2023 CCEA GCSE Physics 1210 模拟试题及答案详解

(a) \( a = \dfrac{v-u}{t} = \dfrac{11-2.0}{6.0} = 1.5 \text{ m/s}^2 \). (b) \( s = \dfrac{(u+v)}{2} \times t = \dfrac{(2.0+11)}{2} \times 6.0 = 39 \text{ m} \) (check: \( s = ut + \tfrac{1}{2}at^2 = 2.0(6.0) + 0.5(1.5)(6.0)^2 = 12 + 27 = 39 \text{ m} \), confirming the answer by a second method). Final answer: a = 1.5 m/s², s = 39 m.

[1] correct equation \( a = \dfrac{v-u}{t} \); [1] a = 1.5 m/s² with unit [2]. [1] correct equation of motion selected (e.g. \( s = \dfrac{(u+v)}{2}t \)); [1] s = 39 m with unit (ecf from (a) not required as independent method) [2].

(a) \( v = u + at = 0 + (10)(2.5) = 25 \text{ m/s} \). (b) \( s = ut + \tfrac{1}{2}at^2 = 0 + 0.5(10)(2.5)^2 = 31.25 \text{ m} \). Self-check by a second route: \( v^2 = u^2 + 2as \Rightarrow s = \dfrac{v^2}{2a} = \dfrac{25^2}{2(10)} = \dfrac{625}{20} = 31.25 \text{ m} \), which agrees. Final answer: v = 25 m/s, height = 31.25 m.

[1] correct equation \( v = u + at \); [1] v = 25 m/s [2]. [1] correct equation of motion selected e.g. \( s = ut + \tfrac{1}{2}at^2 \); [1] height = 31.25 m (accept 31 m) [2].

\( \text{speed} = \dfrac{\text{distance}}{\text{time}} = \dfrac{400}{51.2} = 7.8125 \text{ m/s} \approx 7.8 \text{ m/s} \). Final answer: 7.8 m/s.

[1] correct equation, speed = distance ÷ time; [1] correct substitution \( \dfrac{400}{51.2} \); [1] 7.8 m/s (accept 7.81 m/s).

(a) \( a = \dfrac{F}{m} = \dfrac{60}{12} = 5.0 \text{ m/s}^2 \). (b) \( v = u + at = 0 + (5.0)(3.0) = 15 \text{ m/s} \). Final answer: a = 5.0 m/s², v = 15 m/s.

[1] correct equation \( a = F/m \); [1] a = 5.0 m/s² [2]. [1] correct equation \( v = u+at \); [1] v = 15 m/s (allow ecf from (a)) [2].

(a) \( \text{moment} = F \times d = 40 \times 0.60 = 24 \text{ N m} \). (b) For the beam to balance, the clockwise moment must equal the anticlockwise moment: \( F_2 \times 0.30 = 24 \), so \( F_2 = \dfrac{24}{0.30} = 80 \text{ N} \). Check: 80 N × 0.30 m = 24 N m, which equals the first moment, confirming equilibrium. Final answer: moment = 24 N m, second force = 80 N.

[1] correct equation moment = F × d; [1] 24 N m [2]. [1] states principle of moments (clockwise moment = anticlockwise moment for balance); [1] correct equation \( F_2 = 24 \div 0.30 \); [1] 80 N [3].

(a) \( V = 4.0^3 = 64 \text{ cm}^3 \). (b) \( \rho = \dfrac{m}{V} = \dfrac{512}{64} = 8.0 \text{ g/cm}^3 \); comparing to the three given values, 8.0 g/cm³ is closest to iron (7.9 g/cm³), so the cube is most likely iron. (c) \( 8.0 \text{ g/cm}^3 \times 1000 = 8000 \text{ kg/m}^3 \). Self-check by a second route: \( 512 \text{ g} = 0.512 \text{ kg} \) and \( 64 \text{ cm}^3 = 64 \times 10^{-6} \text{ m}^3 \), so \( \rho = \dfrac{0.512}{64\times10^{-6}} = 8000 \text{ kg/m}^3 \), which agrees. Final answer: volume = 64 cm³, density = 8.0 g/cm³ (iron), 8000 kg/m³.

(a) [1] 64 cm³. (b) [1] correct equation \( \rho = m/V \); [1] 8.0 g/cm³; [1] correctly identifies iron (closest to 7.9 g/cm³). (c) [1] correct conversion method (× 1000, or full unit conversion of g and cm³ to kg and m³); [1] 8000 kg/m³.

(a) \( E_p = mgh = 0.50 \times 10 \times 8.0 = 40 \text{ J} \). (b) All 40 J of g.p.e. becomes k.e.: \( \tfrac{1}{2}mv^2 = 40 \Rightarrow v^2 = \dfrac{2 \times 40}{0.50} = 160 \Rightarrow v = \sqrt{160} = 12.6 \text{ m/s} \). Self-check by a second route (kinematics of free fall): \( v^2 = u^2 + 2gh = 0 + 2(10)(8.0) = 160 \), giving the same v = 12.6 m/s. Final answer: E_p = 40 J, v = 12.6 m/s.

(a) [1] correct equation \( E_p = mgh \); [1] 40 J. (b) [1] correctly equates k.e. to 40 J and rearranges \( \tfrac{1}{2}mv^2 = E_p \); [1] v = 12.6 m/s (accept 12.65 m/s, allow ecf from (a)).

(a) \( E_p = mgh = 2.0 \times 10 \times 15 = 300 \text{ J} \). (b) \( P = \dfrac{\text{work done}}{\text{time taken}} = \dfrac{300}{6.0} = 50 \text{ W} \). (c) \( \text{efficiency} = \dfrac{\text{useful output energy}}{\text{total input energy}} = \dfrac{300}{500} = 0.60 \) (60%). Self-check: 60% of 500 J is 300 J, matching the output energy in (a), confirming consistency. Final answer: E_p = 300 J, power = 50 W, efficiency = 60%.

(a) [1] correct equation \( E_p = mgh \); [1] 300 J. (b) [1] P = 300 ÷ 6.0 = 50 W (allow ecf from (a)). (c) [1] correct equation efficiency = useful output ÷ total input; [1] 60% or 0.60 (allow ecf).

(a) Halving repeatedly from 800: 800 → 400 (1 half-life) → 200 (2) → 100 (3) → 50 (4). It takes 4 half-lives for the count rate to fall from 800 to 50 counts/minute. (b) 4 half-lives occur in 12 minutes, so one half-life \( = \dfrac{12}{4} = 3 \) minutes. Self-check: after 4 half-lives of 3 minutes each (12 minutes total), 800 → 400 → 200 → 100 → 50, which matches the given data. Final answer: 4 half-lives, half-life = 3 minutes.

(a) [1] correct halving sequence shown (800→400→200→100→50); [1] all intermediate values correct; [1] correctly states 4 half-lives. (b) [1] correct method, half-life = 12 ÷ 4; [1] 3 minutes (allow ecf from (a)).

(a) Alpha decay reduces the mass number by 4 and the proton number by 2: mass number = 232 − 4 = 228; proton number = 90 − 2 = 88. (b) Beta-minus decay leaves the mass number unchanged but increases the proton number by 1 (a neutron converts to a proton, emitting an electron): mass number = 228 (unchanged); proton number = 88 + 1 = 89. (c) Alpha particles are stopped by paper or a few centimetres of air, whereas beta particles are more penetrating and require a few millimetres of aluminium to stop them. Self-check: the total change from Th-232 (Z=90) after one alpha and one beta-minus decay is Δmass = −4, Δproton = −2+1 = −1, giving mass 228 and proton 89, consistent with the two-step calculation above. Final answer: X = mass 228, proton 88; Y = mass 228, proton 89; beta particles penetrate further than alpha particles.

