Physics Tutoring —
From Mechanics to Modern Physics
Taught by a PhD physicist who studied at UC Berkeley and AFIT. One-on-one sessions or self-guided community resources.
Classical Mechanics
The bedrock of physics. Classical mechanics governs everything from projectiles to planetary orbits, and mastering it is essential before advancing to any specialized field. The math is tractable but the conceptual traps are real.
Key Concepts
- Newton's three laws and their vector formulations
- Kinematics: displacement, velocity, acceleration in 1D, 2D, and 3D
- Work-energy theorem and conservation of mechanical energy
- Conservation of linear momentum and impulse
- Circular motion: centripetal acceleration and net force
- Rotational dynamics: torque, moment of inertia, angular momentum
- Simple harmonic motion: period, amplitude, phase
- Gravitation and orbital mechanics
Common Student Mistakes
- Sign errors in vector decomposition — always define your coordinate system first
- Forgetting the normal force component on inclined planes (N = mg cosθ, not mg)
- Treating centripetal force as a separate "real" force rather than the net inward force
- Confusing centripetal and centrifugal frames without specifying the reference frame
- Using energy methods when momentum is conserved (inelastic collisions)
- Mixing up rotational inertia values for different geometries
A ball is launched from ground level at 30° above horizontal with an initial speed of 24 m/s. Ignoring air resistance, find: (a) the time of flight, (b) the horizontal range, and (c) the maximum height reached. Use g = 9.8 m/s².
Decompose: vx = 24 cos 30° ≈ 20.78 m/s; vy0 = 24 sin 30° = 12 m/s.
(a) Time of flight: set y = 0, so t = 2vy0/g = 2(12)/9.8 ≈ 2.45 s.
(b) Range: R = vx × t = 20.78 × 2.45 ≈ 50.9 m.
(c) Max height: H = vy0² / (2g) = 144/19.6 ≈ 7.35 m.
A 5 kg block slides from rest down a frictionless incline of angle 37° and travels a distance of 8 m along the surface. (a) What is the work done by gravity? (b) What is the block's speed at the bottom? (c) If instead μk = 0.25, what speed does it reach?
Height dropped: h = 8 sin 37° ≈ 4.81 m.
(a) Wgrav = mgh = 5 × 9.8 × 4.81 ≈ 235.7 J.
(b) Frictionless: ½mv² = 235.7 J → v = √(2 × 235.7/5) ≈ 9.70 m/s.
(c) With friction: Wfriction = μk N × d = 0.25 × 5 × 9.8 × cos 37° × 8 ≈ 78.4 J (negative work). Net W = 235.7 − 78.4 = 157.3 J → v = √(2 × 157.3/5) ≈ 7.93 m/s.
A solid cylinder of mass 3 kg and radius 0.4 m is free to rotate about its central axis. A rope wound around it is pulled with a constant force of 12 N. Starting from rest, what is the angular velocity after 5 seconds? (Moment of inertia of solid cylinder: I = ½mR²)
I = ½ × 3 × (0.4)² = 0.24 kg·m².
Torque: τ = F × R = 12 × 0.4 = 4.8 N·m.
Angular acceleration: α = τ/I = 4.8/0.24 = 20 rad/s².
After 5 s: ω = αt = 20 × 5 = 100 rad/s.
Electromagnetism
From Coulomb's force law to Maxwell's unification of electricity and magnetism — E&M is where calculus and physics converge. Understanding the field picture, not just the formulas, is what separates real mastery from surface-level familiarity.
Key Concepts
- Coulomb's law and the superposition principle for discrete charges
- Electric field and electric potential; their relationship E = −∇V
- Gauss's law: ∮ E·dA = Qenc/ε0
- Capacitors: energy stored, capacitance, dielectrics
- Ohm's law, resistivity, Kirchhoff's current and voltage laws
- Magnetic force on moving charges: F = qv × B
- Faraday's law of induction and Lenz's law
- Maxwell's equations — the complete description of classical E&M
Common Student Mistakes
- Getting B field direction wrong — always apply the right-hand rule carefully to the current element, not the field point
- Treating electric potential as a vector (it is a scalar)
- Sign errors in circuit analysis — establish a consistent loop traversal direction first
- Confusing electric flux with electric field magnitude
- Forgetting that induced EMF opposes the change in flux (Lenz's law direction)
A solid conducting sphere of radius R = 0.10 m carries a total charge Q = +4.0 μC. Find the magnitude of the electric field (a) at r = 0.05 m (inside), (b) at r = 0.10 m (surface), and (c) at r = 0.30 m (outside).
(a) Inside a conductor, E = 0 N/C. All free charge resides on the surface.
