question

Two point charges, q₁ = +4.0 μC and q₂ = -2.0 μC, are placed on the x-axis. q₁ is at the origin (x = 0) and q₂ is at x = 0.30 m.

(a) Calculate the electric field at point P on the x-axis at x = 0.50 m. Give both magnitude and direction.

(b) Calculate the electric potential at point P.

(c) A third charge q₃ = +1.0 μC is placed at point P. Calculate the electric force on q₃.

(d) Calculate the electric potential energy of the three-charge system (q₁, q₂, and q₃).

(e) Sketch the electric field lines in the region around q₁ and q₂. Explain the key features of your diagram.

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Answers to Self-Check Questions from Unit 3 Notes

SCQ 1: The electric field at the midpoint is zero (the fields from the two equal positive charges are equal and opposite, so they cancel). However, the electric potential is NOT zero — potentials add as scalars: V = kQ/r + kQ/r = 2kQ/r. Since both charges are positive, V is positive and nonzero.

SCQ 2: The electron moves from lower potential (20 V) to higher potential (80 V). ΔV = 80 - 20 = +60 V. ΔU = qΔV = (-1.6 × 10⁻¹⁹)(+60) = -9.6 × 10⁻¹⁸ J. Since ΔU is negative, kinetic energy increases by 9.6 × 10⁻¹⁸ J (by conservation of energy: ΔKE = -ΔU).

SCQ 3: F = qE = (-2.0 × 10⁻⁶)(500) = -1.0 × 10⁻³ N. The magnitude is 1.0 × 10⁻³ N, and the negative sign indicates the force is opposite to the field direction (to the left). The negative charge accelerates to the left.

SCQ 4: Electric field lines radiate outward from the positive charge (like spokes from a wheel). Equipotential lines are concentric circles centered on the charge. Field lines are perpendicular to the equipotential lines everywhere they intersect. The field lines point from high to low potential.

SCQ 5: Set the magnitudes equal: kq₁/x² = kq₂/(0.30 - x)², where x is the distance from q₁. Cancel k: q₁/x² = q₂/(0.30 - x)². Substituting: 4.0/x² = 2.0/(0.30 - x)². Taking square roots: 2/x = √2/(0.30 - x). Solving: 2(0.30 - x) = x√2 → 0.60 - 2x = 1.414x → 0.60 = 3.414x → x = 0.176 m. The point where E = 0 is at x = 0.176 m (between the charges). Note that E = 0 is NOT at the midpoint because the charges are unequal.

SCQ 6: In a conductor in electrostatic equilibrium, free electrons redistribute themselves so that the net field inside is zero. If there were a field inside, electrons would move (not equilibrium). All excess charge goes to the surface because like charges repel and push each other as far apart as possible (the surface is the farthest apart they can get). Since E = 0 inside, and the field is perpendicular at the surface, the potential is the same everywhere on and in the conductor.

Question 1. A 12 V battery is connected to two resistors in series: R₁ = 4 Ω and R₂ = 8 Ω. What is the voltage across R₂?

(A) 4 V
(B) 6 V
(C) 8 V
(D) 12 V

Question 2. An RC circuit with R = 10 kΩ and C = 20 μF is connected to a 12 V battery. What is the time constant?

(A) 0.2 s
(B) 2.0 s
(C) 20 s
(D) 200 s

Question 3. Three identical resistors, each of resistance R, are connected in parallel. The equivalent resistance is:

(A) 3R
(B) R
(C) R/3
(D) R/9

Question 4. A battery has an emf of 9.0 V and an internal resistance of 0.5 Ω. When connected to an external resistor of 4.0 Ω, the current in the circuit is:

(A) 1.0 A
(B) 1.5 A
(C) 2.0 A
(D) 2.5 A

Question 5. A capacitor is fully charged and then disconnected from the battery. If the plates are pulled farther apart, what happens to the energy stored in the capacitor?

(A) It decreases
(B) It increases
(C) It stays the same
(D) It depends on the capacitance

Question 6. In the circuit shown, an ammeter measures the current through R₁ and a voltmeter measures the voltage across R₂. The ammeter should be connected in ____ and the voltmeter should be connected in ____.

(A) Series; series
(B) Series; parallel
(C) Parallel; series
(D) Parallel; parallel

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Free Response Question

A circuit consists of a 24 V battery (with negligible internal resistance), a 6.0 Ω resistor (R₁), an 8.0 Ω resistor (R₂) in parallel with R₁, and a switch S that controls the connection to a 100 μF capacitor (C) in series with a 4.0 Ω resistor (R₃). The capacitor and R₃ are in series with the parallel combination of R₁ and R₂.

(a) Calculate the equivalent resistance of the parallel combination of R₁ and R₂.

(b) Calculate the current delivered by the battery immediately after the switch is closed (when the capacitor is uncharged).

(c) Calculate the voltage across the capacitor a long time after the switch is closed.

(d) Calculate the time constant of the RC portion of the circuit.

(e) How much energy is stored in the capacitor when it is fully charged?

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Answers to Self-Check Questions from Unit 4 Notes

SCQ 1: Two 6 Ω resistors in parallel: 1/R_parallel = 1/6 + 1/6 = 2/6 = 1/3 → R_parallel = 3 Ω. In series with the third 6 Ω: R_eq = 3 + 6 = 9 Ω.

SCQ 2: I = ε/(R + r) = 9/(4 + 0.5) = 9/4.5 = 2.0 A. V_terminal = ε - Ir = 9 - 2.0(0.5) = 9 - 1.0 = 8.0 V.

SCQ 3: τ = RC = (100 × 10³)(10 × 10⁻⁶) = 1.0 s. The capacitor is approximately fully charged after 5τ = 5.0 s.

SCQ 4: The 100 W bulb has resistance R₁ = V²/P = 120²/100 = 144 Ω. The 60 W bulb has R₂ = 120²/60 = 240 Ω. In series, the same current flows through both. P = I²R, so the bulb with larger resistance (60 W bulb, 240 Ω) dissipates more power and is brighter. The 60 W bulb is brighter in series.

SCQ 5: Both resistors have 30 V across them (parallel). I₁ = V/R₁ = 30/10 = 3.0 A. I₂ = V/R₂ = 30/20 = 1.5 A. I_total = 3.0 + 1.5 = 4.5 A.

SCQ 6: An ideal ammeter has zero resistance so it doesn't affect the current it measures (in series, any added resistance would change the current). An ideal voltmeter has infinite resistance so no current flows through it (in parallel, it shouldn't draw current away from the component it measures). If an ammeter had significant resistance, placing it in parallel would create a short-circuit path. If a voltmeter had low resistance, it would draw significant current and alter the circuit voltage.

Question 1. An electron moves with velocity v to the right through a uniform magnetic field B directed into the page. What is the direction of the magnetic force on the electron?

(A) To the right
(B) To the left
(C) Upward
(D) Downward

Question 2. A proton and an alpha particle (charge +2e, mass 4m_p) enter a uniform magnetic field perpendicular to their velocities with the same speed. What is the ratio of the radius of the alpha particle's path to the proton's path?

