# A-Level Physics & Mathematics: Mechanics, Electric/Magnetic/Gravitational Fields, Nuclear/Particle Physics & Calculus Integration

> High-yield A-Level Physics and Mathematics across AQA, OCR, and Edexcel: particle physics (quarks/leptons), photoelectric effect, projectile motion, Young modulus, Simple Harmonic Motion, gravitational/electric/magnetic fields, capacitor RC decay, nuclear binding energy, and calculus integration.

- **Canonical URL:** https://flashcards.world/flashcards/sets/50dea812-35bc-472f-a40f-147a2ea56e77/
- **Markdown URL:** https://flashcards.world/flashcards/sets/50dea812-35bc-472f-a40f-147a2ea56e77.md
- **Level:** advanced
- **Languages:** en → en
- **Total Cards:** 32

## Flashcards (32 Cards)

| # | Front (Term / Prompt) | Back (Definition / Answer) | Example | Multiple-Choice Options |
| :--- | :--- | :--- | :--- | :--- |
| 1 | How do you recognize and perform Reverse Chain Rule integration of the form ∫ [f'(x) / f(x)] dx and ∫ f'(x)[f(x)]^n dx? | 1) When the numerator is the derivative of the denominator: ∫ [f'(x) / f(x)] dx = ln\|f(x)\| + C (e.g., ∫ tan(x) dx = ∫ [sin(x)/cos(x)] dx = −ln\|cos(x)\| + C = ln\|sec(x)\| + C). 2) When a function is multiplied by its derivative: ∫ f'(x)[f(x)]^n dx = [f(x)]^(n+1) / (n + 1) + C (for n ≠ −1). |  |  |
| 2 | Explain Einstein's Photoelectric Effect equation (hf = ϕ + Ek(max)) and why classical wave theory could not explain the threshold frequency. | In classical wave theory, light of any frequency should eventually transfer enough energy to eject photoelectrons if bright enough. In reality, no electrons are emitted below a threshold frequency f₀ regardless of intensity, and emission is instantaneous because each electron absorbs a single discrete photon of energy E = hf in a one-to-one interaction. Work function ϕ = hf₀ is the minimum energy to liberate a surface electron; remaining energy becomes maximum kinetic energy: hf = ϕ + ½mv²_max (stopping potential: eVs = Ek(max)). |  |  |
| 3 | State the defining equation for Simple Harmonic Motion (SHM) and the equations for displacement, velocity, maximum acceleration, and period of a mass-spring system and simple pendulum. | Definition of SHM: acceleration is directly proportional to displacement from the equilibrium position and always directed towards the equilibrium position: a = −ω²x (where ω = 2πf = 2π/T). If x = A cos(ωt), then v = ±ω√(A² − x²), v_max = ωA (at equilibrium x=0), and a_max = ω²A (at amplitude x=±A). Mass-spring period: T = 2π√(m / k). Simple pendulum period: T = 2π√(L / g). |  |  |
| 4 | State the Trapezium Rule for numerical integration and explain how to tell if it gives an overestimate or underestimate. | ∫_a^b y dx ≈ ½h [ (y₀ + y_n) + 2(y₁ + y₂ + ... + y_{n−1}) ], where h = (b − a) / n is the strip width for n strips (and n + 1 ordinates). If the curve is concave upwards (d²y/dx² > 0, U-shaped), the chords of the trapezia lie above the curve → OVERESTIMATE. If the curve is concave downwards (d²y/dx² < 0, ∩-shaped), the chords lie below the curve → UNDERESTIMATE. |  |  |
| 5 | In A-Level Pure Mathematics, state the Binomial Expansion formula for (1 + x)^n when n is negative or fractional, and its validity condition. | (1 + x)^n = 1 + nx + [n(n − 1) / 2!]x² + [n(n − 1)(n − 2) / 3!]x³ + ..., which is valid ONLY when \|x\| < 1. For (a + bx)^n, first factor out a^n to write it in the form a^n (1 + (b/a)x)^n, which is valid for \|(b/a)x\| < 1, i.e., \|x\| < \|a/b\|. |  |  |
