2023 CCEA A-Level Physics 1210 模拟试题及答案详解

(a) Nucleon number, A, is the total number of protons and neutrons (nucleons) in the nucleus of an atom. (b) Proton number, Z, is the number of protons in the nucleus of an atom. Isotopes are atoms of the same element (i.e. having the same proton number, Z) but with different nucleon numbers, A (i.e. different numbers of neutrons in the nucleus).

1 mark: correct definition of nucleon number A; 1 mark: correct definition of proton number Z; 1 mark: isotopes correctly described as having the same Z; 1 mark: isotopes correctly described as having different A/different neutron number. [4]

(a) Angular velocity is the rate of change of angular displacement (the angle swept out) with time; \( \omega = \dfrac{\Delta\theta}{\Delta t} = \dfrac{v}{r} \). (b) Centripetal acceleration is the acceleration directed towards the centre of the circular path, required to continuously change the direction of the object's velocity while its speed stays constant; \( a = \dfrac{v^2}{r} = \omega^2 r \).

1 mark: correct definition of angular velocity; 1 mark: correct equation \( \omega = v/r \); 1 mark: correct definition of centripetal acceleration (directed towards centre, changes direction not speed); 1 mark: correct equation \( a = v^2/r \) (or \( \omega^2 r \)). [4]

(a) Activity is the number of nuclear disintegrations (decays) occurring per unit time (per second) within a radioactive sample. (b) Half-life is the time taken for the activity (or the number of undecayed nuclei, or the count rate) of a radioactive sample to fall to half of its original value. Equation: \( A = \lambda N \), where λ is the decay constant and N is the number of undecayed nuclei present.

1 mark: correct definition of activity; 1 mark: correct definition of half-life; 1 mark: correct equation \( A = \lambda N \); 1 mark: symbols/quantities correctly identified (λ = decay constant, N = number of undecayed nuclei). [4]

(a) Stress is the force applied per unit cross-sectional area of the wire: \( \sigma = F/A \), measured in Pa (N m⁻²). (b) Strain is the extension produced per unit original length of the wire: \( \varepsilon = x/L \), a dimensionless ratio (no units). The Young modulus is defined as: \( E = \dfrac{\sigma}{\varepsilon} = \dfrac{F/A}{x/L} = \dfrac{FL}{Ax} \).

1 mark: correct definition and equation for stress; 1 mark: correct definition and equation for strain, noted as dimensionless; 1 mark: correct equation for Young modulus, \( E = \sigma/\varepsilon \); 1 mark: correct expanded form \( E = FL/(Ax) \). [4]

Simple harmonic motion is oscillatory motion in which the acceleration of an object is directly proportional to its displacement from a fixed equilibrium point, and is always directed towards that point (i.e. it is always a restoring acceleration, acting to return the object to equilibrium). This is expressed by the equation \( a = -\omega^2 x \), where the negative sign shows that the acceleration and the displacement are always in opposite directions.

1 mark: acceleration proportional to displacement stated; 1 mark: acceleration always directed towards/restoring towards equilibrium stated; 1 mark: correct equation \( a=-\omega^2x \); 1 mark: correct explanation of the negative sign (opposite directions). [4]

Energy supplied \( = Pt = 50.0 \times 240 = 12000 \text{ J} \). Energy absorbed by the block \( = mc\Delta\theta = 0.150 \times 897 \times (61.2-20.0) = 0.150 \times 897 \times 41.2 = 5544 \text{ J} \). Efficiency \( = \dfrac{5544}{12000}\times100 = 46.2\% \).

1 mark: correct energy supplied by heater (12000 J); 1 mark: correct temperature rise (41.2 K); 1 mark: correct energy absorbed by block (5544 J); 1 mark: correct efficiency formula (useful/supplied ×100); 1 mark: correct final answer 46.2% (accept 45–47%). [5]

\( \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \Rightarrow P_2 = \dfrac{P_1V_1T_2}{T_1V_2} = \dfrac{1.05\times10^5 \times 2.40\times10^{-3}\times350}{290\times1.60\times10^{-3}} = \dfrac{88200}{0.464} = 1.90\times10^5 \text{ Pa} \).

1 mark: correct combined gas law equation quoted; 1 mark: correct numerator substitution/evaluation; 1 mark: correct denominator substitution/evaluation; 1 mark: correct rearrangement for P₂; 1 mark: correct final answer \( 1.90\times10^5 \text{ Pa} \) (accept 1.88–1.92 ×10⁵). [5]

\( c_{rms} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3\times8.31\times300}{0.0280}} = \sqrt{\dfrac{7479}{0.0280}} = \sqrt{267107} = 517 \text{ m s}^{-1} \).

1 mark: correct formula \( c_{rms}=\sqrt{3RT/M} \); 1 mark: correct numerator (3RT = 7479); 1 mark: correct division by M; 1 mark: correct square root evaluation; 1 mark: correct final answer 517 m s⁻¹ (accept 515–519). [5]

Energy supplied \( = Pt = 60.0 \times 300 = 18000 \text{ J} \). Using \( E = mL \): \( m = \dfrac{E}{L} = \dfrac{18000}{2.26\times10^6} = 7.96\times10^{-3} \text{ kg} \). In practice, less steam would be produced than calculated because some of the electrical energy supplied is lost as heat to the surroundings (e.g. by conduction/convection/radiation from the heater and container) rather than all being used to vaporise the water.

1 mark: correct energy supplied (18000 J); 1 mark: correct formula \( m=E/L \); 1 mark: correct substitution; 1 mark: correct final answer \( 7.96\times10^{-3} \text{ kg} \) (accept 7.9–8.0 ×10⁻³); 1 mark: valid reason for heat loss to the surroundings reducing the actual mass produced. [5]

Energy per event: \( E = \Delta mc^2 = 3.20\times10^{-28}\times(3.00\times10^8)^2 = 3.20\times10^{-28}\times9.00\times10^{16} = 2.88\times10^{-11} \text{ J} \). Per mole: \( E_{mol} = E \times N_A = 2.88\times10^{-11}\times6.02\times10^{23} = 1.73\times10^{13} \text{ J mol}^{-1} \).