(A) 1:1
(B) 2:1
(C) 4:1
(D) 1:2

Question 3. A circular loop of wire is in a uniform magnetic field pointing out of the page. The magnetic field strength is increasing. What is the direction of the induced current in the loop?

(A) Clockwise
(B) Counterclockwise
(C) No current is induced
(D) Alternating direction

Question 4. A wire of length 0.5 m carrying 3.0 A of current is perpendicular to a 0.4 T magnetic field. The force on the wire is:

(A) 0.3 N
(B) 0.6 N
(C) 1.2 N
(D) 6.0 N

Question 5. A transformer has 200 turns on the primary coil and 50 turns on the secondary coil. If the primary voltage is 120 V AC, what is the secondary voltage?

(A) 15 V
(B) 30 V
(C) 480 V
(D) 960 V

Question 6. A bar magnet is dropped through a vertical copper tube. Compared to free fall, the magnet:

(A) Falls faster due to magnetic attraction
(B) Falls at the same rate
(C) Falls slower due to electromagnetic braking
(D) Stops midway through the tube

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Free Response Question

A rectangular conducting loop of width w = 0.10 m and height h = 0.20 m has a resistance of 2.0 Ω. The loop is pulled with constant velocity v = 5.0 m/s out of a region of uniform magnetic field B = 0.4 T directed into the page. The field region has a width of 0.30 m. At time t = 0, the right edge of the loop is at the left boundary of the field region.

(a) As the loop exits the field, determine the magnetic flux through the loop as a function of the distance x the loop has been pulled (while the loop is still partially in the field).

(b) Calculate the magnitude of the induced emf while the loop is partially in the field.

(c) What is the direction of the induced current? Explain using Lenz's law.

(d) Calculate the force required to pull the loop at constant velocity while it exits the field.

(e) Calculate the total energy dissipated as heat in the loop during the entire process of exiting the field.

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Answers to Self-Check Questions from Unit 5 Notes

SCQ 1: Using the right-hand rule for a positive charge: fingers point right (v), curl into page (B) → thumb points up. But this is an electron (negative charge), so reverse: force is downward. Magnitude: F = evB = (1.6 × 10⁻¹⁹)(5.0 × 10⁶)(0.3) = 2.4 × 10⁻¹³ N.

SCQ 2: F = BIL sin θ = (0.6)(8.0)(0.5)sin 30° = (0.6)(8.0)(0.5)(0.5) = 1.2 N.

SCQ 3: Φ_initial = BA = (0.2)(π)(0.10)² = 6.28 × 10⁻³ Wb. Φ_final = (0.8)(π)(0.10)² = 2.51 × 10⁻² Wb. ΔΦ = 2.51 × 10⁻² - 6.28 × 10⁻³ = 1.88 × 10⁻² Wb. ε = NΔΦ/Δt = 50(1.88 × 10⁻²)/0.05 = 18.8 V. (The sign depends on Lenz's law direction.)

SCQ 4: As the north pole approaches, the flux through the coil (pointing to the right, away from the north pole) increases. By Lenz's law, the induced current opposes this increase, so it creates a magnetic field pointing to the left (toward the approaching magnet). By the right-hand rule, this means the induced current flows counterclockwise when viewed from the magnet's side.

SCQ 5: V_s/V_p = N_s/N_p → V_s = 120(500/100) = 600 V. For power conservation: V_pI_p = V_sI_s → I_s = V_pI_p/V_s = 120(10)/600 = 2.0 A.

SCQ 6: Work W = F·d·cos θ. The magnetic force is always perpendicular to velocity (θ = 90°), so cos 90° = 0 and W = 0. Since work equals the change in kinetic energy (work-energy theorem), and W = 0, the kinetic energy (and therefore speed) cannot change. The magnetic force only changes the direction of motion, not the speed.

Question 1. Light travels from air (n = 1.00) into glass (n = 1.52) at an angle of incidence of 30°. What is the angle of refraction?

(A) 19.2°
(B) 30.0°
(C) 48.8°
(D) 52.4°

Question 2. An object is placed 30 cm in front of a converging lens with a focal length of 10 cm. The image is:

(A) Real, inverted, and magnified
(B) Real, inverted, and reduced
(C) Virtual, upright, and magnified
(D) Virtual, upright, and reduced

Question 3. In a double-slit experiment, the slit separation is 0.050 mm and the screen is 1.5 m away. The distance between the first and third bright fringes is 3.0 cm. What is the wavelength of the light used?

(A) 400 nm
(B) 500 nm
(C) 600 nm
(D) 700 nm

Question 4. Light of wavelength λ passes through a single slit of width a. The first diffraction minimum occurs at angle θ. If the slit width is doubled, the first minimum occurs at:

(A) θ/4
(B) θ/2
(C) 2θ
(D) 4θ

Question 5. Unpolarized light of intensity I₀ passes through two polarizing filters whose transmission axes are perpendicular to each other. What is the transmitted intensity?

(A) I₀
(B) I₀/2
(C) I₀/4
(D) 0

Question 6. A convex mirror has a focal length of -15 cm. An object is placed 20 cm in front of the mirror. The image distance is:

(A) -8.6 cm
(B) -15 cm
(C) -60 cm
(D) +60 cm

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Free Response Question

Light of wavelength 550 nm is incident normally on a double slit with slit separation d = 0.025 mm. The interference pattern is observed on a screen 2.0 m away.

(a) Calculate the angular position of the second-order bright fringe.

(b) Calculate the distance from the central maximum to the second-order bright fringe on the screen.

(c) If the entire apparatus is submerged in water (n = 1.33), does the spacing between bright fringes increase, decrease, or stay the same? Calculate the new fringe spacing to support your answer.

(d) A thin piece of glass (n = 1.50) of thickness t = 1.0 × 10⁻⁵ m is placed over one of the slits. By how many wavelengths does this glass shift the light from that slit, compared to the uncovered slit? Will the central maximum shift toward the covered slit or away from it?

(e) Explain qualitatively how the pattern would change if the slit separation were increased while keeping all other parameters the same.

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Answers to Self-Check Questions from Unit 6 Notes

SCQ 1: Light goes from higher n (1.52) to lower n (1.33), so it bends away from the normal. From water to glass (lower to higher n), light would bend toward the normal. There is no critical angle when light goes from lower n to higher n — it can always enter the denser medium.

SCQ 2: 1/f = 1/d_o + 1/d_i → 1/10 = 1/15 + 1/d_i → 1/d_i = 1/10 - 1/15 = 3/30 - 2/30 = 1/30 → d_i = 30 cm. m = -d_i/d_o = -30/15 = -2.0. The image is real (d_i > 0), inverted (m < 0), and magnified (|m| > 1).

SCQ 3: Fringe spacing: Δy = λL/d = (500 × 10⁻⁹)(2.0)/(0.040 × 10⁻³) = 0.025 m = 2.5 cm.