(a) [1] mass number 228; [1] proton number 88. (b) [1] mass number 228 (unchanged); [1] proton number 89. (c) [1] any correct comparative statement that beta radiation penetrates further/is stopped by aluminium while alpha is stopped by paper/a few cm of air.

(a) Gradient of the velocity-time graph for 0–6 s: \( a = \dfrac{\Delta v}{\Delta t} = \dfrac{18-0}{6-0} = 3.0 \text{ m/s}^2 \). (b) The area under the graph is a triangle (0–6 s) plus a rectangle (6–12 s): triangle area \( = \tfrac{1}{2} \times 6 \times 18 = 54 \text{ m} \); rectangle area \( = 6 \times 18 = 108 \text{ m} \); total distance \( = 54 + 108 = 162 \text{ m} \). Self-check by a second route: average velocity over the whole 12 s is not simply the mean of endpoints because the motion has two phases, but using \( s = ut + \tfrac{1}{2}at^2 \) for 0–6 s gives \( 0 + 0.5(3.0)(6)^2 = 54 \text{ m} \), matching the triangle area, and constant-velocity distance for 6–12 s is \( 18 \times 6 = 108 \text{ m} \), matching the rectangle area. Final answer: acceleration = 3.0 m/s², total distance = 162 m.

(a) [1] correct method, gradient = Δv/Δt; [1] 3.0 m/s². (b) [1] triangle area (0–6 s) = 54 m; [1] rectangle area (6–12 s) = 108 m; [1] total = 162 m.

(a) \( \text{speed} = \dfrac{\text{distance}}{\text{time}} = \dfrac{25}{10} = 2.5 \text{ m/s} \). (b) Between t = 10 s and t = 20 s the distance stays at 25 m, so the graph is a horizontal (flat) line; this means the walker is stationary (not moving) during this interval. (c) \( \text{average speed} = \dfrac{\text{total distance}}{\text{total time}} = \dfrac{75}{40} = 1.875 \approx 1.9 \text{ m/s} \). Self-check: distance covered 20–40 s is 75 − 25 = 50 m in 20 s, i.e. 2.5 m/s, the same rate as the first 10 s, and the flat middle section correctly contributes zero to the total distance change, consistent with the overall average being below 2.5 m/s. Final answer: 2.5 m/s, stationary (flat graph), average speed 1.9 m/s.

(a) [1] equation speed = distance/time; [1] 2.5 m/s. (b) [1] correctly states the walker is stationary/at rest because the line is horizontal (distance is not changing). (c) [1] method, 75 m ÷ 40 s; [1] 1.9 m/s (accept 1.875 m/s).

(a) Converting extension to metres and finding \( F/x \) for each pair: \( 4/0.02 = 200 \), \( 8/0.04 = 200 \), \( 12/0.06 = 200 \), \( 16/0.08 = 200 \text{ N/m} \) — the ratio is constant, so force is directly proportional to extension (a straight line through the origin), confirming Hooke's law; spring constant k = 200 N/m. (b) The area under a force-extension graph is a triangle: \( E = \tfrac{1}{2} \times F \times x = \tfrac{1}{2} \times 16 \times 0.08 = 0.64 \text{ J} \). Self-check by a second route: \( E = \tfrac{1}{2}kx^2 = 0.5 \times 200 \times (0.08)^2 = 0.5 \times 200 \times 0.0064 = 0.64 \text{ J} \), which agrees. Final answer: k = 200 N/m, elastic potential energy = 0.64 J.

(a) [1] calculates F/x for at least two data points showing a constant value; [1] correctly states this constant ratio (straight line through origin) confirms Hooke's law; [1] k = 200 N/m. (b) [1] correct method, area of triangle = ½ × F × x; [1] 0.64 J.

(a) Time in seconds: \( 15 \times 60 = 900 \text{ s} \). The area under a power-time graph (a rectangle) gives energy: \( E = P \times t = 800 \times 900 = 720\,000 \text{ J} = 7.2 \times 10^5 \text{ J} \). (b) Since 90% of the input energy is usefully transferred: \( \text{input energy} = \dfrac{\text{useful energy}}{0.90} = \dfrac{720\,000}{0.90} = 800\,000 \text{ J} = 8.0 \times 10^5 \text{ J} \). Self-check: 90% of 800 000 J is 720 000 J, matching (a), confirming consistency. Final answer: 7.2 × 10⁵ J useful heat energy; 8.0 × 10⁵ J total electrical energy supplied.

(a) [1] converts 15 minutes to 900 s; [1] correct method E = P × t; [1] 7.2 × 10⁵ J (720 000 J). (b) [1] correct method, input = useful ÷ 0.90 (or equivalent efficiency equation); [1] 8.0 × 10⁵ J.

A strong answer explains that a gas boiler burns natural gas, transferring energy from the fuel's chemical energy store directly to the thermal energy store of the water/air in the house, and is typically around 90% efficient at this transfer. A heat pump instead uses electrical energy to drive a compressor in a refrigeration-type cycle, extracting thermal energy that is already present in the outside air (or ground) and transferring it, together with the electrical energy used, into the house; because it moves existing thermal energy rather than generating it from fuel, it can transfer more energy to the house than the electrical energy it consumes. It compares environmental impact: burning gas is a non-renewable process that releases carbon dioxide, contributing to climate change, whereas a heat pump produces no direct emissions at the point of use (though the electricity used to run it may still be generated partly from fossil fuels, depending on the electricity supply). It compares cost: heat pumps have a much higher installation/capital cost than a gas boiler but can have lower running costs, whereas gas boilers are cheaper to install but have ongoing fuel costs and rising carbon costs. It compares reliability: a gas boiler provides a constant, controllable heat output in any weather, whereas a heat pump's performance (its ability to transfer heat efficiently) decreases as the outside temperature falls, since there is a smaller temperature difference to work with, making it less effective on the coldest days.

Level of response marking (6 marks). Band 3 (5–6 marks): accurately describes the energy transfer for both a gas boiler (chemical → thermal, by combustion) and a heat pump (electrical energy used to move thermal energy from outside), compares efficiency/environmental impact (CO2 emissions vs none at point of use) and installation/running cost, and compares reliability in cold weather, using correct specialist terminology (e.g. energy transfer, thermal energy store, non-renewable, emissions) in clear, well-organised prose. Band 2 (3–4 marks): reasonable coverage of at least two of the three required areas (energy transfer mechanism, efficiency/cost/environmental impact, reliability) with generally correct terminology and communication. Band 1 (1–2 marks): basic, largely descriptive points covering only one area in any depth, weak use of terminology, unclear organisation. Band 0 (0 marks): no creditworthy material.

(a) Acceleration is the rate of change of velocity (the change in velocity per unit time). (b) The base SI unit of acceleration is metres per second squared, m/s². (c) Since \( a = \dfrac{\Delta v}{\Delta t} \) and the cyclist's velocity is constant, the change in velocity, \( \Delta v \), is zero, so the acceleration must also be zero. Final answer: acceleration = rate of change of velocity, unit m/s², zero because Δv = 0.

(a) [1] rate of change of velocity (accept: change in velocity ÷ time taken). (b) [1] m/s². (c) [1] correctly links zero acceleration to velocity not changing (Δv = 0).

(a) The gradient of a displacement-time graph represents velocity. (b) Because the gradient is constant, the object's velocity is constant; since velocity is not changing, the object's acceleration is zero. Final answer: gradient = velocity; the object moves at constant velocity with zero acceleration.

