2024 CCEA GCSE Physics 1210 Practice Paper with Answers

Acceleration describes how quickly an object's velocity changes. It is calculated using \( a = \frac{v-u}{t} \), where \( v \) is final velocity, \( u \) is initial velocity and \( t \) is time taken, giving a result in metres per second squared (m/s²).

[1] Acceleration is the rate of change of velocity; [1] correct equation \( a = \frac{v-u}{t} \); [1] correct unit, m/s².

The resultant force on an object is found by combining (vector-summing) all the individual forces acting on it into a single equivalent force. Newton's First Law states that an object will remain stationary, or continue to move at a constant velocity in a straight line, unless a resultant/unbalanced force acts upon it, causing it to accelerate.

[1] Resultant force defined as the single force equivalent to all forces acting combined; [1] Newton's First Law correctly stated (object stays at rest/constant velocity unless acted on by a resultant force); [1] correct reference to 'unbalanced'/'resultant' force being required to change motion.

Density is calculated using \( \rho = \frac{m}{V} \), with SI units of kg/m³ (or g/cm³). In a solid, particles are held closely together in fixed, regular positions by strong forces and can only vibrate about these positions; in a gas, particles are much further apart, move randomly and at high speed, and the forces between them are negligible.

[1] Correct equation \( \rho = \frac{m}{V} \); [1] correct unit, kg/m³; [1] valid difference in particle arrangement/motion between solid and gas (e.g. closely packed and vibrating vs far apart and moving randomly).

According to kinetic theory, gas particles are in constant random motion; when they collide with the walls of their container, they exert a force on the walls, and this force per unit area is observed as gas pressure. As the temperature of a fixed volume of gas increases, the average kinetic energy (and therefore speed) of the particles increases, so they collide with the container walls more frequently and with greater force per collision, increasing the pressure.

[1] Gas pressure is caused by particles colliding with the container walls; [1] increasing temperature increases the average kinetic energy/speed of the particles; [1] particles collide with the walls more often and with greater force, so pressure increases (at constant volume).

The principle of conservation of energy states that energy cannot be created or destroyed; it can only be transferred usefully, stored, or dissipated (e.g. as heat) from one form/store to another, with the total energy remaining constant. The efficiency of a device measures the proportion of the input energy that is transferred usefully, calculated as \( \text{efficiency} = \frac{\text{useful output energy}}{\text{total input energy}} \times 100\% \).

[1] Energy cannot be created or destroyed, only transferred/transformed between stores; [1] correct efficiency equation, \( \frac{\text{useful output energy}}{\text{total input energy}} \); [1] correctly expressed as a percentage (× 100%).

Isotopes are different forms of the same chemical element; they have the same number of protons (and so the same proton/atomic number and chemical properties) but different numbers of neutrons, giving them different mass/nucleon numbers. A particle made up of 2 protons and 2 neutrons, identical to a helium nucleus, is an alpha particle, emitted during alpha decay.

[1] Isotopes have the same number of protons/same proton number; [1] isotopes have different numbers of neutrons/different mass (nucleon) number; [1] alpha particle correctly named.

(a) \( a = \frac{v-u}{t} = \frac{24-0}{8.0} = 3.0 \text{ m/s}^2 \). (b) \( s = \frac{(u+v)}{2} \times t = \frac{(0+24)}{2} \times 8.0 = 96 \text{ m} \) (equivalently \( s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(3.0)(8.0)^2 = 96 \text{ m} \)).

[1] Correct equation, \( a = \frac{v-u}{t} \); [1] a = 3.0 m/s² with unit; [1] correct equation of motion selected, e.g. \( s = \frac{(u+v)}{2}t \) or \( s = ut + \frac{1}{2}at^2 \); [1] correct substitution; [1] s = 96 m with unit (allow ecf from part (a)).

(a) \( p = mv = 0.40 \times 6.0 = 2.4 \text{ kg m/s} \). (b) The second ball is stationary, so total momentum before collision = 2.4 kg m/s + 0 = 2.4 kg m/s. By conservation of momentum, total momentum after collision = 2.4 kg m/s. Combined mass = 0.40 + 0.20 = 0.60 kg. \( v = \frac{2.4}{0.60} = 4.0 \text{ m/s} \).

[1] Correct equation, p = mv; [1] p = 2.4 kg m/s with unit; [1] momentum is conserved: total momentum before = total momentum after (= 2.4 kg m/s); [1] correct substitution using combined mass 0.60 kg; [1] v = 4.0 m/s with unit (allow ecf).

(a) \( a = \frac{v-u}{t} = \frac{8.0-0}{4.0} = 2.0 \text{ m/s}^2 \). (b) Distance = total area under the graph = area of the first triangle (0-4 s) + area of the rectangle (4-10 s) + area of the final triangle (10-14 s) \( = (\frac{1}{2} \times 4.0 \times 8.0) + (6.0 \times 8.0) + (\frac{1}{2} \times 4.0 \times 8.0) = 16 + 48 + 16 = 80 \text{ m} \).

[1] Correct equation, \( a = \frac{v-u}{t} \); [1] a = 2.0 m/s² with unit; [1] correct method, distance = sum of the areas under the graph (triangle + rectangle + triangle); [1] correct individual areas calculated (16 m, 48 m, 16 m, or equivalent working); [1] total distance = 80 m with unit.

(a) \( F = kx \), so \( x = \frac{F}{k} = \frac{6.0}{40} = 0.15 \text{ m} = 15 \text{ cm} \). (b) \( F = kx = 40 \times 0.030 = 1.2 \text{ N} \).

[1] Correct equation rearranged, \( x = \frac{F}{k} \); [1] correct substitution, \( \frac{6.0}{40} \); [1] x = 0.15 m (15 cm) with unit; [1] correct substitution, F = 40 × 0.030; [1] F = 1.2 N with unit.

