Mechanics, fields, thermal behavior, waves and modern models
Mechanics
What this domain tests
This mechanics review develops kinematics, Newton's laws, momentum, rigid-body equilibrium, circular motion, oscillations and gravitational ideas. State sign conventions, resolve vectors and justify the choice of force, energy or momentum methods. These are shared foundations rather than a claim about any current regional paper.
Kinematics and projectiles
Use the constant-acceleration equations only when acceleration is uniform. For a projectile, resolve the launch velocity into u cosθ (horizontal, constant) and u sinθ (vertical, changing at g). Time of flight comes from the vertical motion; the range comes from horizontal displacement over that time.
Forces, momentum and equilibrium
- Newton's second law uses the resultant force: F = ma. Weight W = mg acts at the centre of gravity.
- Total momentum p = mv is conserved separately in each direction when the net external impulse on the chosen system is zero; elastic collisions also conserve kinetic energy, inelastic ones do not.
- A body in equilibrium has ΣF = 0 and Σmoments about any point = 0. Take moments about the point with the most unknown forces so those unknowns vanish.
- For circular motion the radial resultant force is mv²/r = mω²r toward the centre. Speed is constant only in uniform circular motion; nonuniform motion also has tangential acceleration.
- SHM is defined by a = −ω²x, with T = 2π/ω; energy shuttles between kinetic and potential.
Worked example
In a uniform-gravity model with no air resistance, a ball leaves level ground at 25 m/s at 40° above the horizontal and lands at the same height. Find its range (g = 9.81 m/s²).
Vertical component: 25 sin40° = 16.07 m/s. Time to the top: 16.07/9.81 = 1.638 s, so total flight = 3.276 s. Horizontal component: 25 cos40° = 19.15 m/s. Range = 19.15 × 3.276 = 62.7 m. Cross-check with R = u² sin2θ/g = (625 × sin80°)/9.81 = 62.7 m. ✓
Traps that cost marks
- Swapping sin and cos: the component next to the angle uses cosine.
- Forgetting that at maximum height the vertical velocity is zero while the horizontal velocity is unchanged.
- Mixing g = 10 with data intended for 9.81, or dropping units mid-calculation.
- Sign errors - fix a positive direction and apply it to displacement, velocity and acceleration consistently.
- Calling circular motion "constant velocity", or omitting the direction of the centripetal force.
- Ignoring the mass of a rod or pulley, or treating a rough surface as smooth when friction is asked for.
Electricity and Magnetism
What this domain tests
Electric current, resistance and resistivity, EMF and internal resistance, DC networks, capacitors, electric fields, magnetic forces, electromagnetic induction and alternating current. This practice separates source EMF from terminal potential difference, applies circuit conventions consistently and explains induced-current directions without asserting a particular target paper structure.
Current, resistance and circuits
- I = nAve links current to drift velocity; V = IR; R = ρL/A, so resistance rises with length and falls with cross-sectional area.
- EMF is the energy converted per unit charge by the source; for a source delivering current, terminal p.d. is V = E − Ir in the stated internal-resistance model.
- Kirchhoff: ΣI = 0 at a junction; around any closed loop ΣEMF = Σp.d.
- A potential divider gives Vout = Vin × R₂/(R₁ + R₂) for ideal (high-impedance) meters.
Fields, capacitors and magnetism
- Uniform field: E = V/d; point charges obey Coulomb's law; field lines run from positive to negative.
- Capacitor: Q = CV, energy = ½CV² = ½QV; discharge follows exponential decay with time constant RC.
- Force on a current: F = BIL sinθ; on a moving charge: F = Bqv sinθ, always perpendicular to v.
- Flux Φ = BA cosθ; induced EMF = −N dΦ/dt; Lenz's law says the induced current opposes the change producing it.
- For sinusoidal AC: Irms = I₀/√2, Vrms = V₀/√2; transformers obey Vp/Vs = Np/Ns for an ideal transformer.
