Unit 4, from zero

Momentum and collisions

Block 1≈11 marks a paper2D collision in 16 of 19elastic check in 11 of 19

Learn

• Momentum p = mv, in kg m s⁻¹ (the same unit as N s). It’s a vector, so pick a positive direction and give anything moving the other way a minus sign.
• Conservation: total momentum before = total momentum after, as long as no resultant external force acts (a “closed system”). This holds in every collision and explosion.
• Impulse: FΔt = Δp. The area under a force–time graph is the change in momentum.
• E_k = ½mv² = p²/2m, which rearranges to p = √(2mE_k).
• Elastic means total kinetic energy is also conserved. Inelastic means total kinetic energy falls, though momentum is still conserved. Objects that stick together are always inelastic.
• Resolving (needed for 2D collisions and banked tracks): a vector of size X at angle θ to a direction has a component X cos θ along that direction and X sin θ at 90° to it.

p = mvFΔt = ΔpE_k = p²/2mΣp_before = Σp_after

What they ask

• State the principle (2 marks). You need both halves: total momentum stays the same, and no resultant external force acts.
• 1D collision or explosion: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, with signs for direction.
• 2D collision. Resolve along the original direction (x) and at 90° to it (y). Along x, momentum before = the sum of the x-components after. Along y, 0 = one y-component minus the other. Alternatively, draw a scale vector triangle, tip to tail, where the momentum before is the resultant.
• “Is the collision elastic?” Work out total E_k before and total after (every object), compare them, and write the conclusion in words.
• Impulse. A ball rebounding at the same speed has Δp = 2mv. Crumple zones, rubber strips and air bags make Δt longer for the same Δp, so the force is smaller.
• Newton’s third law (helicopter, flyboard). It pushes air or water down, so the air or water pushes it up with an equal force. If that equals the weight, the resultant is zero and it hovers.

Traps

• Forgetting a minus sign for the object moving backwards.
• Leaving out one object’s kinetic energy “after” in the elastic check.
• Rounding mid-way through a 2D problem. Keep 3–4 significant figures until the end.
• Masses given in grams.

Drill: Jan 2023 Q14, Jan 2024 Q20, Jun 2024 Q15, Oct 2023 Q19

Circular motion

Block 1≈10 marks a paperbanked or conical in 9 of 19vertical circle in 8 of 19

Learn

• Angles in radians: θ (rad) = θ (°) × π/180. One revolution is 2π rad.
• Angular velocity ω = Δθ/Δt = 2π/T = 2πf in rad s⁻¹. From rpm: ω = rpm × 2π ÷ 60.
• v = ωr. The direction of v keeps changing, so there’s an acceleration towards the centre even at constant speed: a = v²/r = rω².
• The centripetal force F = mv²/r = mrω² is the resultant force towards the centre. It’s not an extra force: tension, friction, a contact force, gravity or a magnetic force provides it.

ω = 2π/Tv = ωra = v²/r = rω²F = mv²/rtan θ = v²/rg

What they ask

• Banked track, conical pendulum or banking plane. Resolve the tilted force (contact force N, tension T or lift L). Vertically N cos θ = mg; horizontally N sin θ = mv²/r. Dividing gives tan θ = v²/rg. For a pendulum of length l, r = l sin θ. Check whether θ is measured from the vertical or the horizontal.
• Vertical circle. At the top, N + mg = mv²/r, so the contact force is smallest there. The minimum speed is when N = 0, giving v = √(gr). At the bottom, N − mg = mv²/r, so the force is largest. Over a humpback bridge, mg − N = mv²/r.
• Friction provides the force (coin on a turntable, glass on a rotating tray). Maximum friction = mrω². Mass cancels, so heavy and light objects slide at the same radius.
• Derive a = v²/r (5 marks, 5 of 19 papers). Use the steps below.
• “I was thrown outwards.” Wrong. By Newton’s first law you keep moving in a straight line along the tangent, and the seat or door pushes you inwards to make you turn.

Traps

• Being given the diameter. Halve it.
• Revolutions per minute vs per second, and degrees left unconverted.
• Using g = 10 instead of 9.81.