(a) \( \text{moment} = F \times d = 15 \times 0.40 = 6.0 \text{ N m} \). (b) For the beam to balance, the clockwise moment must equal the anticlockwise moment: \( 20 \times d = 6.0 \), so \( d = \frac{6.0}{20} = 0.30 \text{ m} \).

[1] Correct equation, moment = F × d; [1] moment = 6.0 N m with unit; [1] correct application of the principle of moments, 20 × d = 6.0 (clockwise moment = anticlockwise moment); [1] correct rearrangement; [1] d = 0.30 m with unit (allow ecf).

Volume \( = 5.0 \times 4.0 \times 2.0 = 40 \text{ cm}^3 = 4.0 \times 10^{-5} \text{ m}^3 \). Mass \( = 312 \text{ g} = 0.312 \text{ kg} \). \( \rho = \frac{m}{V} = \frac{0.312}{4.0 \times 10^{-5}} = 7800 \text{ kg/m}^3 \).

[1] Volume calculated as 40 cm³ (5.0 × 4.0 × 2.0); [1] volume correctly converted to 4.0 × 10⁻⁵ m³; [1] mass correctly converted to 0.312 kg; [1] correct substitution into \( \rho = \frac{m}{V} \); [1] final answer, 7800 kg/m³ (accept 7.8 × 10³ kg/m³).

(a) Useful energy gained \( = \text{weight} \times \text{height} = 1200 \times 3.0 = 3600 \text{ J} \). (b) \( \text{efficiency} = \frac{\text{useful output energy}}{\text{total input energy}} \times 100\% = \frac{3600}{5400} \times 100 = 66.7\% \).

[1] Correct equation, useful energy = weight × height; [1] useful energy = 3600 J with unit; [1] correct efficiency equation used; [1] correct substitution, \( \frac{3600}{5400} \times 100 \); [1] efficiency = 66.7% (accept 67%), allow ecf.

(a) Corrected count rate = measured count rate − background count rate \( = 148 - 24 = 124 \) counts/min. (b) The corrected count rate (which is proportional to the activity/number of undecayed nuclei of the source) has fallen to exactly half of its initial value in 15 minutes; since activity halves after exactly one half-life, the half-life of the source is 15 minutes.

[1] Correct equation, corrected count rate = measured − background; [1] corrected count rate = 124 counts/min with unit; [1] recognises the count rate has fallen to half its initial (corrected) value; [1] correctly links one half-life to the count rate halving; [1] half-life = 15 minutes with unit.

A strong answer explains that solar PV panels convert light energy directly into electrical energy using the photovoltaic effect within semiconductor material, while wind turbines convert the kinetic energy store of moving air into the kinetic energy store of a turbine, which drives a generator to produce electrical energy. It compares reliability: fossil fuel power stations can generate electricity on demand at a constant, controllable rate, whereas solar PV output depends on daylight and weather (producing no electricity at night) and wind turbine output depends on wind speed (producing little or no electricity when it is too calm, and none when it is too windy for safety), making both intermittent/unreliable without storage or backup. It compares environmental impact: burning fossil fuels releases carbon dioxide (contributing to climate change/global warming) and other pollutants (e.g. sulfur dioxide, particulates) affecting air quality, and fossil fuels are a finite, non-renewable resource, whereas solar PV and wind produce no emissions during operation, use a renewable energy resource, though manufacturing and installing the equipment does have some environmental impact. It compares costs: solar PV panels and wind turbines have relatively high initial installation/capital costs but very low running costs once installed (the 'fuel' is free), whereas fossil fuel power stations have lower initial cost but ongoing, potentially rising, fuel costs over their lifetime.

Level of response marking (9 marks, scaled from a 6-mark base). Band 3 (7-9 marks): accurately describes how both solar PV and wind generate electricity, compares reliability/intermittency of both against fossil fuels, discusses environmental impact (CO2/pollution/finite resource vs renewable, no operational emissions), compares installation vs running costs, using correct specialist terminology (e.g. photovoltaic effect, kinetic energy store, generator, non-renewable, emissions) in clear, well-organised prose. Band 2 (4-6 marks): reasonable coverage of at least two of the three required areas (generation mechanism, reliability/environmental impact, cost) with generally correct terminology and communication. Band 1 (1-3 marks): basic, largely descriptive points covering only one area in any depth, weak terminology, unclear organisation. Band 0: no creditworthy material.

(a) Within the Hooke's law region, extension is directly proportional to force: each 2.0 N increases extension by 5 mm (2.5 mm per N), so at 6.0 N, X = 15 mm. (b) Extrapolating the linear pattern, 10.0 N would be expected to give an extension of 25 mm, but the actual extension recorded is 28 mm; this disproportionate jump shows that 8.0 N (the last force still following the linear pattern) marks the limit of proportionality. (c) Using data within the Hooke's law region, e.g. at 8.0 N: \( k = \frac{F}{x} = \frac{8.0}{0.020} = 400 \text{ N/m} \).

[1] Recognises the linear pattern (extension increases by 5 mm per 2.0 N, i.e. 2.5 mm per N); [1] X = 15 mm; [1] identifies 8.0 N as the limit of proportionality; [1] valid reason, e.g. the extension at 10.0 N (28 mm) is disproportionately larger than expected from the linear pattern (25 mm); [1] correct substitution, \( k = \frac{8.0}{0.020} \); [1] k = 400 N/m with unit.

In any nuclear equation, the total mass number (A) is conserved: \( 226 = A + 4 \), so \( A = 222 \). The total atomic/proton number (Z) is also conserved: \( 88 = Z + 2 \), so \( Z = 86 \), which correctly identifies the daughter nuclide as radon (Rn), since radon has atomic number 86. This reflects the fact that an alpha particle removes 2 protons and 2 neutrons (4 nucleons in total) from the parent nucleus.