SCQ 4: Total internal reflection requires light to go from higher n to lower n. When light goes from air (n = 1.00) into glass (n = 1.52), it goes from lower to higher n, so there is no critical angle — refraction always occurs regardless of the angle of incidence.

SCQ 5: After the first (vertical) polarizer: I₁ = I₀/2 (unpolarized → polarized). After the second (at 30° from vertical): I₂ = I₁ cos²30° = (I₀/2)(cos 30°)² = (I₀/2)(√3/2)² = (I₀/2)(3/4) = 3I₀/8 = 37.5 W/m².

SCQ 6: For single-slit diffraction, the first minimum is at sin θ = λ/a. The central maximum spans from m = -1 to m = +1, so the angular width = 2θ = 2sin⁻¹(λ/a) = 2sin⁻¹(600 × 10⁻⁹/0.10 × 10⁻³) = 2sin⁻¹(6.0 × 10⁻³) ≈ 2(6.0 × 10⁻³) rad = 0.012 rad (for small angles).

Question 1. Light of wavelength 200 nm is incident on a metal surface with a work function of 4.5 eV. Which of the following is true?

(A) No photoelectrons are emitted
(B) Photoelectrons are emitted with maximum KE of 1.7 eV
(C) Photoelectrons are emitted with maximum KE of 6.2 eV
(D) Photoelectrons are emitted with maximum KE of 4.5 eV

Question 2. A hydrogen atom is in the n = 3 excited state. How many different photon energies can be emitted as the atom returns to the ground state?

(A) 1
(B) 2
(C) 3
(D) 4

Question 3. An electron is accelerated through a potential difference of 100 V. What is its de Broglie wavelength?

(A) 0.012 nm
(B) 0.123 nm
(C) 1.23 nm
(D) 12.3 nm

Question 4. A radioactive isotope has a half-life of 6.0 hours. Starting with 80 g of the isotope, how much remains after 24 hours?

(A) 5.0 g
(B) 10 g
(C) 20 g
(D) 40 g

Question 5. Which of the following nuclear processes decreases the atomic number by 2?

(A) Beta-minus decay
(B) Beta-plus decay
(C) Alpha decay
(D) Gamma decay

Question 6. In the photoelectric effect, the stopping potential is measured for light of various frequencies. A graph of stopping potential versus frequency yields a line with:

(A) Slope = h/e and y-intercept = -φ/e
(B) Slope = e/h and y-intercept = φ/e
(C) Slope = h and y-intercept = -φ
(D) Slope = hf and y-intercept = φ

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Free Response Question

A sample of radioactive phosphorus-32 (³²P) has an initial activity of 8.0 × 10⁸ decays/s. Phosphorus-32 undergoes beta-minus decay with a half-life of 14.3 days.

(a) Write the complete nuclear equation for the decay of ³²P.

(b) What is the decay constant λ for ³²P?

(c) How many ³²P nuclei are present in the initial sample?

(d) How much time must pass before the activity drops to 1.0 × 10⁸ decays/s?

(e) A ³²P nucleus is at rest when it decays. The emitted beta particle (electron) has a kinetic energy of 0.50 MeV. Calculate the speed of the beta particle. (Mass of electron = 9.11 × 10⁻³¹ kg, 1 eV = 1.602 × 10⁻¹⁹ J.)

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Answers to Self-Check Questions from Unit 7 Notes

SCQ 1: E_photon = hc/λ = (6.626 × 10⁻³⁴)(3.0 × 10⁸)/(250 × 10⁻⁹) = 7.95 × 10⁻¹⁹ J = 4.97 eV. Since 4.97 eV > 3.8 eV (work function), photoelectrons ARE emitted. KE_max = 4.97 - 3.8 = 1.17 eV.

SCQ 2: The shortest wavelength corresponds to the largest energy transition: n = 4 → n = 1. ΔE = E₁ - E₄ = -13.6/1 - (-13.6/16) = -13.6 + 0.85 = -12.75 eV. The photon energy is 12.75 eV. λ = hc/ΔE = 1240 eV·nm/12.75 eV = 97.3 nm (ultraviolet).

SCQ 3: The electron has a much smaller mass (9.11 × 10⁻³¹ kg vs. 1.67 × 10⁻²⁷ kg). Since λ = h/(mv), the electron has the longer de Broglie wavelength (by a factor of about 1836).

SCQ 4: Two alpha decays: A decreases by 8, Z decreases by 4. Starting from ²³⁸U: A = 238 - 8 = 230, Z = 92 - 4 = 88 (radium, Ra). One beta-minus decay: A stays 230, Z increases by 1: Z = 89 (actinium, Ac). Final nucleus: ²³⁰₈₉Ac.

SCQ 5: 32 days = 4 half-lives (32/8 = 4). N = N₀(½)⁴ = 100/16 = 6.25 g.

SCQ 6: Nuclear fusion requires extremely high temperatures because protons are both positively charged and strongly repel each other (Coulomb repulsion). To get close enough for the strong nuclear force (which is short-range) to bind them, protons must overcome this enormous electrostatic barrier. High temperatures mean high kinetic energies, giving some protons enough speed to approach within ~1 fm despite the repulsion. Once close enough, the strong force (much stronger than the electromagnetic force at short range) pulls the nuclei together and releases energy. This released energy heats the surrounding material, sustaining the high temperatures needed for further fusion reactions. This is why fusion is self-sustaining in stars once ignited.

Concept	Formula	Notes
Density	ρ = m/V	Water = 1000 kg/m³
Pressure	P = F/A	SI: pascal (Pa)
Hydrostatic pressure	P = P₀ + ρgh	h measured from surface
Pascal's principle	F₁/A₁ = F₂/A₂	Hydraulic press
Buoyant force	F_b = ρ_fluid V_displaced g	Archimedes' principle
Floating fraction	f = ρ_obj/ρ_fluid	Only for floating objects
Continuity equation	A₁v₁ = A₂v₂	Conservation of mass
Bernoulli's equation	P + ½ρv² + ρgh = const	Conservation of energy

Key rules: Pressure is scalar. Pressure depends on depth, not shape. Bernoulli: faster flow = lower pressure.

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Unit 2: Thermodynamics

Concept	Formula	Notes
Temperature conversion	T_K = T_C + 273.15	Always use Kelvin
Ideal gas law	PV = nRT	R = 8.314 J/(mol·K)
Average KE	KE_avg = (3/2)k_BT	Proportional to T
RMS speed	v_rms = √(3RT/M)	Lighter = faster
First law	ΔU = Q - W	W = work BY gas
Work (const P)	W = PΔV	Area under PV curve
Carnot efficiency	η = 1 - T_C/T_H	Max possible efficiency
Entropy change	ΔS = Q/T	Reversible process
COP (refrigerator)	COP = T_C/(T_H - T_C)	Carnot COP

Process summary: Isobaric (const P): W = PΔV. Isothermal (const T): ΔU = 0, Q = W. Isochoric (const V): W = 0, ΔU = Q. Adiabatic (Q = 0): ΔU = -W.

