2025 CCEA GCSE Science Double Award 1370 模拟试题及答案详解

(a) Balancing: one MgCO₃ reacts with two HCl (chlorine and hydrogen atoms must be doubled to balance, since MgCl₂ contains two chlorine atoms and each HCl supplies one), giving MgCl₂, H₂O and CO₂ each with a coefficient of 1: MgCO₃(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l) + CO₂(g). Checking atoms: Mg 1=1, C 1=1, O (3 on left) = (1+2 on right)=3, H 2=2, Cl 2=2 — balanced. (b) Carbon dioxide is identified by bubbling it through limewater: if CO₂ is present the limewater turns milky/cloudy. (c) The general pattern for any acid reacting with a carbonate is: acid + metal carbonate → salt + water + carbon dioxide.
Final answer: MgCO₃(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l) + CO₂(g); limewater turns milky; acid + carbonate → salt + water + carbon dioxide.

(a) 1 mark for correct coefficient '2' before HCl (all other coefficients are 1, allow these to be left blank or written as 1); 1 mark for state symbol (s) after MgCO₃; 1 mark for (aq) after HCl and MgCl₂; 1 mark for (l) after H₂O and (g) after CO₂ (award if all four remaining state symbols correct) — max 4. (b) 1 mark for 'bubble through limewater'; 1 mark for 'turns milky/cloudy' — max 2. (c) 1 mark for 'acid + carbonate → salt + water + carbon dioxide' or equivalent. Reject 'neutralisation' alone for (c), as this specification defines a base specifically as a metal oxide or hydroxide, distinct from a carbonate.

(a) Two magnesium atoms are needed to react with one oxygen molecule (which contains two oxygen atoms) to form two formula units of MgO, so the balanced equation is 2Mg(s) + O₂(g) → 2MgO(s). Checking atoms: Mg 2=2, O 2=2 — balanced. (b) Magnesium (Group 2, electronic structure 2,8,2) must lose its 2 outer electrons to achieve a stable, noble-gas electronic structure (2,8), forming a Mg²⁺ ion. Oxygen (Group 6, electronic structure 2,6) needs to gain 2 electrons to complete its outer shell to 2,8, forming an O²⁻ ion. The 2 electrons lost by magnesium are transferred to the oxygen atom (shown as crosses moving from Mg to become part of O's outer shell in a dot-and-cross diagram). The resulting Mg²⁺ and O²⁻ ions, having opposite charges, attract each other strongly — this electrostatic attraction is the ionic bond. (c) Because ionic bonding extends in all directions between many alternating positive and negative ions, magnesium oxide forms a giant ionic lattice structure.
Final answer: 2Mg(s) + O₂(g) → 2MgO(s); Mg loses 2 electrons to form Mg²⁺ (2,8), O gains 2 electrons to form O²⁻ (2,8), held by electrostatic attraction; giant ionic lattice.

(a) 1 mark for coefficient '2' before Mg; 1 mark for coefficient '2' before MgO; 1 mark for correct state symbols (s), (g), (s) in order — max 3. (b) 1 mark for magnesium losing 2 electrons to form Mg²⁺; 1 mark for oxygen gaining 2 electrons to form O²⁻; 1 mark for both ions reaching a stable/noble-gas (2,8) electron arrangement or for stating the ionic bond is the electrostatic attraction between the oppositely charged ions — max 3. (c) 1 mark for 'giant ionic lattice' (accept 'ionic lattice'). Reject answers describing covalent/molecular structures.

(a) Going down Group 1, each successive alkali metal atom has one more electron shell than the one above it, so its single outer electron is further from the positively charged nucleus and is shielded from the nucleus's attractive pull by an increasing number of inner, filled electron shells. This makes the outer electron progressively easier to remove as the group is descended, so the atom loses its outer electron — and therefore reacts — more readily. Since a Group 1 metal reacts by losing this one outer electron to form a positive ion with a stable electronic configuration, this explains why reactivity (for example, the vigour of reaction with water, producing hydrogen gas and a metal hydroxide) increases from lithium down to potassium and beyond. (b) Group 1 metals share several distinctive physical properties: they have unusually low densities for metals (lithium, sodium and potassium are all less dense than water), they are soft enough to be cut with a knife, and they are shiny immediately after cutting but tarnish rapidly as they react with oxygen and moisture in the air. (c) Group 7 shows the opposite trend to Group 1: reactivity decreases down the group, since halogen atoms react by gaining one electron into their outer shell, and as atoms get larger down the group, the outer shell is further from the nucleus and more shielded, making it harder to attract an extra electron. Fluorine, at the top of the group, is the most reactive halogen.
Final answer: (a) reactivity increases down Group 1 as the outer electron becomes easier to lose (further from nucleus, more shielding); (b) low density and softness/shininess-when-cut are valid properties; (c) reactivity decreases down Group 7; fluorine is the most reactive halogen.