[1] Mass number is conserved in a nuclear equation; [1] A = 226 − 4 = 222; [1] atomic/proton number is conserved in a nuclear equation; [1] Z = 88 − 2 = 86; [1] correctly identifies the daughter nuclide as radon (Rn), consistent with Z = 86; [1] correct explanation that an alpha particle removes 2 protons and 2 neutrons from the parent nucleus.

(a) The count rate halves every 2 hours (800→400→200→100→50), so Y = 100 counts/min. (b) Since the count rate consistently halves every 2 hours, the half-life is 2 hours. (c) From 8 hours (50 counts/min) to 12 hours is a further 4 hours, equal to 2 half-lives, so the count rate halves twice: \( 50 \div 2 \div 2 = 12.5 \) counts/min.

[1] Recognises the count rate halves every 2 hours; [1] Y = 100 counts/min; [1] correctly determines the time taken to halve is consistently 2 hours; [1] half-life = 2 hours with unit; [1] recognises 4 hours (8h to 12h) is equal to 2 half-lives; [1] count rate = 12.5 counts/min (allow ecf).

(a) A ray travelling parallel to the principal axis is refracted by the lens so that it passes through the principal focus on the far side; a ray passing through the centre of the lens continues in a straight line, undeviated. (b) \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{10} - \frac{1}{15} = \frac{3-2}{30} = \frac{1}{30} \), so \( v = 30 \text{ cm} \). (c) \( \text{magnification} = \frac{v}{u} = \frac{30}{15} = 2.0 \).

[1] Ray parallel to the axis refracts through the principal focus; [1] ray through the centre of the lens continues straight/undeviated; [1] correct rearrangement, \( \frac{1}{v} = \frac{1}{10} - \frac{1}{15} \); [1] v = 30 cm; [1] correct equation, magnification = v/u; [1] magnification = 2.0 (no unit), allow ecf.

(a) Total internal reflection occurs when light travelling within the optically denser medium (the fibre) strikes the boundary with a less dense medium (air) at an angle of incidence greater than the critical angle. (b) \( \sin\theta_c = \frac{1}{n} = \frac{1}{1.5} = 0.667 \), so \( \theta_c = \sin^{-1}(0.667) = 41.8° \). (c) Because the angle of incidence at the fibre-air boundary exceeds the critical angle each time the light reaches it, the light undergoes total internal reflection repeatedly along the length of the fibre, with (in principle) no light escaping and very little energy lost at each reflection, allowing the signal to be transmitted over long distances while remaining strong.

[1] Light must be travelling in the denser medium towards a less dense medium; [1] angle of incidence must exceed the critical angle; [1] correct substitution, \( \sin\theta_c = \frac{1}{1.5} \); [1] θc = 41.8° (accept 41-42°); [1] light undergoes repeated total internal reflection along the fibre without escaping; [1] very little energy/signal is lost at each reflection, so the signal remains strong over long distances.

(a) \( v = f\lambda \), with v in m/s. (b) Wavelength = 2.0 cm = 0.020 m, so \( v = f\lambda = 4.0 \times 0.020 = 0.080 \text{ m/s} \). (c) Diffraction is most pronounced when the gap width is similar in size to the wavelength of the wave. The 2.0 cm gap is approximately equal to the wavelength (2.0 cm), so the wave diffracts strongly, spreading out significantly into the region beyond the barrier. The 20 cm gap is much greater than the wavelength, so the wave diffracts only slightly at the edges of the gap and travels through largely undisturbed, in a relatively straight line.

[1] Correct equation, v = fλ; [1] correct substitution using λ = 0.020 m; [1] v = 0.080 m/s with unit; [1] diffraction is greatest when the gap width is comparable to the wavelength; [1] the narrow (2.0 cm) gap causes significant spreading/diffraction; [1] the wide (20 cm) gap causes little diffraction, the wave travels mostly straight through.

(a) Rearranging \( v = f\lambda \) gives \( \lambda = \frac{v}{f} = \frac{340}{680} = 0.5 \text{ m} \). (b) Since \( v = f\lambda \) and v (the speed of sound in air) is constant, f and λ must be inversely proportional; if the frequency doubles, the wavelength must halve to keep v constant, giving a new wavelength of 0.25 m.

[1] Correct rearrangement, \( \lambda = \frac{v}{f} \); [1] correct substitution, \( \frac{340}{680} \); [1] λ = 0.5 m with unit; [1] states the wavelength would halve; [1] new wavelength = 0.25 m; [1] correct explanation, wavelength and frequency are inversely proportional when speed (v = fλ) is constant.

(a) In series, resistances add: \( R_{total} = 4.0 + 8.0 = 12 \text{ Ω} \). (b) \( I = \frac{V}{R} = \frac{12}{12} = 1.0 \text{ A} \). (c) \( V = IR = 1.0 \times 8.0 = 8.0 \text{ V} \).

[1] Total resistance = 12 Ω; [1] correct equation, I = V/R; [1] I = 1.0 A with unit; [1] correct substitution, V = 1.0 × 8.0; [1] V = 8.0 V with unit (allow ecf).

In parallel, the full supply p.d. (24 V) acts across each resistor. (a) \( I = \frac{V}{R} = \frac{24}{6.0} = 4.0 \text{ A} \). (b) \( I = \frac{24}{12} = 2.0 \text{ A} \). (c) Total current = sum of branch currents \( = 4.0 + 2.0 = 6.0 \text{ A} \).

[1] Correct equation, I = V/R; [1] I = 4.0 A through the 6.0 Ω resistor; [1] correct substitution for the 12 Ω resistor; [1] I = 2.0 A through the 12 Ω resistor; [1] total current = 6.0 A (sum of branch currents).