| 6 | State the Radioactive Decay Law (A = λN and N = N₀e^(−λt)) and explain why the nuclear radius obeys R = R₀A^(1/3). | Activity: A = −ΔN/Δt = λN (in Becquerels, Bq), where λ = ln 2 / t₁/₂ is the decay constant (s⁻¹), and N(t) = N₀ e^(−λt). High-energy electron diffraction shows nuclear radius R = R₀ A^(1/3) (with R₀ ≈ 1.05–1.2 fm). Since volume V = (4/3)πR³ = (4/3)πR₀³A and mass m = Au, nuclear density ρ = m / V = 3u / (4πR₀³) ≈ 2 × 10¹⁷ kg m⁻³ is constant and independent of mass number A. |  |  |
| 7 | Explain the Newton–Raphson numerical method (x_{n+1} = x_n − f(x_n)/f'(x_n)) and when it fails. | The Newton–Raphson formula x_{n+1} = x_n − f(x_n) / f'(x_n) finds successively closer approximations to a root of f(x) = 0 by finding where the tangent line at (x_n, f(x_n)) crosses the x-axis. It fails (dividing by zero or diverging wildly) if the starting value x₀ is chosen at or very close to a stationary point (turning point) where the gradient f'(x₀) = 0. |  |  |
| 8 | How do you solve 2D Projectile Motion problems using the SUVAT equations? | Resolve initial velocity u at angle θ above the horizontal into independent perpendicular components: Horizontal velocity u_x = u cos θ (constant velocity since horizontal air resistance = 0, so horizontal range x = (u cos θ)t). Vertical velocity u_y = u sin θ (experiences constant downward acceleration a_y = −g = −9.81 m s⁻²). At maximum height, v_y = 0, so v_y² = u_y² − 2gh_max gives h_max = (u² sin² θ) / (2g). |  |  |
| 9 | In A-Level Statistics, compare the Binomial Distribution B(n, p) and the Normal Distribution N(μ, σ²), and show how to standardize to Z. | Binomial X ~ B(n, p): discrete count of successes in n independent trials with constant probability p (mean = np, variance = np(1−p)). Normal X ~ N(μ, σ²): continuous symmetrical bell-shaped distribution where ~68% lies within μ ± σ, ~95% within μ ± 2σ, and ~99.7% within μ ± 3σ. Standardisation to Z ~ N(0, 1²): Z = (X − μ) / σ; for a sample mean X̄ of size n in hypothesis testing: Z = (X̄ − μ) / (σ / √n). |  |  |
| 10 | Define Tensile Stress (σ), Tensile Strain (ε), and the Young Modulus (E), and state what the area under a force–extension vs stress–strain graph represents. | Tensile Stress: σ = F / A (in Pa or N m⁻², where A = πd²/4 is cross-sectional area). Tensile Strain: ε = ΔL / L (dimensionless ratio). Young Modulus: E = σ / ε = (FL) / (AΔL) (in Pa, equal to the gradient of the linear proportional region of a stress–strain graph). Area under a Force–Extension graph = Elastic strain energy stored (E = ½FΔL); Area under a Stress–Strain graph = Elastic strain energy stored per unit volume (½σε). |  |  |
| 11 | In A-Level Mechanics, how do you resolve forces on an inclined plane at angle θ with coefficient of friction μ? | Resolve weight mg into components parallel and perpendicular to the slope: Parallel down the slope = mg sin θ; Perpendicular into the slope = mg cos θ. Normal reaction R = mg cos θ (if no other force acts at an angle to the slope). When the block is on the point of slipping (limiting equilibrium) or moving, friction acts opposite to motion with maximum value F_max = μR = μ mg cos θ. |  |  |
| 12 | State the de Broglie wavelength equation (λ = h / p) and describe the experimental evidence for wave–particle duality of electrons. | λ = h / p = h / (mv), where h = Planck's constant (6.63 × 10⁻³⁴ J s) and mv = momentum. Evidence for wave nature of electrons: when a beam of accelerated electrons (where eV = ½mv²) passes through a thin polycrystalline graphite film, it diffracts through the atomic lattice spacings and forms concentric bright and dark interference rings on a fluorescent screen. Increasing accelerating voltage increases electron speed, decreasing λ and making the diffraction rings smaller. |  |  |