Worked example
A cell of EMF 12 V and internal resistance 0.5 Ω drives a 5.5 Ω resistor. Find the current, terminal p.d. and efficiency.
I = E/(R + r) = 12/6.0 = 2.0 A. Terminal p.d. = E − Ir = 12 − (2.0 × 0.5) = 11 V. Power in load = I²R = 4 × 5.5 = 22 W; internal power = I²r = 2 W. Efficiency = 22/24 = 91.7%.
Traps that cost marks
- Quoting EMF as the terminal p.d. when current flows.
- Saying current is "used up" around a circuit - energy is transferred, charge is not consumed.
- Using peak values where RMS is required, or the reverse.
- Getting Lenz's law backwards and predicting the wrong induced current direction.
- Writing capacitor energy as CV² instead of ½CV².
- Adding parallel resistances directly instead of using reciprocals.
Thermal Physics
What this domain tests
Thermometry, heat capacity and latent heat, methods of heat transfer, the kinetic theory of gases, ideal gas behaviour and the first law of thermodynamics. Questions test careful accounting of energy in mixtures and correct use of kelvin temperatures in gas calculations.
Heat and temperature
- T(K) = θ(°C) + 273.15. Always use kelvin in pV = nRT and in kinetic theory.
- Q = mcΔθ for a temperature change; Q = mL for a change of state at constant temperature.
- In a mixture, heat lost = heat gained, provided the container is insulated and its heat capacity is either included or negligible.
- A thermometric property needs a reproducible calibrated relationship with temperature over the working range; it need not be linear. Ice and steam points are traditional calibration examples, with pressure and other conditions specified.
- Heat transfer: conduction (lattice vibration and free electrons), convection (bulk fluid motion), thermal radiation (electromagnetic emission over a temperature-dependent spectrum, requiring no material medium and depending on surface properties).
Gases and thermodynamics
- Ideal gas law: pV = nRT, with R = 8.31 J mol⁻¹ K⁻¹.
- Kinetic theory: pV = ⅓Nm⟨c²⟩ and mean translational kinetic energy = (3/2)kT; for a fixed amount of an ideal gas, internal energy depends on temperature rather than pressure and volume independently.
- First law: ΔU = Q − W, where W is work done by the gas. Isothermal: ΔU = 0. Adiabatic: Q = 0. Isobaric: W = pΔV. Isochoric: W = 0.
Worked example
0.25 kg of water at 80 °C is mixed with 0.50 kg of water at 15 °C in an insulated vessel. Find the final temperature (c = 4200 J kg⁻¹ K⁻¹).
Heat lost = 0.25 × 4200 × (80 − θ); heat gained = 0.50 × 4200 × (θ − 15). Cancelling 4200: 0.25(80 − θ) = 0.50(θ − 15) 20 − 0.25θ = 0.50θ − 7.5 → 27.5 = 0.75θ → θ = 36.7 °C.
Traps that cost marks
- Substituting Celsius values into pV = nRT - a classic lost mark.
- Confusing specific heat capacity c with specific latent heat L, or using L during a temperature change.
- Ignoring heat absorbed by the calorimeter (use mcccΔθ when the material and mass are given).
- Forgetting to convert grams to kilograms, or cm³ to m³.
- Sign errors in the first law - state whether W is work done by or on the gas.
- Claiming internal energy of an ideal gas depends on pressure or volume.
Waves and Optics
What this domain tests
This domain covers wave properties, superposition, stationary waves, interference, diffraction, refraction, lenses and optical instruments. Practice relates path difference to phase, uses grating and lens equations consistently, and explains fringe patterns through superposition; the chapter does not reproduce a particular examination format.
Core wave relations
- v = fλ; intensity is proportional to amplitude²; polarisation is possible only for transverse waves.
- Phase difference = 2π × (path difference)/λ.
- Superposition: displacements add vectorially. Two identical waves travelling in opposite directions give a stationary wave with nodes every λ/2 and adjacent node–antinode spacing λ/4. No net energy is transported.