(a) Banded mark scheme (6 marks, QWC assessed): Band A (5–6 marks) — a full, coherent explanation covering: reactivity increases down the group; the outer electron is further from the nucleus down the group; increased shielding by inner shells down the group; and the outer electron is therefore lost more easily — all expressed with accurate use of specialist vocabulary and few errors in spelling, punctuation and grammar. Band B (3–4 marks) — reactivity trend correctly stated with partial or under-developed explanation (e.g. only one of 'further from nucleus' or 'shielding' given), reasonable SPG. Band C (1–2 marks) — reactivity trend stated with little or no valid explanation, or explanation given with the trend direction wrong; weak SPG. Band D (0 marks) — no relevant content. (b) 1 mark each for any two of: low density; soft/easily cut; shiny when cut but tarnishes quickly — max 2. (c) 1 mark for 'decreases' (down the group); 1 mark for 'fluorine'; 1 mark for a valid reason (e.g. outer shell further from nucleus / more shielding down the group, making it harder to gain an electron) — max 3.

(a) The mixture is first filtered: sand is insoluble in water so it is retained as the residue on the filter paper (this can then be rinsed with distilled water and dried to obtain dry sand), while the dissolved salt passes through with the water as the filtrate. To obtain pure, dry salt crystals from the filtrate, the salt solution is evaporated/crystallised — heating the solution gently (for example in an evaporating basin) drives off the water as steam, leaving the dissolved salt behind as dry crystals (heating can be stopped once crystals begin to form and the rest left to evaporate slowly, to produce larger, purer crystals). (b) Anhydrous copper(II) sulfate is white; a small amount is placed in a test tube and a few drops of the liquid to be tested are added — if the liquid contains water, the anhydrous copper(II) sulfate turns blue, forming hydrated copper(II) sulfate. (c) A flame test is carried out by first cleaning a nichrome wire, dipping it into concentrated hydrochloric acid (to clean it and help the sample stick), then dipping the wire into the solid compound being tested, and holding the wire in a roaring/blue Bunsen flame; the colour of the flame produced identifies the metal ion present. The flame colours to recall are: lithium — crimson; sodium — yellow/orange; potassium — lilac; calcium — brick red; copper(II) — blue-green/green-blue. A lilac flame therefore confirms the presence of potassium ions.
Final answer: (a) filtration to separate sand (residue) from salt solution (filtrate), then evaporation/crystallisation of the filtrate to obtain dry salt crystals; (b) anhydrous copper(II) sulfate turns blue in the presence of water; (c) nichrome wire dipped in conc. HCl then the solid, held in a Bunsen flame — potassium gives a lilac flame.

(a) 1 mark for 'filtration' as the first technique; 1 mark for correctly identifying sand as the residue and salt solution as the filtrate; 1 mark for 'evaporation' or 'crystallisation' as the second technique applied to the filtrate; 1 mark for a valid detail of the evaporation/crystallisation process (e.g. heating gently, or stopping heating once crystals form); 1 mark for correctly obtaining 'dry sand' and 'dry salt crystals' as the two final named products — max 5. (b) 1 mark for adding the liquid to (white) anhydrous copper(II) sulfate; 1 mark for 'turns blue' as the positive result — max 2. (c) 1 mark for using a nichrome wire; 1 mark for dipping in concentrated hydrochloric acid (to clean the wire); 1 mark for holding the wire/sample in a Bunsen flame; 1 mark for 'lilac' as the potassium flame colour — max 4. Accept flame colours for other ions if given correctly as additional detail but do not require them.