1 mark: correct equation \( E=\Delta mc^2 \); 1 mark: correct evaluation of \( c^2 \); 1 mark: correct energy per event \( 2.88\times10^{-11} \text{ J} \); 1 mark: correct scaling by \( N_A \); 1 mark: correct final per-mole answer \( 1.73\times10^{13} \text{ J mol}^{-1} \). [5]

Total binding energy before (2 deuterium nuclei, 4 nucleons total): \( 4 \times 1.11 = 4.44 \text{ MeV} \). Total binding energy after (\( ^3He \) has 3 nucleons; the free neutron is unbound and contributes zero binding energy): \( 3 \times 2.57 = 7.71 \text{ MeV} \). Energy released = increase in total binding energy \( = 7.71 - 4.44 = 3.27 \text{ MeV} \).

1 mark: correct total binding energy before (4.44 MeV); 1 mark: correct total binding energy after, with free neutron correctly given zero binding energy (7.71 MeV); 1 mark: correct method (energy released = increase in total binding energy); 1 mark: correct subtraction; 1 mark: correct final answer 3.27 MeV. [5]

\( I = I_0\left(\tfrac12\right)^{x/x_{1/2}} \). \( 0.125 = \left(\tfrac12\right)^{15.0/x_{1/2}} \). Taking logarithms: \( \ln(0.125) = \dfrac{15.0}{x_{1/2}}\ln(0.5) \), so \( x_{1/2} = 15.0\times\dfrac{\ln(0.5)}{\ln(0.125)} = 15.0\times\dfrac{-0.693}{-2.079} = 15.0/3.00 = 5.00 \text{ cm} \). (Equivalently, since \( 0.125 = (0.5)^3 \), 15.0 cm corresponds to exactly 3 half-value thicknesses.)

1 mark: correct exponential attenuation equation; 1 mark: recognising \( 0.125=(0.5)^3 \) or correct use of logarithms; 1 mark: correct number of half-value thicknesses (3); 1 mark: correct division 15.0/3; 1 mark: correct final answer, 5.00 cm. [5]

\( \omega = \sqrt{k/m} = \sqrt{40.0/0.250} = \sqrt{160} = 12.6 \text{ rad s}^{-1} \). (a) \( T = 2\pi/\omega = 2\pi/12.6 = 0.497 \text{ s} \). (b) \( v_{max} = \omega A = 12.6\times0.0500 = 0.632 \text{ m s}^{-1} \). (c) \( a_{max} = \omega^2 A = 160\times0.0500 = 8.00 \text{ m s}^{-2} \).

1 mark: correct ω from \( \sqrt{k/m} \); 1 mark: correct period T = 0.497 s; 1 mark: correct max speed formula \( v_{max}=\omega A \); 1 mark: correct max speed value 0.632 m s⁻¹; 1 mark: correct max acceleration 8.00 m s⁻² (via \( \omega^2A \)). [5]

(a) \( E_{total} = \tfrac12 m\omega^2A^2 = 0.5\times0.0200\times36.0\times(0.0800)^2 = 0.5\times0.0200\times36.0\times0.00640 = 2.30\times10^{-3} \text{ J} \). (b) \( PE = \tfrac12 m\omega^2 x^2 = 0.5\times0.0200\times36.0\times(0.0500)^2 = 0.5\times0.0200\times36.0\times0.00250 = 9.00\times10^{-4} \text{ J} \). \( KE = E_{total}-PE = 2.30\times10^{-3}-9.00\times10^{-4} = 1.40\times10^{-3} \text{ J} \).

1 mark: correct total energy formula \( \tfrac12m\omega^2A^2 \); 1 mark: correct total energy value; 1 mark: correct PE at x = 0.0500 m formula and substitution; 1 mark: correct method KE = E_total − PE; 1 mark: correct final KE value \( 1.40\times10^{-3} \text{ J} \). [5]

\( E = \dfrac{FL}{Ax} = \dfrac{150\times2.500}{1.964\times10^{-6}\times1.20\times10^{-3}} = \dfrac{375}{2.357\times10^{-9}} = 1.59\times10^{11} \text{ Pa} \).

1 mark: correct formula \( E=FL/(Ax) \); 1 mark: correct numerator (375); 1 mark: correct denominator (extension converted to m and multiplied by area); 1 mark: correct final answer \( 1.59\times10^{11} \); 1 mark: correct unit, Pa (or N m⁻²). [5]

\( k = F/x = 18.0/0.120 = 150 \text{ N m}^{-1} \). Elastic strain energy \( = \tfrac12 Fx = 0.5\times18.0\times0.120 = 1.08 \text{ J} \) (equivalently \( \tfrac12kx^2 = 0.5\times150\times(0.120)^2 = 1.08 \text{ J} \)).

1 mark: correct spring constant, 150 N m⁻¹; 1 mark: correct energy formula \( \tfrac12Fx \) (or \( \tfrac12kx^2 \)); 1 mark: correct substitution; 1 mark: correct final energy value, 1.08 J; 1 mark: correct units (N m⁻¹ for k, J for energy) throughout. [5]

\( \lambda = \dfrac{\ln2}{T_{1/2}} = \dfrac{0.693}{12.0} = 0.0578 \text{ day}^{-1} \). \( A = A_0e^{-\lambda t} = 8.00\times10^5 \times e^{-0.0578\times30.0} = 8.00\times10^5\times e^{-1.733} = 8.00\times10^5\times0.1768 = 1.41\times10^5 \text{ Bq} \).