(a) [1] velocity. (b) [1] velocity is constant; [1] acceleration is zero (because velocity is not changing).

(a) Newton's first law states that an object will remain at rest, or continue to move at a constant velocity in a straight line, unless it is acted on by a resultant (unbalanced) force. (b) The two forces acting on the book are its weight (acting downward, due to gravity) and the normal contact/reaction force from the table (acting upward). Since these two forces are equal in size and opposite in direction, they cancel out, giving a resultant force of zero; by Newton's first law, an object with zero resultant force has no change in motion, so the book remains at rest. Final answer: Newton's first law as stated; weight and normal contact force are balanced, giving zero resultant force, so the book stays at rest.

(a) [1] correct statement of Newton's first law (rest or constant velocity unless acted on by a resultant force). (b) [1] correctly identifies weight and the normal contact/reaction force; [1] explains the forces are balanced (resultant force = 0), so by Newton's first law the book remains at rest.

(a) The principle of moments states that for a body in equilibrium (balanced), the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point. (b) For an object to be in equilibrium, the resultant force acting on it must be zero, and the resultant moment (turning effect) about any point must also be zero. Final answer: sum of clockwise moments = sum of anticlockwise moments (about the same point); resultant force = 0 and resultant moment = 0.

(a) [1] sum of clockwise moments = sum of anticlockwise moments; [1] correctly specifies this is about the same point/pivot. (b) [1] resultant force is zero and resultant moment is zero.

(a) \( \rho = \dfrac{m}{V} \), where the SI unit of density is kg/m³. (b) The density of a gas is much lower than the density of the same substance as a solid, because in a gas the particles are far apart and moving randomly, whereas in a solid the particles are closely packed together, so a given volume of gas contains far less mass than the same volume of solid. Final answer: ρ = m/V, unit kg/m³; gas density is lower because gas particles are far apart.

(a) [1] \( \rho = m/V \); [1] kg/m³. (b) [1] lower, with a correct reason referring to particles being further apart/less closely packed in a gas.

(a) Gas particles are in continuous, random motion and repeatedly collide with the walls of their container; each collision exerts a small force on the wall, and the pressure of the gas is the total effect of these very many frequent collisions (force per unit area) on the container walls. (b) Increasing the temperature increases the average kinetic energy of the gas particles, so they move faster; at constant volume this means the particles collide with the walls more frequently and with greater force, so the pressure increases. Final answer: pressure arises from frequent particle collisions with the walls; pressure increases with temperature at constant volume.

(a) [1] particles move randomly and collide with the container walls; [1] these (frequent) collisions exert a force, and pressure is this force per unit area. (b) [1] pressure increases (particles move faster, causing more frequent and/or more forceful collisions).

(a) The principle of conservation of energy states that energy cannot be created or destroyed, only transferred (or transformed) from one energy store to another; the total energy of a system remains constant. (b) As the pendulum bob swings from its highest point to its lowest point, energy is transferred from the gravitational potential energy store to the kinetic energy store, reaching maximum kinetic energy at the lowest point. The bob does not return to exactly the same height because some of this energy is dissipated to the thermal energy store of the surroundings, due to air resistance acting on the bob and friction at the pivot, so less useful mechanical energy is available to raise the bob back up. Final answer: energy is conserved overall; g.p.e. → k.e., with some energy dissipated as heat due to air resistance/friction, reducing the height reached.

(a) [1] correct statement that energy cannot be created or destroyed, only transferred, and total energy is conserved. (b) [1] correctly describes gravitational PE transferring to KE; [1] explains energy dissipated to thermal energy store of surroundings due to air resistance/friction at the pivot.

(a) Efficiency is the proportion (or percentage) of the total input energy to a device that is usefully transferred, given by \( \text{efficiency} = \dfrac{\text{useful output energy}}{\text{total input energy}} \). (b) Energy that is not usefully transferred is not destroyed; it is dissipated to the surroundings, spreading out and becoming less useful. This wasted energy is often transferred to the thermal energy store of the surroundings (as heat), commonly due to friction between moving parts. Final answer: efficiency = useful output ÷ total input (as a fraction or %); wasted energy is dissipated to the surroundings, typically as heat.

(a) [1] correct definition (useful output energy as a proportion/percentage of total input energy). (b) [1] energy is dissipated to the surroundings, not destroyed; [1] correctly identifies this is typically as thermal energy/heat (e.g. due to friction).

(a) An alpha particle consists of 2 protons and 2 neutrons (equivalent to a helium nucleus). (b) One use of gamma radiation is sterilising medical equipment (another accepted example is cancer treatment/radiotherapy, or medical tracers). Gamma radiation is suitable because it is highly penetrating, so it can pass through the packaging of the equipment (or through body tissue, for tracers/treatment) to reach where it is needed, unlike alpha or beta radiation, which have much lower penetrating power and would be absorbed before reaching/passing through the material. Final answer: alpha particle = 2 protons + 2 neutrons; gamma radiation is used because of its high penetrating power.

(a) [1] 2 protons and 2 neutrons. (b) [1] valid use of gamma radiation stated (e.g. sterilising equipment, cancer treatment, medical tracer); [1] correct explanation referring to gamma's high penetrating power.

(a) When a nucleus emits a beta-minus particle, the nucleon (mass) number stays the same (unchanged), while the proton number increases by 1, because a neutron in the nucleus changes into a proton (emitting an electron in the process). (b) A beta-minus particle is a fast-moving (high-energy) electron. Final answer: nucleon number unchanged, proton number increases by 1; a beta-minus particle is a high-energy electron.

(a) [1] nucleon number unchanged; [1] proton number increases by 1. (b) [1] a (fast-moving/high-energy) electron.

(a) \( P = \dfrac{E}{t} = \dfrac{72\,000}{60} = 1200 \text{ W} \). (b) \( P = VI \Rightarrow V = \dfrac{P}{I} = \dfrac{1200}{5.0} = 240 \text{ V} \). Self-check: \( VI = 240 \times 5.0 = 1200 \text{ W} \), matching (a). Final answer: power = 1200 W, p.d. = 240 V.

(a) [1] correct equation \( P = E/t \); [1] 1200 W. (b) [1] correct equation \( V = P/I \); [1] 240 V (allow ecf from (a)).

(a) \( I = \dfrac{V}{R} = \dfrac{6.0}{12} = 0.50 \text{ A} \). (b) In series, resistances add: \( R_{total} = 12 + 12 = 24 \text{ Ω} \); \( I = \dfrac{V}{R_{total}} = \dfrac{6.0}{24} = 0.25 \text{ A} \). Self-check: since resistance doubled and voltage is unchanged, current should halve (0.50 → 0.25 A), which agrees. Final answer: 0.50 A, then 0.25 A.

(a) [1] correct equation \( I = V/R \); [1] 0.50 A. (b) [1] correct total series resistance (24 Ω); [1] 0.25 A.

In parallel, the full 12 V p.d. acts across each resistor. (a) \( I_1 = \dfrac{V}{R_1} = \dfrac{12}{3.0} = 4.0 \text{ A} \). (b) \( I_2 = \dfrac{V}{R_2} = \dfrac{12}{6.0} = 2.0 \text{ A} \). (c) Total current \( = I_1 + I_2 = 4.0 + 2.0 = 6.0 \text{ A} \). Self-check: combined resistance of the parallel pair is \( \dfrac{1}{R} = \dfrac{1}{3.0}+\dfrac{1}{6.0} = 0.5 \Rightarrow R = 2.0 \text{ Ω} \), giving total current \( = 12/2.0 = 6.0 \text{ A} \), which agrees. Final answer: 4.0 A, 2.0 A, total 6.0 A.