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Unit 3: Electric Force, Field, and Potential

Concept	Formula	Notes
Coulomb's law	F = kq₁q₂/r²	k = 8.99 × 10⁹ N·m²/C²
Electric field	E = kQ/r²	Points away from +, toward -
Field of parallel plates	E = V/d	Uniform field
Force from field	F = qE	Vector equation
Electric potential	V = kQ/r	Scalar! Add algebraically
Potential energy (2 charges)	U = kq₁q₂/r	+ for like, - for unlike
Energy from potential	U = qV	Connects U and V
E-V relationship	E = -ΔV/Δx	Field → decreasing V
Work from potential	W = -ΔU = qΔV

Conductor rules: E = 0 inside. Excess charge on surface. E perpendicular to surface. Surface is equipotential.

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Unit 4: Electric Circuits

Concept	Formula	Notes
Current	I = ΔQ/Δt	Conventional: + to -
Ohm's law	V = IR	Ohmic materials only
Resistance	R = ρL/A	ρ = resistivity
Power	P = IV = I²R = V²/R
Series R	R_eq = R₁ + R₂ + ...	Same I, V splits
Parallel R	1/R_eq = 1/R₁ + 1/R₂ + ...	Same V, I splits
Kirchhoff junction	ΣI_in = ΣI_out	Conservation of charge
Kirchhoff loop	ΣV = 0	Conservation of energy
RC charging	Q = Cε(1 - e^(-t/RC))
RC discharging	Q = Q₀e^(-t/RC)
Time constant	τ = RC	63.2% charge in 1τ
Terminal voltage	V = ε - Ir	Internal resistance

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Unit 5: Magnetism and Electromagnetic Induction

Concept	Formula	Notes
Force on charge	F = qvB sin θ	Max at θ = 90°
Circular motion radius	r = mv/(qB)	Period T = 2πm/(qB)
Force on wire	F = BIL sin θ
Torque on loop	τ = NIAB sin θ	Motor principle
Magnetic flux	Φ = BA cos θ	Wb (weber)
Faraday's law	ε = -NΔΦ/Δt
Motional emf	ε = BLv
Lenz's law	Induced I opposes ΔΦ	Conservation of energy
Transformer	V_s/V_p = N_s/N_p	AC only
EM wave speed	c = 3.0 × 10⁸ m/s

Critical notes: Magnetic force does NO work (F ⊥ v). Right-hand rule for + charges; reverse for electrons. Lenz opposes the CHANGE, not the flux.

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Unit 6: Geometric and Physical Optics

Concept	Formula	Notes
Law of reflection	θ_i = θ_r	From normal
Snell's law	n₁ sin θ₁ = n₂ sin θ₂
Critical angle	θ_c = sin⁻¹(n₂/n₁)	n₁ > n₂ required
Mirror/lens equation	1/f = 1/d_o + 1/d_i
Magnification	m = -d_i/d_o	Neg = inverted
Double-slit bright	d sin θ = mλ	m = 0, ±1, ±2, ...
Double-slit dark	d sin θ = (m+½)λ
Single-slit min	a sin θ = mλ	m = ±1, ±2, ...
Thin-film (1 phase shift)	2nt = (m+½)λ	Constructive
Malus's law	I = I₀ cos²θ	Polarization
Wavelength in medium	λ_n = λ₀/n	Frequency unchanged

Sign conventions: Concave mirror & converging lens: f > 0. Convex mirror & diverging lens: f < 0. Real image: d_i > 0. Virtual: d_i < 0.

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Unit 7: Quantum, Atomic, and Nuclear Physics

Concept	Formula	Notes
Photon energy	E = hf = hc/λ	h = 6.626 × 10⁻³⁴ J·s
Photoelectric effect	KE_max = hf - φ	φ = work function
Hydrogen energy levels	E_n = -13.6/n² eV
Bohr transitions	ΔE = hf	Photon emission/absorption
de Broglie wavelength	λ = h/(mv)	Matter waves
Half-life	N = N₀(½)^(t/t₁/₂)
Decay constant	λ = ln2/t₁/₂
Mass-energy	E = mc²	1 u = 931.5 MeV
Alpha decay	A-4, Z-2	Emits ⁴₂He
Beta-minus decay	A, Z+1	n → p + e⁻ + ν̄
Beta-plus decay	A, Z-1	p → n + e⁺ + ν
Gamma decay	A, Z unchanged	Emits photon

Key constants: c = 3 × 10⁸ m/s. h = 6.626 × 10⁻³⁴ J·s. e = 1.602 × 10⁻¹⁹ C. 1 eV = 1.602 × 10⁻¹⁹ J. hc = 1240 eV·nm.

Pacing: 90 Minutes, 50 MCQs

That gives you approximately 1 minute 48 seconds per question. However, not all questions deserve equal time:

• Easy conceptual questions: 30–45 seconds
• Standard calculation questions: 90–120 seconds
• Complex multi-step questions: 150–180 seconds
• Questions you are stuck on: mark and move on in 30 seconds

  Recommended pacing checkpoints:

• After 20 minutes: you should have completed ~12 questions
• After 45 minutes: you should have completed ~28 questions
• After 70 minutes: you should have completed ~42 questions
• Last 20 minutes: review marked questions and guess on any remaining

The Golden Rule: No Penalty for Guessing

AP Physics 2 has no wrong-answer penalty. Never leave a question blank. If you can eliminate even one answer choice, your expected score improves by guessing. If you can eliminate two choices, you have a 33% chance — always guess.

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Section 1: Multiple Choice Strategies

Strategy 1: Process of Elimination

Most MCQs have one or two clearly wrong answers. Eliminate them first:

• Check units: if an answer has wrong units, eliminate it
• Check order of magnitude: if the answer should be ~10 N and an option is ~1000 N, eliminate it
• Check direction: if a force should point left and the answer implies right, eliminate it

Strategy 2: Dimensional Analysis

For calculation questions, check that the answer has the correct units:

• Force: kg·m/s² or N
• Energy: kg·m²/s² or J
• Pressure: N/m² or Pa
• Power: J/s or W
• Potential: J/C or V
• Flux: T·m² or Wb

Strategy 3: Special Cases and Limits

If you are unsure about a general formula, try a special case:

• Set θ = 0° or θ = 90° and see if the formula reduces correctly
• Set r → ∞ and check if the answer approaches zero
• Set equal charges and check symmetry
• Set equal resistors and check that the equivalent resistance makes sense

Strategy 4: Use the Equation Sheet

The equation sheet has every formula you need. Use it as a reference, not a memory test. Spend 30 seconds at the start of the exam orienting yourself to the sheet layout so you can find formulas quickly.