1 mark: correct decay constant λ; 1 mark: correct exponential decay equation used; 1 mark: correct evaluation of the exponent and \( e^{-\lambda t} \); 1 mark: correct final answer \( 1.41\times10^5 \text{ Bq} \) (accept 1.38–1.44 ×10⁵). [4]

A metal block of known mass m has two holes drilled into it: one to hold an electrical immersion heater and one to hold a thermometer (or thermocouple), each smeared with a little oil to ensure good thermal contact. The block's initial temperature is recorded, together with the initial reading on a joulemeter (or, if using an ammeter and voltmeter, the current I and voltage V, together with a stopclock). The heater is switched on for a measured, fixed time t, supplying a known quantity of electrical energy \( E=VIt \) (or read directly from the joulemeter); the block is well insulated (e.g. wrapped in insulating foam or lagging) throughout to reduce heat loss to the surroundings during heating. After switching off, the maximum temperature reached by the block is recorded, giving a temperature rise \( \Delta\theta \). The specific heat capacity is then calculated from \( c = \dfrac{E}{m\Delta\theta} \). To correct for any remaining systematic heat loss to the surroundings, a cooling correction can be applied: the rate at which the block cools is measured over a period after the experiment, and this rate is used to estimate and add back the heat lost during the heating period, giving a more accurate value of \( \Delta\theta \) and hence of c.

Band A (5–6 marks): a full, coherent method including drilled holes with oil for the heater and thermometer, insulation of the block, measurement of electrical energy supplied (joulemeter, or V, I and t), measurement of temperature rise, the correct formula \( c=E/(m\Delta\theta) \), and a valid method for minimising/correcting for heat loss (insulation and/or a cooling correction); fluent scientific language, essentially free from errors in spelling, punctuation and grammar. Band B (3–4 marks): most elements present but less complete, e.g. insulation mentioned but no cooling correction, or vice versa. Band C (1–2 marks): only a fragment of the method, e.g. 'heat the block and measure the temperature rise'. Indicative content: drilled holes with oil for heater/thermometer; insulation/lagging; measurement of V, I, t (or joulemeter reading); measurement of temperature rise; formula \( c=E/(m\Delta\theta) \); cooling correction method. [6]

Since \( F = mr\omega^2 \), a graph of F against \( \omega^2 \) (with m and r constant) is a straight line passing through the origin, with a constant positive gradient. The gradient of this straight line is equal to mr; since r is known, the mass of the object can be found from \( m = \text{gradient}/r \).

1 mark: correct relationship \( F=mr\omega^2 \) identified; 1 mark: correct description of the graph as a straight line through the origin; 1 mark: correct statement that the gradient equals mr; 1 mark: correct method to find m from the gradient (m = gradient/r). [4]

The graph rises steeply from low values for light nuclides (small A) to a maximum at around \( A \approx 56 \) (in the region of iron/nickel), where the binding energy per nucleon is greatest (about 8.8 MeV), and then decreases slowly for heavier nuclides up to around \( A \approx 238 \) (uranium). In fission, a heavy nuclide (large A, to the right of the peak) splits into two medium-sized nuclides closer to the peak; since the binding energy per nucleon of the products is greater than that of the original heavy nuclide, the total binding energy increases, so energy is released. In fusion, light nuclides (small A, to the left of the peak) combine to form a heavier nuclide closer to the peak; since the product again has a greater binding energy per nucleon than the light reactant nuclides, the total binding energy again increases and energy is released.

1 mark: correct shape described, with a peak near A ≈ 56; 1 mark: correct explanation for fission (heavy nuclide moves towards the peak, binding energy per nucleon increases); 1 mark: correct explanation for fusion (light nuclides move towards the peak, binding energy per nucleon increases); 1 mark: correct overall link between increased binding energy and energy release. [4]

The graph is a straight line through the origin, showing that pressure is directly proportional to absolute temperature at constant volume. As temperature increases, the average kinetic energy of the gas molecules increases (mean kinetic energy is proportional to T), so the molecules move faster on average; this increases both the frequency of collisions with the container walls and the average force (rate of change of momentum) delivered per collision, both of which increase the pressure exerted. Since pressure is proportional to the average molecular kinetic energy, which is itself proportional to T, pressure is directly proportional to T, giving a straight line that passes through the origin (P → 0 as T → 0 K, absolute zero, where molecular kinetic energy would be zero).

1 mark: correct shape, straight line through the origin; 1 mark: correct link between temperature and mean molecular kinetic energy; 1 mark: correct explanation of increased collision frequency/force increasing pressure; 1 mark: correct explanation of why the line passes through the origin (P=0 at T=0 K). [4]

The graph is a cosine-shaped curve: it starts at the maximum positive displacement (+A) at t = 0, decreases smoothly through zero displacement, reaches maximum negative displacement (−A), then increases smoothly back through zero to +A again, repeating this pattern periodically. The period T can be determined from the graph as the time interval between two successive peaks (two successive points at which the displacement is at its maximum positive value, +A).

1 mark: correct cosine-shaped curve described, starting at maximum displacement; 1 mark: correct description of the repeating pattern through zero and negative displacement; 1 mark: correct method for finding T (time between successive peaks). [3]

(a) An electron is a lepton, and is a fermion. (b) A proton is a baryon (made of three quarks), and is a fermion. (c) A pion is a meson (made of a quark–antiquark pair), and is a boson. In general, leptons and baryons (built from an odd number of quarks) are fermions, while mesons (built from a quark–antiquark pair) are bosons.

1 mark: electron correctly classified as a lepton, fermion; 1 mark: proton correctly classified as a baryon, fermion; 1 mark: pion correctly classified as a meson, boson; 1 mark: correct general statement linking baryons/leptons to fermions and mesons to bosons. [4]

(a) A linear accelerator accelerates charged particles along a straight path, using a series of alternating electric fields (across a sequence of tubes/electrodes) to increase the particles' kinetic energy at each gap they cross. (b) A synchrotron accelerates charged particles around a fixed circular path many times, using electric fields to accelerate the particles at each pass and magnetic fields (increased in strength in synchronism with the particles' increasing momentum) to keep them on the same circular path. Advantage: because a synchrotron reuses the same accelerating regions repeatedly as the particles loop around many times, it can reach a much higher final energy within a far more compact device than a linear accelerator, which would need to be impractically long to reach a comparable energy in a single straight pass.