Drill: Jan 2023 Q13, Jan 2024 Q17, Oct 2024 Q17(b), Jan 2022 Q15

Electric fields and potential

Block 2≈14 marks a paperfield lines to draw in 12 of 19E = V/d with F = EQ in 13 of 19

Learn

• An electric field is a region where a charged particle experiences a force. E = F/Q, in N C⁻¹ (or V m⁻¹). Its direction is the direction of the force on a positive charge.
• Point charge (radial field): E = kQ/r², where k = 1/4πε₀ = 8.99 × 10⁹. Force between two charges (Coulomb’s law): F = kQ₁Q₂/r². Potential: V = kQ/r, which is a scalar (sign matters) with r, not r².
• Between parallel plates (uniform field): E = V/d. The force F = EQ is the same everywhere between the plates.
• Work to move charge Q through p.d. V: W = QV. An electron accelerated through 1 V gains 1 eV = 1.60 × 10⁻¹⁹ J.
• Field and potential are linked. E is the potential gradient, E = −ΔV/Δr, so the gradient of a V–r graph gives E, and the area under an E–r graph gives the potential difference. Equipotentials bunch together where the field is strongest.
• A charged sphere acts like a point charge at its centre, so measure r from the centre.

E = F/QF = kQ₁Q₂/r²E = kQ/r²V = kQ/rE = V/dW = QV

What they ask

• Draw field lines (2–3 easy marks). For a point charge, draw at least 4 straight radial lines, evenly spaced, with arrows pointing out of + (into −). For plates, draw at least 3 parallel, evenly spaced lines touching both plates, with arrows from + to −. Between two charges, lines curve from + to −; two like charges have a neutral point between them. Equipotentials are always at 90° to field lines. Around a point charge they’re circles that get further apart.
• Coulomb, E and V calculations with nC, μC and cm (convert them). With two charges, fields are vectors: between a + and a − charge they add.
• Closest approach of an alpha to a nucleus. Set its kinetic energy (in J) equal to kQ_αQ_N/r, so r = kQ_αQ_N/E_k. The alpha is 2e and gold is 79e. They may then ask for F and a = F/m.
• A charged drop or sphere between plates. Use E = V/d and F = EQ, and compare with the weight. A stationary Millikan drop has EQ = mg, and the charges come out as whole-number multiples of e. For a hanging charged sphere, tan θ = F/mg.
• A particle fired between plates. Horizontal velocity is constant; vertical acceleration is a = EQ/m. Treat it like a projectile: time between the plates = length ÷ speed, then s = ½at².

Traps

• Squaring r in V, or forgetting to square it in F and E.
• Leaving cm or nC unconverted.
• Dropping the sign of a charge when adding potentials.
• Using a diameter as r.

Drill: Jun 2023 Q17, Jun 2024 Q17, Oct 2023 Q20, Oct 2023 Q15

Capacitors

Block 2≈10 marks a paperexponential calculation in 16 of 19W = ½CV² in 15 of 19

Learn

• A capacitor stores charge (equal and opposite on its plates) and energy. C = Q/V, measured in farads (F).
• Energy stored: W = ½QV = ½CV² = ½Q²/C, the area under a V–Q graph. The battery supplies QV, but only half is stored; the rest is dissipated in the resistance.
• Discharge through R: Q, V and I all fall exponentially, e.g. Q = Q₀e^−t/RC. In log form, ln V = ln V₀ − t/RC, so a graph of ln V against t is a straight line with gradient −1/RC.
• Time constant τ = RC (not on the formula sheet). After one RC, V falls to 37% when discharging or rises to 63% when charging. After about 5RC it’s essentially finished. Bigger R or C means slower.
• Charging (not on the sheet): V_C = V₀(1 − e^−t/RC). The current still decays as I = I₀e^−t/RC, and V_R + V_C = V₀ at every instant.