(a) \( P = VI \Rightarrow I = \frac{P}{V} = \frac{2300}{230} = 10 \text{ A} \). (b) Time = 3.0 minutes = 180 s. \( E = Pt = 2300 \times 180 = 414000 \text{ J} \) (4.14 × 10⁵ J).

[1] Correct rearrangement, I = P/V; [1] I = 10 A with unit; [1] correct equation, E = Pt; [1] time correctly converted to 180 s and substituted; [1] E = 414 000 J (4.14 × 10⁵ J).

(a) The current does not increase in proportion to the potential difference; each successive 1.0 V increase produces a smaller increase in current, so the I-V graph is a curve that becomes less steep (flattens) at higher V. This shows that the resistance of the lamp increases as the potential difference (and current) increases, because as more current flows, the filament gets hotter, and the resistance of a metal filament increases with increasing temperature. (b) \( R = \frac{V}{I} = \frac{4.0}{1.00} = 4.0 \text{ Ω} \).

[1] Graph is a curve that flattens/current increases less steeply at higher V, showing resistance increases; [1] explanation: the filament heats up as current increases, and the resistance of a metal filament increases with temperature; [1] correct equation, R = V/I; [1] correct substitution, 4.0/1.00; [1] R = 4.0 Ω with unit.

(a) Time = 2.0 minutes = 120 s. \( Q = It = 0.50 \times 120 = 60 \text{ C} \). (b) \( E = QV = 60 \times 6.0 = 360 \text{ J} \).

[1] Correct equation, Q = It; [1] time correctly converted to 120 s and substituted; [1] Q = 60 C with unit; [1] correct equation/substitution, E = QV = 60 × 6.0; [1] E = 360 J with unit (allow ecf).

(a) \( F = BIL = 0.50 \times 3.0 \times 0.20 = 0.30 \text{ N} \). (b) The force on a current-carrying wire in a magnetic field depends on the component of the current that is perpendicular to the field; the force is greatest when the wire is at right angles to the field and falls to zero when the wire is parallel to the field lines, since there is then no perpendicular component and no field lines are 'cut' by the current.

[1] Correct equation, F = BIL; [1] correct substitution, 0.50 × 3.0 × 0.20; [1] F = 0.30 N with unit; [1] states the force would be zero (or much reduced) when parallel; [1] correct explanation, force depends on the perpendicular component of current to field, which is zero when the wire is parallel to the field.

(a) \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow V_s = V_p \times \frac{N_s}{N_p} = 230 \times \frac{1000}{200} = 230 \times 5 = 1150 \text{ V} \). (b) Since the secondary coil has more turns than the primary coil, the secondary (output) voltage (1150 V) is greater than the primary (input) voltage (230 V), so this is a step-up transformer.

[1] Correct equation, \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \); [1] correct substitution, \( 230 \times \frac{1000}{200} \); [1] Vs = 1150 V with unit; [1] correctly identifies a step-up transformer; [1] correct reasoning, secondary has more turns / higher output than input voltage.

A strong answer describes a Sun-like star's life cycle: a cloud of dust and gas (nebula) collapses under gravitational attraction to form a protostar; when the core becomes hot and dense enough, nuclear fusion of hydrogen into helium begins, releasing energy that creates an outward radiation pressure balancing the inward pull of gravity, and the star enters a long, stable main sequence phase. Once hydrogen fuel in the core is used up, the star expands to become a red giant; eventually it sheds its outer layers as a planetary nebula, leaving behind a hot, dense core called a white dwarf, which gradually cools and fades over a very long time. It then explains that a star with much greater mass than the Sun follows a similar early sequence (nebula, protostar, main sequence) but progresses much faster because it burns fuel more quickly; instead of becoming a red giant it expands into a red supergiant, and instead of a gentle end, it undergoes a sudden, violent explosion called a supernova, blasting most of its material into space. What remains of the core collapses further, forming an extremely dense neutron star, or, if the original star was massive enough, collapsing further still into a black hole, from which not even light can escape.

Level of response marking (9 marks, scaled from a 6-mark base). Band 3 (7-9 marks): accurate, detailed description of the Sun-like star life cycle (nebula, protostar, main sequence, red giant, planetary nebula, white dwarf) and the massive star life cycle (nebula, protostar, main sequence, red supergiant, supernova, neutron star/black hole), with correct comparison of timescale/outcome and confident, accurate use of specialist terminology throughout, in clear, well-organised prose. Band 2 (4-6 marks): reasonable coverage of both life cycles with most key stages named, generally accurate terminology, some comparison made, generally clear communication. Band 1 (1-3 marks): basic, largely descriptive points on one life cycle only, or both life cycles described with significant gaps/inaccuracy, weak terminology. Band 0: no creditworthy material.

Wave speed is calculated using \( v = f\lambda \), where v is in m/s, f is in Hz and λ is in m. Sound waves are longitudinal waves, meaning that the particles of the medium (e.g. air) vibrate back and forth parallel to the direction in which the wave, and its energy, travels, producing regions of compression and rarefaction.

[1] Correct equation, v = fλ; [1] sound waves correctly identified as longitudinal; [1] correct description of longitudinal, particles vibrate parallel to the direction of energy transfer/wave travel.

The electromagnetic spectrum is a continuous range of waves, all of which travel at the speed of light in a vacuum, arranged in order of increasing frequency (and decreasing wavelength) as: radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays.

[1] Radio waves, microwaves and infrared correctly named and in order; [1] visible light and ultraviolet correctly named and in order; [1] X-rays and gamma rays correctly named and in order (all seven correct, in the correct order, for full marks).