1 mark: correct purpose of a linac; 1 mark: correct purpose of a synchrotron, including the role of increasing magnetic field; 1 mark: valid advantage stated (more compact/higher achievable energy); 1 mark: correct reasoning linking the advantage to repeated use of the same accelerating region. [4]

(a) Gravitational field strength is the force per unit mass experienced by a small test mass placed at that point: \( g = \dfrac{GM}{r^2} \); it is a vector quantity, directed towards the mass M. (b) Gravitational potential is the work done per unit mass in bringing a small test mass from infinity to that point: \( V = -\dfrac{GM}{r} \); it is a scalar quantity, which is negative everywhere and tends to zero as r tends to infinity.

1 mark: correct definition of g; 1 mark: correct equation \( g=GM/r^2 \), noted as a vector; 1 mark: correct definition of V; 1 mark: correct equation \( V=-GM/r \), noted as a scalar with the correct sign convention. [4]

The force on a current-carrying conductor in a magnetic field depends on: the magnetic flux density, B; the current, I, flowing in the conductor; the length, L, of the conductor within the field; and the angle θ between the conductor and the field direction (the force is a maximum when the conductor is perpendicular to the field, and zero when it is parallel to the field). When the conductor is perpendicular to the field: \( F = BIL \).

1 mark: correct factors (B, I, L) stated; 1 mark: correct dependence on the angle between conductor and field (maximum when perpendicular, zero when parallel); 1 mark: correct equation \( F=BIL \) for the perpendicular case. [3]

(a) \( F = \dfrac{kq_1q_2}{r^2} = \dfrac{8.99\times10^9\times3.20\times10^{-6}\times5.00\times10^{-6}}{(0.400)^2} = \dfrac{0.1438}{0.160} = 0.899 \text{ N} \), attractive, since the charges have opposite signs. (b) At the midpoint, distance from each charge = 0.200 m. \( E_1 \) (from +charge) \( = kq_1/r^2 = 8.99\times10^9\times3.20\times10^{-6}/0.0400 = 7.19\times10^5 \text{ N C}^{-1} \); \( E_2 \) (from −charge) \( = kq_2/r^2 = 8.99\times10^9\times5.00\times10^{-6}/0.0400 = 1.12\times10^6 \text{ N C}^{-1} \). Both fields point in the same direction at the midpoint (from + towards −), so they add: \( E = E_1+E_2 = 7.19\times10^5+1.12\times10^6 = 1.84\times10^6 \text{ N C}^{-1} \).

1 mark: correct Coulomb's law formula and force value; 1 mark: correctly identified as attractive (opposite charges); 1 mark: correct E₁ at the midpoint; 1 mark: correct E₂ at the midpoint; 1 mark: correct reasoning that the two fields add (same direction) at the midpoint; 1 mark: correct final total E, \( 1.84\times10^6 \text{ N C}^{-1} \). [6]

(a) \( E = V/d = 500/0.0250 = 2.00\times10^4 \text{ V m}^{-1} \). (b) \( F = qE = 1.60\times10^{-19}\times2.00\times10^4 = 3.20\times10^{-15} \text{ N} \). (c) Work done on the electron \( = qV = 1.60\times10^{-19}\times500 = 8.00\times10^{-17} \text{ J} = \tfrac12mv^2 \). \( v = \sqrt{2\times8.00\times10^{-17}/9.11\times10^{-31}} = \sqrt{1.756\times10^{14}} = 1.33\times10^7 \text{ m s}^{-1} \).

1 mark: correct field strength E; 1 mark: correct force F = qE; 1 mark: correct work done, W = qV; 1 mark: correctly equating W to kinetic energy \( \tfrac12mv^2 \); 1 mark: correct rearrangement for v; 1 mark: correct final speed, \( 1.33\times10^7 \text{ m s}^{-1} \). [6]

\( W = q(V_1-V_2) = 2.00\times10^{-9}\times(150-40.0) = 2.00\times10^{-9}\times110 = 2.20\times10^{-7} \text{ J} \).

1 mark: correct formula \( W=q\Delta V \); 1 mark: correct potential difference (110 V); 1 mark: correct substitution; 1 mark: correct final answer, \( 2.20\times10^{-7} \text{ J} \); 1 mark: correct sign/reasoning that positive work is done as a positive charge moves to a region of lower potential. [5]

\( E = \dfrac{kq}{r^2} = \dfrac{8.99\times10^9\times6.00\times10^{-8}}{(0.0500)^2} = \dfrac{539.4}{0.00250} = 2.16\times10^5 \text{ N C}^{-1} \). Since \( E \propto 1/r^2 \), doubling r (to 0.100 m) reduces E to one quarter of its surface value: \( E = 2.16\times10^5/4 = 5.40\times10^4 \text{ N C}^{-1} \).

1 mark: correct formula \( E=kq/r^2 \); 1 mark: correct surface field value; 1 mark: correct inverse-square reasoning (E ∝ 1/r²); 1 mark: correct scaling factor (falls to one quarter); 1 mark: correct final value at 0.100 m. [5]

(a) \( Q=CV = 470\times10^{-6}\times12.0 = 5.64\times10^{-3} \text{ C} \). (b) \( E = \tfrac12CV^2 = 0.5\times470\times10^{-6}\times(12.0)^2 = 3.38\times10^{-2} \text{ J} \). (c) Combined capacitance \( = 470+470 = 940 \ \mu\text{F} \); total charge is conserved at \( 5.64\times10^{-3} \text{ C} \); new \( V = Q/C_{total} = 5.64\times10^{-3}/940\times10^{-6} = 6.00 \text{ V} \).

1 mark: correct charge Q; 1 mark: correct energy formula and value; 1 mark: correct combined capacitance (940 μF); 1 mark: correct use of charge conservation; 1 mark: correct rearrangement V=Q/C; 1 mark: correct final new voltage, 6.00 V. [6]

(a) \( \tau = RC = 2200\times100\times10^{-6} = 0.220 \text{ s} \). (b) \( V = V_0e^{-t/\tau} = 9.00\times e^{-0.500/0.220} = 9.00\times e^{-2.273} = 9.00\times0.1031 = 0.928 \text{ V} \).

