2025 CCEA GCSE Physics 1210 Practice Paper with Answers

(a) acceleration = (final velocity - initial velocity) ÷ time taken = (20 - 8) ÷ 6 = 12 ÷ 6 = 2 m/s². (b) average velocity = (initial velocity + final velocity) ÷ 2 = (8 + 20) ÷ 2 = 28 ÷ 2 = 14 m/s. (c) distance = average velocity × time = 14 × 6 = 84 m. (d) rate of change of speed = (final speed - initial speed) ÷ time = (0 - 20) ÷ 5 = -20 ÷ 5 = -4 m/s², so the magnitude of the deceleration is 4 m/s². (e) comparing magnitudes: 2 m/s² (part a) versus 4 m/s² (part d) - since 4 > 2, the deceleration in (d) is greater. Final answer: (a) 2 m/s²; (b) 14 m/s; (c) 84 m; (d) 4 m/s²; (e) the deceleration in (d) is greater.

(a) [1] correct formula/substitution (20-8)÷6; [1] correct evaluation 12÷6; [1] final answer 2 m/s² with unit. (b) [1] correct formula (8+20)÷2; [1] correct evaluation 28÷2; [1] final answer 14 m/s with unit. (c) [1] correct formula distance = average velocity × time; [1] correct substitution 14×6; [1] final answer 84 m with unit (error carried forward from (b) accepted). (d) [1] correct formula/substitution (0-20)÷5; [1] correct evaluation; [1] final answer 4 m/s² (magnitude) with unit. (e) [1] correct comparison of the two values; [1] correct conclusion that (d) is greater, with reference to both values (2 m/s² and 4 m/s²).

(a) resultant force = pulling force - frictional force = 9 - 3 = 6 N. (b) resultant force = mass × acceleration, so acceleration = resultant force ÷ mass = 6 ÷ 1.5 = 4 m/s². (c) rearranging acceleration = (final velocity - initial velocity) ÷ time gives final velocity = initial velocity + (acceleration × time) = 0 + (4 × 4) = 16 m/s. (d) Newton's first law states that in the absence of unbalanced forces an object will continue to move in a straight line at constant speed (i.e. with constant velocity), or remain at rest if it was already at rest. (e) F = ke, so e = F ÷ k = 9 ÷ 50 = 0.18 m (18 cm). Check: k × e = 50 × 0.18 = 9 N, which matches the given force, confirming the answer. Final answer: (a) 6 N; (b) 4 m/s²; (c) 16 m/s; (d) Newton's first law as stated; (e) 0.18 m.

(a) [1] correct substitution 9-3; [1] final answer 6 N with unit. (b) [1] correct formula (rearranged F=ma); [1] correct substitution 6÷1.5; [1] final answer 4 m/s² with unit. (c) [1] correct rearranged formula/substitution 0+(4×4); [1] correct evaluation; [1] final answer 16 m/s with unit (ecf from (b) accepted). (d) [1] partial statement referring to constant velocity/no change without unbalanced force; [1] full, accurate statement of Newton's first law. (e) [1] correct formula F=ke rearranged to e=F/k; [1] correct substitution 9÷50; [1] correct evaluation 0.18; [1] final answer 0.18 m or 18 cm with unit.

(a) Ek = ½mv², so v² = 2Ek ÷ m = (2 × 32) ÷ 1 = 64, so v = √64 = 8 m/s. (b) Ep = mgh, and since all kinetic energy converts to potential energy at maximum height, Ep = 32 J, so h = Ep ÷ (mg) = 32 ÷ (1 × 10) = 3.2 m. Check by an independent route: using v² = 2gh (equivalent to the energy relationship for free vertical motion), h = v² ÷ (2g) = 64 ÷ 20 = 3.2 m, which matches, confirming the answer. (c) work = force × distance = 20 × 0.8 = 16 J. (d) power = work done ÷ time taken = 16 ÷ 0.4 = 40 W. (e) The Principle of Conservation of Energy states that energy can be changed from one form to another, but the total amount of energy does not change. Final answer: (a) 8 m/s; (b) 3.2 m; (c) 16 J; (d) 40 W; (e) energy is conserved - it changes form but the total amount stays the same.

(a) [1] correct rearranged formula v²=2Ek/m; [1] correct substitution/evaluation to 64; [1] final answer 8 m/s with unit. (b) [1] correct formula h=Ep/(mg); [1] correct substitution 32÷(1×10); [1] final answer 3.2 m with unit. (c) [1] correct formula work=force×distance; [1] correct substitution 20×0.8; [1] final answer 16 J with unit. (d) [1] correct formula power=work/time; [1] correct substitution 16÷0.4; [1] final answer 40 W with unit (ecf from (c) accepted). (e) [1] partial reference to energy changing form; [1] full accurate statement including that total energy does not change.

(a) Ep = mgh = 5 × 10 × 4 = 200 J. (b) useful power output = useful energy transferred ÷ time = 200 ÷ 8 = 25 W. (c) efficiency = useful output energy ÷ total input energy = 25 ÷ 50 = 0.5 = 50%. (d) total input energy over 8 s = input power × time = 50 × 8 = 400 J; energy wasted = total input energy - useful output energy = 400 - 200 = 200 J. Check by an independent route: wasted power = input power - output power = 50 - 25 = 25 W, so over 8 s energy wasted = 25 × 8 = 200 J, which matches, confirming the answer. Final answer: (a) 200 J; (b) 25 W; (c) 50%; (d) 200 J.

(a) [1] correct formula Ep=mgh; [1] correct substitution 5×10×4; [1] final answer 200 J with unit. (b) [1] correct formula power=energy/time; [1] correct substitution 200÷8; [1] final answer 25 W with unit (ecf from (a)). (c) [1] correct formula efficiency=useful output/total input; [1] correct substitution 25÷50; [2] correct evaluation and final answer expressed as 50% (award 3 if expressed only as decimal 0.5 without converting to %). (d) [1] correct method to find total input energy (50×8=400 J) or equivalent; [1] correct method to find wasted power/energy; [1] correct substitution; [1] final answer 200 J with unit (ecf from earlier parts accepted throughout).