C = Q/VW = ½CV²Q = Q₀e^−t/RCln V = ln V₀ − t/RCτ = RCV = V₀(1 − e^−t/RC)

What they ask

• Find C or R from a graph (almost every paper). Pick one method: (a) read V₀, find 0.37V₀, and read that time, which is RC; (b) take two points and use the ln equation; (c) draw the tangent at t = 0, which meets the time axis at RC; (d) the gradient of a ln graph is −1/RC.
• Time to fall to a value: t = RC ln(V₀/V). “Show that” questions need every line of working.
• Q = CV, then W = ½CV². Energy dissipated = energy at the start − energy at the end. With a ±20% tolerance, use 1.2 × C for the maximum.
• Charge from an I–t graph = the area under it.
• Draw the circuit. Battery, resistor and capacitor in series, with a two-way switch so the capacitor can then discharge through the resistor. Put a voltmeter (or data logger) in parallel with the capacitor, or an ammeter in series for current. A data logger suits fast changes because it takes many readings a second.
• Sketch the graph. Draw exponential decay from I₀ = V₀/R (or from V₀) that never quite reaches zero, and mark 0.37 of the start value at t = RC. A charging V rises from 0 and levels off at V₀.
• Explain charging (the October 2024 6-marker). See the template in the 6-mark answers.

Traps

• μF and nF prefixes.
• Getting the ln ratio upside down, which gives a negative time.
• Using the supply voltage when the question means the capacitor’s voltage.
• Forgetting the ½ in the energy.

Drill: Jan 2024 Q18, Jun 2024 Q12, Jan 2023 Q16, Oct 2024 Q14

Magnetic forces and r = p/BQ

Block 3≈7 marks a paperr = p/BQ calculation in 13 of 19

Learn

• Force on a current-carrying wire: F = BIl sin θ, where θ is the angle between the wire and the field. B is in tesla (T).
• Fleming’s left-hand rule: First finger = Field (N to S), seCond finger = Current (conventional, + to −), thuMb = Motion (force). For electrons, point the current finger opposite to their motion.
• Force on a moving charge: F = Bqv sin θ. It’s always at 90° to the velocity, so it does no work: the speed stays constant and the path is a circle.
• Radius: setting Bqv = mv²/r gives r = mv/Bq = p/BQ. Faster or heavier means a bigger circle; a stronger field or bigger charge means a tighter one.
• The time for one circle, T = 2πm/Bq, doesn’t depend on speed. That’s why a cyclotron can run at a fixed frequency.

F = BIl sin θF = Bqv sin θr = p/BQv = E/B

What they ask

• Derive r = p/BQ (2 marks): equate Bqv = mv²/r, then substitute p = mv. Asked in June 2023 and June 2026.
• Use r = p/BQ with an energy in MeV. Convert to joules (MeV × 1.60 × 10⁻¹³), find p = √(2mE_k), then find r or B. An alpha has Q = 2e.
• Track direction. Use the left-hand rule to find the field direction (into or out of the page) or the sign of the charge. A spiral that tightens means the particle is losing energy, which tells you which way it was travelling.
• Velocity selector (undeflected beam): EQ = BQv, so v = E/B = V/Bd. Mass spectrometer: with the same v, B and Q, r ∝ m, so isotopes separate.
• Motor and balance setups. The change in the balance reading equals BIl. For a coil, moment = force × perpendicular distance, using the component if the coil is tilted.

Traps

• Leaving out sin θ when the field isn’t perpendicular.
• mT left unconverted.
• Using speed where the formula needs momentum.
• Forgetting that an alpha’s charge is 2e.

Drill: Jun 2023 Q12, Oct 2022 Q18, Jan 2023 Q17(a), Jan 2024 Q22(b)(ii)

EM induction

Block 3≈10 marks a paperexplain-in-words in 19 of 19e.m.f. calculation in 15 of 19

Learn

• Magnetic flux φ = BA, in webers (Wb), where A is the area at 90° to B. If the field is tilted, use the component of B perpendicular to the area. Flux linkage = Nφ.
• Faraday’s law: the induced e.m.f. equals the rate of change of flux linkage, ε = −d(Nφ)/dt.
• Lenz’s law (the minus sign): the induced e.m.f. or current always opposes the change that caused it. That’s energy conservation, because work has to be done to make the current.
• Flux linkage changes when you move a magnet or coil, change the current in a nearby coil (a.c.), rotate a coil, or move a rod through a field so it “cuts field lines”.