(a) [1] correct equation \( I=V/R \); [1] 4.0 A. (b) [1] correct equation; [1] 2.0 A. (c) [1] 6.0 A (= sum of branch currents).

(a) \( v = f\lambda \Rightarrow f = \dfrac{v}{\lambda} = \dfrac{340}{0.68} = 500 \text{ Hz} \). (b) \( T = \dfrac{1}{f} = \dfrac{1}{500} = 0.0020 \text{ s} \). Self-check: \( f \times \lambda = 500 \times 0.68 = 340 \text{ m/s} \), matching the given speed. Final answer: f = 500 Hz, T = 2.0 × 10⁻³ s.

(a) [1] correct equation \( f = v/\lambda \); [1] 500 Hz. (b) [1] correct equation \( T = 1/f \); [1] 0.0020 s (allow ecf from (a)).

(a) \( v = f\lambda \Rightarrow \lambda = \dfrac{v}{f} = \dfrac{3.0\times10^8}{1.0\times10^6} = 300 \text{ m} \). (b) A wavelength of hundreds of metres corresponds to radio waves (e.g. typical AM radio broadcast wavelengths), the longest-wavelength region of the electromagnetic spectrum. (c) Radio waves are non-ionising radiation. Self-check: \( f\lambda = 1.0\times10^6 \times 300 = 3.0\times10^8 \text{ m/s} \), matching the given wave speed. Final answer: λ = 300 m, radio waves, non-ionising.

(a) [1] correct equation \( \lambda = v/f \); [1] 300 m. (b) [1] radio waves. (c) [1] non-ionising.

(a) \( \dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} \Rightarrow V_s = 230 \times \dfrac{2000}{400} = 230 \times 5 = 1150 \text{ V} \). (b) For an ideal transformer, input power = output power: \( V_p I_p = V_s I_s \Rightarrow I_s = \dfrac{V_p I_p}{V_s} = \dfrac{230 \times 0.50}{1150} = \dfrac{115}{1150} = 0.10 \text{ A} \). Self-check: output power \( = V_s I_s = 1150 \times 0.10 = 115 \text{ W} \), equal to input power \( = 230 \times 0.50 = 115 \text{ W} \), confirming conservation of power. Final answer: secondary voltage = 1150 V, secondary current = 0.10 A.

(a) [1] correct equation \( V_s/V_p = N_s/N_p \) rearranged; [1] 1150 V. (b) [1] correct equation \( V_pI_p = V_sI_s \); [1] correct substitution; [1] 0.10 A (allow ecf from (a)).

(a) \( F = BIl = 0.25 \times 3.0 \times 0.40 = 0.30 \text{ N} \). (b) Since \( F = BIl \) and B and l are unchanged, F is directly proportional to I; doubling I doubles F, giving \( F = 0.25 \times 6.0 \times 0.40 = 0.60 \text{ N} \). Self-check: 0.60 N is exactly double 0.30 N, consistent with the direct proportionality. Final answer: F = 0.30 N; doubling the current doubles the force to 0.60 N.

(a) [1] correct equation \( F = BIl \); [1] correct substitution; [1] 0.30 N. (b) [1] correctly states the force doubles; [1] correct reasoning (F directly proportional to I) or states new value 0.60 N.

(a) \( \text{time} = \dfrac{\text{distance}}{\text{speed}} = \dfrac{2.28\times10^{11}}{3.00\times10^8} = 760 \text{ s} \). (b) \( \dfrac{760}{60} = 12.67 \approx 12.7 \text{ minutes} \) (12 minutes 40 seconds). Self-check: \( 12.7 \times 60 = 762 \text{ s} \approx 760 \text{ s} \) (small rounding), consistent. Final answer: 760 s, 12.7 minutes.

(a) [1] correct equation, time = distance ÷ speed; [1] 760 s. (b) [1] correct method (÷60); [1] 12.7 minutes (accept 12 min 40 s, allow ecf).

(a) \( n = \dfrac{3.0\times10^8}{2.0\times10^8} = 1.5 \) (no unit). (b) A refractive index greater than 1 means light travels more slowly in the glass than in air (here, 1.5 times slower); when the ray meets the boundary at an angle to the normal, it bends towards the normal as it enters the glass (the denser medium). Final answer: n = 1.5; light slows down and bends towards the normal on entering the glass.

(a) [1] correct substitution into the given equation; [1] n = 1.5. (b) [1] light slows down/travels more slowly in glass; [1] the ray bends towards the normal.

(a) The normal is an imaginary line drawn perpendicular (at 90°) to the boundary (surface) at the point where the ray strikes it; angles of incidence and refraction are always measured from the normal. (b) \( n = \dfrac{\sin i}{\sin r} \Rightarrow \sin r = \dfrac{\sin i}{n} = \dfrac{\sin 50°}{1.5} = \dfrac{0.766}{1.5} = 0.511 \Rightarrow r = \sin^{-1}(0.511) = 30.7° \). Self-check: substituting back, \( \sin(30.7°) = 0.511 \), and \( \dfrac{\sin 50°}{\sin 30.7°} = \dfrac{0.766}{0.511} = 1.5 \), matching the given refractive index. Final answer: the normal is perpendicular to the surface at the point of incidence; angle of refraction = 30.7°.

(a) [1] correct description (line at right angles/perpendicular to the surface at the point of incidence). (b) [1] correct rearrangement, \( \sin r = \sin i / n \); [1] correct substitution \( \sin 50°/1.5 \); [1] r = 30.7° (accept 30–31°).

(a) \( \sin C = \dfrac{1}{n} = \dfrac{1}{1.5} = 0.667 \Rightarrow C = \sin^{-1}(0.667) = 41.8° \). (b) At angles of incidence greater than the critical angle, the ray undergoes total internal reflection: all of the light is reflected back into the glass at the boundary (following the normal law of reflection), and none is transmitted (refracted) into the air. Self-check: substituting back, \( \sin(41.8°) = 0.667 = 1/1.5 \), confirming the value. Final answer: critical angle = 41.8°; total internal reflection occurs.

(a) [1] correct equation \( \sin C = 1/n \); [1] correct substitution; [1] C = 41.8° (accept 41.8°–41.9°). (b) [1] total internal reflection occurs (all the light is reflected back into the glass).

(a) Total internal reflection requires: the light must be travelling within the optically denser medium (the core) towards the boundary with a less dense medium (the cladding); and the angle of incidence at that boundary must be greater than the critical angle for that boundary. (b) A common application is telecommunications, where optical fibres carry data (e.g. broadband internet and telephone signals) as pulses of light over long distances; another accepted application is a medical endoscope, used to see inside the body. Final answer: light travelling core→cladding at an angle greater than the critical angle; used in telecommunications/endoscopy.

(a) [1] light travelling from a denser to a less dense medium (core to cladding); [1] angle of incidence greater than the critical angle. (b) [1] valid, correctly explained application (e.g. telecommunications/broadband, medical endoscopy).

(a) The displacement pattern (0 → 5 → 0 → −5 → 0) repeats every 2.0 m along the rope, so the wavelength is 2.0 m. (b) The amplitude is the maximum displacement from the undisturbed (rest) position, which is 5 cm. (c) In a transverse wave, the particles oscillate at right angles (perpendicular) to the direction the wave travels, whereas in a longitudinal wave the particles oscillate parallel to the direction of travel; also, transverse waves can be polarised, while longitudinal waves cannot (or: longitudinal waves form compressions and rarefactions, transverse waves form crests and troughs). Final answer: wavelength = 2.0 m, amplitude = 5 cm; transverse waves oscillate perpendicular to the direction of travel and can be polarised, unlike longitudinal waves.

