2025 CCEA A-Level Physics 1210 Practice Paper with Answers

(a) \( E_1 = mc_{ice}\Delta T = 0.40\times2100\times10 = 8400\ \text{J} \).
(b) \( E_2 = mL_f = 0.40\times3.34\times10^{5} = 1.336\times10^{5}\ \text{J} \) (133600 J).
(c) \( E_3 = mc_{water}\Delta T = 0.40\times4200\times25 = 42000\ \text{J} \). Total energy \( E = E_1+E_2+E_3 = 8400+133600+42000 = 1.84\times10^{5}\ \text{J} \).
(d) \( t = E/P = 1.84\times10^{5}/150 = 1227\ \text{s} \approx 1230\ \text{s} \) (about 20.5 minutes).
(e) In practice, some of the thermal energy supplied by the heater is lost to the surroundings by conduction, convection and radiation from the container and the ice/water itself, rather than all of it going into heating and melting the ice; the heater may also not convert 100% of its electrical energy input into useful thermal energy. Both effects mean more total energy — and therefore more time — is needed than the ideal calculated value, which represents a minimum.

(a) [1] correct formula \( E=mc\Delta T \); [1] correct substitution; [1] correct answer 8400 J with unit.
(b) [1] correct formula \( E=mL \); [1] correct substitution; [1] correct answer 1.336×10^5 J (accept 133600 J) with unit.
(c) [1] correct formula and substitution for the water-heating stage; [1] correct value 42000 J; [1] correct total of all three stages (allow error carried forward); [1] correct final total 1.84×10^5 J with unit.
(d) [1] correct use of \( t=E/P \); [1] correct substitution (allow ECF from (c)); [1] correct final value ≈1230 s (or equivalent in minutes) with unit.
(e) [1] identifies heat loss to the surroundings; [1] identifies heater inefficiency or a second distinct valid loss mechanism; [1] correctly links either/both to the calculated time being an underestimate/minimum.

(a) \( T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{0.50}{20}} = 2\pi\sqrt{0.025} = 2\pi \times 0.1581 = 0.993\ \text{s} \approx 1.0\ \text{s} \).
(b) Angular frequency \( \omega = \sqrt{k/m} = \sqrt{20/0.50} = \sqrt{40} = 6.32\ \text{rad s}^{-1} \). \( v_{max} = \omega A = 6.32\times0.12 = 0.76\ \text{m s}^{-1} \).
(c) \( a_{max} = \omega^2 A = 40\times0.12 = 4.8\ \text{m s}^{-2} \). This maximum acceleration occurs at maximum displacement (the extremes of the oscillation), where the spring's extension/compression, and hence the restoring force, is greatest.
(d) \( E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}\times20\times0.12^2 = \tfrac{1}{2}\times20\times0.0144 = 0.144\ \text{J} \).
(e) When KE = PE, each equals half the total energy: \( \tfrac{1}{2}kx^2 = E/2 = 0.072\ \text{J} \), so \( x^2 = \dfrac{2\times0.072}{20} = 0.0072 \), giving \( x = \pm0.0849\ \text{m} \approx \pm0.085\ \text{m} \) (equivalently \( x = A/\sqrt{2} \)).

(a) [1] correct formula \( T=2\pi\sqrt{m/k} \) quoted; [1] correct substitution shown in full; [1] concluding value ≈0.99 s clearly shown to justify 'approximately 1.0 s'.
(b) [1] correct value of \( \omega \) (or \( 2\pi/T \)); [1] correct use of \( v_{max}=\omega A \); [1] correct final value ≈0.76 m/s with unit.
(c) [1] correct formula \( a_{max}=\omega^2A \); [1] correct value 4.8 m/s² with unit; [1] correctly identifies maximum displacement/extremes of oscillation as the location; [1] correct physical reasoning (restoring force/extension greatest there).
(d) [1] correct formula \( E=\tfrac12kA^2 \); [1] correct substitution; [1] correct final value 0.144 J with unit.
(e) [1] correct statement that KE=PE=E/2; [1] correct rearrangement of \( \tfrac12kx^2=E/2 \); [1] correct final value ≈0.085 m (accept ± sign, or A/√2).

(a) \( T_{1/2} = 8.0\ \text{days} = 8.0\times24\times3600 = 6.912\times10^{5}\ \text{s} \). \( \lambda = \dfrac{\ln2}{T_{1/2}} = \dfrac{0.693}{6.912\times10^{5}} = 1.00\times10^{-6}\ \text{s}^{-1} \).
(b) \( A_0 = \lambda N_0 \Rightarrow N_0 = \dfrac{A_0}{\lambda} = \dfrac{6.4\times10^{10}}{1.00\times10^{-6}} = 6.4\times10^{16} \) nuclei.
(c) 24 days is exactly 3 half-lives (\( 24/8 = 3 \)). \( A = A_0\left(\tfrac12\right)^3 = 6.4\times10^{10}\times0.125 = 8.0\times10^{9}\ \text{Bq} \).
(d) Using \( A = A_0\left(\tfrac12\right)^{t/T_{1/2}} \): \( \dfrac{1.0\times10^9}{6.4\times10^{10}} = \left(\tfrac12\right)^{t/8} \Rightarrow \dfrac{1}{64} = \left(\tfrac12\right)^{t/8} \). Since \( \tfrac{1}{64} = \left(\tfrac12\right)^6 \), \( t/8 = 6 \Rightarrow t = 48\ \text{days} \).
(e) In beta-minus decay, a neutron within the nucleus is transformed into a proton, with the simultaneous emission of a fast-moving electron (the beta particle) and an electron antineutrino. Because a neutron becomes a proton, the atomic (proton) number of the nucleus increases by 1, while the mass number stays the same, since the total number of nucleons is unchanged.