φ = BAflux linkage = Nφε = −d(Nφ)/dtε = Blv

What they ask

• Explain it in words, which every one of the 19 papers did. Use the induction chain in the 6-mark answers; for a 2–4 mark part, give the first 2–4 links of that chain.
• ε = NΔ(BA)/Δt. From a B–t graph, the steepest gradient gives the maximum e.m.f. A coil’s area is πr², with the radius in metres.
• Moving rod or aircraft wing. The area swept per second is l × v, so ε = Blv. Where the field has no vertical component (at the equator) a horizontal wing gets no e.m.f.
• Generator or rotating coil. The e.m.f. is sinusoidal and is largest when the coil’s plane is parallel to B, where the flux is zero but changing fastest. Spinning twice as fast doubles the peak e.m.f. and halves the period; power (∝ ε²) goes up four times.
• Magnet dropped through a coil. You get two pulses of opposite sign. The second is taller and narrower because the magnet is moving faster by then.
• Transformers, induction hobs, wireless chargers. Alternating current makes an alternating field, so the flux linking the second coil (or the pan) keeps changing and an e.m.f. is induced. Without an iron core, or with the coils far apart, not all the flux links. Only conductors heat on an induction hob.

Traps

• Forgetting the number of turns N.
• Using the diameter to find the coil area.
• Writing “the magnet cuts the field lines”. It’s the conductor that cuts them.
• Giving a direction that doesn’t oppose the change.

Drill: Jun 2023 Q16, Oct 2023 Q16 and Q17, Jan 2024 Q21, Jun 2024 Q18

Rutherford and nuclear notation

Block 4≈4 marks a paper6-marker 4 timesMCQ in most papers

Learn

• In the notation ^A_ZX, A is the nucleon number (protons + neutrons) and Z is the proton number. Neutrons: N = A − Z. Isotopes have the same Z but different N.
• Alpha decay: A falls by 4 and Z by 2 (an alpha is a helium nucleus, charge +2e). β⁻ decay: n → p + e⁻ + ν̄_e, so Z goes up by 1. β⁺ decay: p → n + e⁺ + ν_e, so Z goes down by 1.
• The experiment: alphas fired at thin gold foil in a vacuum. The vacuum stops air absorbing or deflecting the alphas; the thin foil means each alpha is scattered by at most one nucleus.
• What the results showed is the Rutherford template in the 6-mark answers. Learn it word for word.

What they ask

• Count neutrons or protons, or find a decay product. An MCQ in nearly every recent paper, and a free mark.
• “Which is NOT a valid conclusion?” The usual wrong options are “the nucleus contains protons and neutrons” (neutrons weren’t discovered until Chadwick in 1932) and “electrons orbit in shells”. Valid ones: the atom is mostly empty space, the nucleus is tiny, it holds most of the mass, and it is charged.
• Why alphas, not beta or gamma? Beta particles are far lighter, so the atom’s electrons deflect them, and they’re more penetrating. Gamma rays are uncharged, so there’s no electrostatic deflection at all.
• Closest approach to the nucleus. That calculation is in the electric fields section.

Drill: Jan 2024 Q14, Oct 2021 Q13, Jun 2022 Q16

Accelerators, electron beams and de Broglie

Block 4≈7 marks a paperlinac or cyclotron in 18 of 19none in June 2026

Learn

• Thermionic emission: a heated filament gives electrons enough energy to escape the metal. Accelerated through a p.d. V, they gain eV = ½mv².
• Electron diffraction shows electrons behave as waves, with λ = h/p (on the sheet). A higher accelerating voltage gives more momentum, a shorter wavelength and smaller rings.
• Why high energies are needed to probe nucleons: high energy means high momentum, so the de Broglie wavelength is short enough (about 10⁻¹⁵ m, the size of a nucleon) to resolve structure. Creating new heavy particles through ΔE = c²Δm is a different reason, and MCQs test that difference.
• Linac: a straight line of drift tubes on an a.c. supply. Particles speed up in the gaps between tubes.
• Cyclotron: two D-shaped “dees” with an a.c. p.d. across the gap and a magnetic field that bends the path. Particles spiral outwards.
• Colliding beams beat a fixed target. Their total momentum is zero, so all the kinetic energy can become the mass of new particles. With a fixed target the products must keep moving to conserve momentum, which wastes energy.