(a) Applying the given pH classification directly: pH 1 falls in the 0–2 range, so W is a strong acid; pH 7 is exactly neutral, so X is neutral; pH 9 falls in the 8–11 range, so Y is a weak alkali; pH 13 falls in the 12–14 range, so Z is a strong alkali. (b) All acidic solutions are acidic because they dissolve in water to release hydrogen ions, H⁺(aq); all alkaline solutions are alkaline because they dissolve in water to release hydroxide ions, OH⁻(aq). The higher the concentration of H⁺(aq) ions, the lower the pH. (c) Sulfuric acid, H₂SO₄, has two replaceable hydrogen ions, so it reacts with two formula units of sodium hydroxide, NaOH, to form sodium sulfate, Na₂SO₄, and water: H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l). Checking atoms: H (2+2=4 on left) = (2×2=4 on right); S 1=1; O (4+2=6 on left) = (4+2=6 on right); Na 2=2 — balanced. In every neutralisation reaction between a strong acid and a strong alkali, the essential reaction occurring is between the hydrogen ions from the acid and the hydroxide ions from the alkali, which combine to form water: H⁺(aq) + OH⁻(aq) → H₂O(l).
Final answer: (a) W strong acid, X neutral, Y weak alkali, Z strong alkali; (b) H⁺(aq) for acidic, OH⁻(aq) for alkaline; (c) H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l), ionic equation H⁺(aq) + OH⁻(aq) → H₂O(l).

(a) 1 mark each for correctly classifying W, X, Y and Z — max 4. (b) 1 mark for 'H⁺(aq)'/'hydrogen ions' for acidic; 1 mark for 'OH⁻(aq)'/'hydroxide ions' for alkaline — max 2. (c) 1 mark for correct formulae of all four species (H₂SO₄, NaOH, Na₂SO₄, H₂O); 1 mark for correct balancing (coefficient 2 before NaOH and before H₂O); 1 mark for correct state symbols throughout; 1 mark for the ionic equation H⁺(aq) + OH⁻(aq) → H₂O(l) with correct species; 1 mark for correct state symbols in the ionic equation — max 5.

(a) The atomic number (17) gives the number of protons, which equals the number of electrons in a neutral atom. The mass number (35) is the total number of protons and neutrons, so the number of neutrons = 35 − 17 = 18. (b) Isotopes are atoms of the same element — meaning they have the same atomic number and therefore the same number of protons (and electrons) — but they have different numbers of neutrons, giving them different mass numbers. Chlorine-35 (17 protons, 18 neutrons) and chlorine-37 (17 protons, 20 neutrons) are both chlorine because they both have 17 protons, but they are different isotopes because they have different numbers of neutrons. (c) Relative atomic mass is a weighted mean of the mass numbers of the isotopes, weighted by their natural abundance: \( A_r = (0.75 \times 35) + (0.25 \times 37) = 26.25 + 9.25 = 35.5 \). This matches the accepted relative atomic mass of chlorine found on the Periodic Table.
Final answer: (a) 17 protons, 18 neutrons, 17 electrons; (b) isotopes have the same number of protons but different numbers of neutrons (Cl-35 has 18, Cl-37 has 20); (c) \( A_r = 35.5 \).

(a) 1 mark for 17 protons; 1 mark for 18 neutrons; 1 mark for 17 electrons — max 3. (b) 1 mark for 'same number of protons/same atomic number'; 1 mark for 'different number of neutrons' (applied correctly to the Cl-35/Cl-37 example) — max 2. (c) 1 mark for correct method, i.e. \( (0.75 \times 35) + (0.25 \times 37) \); 1 mark for the correct final answer 35.5 to 1 decimal place — max 2. No mark for the final answer if given to the wrong number of decimal places without the correct value shown in working.

(a) Calcium (atomic number 20) has the electronic structure 2,8,8,2 (2 electrons in the first shell, 8 in the second, 8 in the third, 2 in the fourth/outer shell). Chlorine (atomic number 17) has the electronic structure 2,8,7 (2, then 8, then 7 in its outer shell). (b) Because calcium chloride has the formula CaCl₂, one calcium atom must react with two chlorine atoms. Calcium, with 2 electrons in its outer shell, loses both of these electrons to achieve the stable, noble-gas-like configuration 2,8,8, forming a Ca²⁺ ion. Each chlorine atom, with 7 electrons in its outer shell, needs only 1 more electron to complete its outer shell to 2,8,8, so each gains one of the two electrons lost by calcium, forming a Cl⁻ ion. In a dot-and-cross diagram, the two electrons originally shown as calcium's (crosses) would each be redrawn in the outer shell of a separate chlorine ion (as crosses among the chlorine atom's own dots), with square brackets and charges (2+ and −) around each resulting ion. The one Ca²⁺ ion and two Cl⁻ ions formed are then held together in a lattice by strong electrostatic attraction between the oppositely charged ions — this is the ionic bond. (c) The calcium ion formed is Ca²⁺ and the chloride ion formed is Cl⁻.
Final answer: Ca is 2,8,8,2; Cl is 2,8,7; Ca loses 2 electrons (one to each of two Cl atoms) to form Ca²⁺ (2,8,8), each Cl gains 1 electron to form Cl⁻ (2,8,8); ions are Ca²⁺ and Cl⁻.