(a) [1] correct conversion of half-life into seconds; [1] correct formula \( \lambda=\ln2/T_{1/2} \); [1] correct final value ≈1.0×10^-6 s⁻¹.
(b) [1] correct formula \( N_0=A_0/\lambda \); [1] correct substitution (allow ECF from (a)); [1] correct final value ≈6.4×10^16.
(c) [1] correct identification that 24 days = 3 half-lives; [1] correct use of \( (\tfrac12)^n \) or the exponential decay equation; [1] correct substitution; [1] correct final value 8.0×10^9 Bq.
(d) [1] correct decay equation set up with the given activity ratio; [1] correct identification that the ratio equals \( (\tfrac12)^6 \) (or equivalent logarithmic method); [1] correct number of half-lives (6); [1] correct final value 48 days.
(e) [1] correct statement that a neutron converts to a proton; [1] correct statement that a beta particle (electron) and antineutrino are emitted; [1] correct statement of the effect on atomic number (increases by 1) and mass number (unchanged).

(a) Total mass before: \( 2.014102+3.016049 = 5.030151\ \text{u} \). Total mass after: \( 4.002602+1.008665 = 5.011267\ \text{u} \). \( \Delta m = 5.030151-5.011267 = 0.018884\ \text{u} = 0.018884\times1.661\times10^{-27} = 3.14\times10^{-29}\ \text{kg} \).
(b) \( E = \Delta m c^2 = 3.14\times10^{-29}\times(3.00\times10^{8})^2 = 3.14\times10^{-29}\times9.00\times10^{16} = 2.82\times10^{-12}\ \text{J} \).
(c) \( E(\text{eV}) = \dfrac{2.82\times10^{-12}}{1.60\times10^{-19}} = 1.76\times10^{7}\ \text{eV} = 17.6\ \text{MeV} \).
(d) The binding energy per nucleon of light nuclei such as deuterium and tritium is lower than that of helium-4, which lies much closer to the peak of the binding-energy-per-nucleon curve. When the light nuclei fuse to form the more tightly bound helium-4 nucleus, the nucleons end up in a lower (more negative) energy, more stable configuration; the difference in total binding energy between the reactants and the more strongly-bound product is released as the kinetic energy of the helium nucleus and neutron.
(e) An extremely high temperature (of the order of \( 10^{8}\ \text{K} \)) is required, so that the nuclei have enough kinetic energy to overcome the electrostatic (Coulomb) repulsion between their positive charges and approach closely enough for the short-range strong nuclear force to bind them together.

(a) [1] correct total masses before and after; [1] correct mass defect in u; [1] correct conversion to kg with correct value ≈3.14×10^-29 kg.
(b) [1] correct formula \( E=\Delta mc^2 \); [1] correct substitution; [1] correct final value ≈2.82×10^-12 J.
(c) [1] correct method (division by 1.60×10^-19); [1] correct value in eV; [1] correctly expressed in MeV ≈17.6 MeV.
(d) [1] correct statement that helium-4 has higher/greater binding energy per nucleon than the reactants; [1] correct reference to the shape/peak of the binding-energy-per-nucleon curve or greater stability; [1] correct statement that the products are in a lower-energy/more stable configuration; [1] correct link to energy being released as kinetic energy of the products.
(e) [1] correctly identifies very high temperature as the condition; [1] correct reference to overcoming Coulomb/electrostatic repulsion; [1] correct reference to the strong nuclear force acting only at very short range once nuclei are close enough.

(a) On a flat track, the maximum available friction supplies the centripetal force: \( F = \dfrac{mv^2}{r} \Rightarrow v^2 = \dfrac{Fr}{m} = \dfrac{4800\times50}{1200} = 200 \Rightarrow v = \sqrt{200} = 14.1\ \text{m s}^{-1} \).
(b) \( \omega = \dfrac{v}{r} = \dfrac{14.1}{50} = 0.283\ \text{rad s}^{-1} \).
(c) \( T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{0.283} = 22.2\ \text{s} \) (equivalently \( T = 2\pi r/v \)).
(d) On a frictionless banked track, the horizontal component of the normal force provides the centripetal force: \( \tan\theta = \dfrac{v^2}{rg} = \dfrac{200}{50\times9.81} = \dfrac{200}{490.5} = 0.408 \Rightarrow \theta = \tan^{-1}(0.408) = 22.2\,^{\circ} \).
(e) On a flat track, only friction between the tyres and road can provide the centripetal force needed to keep the car moving in a circle, and this is limited to a maximum value. On a banked track, the normal contact force from the road no longer acts vertically; it has a horizontal component directed toward the centre of the circle, which itself supplies some or all of the required centripetal force. Because banking allows the normal force (rather than only friction) to contribute to the centripetal force, a higher speed can be reached before the total force required would exceed what the road can supply (or before the car would tend to slide).

(a) [1] correct formula \( F=mv^2/r \) rearranged for v; [1] correct substitution; [1] correct final value ≈14.1 m/s.
(b) [1] correct formula \( \omega=v/r \); [1] correct substitution (allow ECF); [1] correct final value ≈0.283 rad/s.
(c) [1] correct formula \( T=2\pi/\omega \) (or equivalent); [1] correct substitution (allow ECF); [1] correct final value ≈22.2 s.
(d) [1] correct identification that \( \tan\theta=v^2/(rg) \) (or full derivation from resolving forces); [1] correct substitution; [1] correct value of \( \tan\theta \); [1] correct final angle ≈22.2°.
(e) [1] correctly identifies friction as the sole centripetal-force source on a flat track, with a maximum limit; [1] correctly identifies the normal force gains a horizontal component when banked; [1] correctly explains this horizontal component supplies/contributes to the centripetal force; [1] correctly links this to a higher achievable speed before sliding occurs.

Nuclear fusion releases energy by combining light nuclei, such as deuterium and tritium (both isotopes of hydrogen), into a heavier nucleus (helium) that has a greater binding energy per nucleon; the difference in binding energy is released as kinetic energy of the products, which can ultimately be used to generate heat and, via a conventional turbine/generator system, electricity.

To sustain a fusion reaction, the fuel must be raised to an extremely high temperature (of the order of \( 10^{8}\ \text{K} \)), at which point it exists as a plasma (a fully ionised gas of nuclei and free electrons), so that the nuclei have enough kinetic energy to overcome their mutual electrostatic (Coulomb) repulsion and approach closely enough for the strong nuclear force to bind them. In addition to temperature, a sufficiently high plasma density and a long enough confinement time are needed for enough fusion reactions to occur to produce more energy than is put in; these three requirements together are captured by the Lawson criterion.