(a) The student should partially fill a measuring cylinder with water and record the initial volume reading, then carefully lower the steel into the cylinder so it is fully submerged, and record the new (higher) volume reading; the volume of the steel is the difference between the two readings (the volume of water displaced). (b) volume = final reading - initial reading = 80 - 60 = 20 cm³. (c) density = mass ÷ volume = 158 ÷ 20 = 7.9 g/cm³. (d) pressure = force ÷ area = 1.58 ÷ 4 = 0.395 N/cm². Check: mass 158 g = 0.158 kg, so weight = mg = 0.158 × 10 = 1.58 N, which matches the force given, confirming the data is self-consistent; and density × volume = 7.9 × 20 = 158 g, matching the given mass, further confirming part (c). (e) In a solid, the particles are held in fixed positions by strong forces between them and can only vibrate, giving a solid a fixed shape and volume; in a liquid, the particles are mainly touching but some gaps have appeared, allowing the particles enough energy and space to move past each other while forces still hold them together, giving a liquid a fixed volume but no fixed shape. Final answer: (a) as described; (b) 20 cm³; (c) 7.9 g/cm³; (d) 0.395 N/cm²; (e) solid particles are fixed and only vibrate, liquid particles have gaps and can move while remaining mainly in contact.

(a) [1] record initial volume of water; [1] fully submerge the steel and record new volume; [1] volume of steel = difference between the two readings. (b) [1] correct subtraction 80-60; [1] final answer 20 cm³ with unit. (c) [1] correct formula density=mass/volume; [1] correct substitution 158÷20; [1] final answer 7.9 g/cm³ with unit. (d) [1] correct formula pressure=force/area; [1] correct substitution 1.58÷4; [1] final answer 0.395 N/cm² with unit (accept equivalent working in other consistent units). (e) [1] mark for describing solid particle arrangement (fixed positions/vibrate only); [1] mark for describing liquid particle arrangement (gaps, can move, still mainly touching); [1] mark for a clear, explicit comparison/contrast linking the two states.

Nuclear fission: for fission to occur, a uranium nucleus must first absorb a neutron; the nucleus then splits into two smaller nuclei, releasing energy and several further neutrons. These fission neutrons go on to cause further fissions in other uranium nuclei, creating a self-sustaining chain reaction that releases a large, continuous amount of energy, which is used to generate electricity in a power station. Issue 1: although nuclear power provides employment opportunities, many people remain concerned about living close to nuclear power plants and to the facilities used to store radioactive waste, due to perceived health and safety risks. Issue 2: incidents at nuclear power plants, such as those in Ukraine (Chornobyl) and Japan (Fukushima), have caused huge economic, health and environmental damage to the surrounding area, raising ongoing safety concerns about the technology. A further issue that could be credited: although nuclear fission itself does not release carbon dioxide, the mining, transport and purification of uranium ore releases significant amounts of greenhouse gases into the atmosphere. Final answer: fission involves a uranium nucleus absorbing a neutron, splitting to release energy and further neutrons that sustain a chain reaction; associated issues include public concern about living near plants/waste storage and the risk of major, damaging incidents.

Levels of response grid (6 marks). Level C (1-2 marks): basic, list-like points about fission or nuclear issues, minimal specialist vocabulary, weak organisation. Level B (3-4 marks): sound explanation of the fission process (neutron absorption, splitting, chain reaction) OR of the issues, with some development; reasonably organised with some specialist vocabulary. Level A (5-6 marks): detailed, accurate explanation of the fission process AND two distinct, developed social/environmental/ethical issues, with excellent syntax, spelling, punctuation and specialist terminology throughout.

average speed = distance moved ÷ time taken = 450 ÷ 30 = 15 m/s. Final answer: 15 m/s.

[1] correct formula average speed = distance/time; [1] correct substitution 450÷30; [1] final answer 15 m/s with unit.

acceleration = (final velocity - initial velocity) ÷ time taken = (9 - 0) ÷ 3 = 3 m/s². Final answer: 3 m/s².

[1] correct formula/substitution (9-0)÷3; [1] correct evaluation; [1] final answer 3 m/s² with unit.

resultant force = mass × acceleration, so acceleration = resultant force ÷ mass = 15 ÷ 3 = 5 m/s². Final answer: 5 m/s².

[1] correct rearranged formula acceleration=force/mass; [1] correct substitution 15÷3; [1] final answer 5 m/s² with unit.

W = mg = 12 × 10 = 120 N. Final answer: 120 N.

[1] correct formula W=mg; [1] correct substitution 12×10; [1] final answer 120 N with unit.

Alpha radiation is stopped by a few centimetres of air or a thin sheet of paper, because alpha particles are large and heavily ionising with a very short range. Gamma radiation easily passes through paper and several metres of air and even aluminium, and can only be blocked by a thick sheet of lead (or a similarly dense material), because it is highly penetrating electromagnetic radiation. Final answer: alpha radiation is stopped by paper; gamma radiation requires lead to stop it.

[1] mark for correctly identifying alpha as the type stopped by paper; [1] mark for correctly identifying gamma as the type requiring lead; [1] further mark for both correctly matched together with no confusion between the two types.

In an alpha decay equation, the mass numbers (top) must balance on both sides, and the atomic numbers (bottom) must balance on both sides. Mass number: \( 238 = A + 4 \), so \( A = 238 - 4 = 234 \). Atomic number: \( 92 = Z + 2 \), so \( Z = 92 - 2 = 90 \). This matches the real identity of thorium, which has atomic number 90, confirming the balancing is correct. Final answer: A = 234, Z = 90 (i.e. \( {}^{238}_{92}\text{U} \rightarrow {}^{234}_{90}\text{Th} + {}^{4}_{2}\text{He} \)).

[1] mark for correctly balancing the mass number, A = 234; [1] mark for correctly balancing the atomic number, Z = 90; [1] mark for both values stated correctly together with correct working shown (238-4 and 92-2).

density = mass ÷ volume = 250 ÷ 200 = 1.25 g/cm³. Final answer: 1.25 g/cm³.