| 13 | How do stationary (standing) waves form on a stretched string, and what is the equation for the first harmonic (fundamental) frequency? | A stationary wave is formed by the superposition of two progressive waves of the same frequency, wavelength, and amplitude travelling in opposite directions. Nodes are points of zero amplitude (destructive interference, phase difference π rad); antinodes are points of maximum amplitude (constructive interference). For the first harmonic on a string of length L (with nodes at both ends, L = λ/2): f₁ = (1 / 2L) √(T / μ), where T is tension (N) and μ is mass per unit length (kg m⁻¹). |  |  |
| 14 | State the magnetic force on a moving charged particle (F = BQv) and derive the radius of its circular path in a uniform magnetic field. | A charge Q moving at speed v perpendicular to a magnetic field of flux density B experiences a magnetic force F = BQv perpendicular to both v and B (by Fleming's Left-Hand Rule). Because the force is always perpendicular to the velocity, it does no work (speed and Ek stay constant) and acts as a centripetal force: BQv = mv² / r → r = mv / (BQ). |  |  |
| 15 | Define Specific Charge of a nucleus or ion, and compare the quark compositions of a proton, neutron, and pion. | Specific charge = Total Charge ÷ Total Mass (Q / m, in C kg⁻¹). Hadrons experience the strong nuclear force and are subdivided into Baryons (3 quarks, baryon number B = +1) and Mesons (1 quark + 1 antiquark, B = 0). Proton = uud (+⅔ + ⅔ − ⅓ = +1); Neutron = udd (+⅔ − ⅓ − ⅓ = 0). Pions (π⁺ = u d̄, π⁻ = d ū, π⁰ = u ū or d d̄) are the exchange particles of the strong nuclear force; W⁺/W⁻ bosons mediate the weak nuclear force (β decay and electron capture). |  |  |
| 16 | State the Integration by Parts formula and the 'LATE' rule for choosing u (especially for ∫ ln(x) dx and ∫ x e^x dx). | ∫ u (dv/dx) dx = u v − ∫ v (du/dx) dx. Choose u using the priority acronym LATE: Logarithms (ln x) → Algebraic powers (x^n) → Trigonometric → Exponential (e^x). To integrate ∫ ln(x) dx, write it as ∫ (ln x) × 1 dx with u = ln(x) and dv/dx = 1, giving x ln(x) − ∫ x(1/x) dx = x ln(x) − x + C. |  |  |
| 17 | Define Mass Defect (Δm) and Nuclear Binding Energy, and explain why both nuclear fission of U-235 and nuclear fusion of H-2/H-3 release energy. | Mass defect Δm is the difference between the total mass of the separated nucleons (Z m_p + (A−Z) m_n) and the mass of the nucleus; Binding Energy = Δm c² (or Δm in u × 931.5 MeV). The curve of Binding Energy per nucleon vs Mass Number A peaks at Iron-56 (~8.8 MeV/nucleon, the most stable nucleus). Both fusion of light nuclei (A < 56) and fission of heavy nuclei (A > 56) move products UP the curve towards higher binding energy per nucleon, releasing the difference as kinetic energy/photons. |  |  |
| 18 | State the equations for Centripetal Acceleration and Centripetal Force, and explain the forces on a vehicle over a curved bridge or banked track. | Angular speed: ω = v / r = 2πf (rad s⁻¹). Centripetal acceleration: a = v² / r = ω²r (directed towards the centre of the circle). Centripetal force: F = mv² / r = mω²r. Over the top of a convex humpback bridge of radius r: mg − R = mv² / r, so the vehicle loses contact with the road (R = 0) when v ≥ √(gr). |  |  |
| 19 | How do you decompose a rational function into Partial Fractions for integration or binomial expansion? | 1) Distinct linear factors: (px + q) / [(x − a)(x − b)] ≡ A / (x − a) + B / (x − b). 2) Repeated linear factor: (px² + qx + r) / [(x − a)(x − b)²] ≡ A / (x − a) + B / (x − b) + C / (x − b)². (If the degree of the numerator ≥ degree of the denominator, perform polynomial long division FIRST). |  |  |