Because no solid material could survive direct contact with plasma at these temperatures, the plasma must be confined without touching the walls of the reactor. Magnetic confinement, used in devices such as a tokamak, uses strong magnetic fields to hold the charged plasma particles in a torus (ring)-shaped path away from the walls. An alternative approach, inertial confinement, uses powerful lasers to rapidly compress and heat a small fuel pellet so that fusion occurs before the fuel has time to disperse.

Several major challenges remain before fusion becomes commercially viable. It has proved extremely difficult to sustain the required temperature, density and confinement time simultaneously for long enough to achieve a net energy gain (more energy output than input), because the hot plasma is inherently unstable and tends to develop turbulence that allows it to lose heat and touch the reactor walls. The materials lining the reactor must also withstand intense bombardment by high-energy neutrons produced by the reactions, which causes damage and induced radioactivity over time, requiring the development of new, more resilient materials. Finally, the engineering complexity and cost of building and maintaining a fusion reactor capable of continuous operation at a commercial scale remain substantial obstacles that current experimental reactors are still working to overcome.

Level-of-response mark scheme (8 marks). Level 0 [0]: Not worthy of credit. Level 1 [1–2]: Basic, limited description of fusion with little or no reference to conditions or challenges; poor use of specialist terms; weak grammar/spelling/organisation. Level 2 [3–4]: Basic clarity — some correct description of the fusion process and at least one condition or challenge mentioned, but underdeveloped; limited specialist terminology. Level 3 [5–6]: Good QWC — satisfactory description of the fusion process, at least one correctly explained condition (temperature/density/confinement) and at least one genuine challenge, with reasonably accurate specialist terminology and organisation. Level 4 [7–8]: Excellent QWC — comprehensive, accurate description of the fusion process, the conditions required (temperature, density, confinement time / Lawson criterion), at least one confinement method, and well-explained challenges (achieving net energy gain, plasma stability, material damage); accurate and consistent use of specialist terminology; clear, well-structured response.

(a) The wire should be checked to be straight and under a small initial tension (not slack) before the zero-extension reading is taken, so that all subsequent extensions are measured from a consistent, taut starting point. The marker/scale reading should be taken with the eye level with the marker (or using a set-square against the scale) to avoid parallax error, which would otherwise introduce a systematic error into every extension reading.
(b) Using the extremes of the data: gradient \( = \dfrac{x}{F} = \dfrac{(3.32-0.66)\times10^{-3}}{20.0-4.0} = \dfrac{2.66\times10^{-3}}{16.0} = 1.66\times10^{-4}\ \text{m N}^{-1} \).
Cross-sectional area: radius \( r = 0.20\ \text{mm} = 2.0\times10^{-4}\ \text{m} \), \( A = \pi r^2 = \pi\times(2.0\times10^{-4})^2 = 1.257\times10^{-7}\ \text{m}^2 \).
Young's modulus: \( E = \dfrac{\text{stress}}{\text{strain}} = \dfrac{F/A}{x/L_0} = \dfrac{L_0}{A\times(\text{gradient})} = \dfrac{2.500}{1.257\times10^{-7}\times1.66\times10^{-4}} = \dfrac{2.500}{2.087\times10^{-11}} = 1.20\times10^{11}\ \text{Pa} \).
(c) The extensions being measured are of the order of a millimetre, which is difficult to read precisely (to better than about ±0.1–0.5 mm) using a simple ruler and marker, introducing a significant percentage uncertainty especially at small loads. This could be reduced by using a travelling microscope (or a vernier/digital extensometer) to read the marker's position with much greater precision, or by using a longer wire so that, for the same loads, the extensions produced are proportionally larger and therefore easier to measure accurately.

(a) [1] first valid precaution (e.g. ensure wire taut before zeroing, avoid parallax, check wire hangs freely over the pulley); [1] second, distinct valid precaution; [1] for a developed reason linking a precaution to improved reliability/accuracy.
(b) [1] correct gradient (x/F) with correct value and units, from a valid method (e.g. using two well-separated points or an equivalent best-fit method); [1] correct calculation of cross-sectional area A; [1] correct substitution into \( E=L_0/(A\times\text{gradient}) \) or equivalent rearrangement; [1] correct final value ≈1.2×10^11 Pa with correct unit.
(c) [1] valid, significant source of uncertainty identified and correctly explained; [1] valid, relevant improvement suggested; [1] for a developed explanation of why the improvement reduces the uncertainty.

\( r = 0.15\ \text{m} \) (distance from each charge to the midpoint).
(a) \( E_1 = \dfrac{kQ_1}{r^2} = \dfrac{8.99\times10^{9}\times4.0\times10^{-6}}{0.15^2} = \dfrac{3.596\times10^{4}}{0.0225} = 1.60\times10^{6}\ \text{N C}^{-1} \), directed away from \( Q_1 \) (from A towards B), since \( Q_1 \) is positive.
(b) \( E_2 = \dfrac{k|Q_2|}{r^2} = \dfrac{8.99\times10^{9}\times2.0\times10^{-6}}{0.0225} = 7.99\times10^{5}\ \text{N C}^{-1} \), directed towards \( Q_2 \) (also from A towards B, since \( Q_2 \) is negative).
(c) Both fields point in the same direction at M (from A towards B), so they add: \( E = E_1+E_2 = 1.60\times10^{6}+7.99\times10^{5} = 2.40\times10^{6}\ \text{N C}^{-1} \), directed from A towards B.
(d) Electric potential is a scalar, so the potentials due to each charge simply add (with sign): \( V = \dfrac{kQ_1}{r}+\dfrac{kQ_2}{r} = \dfrac{k}{r}(Q_1+Q_2) = \dfrac{8.99\times10^{9}}{0.15}\times(4.0\times10^{-6}-2.0\times10^{-6}) = 5.993\times10^{10}\times2.0\times10^{-6} = 1.20\times10^{2}\ \text{V} \).