(a) 1 mark for calcium 2,8,8,2; 1 mark for chlorine 2,8,7 — max 2. (b) 1 mark for calcium losing 2 electrons; 1 mark for correctly identifying that ONE calcium atom transfers one electron to EACH of two separate chlorine atoms (ratio 1 Ca : 2 Cl); 1 mark for both ions reaching a stable/noble-gas (2,8,8) configuration; 1 mark for describing the ionic bond as electrostatic attraction between oppositely charged ions — max 4. (c) 1 mark for 'Ca²⁺' AND 'Cl⁻' both correct with charges shown.

(a) \( M_r(\text{CaCO}_3) = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100 \). (b) The number of moles is found using \( \text{moles} = \dfrac{\text{mass}}{M_r} = \dfrac{10.0}{100} = 0.1 \text{ mol} \). (c) From the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂ (a 1:1 ratio), so 0.1 mol CaCO₃ produces 0.1 mol CO₂. Using \( \text{mass} = \text{moles} \times M_r \): \( \text{mass of CO}_2 = 0.1 \times 44 = 4.4 \text{ g} \).
Final answer: (a) \( M_r = 100 \); (b) 0.1 mol; (c) 4.4 g of CO₂.

(a) 1 mark for \( M_r = 100 \) (working not essential if correct). (b) 1 mark for correct method (mass ÷ Mr); 1 mark for correct answer 0.1 mol — max 2. (c) 1 mark for identifying the 1:1 mole ratio between CaCO₃ and CO₂ (i.e. moles CO₂ = 0.1 mol); 1 mark for correct method (moles × Mr of CO₂); 1 mark for correct final answer 4.4 g — max 3. Award full credit for a correct final answer with clearly correct working even if not laid out in exactly these steps.

(a) An allotrope is one of two or more different structural forms in which an element can exist in the same physical state — for example, carbon can exist as diamond, graphite or graphene, which differ in how their carbon atoms are bonded together, even though all three are the same element. (b) In graphite, each carbon atom is covalently bonded to only 3 neighbouring carbon atoms, arranged in flat hexagonal layers; this leaves one outer electron per carbon atom that is not used in bonding, and these electrons become delocalised, free to move along the layers, which allows graphite to conduct electricity. In diamond, by contrast, each carbon atom uses all 4 of its outer electrons to form 4 strong covalent bonds to 4 other carbon atoms in a rigid 3-dimensional giant covalent structure; because every outer electron is held tightly in a covalent bond, there are no delocalised or free electrons (and no ions) able to carry charge, so diamond cannot conduct electricity. (c) Graphene is a single, one-atom-thick layer of graphite; its structure gives it very high electrical conductivity (because, like graphite, it has delocalised electrons free to move across the layer) and great strength, which is why it is used in applications such as batteries and solar cells.
Final answer: (a) different structural forms of the same element in the same physical state; (b) graphite has 1 delocalised electron per carbon atom (only 3 bonds each) that can move and carry charge, diamond has all 4 outer electrons used in bonds so none are free; (c) graphene is used in batteries/solar cells because of its high electrical conductivity.

(a) 1 mark for a correct definition of allotrope. (b) 1 mark for graphite having 3 covalent bonds per carbon atom (layered structure); 1 mark for graphite having 1 delocalised/free electron per atom able to move and carry charge; 1 mark for diamond having 4 covalent bonds per carbon atom using all outer electrons, so no free/delocalised electrons — max 3 (must explain both graphite AND diamond to gain full marks). (c) 1 mark for a valid use (e.g. batteries, solar cells) AND a correctly linked property (e.g. high electrical conductivity, or great strength/thinness) — award the mark only if both the use and a matching property are given.