Standard UK three-pin plugs include several safety features, including a fuse, an earth wire/pin, a cable grip, and (in some appliances) double insulation. The fuse contains a thin wire of a specific rating that melts (and so breaks the circuit) if the current flowing through it exceeds a safe value, for example due to a fault; this immediately disconnects the appliance from the mains supply, preventing overheating of the wiring and reducing the risk of fire or electric shock.

[1] Two valid safety features named (e.g. fuse, earth wire, cable grip, double insulation); [1] correct explanation of the purpose of the chosen feature; [1] correct additional detail of the mechanism (e.g. the fuse wire melts/breaks the circuit when current exceeds its rating, or the earth wire provides a low-resistance path to earth causing a fault current to blow the fuse).

The strength of the magnetic field produced by a solenoid can be increased by increasing the current flowing through it, by increasing the number of turns of wire in the solenoid (for a given length), or by placing a soft iron core inside the solenoid, which concentrates and strengthens the magnetic field (this is the principle used in an electromagnet).

[1] Size of the current; [1] number of turns on the solenoid; [1] presence of a (soft iron) core.

Fleming's left-hand rule is used to predict the direction of the force (motor effect) on a current-carrying wire in a magnetic field. Using the left hand, with the First finger, seCond finger and thuMb held at right angles to each other: the First finger points in the direction of the magnetic Field, the seCond finger points in the direction of the conventional Current, and the thuMb then points in the direction of the resultant force/Motion of the wire.

[1] Fleming's left-hand rule correctly named; [1] correctly states it relates the directions of magnetic field, current and force/motion; [1] correct description of the finger assignment (First finger = Field, seCond finger = Current, thuMb = Motion/force).

Observations show that light from almost all distant galaxies is redshifted, meaning its wavelength has been stretched towards the red end of the spectrum. This redshift indicates that these galaxies are moving away from us (and from each other); the further away a galaxy is, the greater its redshift and the faster it is moving away, showing that the universe is expanding. This evidence is consistent with the Big Bang theory, in which the universe originated from a single point and has been expanding ever since.

[1] Redshift of light from (most) distant galaxies observed; [1] this shows galaxies are moving away from us / the universe is expanding; [1] more distant galaxies show greater redshift/move away faster, consistent with an expanding universe originating from a single point.

(a) The weight (force) applied by each mass can be calculated using \( W = mg \), with mass converted to kilograms and g taken as approximately 10 N/kg (or 9.8 N/kg), or it can be measured directly by suspending the mass from a newton-meter. (b) The natural (unstretched) length of the spring is measured with a metre ruler before any mass is attached, with the ruler and spring viewed at eye level to avoid parallax error. As each mass is added, the new total length of the spring is measured in the same way, and the extension for that mass is calculated as extension = new (stretched) length − natural (unstretched) length.

[1] Weight calculated using W = mg (with g stated, e.g. 10 N/kg), or measured directly with a newton-meter; [1] mass correctly converted to kg before use (if calculating); [1] natural/unstretched length measured with no mass attached, viewed at eye level to avoid parallax error; [1] stretched length measured in the same way for each mass added; [1] extension = stretched length − natural (unstretched) length.

(a) The student should read the thermometer scale at eye level, directly in line with the liquid column, to avoid parallax error, and should wait until the temperature reading has stabilised before recording it. (b) Using a temperature probe (sensor) connected to a data logger and computer, instead of a thermometer and stopwatch, would increase precision: the data logger can record temperature automatically, at precisely regular time intervals and to a greater number of decimal places, removing the human reaction-time error involved in starting/stopping a stopwatch and reading a thermometer manually.

[1] Read the thermometer at eye level to avoid parallax error; [1] wait for the reading to stabilise before recording; [1] suggests using a temperature probe/data logger; [1] explains this removes human reaction-time/reading error; [1] explains it records more precisely/at more regular, exact intervals.

(a) The ammeter must be connected in series with the resistor, so that the same current flows through both; the voltmeter must be connected in parallel across the resistor, so that it measures the potential difference across it without significantly affecting the current in the circuit. (b) The potential difference can be varied using a variable resistor (rheostat) connected in series in the circuit, or by using a variable power supply; to ensure the readings are reliable, each current and p.d. reading should be repeated (e.g. two or three times) at each setting and a mean value calculated.

[1] Ammeter connected in series with the resistor; [1] voltmeter connected in parallel across the resistor; [1] use of a variable resistor/rheostat or variable power supply to vary the p.d.; [1] readings repeated at each setting; [1] mean value calculated to improve reliability.

(a) The length of the pendulum should be measured using a ruler or measuring tape, from the fixed point of suspension (where the string is attached to its support) to the centre of the pendulum bob (mass). (b) Human reaction time when starting and stopping the stopwatch introduces a small, roughly fixed absolute error into any timed reading; if only a single oscillation is timed, this fixed error represents a large percentage of the (short) total time, giving a large percentage uncertainty in the period. By timing 20 complete oscillations and dividing the total time by 20, the same fixed reaction-time error is spread over a much longer total time, so it represents a much smaller percentage of the time for one oscillation, considerably reducing the percentage uncertainty in the calculated period.

[1] Length measured from the point of suspension/support; [1] to the centre of the pendulum bob (mass); [1] explains reaction-time error is a small/fixed absolute error that becomes a large percentage error if only one oscillation is timed; [1] timing many oscillations spreads this fixed error over a longer time, reducing the percentage/relative error; [1] period = total time ÷ number of oscillations (20).

(a) Since \( F = kx \), plotting force (y-axis) against extension (x-axis) gives a gradient equal to k, the spring constant. (b) The graph should be a straight line passing through the origin; this confirms that force is directly proportional to extension over this range, which is the definition of Hooke's law being obeyed. (c) Using the extremes, converting extension to metres: \( \text{gradient} = \frac{\Delta F}{\Delta x} = \frac{4.0 - 0}{0.10 - 0} = \frac{4.0}{0.10} = 40 \text{ N/m} \); since gradient = F/x = k, the spring constant is 40 N/m. (d) Any point that does not lie close to (on) the straight line of best fit is identified as anomalous; such a point should be circled and, where possible, excluded from the line of best fit or the measurement repeated to check it.