1 mark: correct time constant formula; 1 mark: correct value of τ; 1 mark: correct exponential decay formula; 1 mark: correct evaluation of the exponent and \( e^{-t/\tau} \); 1 mark: correct final voltage, 0.928 V. [5]

(a) \( \dfrac{1}{C_{total}} = \dfrac{1}{200}+\dfrac{1}{300} = \dfrac{3+2}{600} = \dfrac{5}{600} \Rightarrow C_{total} = 120 \ \mu\text{F} \). (b) In series, the charge is the same on every capacitor: \( Q = C_{total}V = 120\times10^{-6}\times12.0 = 1.44\times10^{-3} \text{ C} \), the same on both the 200 μF and 300 μF capacitors.

1 mark: correct series capacitance formula; 1 mark: correct combined capacitance, 120 μF; 1 mark: correct charge formula using combined C; 1 mark: correct charge value, \( 1.44\times10^{-3} \text{ C} \); 1 mark: correct statement that this charge is the same on both capacitors in series. [5]

(a) \( F = BIL = 0.800\times3.50\times0.250 = 0.700 \text{ N} \). (b) The magnetic force provides the centripetal force: \( Bqv = \dfrac{mv^2}{r} \Rightarrow r = \dfrac{mv}{Bq} = \dfrac{1.67\times10^{-27}\times2.00\times10^6}{0.800\times1.60\times10^{-19}} = \dfrac{3.34\times10^{-21}}{1.28\times10^{-19}} = 0.0261 \text{ m} \).

1 mark: correct formula F=BIL; 1 mark: correct force value, 0.700 N; 1 mark: correct centripetal force = magnetic force equation set up; 1 mark: correct rearrangement for r; 1 mark: correct substitution; 1 mark: correct final radius, 0.0261 m. [6]

\( \varepsilon = N\dfrac{\Delta\Phi}{\Delta t} = N\dfrac{A\Delta B}{\Delta t} = 250\times\dfrac{4.00\times10^{-3}\times(0.200-0.0500)}{0.400} = 250\times\dfrac{4.00\times10^{-3}\times0.150}{0.400} = 0.375 \text{ V} \).

1 mark: correct Faraday's law formula; 1 mark: correct ΔB; 1 mark: correct flux change ΔΦ = AΔB; 1 mark: correct substitution including N; 1 mark: correct final EMF, 0.375 V. [5]

(a) \( KE = qV = 1.60\times10^{-19}\times2.50\times10^6 = 4.00\times10^{-13} \text{ J} \). Since the proton's charge is 1e and it is accelerated through \( 2.50\times10^6 \text{ V} \), its energy gain is \( 2.50\times10^6 \text{ eV} = 2.50 \text{ MeV} \). (b) \( KE=\tfrac12mv^2 \Rightarrow v=\sqrt{2KE/m} = \sqrt{2\times4.00\times10^{-13}/1.67\times10^{-27}} = \sqrt{4.79\times10^{14}} = 2.19\times10^7 \text{ m s}^{-1} \).

1 mark: correct KE in joules; 1 mark: correct KE in MeV; 1 mark: correct rearrangement of \( KE=\tfrac12mv^2 \) for v; 1 mark: correct substitution; 1 mark: correct final speed; 1 mark: valid comment that this speed (~7% of c) justifies the non-relativistic assumption. [6]

The total rest mass energy of the pion, 135 MeV, is entirely converted into the total energy of the two photons (the pion has zero kinetic energy, being at rest). By symmetry — since the two photons are identical particles — each photon carries exactly half of this total energy: \( E_{\text{each}} = 135/2 = 67.5 \text{ MeV} \). Since the pion is at rest before decay, its total momentum is zero; by conservation of momentum, the total momentum of the two photons after decay must also be zero. This requires the photons' momenta to be equal in magnitude but opposite in direction, so that their vector sum is zero — hence the two photons must be emitted in exactly opposite directions (180° apart).

1 mark: correct total energy (135 MeV) from rest mass energy; 1 mark: correct reasoning that identical photons share the energy equally; 1 mark: correct energy per photon, 67.5 MeV; 1 mark: correct statement that initial momentum is zero (pion at rest); 1 mark: correct conservation of momentum reasoning; 1 mark: correct conclusion that momenta must be equal and opposite, requiring opposite directions. [6]

As a satellite moves to a higher orbit, its distance r from the Earth's centre increases. Gravitational potential energy, \( E_p = -\dfrac{GMm}{r} \), is negative and becomes less negative (i.e. increases) as r increases, since dividing by a larger r brings the negative value closer to zero. The orbital speed of a satellite in a higher circular orbit is lower, since \( v=\sqrt{GM/r} \) decreases as r increases, so kinetic energy, \( E_k=\tfrac12mv^2=\dfrac{GMm}{2r} \), decreases as the satellite moves to a higher orbit. The total energy of a satellite in a circular orbit, \( E_{total}=E_k+E_p=-\dfrac{GMm}{2r} \), also increases (becomes less negative) as r increases, because the increase in potential energy is twice as large as the decrease in kinetic energy. Because the total energy must increase for the satellite to move to a higher orbit, positive work must be done on the satellite — for example, by firing its rocket engines — to raise it from the lower orbit to the higher orbit; this external work supplies the necessary increase in total mechanical energy.