[1] correct formula density=mass/volume; [1] correct substitution 250÷200; [1] final answer 1.25 g/cm³ with unit.

Studies of light from stars, including the Sun, show they are composed mainly of hydrogen and helium, and their energy is supplied by the fusion of hydrogen nuclei into helium. This process is called nuclear fusion. Final answer: nuclear fusion; hydrogen combines (fuses) to form helium.

[1] mark for correctly naming the process as nuclear fusion; [1] mark for correctly naming hydrogen as the element that fuses; [1] mark for correctly naming helium as the element formed.

(a) for resistors in series, total resistance = sum of the individual resistances = 4 + 8 = 12 Ω. (b) voltage = current × resistance, so current = voltage ÷ total resistance = 12 ÷ 12 = 1 A. (c) the current through each component in series is the same (1 A), so voltage across the 8 Ω resistor = current × resistance = 1 × 8 = 8 V. Check: voltage across the 4 Ω resistor = 1 × 4 = 4 V, and 4 + 8 = 12 V, which equals the supply voltage, confirming the series voltage rule and the answer. (d) for resistors in parallel, the combined resistance is always less than the smallest individual resistance, because connecting resistors in parallel provides additional paths for current to flow, increasing the total current for a given voltage and therefore reducing the overall resistance. Final answer: (a) 12 Ω; (b) 1 A; (c) 8 V; (d) less than 4 Ω, because parallel paths reduce overall resistance below the smallest branch resistance.

(a) [1] correct method (addition of resistances); [1] final answer 12 Ω with unit. (b) [1] correct rearranged formula current=voltage/resistance; [1] correct substitution 12÷12; [1] final answer 1 A with unit. (c) [1] correct recall that current is the same throughout a series circuit; [1] correct substitution 1×8; [1] final answer 8 V with unit (ecf from (b) accepted). (d) [1] correct statement that the combined resistance is less than 4 Ω; [1] correct reasoning referring to an additional current path/route; [1] further mark for a clear, complete explanation linking the additional path to reduced total resistance.

(a) power = current × voltage, so current = power ÷ voltage = 2300 ÷ 230 = 10 A. (b) 3 minutes = 3 × 60 = 180 s; energy = power × time = 2300 × 180 = 414000 J (414 kJ). (c) power rating in kilowatts = 2300 ÷ 1000 = 2.3 kW; energy in kWh = power (kW) × time (hours) = 2.3 × 0.5 = 1.15 kWh. (d) appliances with metal cases are usually earthed: the earth wire connects the metal case to earth, providing a very low-resistance path so that if a fault causes the live wire to touch the case, a large current flows to earth rather than through anyone touching the case, and this surge of current blows the fuse, cutting off the supply and protecting the user from electric shock and the appliance from further damage. Final answer: (a) 10 A; (b) 414000 J; (c) 1.15 kWh; (d) as described above.

(a) [1] correct rearranged formula current=power/voltage; [1] correct substitution 2300÷230; [1] final answer 10 A with unit. (b) [1] correct conversion of time to seconds (180 s); [1] correct formula energy=power×time; [1] correct substitution 2300×180; [1] final answer 414000 J (or 414 kJ) with unit. (c) [1] correct conversion of power to kW (2.3 kW) and/or correct method; [1] final answer 1.15 kWh with unit. (d) [1] mark for stating the earth wire connects the case to earth/provides a low-resistance path; [1] further mark for explaining this (with the fuse) prevents electric shock by causing a large current to blow the fuse.

(a) turns-ratio equation: Ns/Np = Vs/Vp, so Vs = Vp × (Ns/Np) = 230 × (100/2000) = 230 × 0.05 = 11.5 V. (b) since the transformer is treated as 100% efficient, input power = output power, so Vp × Ip = Vs × Is, giving Ip = (Vs × Is) ÷ Vp = (11.5 × 4.6) ÷ 230 = 52.9 ÷ 230 = 0.23 A. (c) power delivered by the secondary = Vs × Is = 11.5 × 4.6 = 52.9 W; since the transformer is 100% efficient, the power taken from the primary supply equals this output power, 52.9 W (check: Vp × Ip = 230 × 0.23 = 52.9 W, which matches, confirming the answer). (d) real transformers lose some energy, for example through resistive heating in the coils and through eddy currents induced in the core; using a laminated (rather than solid) iron core reduces eddy current losses, and using thick, low-resistance copper wire for the coils reduces resistive heating losses. Final answer: (a) 11.5 V; (b) 0.23 A; (c) 52.9 W (both primary and secondary, since efficiency is 100%); (d) laminated core and low-resistance coil wire (or another valid design feature).

(a) [1] correct turns-ratio equation; [1] correct substitution 230×(100/2000); [1] final answer 11.5 V with unit. (b) [1] correct statement that input power=output power (100% efficiency); [1] correct rearranged formula Ip=(Vs×Is)/Vp; [1] correct substitution/evaluation; [1] final answer 0.23 A with unit (ecf from (a) accepted). (c) [1] correct calculation of secondary power (Vs×Is=52.9 W); [1] correct statement that the primary power equals this value, with brief reasoning (100% efficiency). (d) [1] mark for each valid design feature stated, to a maximum of 2: laminated core [1]; low-resistance/thick copper coil wire [1] (accept other valid features, e.g. soft iron core to allow easy magnetisation/demagnetisation).

(a) resistance is proportional to length at constant temperature, so resistance per metre = 5 ÷ 2 = 2.5 Ω/m; for a 6 m wire, resistance = 2.5 × 6 = 15 Ω (equivalently, since length is tripled from 2 m to 6 m, resistance is also tripled: 5 × 3 = 15 Ω, confirming the answer). (b) charge = current × time, so current = charge ÷ time = 15 ÷ 5 = 3 A. (c) voltage = current × resistance = 3 × 5 = 15 V. (d) as the cross-sectional area of a metallic conductor increases, its resistance decreases, because a wider conductor gives the charge carriers more space to flow through with less obstruction. Final answer: (a) 15 Ω; (b) 3 A; (c) 15 V; (d) resistance decreases as cross-sectional area increases.

