Physics, step by step

Newton’s laws and conservation

• Community college
• Undergraduate (freshman–senior)

Track: Physics · About 25 min · Numbers verified in build: 2026-10-16

Step by step

1. $$\mathbf F=\frac{d\mathbf p}{dt},\qquad \mathbf p=m\mathbf v$$
   Second law in momentum form.
2. $$\mathbf F_{12}=-\mathbf F_{21}\;\Rightarrow\;\frac{d}{dt}(\mathbf p_1+\mathbf p_2)=0$$
   Third law: internal forces cancel, so total momentum is conserved.
3. $$W=\int\mathbf F\cdot d\mathbf r=\Delta\left(\tfrac12mv^2\right)$$
   Work–energy theorem.
4. $$v_1=\frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2},\qquad v_2=\frac{(m_2-m_1)u_2+2m_1u_1}{m_1+m_2}$$
   Solve momentum and kinetic-energy conservation together for an elastic collision.

m1, u1, m2, u2 = 2.0, 3.0, 1.0, 0.0
v1 = ((m1-m2)*u1 + 2*m2*u2) / (m1+m2)
v2 = ((m2-m1)*u2 + 2*m1*u1) / (m1+m2)
print(v1, v2, m1*v1 + m2*v2, 0.5*m1*v1**2 + 0.5*m2*v2**2)

Auto-checked practice

1. A 1 kg cart at 4 m/s hits a 3 kg cart at rest (elastic). Speed of the 3 kg cart afterwards, in m/s?

   m/s Check Show a hint
   Full solution

   v₂ = 2·1·4/4 = 2.0 m/s; the light cart bounces back at -2.0 m/s.

Practice

1. A ball falls 20 m from rest (no air). How fast is it moving at the bottom?

   Show solution

   mgh = ½mv², so v = √(2gh) = 19.81 m/s, independent of mass.

References

• D. Halliday, R. Resnick, J. Walker, Fundamentals of Physics, 12th ed., Wiley, 2021.
• J. R. Taylor, Classical Mechanics, University Science Books, 2005.
• MIT OpenCourseWare, 8.01 Classical Mechanics and 8.02 Electricity and Magnetism.

Projectiles with air drag (numerical ODE)

• Undergraduate (freshman–senior)
• Graduate

Track: Physics · About 40 min · Numbers verified in build: 2026-10-16

Step by step

1. $$m\dot{\mathbf v}=m\mathbf g-\tfrac12\rho C_DA\,|\mathbf v|\,\mathbf v$$
   Quadratic drag, opposite to the velocity.
2. $$\dot v_x=-k|\mathbf v|v_x,\quad \dot v_y=-g-k|\mathbf v|v_y,\qquad k=\frac{\rho C_DA}{2m}$$
   Coupled through |v|: no elementary closed form.
3. $$v_t=\sqrt{\frac{2mg}{\rho C_DA}}$$
   Terminal velocity: drag balances weight.
4. $$\mathbf s_{n+1}=\mathbf s_n+\tfrac{\Delta t}{6}(\mathbf k_1+2\mathbf k_2+2\mathbf k_3+\mathbf k_4)$$
   RK4: global error shrinks like Δt⁴.

import math
m, d, cd, rho, g = 0.145, 0.074, 0.35, 1.225, 9.80665
k = 0.5*rho*cd*math.pi*d*d/4/m
def f(s):
    x, y, vx, vy = s; v = math.hypot(vx, vy)
    return [vx, vy, -k*v*vx, -g - k*v*vy]
s, dt = [0, 0, 40*math.cos(math.radians(35)), 40*math.sin(math.radians(35))], 1e-3
while s[1] >= 0:
    a = f(s); b = f([s[i]+dt/2*a[i] for i in range(4)])
    c = f([s[i]+dt/2*b[i] for i in range(4)]); e = f([s[i]+dt*c[i] for i in range(4)])
    s = [s[i]+dt/6*(a[i]+2*b[i]+2*c[i]+e[i]) for i in range(4)]
print(s[0])

Auto-checked practice

1. Same ball and speed, launched at 45°: range with drag, in m? (Use the Physics Lab or the Python above.)

   m Check Show a hint
   Full solution

   R(45°) = 94.8 m, versus 93.9 m at 35° and 163.2 m in vacuum at 45°.

Practice

1. Why does the ball come down more steeply than it went up?

   Show solution

   Drag removes horizontal speed the whole time, while gravity keeps adding vertical speed on the way down, so the descent angle is steeper.

References

• J. R. Taylor, Classical Mechanics, University Science Books, 2005.
• W. H. Press et al., Numerical Recipes, 3rd ed., Cambridge University Press, 2007.
• F. M. White, Fluid Mechanics, 8th ed., McGraw-Hill, 2016.