Strategy 5: Read Carefully

AP Physics 2 MCQs often hinge on a single word:

• "Gauge pressure" vs. "absolute pressure"
• "Electric potential" vs. "electric potential energy"
• "Work done by the gas" vs. "work done on the gas"
• "Real image" vs. "virtual image"
• "Electron" vs. "proton" (direction of force reverses)
• "Constant pressure" vs. "constant temperature"

Section 2: Free Response Strategies

Understanding the Four FRQ Types

1. Mathematical Routines (1 question):

• Focus on clear algebraic steps
• Show every calculation, including formula substitution
• Include units in every step and box your final answer with units
• If you make a calculation error, you can still earn full points for correct physics if your method is right

  2. Translation Between Representations (1 question):

• You may be asked to convert between:
  • Verbal descriptions and equations
  • Graphs and verbal descriptions
  • Diagrams and equations
  • Data tables and mathematical relationships
• Be explicit about the connections you are making
• Label all axes, points, and features on any graph you draw

  3. Experimental Design and Analysis (1 question):

• Include: hypothesis, procedure, measurements, analysis, and error analysis
• Control variables explicitly
• Describe HOW you measure each quantity (not just what)
• Explain how you would use the data to reach a conclusion
• Discuss sources of error and their effects (direction, not just "human error")
• A good procedure is detailed enough that another student could replicate it

  4. Qualitative/Quantitative Translation (1 question):

• Combines conceptual explanation with mathematical justification
• Often asks "explain why" or "justify your answer"
• Use physics vocabulary precisely (not vague language)
• Support every claim with an equation, principle, or law

Universal FRQ Best Practices

1. Show all work. Partial credit is generous. Even if the final answer is wrong, correct setup earns significant points.
2. Box or circle final answers. Make it easy for the grader to find your answer.
3. Use correct significant figures. Usually 2–3 significant figures are appropriate.
4. Include units in every answer. A number without units is incomplete.
5. Draw diagrams. Free-body diagrams, circuit diagrams, field maps, and ray diagrams earn points and help your reasoning.
6. Use the equation sheet. Write down the general formula first, then substitute. This shows the grader which equation you are using.
7. Answer every part. Even if you cannot do part (a), you can often do part (b) using the answer to (a) as a variable. Write: "Using the result from (a)..."
8. Explain in complete sentences. The FRQ is not just about numbers. Graders look for physics understanding expressed in words.

Time Management for FRQs

90 minutes for 4 questions = 22.5 minutes per question. Budget:

• 1–2 min: Read the entire question carefully
• 1 min: Identify what each part is asking
• 15–18 min: Solve all parts
• 2–3 min: Review and check answers

  Do NOT spend more than 25 minutes on any single FRQ. If you are stuck, write down relevant formulas and move on.

Topic-Specific Strategies

Fluids

• Always draw a free-body diagram for submerged objects (gravity down, buoyancy up)
• For Bernoulli problems, identify the two points and write out all three terms at each
• Remember: faster flow = lower pressure (this alone answers many conceptual questions)

Thermodynamics

• Always convert to Kelvin before using gas laws
• On PV diagrams: work = area under the curve (positive for expansion, negative for compression)
• Clockwise cycle = heat engine; counterclockwise = refrigerator
• For the first law, be meticulous about sign conventions: Q into system = positive, W by gas = positive

Electric Force, Field, and Potential

• Potential is a scalar; add algebraically. Field is a vector; add component-wise.
• E points from high V to low V. Equipotentials are perpendicular to field lines.
• For conductor problems, always state: E = 0 inside, charge on surface, field perpendicular to surface.

Electric Circuits

• For complex circuits, assign current directions, write KCL and KVL equations, and solve the system.
• RC circuit tip: an uncharged capacitor initially acts as a wire (I = V/R); a fully charged capacitor acts as an open circuit (I = 0).
• Ammeters in series (low R), voltmeters in parallel (high R).

Magnetism

• Right-hand rule for positive charges; LEFT-hand rule (or reverse) for electrons.
• Magnetic force does no work (F ⊥ v always).
• Lenz's law: oppose the CHANGE, not the flux itself.

Optics

• All angles are measured from the NORMAL, not the surface.
• Real images: d_i > 0, inverted, can be projected. Virtual: d_i < 0, upright, cannot be projected.
• TIR requires going from higher n to lower n.
• Wavelength changes in a medium; frequency does not.

Quantum Physics

• Photoelectric effect: frequency determines whether electrons are emitted AND their KE. Intensity only determines the number of electrons.
• For hydrogen: E_n = -13.6/n² eV. Lower n = more negative (lower energy).
• Alpha decay: A-4, Z-2. Beta-minus: A same, Z+1. Gamma: A and Z same.

The Night Before and Morning Of

Night before:

• Review the summary sheet (not the full notes)
• Do 3–5 practice problems from your weakest unit
• Get 7–8 hours of sleep
• Pack your calculator (with fresh batteries), pencils, and a watch

  Morning of:

• Eat a balanced meal
• Arrive early
• During the briefing, flip through the equation sheet to orient yourself
• Take deep breaths. You have prepared thoroughly.

Final Score Estimation

Raw MCQ (out of 50)	Raw FRQ (out of 40)	Approx. AP Score
35–40	28–35	5
28–34	22–27	4
20–27	16–21	3
14–19	10–15	2
0–13	0–9	1

These are approximate. The College Board applies statistical adjustments. Aiming for 30+ correct MCQs and 20+ FRQ points gives a strong chance at a 4 or 5.

AP Physics 2: Complete Course Review

• Algebra-Based Physics
• 7 Units | 3-Hour Exam | 50% MCQ + 50% FRQ
• Presenter name / Date

Slide 2: Exam Format Overview

• Section 1: 50 MCQ, 90 minutes (50%)
• Section 2: 4 FRQ, 90 minutes (50%)
• Calculator and equation sheet allowed
• No penalty for guessing on MCQ

Slide 3: Unit Weightings

• Unit 1 (Fluids): 10–12%
• Unit 2 (Thermo): 12–18%
• Unit 3 (E-Force/Field/Potential): 18–22% ← HIGHEST
• Unit 4 (Circuits): 10–14%
• Unit 5 (Magnetism): 10–14%
• Unit 6 (Optics): 12–16%
• Unit 7 (Quantum/Nuclear): 10–14%

UNIT 1: FLUIDS (Slides 4–9)

Slide 4: Density and Pressure

• Density: ρ = m/V (water = 1000 kg/m³)
• Pressure: P = F/A (SI: pascal)
• Hydrostatic: P = P₀ + ρgh
• Gauge vs. absolute pressure

Slide 5: Pascal's Principle

• Pressure transmits undiminished in enclosed fluid
• F₁/A₁ = F₂/A₂
• Hydraulic press: small force → large force
• Mechanical advantage = A₂/A₁

Slide 6: Buoyancy — Archimedes' Principle

• F_buoy = ρ_fluid × V_displaced × g
• Floating: ρ_obj < ρ_fluid (fraction submerged = ρ_obj/ρ_fluid)
• Sinking: ρ_obj > ρ_fluid
• Apparent weight = mg - F_buoy

Slide 7: Fluid Dynamics — Continuity Equation

• A₁v₁ = A₂v₂ (conservation of mass)
• Narrower pipe → faster flow
• Volume flow rate = Av = constant

