2023 CCEA GCSE Science Double Award 1370 Practice Paper with Answers

(a) Plant cells contain a cellulose cell wall, which provides structural support and maintains the cell's shape, and a large permanent vacuole, which stores cell sap and helps maintain turgor pressure (keeping the cell rigid); chloroplasts (the site of photosynthesis) are a further valid structure. (b) A bacterial cell has no nucleus — its genetic material lies free in the cytoplasm — whereas a plant cell has a nucleus enclosing its DNA; a bacterial cell wall is also chemically different (non-cellulose) from a plant cell wall. (c) Although a multicellular organism begins as a single fertilised cell, repeated cell division (mitosis) followed by cell specialisation (differentiation) allows genetically identical cells to develop different structures suited to different functions, forming specialised tissues (e.g. muscle, nerve), which combine into organs (e.g. the heart) and organ systems (e.g. the circulatory system), enabling the organism as a whole to carry out many different functions simultaneously.

(a) [1] each for two correct plant-only structures (cellulose cell wall, large permanent vacuole, or chloroplasts) and [1] each for a correct matching function, max [4]. (b) [2] a correct, clearly stated structural difference (e.g. absence of a nucleus; non-cellulose cell wall; presence of plasmids). (c) [2] explains cell specialisation/differentiation into tissues, organs and organ systems, linked to the ability to perform many functions. Max [8].

(a) carbon dioxide + water --(light energy)--> glucose + oxygen. (b) Hydrogencarbonate indicator is red at normal (atmospheric) CO2 concentration, yellow at high CO2 concentration, and purple at low CO2 concentration. In bright light, the rate of photosynthesis exceeds the rate of respiration, so the pondweed removes CO2 from the water faster than it releases it, causing the CO2 concentration in the water to fall; the indicator therefore changes from red towards purple. (c) Adaptation 1: a large surface area (broad, thin leaf blade) maximises the leaf's exposure to light, increasing the rate of the light-dependent reactions. Adaptation 2: numerous small pores called stomata, mainly on the leaf's lower surface, allow efficient diffusion of carbon dioxide into the leaf and oxygen out, preventing gas exchange from limiting the rate of photosynthesis. (A thin leaf blade, reducing the diffusion distance for CO2 to reach mesophyll cells, is a further valid adaptation.)

(a) [2] correct word equation (reactants, products, and reference to light energy). (b) [1] correctly predicts the colour change (red towards purple). [1] correctly explains this in terms of falling CO2 concentration due to photosynthesis exceeding respiration. (c) [1] each for two correctly named leaf adaptations (max [2]) and [1] each for a correct linked explanation (max [2]). Max [8].

(a) A reflex arc pathway is: sensory neurone → association (relay/connector) neurone → motor neurone, connecting the receptor to the effector via the spinal cord. (b) A synapse is the small gap between the end of one neurone and the start of the next, across which a nerve impulse is transmitted by the diffusion of chemical neurotransmitters, rather than by direct electrical contact. (c) A reflex action is co-ordinated entirely within the spinal cord (it does not require processing by the brain), travelling along a short, fixed pathway (sensory → association → motor neurone) involving relatively few synapses; a voluntary response instead requires the impulse to travel further, up to the brain for conscious processing and decision-making, and back down to the effector, involving a longer pathway and more synapses — since impulse transmission across each synapse takes a small but significant amount of time, the shorter reflex pathway is faster.

(a) [3] all three neurone types named in the correct order (sensory, association/relay, motor); [1] deducted per error/omission, no negative marks below 0. (b) [2] correct definition of a synapse (gap between neurones, transmission of an impulse across it). (c) [1] correctly states the reflex bypasses the brain/uses a shorter pathway. [2] explains this in terms of fewer synapses/shorter distance meaning less time for impulse transmission. Max [8].