(a) [1] 2.0 m. (b) [1] 5 cm. (c) [1] correct statement re perpendicular vs parallel oscillation direction; [1] correct statement re polarisation, or crests/troughs vs compressions/rarefactions.

(a) The law of reflection states that the angle of incidence equals the angle of reflection, both measured from the normal to the boundary. (b) When a wave pulse reflects off a fixed end, the reflected pulse is inverted (flipped upside down) relative to the incoming pulse. (c) A wavefront is a line (or surface) joining points on a wave that are in phase with each other (e.g. all momentarily at a crest), showing the shape of the wave and the direction in which it is travelling. Final answer: angle of incidence = angle of reflection; the pulse is inverted; a wavefront joins points in phase.

(a) [1] angle of incidence = angle of reflection. (b) [1] inverted/flipped. (c) [1] correct definition of wavefront.

A strong answer explains that in a series circuit there is only one path for current, so the same current flows through every component in the circuit, while the total potential difference supplied by the source is shared (split) between the components. In a parallel circuit there is more than one path (branch) for current, so the current splits between the branches (the branch currents add up to equal the total supply current), while every branch has the same potential difference across it, equal to the full supply voltage. It explains that if a component fails in a series circuit (e.g. a bulb filament breaks), the single loop is broken, so no current can flow anywhere in that circuit and everything connected in that series loop stops working; but if a component fails in a parallel circuit, the other branches still form complete loops, so current continues to flow through them and they continue to operate normally — only the failed branch is affected. It concludes that household circuits are wired in parallel so that each light or appliance can be switched on and off independently of the others, and so that every appliance receives the full supply voltage (e.g. 230 V) regardless of how many other appliances are connected or switched on, allowing them all to work correctly and independently.

Level of response marking (6 marks). Band 3 (5–6 marks): correctly and fully describes current/p.d. behaviour in both series and parallel circuits, correctly explains the effect of a component failing in each case, and gives a complete, correct explanation of why household circuits are wired in parallel (independent operation, full supply voltage to each device), using correct specialist terminology in clear, well-organised prose. Band 2 (3–4 marks): reasonable coverage of at least two of the three required areas, with generally correct terminology. Band 1 (1–2 marks): basic, largely descriptive coverage of only one area, weak terminology or unclear organisation. Band 0 (0 marks): no creditworthy material.

(a) \( R = \dfrac{V}{I} = \dfrac{4.0}{0.80} = 5.0 \text{ Ω} \) (this ratio is constant for all data pairs, confirming it is a fixed/ohmic resistor). (b) At low voltage the lamp's resistance is \( R = 1.0/0.30 = 3.3 \text{ Ω} \), while at 4.0 V it is \( R = 4.0/0.70 = 5.7 \text{ Ω} \), so resistance increases as voltage (and current) increases; this happens because a larger current heats the filament to a higher temperature, and the increased temperature causes the metal ions in the filament to vibrate more, increasing the frequency of collisions with the charge carriers (electrons) and so increasing resistance. (c) Both graphs pass through the origin — at zero voltage, the current in each component is zero. Final answer: resistor resistance = 5.0 Ω (constant); lamp resistance increases with temperature; both graphs pass through the origin.

(a) [1] correct equation \( R = V/I \); [1] 5.0 Ω. (b) [1] correctly states resistance increases as voltage/current increases; [1] correct explanation referring to the filament heating up and increased ion vibration/collisions increasing resistance. (c) [1] both graphs pass through the origin (zero current at zero voltage).

(a) Using \( v = f\lambda \) for each pair: \( 2.0 \times 1.20 = 2.4 \); \( 3.0 \times 0.80 = 2.4 \); \( 4.0 \times 0.60 = 2.4 \); \( 6.0 \times 0.40 = 2.4 \text{ m/s} \). All four data points give the same wave speed of 2.4 m/s, confirming the speed is constant. (b) As frequency increases, wavelength decreases in the same proportion (e.g. tripling the frequency, from 2.0 to 6.0 Hz, reduces the wavelength to a third, from 1.20 to 0.40 m), so frequency and wavelength are inversely proportional; their product, the wave speed, stays constant. Final answer: f × λ = 2.4 m/s for all points; frequency and wavelength are inversely proportional.

(a) [1] calculates f × λ correctly for at least two data points; [1] correctly concludes all four pairs give 2.4 m/s, confirming constant speed. (b) [1] correctly states frequency and wavelength are inversely proportional; [1] correct justification (their product/the wave speed stays constant).

(a) \( v = \dfrac{2\pi r}{T} = \dfrac{2\pi \times (60\times10^6)}{90} = \dfrac{3.77\times10^8}{90} = 4.19\times10^6 \text{ km/day} \). (b) As orbital radius increases (60 → 110 → 230 × 10⁶ km), orbital period also increases (90 → 220 → 690 days), and the period increases proportionally faster than the radius; using \( v = 2\pi r/T \), calculating the speeds of Q and R in the same way gives approximately \( 3.14\times10^6 \) km/day and \( 2.09\times10^6 \) km/day respectively, both lower than planet P's speed — so orbital speed decreases as distance from the star increases. Final answer: v(P) ≈ 4.19 × 10⁶ km/day; period increases with radius, and orbital speed decreases with distance from the star.

(a) [1] correct substitution into \( v = 2\pi r/T \); [1] v ≈ 4.19 × 10⁶ km/day (accept 4.2 × 10⁶). (b) [1] correctly states orbital period increases as orbital radius increases; [1] correctly deduces/states that orbital speed decreases with increasing distance from the star.

(a) Refraction is the change in direction (bending) of light as it crosses a boundary between two transparent media of different optical density, caused by a change in the speed of the light. (b) On entering a denser medium such as glass, the light slows down; because it slows down while crossing the boundary at an angle to the normal, the ray bends towards the normal. Final answer: refraction is the bending of light due to a change in speed at a boundary; entering a denser medium, light slows down and bends towards the normal.

(a) [1] correct definition (change of direction/bending of light due to a change of speed at a boundary). (b) [1] light slows down entering the denser medium; [1] the ray bends towards the normal.

(a) Violet light has a wavelength of approximately 400 nm, and red light has a wavelength of approximately 700 nm. (b) Since \( v = f\lambda \) and all visible light travels at the same speed in a given medium, a longer wavelength corresponds to a lower frequency; as red light has the longer wavelength, it has a lower frequency than violet light. Final answer: violet ≈ 400 nm, red ≈ 700 nm; red light has a lower frequency than violet light.

(a) [1] violet ≈ 400 nm; [1] red ≈ 700 nm. (b) [1] correctly states red light has a lower frequency than violet light, with correct reasoning (same speed, longer wavelength means lower frequency).

(a) In order of increasing frequency (decreasing wavelength): radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. (b) A common use of microwaves is cooking/heating food in a microwave oven (another accepted use is mobile phone or satellite communication signals). Final answer: radio waves → microwaves → infrared → visible → ultraviolet → X-rays → gamma rays; microwaves are used for cooking food or communications.

(a) [1] correct order for at least the first four regions; [1] fully correct order for all seven regions. (b) [1] valid correct use of microwaves.