(a) [1] correct use of the proportionality relationship (e.g. resistance per unit length or ratio method); [1] correct substitution/evaluation; [1] final answer 15 Ω with unit. (b) [1] correct rearranged formula current=charge/time; [1] correct substitution 15÷5; [1] final answer 3 A with unit. (c) [1] correct formula voltage=current×resistance; [1] correct substitution 3×5; [1] final answer 15 V with unit. (d) [1] mark for stating resistance decreases as area increases (or the inverse); [1] further mark for a correct qualitative reason (more space/paths for charge carriers).

(a) turns-ratio equation: Ns/Np = Vs/Vp, so Vs = Vp × (Ns/Np) = 1000 × (200000/500) = 1000 × 400 = 400000 V. (b) power = current × voltage, so current = power ÷ voltage = 2000000 ÷ 400000 = 5 A. (c) transmitting the same 2 MW of power at a high voltage of 400000 V requires only a small current of 5 A (from part b), whereas transmitting it at a much lower voltage would require a correspondingly larger current; since the heating effect in a cable depends on the current flowing through it, a smaller current produces far less energy wasted as heat in the transmission cables, making high-voltage transmission much more efficient. (d) mains electricity in homes is used at a much lower, safer voltage (in the UK, 230 V) suited to household wiring and appliances; transmitting electricity to homes at 400000 V would be extremely dangerous to the public and would destroy ordinary household appliances, so step-down transformers are used to reduce the voltage before it reaches consumers. Final answer: (a) 400000 V; (b) 5 A; (c) high voltage means low current for the same power, and lower current causes much less heating/energy loss in the cables; (d) household appliances and wiring need a much lower, safer voltage than the transmission voltage.

(a) [1] correct turns-ratio equation; [1] correct substitution 1000×(200000/500); [1] final answer 400000 V with unit. (b) [1] correct rearranged formula current=power/voltage; [1] correct substitution 2000000÷400000; [1] final answer 5 A with unit (ecf from (a)). (c) [1] mark for recognising that high voltage means lower current for the same power; [1] mark for linking lower current to reduced heating/energy loss in the cables; [1] further mark for a clear, complete explanation connecting both ideas. (d) [1] mark for stating homes need a much lower/safer voltage; [1] further mark for explaining the danger/damage that would result without stepping down.

(a) for each pair of readings, voltage ÷ current = 1.0÷0.5 = 2.0; 2.0÷1.0 = 2.0; 3.0÷1.5 = 2.0; 4.0÷2.0 = 2.0. Since the ratio of voltage to current is constant (2.0 Ω) for every reading, the graph of voltage against current would be a straight line through the origin, showing that current and voltage are proportional for this metal wire at constant temperature - this is Ohm's law. (b) resistance = voltage ÷ current = constant ratio found above = 2 Ω (using any pair, e.g. 4.0 ÷ 2.0 = 2 Ω). (c) voltage = current × resistance = 3.5 × 2 = 7 V. (d) unlike a metal wire at constant temperature, the resistance of a filament lamp increases as the current through it increases; this is because the increasing current causes the filament to heat up, and the resistance of the filament increases as its temperature rises. Final answer: (a) constant V/I ratio of 2.0 Ω for all readings confirms proportionality (Ohm's law); (b) 2 Ω; (c) 7 V; (d) filament lamp resistance increases with current, because the filament gets hotter.

(a) [1] correct calculation of the V/I ratio for at least two pairs of readings; [1] recognition that the ratio is constant across all readings; [1] correct conclusion that this constant ratio (straight line through the origin) demonstrates Ohm's law. (b) [1] correct formula resistance=voltage/current; [1] correct substitution using data from the table; [1] final answer 2 Ω with unit. (c) [1] correct formula voltage=current×resistance; [1] correct substitution 3.5×2; [1] final answer 7 V with unit (ecf from (b)). (d) [1] mark for correctly stating resistance increases with current; [1] mark for the correct reason (filament heats up/temperature rises).

In a ray diagram of reflection at a plane mirror, a dashed line called the normal is drawn at right angles (90°) to the mirror surface at the point where the incident ray strikes it. The incident ray is drawn approaching the mirror, and the angle between this ray and the normal is the angle of incidence. The reflected ray is then drawn leaving the mirror on the same side as the incident ray arrived, such that the angle between the reflected ray and the normal (the angle of reflection) is drawn equal in size to the angle of incidence, since the angle of incidence always equals the angle of reflection. Final answer: draw the normal at right angles to the mirror at the point of incidence, then draw the incident and reflected rays so that the angle of incidence equals the angle of reflection, both measured from the normal.

[1] mark for drawing/describing the normal at right angles to the mirror at the point of incidence; [1] mark for correctly identifying the angle of incidence as being measured from the normal to the incident ray; [1] mark for correctly identifying the angle of reflection as being measured from the normal to the reflected ray; [1] mark for stating that the angle of incidence equals the angle of reflection.

In a short-sighted eye, the eyeball has grown slightly too long, so when parallel rays of light from a distant object enter the eye, they are refracted (bent) too strongly by the eye's lens and cornea and come to a focus in front of the light-sensitive retina rather than on it; the rays then continue to spread out again before reaching the retina, so distant objects appear blurred. To correct this, a diverging (concave) lens is placed in front of the eye; this lens spreads the parallel rays out slightly before they reach the eye's own lens, so that the eye's stronger-than-needed refraction now brings the rays to a focus exactly on the retina, producing a clear image. Final answer: a short-sighted eye focuses light in front of the retina because it refracts too strongly; a diverging lens placed in front of the eye spreads the light slightly before it enters, moving the focus back onto the retina.

[1] mark for correctly describing that in a short-sighted eye, light focuses in front of the retina; [1] mark for correctly attributing this to the eye refracting too strongly / being too long; [1] mark for correctly stating a diverging (concave) lens is used for correction; [1] mark for explaining that the diverging lens spreads the rays before they enter the eye so they now focus on the retina.