Band A (5–6 marks): comprehensive, correct explanation covering all three energy changes (PE increases, KE decreases, total energy increases) with correct supporting reasoning/equations, and a correct link between the total energy increase and the positive work that must be done on the satellite; fluent scientific language, essentially free from errors in spelling, punctuation and grammar. Band B (3–4 marks): most elements present but less complete, e.g. correctly identifies the PE and KE changes but does not fully link this to total energy or work done. Band C (1–2 marks): only a fragment, e.g. states PE increases with height without further reasoning. Indicative content: \( E_p=-GMm/r \) becomes less negative (increases) with r; orbital speed and KE decrease as r increases; total energy \( E=-GMm/2r \) increases (less negative) with r; positive work must be done (e.g. by engines) to raise the satellite to a higher orbit. [6]

Since \( E = \dfrac{kQ}{r^2} \), E is directly proportional to \( 1/r^2 \), so the graph is a straight line through the origin, with gradient \( = kQ \). \( Q = \dfrac{\text{gradient}}{k} = \dfrac{3.60\times10^2}{8.99\times10^9} = 4.00\times10^{-8} \text{ C} \). Plotting E against \( 1/r^2 \) (rather than directly against r, which would give a curve) linearises the relationship into a straight line; a straight-line graph allows Q to be determined more accurately and reliably from a single gradient calculated using the whole data set (e.g. by a line of best fit), rather than by reading and averaging individual, less precise points from a curve.

1 mark: correct relationship/shape, straight line through the origin; 1 mark: gradient correctly identified as kQ; 1 mark: correct rearrangement Q = gradient/k; 1 mark: correct final value of Q; 1 mark: valid advantage of linearising the graph; 1 mark: correct reasoning that a straight line allows a more reliable/accurate determination using all data points. [6]

(a) The graph rises from zero, steeply at first, with a continuously decreasing gradient, approaching the EMF value of 6.00 V asymptotically (getting closer and closer but never quite reaching it) — an exponential growth/rising curve. (b) \( \tau = RC \Rightarrow C = \tau/R = 0.150/3000 = 5.00\times10^{-5} \text{ F} = 50.0 \ \mu\text{F} \). (c) \( V_C = V_0(1-e^{-t/\tau}) \); after one time constant, \( t=\tau \), so \( V_C = 6.00\times(1-e^{-1}) = 6.00\times(1-0.368) = 6.00\times0.632 = 3.79 \text{ V} \); this represents 63.2% of the final (maximum) voltage.

1 mark: correct description of the graph shape (rising exponential curve, decreasing gradient, approaching EMF asymptotically); 1 mark: correct time constant formula; 1 mark: correct capacitance value, 50.0 μF; 1 mark: correct charging equation \( V_C=V_0(1-e^{-t/\tau}) \); 1 mark: correct evaluation of \( e^{-1} \); 1 mark: correct value of \( V_C \), 3.79 V; 1 mark: correct percentage, 63.2%, correctly identified as the standard fraction reached after one time constant. [7]

Since \( B = \dfrac{\mu_0I}{2\pi r} \), a graph of B against \( 1/r \) is a straight line through the origin, with gradient \( = \dfrac{\mu_0I}{2\pi} \). \( I = \text{gradient} \times \dfrac{2\pi}{\mu_0} = 4.00\times10^{-7}\times\dfrac{2\pi}{4\pi\times10^{-7}} = \dfrac{8.00\times10^{-7}}{4.00\times10^{-7}} = 2.00 \text{ A} \).

1 mark: correct shape/relationship, straight line through origin; 1 mark: gradient correctly identified as \( \mu_0I/2\pi \); 1 mark: correct rearrangement for I; 1 mark: correct substitution; 1 mark: correct final answer, 2.00 A. [5]

A pendulum bob is attached to a string of measured length L (measured with a metre rule from the point of suspension to the centre of the bob, using a set square to ensure the length is measured vertically), and set swinging through a small angle (less than about 10°) so that simple harmonic motion applies. Using a stopwatch, the time for a large number of oscillations (e.g. 20 complete oscillations) is measured, starting the stopwatch as the pendulum passes through the equilibrium (lowest) point, where it moves fastest, reducing the effect of reaction-time error; timing a single oscillation would give a much larger percentage timing error. This is repeated at least three times for the same length to check consistency and to allow a mean to be calculated.

Trial | Time for 20 oscillations, t / s | Period, T = t/20 / s
1 | ... | ...
2 | ... | ...
3 | ... | ...
Mean T / s | | ...

1 mark: correct method for measuring the pendulum length (metre rule, set square, to centre of bob); 1 mark: small-angle condition stated with reason; 1 mark: timing many oscillations rather than one, with correct reasoning; 1 mark: correct starting point (equilibrium position) with reasoning about reaction time; 1 mark: at least 3 repeats specified for reliability/mean; 1 mark: correctly headed table with units for t and T; 1 mark: correct method for calculating T from total time (T = t/20). [7]

The spring is clamped vertically at its upper end using a stand, boss and clamp, with a pointer or reference mark attached at its lower (free) end, and a metre rule clamped vertically alongside the spring, with its scale close to the pointer and the observer's eye level with the pointer when taking readings, to minimise parallax error. The unstretched (natural) length of the spring, \( l_0 \), is first measured and recorded with no load attached. A range of known masses is then added one at a time (at least 5 different loads, in equal increments), and after each addition the new length of the spring is recorded once it has settled; the extension for each load is calculated as \( x = l - l_0 \).

Load, F / N | Length, l / cm | Extension, x = l − l₀ / cm
0 (no load) | ... | 0
(5 further rows) | ... | ...

1 mark: correct clamping arrangement (stand, boss, clamp, spring vertical); 1 mark: correct use of a pointer and ruler with parallax reduction; 1 mark: unstretched length measured first, before loading; 1 mark: at least 5 incremental loads specified; 1 mark: correct method for calculating extension (x = l − l₀); 1 mark: correctly headed table with units; 1 mark: clear, logical column structure covering all required quantities. [7]

\( T=2\pi\sqrt{L/g} \Rightarrow T^2 = \dfrac{4\pi^2}{g}L \). This is of the form \( y=mx \) (with \( y=T^2 \), \( x=L \), and \( c=0 \)), so a graph of \( T^2 \) against L gives a straight line through the origin with gradient \( m = \dfrac{4\pi^2}{g} \).