(a) The magnetic field around a long straight current-carrying wire forms a pattern of concentric circles, centred on the wire, in a plane perpendicular to the wire. (b) The strength of a solenoid's magnetic field can be increased by increasing the current flowing through it, or by increasing the number of turns on the coil (or by placing a soft-iron core inside the solenoid). Final answer: concentric circles around the wire; increase current and/or number of turns (or add an iron core).

(a) [1] concentric circles centred on the wire. (b) [1] increase the current; [1] increase the number of turns (or add a soft-iron core).

The secondary coil has more turns (1000) than the primary coil (200), and since \( V_s/V_p = N_s/N_p \), a greater number of secondary turns means the secondary (output) voltage is greater than the primary (input) voltage — this is a step-up transformer. Final answer: step-up transformer, because the secondary has more turns than the primary, giving a higher output voltage.

[1] correctly identifies step-up; [1] correct explanation (more secondary turns than primary turns means higher output than input voltage).

Red-shift is the increase in the wavelength of light from a distant galaxy (a shift of its spectral lines towards the red end of the visible spectrum), observed because the galaxy is moving away from (receding from) the observer. Final answer: red-shift = increased wavelength of light from a receding galaxy, shifted towards the red end of the spectrum.

[1] wavelength of light from the galaxy is increased/shifted towards the red end of the spectrum; [1] because the galaxy is moving away from (receding from) the observer/Earth.

This evidence suggests that all galaxies are moving apart from one another, i.e. that the universe is expanding, with more distant galaxies moving away faster than closer ones. This supports the Big Bang theory, which states that the universe began from a very hot, dense point and has been expanding ever since. Final answer: the universe is expanding; this supports the Big Bang theory.

[1] correctly states the universe is expanding (galaxies moving apart, more distant ones faster); [1] correctly names the Big Bang theory.

During the main sequence stage, a star fuses hydrogen nuclei into helium in its core (nuclear fusion), releasing the energy that makes the star shine; the star is stable (in equilibrium), because the outward pressure produced by this fusion balances the inward pull of gravity trying to collapse the star. A star spends most of its life in this stable, main-sequence stage. Final answer: hydrogen fuses into helium in the core, releasing energy; the star is stable because fusion pressure balances gravitational collapse.

[1] hydrogen nuclei fuse into helium in the core, releasing energy (nuclear fusion); [1] the star is stable/in equilibrium because the outward pressure from fusion balances the inward pull of gravity.

Diameter = main scale reading + (thimble division × 0.01 mm) = 4.5 + (23 × 0.01) = 4.5 + 0.23 = 4.73 mm. Final answer: 4.73 mm.

[1] correct method (4.5 + 0.23 shown); [1] 4.73 mm.

To the nearest whole gram, 127.4 g rounds to 127 g. However, recording the full reading (127.4 g) retains more precision; rounding early to 127 g discards information and introduces an additional rounding error that then carries through (and can be magnified) in any later calculation using this mass, such as a density calculation. Final answer: 127 g; recording to one decimal place preserves precision and avoids extra rounding error in subsequent calculations.

[1] 127 g; [1] correct explanation that retaining more decimal places preserves precision/reduces rounding error carried into later calculations.

Period \( T = \dfrac{24.6}{20} = 1.23 \text{ s} \). Timing 20 oscillations and dividing by 20 gives a more accurate value for the period because the fixed absolute error introduced by the student's reaction time (in starting/stopping the stopwatch) is spread over 20 oscillations rather than 1, so it has a much smaller effect (percentage impact) on the final value of the period. Final answer: T = 1.23 s; timing many oscillations reduces the effect of reaction-time error on the calculated period.

[1] T = 1.23 s; [1] correct explanation that timing more oscillations reduces the effect/percentage impact of reaction-time error on the period.

An ammeter must be connected in series in the circuit, so that the same current flowing through the component also flows through the ammeter. A voltmeter must be connected in parallel across the component being measured, so that it reads the potential difference across that component without significantly affecting the current in the main circuit. Final answer: ammeter connected in series; voltmeter connected in parallel across the component.

[1] ammeter connected in series; [1] voltmeter connected in parallel (across the component).

This avoids parallax error. Parallax error occurs when the observer's eye is not positioned directly in line with (perpendicular to) the scale and the point being measured; viewing the scale from an angle makes the reading appear shifted from its true value. Final answer: parallax error, caused by viewing a scale at an angle rather than directly in line with it.

[1] parallax error; [1] correct explanation (viewing the scale at an angle causes the reading to appear incorrect/shifted from its true value).

(a) Mean \( = \dfrac{2.31+2.34+2.28+2.35+2.30}{5} = \dfrac{11.58}{5} = 2.316 \approx 2.32 \text{ s} \). (b) Taking repeat readings and finding a mean reduces the effect of random error on the result (any readings that are too high tend to be balanced out by readings that are too low), giving a value that is more precise and more likely to be close to the true value. Final answer: mean = 2.32 s; repeats reduce the effect of random error, improving precision/reliability.

(a) [1] correct sum of the five readings (11.58); [1] mean = 2.32 s (accept 2.316 s). (b) [1] correct advantage: reduces the effect of random error, improving precision/reliability of the result.

Volume of stone (by displacement) \( = 68.0 - 50.0 = 18.0 \text{ cm}^3 \). \( \rho = \dfrac{m}{V} = \dfrac{45.6}{18.0} = 2.53 \text{ g/cm}^3 \). Final answer: 2.53 g/cm³.

[1] correct volume by displacement (18.0 cm³); [1] correct equation ρ = m/V; [1] 2.53 g/cm³ (accept 2.5 g/cm³).

Mean \( = \dfrac{24.8+25.3+24.6}{3} = \dfrac{74.7}{3} = 24.9 \text{ N/m} \). Final answer: 24.9 N/m (to 3 significant figures, matching the precision of the data).

[1] correct sum of the three values (74.7); [1] correctly divides by 3; [1] 24.9 N/m.

\( \text{percentage uncertainty} = \dfrac{\text{absolute uncertainty}}{\text{measured value}} \times 100 = \dfrac{0.1}{24.0} \times 100 = 0.417\% \approx 0.42\% \). Final answer: 0.42%.

[1] correct equation, % uncertainty = (absolute uncertainty ÷ measured value) × 100; [1] correct substitution (0.1 ÷ 24.0); [1] 0.42% (accept 0.4%).

(a) \( \Delta\theta = 38-20 = 18°C \). Rearranging \( Q = mc\Delta\theta \): \( c = \dfrac{Q}{m\Delta\theta} = \dfrac{4500}{0.50 \times 18} = \dfrac{4500}{9.0} = 500 \text{ J/(kg °C)} \). (b) In practice, some of the electrical energy supplied is lost to the surroundings (e.g. warming the surrounding air rather than the block), so the block's actual temperature rise for a given amount of useful energy is smaller than it would be with no losses; since the calculation assumes all the supplied energy (Q) went into the block, using this measured (smaller) Δθ with the full Q gives a calculated c that is higher than the true value. Final answer: c = 500 J/(kg °C); the experimental value is high because of heat losses to the surroundings.

(a) [1] correct Δθ = 18°C; [1] correct rearrangement c = Q/(mΔθ); [1] c = 500 J/(kg °C). (b) [1] correct reason: heat losses to the surroundings mean not all the supplied energy raises the block's temperature, giving an inflated value of c.

(a) The data show current increasing steadily and roughly in proportion with voltage, so the graph is (approximately) a straight line through the origin, showing ohmic behaviour. (b) Gradient \( = \dfrac{I_{last}-I_{first}}{V_{last}-V_{first}} = \dfrac{1.02-0.21}{5.0-1.0} = \dfrac{0.81}{4.0} = 0.20 \text{ A/V} \). Since \( I = \dfrac{V}{R} \), the gradient of an I-V graph equals \( \dfrac{1}{R} \), the reciprocal of the resistance. Final answer: straight line through the origin; gradient ≈ 0.20 A/V = 1/R.