(b) At the surface: E = kQ/R² = (8.99×10&sup9;)(4.0×10−&sup6;)/(0.10)² ≈ 3.60 × 10&sup6; N/C (radially outward).
(c) Outside: E = kQ/r² = (8.99×10&sup9;)(4.0×10−&sup6;)/(0.30)² ≈ 4.00 × 10&sup5; N/C.
A circuit has a 12 V battery and three resistors: R1 = 4 Ω in series with a parallel combination of R2 = 6 Ω and R3 = 12 Ω. Find: (a) the equivalent resistance, (b) the total current from the battery, (c) the voltage across the parallel combination.
(a) Parallel: R23 = (6 × 12)/(6 + 12) = 72/18 = 4 Ω. Total: Req = 4 + 4 = 8 Ω.
(b) Itotal = V/Req = 12/8 = 1.5 A.
(c) Vparallel = I × R23 = 1.5 × 4 = 6 V.
A circular loop of radius 0.15 m lies in a uniform magnetic field that decreases from B0 = 0.80 T to 0 T over exactly 0.40 s. The loop has resistance R = 2.5 Ω. Calculate (a) the induced EMF and (b) the induced current magnitude.
Area: A = πr² = π(0.15)² ≈ 0.0707 m².
Change in flux: ΔΦ = A × ΔB = 0.0707 × (0 − 0.80) = −0.0566 Wb.
(a) |EMF| = |ΔΦ/Δt| = 0.0566/0.40 ≈ 0.141 V.
(b) I = EMF/R = 0.141/2.5 ≈ 0.057 A. By Lenz's law, the current opposes the decreasing flux, flowing to maintain B in the original direction.
Thermodynamics
Thermodynamics governs energy conversion, heat engines, and the statistical behavior of matter. The four laws are deceptively simple; the real work is in applying them correctly to processes where the sign conventions can make or break your answer.
Key Concepts
- Zeroth law: thermal equilibrium and temperature
- First law: ΔU = Q − W (internal energy, heat, work)
- Second law: entropy never decreases in an isolated system
- Third law: absolute zero is unattainable
- Ideal gas law: PV = nRT
- Process types: isothermal, adiabatic, isobaric, isochoric
- Carnot efficiency: η = 1 − TC/TH
- Entropy and the statistical interpretation of the second law
Common Student Mistakes
- Sign convention errors: physics convention is W = work done by the gas; chemistry sometimes flips this
- Not distinguishing between isothermal (T constant, use PV = nRT) and adiabatic (no heat exchange, use PVγ = const) processes
- Using Celsius instead of Kelvin in absolute temperature calculations
- Confusing heat capacity at constant pressure Cp with Cv
An ideal gas expands isothermally at T = 300 K. During the process, the gas absorbs Q = 500 J of heat. (a) What is the change in internal energy? (b) How much work does the gas do on its surroundings?
(a) For an ideal gas at constant temperature, internal energy depends only on T. Since T is constant, ΔU = 0 J.
(b) First law: ΔU = Q − W → 0 = 500 − W → W = 500 J. All absorbed heat converts directly to work in an isothermal expansion of an ideal gas.
A Carnot heat engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. If the engine absorbs QH = 800 J per cycle, find: (a) the Carnot efficiency, (b) the net work output per cycle, and (c) the heat rejected to the cold reservoir.
(a) ηCarnot = 1 − TC/TH = 1 − 300/600 = 50%.
(b) W = η × QH = 0.50 × 800 = 400 J.
(c) QC = QH − W = 800 − 400 = 400 J rejected.
2.0 moles of ideal monatomic gas (γ = 5/3) at initial state P1 = 2.0 atm, V1 = 10 L undergoes an adiabatic expansion to V2 = 20 L. Find the final pressure P2. (1 atm = 101,325 Pa)
Adiabatic relation: P1V1γ = P2V2γ.
P2 = P1(V1/V2)γ = 2.0 × (10/20)5/3 = 2.0 × (0.5)1.667.
(0.5)5/3 = e(5/3)ln(0.5) = e−1.155 ≈ 0.315.
P2 ≈ 2.0 × 0.315 ≈ 0.63 atm. The gas cools as it expands adiabatically.
Quantum Mechanics
Quantum mechanics is the most accurate physical theory ever developed, describing matter and light at the atomic scale with extraordinary precision. The mathematics requires letting go of classical intuitions — particles do not have definite positions and momenta simultaneously, and that is not a limitation of measurement, it is fundamental to nature.