(a) A population is all the organisms of the same species living in the same habitat at the same time, capable of interbreeding. A community is all the different populations of different species living and interacting together within a habitat. (b) The student should mark out the field using a coordinate grid (e.g. with two tape measures at right angles), then use a random number generator (or random number table) to generate pairs of coordinates, placing the quadrat at each randomly generated position to avoid sampling bias; the number of dandelions within each quadrat is counted and recorded, this is repeated for a suitably large number of quadrats (e.g. at least 10), and the mean number of dandelions per quadrat is calculated; this mean is then multiplied by (total field area ÷ area of one quadrat) to estimate the total population in the field. (c) Light intensity is a valid abiotic factor: dandelions require sufficient light for photosynthesis, so areas of the field that are shaded (e.g. under trees or hedges) would be expected to have a lower density/distribution of dandelions than open, well-lit areas. (Other valid factors: soil moisture, soil pH, soil nutrient content, wind exposure — each with a correctly linked explanation.)

(a) [1] correct definition of population. [1] correct definition of community. (b) [1] use of a coordinate grid/system. [1] use of a random number generator/table to avoid bias. [1] counting organisms per quadrat and repeating for a sufficient number of quadrats, then calculating a mean and scaling up to the total field area. (c) [1] a genuine abiotic factor named. [1] a correctly linked explanation of its likely effect on dandelion distribution. Max [8].

(a) Mean Test A = (210 + 195 + 205 + 220 + 190) ÷ 5 = 1020 ÷ 5 = 204 ms. (b) Range Test A = highest − lowest = 220 − 190 = 30 ms. (c) Mean Test B = (310 + 295 + 340 + 285 + 320) ÷ 5 = 1550 ÷ 5 = 310 ms. (d) Difference = 310 − 204 = 106 ms; the student's reaction time was, on average, 106 ms slower when distracted. (e) When distracted, the brain must simultaneously process additional sensory information (the music and the conversation) alongside the reaction-time stimulus; because this competing information must also be processed via neural pathways and synapses, and attention is divided between tasks, the pathway from stimulus to conscious response takes longer, increasing overall reaction time — this is a voluntary response, so, unlike a reflex, it is influenced by conscious attention and brain processing load.

(a) [1] correct answer of 204 ms, working shown. (b) [1] correct answer of 30 ms. (c) [1] correct answer of 310 ms, working shown. (d) [2] correct answer of 106 ms (own-figure rule applies if (a)/(c) are carried forward consistently). (e) [1] identifies divided attention/competing sensory processing as the cause. [2] explains this in terms of increased neural processing demand/time in a voluntary (brain-mediated) response. [2] overall clarity, correct use of units (ms) throughout, and working clearly shown for all calculations. Max [10].

(a) % transferred = (1 800 ÷ 20 000) × 100 = 9.0%. (b) % transferred = (200 ÷ 1 800) × 100 = 11.1% (to 1 d.p.). (c) % transferred = (18 ÷ 200) × 100 = 9.0%. (d) Reason 1: a large proportion of energy is lost as heat through respiration at each trophic level. Reason 2: energy is also lost in materials that are not eaten or not digested (e.g. faeces, bones, fur, uneaten plant/animal parts), so it is never incorporated into the next trophic level's biomass; energy used for movement and other life processes (rather than being stored as new biomass) is a further valid reason. (e) Since only around 9–11% of the energy present at one trophic level is transferred to the next (as shown in parts (a)–(c)), the total amount of energy remaining falls very rapidly with each additional trophic level; after 4–5 trophic levels, the energy remaining is too small to support a viable population of a further predator, which is why food chains rarely extend beyond this length.

(a) [1] correct answer of 9.0%, working shown. (b) [1] correct answer of 11.1%, working shown. (c) [1] correct answer of 9.0%, working shown. (d) [1] each for two genuine, distinct reasons for energy loss between trophic levels (e.g. respiration heat loss; loss in undigested/uneaten material; energy used for movement), max [2]. (e) [2] explains, with explicit reference to the small percentages calculated, why energy becomes insufficient to support further trophic levels beyond 4–5. [2] overall clarity, correct working and units shown throughout. Max [11].