Oscillators: simple, damped and eigenvalues

• Undergraduate (freshman–senior)
• Graduate

Track: Physics · About 35 min · Numbers verified in build: 2026-10-16

Step by step

1. $$m\ddot x+c\dot x+kx=0$$
   Newton’s law with a spring force and viscous damping.
2. $$x=e^{st}\;\Rightarrow\;s^2+2\zeta\omega_ns+\omega_n^2=0,\quad \omega_n=\sqrt{k/m},\ \zeta=\frac{c}{2\sqrt{km}}$$
   Characteristic equation.
3. $$s=-\zeta\omega_n\pm i\,\omega_n\sqrt{1-\zeta^2}\quad(\zeta<1)$$
   Underdamped: complex pair; real part = decay rate, imaginary part = damped frequency.
4. $$\dot{\mathbf x}=\begin{pmatrix}0&1\\-\omega_n^2&-2\zeta\omega_n\end{pmatrix}\mathbf x,\qquad Q=\frac{1}{2\zeta}$$
   Same thing as a first-order system: the eigenvalues of the matrix are the roots s. This is exactly the aircraft short-period analysis.

import numpy as np
m, c, k = 1.0, 2.0, 16.0
A = np.array([[0, 1], [-k/m, -c/m]])
s = np.linalg.eigvals(A)
wn = np.sqrt(k/m); z = c/(2*np.sqrt(k*m))
print(s, wn, z, 1/(2*z))

Auto-checked practice

1. m = 0.5 kg, k = 8 N/m, c = 1.2 N·s/m. Damping ratio ζ?

   Check Show a hint
   Full solution

   ζ = 1.2 / (2·2) = 0.30; ω_n = 4 rad/s.

Practice

1. What does critical damping (ζ = 1) look like, and where is it used?

   Show solution

   A double real root: the fastest return to rest without overshoot. Door closers and instrument needles are designed near it.

References

• J. R. Taylor, Classical Mechanics, University Science Books, 2005.
• D. Halliday, R. Resnick, J. Walker, Fundamentals of Physics, 12th ed., Wiley, 2021.

Lagrangian and Hamiltonian mechanics

• Graduate
• PhD / research

Track: Physics · About 45 min · Numbers verified in build: 2026-10-16

Step by step

1. $$S=\int L(q,\dot q,t)\,dt,\quad \delta S=0\;\Rightarrow\;\frac{d}{dt}\frac{\partial L}{\partial\dot q}-\frac{\partial L}{\partial q}=0$$
   Hamilton’s principle gives the Euler–Lagrange equation.
2. $$L=\tfrac12m\ell^2\dot\theta^2+mg\ell\cos\theta\;\Rightarrow\;\ddot\theta=-\frac{g}{\ell}\sin\theta$$
   Pendulum, using the angle as the coordinate.
3. $$T=4\sqrt{\frac{\ell}{g}}\,K(k),\quad k=\sin\frac{\theta_0}{2}\;\Longleftrightarrow\;T=\frac{2\pi\sqrt{\ell/g}}{\operatorname{AGM}(1,\cos\frac{\theta_0}{2})}$$
   Exact period from energy conservation; the arithmetic–geometric mean evaluates the elliptic integral K quickly.
4. $$H=\sum_i p_i\dot q_i-L,\quad \dot q=\frac{\partial H}{\partial p},\ \dot p=-\frac{\partial H}{\partial q};\qquad \frac{\partial L}{\partial q}=0\Rightarrow p=\text{const}$$
   Hamilton’s equations. Noether: an ignorable coordinate means a conserved momentum (e.g. angular momentum in a central force).

import math
def agm(a, b):
    while abs(a-b) > 1e-15: a, b = (a+b)/2, math.sqrt(a*b)
    return a
L, g, th0 = 1.0, 9.80665, math.radians(60)
T0 = 2*math.pi*math.sqrt(L/g)
print(T0, T0/agm(1, math.cos(th0/2)))

Auto-checked practice

1. Pendulum ℓ = 2 m released at 90°. Exact period, in s?

   s Check Show a hint
   Full solution

   T₀ = 2.8375 s; T = 3.3492 s (18 % longer).

Practice

1. In a central-force problem written in polar coordinates, which momentum is conserved and why?

   Show solution

   L does not depend on the angle φ (rotational symmetry), so p_φ = mr²φ̇, the angular momentum, is constant: this is Kepler’s second law.

References

• H. Goldstein, C. Poole, J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2002.
• J. R. Taylor, Classical Mechanics, University Science Books, 2005.