(a) [1] straight line through the origin (approximately directly proportional/ohmic behaviour). (b) [1] correct gradient calculation (0.81 ÷ 4.0); [1] gradient ≈ 0.20 A/V, correctly identified as representing 1/R.

(a) Gradient \( = \dfrac{\Delta y}{\Delta x} = \dfrac{0.16-0}{8.0-0} = 0.02 \text{ m/N} \). (b) Since \( F = kx \), rearranging gives \( x = \dfrac{1}{k}F \), so a graph of extension (x) against force (F) has a gradient equal to \( \dfrac{1}{k} \); here, \( k = \dfrac{1}{0.02} = 50 \text{ N/m} \). Final answer: gradient = 0.02 m/N = 1/k, giving k = 50 N/m.

(a) [1] correct method (0.16 ÷ 8.0); [1] 0.02 m/N. (b) [1] correctly identifies the gradient as 1/k (reciprocal of the spring constant).

(a) A straight line through the origin shows that T² is directly proportional to l. (b) Comparing \( T^2 = \dfrac{4\pi^2}{g}l \) with \( y = (\text{gradient}) \times x \), the gradient equals \( \dfrac{4\pi^2}{g} \), so \( g = \dfrac{4\pi^2}{\text{gradient}} = \dfrac{4\pi^2}{4.0} = \dfrac{39.48}{4.0} = 9.87 \text{ m/s}^2 \). Self-check: this is very close to the accepted value of g (9.81 m/s²), confirming the result is physically sensible. Final answer: T² ∝ l; g ≈ 9.87 m/s².

(a) [1] T² is directly proportional to l. (b) [1] correct rearrangement, g = 4π² ÷ gradient; [1] g = 9.87 m/s² (accept 9.8–9.9 m/s²).

(a) Since \( E_p = mgh \), plotting E_p against h gives a gradient equal to mg, the weight of the ball (in newtons). (b) \( \text{gradient} = mg \Rightarrow m = \dfrac{\text{gradient}}{g} = \dfrac{4.9}{9.8} = 0.50 \text{ kg} \). Final answer: gradient = weight (mg); mass = 0.50 kg.

(a) [1] correctly identifies the gradient as the weight of the ball, mg. (b) [1] correct method m = gradient ÷ g; [1] 0.50 kg.

(a) Using scales that make the plotted points cover a large proportion (at least half) of both axes reduces the reading error (and so the percentage error) when values, such as the gradient, are taken from the graph, improving the precision of the result. (b) A line of best fit should be a single, smooth straight line drawn through the points so that it passes as close as possible to all of them, with a roughly equal number of points scattered above and below the line (not simply joined point-to-point). (c) Gradient \( = \dfrac{10.1-2.1}{5.0-1.0} = \dfrac{8.0}{4.0} = 2.0 \text{ cm/N} \). (d) Converting to metres: \( 2.0 \text{ cm/N} = 0.02 \text{ m/N} \); since extension \( x = \dfrac{1}{k}F \), the gradient equals \( 1/k \), so \( k = \dfrac{1}{0.02} = 50 \text{ N/m} \). Self-check: checking the ratio extension/load for other data points (e.g. 4.0/2.0 = 2.0 cm/N, 7.9/4.0 ≈ 1.98 cm/N) gives values close to 2.0 cm/N, consistent with the calculated gradient. Final answer: gradient = 2.0 cm/N; spring constant k = 50 N/m.

(a) [1] correct reason (reduces reading/percentage error, improving precision). (b) [1] correct description of a line of best fit (smooth straight line, close to all points, roughly equal points above/below). (c) [1] correct method using two points on/near the line; [1] correct substitution; [1] 2.0 cm/N. (d) [1] correct unit conversion (cm/N to m/N); [1] k = 50 N/m (allow ecf from (c)).

(a) Since \( P = I^2R \), a graph of P against I² is a straight line through the origin with gradient R; a graph of P against I would instead be a curve, making it much harder to determine R accurately, so plotting against I² linearises the relationship. (b) Gradient \( = \dfrac{10.0-2.0}{0.91-0.18} = \dfrac{8.0}{0.73} = 11.0 \text{ Ω} \) (to 3 s.f., 10.96 Ω); since the gradient of P against I² equals R, the resistance is approximately 11.0 Ω. (c) The student could take repeat readings at each setting and calculate a mean value, and use a wider range of current/power values spread evenly across the range, to obtain a more reliable line of best fit. (d) A possible source of random error is small fluctuations in the ammeter or voltmeter readings (e.g. due to reading the scale/display or slight variations in the supply). Final answer: gradient ≈ 11.0 Ω = R.

(a) [1] correct reason (P vs I² gives a straight line allowing R to be read directly from the gradient, whereas P vs I would be a curve). (b) [1] correct method (differences used); [1] correct substitution (8.0 ÷ 0.73); [1] R ≈ 11.0 Ω (accept 10.9–11.0 Ω). (c) [1] repeat readings and take a mean; [1] use a wider/more evenly spread range of values. (d) [1] valid random error source (e.g. fluctuations/parallax in reading the meters).

(a) x-axis = length, l; y-axis = resistance, R. (b) A straight line through the origin would confirm that R is directly proportional to l. (c) Gradient \( = \dfrac{4.9-1.1}{100-20} = \dfrac{3.8}{80} = 0.0475 \text{ Ω/cm} \). (d) Since \( R = \dfrac{\rho}{A}\times l \), comparing with \( y = (\text{gradient})\times x \), the gradient of R against l equals \( \dfrac{\rho}{A} \) — the resistivity of the wire's material divided by its cross-sectional area. Final answer: x = length, y = resistance; straight line through the origin confirms proportionality; gradient = 0.0475 Ω/cm = ρ/A.

(a) [1] correctly identifies x = length, y = resistance. (b) [1] straight line through the origin. (c) [1] correct method (3.8 ÷ 80); [1] 0.0475 Ω/cm (accept 0.047–0.048). (d) [1] gradient = ρ/A; [1] correctly explains this is resistivity divided by cross-sectional area.

Gradient \( = \dfrac{23.0-5.0}{8.0-2.0} = \dfrac{18.0}{6.0} = 3.0 \text{ m/s}^2 \). The gradient of a velocity-time graph represents acceleration. Final answer: gradient = 3.0 m/s² (acceleration).

[1] correct method (differences used); [1] 3.0 m/s²; [1] correctly identifies this as the acceleration.

Gradient \( = \dfrac{12.5-0}{0.050-0} = 250 \text{ N/m} \). The gradient of a force-extension graph (within the elastic limit) represents the spring constant, k, since \( F = kx \). Final answer: gradient = 250 N/m = spring constant k.

[1] correct method; [1] 250 N/m; [1] correctly identifies this as the spring constant, k.

(a) Gradient \( = \dfrac{4.4-6.0}{2.0-0} = \dfrac{-1.6}{2.0} = -0.8 \text{ V/A} \). (b) Comparing \( V = E - Ir \) with \( y = c + (\text{gradient})x \), the gradient equals \( -r \) (the negative of the internal resistance); so the internal resistance \( r = 0.8 \text{ Ω} \) (the size/magnitude of the gradient). Final answer: gradient = −0.8 V/A; internal resistance r = 0.8 Ω.