(a) [1] correct formula \( E=kQ/r^2 \); [1] correct substitution; [1] correct value ≈1.60×10^6 N/C.
(b) [1] correct substitution using \( Q_2 \)'s magnitude; [1] correct value ≈7.99×10^5 N/C; [1] correct direction stated (towards B).
(c) [1] correct recognition that the two field contributions point in the same direction at M and should be added (not subtracted); [1] correct sum; [1] correct final value ≈2.40×10^6 N/C with direction stated.
(d) [1] correct recognition that potential is a scalar and potentials add algebraically (with sign); [1] correct substitution; [1] correct final value ≈120 V.

(a) \( V = Ed = 2.5\times10^{4}\times0.080 = 2.0\times10^{3}\ \text{V} \).
(b) \( W = qV = 1.60\times10^{-19}\times2.0\times10^{3} = 3.20\times10^{-16}\ \text{J} \).
(c) By the work–energy theorem, \( W = \tfrac12mv^2 \Rightarrow v = \sqrt{\dfrac{2W}{m}} = \sqrt{\dfrac{2\times3.20\times10^{-16}}{9.11\times10^{-31}}} = \sqrt{7.03\times10^{14}} = 2.65\times10^{7}\ \text{m s}^{-1} \).
(d) The acceleration remains constant throughout the electron's journey between the plates. This is because the electric field between two oppositely charged parallel plates is uniform (equal in magnitude and direction at every point between them), so the force \( F=qE \) on the electron does not change, and hence, by \( a=F/m \), the acceleration does not change either.

(a) [1] correct formula \( V=Ed \); [1] correct substitution (with d converted to metres); [1] correct value 2000 V.
(b) [1] correct formula \( W=qV \); [1] correct substitution (allow ECF); [1] correct value 3.20×10^-16 J.
(c) [1] correct use of the work–energy theorem \( W=\tfrac12mv^2 \); [1] correct rearrangement/substitution; [1] correct final value ≈2.65×10^7 m/s.
(d) [1] correctly states the acceleration is constant; [1] correctly identifies the field between the plates is uniform; [1] correctly links this (via F=qE and a=F/m) to constant force and hence constant acceleration.

(a) \( F = BIL = 0.40\times3.0\times0.50 = 0.60\ \text{N} \).
(b) The force acts vertically upward, in the plane of the page. This is found using Fleming's left-hand rule: with the First finger pointing in the direction of the field (into the page), the seCond finger pointing in the direction of the conventional current (left to right), the thuMb then gives the direction of the force/motion — upward in this case.
(c) \( F = qvB = 1.60\times10^{-19}\times2.0\times10^{6}\times0.40 = 1.28\times10^{-13}\ \text{N} \).
(d) The magnetic force on a moving charged particle acts at right angles to its velocity at all times. A force that is always perpendicular to velocity changes only the direction of motion, not the speed, and has constant magnitude for a constant speed and field — this is exactly the condition for circular motion, with the magnetic force providing the centripetal force. Equating the magnetic force to the centripetal force: \( qvB = \dfrac{mv^2}{r} \Rightarrow r = \dfrac{mv}{qB} \).

(a) [1] correct formula \( F=BIL \); [1] correct substitution; [1] correct value 0.60 N.
(b) [1] correct direction (upward); [1] correct reference to Fleming's left-hand rule; [1] correct explanation of how the rule is applied (field/current/force fingers correctly assigned).
(c) [1] correct formula \( F=qvB \); [1] correct substitution; [1] correct value 1.28×10^-13 N.
(d) [1] correct explanation that the force is always perpendicular to velocity, changing direction but not speed; [1] correct identification that this force acts as the centripetal force; [1] correct derivation \( qvB=mv^2/r \Rightarrow r=mv/(qB) \).

(a) \( r = \dfrac{mv}{qB} = \dfrac{1.67\times10^{-27}\times2.0\times10^{6}}{1.60\times10^{-19}\times0.40} = \dfrac{3.34\times10^{-21}}{6.40\times10^{-20}} = 5.22\times10^{-2}\ \text{m} \).
(b) \( T = \dfrac{2\pi r}{v} = \dfrac{2\pi\times5.22\times10^{-2}}{2.0\times10^{6}} = 1.64\times10^{-7}\ \text{s} \) (equivalently, \( T=2\pi m/(qB) \), giving the same value independent of v).
(c) The period would stay the same. Since \( T = \dfrac{2\pi m}{qB} \), the period depends only on the particle's mass, charge and the field strength, not on its speed; doubling the speed increases the radius of the circular path proportionally (so the particle still takes the same time to complete one full revolution).
(d) \( r = \dfrac{mv}{qB} \), so with v, q and B all unchanged, the radius is directly proportional to mass. Doubling the mass therefore doubles the radius: the new particles travel in a circle of twice the radius of the original protons' path.

(a) [1] correct formula \( r=mv/(qB) \); [1] correct substitution; [1] correct final value ≈5.2×10^-2 m.
(b) [1] correct formula \( T=2\pi r/v \) (or \( T=2\pi m/(qB) \)); [1] correct substitution (allow ECF from (a)); [1] correct final value ≈1.6×10^-7 s.
(c) [1] correctly states the period is unchanged; [1] correct justification referencing \( T=2\pi m/(qB) \) being independent of v.
(d) [1] correctly identifies \( r\propto m \) from the formula (with v, q, B constant); [1] correctly concludes the radius doubles; [1] for a clear, complete comparative statement.

(a) \( r = 6.37\times10^{6}+400\times10^{3} = 6.77\times10^{6}\ \text{m} \).
(b) The gravitational force provides the centripetal force: \( \dfrac{GMm}{r^2} = \dfrac{mv^2}{r} \Rightarrow v^2 = \dfrac{GM}{r} \Rightarrow v = \sqrt{\dfrac{GM}{r}} \). Substituting: \( v = \sqrt{\dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{6.77\times10^{6}}} = \sqrt{\dfrac{3.982\times10^{14}}{6.77\times10^{6}}} = \sqrt{5.88\times10^{7}} = 7.67\times10^{3}\ \text{m s}^{-1} \).
(c) \( T = \dfrac{2\pi r}{v} = \dfrac{2\pi\times6.77\times10^{6}}{7.67\times10^{3}} = \dfrac{4.254\times10^{7}}{7.67\times10^{3}} = 5547\ \text{s} \). Converting to minutes: \( 5547/60 = 92.5\ \text{minutes} \).
(d) A geostationary orbit is one in which a satellite orbits directly above the Earth's equator with an orbital period of exactly 24 hours, matching the Earth's own rotation, so that the satellite remains permanently above the same point on the Earth's surface. This satellite's period is only about 92.5 minutes, far shorter than 24 hours, so it does not remain above a fixed point and is therefore not in a geostationary orbit — it is a low-Earth-orbit satellite, orbiting at a much smaller radius than the ~42,000 km (from Earth's centre) required for a geostationary orbit.