1 mark: correct squaring/rearrangement to \( T^2 = (4\pi^2/g)L \); 1 mark: gradient correctly identified as \( 4\pi^2/g \). [2]

\( T=2\pi\sqrt{m/k} \Rightarrow T^2 = \dfrac{4\pi^2}{k}m \). This is of the form \( y=mx \) (with \( y=T^2 \), \( x=m \), and \( c=0 \)), giving a straight line through the origin with gradient \( = \dfrac{4\pi^2}{k} \).

1 mark: correct squaring/rearrangement to \( T^2=(4\pi^2/k)m \); 1 mark: gradient correctly identified as \( 4\pi^2/k \). [2]

Scales should be chosen for both axes so that the plotted points occupy at least half the width and half the height of the graph grid, using sensible scale ratios (e.g. 1, 2 or 5 units per large square, avoiding awkward multiples such as 3 or 7); both axes should be clearly labelled with the quantity and its unit (e.g. '\( T^2 \) / s²' and 'L / m'). Each data point should be plotted accurately, to within about half a small square. A single, thin, straight line of best fit should then be drawn through the points using a clear plastic ruler, positioned so that there is a roughly equal number of points scattered on either side of the line along its length, rather than simply joining the first and last points.

1 mark: correct rule on scale size (points occupy at least half the grid, sensible scale ratios); 1 mark: correct axis labelling with units; 1 mark: correct accurate plotting of points; 1 mark: correct method for drawing a genuine line of best fit (balanced scatter, not joining end points). [4]

Two points that lie on the drawn best-fit line itself (not necessarily two of the original data points) are selected, chosen to be as far apart as possible along the line, to construct a large gradient triangle. The vertical (Δy) and horizontal (Δx) sides of this triangle are read off the axes as precisely as possible, and the gradient is calculated as \( \Delta y/\Delta x \), including the correct units. A large gradient triangle should be used, rather than two adjacent data points, because the absolute uncertainty in reading each coordinate from the graph is fixed (related to the smallest graph division); using a large triangle makes this fixed absolute reading uncertainty a much smaller proportion of the (larger) Δy and Δx values used, significantly reducing the percentage uncertainty in the calculated gradient.

1 mark: correct method — two points chosen on the line itself, not necessarily original data points; 1 mark: points chosen as far apart as possible; 1 mark: correct calculation method, gradient = Δy/Δx with units; 1 mark: correct reasoning that reading uncertainty is a fixed absolute value; 1 mark: correct explanation that a larger triangle reduces the percentage uncertainty in the gradient. [5]

\( \text{gradient} = \dfrac{4\pi^2}{g} \Rightarrow g = \dfrac{4\pi^2}{\text{gradient}} = \dfrac{39.48}{4.02} = 9.82 \text{ m s}^{-2} \). Percentage difference \( = \dfrac{|9.82-9.81|}{9.81}\times100 = 0.10\% \). This is excellent agreement, well within typical experimental uncertainty, supporting the validity of the pendulum method for determining g.

1 mark: correct rearrangement, g = 4π²/gradient; 1 mark: correct substitution; 1 mark: correct value of g, 9.82 m s⁻²; 1 mark: correct percentage difference formula; 1 mark: correct percentage difference value, 0.10%; 1 mark: valid comparative conclusion (close agreement with accepted value). [6]

(a) \( \text{gradient} = \dfrac{4\pi^2}{k} \Rightarrow k = \dfrac{4\pi^2}{\text{gradient}} = \dfrac{39.48}{0.395} = 99.9 \text{ N m}^{-1} \). (b) \( \text{intercept} = \dfrac{4\pi^2m_s}{3k} \Rightarrow m_s = \dfrac{3k\times\text{intercept}}{4\pi^2} = \dfrac{3\times99.9\times0.020}{39.48} = \dfrac{5.994}{39.48} = 0.152 \text{ kg} \).

1 mark: correct gradient–k relationship; 1 mark: correct value of k, 99.9 N m⁻¹; 1 mark: correct intercept–mₛ relationship identified; 1 mark: correct rearrangement for mₛ; 1 mark: correct substitution; 1 mark: correct final value of mₛ, 0.152 kg; 1 mark: correct units throughout (N m⁻¹ for k, kg for mₛ). [7]

Matching base units on both sides: \( \text{kg}^0\text{m}^0\text{s}^1 = (\text{kg m}^{-3})^a(\text{kg s}^{-2})^b(\text{m})^c = \text{kg}^{a+b}\text{m}^{-3a+c}\text{s}^{-2b} \). Equating powers of kg: \( a+b=0 \). Equating powers of m: \( -3a+c=0 \). Equating powers of s: \( -2b=1 \Rightarrow b=-\tfrac12 \). From the kg equation: \( a=-b=\tfrac12 \). From the m equation: \( c=3a=\tfrac32 \).

1 mark: correct unit equations set up for kg and m powers; 1 mark: correct equation for s powers, solved to give b = −1/2; 1 mark: correct value of a = 1/2 from the kg equation; 1 mark: correct value of c = 3/2 from the m equation. [4]

Taking logarithms (base 10) of both sides of \( y=Ax^n \): \( \lg y = \lg A + \lg(x^n) = \lg A + n\lg x \). This is of the form \( Y=mX+c \), with \( Y=\lg y \), \( X=\lg x \), gradient \( m=n \), and intercept \( c=\lg A \). Plotting \( \lg y \) (y-axis) against \( \lg x \) (x-axis) therefore gives a straight line whose gradient equals the power n, and whose y-intercept equals \( \lg A \), so that \( A=10^{c} \).

1 mark: correct application of log rules to both sides; 1 mark: correct identification that Y = lg y and X = lg x; 1 mark: correct axes stated (lg y vs lg x); 1 mark: gradient correctly identified as n; 1 mark: intercept correctly identified as lg A; 1 mark: correct final step, A = 10^c. [6]

\( \lg(2.00)=0.301,\ \lg(15.8)=1.199 \); \( \lg(4.00)=0.602,\ \lg(63.5)=1.803 \); \( \lg(8.00)=0.903,\ \lg(254)=2.405 \). Using the first and last points: \( n = \dfrac{\Delta(\lg y)}{\Delta(\lg x)} = \dfrac{2.405-1.199}{0.903-0.301} = \dfrac{1.206}{0.602} = 2.00 \).