When a ray of light travelling in air meets the surface of a glass block, part of it is refracted into the glass. In a ray diagram, the normal is drawn at right angles to the glass surface at the point where the ray meets it; the refracted ray inside the glass is drawn bent towards the normal, so the angle of refraction (measured from the normal) is smaller than the angle of incidence in the air. This bending occurs because light travels more slowly in glass than in air; when light slows down on entering a denser medium, it bends towards the normal (the converse is also true - light speeds up and bends away from the normal when leaving a denser medium for a less dense one). Final answer: the ray bends towards the normal on entering the glass, because light slows down when passing from the less dense air into the denser glass.

[1] mark for correctly stating the ray bends towards the normal on entering the glass; [1] mark for correctly identifying that light travels more slowly in glass than in air; [1] mark for correctly linking the slowing of light to bending towards the normal; [1] mark for reference to the normal being drawn at right angles to the surface, or correct description of the angle of refraction being smaller than the angle of incidence.

To use a converging lens as a magnifying glass, the object is placed closer to the lens than its focal length. Two construction rays are drawn from the top of the object: one travelling parallel to the principal axis, which is refracted by the lens to pass through the focal point on the far side; and one travelling through the centre of the lens, which passes through undeviated. Because the object is inside the focal length, these two refracted rays continue to diverge on the far side of the lens and never actually meet, so they must be extended backwards (as dashed construction lines) on the same side of the lens as the object; the point where these dashed lines cross is where the image appears to be. The image formed is virtual (it cannot be projected onto a screen because no real light rays meet there), upright, and magnified (larger than the object). Final answer: with the object inside the focal length, the refracted rays diverge and appear (when traced backwards) to come from a point on the same side as the object, forming a magnified, upright, virtual image.

[1] mark for correctly stating the object must be placed closer to the lens than the focal length; [1] mark for correctly describing at least one construction ray (parallel ray refracting through the focal point, or ray through the centre passing straight through); [1] mark for correctly stating the rays must be extended backwards to locate the image, since they do not actually meet; [1] mark for correctly describing the image as virtual, upright and magnified.

In a wavefront diagram, plane wavefronts are shown as a series of parallel straight lines moving in the direction of travel. As the wavefronts cross the boundary from deep water into shallower water at an angle, the part of each wavefront that reaches the shallow water first slows down, while the rest of the wavefront (still in deep water) continues at its original, faster speed; this causes the wavefronts to bend, changing the direction of travel towards the normal to the boundary, analogous to the refraction of light on entering a denser medium. The wave speed decreases in the shallower water, and since v = fλ and the frequency of the wave stays the same (it is set by the source), the wavelength must decrease to match the reduced speed, so the wavefronts also appear closer together in the shallow water. Final answer: the wavefronts bend towards the normal on entering shallow water because the wave slows down there, and the wavelength decreases (with frequency unchanged) to match the lower speed.

[1] mark for correctly describing the wave slowing down in shallower water; [1] mark for correctly describing the wavefronts bending/changing direction (towards the normal) at the boundary; [1] mark for correctly stating the wavelength decreases in the shallower water (wavefronts closer together); [1] mark for correctly stating the frequency remains unchanged.

A star forms when enough dust and gas from space is pulled together by gravitational attraction, first forming a protostar. Once the core becomes hot and dense enough for nuclear fusion of hydrogen into helium to begin, the star enters the stable main sequence period of its life, which lasts for most of the star's lifetime. A star of about the Sun's mass will then, once its hydrogen fuel is used up, expand and cool at its surface to become a red giant. After this, the outer layers of the star drift away into space, leaving behind a hot, dense core called a white dwarf, which gradually cools over a very long time to eventually become a cold, dark black dwarf. Stability during the main sequence: the star remains a stable size during this period because the outward force caused by thermal expansion (the pressure generated by the energy released through nuclear fusion in the core, pushing outward) is exactly balanced by the inward force of gravity (the star's own mass pulling material inward); as long as these two forces remain in balance, the star neither expands nor collapses. Final answer: protostar → main sequence → red giant → white dwarf → black dwarf, and the star is stable on the main sequence because outward thermal expansion from fusion balances inward gravitational pull.

Levels of response grid (6 marks). Level C (1-2 marks): basic, list-like naming of one or two stages, with limited explanation of stability, minimal specialist vocabulary. Level B (3-4 marks): sound description of most/all life cycle stages in correct order OR a sound explanation of main sequence stability, with some development; reasonably organised, some specialist vocabulary. Level A (5-6 marks): detailed, accurate description of all five stages (protostar, main sequence, red giant, white dwarf, black dwarf) in correct order AND a clear, accurate explanation of main sequence stability (balance of thermal expansion and gravity), with excellent syntax, spelling, punctuation and specialist terminology throughout.

Similarity: both sound waves and water waves are waves that transfer energy from one point to another through vibrations, without transferring matter overall. Difference: sound (and ultrasound) is a longitudinal wave, in which the particles of the medium vibrate parallel to (in the same direction as) the direction of energy transfer, whereas water waves are transverse waves, in which the particles vibrate perpendicular to (at right angles to) the direction of energy transfer. Final answer: both transfer energy via vibrations, but sound is longitudinal while water waves are transverse.

[1] mark for a valid similarity (e.g. both transfer energy through vibrations/are waves); [1] mark for a valid, correctly-classified difference (sound = longitudinal, water waves = transverse).

The full electromagnetic spectrum, in order of increasing wavelength (decreasing frequency), is: gamma rays, X-rays, ultraviolet, visible light, infrared, microwaves, radio waves. Any two regions other than visible light can be chosen and correctly ordered, for example: X-rays have a shorter wavelength than infrared, so the order of increasing wavelength would be X-rays, then infrared. Final answer: any two valid regions (e.g. X-rays and infrared) correctly ordered by increasing wavelength according to the full spectrum sequence gamma rays < X-rays < ultraviolet < visible light < infrared < microwaves < radio waves.

[1] mark for stating two valid regions of the electromagnetic spectrum (excluding visible light); [1] mark for arranging the two chosen regions correctly in order of increasing wavelength, consistent with the full spectrum order (radio waves longest, gamma rays shortest).