Electromagnetism, Maxwell’s equations and space plasma

• Undergraduate (freshman–senior)
• Graduate

Track: Physics · About 40 min · Numbers verified in build: 2026-10-16

Step by step

1. $$\nabla\!\cdot\mathbf E=\frac{\rho}{\varepsilon_0},\quad \nabla\!\cdot\mathbf B=0,\quad \nabla\!\times\mathbf E=-\frac{\partial\mathbf B}{\partial t},\quad \nabla\!\times\mathbf B=\mu_0\mathbf J+\mu_0\varepsilon_0\frac{\partial\mathbf E}{\partial t}$$
   Maxwell’s equations (SI).
2. $$\nabla^2\mathbf E=\mu_0\varepsilon_0\frac{\partial^2\mathbf E}{\partial t^2}\;\Rightarrow\;c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}$$
   In vacuum, take the curl of Faraday’s law: a wave equation with speed c.
3. $$m\dot{\mathbf v}=q(\mathbf E+\mathbf v\times\mathbf B)\;\Rightarrow\;r_g=\frac{mv_\perp}{|q|B},\quad \omega_g=\frac{|q|B}{m}$$
   Lorentz force: circular gyration perpendicular to B.
4. $$f_{pe}=\frac{1}{2\pi}\sqrt{\frac{n e^2}{\varepsilon_0 m_e}}$$
   Electron plasma frequency: radio waves below it cannot pass through the plasma.

import math
mp, e, me, eps0 = 1.67262192369e-27, 1.602176634e-19, 9.1093837015e-31, 8.8541878128e-12
v, B, n = 4.0e5, 5e-9, 5e6
rg = mp*v/(e*B); wg = e*B/mp
fpe = math.sqrt(n*e*e/(eps0*me))/(2*math.pi)
print(rg/1e3, 2*math.pi/wg, fpe/1e3)

Auto-checked practice

1. An electron at 1000 km/s perpendicular to a 30 μT field (roughly Earth’s surface field). Gyroradius, in m?

   m Check Show a hint
   Full solution

   r_g = 9.109e-31 · 1e6 / (1.602e-19 · 3e-5) = 0.190 m.

Practice

1. Why does ∇·B = 0 matter physically?

   Show solution

   There are no magnetic monopoles: field lines never start or end, so they close on themselves. Numerical plasma codes must preserve it or they produce false forces.

References

• D. J. Griffiths, Introduction to Electrodynamics, 4th ed., Cambridge University Press, 2017.
• F. F. Chen, Introduction to Plasma Physics and Controlled Fusion, 3rd ed., Springer, 2016.
• E. Tiesinga et al., “CODATA Recommended Values of the Fundamental Physical Constants: 2018”, Rev. Mod. Phys. 93, 025010 (2021).

Thermodynamics and statistical mechanics

• Community college
• Undergraduate (freshman–senior)
• Graduate

Track: Physics · About 35 min · Numbers verified in build: 2026-10-16

Step by step

1. $$f(v)=4\pi\left(\frac{m}{2\pi k_BT}\right)^{3/2}v^2e^{-mv^2/(2k_BT)}$$
   Maxwell–Boltzmann speed distribution.
2. $$\tfrac12m\langle v^2\rangle=\tfrac32k_BT\;\Rightarrow\;v_{rms}=\sqrt{\frac{3k_BT}{m}},\qquad \bar v=\sqrt{\frac{8k_BT}{\pi m}}$$
   Moments of the distribution (equipartition).
3. $$dU=\delta Q-\delta W,\qquad dS\ge\frac{\delta Q}{T}$$
   First law (energy) and second law (entropy).
4. $$\eta\le\eta_C=1-\frac{T_c}{T_h}$$
   No engine between two reservoirs beats Carnot.

import math
kB, NA = 1.380649e-23, 6.02214076e23
R, M, T = kB*NA, 0.0280134, 300.0
print(math.sqrt(3*R*T/M), math.sqrt(8*R*T/(math.pi*M)))

Auto-checked practice

1. RMS speed of oxygen O₂ (31.9988 g/mol) at 300 K, in m/s?

   m/s Check Show a hint
   Full solution

   v_rms = √(3RT/M) = 483.6 m/s (vs 516.8 m/s for N₂).

Practice

1. An engine runs between 600 K and 300 K. What is the most efficiency it can ever have?

   Show solution

   η_C = 1 − 300/600 = 0.50, i.e. 50 %. Real engines get noticeably less because of friction and finite-speed heat transfer.

References

• D. V. Schroeder, An Introduction to Thermal Physics, Addison-Wesley, 2000.
• D. Halliday, R. Resnick, J. Walker, Fundamentals of Physics, 12th ed., Wiley, 2021.
• E. Tiesinga et al., “CODATA Recommended Values of the Fundamental Physical Constants: 2018”, Rev. Mod. Phys. 93, 025010 (2021).