1 mark: all three lg x values correct; 1 mark: all three lg y values correct; 1 mark: correct gradient formula using the first and last points; 1 mark: correct Δ(lg y); 1 mark: correct Δ(lg x); 1 mark: correct final value, n = 2.00. [6]

A line of best fit is first drawn through the data points, balancing points on either side of it. Using the error bars on each point, a 'maximum' line is then drawn as the steepest straight line that still passes through all (or the great majority) of the error bars/uncertainty ranges; similarly, a 'minimum' line is drawn as the least steep straight line that still passes through all (or the great majority) of the error bars. The gradient of each of these two lines is calculated using a large gradient triangle spanning most of the line's length, as for the best-fit line. Since the gradient of this graph equals n, the uncertainty in n is estimated as half the difference between the maximum and minimum gradients: \( \Delta n = \dfrac{|\text{gradient}_{max}-\text{gradient}_{min}|}{2} \).

1 mark: best-fit line drawn first, correctly described; 1 mark: correct role of the error bars in constraining the lines; 1 mark: correct definition of the maximum (steepest) line; 1 mark: correct definition of the minimum (least steep) line; 1 mark: correct method for finding each gradient (large triangle); 1 mark: correct uncertainty formula, Δn = half the difference in gradients; 1 mark: correct reasoning that n equals the gradient so this uncertainty transfers directly; 1 mark: overall clear, correctly sequenced explanation. [8]

(a) The best estimate of n is taken as the gradient of the best-fit line: \( n = 1.98 \). The uncertainty is half the difference between the maximum and minimum gradients: \( \Delta n = \dfrac{2.15-1.83}{2} = \dfrac{0.32}{2} = 0.16 \). Result: \( n = 1.98 \pm 0.16 \). (b) The theoretical value n = 2 lies within the range covered by the experimental result (1.98 − 0.16 = 1.82 up to 1.98 + 0.16 = 2.14, and 2 lies between 1.82 and 2.14), so the experimental result is consistent with, and does not rule out, the theoretical prediction of n = 2.

1 mark: n correctly taken as the best-fit gradient, 1.98; 1 mark: correct formula for Δn (half the difference of max/min gradients); 1 mark: correct value, Δn = 0.16; 1 mark: correct final expression, n = 1.98 ± 0.16; 1 mark: consistent rounding/sig figs between n and Δn; 1 mark: correct range calculated (1.82 to 2.14); 1 mark: n = 2 correctly identified as lying within this range; 1 mark: correct overall conclusion that the result is consistent with the theoretical prediction. [8]

\( \%\text{unc}(m) = \dfrac{0.002}{0.250}\times100 = 0.80\% \). \( \%\text{unc}(L) = \dfrac{0.005}{0.800}\times100 = 0.625\% \). \( \%\text{unc}(T) = \dfrac{0.02}{1.20}\times100 = 1.67\% \); since T is squared in the formula, its contribution doubles: \( \%\text{unc}(T^2) = 2\times1.67 = 3.33\% \). Total percentage uncertainty (uncertainties in a product/quotient add): \( \%\text{unc}(Q) = 0.80+0.625+3.33 = 4.76\% \). Absolute uncertainty: \( \Delta Q = 0.0476\times21.9 = 1.04 \).

1 mark: correct %unc(m); 1 mark: correct %unc(L); 1 mark: correct %unc(T) doubled for the square term; 1 mark: correct total %unc(Q) by addition; 1 mark: correct method converting % to absolute uncertainty; 1 mark: correct final ΔQ ≈ 1.04 (accept 1.0–1.1). [6]

\( R = V/I = 6.00/0.250 = 24.0 \ \Omega \). \( \%\text{unc}(V) = 0.10/6.00\times100 = 1.67\% \). \( \%\text{unc}(I) = 0.008/0.250\times100 = 3.20\% \). Total \( \%\text{unc}(R) = 1.67+3.20 = 4.87\% \). \( \Delta R = 0.0487\times24.0 = 1.17 \approx 1.2 \ \Omega \). Final: \( R = 24.0\pm1.2 \ \Omega \).

1 mark: correct value of R, 24.0 Ω; 1 mark: correct %unc(V); 1 mark: correct %unc(I); 1 mark: correct total %unc(R) by addition; 1 mark: correct final expression, R = 24.0 ± 1.2 Ω. [5]

(a) \( r=1.00 \text{ mm}: t\times r^2 = 8.20\times(1.00)^2 = 8.20 \). \( r=2.00 \text{ mm}: t\times r^2 = 2.05\times(2.00)^2 = 2.05\times4.00 = 8.20 \). \( r=3.00 \text{ mm}: t\times r^2 = 0.91\times(3.00)^2 = 0.91\times9.00 = 8.19 \). All three values of \( t\times r^2 \) are constant to within about 0.1% (≈8.20 mm²s), strongly supporting the proposed relationship \( t\propto r^{-2} \), i.e. \( t=k/r^2 \) with \( k\approx8.20 \text{ mm}^2\text{s} \); such close agreement across three independent points is well within any reasonable experimental uncertainty. (b) Take logarithms of both sides of \( t=kr^n \): \( \lg t = \lg k + n\lg r \); plotting \( \lg t \) against \( \lg r \) for a wider range of r values gives a straight line whose gradient equals n, allowing the power to be confirmed rigorously (it should be found to equal exactly −2) using a full data set rather than relying on just three points.

1 mark: correct t×r² for r=1.00 mm; 1 mark: correct t×r² for r=2.00 mm; 1 mark: correct t×r² for r=3.00 mm; 1 mark: correct conclusion that the values are constant, supporting the proposed relationship; 1 mark: valid comment on the closeness/precision of agreement; 1 mark: correct log-linearisation method described for (b); 1 mark: correct statement that the gradient of the log-log graph equals n. [7]