Direct current (d.c.) flows in one direction only around a circuit, with a constant polarity, and is produced by sources such as batteries and cells. Alternating current (a.c.) regularly reverses its direction of flow, continuously switching between positive and negative, and is the type of current supplied to homes from the mains. On a cathode ray oscilloscope (CRO), d.c. appears as a constant, unchanging horizontal line, while a.c. appears as a regularly repeating wave that oscillates above and below the zero line. Final answer: d.c. flows in one direction only; a.c. regularly reverses direction.

[1] mark for correctly describing d.c. as flowing in one direction only; [1] mark for correctly describing a.c. as regularly reversing/changing direction.

The strength of the magnetic field produced by a current-carrying coil (electromagnet) depends on: the size of the current flowing through the coil (a larger current produces a stronger field); the number of turns in the coil (more turns produce a stronger field); and the material used as the core of the coil (a soft iron core produces a much stronger field than an air core). Final answer: any two of current in the coil, number of turns in the coil, or the material used as the core.

[1] mark for each valid factor stated, to a maximum of 2: current in the coil [1]; number of turns in the coil [1]; material used as the core [1] (any two of these three).

The spring should be hung vertically from the clamp, with the metre rule fixed vertically alongside it (or clamped separately) so the position of the bottom of the spring (or a pointer attached to it) can be read against the scale. With no masses attached, the unstretched (zero-load) reading is recorded first, as the reference position. Masses are then added to the spring one 100 g slotted mass at a time, and after each addition the new scale reading is recorded once the spring has settled. The extension for each load is calculated as the new reading minus the original unstretched reading. All results should be recorded in a table with a column for the applied force (in N, found from the mass added) and a column for the corresponding extension (in cm or m), for each of the five loads. Final answer: hang the spring, record the unstretched length, add masses in 100 g steps recording the new length each time, then tabulate force against extension (new length minus unstretched length) for all five loads.

[1] mark for hanging the spring vertically with the ruler positioned to measure its length/position; [1] mark for recording the unstretched (zero-load) reading as a reference; [1] mark for adding masses in known steps (e.g. 100 g at a time) and recording the new reading after each; [1] mark for correctly describing how extension is calculated (new reading minus unstretched reading); [1] mark for describing a suitable table with force/load and extension columns covering all five loads.

Using the 30 cm ruler, the length, width and height of the rectangular block should each be measured, with the eye positioned directly above/level with the scale to avoid parallax error, and each dimension recorded in a suitable table (in cm). The electronic balance should be zeroed (tared) before the block is placed on it, and the mass of the block recorded in grams. The volume of the block can then be calculated by multiplying the three dimensions together (length × width × height). The mass and calculated volume are recorded together so that the density can subsequently be calculated as mass divided by volume. Final answer: measure length, width and height with the ruler and record them; measure mass with the zeroed balance; calculate volume as length × width × height, ready to compute density = mass ÷ volume.

[1] mark for measuring length, width and height with the ruler; [1] mark for reference to avoiding parallax error (eye level with the scale); [1] mark for zeroing/taring the balance before measuring mass; [1] mark for correctly describing how volume is calculated (length × width × height); [1] mark for recording mass and volume in a suitable table ready for the density calculation.

Label the y-axis 'Force (N)' and the x-axis 'Extension (cm)'. Choose a linear scale for each axis that uses at least half of the available grid (for example, 1 large square = 1.0 N on the y-axis, and 1 large square = 4.0 cm on the x-axis). Plot each of the six data points accurately: (0,0), (4.0,1.0), (8.0,2.0), (12.0,3.0), (16.0,4.0), (20.0,5.0). Since the extension is directly proportional to the force applied (within the limit of proportionality), the points should lie on a straight line, so draw a single straight line of best fit through the origin and the plotted points. Final answer: a straight line through the origin from (0,0) to (20.0,5.0), correctly labelled axes with units, and all six points accurately plotted.

[1] each axis correctly labelled with quantity and unit, to a maximum of [2]; [1] suitable linear scale chosen for both axes, using at least half the available grid; [1] all six points plotted accurately (within half a small square); [1] a single smooth straight line of best fit drawn through the origin and the plotted points (not a dot-to-dot join). Total 5 marks.

Label the y-axis 'Voltage (V)' and the x-axis 'Current (A)'. Choose a linear scale for each axis that uses at least half of the available grid (for example, 1 large square = 2.0 V on the y-axis, and 1 large square = 0.5 A on the x-axis). Plot each of the four data points accurately: (0.5,2.0), (1.0,4.0), (1.5,6.0), (2.0,8.0), and include the origin (0,0) since a metal wire at constant temperature obeys Ohm's law. Since current and voltage are proportional for a metal wire at constant temperature, draw a single straight line of best fit through the origin and the plotted points. Final answer: a straight line through the origin from (0,0) to (2.0,8.0), correctly labelled axes with units, and all four points accurately plotted.

[1] each axis correctly labelled with quantity and unit, to a maximum of [2]; [1] suitable linear scale chosen for both axes, using at least half the available grid; [1] all four points plotted accurately (within half a small square); [1] a single smooth straight line of best fit drawn through the origin and the plotted points. Total 5 marks.

Choosing two points on the line that are far apart to maximise accuracy, for example (0,0) and (20.0 cm, 5.0 N): the change in force is 5.0 - 0 = 5.0 N, and the change in extension is 20.0 - 0 = 20.0 cm = 0.20 m (converting to metres). Gradient = change in force ÷ change in extension = 5.0 ÷ 0.20 = 25 N/m. Since the gradient of a force-extension graph is numerically equal to the spring constant (F = ke), the spring constant k = 25 N/m. Check using a different pair of points: (8.0 cm, 2.0 N) to (16.0 cm, 4.0 N) gives gradient = (4.0-2.0) ÷ (0.16-0.08) = 2.0 ÷ 0.08 = 25 N/m, which matches, confirming the answer. Final answer: k = 25 N/m.