(a) [1] correct conversion of altitude to metres; [1] correct addition to Earth's radius; [1] correct final value 6.77×10^6 m.
(b) [1] correct equating of gravitational force to centripetal force; [1] correct rearrangement to \( v=\sqrt{GM/r} \); [1] correct substitution; [1] correct final value ≈7.67×10^3 m/s.
(c) [1] correct formula \( T=2\pi r/v \); [1] correct substitution (allow ECF); [1] correct final value ≈92.5 minutes (accept equivalent in seconds if correctly converted).
(d) [1] correct definition of geostationary orbit (24-hour period, above the equator, stays above a fixed point); [1] correct explanation that this satellite's much shorter period means it is not geostationary.

(a) \( Q_0 = CV = 470\times10^{-6}\times12 = 5.64\times10^{-3}\ \text{C} \).
(b) \( E = \tfrac12CV^2 = \tfrac12\times470\times10^{-6}\times12^2 = \tfrac12\times470\times10^{-6}\times144 = 3.38\times10^{-2}\ \text{J} \).
(c) \( \tau = RC = 2200\times470\times10^{-6} = 1.03\ \text{s} \approx 1.0\ \text{s} \).
(d) \( V = V_0 e^{-t/\tau} = 12\times e^{-2.0/1.03} = 12\times e^{-1.94} = 12\times0.144 = 1.73\ \text{V} \approx 1.7\ \text{V} \).

(a) [1] correct formula \( Q=CV \); [1] correct substitution; [1] correct value 5.64×10^-3 C.
(b) [1] correct formula \( E=\tfrac12CV^2 \); [1] correct substitution; [1] correct value ≈3.38×10^-2 J.
(c) [1] correct formula \( \tau=RC \); [1] correct substitution; [1] correct value ≈1.0 s.
(d) [1] correct use of \( V=V_0e^{-t/\tau} \) with correct substitution; [1] correct final value ≈1.7 V.

(a) \( E = \dfrac{V}{d} = \dfrac{300}{0.020} = 1.5\times10^{4}\ \text{N C}^{-1} \).
(b) \( F = eE = 1.60\times10^{-19}\times1.5\times10^{4} = 2.40\times10^{-15}\ \text{N} \). \( a = F/m = 2.40\times10^{-15}/9.11\times10^{-31} = 2.64\times10^{15}\ \text{m s}^{-2} \).
(c) Time between the plates: \( t = \dfrac{L}{v_x} = \dfrac{0.050}{3.0\times10^{7}} = 1.67\times10^{-9}\ \text{s} \). Vertical deflection: \( y = \tfrac12at^2 = \tfrac12\times2.64\times10^{15}\times(1.67\times10^{-9})^2 = \tfrac12\times2.64\times10^{15}\times2.78\times10^{-18} = 3.7\times10^{-3}\ \text{m} \) (3.7 mm).
(d) If the horizontal speed were doubled, the time spent between the plates would halve, since \( t=L/v_x \); the vertical acceleration a is unaffected by horizontal speed, as it depends only on the (unchanged) electric field. Since \( y=\tfrac12at^2 \), halving t means y is multiplied by \( (\tfrac12)^2=\tfrac14 \); the vertical deflection would fall to one quarter of its original value.

(a) [1] correct formula \( E=V/d \); [1] correct substitution with d in metres; [1] correct value 1.5×10^4 N/C.
(b) [1] correct force \( F=eE \); [1] correct use of \( a=F/m \); [1] correct final value ≈2.64×10^15 m/s².
(c) [1] correct time \( t=L/v_x \); [1] correct use of \( y=\tfrac12at^2 \) (allow ECF); [1] correct final deflection ≈3.7×10^-3 m (3.7 mm).
(d) [1] correctly identifies t halves; [1] correctly identifies a is unchanged; [1] correctly concludes y falls to one quarter, with valid reasoning from \( y\propto t^2 \).

A cyclotron consists of two hollow, D-shaped electrodes (called 'dees'), placed in a strong, uniform magnetic field directed perpendicular to the plane of the dees, with a small gap between them. An alternating high-frequency voltage is applied across this gap. A charged particle, injected near the centre, is accelerated each time it crosses the gap between the dees, gaining kinetic energy from the electric field there; while inside a dee (where there is no electric field), the magnetic field causes it to move in a semicircular path due to the magnetic force providing a centripetal force. As the particle's speed increases with each crossing of the gap, the radius of its circular path (\( r=mv/(qB) \)) increases, so it follows an outward spiral, gaining more energy on each successive half-circle until it exits at the outer edge of the dees with high kinetic energy.

A key limitation of the basic cyclotron design becomes significant as the particle's speed approaches a substantial fraction of the speed of light. At these relativistic speeds, the particle's relativistic mass increases, which increases the time taken to complete each semicircular path (since \( T=2\pi m/(qB) \) increases as m increases). Because the accelerating voltage alternates at a fixed frequency, the particle gradually falls out of step (out of synchronisation) with the alternating voltage, meaning it may arrive at the gap when the field is in the wrong phase to accelerate it further, ultimately limiting the maximum energy the cyclotron can impart.

The synchrotron was developed to overcome this limitation. Rather than allowing the particle to spiral outward, a synchrotron confines the particle to a fixed-radius circular path using an electromagnet whose field strength is increased in step with the particle's increasing momentum, keeping the radius constant; the frequency of the accelerating voltage is also adjusted (synchronised) to match the particle's changing revolution time as it approaches relativistic speeds. This synchronisation of both field strength and accelerating frequency allows synchrotrons to accelerate particles to much higher, fully relativistic energies than a cyclotron can achieve.