Fluids: Navier–Stokes and the Reynolds number

• Undergraduate (freshman–senior)
• Graduate
• PhD / research

Track: Physics · About 40 min · Numbers verified in build: 2026-10-16

Step by step

1. $$\rho\left(\frac{\partial\mathbf u}{\partial t}+\mathbf u\cdot\nabla\mathbf u\right)=-\nabla p+\mu\nabla^2\mathbf u,\qquad \nabla\!\cdot\mathbf u=0$$
   Incompressible Navier–Stokes and continuity.
2. $$\mathbf u^*=\frac{\mathbf u}{U},\ \mathbf x^*=\frac{\mathbf x}{L}\;\Rightarrow\;\ldots=-\nabla^*p^*+\frac{1}{Re}\nabla^{*2}\mathbf u^*,\quad Re=\frac{\rho UL}{\mu}$$
   Scaling leaves one parameter: the Reynolds number.
3. $$u(r)=\frac{\Delta p}{4\mu L}(R^2-r^2)\;\Rightarrow\;Q=\frac{\pi R^4\Delta p}{8\mu L}$$
   Steady pipe flow: an exact solution (Hagen–Poiseuille), valid while Re is below about 2300.
4. $$\mu(T)=\frac{1.458\times10^{-6}\,T^{3/2}}{T+110.4}\ \ \text{Pa·s}$$
   Sutherland’s law for air viscosity (as used in the 1976 Standard Atmosphere).

rho, U, L, T = 0.363918, 250.0, 5.0, 216.65
mu = 1.458e-6*T**1.5/(T + 110.4)
print(mu, rho*U*L/mu)

Auto-checked practice

1. Sea-level air (ρ = 1.225 kg/m³, T = 288.15 K) at 30 m/s over a 0.2 m chord model. Reynolds number? (Enter the full number, e.g. 123000.)

   Check Show a hint
   Full solution

   Re = 1.225 · 30 · 0.2 / 1.789×10⁻⁵ ≈ 410757: about 80 times smaller than the cruise wing.

Practice

1. Doubling a pipe’s radius at the same pressure drop multiplies the laminar flow rate by how much?

   Show solution

   Q ∝ R⁴, so by 16. Example: R = 0.5 mm, L = 10 cm, Δp = 1 kPa, water μ = 1.0 mPa·s gives Q = 0.245 mL/s.

References

• F. M. White, Fluid Mechanics, 8th ed., McGraw-Hill, 2016.
• J. D. Anderson Jr., Fundamentals of Aerodynamics, 6th ed., McGraw-Hill, 2017.

Relativity for orbits: GPS clocks and Mercury

• Graduate
• PhD / research

Track: Physics · About 45 min · Numbers verified in build: 2026-10-16

Step by step

1. $$\frac{d\tau}{dt}\approx1+\frac{\Phi}{c^2}-\frac{v^2}{2c^2},\qquad \Phi=-\frac{GM}{r}$$
   Weak-field clock rate: potential term (GR) plus velocity term (SR).
2. $$\left.\frac{\Delta f}{f}\right|_{GR}=\frac{GM}{c^2}\left(\frac1{R_\oplus}-\frac1a\right),\qquad \left.\frac{\Delta f}{f}\right|_{SR}=-\frac{GM}{2ac^2}$$
   Satellite vs ground, circular orbit v² = GM/a (ground clock motion neglected here).
3. $$\Delta\varpi=\frac{6\pi GM_\odot}{c^2a(1-e^2)}\ \text{per orbit}$$
   GR perihelion advance from the Schwarzschild orbit equation.

GM, c, R, a = 3.986004418e14, 299792458.0, 6378137.0, 26561750.0
gr = GM*(1/R - 1/a)/c**2*86400e6
sr = -GM/(2*a*c**2)*86400e6
print(gr, sr, gr + sr, (gr + sr)*1e-6*c/1e3)

Auto-checked practice

1. Relativistic perihelion advance of Venus (a = 1.0821×10¹¹ m, e = 0.006772, period 224.701 days), in arcseconds per century?

   ″/century Check Show a hint
   Full solution

   Δϖ ≈ 8.62″ per century for Venus, versus 42.98″ for Mercury (Will 2014 quotes 42.98″ for Mercury).

Practice

1. Why does GR change Mercury’s orbit more than Earth’s?

   Show solution

   The advance per orbit scales as 1/(a(1−e²)), and Mercury also completes about 4 times as many orbits per century, so the effect per century is much larger close to the Sun.

References

• N. Ashby, “Relativity in the Global Positioning System”, Living Reviews in Relativity 6, 1 (2003).
• C. M. Will, “The Confrontation between General Relativity and Experiment”, Living Reviews in Relativity 17, 4 (2014).
• E. Tiesinga et al., “CODATA Recommended Values of the Fundamental Physical Constants: 2018”, Rev. Mod. Phys. 93, 025010 (2021).

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