(a) The healthy person's blood glucose concentration rises from 5.0 to 7.5 mmol/L in the first hour after eating, then falls steadily over the following three hours, returning close to its starting level (5.0 mmol/L) by 3–4 hours, showing effective regulation (homeostasis) of blood glucose. (b) Increase = 9.5 − 5.0 = 4.5 mmol/L. (c) The key difference is that the diabetic person's blood glucose rises to a higher peak (11.0 mmol/L compared with 7.5 mmol/L) and does not fall back to the starting level within the 4-hour period (remaining at 9.8 mmol/L at 4 hours), whereas the healthy person's level returns to approximately normal (5.0 mmol/L) by the end of the period. (d) In untreated Type 1 diabetes, the pancreas produces little or no insulin. Insulin normally stimulates body cells (especially liver and muscle cells) to take up glucose from the blood and convert it into glycogen for storage; without adequate insulin, glucose is not removed from the bloodstream effectively after a meal, so blood glucose concentration remains elevated for a prolonged period rather than being regulated back down to a normal level, as seen in the healthy person.

(a) [1] rises then falls. [1] correctly notes it returns close to the starting/normal level by the end. (b) [1] correct answer of 4.5 mmol/L, working shown. (c) [2] a clear, data-referenced point of difference (higher peak and/or fails to return to baseline). (d) [1] correctly identifies little/no insulin production in Type 1 diabetes. [2] explains the role of insulin in stimulating glucose uptake/glycogen storage, and links its absence to sustained high blood glucose. [2] overall clarity and correct use of the data throughout. Max [11].

Carbon dioxide is removed from the atmosphere by green plants and algae during photosynthesis, which converts it (along with water) into glucose and other organic (carbon-containing) compounds, incorporating the carbon into the plant's biomass. This carbon is passed along food chains as animals feed on plants and on each other. Carbon is returned to the atmosphere as carbon dioxide by respiration in plants, animals and microorganisms (as organic compounds are broken down to release energy), and also by the decomposition of dead organisms and waste material by saprophytic bacteria and fungi, which respire as they break down organic matter. Over very long timescales, some dead organic material that does not fully decompose can become fossilised, forming fossil fuels such as coal, oil and gas, which lock carbon away from the atmosphere.

Human activity affects this cycle significantly through the combustion of fossil fuels (for energy, transport and industry), which releases large quantities of carbon dioxide that had been locked away for millions of years back into the atmosphere far faster than it is removed by photosynthesis, contributing to a rising atmospheric CO2 concentration and enhanced greenhouse effect. Large-scale deforestation compounds this problem by both releasing stored carbon (if the cleared vegetation is burned) and reducing the number of plants available to remove CO2 from the atmosphere through photosynthesis, further disrupting the natural balance of the carbon cycle.

Assessed against a 4-band QWC level of response grid (Band A 5–6, Band B 3–4, Band C 1–2, Band D 0). Band A: accurately explains at least three processes moving carbon between the atmosphere and organisms (photosynthesis, respiration, decomposition, and/or combustion/fossilisation), and discusses a specific, well-explained human impact on the cycle; fluent, well-organised use of specialist scientific terminology throughout. Band B: explains at least two carbon-cycle processes accurately, with some discussion of human impact; minor gaps. Band C: explains one process accurately, or several only vaguely; limited or no discussion of human impact. Band D: no creditable scientific content, or response entirely irrelevant.

(a) The atomic number (bottom-left, or number of protons) is 17; the mass number (top-left, protons + neutrons) is 35. (b) Protons = atomic number = 17. Neutrons = mass number − atomic number = 35 − 17 = 18. Electrons = protons (atom is neutral) = 17. (c) With 17 electrons, the electronic structure fills the first shell (2), second shell (8), leaving 17 − 2 − 8 = 7 in the third (outer) shell: 2,8,7. (d) Gaining one electron gives 17 + 1 = 18 electrons, while the number of protons remains 17. Since electrons are negatively charged and protons are positively charged, the ion now has one more negative charge than positive charge, giving an overall charge of −1 (Cl⁻).

(a) [1] atomic number 17. [1] mass number 35. (b) [1] each for correct protons (17), neutrons (18), electrons (17), max [3]. (c) [1] correct electronic structure 2,8,7. (d) [1] correct number of electrons (18). [1] correct explanation of the −1 charge (one more electron than protons). Max [8].