Slide 8: Bernoulli's Equation

• P + ½ρv² + ρgh = constant
• Conservation of energy per unit volume
• Key insight: faster flow = lower pressure
• Applications: airplane lift, Venturi effect, spray bottles

Slide 9: Unit 1 — Common Pitfalls

• Measuring depth from wrong reference
• Confusing V_displaced with V_total for floating objects
• Applying Bernoulli without ρgh terms in non-horizontal pipes
• Pressure is scalar, not vector

UNIT 2: THERMODYNAMICS (Slides 10–16)

Slide 10: Temperature and Gas Laws

• Three temperature scales; always use Kelvin
• Ideal gas law: PV = nRT
• Boyle (const T): P₁V₁ = P₂V₂
• Charles (const P): V₁/T₁ = V₂/T₂
• Gay-Lussac (const V): P₁/T₁ = P₂/T₂

Slide 11: Kinetic Molecular Theory

• Gas = particles in random motion, elastic collisions
• KE_avg = (3/2)k_BT → proportional to temperature
• v_rms = √(3RT/M) → lighter molecules move faster

Slide 12: First Law of Thermodynamics

• ΔU = Q - W (conservation of energy)
• Q: heat into system (positive in)
• W: work by system (positive out)
• Work = area under PV curve

Slide 13: PV Diagrams and Processes

• Isobaric (const P): horizontal line, W = PΔV
• Isothermal (const T): hyperbola, ΔU = 0
• Isochoric (const V): vertical line, W = 0
• Adiabatic (Q = 0): steepest curve, ΔU = -W
• Cyclic: ΔU = 0, W_net = area enclosed

Slide 14: Second Law and Entropy

• Entropy of isolated system never decreases
• ΔS = Q/T
• Natural processes: irreversible, ΔS > 0
• Heat flows hot → cold, never reverses spontaneously

Slide 15: Heat Engines and Refrigerators

• Engine: η = W/Q_H = 1 - Q_C/Q_H
• Carnot: η_max = 1 - T_C/T_H
• Refrigerator: COP = Q_C/W
• Clockwise cycle = engine; counterclockwise = refrigerator

Slide 16: Unit 2 — Common Pitfalls

• Forgetting Kelvin conversion
• Sign convention errors (Q and W)
• Assuming ΔU = 0 for all processes
• Exceeding Carnot efficiency in calculations

UNIT 3: ELECTRIC FORCE, FIELD, AND POTENTIAL (Slides 17–23)

Slide 17: Electric Charge and Coulomb's Law

• Charge is quantized: Q = ne
• Conduction vs. induction
• F = kq₁q₂/r² (attractive for unlike, repulsive for like)
• Superposition: vector sum of forces

Slide 18: Electric Field

• E = F/q₀ = kQ/r² (N/C or V/m)
• Points away from +, toward −
• Field lines: density = strength, never cross
• Uniform field between parallel plates: E = V/d

Slide 19: Conductors in Electrostatic Equilibrium

• E = 0 inside the conductor
• Excess charge on the surface
• E perpendicular to surface at the surface
• Entire conductor is equipotential

Slide 20: Electric Potential Energy

• U = kq₁q₂/r (two point charges)
• Positive for like charges, negative for unlike
• W = -ΔU
• For uniform field: U = qEd

Slide 21: Electric Potential

• V = U/q = kQ/r (volts, scalar!)
• Add potentials algebraically (NOT as vectors)
• E = -ΔV/Δx: field points toward decreasing V
• Equipotential lines ⊥ to field lines

Slide 22: Key Relationships in Unit 3

• F = qE (force from field)
• E = -ΔV/Δx (field from potential)
• U = qV (energy from potential)
• W = -ΔU = qΔV (work from potential)

Slide 23: Unit 3 — Common Pitfalls

• Confusing field (vector) and potential (scalar)
• Adding potentials as vectors
• Forgetting negative sign in E = -ΔV/Δx
• Confusing potential energy (J) with potential (V)

UNIT 4: ELECTRIC CIRCUITS (Slides 24–28)

Slide 24: Current, Resistance, and Power

• I = ΔQ/Δt; V = IR; P = IV = I²R = V²/R
• R = ρL/A
• Conventional current: + to −

Slide 25: Series and Parallel Circuits

• Series: same I, V splits, R_eq = R₁ + R₂ + ...
• Parallel: same V, I splits, 1/R_eq = 1/R₁ + 1/R₂ + ...
• Series: larger R_eq. Parallel: smaller R_eq

Slide 26: Kirchhoff's Laws and Meters

• Junction rule: ΣI_in = ΣI_out (charge conservation)
• Loop rule: ΣV = 0 (energy conservation)
• Ammeter: series, low R. Voltmeter: parallel, high R.
• Internal resistance: V_term = ε − Ir

Slide 27: RC Circuits

• Charging: Q(t) = Cε(1 − e^(−t/RC))
• Discharging: Q(t) = Q₀e^(−t/RC)
• Time constant τ = RC (63.2% in 1τ, ~100% in 5τ)
• Uncharged capacitor = wire; fully charged = open circuit

Slide 28: Unit 4 — Common Pitfalls

• Mixing series/parallel rules
• Forgetting reciprocal in parallel resistance
• Ohm's law doesn't apply to capacitors
• Incorrect ammeter/voltmeter placement

UNIT 5: MAGNETISM (Slides 29–33)

Slide 29: Magnetic Force on Charges and Wires

• F = qvB sin θ (max at 90°, zero at 0°)
• Circular motion: r = mv/(qB)
• Force on wire: F = BIL sin θ
• RIGHT HAND for + charges; REVERSE for electrons
• Magnetic force does NO work (F ⊥ v)

Slide 30: Torque on a Current Loop

• τ = NIAB sin θ
• Maximum when loop parallel to B
• Zero when loop perpendicular to B (equilibrium)
• Basis for electric motors

Slide 31: Faraday's Law and Lenz's Law

• Φ = BA cos θ (magnetic flux)
• ε = −NΔΦ/Δt (Faraday's law)
• Lenz's law: induced current opposes the CHANGE in flux
• Motional emf: ε = BLv

Slide 32: Transformers and EM Waves

• V_s/V_p = N_s/N_p (AC only!)
• Power conservation: V_pI_p = V_sI_s
• Maxwell: accelerating charges → EM waves
• c = 3.0 × 10⁸ m/s; E ⊥ B ⊥ direction

Slide 33: Unit 5 — Common Pitfalls

• Wrong hand for negative charges
• Confusing angle θ in F = qvB sin θ
• Thinking magnetic force does work
• Misapplying Lenz's law (oppose change, not flux)
• Transformers only work with AC

UNIT 6: OPTICS (Slides 34–39)

Slide 34: Reflection and Refraction

• Law of reflection: θ_i = θ_r (from normal!)
• Snell's law: n₁ sin θ₁ = n₂ sin θ₂
• Toward normal (lower to higher n), away (higher to lower n)
• Wavelength in medium: λ = λ₀/n (f unchanged)