Level-of-response mark scheme (8 marks). Level 0 [0]: Not worthy of credit. Level 1 [1–2]: Basic, limited description of the cyclotron with little or no reference to its limitation; poor specialist terminology; weak grammar/spelling/organisation. Level 2 [3–4]: Basic clarity — some correct description of the dees and accelerating voltage, with a vague or underdeveloped reference to a limitation. Level 3 [5–6]: Good QWC — satisfactory, largely accurate description of the cyclotron's operation (dees, alternating voltage, magnetic field causing circular motion) and a valid, reasonably explained limitation; some accurate specialist terminology. Level 4 [7–8]: Excellent QWC — comprehensive, accurate description of the cyclotron's operating principle including the spiral path and increasing radius, a clearly explained relativistic desynchronisation limitation, and a correct explanation of how the synchrotron addresses it (fixed radius, synchronised frequency and/or field); accurate, consistent specialist terminology; clear, well-structured response.

(a) (1) A baryon is any hadron made of three quarks; a proton (uud) or neutron (udd) are correct examples. (2) A meson is any hadron made of a quark and an antiquark; a pion (e.g. \( \pi^{+} \)) or kaon are correct examples. (3) The electron is the lightest, most familiar charged lepton. (4) The muon (or the tau) is a second, heavier charged lepton.
(b) Leptons, such as the electron and muon and their associated neutrinos, are believed to be fundamental particles — they have no known internal structure and are not composed of smaller constituent particles — and they do not experience the strong nuclear force. Hadrons, such as protons, neutrons and pions, are not fundamental; they are composite particles, made up of quarks bound together, and it is because they contain quarks (which do interact via the strong force) that hadrons as a whole experience the strong nuclear force.
(c) A proton consists of two up quarks and one down quark, written uud. A neutron consists of one up quark and two down quarks, written udd.

(a) [1] each for correctly completing (1) a valid baryon; (2) a valid meson; (3) electron; (4) muon or tau — 4 marks total.
(b) [1] correctly states leptons are fundamental particles (no substructure); [1] correctly states hadrons are composite, made of quarks; [1] correctly explains hadrons experience the strong force via their constituent quarks, while leptons do not.
(c) [1] correct proton composition uud; [1] correct neutron composition udd; [1] for a correct additional comparative statement (e.g. the difference is one up quark exchanged for one down quark, giving the neutron zero net charge).

(a) The student should measure L (from the point of suspension to the centre of the pendulum bob) using a metre ruler, then displace the bob through a small angle (less than about 10°) and release it. Rather than timing a single oscillation, the student should time 20 complete oscillations with a stopwatch and divide the total time by 20 to find T; this greatly reduces the effect of the student's reaction time on the calculated period, since any fixed timing error is spread over 20 oscillations rather than 1. A fixed reference point (a fiducial marker) at the lowest point of the swing should be used to judge when each oscillation is complete, viewed at eye level to avoid parallax error, and the measurement should be repeated (e.g. three times) and averaged to reduce the effect of random error further.

(b) \( T = \text{time}/20 \): for L=0.200 m, \( T=17.9/20=0.895\ \text{s} \), \( T^2=0.801\ \text{s}^2 \); for L=0.400 m, \( T=25.3/20=1.265\ \text{s} \), \( T^2=1.600\ \text{s}^2 \); for L=0.600 m, \( T=31.0/20=1.550\ \text{s} \), \( T^2=2.403\ \text{s}^2 \); for L=0.800 m, \( T=35.8/20=1.790\ \text{s} \), \( T^2=3.204\ \text{s}^2 \); for L=1.000 m, \( T=40.0/20=2.000\ \text{s} \), \( T^2=4.000\ \text{s}^2 \).

(c) Squaring \( T=2\pi\sqrt{L/g} \) gives \( T^2 = \dfrac{4\pi^2}{g}L \), which has the form \( y=mx \) (with \( y=T^2 \), \( x=L \)) — a straight line passing through the origin, with gradient \( m=\dfrac{4\pi^2}{g} \).

(d) Rearranging, \( g = \dfrac{4\pi^2 L}{T^2} = \dfrac{4\pi^2\times1.000}{4.000} = \dfrac{39.48}{4.000} = 9.87\ \text{m s}^{-2} \approx 9.9\ \text{m s}^{-2} \).

(e) Rather than calculating g from a single (L, T) pair, the student should plot \( T^2 \) against L for all five data points and draw a straight line of best fit through them; g is then calculated from the gradient of this line (\( g=4\pi^2/\text{gradient} \)). Using the gradient of a best-fit line through all the data averages out the effect of random error in each individual reading, and also allows any anomalous data point to be identified and excluded, giving a more accurate value of g than relying on a single measurement.

(f) If the amplitude of the pendulum's swing is not kept small (well below about 10°), the simple harmonic motion approximation underlying \( T=2\pi\sqrt{L/g} \) becomes progressively less accurate, and the true period of the pendulum becomes systematically longer than this formula predicts. Using these (too large) measured values of T in the formula \( g=4\pi^2L/T^2 \) would therefore cause the calculated value of g to be systematically too small — an underestimate of the true value.

(a) [1] valid method for measuring the period (time many oscillations, e.g. 20, rather than one); [1] correct explanation that this reduces the percentage effect of reaction-time error; [1] valid additional precaution (fiducial marker at lowest point / eye-level viewing to avoid parallax); [1] repeats and averaging for reliability.
(b) [1] all five T values correct; [1] all five \( T^2 \) values correct (allow error carried forward from candidate's own T values); [1] values given to a consistent, sensible number of significant figures; [1] correct units shown throughout.
(c) [1] correct squaring of the equation to \( T^2=(4\pi^2/g)L \); [1] correct identification that this is straight-line form (y=mx) through the origin; [1] correct statement that the gradient equals \( 4\pi^2/g \).
(d) [1] correct rearrangement \( g=4\pi^2L/T^2 \); [1] correct substitution (allow ECF from (b)); [1] correct final value ≈9.9 m/s² with unit.
(e) [1] valid method (gradient of best-fit line through all data, not a single point); [1] correct explanation that this averages random error across all readings; [1] further valid point (anomalous points can be identified/excluded).
(f) [1] valid systematic error identified; [1] correct explanation of its physical cause; [1] correct explanation of its directional effect on the calculated g (too large or too small), with valid reasoning.