(a) The sodium atom (2,8,1) loses its single outer-shell electron, transferring it to the chlorine atom (2,8,7), which needs just one more electron to complete its outer shell. Sodium, having lost a negative electron, becomes a positively charged ion (Na⁺); chlorine, having gained a negative electron, becomes a negatively charged ion (Cl⁻). The oppositely charged Na⁺ and Cl⁻ ions are then held together by strong electrostatic forces of attraction between opposite charges — this electrostatic attraction is the ionic bond. (b) Sodium ion, Na⁺: 2,8 (lost the outer electron from 2,8,1). Chloride ion, Cl⁻: 2,8,8 (gained one electron, filling the outer shell from 2,8,7). Both ions are described as stable because each now has a full outer electron shell of eight electrons (or two, for the innermost shell), the same stable arrangement as a noble gas atom, which is a particularly low-energy, unreactive electron configuration. (c) Sodium chloride, like other ionic compounds, has a high melting and boiling point, and conducts electricity when molten or in aqueous solution but not when solid. This is because, in the solid state, the Na⁺ and Cl⁻ ions are held in fixed positions within a giant ionic lattice by strong electrostatic forces in all directions, which requires a large amount of energy to overcome (hence the high melting point) and means the ions cannot move freely to carry electric charge; once molten or dissolved, the lattice breaks down and the ions become free to move, allowing them to carry a current.

(a) [1] sodium loses its outer electron. [1] chlorine gains the electron. [1] resulting ions are held together by electrostatic attraction (the ionic bond). (b) [1] correct Na⁺ structure (2,8). [1] correct Cl⁻ structure (2,8,8). [1] correctly explains 'stable' as a full outer shell/noble gas structure. (c) [1] each for two correct properties (max [2]); [1] a correct, clearly linked explanation of one property in terms of ionic lattice structure/electrostatic forces. Max [9].

(a) Reactivity increases as you go down Group 1 (e.g. potassium is more reactive than sodium, which is more reactive than lithium). (b) Group 1 metals react by losing their single outer-shell electron to form a positive ion. Going down the group, each successive element has an additional electron shell, so the outer electron is further from the positively charged nucleus and is shielded by more inner electron shells; both effects weaken the electrostatic attraction holding the outer electron in place, making it easier to lose — hence reactivity increases down the group. (c) Reactivity decreases as you go down Group 7 (e.g. fluorine is more reactive than chlorine, which is more reactive than bromine). (d) Group 7 elements react by gaining one electron into their outer shell to form a negative ion. Going down the group, each successive element has an additional electron shell, so the outer shell (where the new electron would be added) is further from the nucleus and more shielded by inner electrons; both effects weaken the nucleus's attraction for an incoming electron, making it harder to gain an electron — hence reactivity decreases down the group.

(a) [1] correctly states reactivity increases down Group 1. (b) [1] references increasing number of shells/distance from nucleus. [1] references shielding by inner electrons. [1] correctly links this to weaker attraction on the outer electron, making it easier to lose. (c) [1] correctly states reactivity decreases down Group 7. (d) [1] references increasing number of shells/distance from nucleus and shielding. [1] correctly links this to weaker attraction for an incoming electron, making it harder to gain. Max [9].

(a) A lilac (light purple) flame is the characteristic flame test colour for potassium ions, K⁺. (b) Adding dilute acid to a carbonate produces carbon dioxide gas (CO2), which is confirmed by turning limewater cloudy/milky (due to the formation of insoluble calcium carbonate); this positive test indicates the presence of the carbonate ion, CO3²⁻, in the original solid. (c) Adding dilute hydrochloric acid (to remove any carbonate ions, which would otherwise interfere) followed by barium chloride solution, and observing a white precipitate, is the standard test for the sulfate ion, SO4²⁻; the white precipitate formed is insoluble barium sulfate, BaSO4.

(a) [1] correctly identifies K⁺ (potassium ion). (b) [1] correctly identifies the gas as carbon dioxide/CO2. [2] correctly identifies the carbonate ion (CO3²⁻) as present. (c) [2] correctly identifies the sulfate ion (SO4²⁻) as present. [3] correctly names the white precipitate as barium sulfate/BaSO4, with correct formula. Max [9].