Slide 35: Total Internal Reflection

• Requires: higher n to lower n, angle > θ_c
• θ_c = sin⁻¹(n₂/n₁)
• Applications: fiber optics, diamonds
• No TIR from lower to higher n

Slide 36: Mirrors and Lenses

• 1/f = 1/d_o + 1/d_i; m = −d_i/d_o
• Concave mirror / converging lens: f > 0
• Convex mirror / diverging lens: f < 0
• Real (d_i > 0, inverted) vs. virtual (d_i < 0, upright)
• Three principal rays for ray diagrams

Slide 37: Interference

• Double slit: d sin θ = mλ (bright), d sin θ = (m+½)λ (dark)
• Fringe spacing: Δy = λL/d
• Diffraction grating: same equation, sharper fringes
• Single-slit minimum: a sin θ = mλ (m = ±1, ±2, ...)
• Central maximum is twice as wide as secondary maxima

Slide 38: Thin Films and Polarization

• Phase shift when reflecting off higher n medium
• Constructive/destructive conditions depend on # of phase shifts
• Malus's law: I = I₀ cos²θ
• Perpendicular polarizers block all light

Slide 39: Unit 6 — Common Pitfalls

• Measuring angles from surface instead of normal
• Forgetting wavelength changes in medium
• Mixing up single-slit and double-slit conditions
• Wrong sign conventions for mirrors/lenses
• Thin-film phase shift errors

UNIT 7: QUANTUM PHYSICS (Slides 40–44)

Slide 40: Photoelectric Effect

• E = hf = hc/λ (photon energy)
• KE_max = hf − φ (φ = work function)
• Frequency determines if emission occurs AND max KE
• Intensity determines number of electrons only
• Threshold frequency: f₀ = φ/h

Slide 41: Atomic Spectra and Bohr Model

• E_n = −13.6/n² eV (hydrogen)
• Emission: electron drops, photon released
• Absorption: electron jumps up, photon absorbed
• ΔE = hf = E_higher − E_lower
• Lyman (UV), Balmer (visible), Paschen (IR)

Slide 42: Wave-Particle Duality

• de Broglie: λ = h/(mv)
• All matter has wave properties
• Significant for small masses (electrons)
• Explains electron diffraction

Slide 43: Nuclear Physics

• Alpha: A−4, Z−2. Beta-minus: A, Z+1. Gamma: A, Z same.
• Half-life: N = N₀(½)^(t/t₁/₂)
• E = mc²; 1 u = 931.5 MeV
• Fission: heavy splits. Fusion: light combines.
• Binding energy per nucleon peaks at Fe-56

Slide 44: Unit 7 — Common Pitfalls

• Confusing intensity and frequency in photoelectric effect
• Forgetting eV ↔ J conversion
• Wrong transition for energy level calculations
• Mixing up alpha/beta/gamma decay products
• Misinterpreting half-life as total decay time

Slide 45: Final Review and Study Plan

• Prioritize Units 2 and 3 (highest weight)
• Practice all four FRQ types
• Memorize sign conventions
• Master right-hand rules
• Take the full practice exam under timed conditions
• Review the summary sheet the night before
• Trust your preparation. Good luck!

This script is designed to be read aloud as an audio study guide. Estimated duration: 35–40 minutes at a natural speaking pace. Pause briefly between sections.

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Introduction (2 minutes)

Welcome to your AP Physics 2 audio study guide. This recording covers all seven units of the AP Physics 2 exam: fluids, thermodynamics, electric force and fields, electric circuits, magnetism, optics, and quantum physics. I will walk you through the key concepts, most important formulas, and common mistakes for each unit.

The AP Physics 2 exam is three hours long. Section one has fifty multiple-choice questions in ninety minutes, worth fifty percent of your score. Section two has four free-response questions in ninety minutes, worth the other fifty percent. You get an equation sheet and a calculator. There is no penalty for guessing on the multiple-choice section, so never leave a question blank.

Let's begin.

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Unit 1: Fluids (5 minutes)

Fluids are substances that flow, including both liquids and gases. We start with density, which is mass per unit volume: rho equals m over V. Water has a density of one thousand kilograms per cubic meter. Mercury is about thirteen point six times denser.

Pressure is force per unit area: P equals F over A. The SI unit is the pascal. In a fluid at rest, pressure increases with depth according to P equals P-naught plus rho g h, where P-naught is the pressure at the surface, rho is the fluid density, g is nine point eight meters per second squared, and h is the depth below the surface. A critical point: pressure depends on depth, not on the shape or total volume of the container.

Pascal's principle says that a pressure change applied to an enclosed fluid is transmitted equally throughout. This is how hydraulic systems work: F-one over A-one equals F-two over A-two. A small force on a small piston creates a large force on a large piston.

Archimedes' principle states that the buoyant force equals the weight of fluid displaced: F-buoy equals rho-fluid times V-displaced times g. For a floating object, the fraction submerged equals the density of the object divided by the density of the fluid. A common mistake: using the object's density instead of the fluid's density in the buoyant force formula.

For fluids in motion, we use the continuity equation and Bernoulli's equation. The continuity equation, A-one V-one equals A-two V-two, says that fluid flows faster through narrower pipes. Bernoulli's equation states that P plus one-half rho V-squared plus rho g h equals a constant. This tells us that where a fluid flows faster, the pressure is lower. This explains airplane lift: air moves faster over the curved top of the wing, creating lower pressure above, which produces an upward force.

Key takeaways for fluids: pressure is a scalar, not a vector. Measure depth from the surface, not the bottom. And remember, faster flow means lower pressure.

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Unit 2: Thermodynamics (6 minutes)

Thermodynamics deals with heat, work, and energy. Always use Kelvin for temperature. Convert Celsius to Kelvin by adding 273.15.

The ideal gas law, PV equals nRT, combines all the gas laws into one equation. From it, we derive the individual laws: Boyle's law for constant temperature, Charles's law for constant pressure, and Gay-Lussac's law for constant volume.

Kinetic molecular theory tells us that the average kinetic energy of gas molecules is proportional to temperature: KE-average equals three-halves k-B T. The root-mean-square speed is v-rms equals the square root of three RT over M. Lighter molecules move faster at the same temperature.

The first law of thermodynamics is conservation of energy: delta U equals Q minus W. Delta U is the change in internal energy, Q is heat added to the system, and W is work done by the system. The sign convention matters enormously on the exam: heat into the system is positive, and work done by the gas is positive.

On a PV diagram, work equals the area under the curve. For the four basic processes: isobaric means constant pressure with a horizontal line; isothermal means constant temperature with a hyperbolic curve and zero change in internal energy; isochoric means constant volume with a vertical line and zero work; and adiabatic means no heat exchange, so delta U equals negative W. The adiabatic curve is the steepest on a PV diagram.

For a cyclic process, the net work equals the area enclosed. A clockwise cycle is a heat engine. A counterclockwise cycle is a refrigerator.