Key Concepts
- Wave-particle duality: de Broglie relation λ = h/p
- The Schrödinger equation: time-dependent and time-independent forms
- Heisenberg uncertainty principle: Δx Δp ≥ ℏ/2
- Energy quantization: particle in a box, harmonic oscillator
- Wave function normalization and probability interpretation
- Hydrogen atom: quantum numbers n, l, ml, ms
- Electron spin and the Pauli exclusion principle
- Operators, eigenvalues, and expectation values
Common Student Mistakes
- Treating quantum states as having definite values before measurement — the wave function describes a superposition, not ignorance
- Forgetting normalization when computing probabilities: ∫|ψ|²dx = 1 must hold
- Confusing the n quantum number energy levels with orbital shapes (l determines shape)
- Applying classical momentum in regions where V > E (tunneling is possible quantum mechanically)
An electron is confined to a one-dimensional box of length L = 0.50 nm. Calculate the ground state energy E1 and the first excited state energy E2. Express your answers in eV. (me = 9.11 × 10−31 kg, ℏ = 1.055 × 10−34 J·s, 1 eV = 1.602 × 10−19 J)
En = n²π²ℏ²/(2meL²).
E1 = π²(1.055×10−34)² / [2 × 9.11×10−31 × (0.50×10−9)²]
≈ 2.41 × 10−19 J ≈ 1.51 eV.
E2 = 4 × E1 ≈ 6.02 eV. The n² scaling of energy levels is a signature of confinement.
An electron is localized in a region Δx = 0.10 nm. What is the minimum uncertainty in its momentum, and what kinetic energy does this imply? Express KE in eV.
Δpmin = ℏ/(2Δx) = (1.055×10−34)/(2 × 0.10×10−9) = 5.28 × 10−25 kg·m/s.
KEmin = (Δp)²/(2me) = (5.28×10−25)² / (2 × 9.11×10−31)
≈ 1.53 × 10−19 J ≈ 0.95 eV. This zero-point kinetic energy is why electrons cannot be at rest inside atoms.
A neutron moves with kinetic energy KE = 0.025 eV (thermal neutron energy). Calculate its de Broglie wavelength. Compare this to the typical atomic spacing of ~0.2 nm. (mn = 1.675 × 10−27 kg)
KE = 0.025 × 1.602×10−19 = 4.0 × 10−21 J.
p = √(2mn × KE) = √(2 × 1.675×10−27 × 4.0×10−21) = 3.66 × 10−24 kg·m/s.
λ = h/p = 6.626×10−34 / 3.66×10−24 ≈ 1.81 × 10−10 m = 0.181 nm.
This is comparable to interatomic spacing — which is exactly why thermal neutrons produce diffraction patterns in crystals (neutron crystallography).
Nuclear Physics
Nuclear physics addresses the structure of atomic nuclei, the forces that bind protons and neutrons together, and the reactions — fission, fusion, and decay — that release enormous amounts of energy. This is the subject area where Dr. Preston's research expertise is deepest, having studied it at Berkeley, AFIT, LLNL, and LBNL.
Key Concepts
- Nuclear binding energy and the mass defect: BE = (Δm)c²
- Semi-empirical mass formula (liquid drop model)
- Types of radioactive decay: alpha, beta, gamma
- Radioactive decay law: N(t) = N0e−λt
- Half-life: t1/2 = ln(2)/λ
- Q-value: energy released in a nuclear reaction
- Nuclear fission: chain reactions, criticality, multiplication factor k
- Nuclear fusion: Coulomb barrier, stellar nucleosynthesis
Common Student Mistakes
- Mass defect direction: the bound nucleus is less massive than its constituents, not more — that missing mass is the binding energy
- Incorrect decay chain notation: always check that both mass number A and atomic number Z balance on both sides
- Confusing activity (decays per second) with number of nuclei N — activity A = λN, not N itself
- Using mass in kg instead of atomic mass units (u) when calculating Q-values; 1 u = 931.5 MeV/c²
A sample initially contains 8.0 × 1012 atoms of ¹³&sup7;Cs, which has a half-life of 30.1 years. (a) What is the decay constant λ? (b) How many atoms remain after 90.3 years? (c) What is the initial activity in Bq?
(a) λ = ln(2)/t1/2 = 0.693/(30.1 yr × 3.156×107 s/yr) = 7.29 × 10−10 s−1.
(b) 90.3 years = 3 × t1/2, so N = N0/2³ = 8.0×1012/8 = 1.0 × 1012 atoms.
(c) Activity A0 = λN0 = 7.29×10−10 × 8.0×1012 ≈ 5,830 Bq.
Calculate the Q-value for the alpha decay: 238U → 234Th + 4He. Atomic masses: M(238U) = 238.050788 u, M(234Th) = 234.043601 u, M(4He) = 4.002602 u. (1 u = 931.5 MeV/c²)
Δm = Mparent − (MTh + MHe)
= 238.050788 − (234.043601 + 4.002602)
= 238.050788 − 238.046203 = 0.004585 u.