(a) \( R = \dfrac{\rho l}{A} \). A graph of R (y-axis) against l (x-axis) is therefore a straight line through the origin, of the form \( y=mx \) with gradient \( m=\rho/A \). Since A can be found independently from the wire's measured diameter, the resistivity can be calculated as \( \rho = \text{gradient}\times A \).

(b) Using the two extreme data points: gradient \( = \dfrac{6.40-1.30}{1.000-0.200} = \dfrac{5.10}{0.800} = 6.375\ \Omega\,\text{m}^{-1} \approx 6.4\ \Omega\,\text{m}^{-1} \).

(c) Diameter \( d = 0.315\ \text{mm} = 3.15\times10^{-4}\ \text{m} \), so radius \( r = 1.575\times10^{-4}\ \text{m} \). \( A = \pi r^2 = \pi\times(1.575\times10^{-4})^2 = 7.79\times10^{-8}\ \text{m}^2 \). Percentage uncertainty in d: \( \dfrac{0.005}{0.315}\times100 = 1.59\% \); percentage uncertainty in A is double this (see part (e)), \( \approx3.17\% \); absolute uncertainty in A \( = 0.0317\times7.79\times10^{-8} = 2.47\times10^{-9}\ \text{m}^2 \approx0.25\times10^{-8}\ \text{m}^2 \). So \( A = (7.8\pm0.2)\times10^{-8}\ \text{m}^2 \).

(d) \( \rho = \text{gradient}\times A = 6.375\times7.79\times10^{-8} = 4.97\times10^{-7}\ \Omega\,\text{m} \approx5.0\times10^{-7}\ \Omega\,\text{m} \).

(e) Percentage uncertainty in \( d = \dfrac{0.005}{0.315}\times100 = 1.59\%\approx1.6\% \). Since \( A=\pi r^2=\pi d^2/4 \), A depends on the square of d. When a quantity is raised to a power n, its percentage uncertainty is multiplied by n; since A depends on \( d^2 \) (power 2), the percentage uncertainty in A is twice the percentage uncertainty in d, i.e. \( \approx3.2\% \), consistent with the value used in part (c).

(f) Entering the length for each reading manually, from memory, after the resistance values were already logged risks the student pairing a resistance reading with an incorrect length (mismatching rows), or misremembering the exact length used, introducing an error into specific data points that does not reflect the actual conditions under which that resistance was measured. This could be improved by having the data-logging software record (or prompt the student to enter) the length value at the same time as each resistance reading is taken, removing any reliance on memory and ensuring every resistance value is correctly paired with its corresponding length.

(a) [1] correct equation \( R=\rho l/A \); [1] correct statement that the graph is a straight line through the origin with gradient \( \rho/A \); [1] correct statement that \( \rho \) is then found from gradient × A.
(b) [1] valid method (e.g. using two well-separated points, or an equivalent best-fit approach); [1] correct substitution; [1] correct value ≈6.4 Ω/m with unit.
(c) [1] correct conversion of diameter to radius in metres; [1] correct area formula and substitution giving ≈7.79×10^-8 m²; [1] correct calculation of the absolute uncertainty in A (allow ECF from (e)); [1] correctly expressed final answer with matching sig figs and unit.
(d) [1] correct method \( \rho=\text{gradient}\times A \); [1] correct substitution (allow ECF); [1] correct final value ≈5.0×10^-7 Ω m with unit.
(e) [1] correct percentage uncertainty in d ≈1.6%; [1] correct statement that \( A\propto d^2 \); [1] correct general rule that percentage uncertainty is multiplied by the power when a quantity is raised to a power; [1] correctly stated resulting percentage uncertainty in A ≈3.2%.
(f) [1] valid explanation of how manual/memory-based length entry could introduce error (mismatched or misremembered values); [1] valid, relevant improvement to the logging procedure; [1] for a developed link between the improvement and more reliable/accurate data.

(a) \( d = 4.82\ \text{mm} = 4.82\times10^{-3}\ \text{m} \); uncertainty \( = 0.02\ \text{mm} = 2\times10^{-5}\ \text{m} \).
(b) Radius \( r = d/2 = 2.41\times10^{-3}\ \text{m} \). \( V = \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi\times(2.41\times10^{-3})^3 = \dfrac{4}{3}\pi\times1.400\times10^{-8} = 5.86\times10^{-8}\ \text{m}^3 \). To 3 significant figures (matching the precision of the original data): \( V \approx 5.86\times10^{-8}\ \text{m}^3 \).
(c) Percentage uncertainty \( = \dfrac{0.02}{4.82}\times100 = 0.415\% \approx 0.42\% \).
(d) Since \( V=\tfrac{4}{3}\pi r^3 \propto d^3 \), volume depends on the cube of the diameter. When a quantity is raised to a power n, its percentage uncertainty is multiplied by n; here n=3, so the percentage uncertainty in V is three times that in d: \( 3\times0.415\% = 1.245\% \approx1.2\% \).

(a) [1] correct diameter in m, standard form; [1] correct uncertainty in m, standard form.
(b) [1] correct radius; [1] correct formula \( V=\tfrac43\pi r^3 \); [1] correct substitution; [1] correct final value ≈5.86×10^-8 m³ to an appropriate number of significant figures with correct unit.
(c) [1] correct method (uncertainty/value ×100); [1] correct value ≈0.42%.
(d) [1] correct statement that \( V\propto d^3 \) so percentage uncertainty is multiplied by 3; [1] correct final percentage uncertainty in V ≈1.2%.

(a) For a cell of EMF \( \varepsilon \) and internal resistance r, the terminal potential difference is given by \( V = \varepsilon - Ir \). Comparing this with \( V=1.482-0.86I \): the y-intercept (the value of V when I=0, i.e. when no current is drawn) corresponds to \( \varepsilon \), the EMF of the cell; the gradient (−0.86 V/A) has magnitude equal to the internal resistance r, with the negative sign reflecting that V decreases as I increases due to the 'lost volts' across the internal resistance.
(b) \( \varepsilon = 1.482\ \text{V} \approx 1.48\ \text{V} \); \( r = 0.86\ \Omega \).
(c) When short-circuited, the external resistance is negligible, so \( V\approx0 \). Setting \( 0=\varepsilon-Ir \Rightarrow I=\varepsilon/r = 1.482/0.86 = 1.72\ \text{A} \approx1.7\ \text{A} \). This assumes that the internal resistance r remains constant (i.e. the linear \( V=\varepsilon-Ir \) model continues to apply) even at this higher current, which may not hold exactly in practice as significant heating effects inside the cell at high current can change its internal resistance.