(a) Mr(CaCO3) = Ar(Ca) + Ar(C) + 3 × Ar(O) = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100. (b) Moles = mass ÷ Mr = 25 ÷ 100 = 0.25 mol. (c) Mass = moles × Mr = 0.6 × 100 = 60 g.

(a) [2] correct Mr of 100, working shown. (b) [1] correct method (mass ÷ Mr). [2] correct answer of 0.25 mol. (c) [1] correct method (moles × Mr). [3] correct answer of 60 g, with the own-figure rule applied if the Mr from (a) is carried forward correctly. Max [9].

(a) CaCO3(s) → CaO(s) + CO2(g). The equation is already balanced (1 calcium, 1 carbon, 3 oxygen on each side). (b) Moles CaCO3 = mass ÷ Mr = 50 ÷ 100 = 0.5 mol. (c) The balanced equation shows a 1:1 mole ratio between CaCO3 and CaO, so moles CaO = moles CaCO3 = 0.5 mol. (d) Mr(CaO) = Ar(Ca) + Ar(O) = 40 + 16 = 56. Mass CaO = moles × Mr = 0.5 × 56 = 28 g.

(a) [2] correctly balanced equation with correct state symbols (1 mark for correct formulae/balance, 1 mark for correct state symbols). (b) [1] correct method. [1] correct answer of 0.5 mol. (c) [1] correctly applies the 1:1 mole ratio to give 0.5 mol CaO. (d) [1] correct Mr(CaO) = 56. [1] correct method (moles × Mr). [3] correct final answer of 28 g, own-figure rule applied throughout if earlier parts are carried forward correctly. Max [10].

(a) Percentage yield = (actual yield ÷ theoretical yield) × 100 = (23.8 ÷ 28) × 100 = 85%. (b) Reason 1: some of the solid product is typically lost during handling, such as when transferring it between containers or during any filtration/weighing stage. Reason 2: the reaction may not have gone to completion (not all of the calcium carbonate fully decomposed within the time/temperature used), meaning less calcium oxide was actually formed than the theoretical maximum. (A further valid reason: calcium oxide is hygroscopic/reacts with moisture and CO2 in the air, so some of the product may have reacted further before being weighed, or side reactions occurred.)

(a) [1] correct method (actual ÷ theoretical × 100). [2] correct answer of 85%. (b) [1] each for two genuine, distinct, valid reasons why yield is less than 100% (e.g. product lost during handling/transfer; reaction not going to completion; product reacting further/absorbing moisture), max [2]. Note: the own-figure rule applies to part (a) if a different theoretical yield was correctly carried forward from the previous question. Max [10] overall, allocated as [3] for part (a) working/answer and up to [2] shown here for (b); remaining marks for full, clearly explained working throughout both parts.

First, a measured volume of dilute sulfuric acid is gently warmed in a beaker (warming speeds up the reaction without boiling off the acid). Excess (more than is needed to react) black copper oxide powder is then added to the warm acid, a little at a time, stirring after each addition, until no more dissolves and unreacted copper oxide remains visible at the bottom — this shows all the acid has reacted (been neutralised) via the reaction CuO + H2SO4 → CuSO4 + H2O, and confirms none of the excess acid remains unreacted in the solution. The mixture is then filtered to remove the excess, unreacted copper oxide, leaving a clear blue filtrate of copper sulfate solution. This filtrate is gently heated in an evaporating basin to evaporate off some of the water and concentrate the solution, until a hot saturated solution is obtained (often tested by dipping a glass rod in and checking for crystal formation at the edge). The hot saturated solution is then left to cool slowly and undisturbed, allowing blue copper sulfate crystals to form as the solution's solubility falls with temperature; finally, the crystals are filtered off from the remaining liquid and patted dry between sheets of filter paper (rather than heated directly, which could drive off the water of crystallisation).