Q = Δm × 931.5 MeV/u = 0.004585 × 931.5 ≈ 4.27 MeV.
This energy appears primarily as kinetic energy of the alpha particle, which is why uranium alpha particles carry ~4.2 MeV.
Calculate the binding energy per nucleon for 4He (the alpha particle). Proton mass: 1.007276 u, neutron mass: 1.008665 u, 4He nuclear mass: 4.001506 u. (1 u = 931.5 MeV/c²)
4He has Z=2 protons and N=2 neutrons.
Mass of free nucleons: 2(1.007276) + 2(1.008665) = 2.014552 + 2.017330 = 4.031882 u.
Mass defect: Δm = 4.031882 − 4.001506 = 0.030376 u.
Total BE = 0.030376 × 931.5 = 28.3 MeV.
BE per nucleon = 28.3/4 = 7.07 MeV/nucleon. This is one of the highest BE/A values for light nuclei, explaining the stability of alpha particles and their prevalence in radioactive decay.
Modern Physics
Modern physics encompasses the revolutionary discoveries of the early twentieth century: special relativity, which redefined space and time, and the quantum hypothesis, which revealed the discrete nature of energy and light. These are not abstract ideas — they underpin GPS satellites, nuclear reactors, lasers, and semiconductor devices.
Key Concepts
- Special relativity postulates: constant speed of light, equivalence of inertial frames
- Lorentz factor: γ = 1/√(1 − v²/c²)
- Time dilation: Δt = γΔt0
- Length contraction: L = L0/γ
- Mass-energy equivalence: E = γmc², rest energy E0 = mc²
- Photoelectric effect: photon energy E = hf, work function φ
- Compton scattering: photon wavelength shift upon scattering from electrons
- Bohr model of hydrogen and its limitations
Common Student Mistakes
- Confusing rest mass m with relativistic mass γm — modern physics uses invariant rest mass; the term "relativistic mass" is largely deprecated
- Lorentz factor direction: γ ≥ 1 always, moving clocks run slow, moving lengths contract
- Applying non-relativistic KE = ½mv² at speeds v > 0.1c — use KE = (γ − 1)mc²
- Photoelectric effect: intensity does not increase photon energy — only frequency does
A muon moving at v = 0.98c is created in Earth's upper atmosphere 60 km above the surface. In its own rest frame, the muon's proper lifetime is 2.2 μs. (a) What is the Lorentz factor γ? (b) What is the muon's lifetime in the Earth frame? (c) Does the muon reach Earth's surface?
(a) γ = 1/√(1 − 0.98²) = 1/√(0.0396) ≈ 5.03.
(b) Δt = γΔt0 = 5.03 × 2.2 μs ≈ 11.1 μs.
(c) Distance traveled = v × Δt = 0.98 × 3×108 m/s × 11.1×10−6 s ≈ 3.26 km. This is well short of 60 km, so the muon does not reach the surface from that altitude — but muons actually created around 10–15 km altitude do reach the ground. This is the classic experimental confirmation of time dilation.
Light of wavelength 220 nm strikes a metal surface with work function φ = 4.0 eV. (a) What is the photon energy in eV? (b) What is the maximum kinetic energy of the ejected electrons? (c) What is the stopping potential? (h = 6.626 × 10−34 J·s, c = 3×108 m/s)
(a) Ephoton = hc/λ = (6.626×10−34 × 3×108)/(220×10−9) = 9.04×10−19 J ≈ 5.64 eV.
(b) KEmax = Ephoton − φ = 5.64 − 4.0 = 1.64 eV.
(c) Stopping potential Vs = KEmax/e = 1.64 V. Electrons with exactly this KE are stopped by the retarding potential.
A proton (rest mass energy 938.3 MeV) is accelerated to kinetic energy KE = 2000 MeV in a particle accelerator. (a) What is its total energy? (b) What is its Lorentz factor γ? (c) What fraction of the speed of light does it travel at?
(a) Etotal = KE + mpc² = 2000 + 938.3 = 2938.3 MeV.
(b) γ = Etotal/(mpc²) = 2938.3/938.3 ≈ 3.13.
(c) γ = 1/√(1−v²/c²) → v/c = √(1 − 1/γ²) = √(1 − 1/9.80) = √(0.8980) ≈ 0.948c. At this energy the proton travels at 94.8% of the speed of light — a regime where non-relativistic approximations completely fail.
Textbooks Dr. Preston Recommends
These are the actual textbooks used in serious undergraduate and graduate physics programs. Every title below has been field-tested in courses, research environments, or both.
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