(a) [1] correct method (differences used); [1] −0.8 V/A. (b) [1] correctly identifies gradient = −r; [1] r = 0.8 Ω.

(a) This constant time interval is called the half-life. (b) In 15 minutes, the number of half-lives that have passed is \( 15 \div 5.0 = 3 \). Halving repeatedly: 800 → 400 (1 half-life) → 200 (2) → 100 (3). Final answer: half-life; count rate at t = 15 min = 100 counts/min.

(a) [1] half-life. (b) [1] correctly identifies 3 half-lives have passed; [1] correct halving sequence shown (800→400→200→100); [1] 100 counts/min.

The independent variable (the one the student deliberately changes) is the length of the wire; the dependent variable (the one measured, which may change as a result) is the resistance of the wire. Final answer: independent variable = length; dependent variable = resistance.

[1] independent variable = length of wire; [1] dependent variable = resistance.

To make this a fair test, variables that could also affect resistance must be kept constant, such as the cross-sectional area (diameter/thickness) of the wire, and the material the wire is made from (and its temperature). Final answer: any two of — cross-sectional area, material, temperature of the wire kept constant.

[1] cross-sectional area/diameter of the wire kept constant; [1] material of the wire (or temperature) kept constant (any two valid controlled variables credited).

Beyond the elastic limit, the spring becomes permanently (plastically) deformed and will not return to its original length when the force is removed; also, Hooke's law — the direct proportionality between force and extension — no longer applies beyond this point, so the linear relationship the experiment relies on to find the spring constant would break down. Final answer: beyond the elastic limit the spring is permanently deformed and no longer obeys Hooke's law.

[1] beyond the elastic limit the spring is permanently deformed (does not return to its original length); [1] Hooke's law (direct proportionality between force and extension) no longer applies.

(a) A hazard is that components and wires can become hot at higher currents, risking burns; a suitable precaution is to switch the circuit off between readings and avoid touching the lamp or wires while current is flowing. (b) A suitable range would be from 0 V to 6.0 V in steps of 1.0 V, giving 7 evenly spaced readings across a safe, low-voltage school-laboratory range. Final answer: hazard = hot components (burns), precaution = switch off between readings; range = 0–6.0 V in 1.0 V steps.

(a) [1] valid hazard (e.g. hot components/burns); [1] valid corresponding precaution. (b) [1] sensible range and even interval suggested giving several data points (accept any reasonable range/interval, e.g. 0–6 V in 1.0 V steps).

(a) Possible sources of random error include the trolley not being released from exactly the same starting position each time, and slight, unpredictable variations in friction or air resistance between runs (e.g. dust on the ramp). (b) Taking repeat readings and calculating a mean reduces the effect of these random errors, giving a more precise, reliable result. Final answer: e.g. inconsistent release point, varying friction; repeat and average readings.

(a) [1] valid source 1 (e.g. inconsistent release point); [1] valid source 2 (e.g. varying friction/air resistance). (b) [1] repeat readings and calculate a mean.

(a) The range should be chosen so that it covers the expected range of currents in the circuit without going off-scale, while still giving precise, easily readable values (not too small a range of the scale being used). (b) An ammeter is connected in series, so if it had significant resistance it would add to the total circuit resistance and reduce the current, giving a falsely low reading; it should therefore have a very low resistance so it does not affect the current it is measuring. A voltmeter is connected in parallel, so if it had low resistance a significant current would flow through it, drawing current away from the component and giving a falsely low p.d. reading; it should therefore have a very high resistance so it draws a negligible current and does not affect the circuit. Final answer: ammeter range should suit expected currents; ammeter needs low resistance (series), voltmeter needs high resistance (parallel).

(a) [1] correct factor (range should cover the expected current values while allowing precise readings). (b) [1] ammeter needs low resistance because it is in series (to avoid reducing the current); [1] voltmeter needs high resistance because it is in parallel (to avoid drawing significant current).

(a) The extension values increase in a fairly regular pattern except at 3.0 N, where the recorded extension (8.9 cm) is much higher than the surrounding pattern suggests (roughly 6 cm would fit the trend), making this result anomalous. (b) A likely cause is a measurement error (e.g. a misreading of the ruler/parallax error), or the spring may have been disturbed or knocked while this reading was taken. (c) An anomalous result should be excluded (not included) when calculating a mean, and should be ignored (not used to help position the line) when drawing a line of best fit — ideally the measurement should also be repeated to check it. Final answer: 8.9 cm at 3.0 N is anomalous; likely a reading error; it should be excluded from the mean and from the line of best fit.

(a) [1] correctly identifies 8.9 cm (at 3.0 N) as anomalous. (b) [1] valid plausible cause (e.g. reading/measurement error, spring disturbed). (c) [1] excluded when calculating the mean; [1] excluded/ignored when drawing the line of best fit (or repeated to check/replace it).

(a) \( R = \dfrac{V}{I} = \dfrac{6.0}{1.22} = 4.92 \text{ Ω} \). (b) The other three rows give resistances of 5.0 Ω, 5.1 Ω and 5.0 Ω; the calculated value of 4.92 Ω is close to this range (within about 2–3%), so it is reasonably consistent with the pattern shown by the other rows. This suggests the third measurement is a normal, reliable result showing typical small random variation, rather than an anomaly that should be excluded. Final answer: R = 4.92 Ω, which fits the pattern of the other rows (no anomaly).

(a) [1] correct equation R = V/I; [1] R = 4.92 Ω (accept 4.9 Ω). (b) [1] correctly states this value is reasonably consistent with (close to) the other rows; [1] correctly concludes this is a normal/reliable measurement rather than an anomaly.

(a) No, this conclusion is not fully justified by the data described. (b) The results show a correlation — that resistance increases as length increases — but the student did not measure temperature, so the data cannot show that a temperature change is the cause; in fact, the standard explanation for why a longer wire (of the same material and cross-section) has greater resistance is simply that there is more material for charge carriers to pass through, using \( R = \rho l/A \), which does not require any change in temperature at all. (c) To properly justify the temperature explanation, the student would need to directly measure the temperature of the wire during each reading (e.g. using a thermometer, or by controlling/minimising self-heating, such as passing current only briefly), to check whether temperature actually does change with length. Final answer: not justified; the data shows only a correlation, and resistivity/length (R = ρl/A) explains the trend without any temperature change; temperature would need to be measured to test the temperature explanation.

(a) [1] correctly states the conclusion is not fully justified. (b) [1] correctly notes the data shows only a correlation, with temperature not measured; [1] correctly notes the standard R = ρl/A explanation does not require a temperature change. (c) [1] correctly suggests measuring/controlling temperature to test the explanation.

(a) The student should consider repeating this particular measurement, to check whether the point is a genuine result or the result of an error, before deciding whether to exclude it or finalising a conclusion based on it. (b) Precision can be assessed by looking at how closely the plotted points lie to the line of best fit (how much scatter there is around the line), or by comparing the spread (range) of repeated readings taken at the same value. (c) A precise result is one where repeated measurements are close to one another (little scatter/spread), regardless of whether they are close to the true value; an accurate result is one where a measurement (or the mean of measurements) is close to the true (accepted) value. Final answer: repeat the anomalous reading to check it; assess precision from scatter around the line of best fit or spread of repeats; precision = closeness of readings to each other, accuracy = closeness to the true value.

(a) [1] repeat the anomalous reading to check it, before deciding whether to exclude it. (b) [1] assess how close the points lie to the line of best fit (scatter); [1] compare the spread/range of repeat readings at a given value. (c) [1] correct distinction between precision (closeness of readings to each other) and accuracy (closeness to the true value).