[1] mark for correctly identifying two points on the line, well separated, to calculate the gradient; [1] mark for correct calculation of the change in force; [1] mark for correct calculation of the change in extension, converted to metres; [1] mark for correct division to find the gradient; [1] mark for correctly stating the spring constant k = 25 N/m, with reference to the gradient being numerically equal to k.

Choosing two points on the line that are far apart to maximise accuracy, for example (0,0) and (2.0 A, 8.0 V): the change in voltage is 8.0 - 0 = 8.0 V, and the change in current is 2.0 - 0 = 2.0 A. Gradient = change in voltage ÷ change in current = 8.0 ÷ 2.0 = 4.0 Ω. Since voltage = current × resistance, the gradient of a voltage-current graph is numerically equal to the resistance, so R = 4 Ω. Check using a different pair of points: (0.5 A, 2.0 V) to (1.5 A, 6.0 V) gives gradient = (6.0-2.0) ÷ (1.5-0.5) = 4.0 ÷ 1.0 = 4 Ω, which matches, confirming the answer. Final answer: R = 4 Ω.

[1] mark for correctly identifying two points on the line, well separated, to calculate the gradient; [1] mark for correct calculation of the change in voltage; [1] mark for correct calculation of the change in current; [1] mark for correct division to find the gradient; [1] mark for correctly stating the resistance R = 4 Ω, with reference to the gradient being numerically equal to the resistance.

(a) with time (s) on the x-axis and distance (m) on the y-axis, correctly labelled with units, and a suitable linear scale, the five points (0,0), (1,0.3), (2,0.6), (3,0.9) and (4,1.2) are plotted and a single straight line of best fit is drawn through the origin and the points. (b) the slope of a distance-time graph is the speed; using the two end points (0,0) and (4,1.2), gradient = change in distance ÷ change in time = (1.2 - 0) ÷ (4 - 0) = 1.2 ÷ 4 = 0.3 m/s. Check using a different pair of points: (1,0.3) to (3,0.9) gives (0.9-0.3)÷(3-1) = 0.6÷2 = 0.3 m/s, which matches, confirming the answer. (c) since the distance-time graph is a straight line (constant gradient), the trolley is moving at a constant (uniform) speed - equal distances (0.3 m) are covered in each equal time interval (1 s). Final answer: (a) straight line through the origin; (b) 0.3 m/s; (c) the straight line shows the trolley moves at a constant, uniform speed.

(a) [1] each axis labelled with quantity and unit, to a max of [2]; [1] suitable linear scale; [1] all five points plotted accurately; [1] single straight line of best fit through the origin. (b) [1] correct identification that gradient of a distance-time graph = speed; [1] correct substitution using two points on the line; [1] final answer 0.3 m/s with unit. (c) [1] mark for identifying the motion as constant/uniform speed; [1] mark for correctly linking this to the straight-line shape; [1] further mark for reference to equal distances being covered in equal time intervals.

(a) with angle of incidence (°) on the x-axis and angle of refraction (°) on the y-axis, correctly labelled with units, and a suitable scale, the five points are plotted accurately and a single smooth curve is drawn through them, starting at the origin and curving so that the angle of refraction increases more slowly than the angle of incidence at larger angles. (b) the graph is a smooth curve rather than a straight line; this shows that although the angle of refraction always increases as the angle of incidence increases, the two quantities are not directly proportional to one another (doubling the angle of incidence does not double the angle of refraction) - they are related, but the relationship is not a simple proportional one. (c) reading from the curve, an angle of incidence of 50° lies between the plotted points at 40° (giving 25°) and 60° (giving 35°); interpolating along the curve gives an estimated angle of refraction of approximately 30°. Final answer: (a) a smooth curve from the origin; (b) the curved shape shows the angles are related but not proportional; (c) approximately 30° (accept a sensible reading in the range 28°-33°).

(a) [1] each axis labelled with quantity and unit, to a max of [2]; [1] suitable scale; [1] all five points plotted accurately; [1] single smooth curve drawn through the points from the origin. (b) [1] mark for correctly describing the graph as a curve, not a straight line; [1] mark for stating the angles are related (refraction increases as incidence increases); [1] mark for correctly concluding the relationship is not one of direct proportion. (c) [1] mark for correct method of interpolation between the 40° and 60° data points; [1] mark for a sensible read-off value; [1] mark for a final answer within the accepted range of 28°-33°.

(a) with force (N) on the y-axis and extension (cm) on the x-axis, correctly labelled with units and a suitable scale, all seven points are plotted accurately: (0,0), (5,1), (10,2), (15,3), (20,4), (28,5), (38,6). (b) from 0 N to 4 N, each additional 1 N increases the extension by a constant 5 cm, so the graph is a straight line through the origin over this range, showing the spring obeys Hooke's law (extension directly proportional to force) up to 4 N. Beyond 4 N, however, the extension increases by 8 cm (to 28 cm) for the next 1 N, and then by 10 cm (to 38 cm) for the following 1 N - larger increases than the constant 5 cm seen before - so the graph curves upward more steeply; this shows that beyond 4 N the spring has been stretched past its limit of proportionality, and Hooke's law (F = ke) no longer applies. (c) the student could repeat each extension measurement (for example, three times at each force) and calculate a mean extension for each force value, which reduces the effect of random errors in reading the ruler and produces a more reliable set of results; alternatively, using a set square against the ruler when taking each reading would reduce parallax error. Final answer: (a) as plotted; (b) straight line to 4 N (Hooke's law obeyed), then curving beyond 4 N as the limit of proportionality is exceeded; (c) repeat readings and take a mean (or reduce parallax error using a set square).

(a) [1] each axis labelled with quantity and unit, to a max of [2]; [1] suitable scale; [1] all seven points plotted accurately; [1] a line drawn correctly showing a straight section to 4 N followed by a curve. (b) [1] mark for correctly describing the straight-line, proportional behaviour up to 4 N; [1] mark for correctly describing the curve/steeper increase beyond 4 N; [1] mark for correctly concluding this shows the limit of proportionality has been exceeded. (c) [1] mark for a valid improvement (e.g. repeat readings and take a mean, reduce parallax error); [1] further mark for explaining how it improves reliability/reduces error; [1] further mark for a fully clear, practical description of how it would be implemented.