[1] Extension on the x-axis, force on the y-axis; [1] straight line through the origin; [1] this shows force directly proportional to extension, confirming Hooke's law; [1] correct method for gradient using two points well separated on the line; [1] extension correctly converted to metres; [1] gradient correctly calculated as 40; [1] correct unit, N/m, correctly identified as the spring constant; [1] an anomalous point is one lying noticeably off the line of best fit; [1] correct action described (circled/excluded from the line, or the reading repeated/checked).

(a) Since \( V = IR \), plotting p.d. (y-axis) against current (x-axis) gives a gradient equal to R. (b) The graph should be a straight line through the origin, showing that p.d. is directly proportional to current, i.e. the resistor has a constant resistance and obeys Ohm's law over this range. (c) \( \text{gradient} = \frac{\Delta V}{\Delta I} = \frac{4.0-0}{0.80-0} = \frac{4.0}{0.80} = 5.0 \), with units V/A = Ω; this gradient represents the resistance of the resistor, R = 5.0 Ω. (d) Repeating each current and p.d. reading (e.g. two or three times) and calculating a mean before plotting would reduce the effect of random error, improving reliability.

[1] Current on the x-axis, p.d. on the y-axis; [1] straight line through the origin; [1] this shows p.d. directly proportional to current, the resistor obeys Ohm's law/has constant resistance; [1] correct method using two well-separated points; [1] gradient correctly calculated as 5.0; [1] correct unit, V/A = Ω; [1] correctly identifies the gradient as the resistance, R = 5.0 Ω; [1] repeat readings and calculate a mean; [1] correct explanation that this reduces random error/improves reliability.

Because the newton-meter reads 0.20 N with no force applied, this is a systematic (zero) error: every subsequent force reading taken with it will be too high by this same fixed amount of 0.20 N, regardless of the true force applied. This does not affect the trend/pattern of the results (e.g. the gradient of a graph), only their absolute values. It can be corrected either by subtracting 0.20 N from every recorded force reading during data processing, or, preferably, by re-zeroing/recalibrating the newton-meter before starting the investigation.

[1] Identifies this as a systematic (zero) error that makes every reading too high by 0.20 N; [1] explains this affects every reading by the same fixed amount (does not affect the trend/gradient, only absolute values); [1] correction described (subtract 0.20 N from each reading, or re-zero the instrument before use).

If temperature is not controlled, it becomes an uncontrolled variable that could independently affect the rate of photosynthesis (a higher temperature generally increases enzyme activity and reaction rate up to an optimum). Any change observed in the rate of photosynthesis could therefore be caused, wholly or partly, by a change in temperature rather than by the change in light intensity being tested, meaning the investigation no longer validly tests the effect of light intensity alone (it is not a fair test). This could be improved by carrying out the investigation with the apparatus in a water bath maintained at a constant temperature, checked regularly with a thermometer throughout.

[1] Explains temperature is an uncontrolled variable that could independently affect the rate of photosynthesis, confounding the results; [1] explains this means the investigation is no longer a fair test/does not validly test the effect of light intensity alone; [1] valid improvement suggested, e.g. use of a water bath to maintain a constant, checked temperature.

(a) Since \( I = \frac{V}{R} \), plotting I (y-axis) against V (x-axis) gives a gradient of 1/R. (b) All data points should be plotted accurately as small crosses (×) or encircled dots; a single straight line should then be drawn passing as close as possible to all the points, with a roughly equal number of points above and below the line, ignoring any clearly anomalous point. (c) Using two well-separated points on the line of best fit, e.g. (0.5, 0.10) and (3.0, 0.60): \( \text{gradient} = \frac{0.60-0.10}{3.0-0.5} = \frac{0.50}{2.5} = 0.20 \text{ A/V} \). (d) Since gradient \( = \frac{1}{R} \), \( R = \frac{1}{0.20} = 5.0 \text{ Ω} \).

[1] V on the x-axis, I on the y-axis; [1] points plotted accurately as small crosses/encircled dots; [1] single straight line drawn close to all points, roughly equal points either side, ignoring anomalies; [1] uses two points well separated on the line of best fit; [1] correct method, gradient = Δy/Δx; [1] gradient correctly calculated ≈ 0.20; [1] correct unit, A/V; [1] correctly states gradient = 1/R (or equivalent reasoning); [1] R = 5.0 Ω correctly calculated (allow ecf).

(a) The graph is a curve (not a straight line), which decreases steeply at low temperatures and becomes progressively less steep as temperature increases, approaching (but not necessarily reaching) zero. (b) A smooth curve of best fit should be drawn through the plotted points; a vertical line is then drawn upward from 50°C on the temperature axis to meet the curve, and a horizontal line is drawn across from this point to the resistance axis, reading off the corresponding resistance value (interpolation). (c) As temperature increases, the resistance of the thermistor decreases (a negative correlation); this is because increasing temperature gives more charge carriers (electrons) within the semiconductor material of the thermistor enough energy to become freed and mobile, increasing the number of charge carriers available to carry a current, which decreases the material's resistance. (d) Random experimental error in measuring either the temperature (e.g. thermometer reading error) or the resistance (e.g. from the ohmmeter/multimeter, or contact resistance) would cause small variations in the plotted points, so not every point lies exactly on a perfectly smooth curve.