| 20 | State the Differentiation from First Principles formula and the derivatives of e^(kx), ln(x), sin(x), cos(x), and tan(x) (in radians). | First principles: f'(x) = lim(h→0) [f(x + h) − f(x)] / h. Standard derivatives (x in radians!): d/dx[e^(kx)] = k e^(kx); d/dx[a^x] = a^x ln(a); d/dx[ln(x)] = 1 / x; d/dx[sin(kx)] = k cos(kx); d/dx[cos(kx)] = −k sin(kx); d/dx[tan(kx)] = k sec²(kx). |  |  |
| 21 | How do you solve connected particles problems (pulleys and tow-bars) and variable acceleration problems in A-Level Mechanics? | For connected particles (inextensible string = same acceleration a and same tension T throughout; smooth pulley = no friction): write Newton's Second Law (F_net = ma) separately for each particle in its direction of motion and add the two simultaneous equations to eliminate tension T. For variable acceleration (when a is a function of time t): use calculus (r → differentiate → v = dr/dt → differentiate → a = dv/dt; reverse by integrating with +C). |  |  |
| 22 | Define Resistivity (ρ), state the equation for the terminal potential difference of a cell with internal resistance r, and explain how to find ε and r from a V–I graph. | Resistivity: ρ = RA / L (in Ω m). For a battery with Electromotive Force (EMF, ε) and internal resistance r connected to external load R: ε = I(R + r) = V + Ir, which rearranges to V = −rI + ε. Plotting terminal p.d. (V) on the y-axis against current (I) on the x-axis gives a straight line with y-intercept = ε (EMF) and gradient = −r. |  |  |
| 23 | Compare Newton's Law of Gravitation, Gravitational Field Strength (g), Gravitational Potential (V), and Kepler's Third Law for circular orbits. | Gravitational Force: F = Gm₁m₂ / r². Radial Gravitational Field Strength: g = F / m = GM / r² (N kg⁻¹). Gravitational Potential (work done per unit mass to move from infinity to a point, always negative): V = −GM / r (J kg⁻¹; Escape velocity v_esc = √(2GM / R)); also g = −ΔV / Δr. Equating gravitational force to centripetal force (GMm / r² = mv² / r where v = 2πr / T) gives Kepler's 3rd Law: T² = (4π² / GM) r³. |  |  |
| 24 | How do you solve a First-Order Separable Differential Equation of the form dy/dx = f(x)g(y)? | Separate the variables by collecting all y terms with dy on the left and all x terms with dx on the right: ∫ [1 / g(y)] dy = ∫ f(x) dx + C. Substitute the boundary condition (x₀, y₀) to find the constant of integration C (or ln A = C) and rearrange for the particular solution y = f(x). |  |  |
| 25 | State the Three Ideal Gas Laws, the Ideal Gas Equation (pV = NkT), and the Kinetic Theory equation (pV = ⅓Nm⟨c²⟩ = ⅔N Ek). | pV = nRT = NkT (where k = R / NA = 1.38 × 10⁻²³ J K⁻¹ is the Boltzmann constant). From kinetic theory of gases: pV = ⅓Nm(c_rms)², where c_rms is the root-mean-square speed. Combining these gives the average translational kinetic energy of a single gas molecule: ½m(c_rms)² = (3/2)kT = (3RT) / (2NA), proving that average molecular kinetic energy is directly proportional ONLY to absolute temperature T in Kelvin. |  |  |
| 26 | State the equations for Young's Double-Slit Interference (w = λD / s) and a Diffraction Grating (d sin θ = nλ). | Double-slit fringe spacing: w = λD / s (where w = distance between adjacent bright fringes, λ = wavelength, D = distance from slits to screen, s = slit separation, valid when s ≪ D). Diffraction grating: d sin θ = nλ (where d = 1 / N is the slit spacing in metres for a grating with N lines per metre, θ = angle of diffraction, and n = order of maximum; maximum order is found by setting sin θ = 1 and rounding n down to the nearest integer). |  |  |