The second law says the entropy of an isolated system never decreases. This means heat flows spontaneously from hot to cold, and no heat engine can be one hundred percent efficient. The maximum possible efficiency is the Carnot efficiency: one minus T-cold over T-hot, using Kelvin temperatures.

Common mistakes: forgetting to convert to Kelvin, messing up the sign convention for Q and W, and calculating an efficiency higher than Carnot.

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Unit 3: Electric Force, Field, and Potential (6 minutes)

This is the highest-weighted unit on the exam, so pay close attention.

Coulomb's law gives the force between two point charges: F equals k q-one q-two over r-squared. Like charges repel; unlike charges attract. The superposition principle says the net force is the vector sum of all individual forces.

The electric field E equals the force per unit charge: F over q-naught, or kQ over r-squared for a point charge. The field points away from positive charges and toward negative charges. Electric field lines show the direction a positive test charge would move. They never cross, and their density indicates field strength. Between parallel plates, the field is uniform: E equals V over d.

For conductors in electrostatic equilibrium, remember four rules: the electric field inside is zero, excess charge resides on the surface, the field at the surface is perpendicular to the surface, and the entire conductor is an equipotential.

Electric potential is a scalar quantity: V equals kQ over r. This is crucial: you add potentials algebraically, not as vectors. The relationship between field and potential is E equals negative delta V over delta x. The electric field points in the direction of decreasing potential. Equipotential lines are always perpendicular to electric field lines.

The four key connections in this unit are: F equals qE connects force and field. E equals negative delta V over delta x connects field and potential. U equals qV connects energy and potential. And W equals negative delta U connects work and energy.

The most common mistakes here are confusing field and potential, adding potentials as vectors, and forgetting the negative sign in the E-V relationship.

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Unit 4: Electric Circuits (4 minutes)

Current I equals delta Q over delta t. Resistance R relates voltage and current through Ohm's law: V equals IR. Power: P equals IV, which also equals I-squared R and V-squared over R.

In series circuits, current is the same through all components, voltage splits, and equivalent resistance is the sum. In parallel circuits, voltage is the same across all branches, current splits, and one over R-equivalent equals the sum of one over each R. Series increases resistance; parallel decreases it.

Kirchhoff's junction rule says current in equals current out at any node. His loop rule says the sum of voltage changes around any closed loop is zero.

RC circuits involve a resistor and capacitor. The time constant tau equals RC. During charging, the current starts at a maximum and decays exponentially. During discharging, the charge starts at a maximum and decays. After one time constant, you reach about sixty-three percent of the final value. After five time constants, the capacitor is essentially fully charged or discharged.

A key practical tip: an uncharged capacitor initially acts like a wire, and a fully charged capacitor acts like an open circuit.

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Unit 5: Magnetism and Electromagnetic Induction (5 minutes)

A charge moving through a magnetic field experiences a force: F equals qvB sine theta. The force is maximum when velocity is perpendicular to the field, and zero when parallel. The magnetic force is always perpendicular to velocity, which means it does no work. It changes direction but not speed.

For circular motion in a magnetic field: r equals mv over qB. Heavier or faster particles make larger circles.

Use the right-hand rule for positive charges: point your fingers in the direction of velocity, curl them toward the magnetic field, and your thumb gives the force direction. For electrons, the force is in the opposite direction. Getting this wrong is the single most common mistake in this unit.

For a current-carrying wire: F equals BIL sine theta. The torque on a current loop is tau equals NIAB sine theta, which is the basis for electric motors.

Faraday's law says a changing magnetic flux induces an emf: emf equals negative N times delta phi over delta t. Magnetic flux phi equals BA cosine theta. Lenz's law says the induced current opposes the change in flux that created it. This is a consequence of conservation of energy.

Transformers change voltage levels using V-s over V-p equals N-s over N-p. They only work with alternating current because they require changing flux.

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Unit 6: Optics (5 minutes)

Optics is divided into geometric optics, which uses rays, and physical optics, which uses the wave nature of light.

The law of reflection says the angle of incidence equals the angle of reflection, both measured from the normal. Snell's law governs refraction: n-one sine theta-one equals n-two sine theta-two. All angles in optics are measured from the normal, never from the surface.

Total internal reflection occurs when light travels from a higher-index medium to a lower-index medium at an angle exceeding the critical angle, theta-c equals the inverse sine of n-two over n-one.

For mirrors and lenses, the thin lens equation applies: one over f equals one over d-o plus one over d-i. Magnification m equals negative d-i over d-o. Concave mirrors and converging lenses have positive focal length. Convex mirrors and diverging lenses have negative focal length. Real images are inverted with positive d-i; virtual images are upright with negative d-i.

For interference, the double-slit bright fringe condition is d sine theta equals m lambda, and the dark fringe condition is d sine theta equals (m plus one-half) lambda. Fringe spacing is lambda L over d. For single-slit diffraction, the minimum condition is a sine theta equals m lambda, where m starts at one, not zero.

Remember that wavelength decreases inside a medium: lambda equals lambda-naught over n. The frequency stays the same.

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Unit 7: Quantum, Atomic, and Nuclear Physics (4 minutes)

The photoelectric effect demonstrated that light consists of photons. The energy of a photon is E equals hf equals hc over lambda. Einstein's equation: KE-max equals hf minus phi, where phi is the work function. If the photon energy is less than the work function, no electrons are emitted, regardless of intensity. Intensity only affects the number of electrons. Frequency determines whether emission occurs and the maximum kinetic energy.

The Bohr model gives hydrogen energy levels: E-n equals negative thirteen point six over n-squared electron-volts. When an electron drops from a higher level to a lower one, it emits a photon with energy equal to the difference. The Lyman series goes to n equals one in the ultraviolet, the Balmer series to n equals two in the visible, and the Paschen series to n equals three in the infrared.

De Broglie proposed that all matter has wave properties: lambda equals h over mv. This is significant for small particles like electrons.

In nuclear physics, know the three types of decay. Alpha decay emits a helium nucleus, reducing mass number by four and atomic number by two. Beta-minus decay converts a neutron to a proton, increasing atomic number by one with no change in mass number. Gamma decay emits a photon with no change in A or Z.

Half-life: N equals N-naught times one-half to the power of t over t-one-half. After n half-lives, the fraction remaining is one over two to the n. Mass and energy are related by E equals mc-squared. One atomic mass unit equals 931.5 MeV.

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Closing Remarks (2 minutes)

Let me leave you with some final advice. First, understand, don't memorize. The equation sheet has every formula. Your job is knowing when and why to use each one. Second, draw diagrams constantly: free-body diagrams, field maps, circuit diagrams, ray diagrams, PV diagrams. Third, practice the four FRQ types: mathematical routines, representation translation, experimental design, and qualitative-quantitative translation.

Fourth, show all your work on free-response questions. Partial credit is generous. Fifth, master your right-hand rules for magnetism. Sixth, check units on every calculation. And seventh, get good sleep the night before the exam.

You have put in the work. Trust your preparation, stay calm, and do your best. Good luck on your AP Physics 2 exam.