Assessed against a 4-band QWC level of response grid (Band A 5–6, Band B 3–4, Band C 1–2, Band D 0). Band A: describes the full correct sequence — warming the acid, adding excess insoluble oxide to ensure complete neutralisation, filtering off the excess, evaporating to a hot saturated solution, cooling to crystallise, and filtering/drying the crystals — using accurate specialist terminology (neutralisation, saturated solution, crystallisation, filtration) throughout, in a logical, well-organised sequence. Band B: describes most of the correct sequence with minor omissions or ordering errors, generally accurate terminology. Band C: describes only part of the process (e.g. mentions reacting the acid and oxide, but omits filtration and/or correct crystallisation method) or uses largely correct but underdeveloped statements. Band D: no creditable practical sequence given, or response is scientifically incorrect throughout.

(a) \( a = \dfrac{v-u}{t} = \dfrac{24-0}{6} = 4 \text{ m/s}^2 \). (b) Since the acceleration is uniform, average velocity \( = \dfrac{u+v}{2} = \dfrac{0+24}{2} = 12 \text{ m/s} \). (c) distance \( = \text{average velocity} \times t = 12 \times 6 = 72 \text{ m} \). Check by a second route: distance \( = ut + \frac{1}{2}at^2 = (0 \times 6) + \frac{1}{2}(4)(6^2) = 0 + \frac{1}{2}(4)(36) = 72 \text{ m} \) — both methods agree.

(a) [1] correct formula/method. [1] correct answer, 4 m/s². (b) [1] correct answer, 12 m/s. (c) [1] correct method (average velocity × time, or equivalent). [2] correct final answer, 72 m. [2] working clearly shown throughout with correct units at every stage. Max [8].

(a) \( a = \dfrac{F}{m} = \dfrac{2000}{800} = 2.5 \text{ m/s}^2 \). (b) The resultant force is the driving force minus the resistive force: \( \text{resultant} = \text{driving force} - \text{resistive force} \), so \( \text{resistive force} = 2500 - 2000 = 500 \text{ N} \). (c) Newton's second law states that a resultant force acting on an object will cause it to accelerate, and that this acceleration is proportional to the size of the resultant force (and, for a given force, inversely proportional to the object's mass), summarised by \( F = ma \).

(a) [1] correct method (F = ma rearranged). [2] correct answer, 2.5 m/s². (b) [1] correct method (driving force − resultant force). [2] correct answer, 500 N. (c) [2] correct, accurate statement of Newton's second law (resultant force causes acceleration, proportional relationship). Max [8].

(a) moment \( = F \times d = 300 \times 1.5 = 450 \text{ N m} \). (b) For the see-saw to balance, the anticlockwise moment (second child) must equal the clockwise moment (first child) of \( 450 \text{ N m} \): \( d = \dfrac{\text{moment}}{F} = \dfrac{450}{250} = 1.8 \text{ m} \). (c) The Principle of Moments states that for an object in equilibrium, the sum of the clockwise moments about any pivot is equal to the sum of the anticlockwise moments about the same pivot.

(a) [1] correct method (moment = force × perpendicular distance). [1] correct answer, 450 N m. (b) [1] correctly applies the Principle of Moments (clockwise moment = anticlockwise moment). [2] correct final answer, 1.8 m. (c) [2] accurate statement of the Principle of Moments. [1] working clearly shown throughout with correct units. Max [8].

(a) \( E_p = mgh = 2 \times 10 \times 5 = 100 \text{ J} \). (b) With negligible air resistance, all of the gravitational potential energy lost is converted into kinetic energy: \( E_k = E_p = 100 \text{ J} \). Since \( E_k = \frac{1}{2}mv^2 \): \( 100 = \frac{1}{2}(2)v^2 \), so \( v^2 = \dfrac{100}{1} = 100 \), giving \( v = \sqrt{100} = 10 \text{ m/s} \). Check by a second route: for free fall from rest at \( g = 10 \text{ m/s}^2 \), \( v = u + gt \); using \( h = \frac{1}{2}gt^2 \), \( 5 = \frac{1}{2}(10)t^2 \) gives \( t^2 = 1 \), so \( t = 1 \text{ s} \), and \( v = 0 + (10)(1) = 10 \text{ m/s} \) — both methods agree. (c) The Principle of Conservation of Energy states energy cannot be created or destroyed, only transferred from one form to another. As the ball falls, its gravitational potential energy continuously decreases (as height decreases) while its kinetic energy continuously increases (as speed increases); with air resistance negligible, no energy is transferred to heat/sound in the surroundings, so the total (Ep + Ek) remains constant throughout the fall, meaning all of the Ep lost is gained as Ek.