- Young's double slit: fringe spacing x = λD/a, where a is the slit separation and D the slit-to-screen distance.
- Diffraction grating: d sinθ = nλ, with d = 1/N and N the number of lines per metre.
Refraction and lenses
Snell's law n₁ sinθ₁ = n₂ sinθ₂; refractive index n = c/v; critical angle sinθc = n₂/n₁ for n₁ > n₂. The special form 1/n applies when the second medium has index approximately one. For lenses use 1/f = 1/u + 1/v with a consistent real-is-positive convention, and signed transverse magnification m = -v/u = image height/object height. For ordinary thin glass lenses in air, a convex lens converges; a concave lens gives a virtual diminished image for a real object.
Worked example
A grating has 500 lines per mm. Find the angle of the second-order maximum for light of wavelength 589 nm.
d = 1/(500 × 1000) = 2.0 × 10⁻⁶ m. sinθ = nλ/d = (2 × 589 × 10⁻⁹)/(2.0 × 10⁻⁶) = 0.589. θ = 36.1°.
Traps that cost marks
- Using degrees inside small-angle formulae that require radians.
- Confusing the order n in d sinθ = nλ with a refractive index.
- Mixing sign conventions in 1/f = 1/u + 1/v, then reporting a virtual image as real.
- Claiming a stationary wave transfers energy along the string.
- Confusing path difference with phase difference (a factor of 2π/λ).
- Forgetting that sinθ ≤ 1 sets a maximum observable order.
Modern Physics
What this domain tests
The photoelectric effect, photon energy, atomic energy levels and spectra, wave–particle duality, nuclear binding energy, radioactive decay and half-life. Questions test whether you keep energies in the correct unit (J or eV) and whether you can write balanced nuclear equations.
Photons and the photoelectric effect
- Photon energy: E = hf = hc/λ.
- Einstein's equation: hf = φ + KEmax, where φ is the work function and KEmax = eVs (Vs is the stopping potential).
- Threshold frequency f₀ = φ/h; below it, ordinary single-photon emission does not occur in the specified weak-illumination model.
- Increasing intensity increases the number of photoelectrons per second, not their maximum kinetic energy.
- Ordinary photoemission does not exhibit the long energy-accumulation delay predicted by the elementary classical picture; this supports the photon description.
Energy levels and nuclear physics
- Spectral lines arise from hf = E₂ − E₁; line spectra are evidence for discrete levels.
- de Broglie wavelength: λ = h/p; for a nonrelativistic massive particle, p = mv.
- Mass defect Δm gives binding energy E = Δmc²; 1 u c² is approximately 931.5 MeV.
- α decay: A falls by 4, Z falls by 2. β⁻ decay: Z rises by 1, A unchanged. γ emission changes neither A nor Z.
- Decay law: N = N₀e^(−λt); half-life t½ = ln2/λ; activity A = λN.
Worked example
Light of wavelength 400 nm falls on sodium (work function 2.28 eV). Find the maximum kinetic energy of the emitted electrons.
Photon energy: E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(4.00 × 10⁻⁷) = 4.97 × 10⁻¹⁹ J = 3.11 eV. KEmax = 3.11 − 2.28 = 0.83 eV (= 1.3 × 10⁻¹⁹ J). Since 0.83 eV > 0, emission occurs; the stopping potential is 0.83 V.
Traps that cost marks
- Claiming brighter light gives faster photoelectrons.
- Mixing eV and J in the same calculation - convert once, then stay consistent.
- Using the count of half-lives incorrectly (20 days with a 5-day half-life is four half-lives, not twenty).
- Changing the wrong number in a nuclear equation: mass number and proton number must both balance.
- Using p = mv outside its nonrelativistic regime, or confusing de Broglie momentum with the separate kinetic-energy expression mv²/2.
- Forgetting that γ emission accompanies α or β decay and does not change A or Z.