Apparatus required: a length of resistance wire (such as constantan) mounted alongside a metre rule so different lengths can be selected using a crocodile clip, an ammeter to measure current, a voltmeter to measure the voltage across the length of wire in the circuit, a cell or power supply, a switch, and connecting leads with crocodile clips. To keep the wire's temperature constant, the student should only close the switch briefly while taking each pair of voltage and current readings, then open it again, rather than leaving the circuit connected continuously; using small currents also helps prevent the wire heating up significantly, since it is the heating effect of the current that would otherwise raise the wire's temperature and change its resistance. Final answer: apparatus includes the test wire, ammeter, voltmeter, power supply, switch and crocodile clip leads; temperature is kept constant by using small currents and only closing the switch briefly to take each reading.

[1] mark for stating the test wire (mounted with a metre rule/crocodile clip to vary length); [1] mark for stating an ammeter and a voltmeter; [1] mark for stating a power supply/cell and a switch; [1] mark for correctly describing use of a switch to close the circuit only briefly for each reading; [1] mark for correctly describing use of small currents to minimise heating; [1] further mark for a complete, coherent description linking both temperature-control methods to preventing a change in resistance.

The independent variable - the one deliberately changed by the student - is the number of turns on the coil. The dependent variable - the one measured to see the effect - is the strength of the magnetic field produced, which could be measured, for example, as the number of paperclips the electromagnet can pick up, or as a force reading on a newton meter. To make this a fair test, variables that could also affect the strength of the field must be controlled (kept the same) between trials, most importantly the current flowing through the coil (since a larger current also increases field strength) and the material used as the core (e.g. always using the same soft iron core). Final answer: independent variable = number of turns; dependent variable = electromagnet strength (e.g. paperclips picked up); controlled variable = current in the coil (and/or the core material).

[1] mark for correctly identifying the number of turns as the independent variable; [1] mark for correctly identifying the strength of the magnetic field (or a valid measurable proxy, e.g. paperclips picked up) as the dependent variable; [1] mark for correctly identifying a valid controlled variable (current in the coil, or core material); [1] further mark for justifying why the controlled variable would otherwise affect the results; [2] further marks for a clear, complete, methodologically sound description of all three variables together, correctly distinguishing their roles.

The student's mass should first be measured using bathroom scales, and converted to a weight in newtons using W = mg (with g = 10 N/kg). The height of the platform/step should be measured accurately using a ruler or metre rule. The student then performs a fixed, counted number of step-ups onto the platform at a steady pace, while a partner times the total duration using a stopwatch. The total work done can then be calculated as weight × height × number of step-ups (since each step-up raises the student's full weight through the height of the step), and the power is calculated by dividing this total work done by the total time taken. Final answer: measure mass (convert to weight), measure platform height, count and time a fixed number of step-ups, then calculate power = total work done (weight × height × number of steps) ÷ time taken.

[1] mark for measuring the student's mass and converting to weight (W=mg); [1] mark for measuring the height of the platform/step; [1] mark for counting a fixed number of step-ups; [1] mark for timing the total duration with a stopwatch; [1] mark for correctly describing the calculation of total work done (weight × height × number of steps); [1] mark for correctly describing power = work done ÷ time taken.

Improvement 1: for each height of the ramp tested, the student should repeat the timing measurement several times (for example, three repeats) and calculate a mean average speed from the repeats, rather than relying on a single reading; this reduces the effect of random errors, such as small variations in reaction time when starting or stopping a stopwatch. Improvement 2: the student should ensure the trolley is released from rest in exactly the same way each time, for example by using a barrier or gate that is removed at the start rather than pushing the trolley by hand, since an inconsistent push would introduce an uncontrolled extra variable that could make the results unreliable. Final answer: repeat readings at each height and take a mean; and use a consistent, controlled method of releasing the trolley from rest (e.g. a barrier/gate) each time.

[1] mark for stating repeating the readings; [1] further mark for correctly explaining this as taking a mean to reduce random error; [1] mark for stating a consistent/controlled release method; [1] further mark for correctly explaining why this improves reliability (removes an uncontrolled variable); [2] further marks for two fully-developed, distinct, valid improvements given together with clear justification.

mean = sum of readings ÷ number of readings = (2.1 + 2.3 + 2.2) ÷ 3 = 6.6 ÷ 3 = 2.2 s. Final answer: 2.2 s.

[1] correct sum of the three readings (6.6); [1] correct division by 3; [1] final answer 2.2 s with unit.

average speed = distance moved ÷ time taken = 1.1 ÷ 2.2 = 0.5 m/s. Final answer: 0.5 m/s.

[1] correct formula average speed=distance/time; [1] correct substitution 1.1÷2.2; [1] final answer 0.5 m/s with unit.

The reading of 18.5 cm is much larger than the other four readings (11.9, 12.0, 12.1, 12.0 cm), which are all close together, so 18.5 cm is identified as the anomalous result and should be excluded from the mean. Mean of the remaining four readings = (11.9 + 12.0 + 12.1 + 12.0) ÷ 4 = 48.0 ÷ 4 = 12.0 cm. Final answer: the anomaly is 18.5 cm; the mean of the remaining readings is 12.0 cm.

[1] mark for correctly identifying 18.5 cm as the anomalous result; [1] mark for correctly summing the remaining four readings (48.0); [1] mark for the final mean, 12.0 cm, correctly excluding the anomaly.

The gradient of a force-extension graph is numerically equal to the spring constant, so k = 40 N/m. Using F = ke, rearranged: e = F ÷ k = 6 ÷ 40 = 0.15 m = 15 cm. The new total length of the spring = natural (unstretched) length + extension = 15 + 15 = 30 cm. Final answer: 30 cm.

[1] mark for correctly rearranging F=ke to e=F/k; [1] mark for correct substitution and evaluation, 6÷40=0.15 m (15 cm); [1] mark for correctly adding the natural length to the extension; [1] mark for the final answer, 30 cm, with unit.