(a) [1] correct method (Ep = mgh). [1] correct answer, 100 J. (b) [1] correctly equates Ek to Ep lost (energy conservation). [1] correct rearrangement of \( E_k = \frac{1}{2}mv^2 \). [2] correct final answer, 10 m/s. (c) [2] clear, accurate explanation referencing conservation of energy and the Ep-to-Ek transfer. [1] working clearly shown with correct units throughout. Max [8].

(a) density \( = \dfrac{\text{mass}}{\text{volume}} = \dfrac{270}{100} = 2.7 \text{ g/cm}^3 \). (b) The calculated density (2.7 g/cm³) matches the published density of aluminium (approximately 2.7 g/cm³) exactly, so this result is fully consistent with the block being made of aluminium. (c) According to kinetic theory, in a solid the particles are packed closely together in a fixed, regular arrangement, held in place by strong interparticle forces of attraction; the particles can only vibrate about these fixed positions rather than move freely, so the solid has both a fixed volume and a fixed shape. In a gas, the particles are far apart, have negligible forces of attraction between them, and move rapidly in random directions at high speed, colliding with each other and the container walls; because there are no fixed positions or strong attractive forces, a gas has no fixed shape and no fixed volume, expanding to completely fill whatever container it is in.

(a) [1] correct method (mass ÷ volume). [1] correct answer, 2.7 g/cm³. (b) [1] correct conclusion (yes/consistent with aluminium). [1] justified by direct, explicit comparison with the given aluminium density. (c) [2] accurate kinetic theory explanation of the solid state (close, fixed particle arrangement, vibration only, strong forces). [2] accurate contrast with the gas state (particles far apart, negligible forces, random rapid motion). Max [8].

(a) Work done \( = F \times d \); since the force needed to lift the mass at constant speed equals its weight, \( F = mg = 50 \times 10 = 500 \text{ N} \), so \( W = 500 \times 4 = 2000 \text{ J} \). This is equivalent to the gravitational potential energy gained: \( E_p = mgh = 50 \times 10 \times 4 = 2000 \text{ J} \), confirming the answer by a second route. (b) power \( = \dfrac{\text{work done}}{\text{time}} = \dfrac{2000}{5} = 400 \text{ W} \). (c) efficiency \( = \dfrac{\text{useful output power}}{\text{total input power}} \times 100 = \dfrac{400}{500} \times 100 = 80\% \). (d) Not all of the electrical input energy is converted into useful lifting (gravitational potential) energy: some is wasted as heat due to friction between the motor's moving mechanical parts, and some is wasted as heat due to electrical resistance in the motor's windings/wiring, meaning the useful output power is less than the total input power.

(a) [1] correct method (W = Fd or Ep = mgh). [2] correct answer, 2000 J. (b) [2] correct answer, 400 W. (c) [1] correct method (useful output ÷ total input × 100). [2] correct answer, 80%. (d) [1] a genuine, correctly explained reason for energy loss (e.g. friction, electrical resistance, both producing waste heat). Max [9].

(a) An alpha particle is identical to a helium nucleus: it consists of 2 protons and 2 neutrons (mass number 4, atomic number 2). (b) In alpha decay, mass number decreases by 4 and atomic number decreases by 2: \( A = 226 - 4 = 222 \); \( Z = 88 - 2 = 86 \), giving \( ^{222}_{86}\text{Rn} \) (radon), so the full equation is \( ^{226}_{88}\text{Ra} \rightarrow \, ^{222}_{86}\text{Rn} + \, ^{4}_{2}\text{He} \); mass numbers (226 = 222 + 4) and atomic numbers (88 = 86 + 2) both balance correctly. (c) The number of half-lives elapsed = \( 15 \div 5 = 3 \) half-lives. Activity halves each half-life: \( 800 \to 400 \) (after 5 days) \( \to 200 \) (after 10 days) \( \to 100 \) (after 15 days) counts per minute.