(a) [1] correct equation \( V=\varepsilon-Ir \) stated/recalled; [1] correct identification that the y-intercept equals \( \varepsilon \) (EMF) when I=0; [1] correct identification that the magnitude of the gradient equals r; [1] correct explanation of the physical meaning of the negative gradient (V falls as I rises due to internal resistance).
(b) [1] correct EMF ≈1.48 V; [1] correct internal resistance ≈0.86 Ω.
(c) [1] correct recognition that V=0 at short circuit; [1] correct rearrangement \( I=\varepsilon/r \); [1] correct final value ≈1.7 A; [1] valid, clearly stated assumption (r/model remains constant/linear at this current).

(a) \( m = 154.2\ \text{g} = 0.1542\ \text{kg} \); \( V = 18.4\ \text{cm}^3 = 18.4\times10^{-6}\ \text{m}^3 = 1.84\times10^{-5}\ \text{m}^3 \). \( \rho = \dfrac{m}{V} = \dfrac{0.1542}{1.84\times10^{-5}} = 8380\ \text{kg m}^{-3} \).
(b) Percentage uncertainty in mass \( = \dfrac{0.1}{154.2}\times100 = 0.0648\% \approx 0.065\% \). Percentage uncertainty in volume \( = \dfrac{0.3}{18.4}\times100 = 1.630\% \approx1.63\% \).
(c) Since density is found by dividing two independently measured quantities, their percentage uncertainties are added: total percentage uncertainty \( = 0.065\%+1.63\% = 1.695\% \approx1.7\% \). Absolute uncertainty in \( \rho = 0.017\times8380 = 142\ \text{kg m}^{-3} \approx140\ \text{kg m}^{-3} \) (2 significant figures, matching the precision implied by the percentage uncertainty). Final result: \( \rho = (8380\pm140)\ \text{kg m}^{-3} \), i.e. \( (8.38\pm0.14)\times10^{3}\ \text{kg m}^{-3} \).

(a) [1] correct unit conversions of mass and volume into kg and m³; [1] correct formula and substitution; [1] correct final value ≈8380 kg/m³.
(b) [1] correct percentage uncertainty in mass ≈0.065%; [1] correct percentage uncertainty in volume ≈1.6%.
(c) [1] correct rule that percentage uncertainties are added for a quantity found by division; [1] correct total percentage uncertainty ≈1.7%; [1] correct conversion to an absolute uncertainty; [1] correctly expressed final density with matching significant figures; [1] uncertainty sensibly rounded (1–2 significant figures).

(a) A line of worst fit is a straight line, drawn through the plotted data, that still passes through (or reasonably close to) the error bars of the data points, but has the steepest (or shallowest) gradient that can still be justified by the data — as distinct from the line of best fit, which is drawn to lie as close as possible to all the points on average. The uncertainty in the gradient is estimated by finding the difference between the gradient of the best-fit line and the gradient of the worst-fit line.
(b) Difference in gradients \( = 0.0271-0.0246 = 0.0025\ \text{m N}^{-1} \). Percentage uncertainty \( = \dfrac{0.0025}{0.0246}\times100 = 10.16\% \approx10\% \).
(c) \( k = \dfrac{1}{\text{gradient (best fit)}} = \dfrac{1}{0.0246} = 40.7\ \text{N m}^{-1} \). Absolute uncertainty \( = 0.1016\times40.7 = 4.13\approx4\ \text{N m}^{-1} \). So \( k = (41\pm4)\ \text{N m}^{-1} \).

(a) [1] correct definition of a line of worst fit (steepest/shallowest line still consistent with the error bars); [1] correct distinction from the line of best fit; [1] correct statement that the uncertainty is estimated from the difference between the two gradients.
(b) [1] correct difference in gradients calculated; [1] correct method (difference divided by best-fit gradient); [1] correct value ≈10%; [1] appropriately rounded/expressed.
(c) [1] correct value of k from the best-fit gradient ≈40.7 N/m; [1] correct absolute uncertainty ≈4 N/m using the percentage from (b); [1] correctly expressed final answer with matching significant figures/rounding.

(a) Percentage error \( = \dfrac{|336-343|}{343}\times100 = \dfrac{7}{343}\times100 = 2.04\% \approx2.0\% \).
(b) \( 336\ \text{m s}^{-1}\times3.6 = 1209.6\ \text{km h}^{-1} \approx1210\ \text{km h}^{-1} \) (since \( 1\ \text{m s}^{-1} = 3.6\ \text{km h}^{-1} \)).
(c) \( v = \dfrac{\text{distance}}{\text{time}} = \dfrac{136.0}{0.405} = 335.8\ \text{m s}^{-1} \approx 336\ \text{m s}^{-1} \), which matches the student's quoted result, confirming consistency.
Percentage uncertainty in distance \( = \dfrac{0.5}{136.0}\times100 = 0.368\% \approx0.37\% \). Percentage uncertainty in time \( = \dfrac{0.006}{0.405}\times100 = 1.481\% \approx1.48\% \). Since v is found by dividing distance by time, the percentage uncertainties are added: total percentage uncertainty \( = 0.37\%+1.48\% = 1.85\% \approx1.9\% \).

(a) [1] correct method (difference/accepted value ×100); [1] correct substitution; [1] correct final value ≈2.0%.
(b) [1] correct conversion factor (×3.6) identified; [1] correct final value ≈1210 km/h.
(c) [1] correct substitution showing v=136.0/0.405≈336 m/s, confirming consistency with the quoted result; [1] correct percentage uncertainty in distance ≈0.37%; [1] correct percentage uncertainty in time ≈1.48%; [1] correct rule that percentages are added for a quantity found by division; [1] correct total percentage uncertainty ≈1.9%.