λ = h/peV = ½mv²f = Bq/2πm

What they ask

• Explain a linac or cyclotron. One or the other was in 18 of 19 papers, and it was the 6-marker 5 times. The templates are in the 6-mark answers.
• Cyclotron numbers. Frequency f = Bq/2πm; time for each half-circle πm/Bq; energy gained per gap crossing QV; number of crossings = total energy ÷ QV.
• Speed check. Work out v from ½mv². If it comes out above 3 × 10⁸ m s⁻¹, the particle must be relativistic and the simple formula has failed. The last tubes of a linac are equal in length for the same reason: v is close to c and barely increases.
• λ = h/p from an energy. Use p = √(2mE_k); if you’re given a voltage, E_k = eV first.

Drill: Jun 2024 Q16, Jun 2023 Q13, Jan 2024 Q22(b), Jun 2022 Q13

Particle physics

Block 4≈14 marks a paperconservation laws and MeV ↔ kg in every papermuon lifetime in 11 of 19

Learn

• Every quark has baryon number +⅓. Antiquarks have the opposite charge and baryon number −⅓. The symmetry of this table is what predicted the top quark as the bottom quark’s partner.
• Baryons are 3 quarks (B = +1): proton uud, neutron udd. Antibaryons are 3 antiquarks (B = −1).
• Mesons are a quark plus an antiquark (B = 0): π⁺ = u d̄, π⁻ = ū d, π⁰ = uū or dd̄, K⁺ = u s̄, K⁻ = ū s. Baryons and mesons are hadrons, and hadrons are not fundamental.
• Leptons are fundamental with lepton number L = +1: e⁻, μ⁻, τ⁻ and their neutrinos. Antileptons (e⁺, μ⁺, τ⁺, antineutrinos) have L = −1. Leptons have B = 0 and hadrons have L = 0.
• An antiparticle has the same mass, the opposite charge, and opposite baryon and lepton numbers.
• Every interaction conserves charge, baryon number, lepton number, energy and momentum.
• Units: 1 eV = 1.60 × 10⁻¹⁹ J, 1 MeV = 1.60 × 10⁻¹³ J, 1 GeV = 1.60 × 10⁻¹⁰ J. To turn a mass in MeV/c² into kg, multiply by 1.60 × 10⁻¹³ and divide by (3.00 × 10⁸)². A momentum in GeV/c becomes kg m s⁻¹ by multiplying by 1.60 × 10⁻¹⁰ and dividing by 3.00 × 10⁸.

ΔE = c²Δm1 MeV = 1.60 × 10⁻¹³ Jm = E/c²

What they ask

• The conservation table (every paper). Write the charge, B and L under each particle, add up each side, and state “conserved” or “not conserved, so not possible”. For a missing particle X, pick whatever balances all three. It’s often a neutrino or antineutrino (neutral, L = ±1).
• Quark structure of a given hadron from its charge, plus an s quark if it’s “strange”.
• MeV/c² ↔ kg, and “how many times heavier than an electron or proton”.
• Annihilation: particle + antiparticle → two photons (two, to conserve momentum). Each photon has E = mc² plus its share of any kinetic energy, then f = E/h.
• Pair production: photon → particle + antiparticle. The minimum photon energy is 2mc² (an electron–positron pair needs 1.02 MeV). The photon leaves no track, and the pair curve opposite ways with equal radii.
• Making heavy particles: the colliding particles’ kinetic energy becomes rest-mass energy. Total energy = rest energy + kinetic energy.
• Relativistic lifetime. Particles moving close to c experience time dilation, so the lifetime we measure is longer and they travel further before decaying. That’s why muons reach the ground. Show that the non-relativistic distance (speed × lifetime, about 650 m) is far too small.
• Reading tracks. Charged particles ionise and leave tracks; neutral ones leave gaps (a V appears where a neutral particle decays). The curve direction plus the field gives the sign of the charge, and a tightening spiral means the particle is losing energy.

Traps

• Mixing up MeV and GeV.
• Forgetting to square c.
• Saying annihilation makes one photon.
• Giving an antiparticle the same lepton or baryon number.

Drill: Oct 2023 Q21, Jun 2024 Q14, Jan 2023 Q11 and Q18, Oct 2024 Q13 and Q18