| 27 | State Faraday's Law and Lenz's Law of Electromagnetic Induction (ε = −N ΔΦ / Δt) and explain how Lenz's Law exemplifies conservation of energy. | Magnetic Flux: Φ = BA cos θ (in Webers, Wb); Flux Linkage: NΦ = BAN cos θ (Wb turns). Faraday's Law: The magnitude of induced EMF is directly proportional to the rate of change of magnetic flux linkage. Lenz's Law: The direction of the induced EMF/current is always such as to oppose the change in flux that produces it (ε = −N ΔΦ / Δt). Work must be done against the opposing magnetic force to move the magnet/coil, converting mechanical work into electrical energy. |  |  |
| 28 | Define Capacitance (C), state the energy stored in a capacitor, and give the RC exponential discharge equations and time constant τ = RC. | Capacitance: C = Q / V, and for parallel plates C = ε₀εᵣA / d (in Farads, F). Energy stored: E = ½QV = ½CV² = ½Q²/C. When a capacitor discharges through a resistor R: X(t) = X₀ e^(−t/RC), where X can be Charge Q, Voltage V, or Current I. The Time Constant τ = RC is the time taken for Q, V, or I to fall to 1/e (≈ 37%) of its initial value, and the half-life is t₁/₂ = RC ln 2 ≈ 0.693 RC. |  |  |
| 29 | Compare Coulomb's Law, Electric Field Strength (E) in radial vs uniform fields, and Electric Potential (V). | Coulomb's Law in free space: F = Q₁Q₂ / (4πε₀r²). Radial Electric Field Strength: E = F / q = Q / (4πε₀r²) (N C⁻¹ or V m⁻¹). Uniform Electric Field between parallel plates: E = V / d (a charged particle enters horizontally and follows a parabolic trajectory, just like a projectile in uniform gravity). Radial Electric Potential: V = Q / (4πε₀r) (positive for +Q, negative for −Q). |  |  |
| 30 | State the key Trigonometric Identities required in A2 Mathematics: Pythagorean identities, Compound-Angle, Double-Angle, and R sin(x ± α). | Pythagorean: sin²x + cos²x ≡ 1; 1 + tan²x ≡ sec²x; 1 + cot²x ≡ cosec²x. Double-Angle: sin(2x) ≡ 2 sin(x) cos(x); cos(2x) ≡ cos²x − sin²x ≡ 2cos²x − 1 ≡ 1 − 2sin²x (rearranged to integrate sin²x = ½(1 − cos 2x) and cos²x = ½(1 + cos 2x)); tan(2x) ≡ 2tan(x) / (1 − tan²x). Harmonic form: a sin(x) ± b cos(x) ≡ R sin(x ± α), where R = √(a² + b²) and tan α = b / a. |  |  |
| 31 | State the Chain Rule, Product Rule, Quotient Rule, and Parametric Differentiation formula in A-Level Calculus. | Chain Rule: dy/dx = (dy/du) × (du/dx). Product Rule for y = u(x)v(x): dy/dx = u(dv/dx) + v(du/dx). Quotient Rule for y = u(x)/v(x): dy/dx = [v(du/dx) − u(dv/dx)] / v². Parametric differentiation for x = f(t), y = g(t): dy/dx = (dy/dt) ÷ (dx/dt). For implicit differentiation, differentiate y terms with respect to y and multiply by dy/dx. |  |  |
| 32 | State Snell's Law of Refraction (n₁ sin θ₁ = n₂ sin θ₂) and the two conditions required for Total Internal Reflection (TIR) in an optical fibre. | Refractive index of a medium: n = c / v. Snell's Law: n₁ sin θ₁ = n₂ sin θ₂. At the critical angle θc (where θ₂ = 90°): sin θc = n₂ / n₁ (for n₁ > n₂). Total Internal Reflection occurs when: 1) Light travels from a more optically dense core (higher n₁) towards a less optically dense cladding (lower n₂); AND 2) The angle of incidence θ₁ exceeds the critical angle θc. Step-index optical fibres use a narrow core and monochromatic light to minimise modal and material pulse dispersion. |  |  |

---
Study this deck interactively with FSRS-6 spaced repetition, audio review, and multiple-choice mode at [Flashcards World](https://flashcards.world/flashcards/sets/50dea812-35bc-472f-a40f-147a2ea56e77/).