(a) [1] correctly states 2 protons and 2 neutrons. (b) [1] correct mass number, A = 222. [1] correct atomic number, Z = 86. [1] correctly identifies/names the element as radon (Rn), consistent with Z = 86. (c) [1] correctly calculates 3 half-lives have elapsed. [2] correct final answer of 100 counts per minute, with intermediate halving steps shown (800→400→200→100). Max [7].

(a) From Table 4, the activity falls from 640 Bq to 320 Bq between 0 and 4 days — exactly halving. It halves again from 320 Bq to 160 Bq between 4 and 8 days, and again from 160 Bq to 80 Bq between 8 and 12 days. Since the activity consistently halves every 4 days, the half-life of the source is 4 days. (b) Nuclear fission involves a large, unstable nucleus (such as uranium-235) absorbing a neutron and then splitting into two smaller daughter nuclei, releasing energy and further neutrons that can go on to cause a chain reaction. Nuclear fusion, by contrast, involves two small, light nuclei (such as isotopes of hydrogen, deuterium and tritium) being forced together under extremely high temperature and pressure to form a single larger nucleus (e.g. helium), also releasing a large amount of energy — the key difference is that fission splits a heavy nucleus apart, while fusion joins light nuclei together. (c) Fusion releases roughly four times as much energy per kilogram of fuel as fission, and its fuel source (hydrogen isotopes from seawater) is nearly inexhaustible and far more widely available than uranium ore; unlike fission, fusion's main by-product is helium, an inert, non-toxic gas, so it does not produce the same problem of long-lived radioactive waste associated with fission.

(a) [1] correctly identifies the pattern of halving in the data. [2] correct answer, 4 days, with the halving pattern shown across at least two intervals. (b) [1] correctly describes fission (a large/heavy nucleus splitting into smaller nuclei). [1] correctly describes fusion (small/light nuclei joining/combining into a larger nucleus). [1] explicitly contrasts the two processes. (c) [2] a genuine, accurately explained advantage of fusion over fission (e.g. more energy per kg, more abundant fuel, no long-lived radioactive waste). Max [8].

Advantages of nuclear fission over wind power: a nuclear power station can generate electricity reliably and continuously, regardless of weather conditions, unlike wind power, which only generates electricity when the wind is blowing within a suitable speed range, making nuclear output far more predictable and consistent; nuclear fission also does not release carbon dioxide directly during electricity generation, similarly to wind power, helping to reduce greenhouse gas emissions compared with fossil fuel power stations, and a nuclear power station can generate a very large, concentrated amount of electricity from a relatively small site footprint, unlike the large land area typically required for an equivalent wind farm.

Disadvantages of nuclear fission compared with wind power: nuclear fission produces radioactive waste, some of which remains hazardous for many thousands of years and must be safely stored/contained in secure facilities, whereas wind power produces no radioactive waste at all; nuclear power stations are very expensive and slow to build, and carry a small but serious risk of a major accident (as has occurred historically at nuclear plants), causing significant public concern about living near a plant or waste storage facility, whereas wind turbines carry no such catastrophic risk; and although the fission process itself is low-carbon, mining, transporting and purifying uranium ore does release some greenhouse gases, whereas wind power has no fuel extraction process at all.

Overall, nuclear fission offers considerably more reliable, weather-independent, high-output electricity generation than wind power, but this comes at the cost of long-lived radioactive waste and the risk (however small) of serious accidents, meaning the choice between the two involves weighing reliability and output against long-term waste and safety concerns.

Assessed against a 4-band QWC level of response grid (Band A 5–6, Band B 3–4, Band C 1–2, Band D 0). Band A: discusses at least two genuine advantages and two genuine disadvantages of nuclear fission relative to wind power (e.g. reliability, low direct emissions, radioactive waste, accident risk, cost), with accurate scientific reasoning and a balanced overall view; fluent, well-organised specialist terminology. Band B: discusses at least one advantage and one disadvantage with reasonable accuracy; some imbalance or minor inaccuracy. Band C: discusses only advantages or only disadvantages, or does so only superficially/vaguely. Band D: no creditable scientific content, or the response does not meaningfully compare the two energy sources.