[1] Curve identified (not a straight line); [1] correct overall shape, decreasing and becoming less steep as temperature increases; [1] draw a smooth curve of best fit; [1] correct interpolation method (vertical then horizontal line from 50°C) to read the resistance; [1] resistance decreases as temperature increases (negative correlation) stated; [1] explanation that higher temperature frees/releases more charge carriers in the thermistor; [1] more charge carriers available leads to lower resistance; [1] random error in temperature/resistance measurement identified; [1] correct explanation that this causes small deviations from a perfectly smooth curve.

(a) A spring obeying Hooke's law produces a straight-line graph of force against extension, passing through the origin, showing that force is directly proportional to extension. The rubber band's curved graph shows that force is not directly proportional to extension; because the gradient of the curve changes, the effective stiffness ('spring constant') of the rubber band is not constant but changes as it is stretched further. (b) Hooke's law specifically requires a straight-line, directly proportional relationship between force and extension (a straight line through the origin); since the rubber band's graph is a curve, the data do not support the conclusion that it obeys Hooke's law. A more accurate conclusion would be that the rubber band does not obey Hooke's law (or obeys it only approximately, over a small initial range of extension, if at all). (c) The student could repeat the investigation using several different rubber bands (or repeat readings at each extension) to check whether the same curved, non-proportional relationship is consistently obtained; this would increase confidence that the conclusion is generally true of rubber bands, rather than being due to an unusual sample or a measurement error.

[1] Spring shows a straight-line/directly proportional relationship, rubber band shows a curve; [1] force is not directly proportional to extension for the rubber band; [1] the gradient (effective stiffness) changes as extension increases; [1] explains the curved graph does not show direct proportionality, which Hooke's law requires; [1] conclusion is therefore not supported by the data/is an overgeneralisation; [1] improved conclusion suggested, e.g. the rubber band does not obey Hooke's law (or only approximately); [1] suggests repeating the investigation (e.g. with other rubber bands, or repeat readings); [1] explains this checks whether the result is reproducible/consistent; [1] explains this increases confidence the conclusion is generally true and not due to error/an unusual sample.

(a) The independent variable, deliberately changed by the student, is the length of the wire. (b) The dependent variable, measured as the outcome, is the resistance of the wire. (c) To make this a fair test, variables such as the diameter/cross-sectional area of the wire, the material the wire is made from, its temperature, and the current used to test it should all be kept the same (standardised) for every length tested, since each of these could independently affect resistance.

[1] Independent variable = length of wire; [1] dependent variable = resistance of the wire; [1] each for any two valid standardised variables (e.g. diameter/thickness of wire, material of wire, temperature, current used). Max [2] for part (c).

(a) The student could use a series of containers of different diameters or shapes (for example, a wide, shallow dish compared with a narrow, tall container), while adding an identical, measured volume of water to each container, so that different surface areas of water are exposed to the air while the volume of water is kept the same. (b) The initial temperature of the water should be controlled (kept the same at the start of each trial); if it were not controlled, any observed difference in the rate of cooling between trials could be caused, wholly or partly, by a difference in starting temperature rather than by the difference in surface area being investigated, making the comparison invalid.

[1] Valid method suggested for varying surface area at constant volume; [1] correct detail (e.g. different container shapes/diameters, with the same volume of water added each time); [1] valid control variable identified (e.g. initial temperature, volume of water, room temperature/draughts); [1] correct explanation of why controlling it matters for validity.

(a) The independent variable is the angle of the ramp, and the dependent variable is the acceleration of the trolley. (b) The student could use a trolley with low-friction wheels/bearings, or compensate for friction by tilting the ramp very slightly before the main investigation begins, so that the trolley moves at a constant velocity with no push (cancelling out the effect of friction). This is important because friction would otherwise act against the trolley's motion, systematically reducing its measured acceleration by an amount unrelated to the ramp angle being tested, introducing error into the investigation.

[1] Independent variable = angle of the ramp; [1] dependent variable = acceleration of the trolley; [1] valid method to reduce/compensate for friction (e.g. low-friction wheels, friction compensation by pre-tilting the ramp); [1] correct explanation that friction would otherwise systematically reduce/affect the measured acceleration, introducing error unrelated to the variable tested.

(a) Variables that should be kept constant include the current flowing through the solenoid, the presence and type of any core material (e.g. a soft iron core) inside the solenoid, and the type/length of wire used. (b) Because the strength of a solenoid's magnetic field depends on both the current flowing through it and the number of turns, if the current were allowed to change between trials, any observed change in field strength could not be attributed solely to the change in the number of turns; the current must be kept constant so that the number of turns is the only variable being changed, making it a valid, fair test.

[1] each for any two valid controlled variables (e.g. current, core material/presence, wire type/length); [1] explains current independently affects field strength; [1] explains current must therefore be constant to isolate the effect of the number of turns, so the investigation is a fair test.

(a) \( \text{mean} = \frac{2.1+1.9+2.0}{3} = \frac{6.0}{3} = 2.0 \text{ s} \). (b) \( \text{speed} = \frac{\text{distance}}{\text{time}} = \frac{1.5}{2.0} = 0.75 \text{ m/s} \).

[1] Correct sum of times, 6.0 s; [1] mean = 2.0 s; [1] correct equation, speed = distance/time; [1] correct substitution, 1.5/2.0; [1] speed = 0.75 m/s with unit.

(a) \( \text{mean} = \frac{0.82+0.86+0.84}{3} = \frac{2.52}{3} = 0.84 \text{ mm} \). (b) Range \( = 0.86 - 0.82 = 0.04 \text{ mm} \). \( \% \text{ uncertainty} = \frac{0.04}{2 \times 0.84} \times 100 = \frac{0.04}{1.68} \times 100 = 2.4\% \).

[1] Correct sum, 2.52 mm; [1] mean = 0.84 mm; [1] correct range = 0.04 mm; [1] correct substitution into the given formula; [1] percentage uncertainty ≈ 2.4% (accept 2.3-2.4%).

(a) The volume of water displaced equals the volume of the stone: \( 76 - 50 = 26 \text{ cm}^3 \). (b) \( \rho = \frac{m}{V} = \frac{68}{26} = 2.6 \text{ g/cm}^3 \) (2 s.f.).

[1] Correct method, final volume − initial volume; [1] volume = 26 cm³; [1] correct equation, ρ = m/V; [1] correct substitution, 68/26; [1] density = 2.6 g/cm³ with unit.

(a) \( \Delta\theta = 45 - 20 = 25°C \). (b) Rearranging \( E = mc\Delta\theta \) gives \( c = \frac{E}{m\Delta\theta} \); with mass converted to kg (0.200 kg), \( c = \frac{21000}{0.200 \times 25} = \frac{21000}{5.0} = 4200 \text{ J/(kg°C)} \).

[1] Δθ = 25°C; [1] correct rearrangement, \( c = \frac{E}{m\Delta\theta} \); [1] mass correctly converted to kg (0.200 kg); [1] correct substitution, \( \frac{21000}{0.200 \times 25} \); [1] c = 4200 J/(kg°C) (J kg⁻¹ °C⁻¹) with unit.

(a) The block is most likely to be copper, since the calculated density (8.90 g/cm³) matches the reference density given for copper exactly. (b) Possible reasons for a discrepancy include: the block may not be pure copper (it could be an alloy, or contain impurities or trapped air pockets, which would affect its density); there could be random measurement error in the mass reading (e.g. an uncalibrated balance) or the volume reading (e.g. parallax error when reading the level in a measuring cylinder, or trapped air bubbles when using water displacement for an irregular shape).

[1] Copper correctly identified; [1] correct reasoning, the calculated value matches the reference density for copper; [1] valid reason suggested (e.g. impurities/alloying, trapped air bubbles, measurement error in mass or volume); [1] further relevant detail/explanation of how this would affect the calculated density; [1] additional distinct valid point or fuller explanation (e.g. a specific measurement technique issue).

(a) \( \% \text{ difference} = \frac{|9.3-9.8|}{9.8} \times 100 = \frac{0.5}{9.8} \times 100 = 5.1\% \). (b) Human reaction time when starting and stopping the stopwatch tends to make the measured falling time longer than the true value. Using an equation such as \( s = \frac{1}{2}gt^2 \), a larger measured time t (for the same fall distance s) gives a smaller calculated value of g, which would explain why the student's calculated value is lower than the accepted value.

[1] Correct method, \( \frac{|9.3-9.8|}{9.8} \times 100 \); [1] percentage difference ≈ 5.1% (accept 5.0-5.2%); [1] valid source of error suggested (e.g. reaction time in timing); [1] correct explanation that this makes the measured time too long; [1] correct link explaining why a longer measured time leads to a lower calculated value of g.

(a) Two pins are placed upright, some distance apart, to define the path of the incident ray before it enters the glass block. Looking through the glass block from the opposite side, two further pins are placed so that they appear to line up exactly with the images of the first two pins as seen through the block; these mark the path of the emergent (refracted) ray. Once the pins and block are removed, the pin positions and the outline of the block (drawn beforehand) are used to draw and measure the ray paths. (b) Angles of incidence and refraction must be measured from the normal (a line drawn at right angles to the glass surface at the point the ray meets it), using a protractor and aligning it carefully; in addition, the outline of the glass block should be traced accurately in sharp pencil before it is removed, so that the entry and exit points of the ray are marked precisely.

[1] Two pins used to define/sight the incident ray before the block; [1] two further pins placed in line with the images of the first pins, viewed through the block, to mark the emergent ray; [1] angles measured from the normal (line perpendicular to the surface) using a protractor; [1] the outline of the glass block is traced accurately (in sharp pencil) before removal, so entry/exit points are marked precisely.

(a) The student draws a straight line on paper to represent the mirror, with a normal line at right angles to it at the point where the light ray will strike. A ray from the ray box is directed at this point at a chosen angle of incidence, measured from the normal using a protractor; the path of the reflected ray is then marked and its angle from the normal (the angle of reflection) is measured, also using the protractor. (b) The law of reflection states that the angle of incidence should equal the angle of reflection. The student could check that this holds generally by repeating the investigation for several different angles of incidence (e.g. 20°, 40°, 60°) and confirming that the measured angle of reflection equals the angle of incidence each time.

[1] Normal drawn at right angles to the mirror line at the point of incidence, and ray directed at a chosen, measured angle of incidence; [1] angle of reflection measured (from the normal) using a protractor; [1] angle of incidence = angle of reflection correctly stated; [1] repeating for several different angles of incidence and confirming equality each time.

(a) The lens is held so that light from the very distant object passes through it and forms an image on a screen positioned on the other side of the lens; the screen is moved back and forth until the sharpest possible image is obtained, and the distance from the lens to the screen at this point is measured and taken as an estimate of the focal length, since light rays from a very distant object are approximately parallel when they reach the lens and so converge close to the focal point. (b) In reality the object, although distant, is at a large but finite distance rather than truly at infinity, so the rays reaching the lens are not perfectly parallel; this means the image forms slightly further from the lens than the true focal point, giving a slight overestimate of the focal length. The student could improve the estimate by repeating the measurement several times and calculating a mean lens-to-screen distance, or by using an object as far away as practically possible so the rays are closer to parallel.

[1] Lens used to focus light from a distant object onto a screen, moving the screen to find the sharpest image; [1] lens-to-screen distance at best focus measured and taken as the focal length; [1] correct explanation that the object is only at a large, finite distance (not true infinity), so rays are not perfectly parallel, giving only an approximate (slightly overestimated) value; [1] valid improvement suggested (e.g. repeat and take a mean, use as distant an object as possible).