(a) Using the Principle of Moments, the plank balances when the clockwise moment equals the anticlockwise moment about the pivot: \( \text{moment} = F \times d \), so \( 300 \times 1.5 = 250 \times d \). Rearranging: \( d = \dfrac{300 \times 1.5}{250} = \dfrac{450}{250} = 1.8 \text{ m} \). (b) The Principle of Moments states that when an object is balanced (in equilibrium), the sum of the clockwise moments about a pivot is equal to the sum of the anticlockwise moments about that same pivot. (c) The centre of gravity is the single point at which the weight of an object can be considered to act. If an object is tilted, it will topple over once the line of action of its weight (a vertical line down from the centre of gravity) falls outside its base, because the weight then creates a turning moment about the edge of the base that continues the tilt. An object with a wide base and a low centre of gravity needs to be tilted through a much larger angle before the vertical line from its centre of gravity moves outside its base, so it is more stable; an object with a narrow base and a high centre of gravity only needs a small tilt before this happens, making it easier to topple and therefore less stable.
Final answer: (a) 1.8 m; (b) clockwise moments = anticlockwise moments for a balanced object; (c) a wide base and low centre of gravity require a much larger tilt before the weight's line of action falls outside the base, so the object is more stable.

(a) 1 mark for the correct moments equation \( 300 \times 1.5 = 250 \times d \); 1 mark for correct rearrangement; 1 mark for the correct final answer 1.8 m — max 3 (award full marks for a correct final answer with clear working even if not laid out in exactly this order). (b) 1 mark for 'clockwise moments = anticlockwise moments'; 1 mark for reference to the object being balanced/in equilibrium about the pivot — max 2. (c) 1 mark for reference to the line of action of the weight from the centre of gravity; 1 mark for explaining that toppling occurs once this line falls outside the base; 1 mark for correctly linking a wider base/lower centre of gravity to requiring a greater tilt before this happens — max 3.

To carry out this investigation, the student first needs to find their weight: they measure their own mass, m, using a set of bathroom or electronic scales, then calculate their weight using \( W = mg \), taking \( g = 10 \text{ N/kg} \). To find the height climbed, the student measures the height of a single step using a ruler or metre rule, and counts the total number of steps in the staircase (or measures the total vertical height of the staircase directly using a long tape measure); multiplying the height of one step by the number of steps gives the total height, h, climbed. The student then uses a stopwatch to time, t, how long it takes them to climb from the bottom to the top of the staircase, moving as quickly as they safely can. The energy transferred in climbing the stairs is equal to the work done against gravity, \( E = mgh \) (equivalently, weight × height climbed). Power is then calculated using \( P = \dfrac{E}{t} \) (energy transferred per second). To improve the reliability of the result, the student should repeat the climb several times and calculate an average time, and could also repeat the whole experiment on different days.
Final answer: measure mass (for weight, W = mg), height of stairs (h), and time taken (t) to climb; calculate energy transferred E = mgh, then power P = E / t; repeat for reliability.

Banded mark scheme (6 marks, QWC assessed). Indicative content: measure mass using a balance/scales and calculate weight using W = mg; measure the height of a step and count the number of steps (or measure total height directly) to find total height climbed, h; use a stopwatch to time how long it takes to climb the stairs, t; calculate energy transferred using E = mgh (or weight × height); calculate power using P = E ÷ t; repeat and average for reliability. Band A (5–6 marks): at least 5 of the indicative points given, in a clear, logically sequenced method, with accurate use of specialist vocabulary and few errors in spelling, punctuation and grammar. Band B (3–4 marks): 3–4 indicative points given, method reasonably clear, some errors in SPG. Band C (1–2 marks): 1–2 indicative points given, method poorly organised or with significant gaps, weak SPG. Band D (0 marks): no relevant content, or a description that does not amount to a workable method.

(a) A proton has a relative charge of +1 and a relative mass of 1; a neutron has a relative charge of 0 (it is uncharged) and a relative mass of 1 (approximately the same mass as a proton); an electron has a relative charge of −1 and a relative mass that is very small/negligible compared with that of a proton or neutron. (b) The top number in ₂⁴He is the mass number (4) and the bottom number is the atomic number (2). The atomic number gives the number of protons, so there are 2 protons; the number of neutrons is found by subtracting the atomic number from the mass number, \( 4 - 2 = 2 \) neutrons; since the atom is neutral overall, the number of electrons equals the number of protons, so there are 2 electrons. (c) Inside the Sun, the extremely high temperature and pressure force light nuclei (hydrogen nuclei) close enough together to overcome their mutual repulsion and fuse together, forming larger nuclei; this process releases a very large amount of energy, which is the source of the Sun's energy output. The main by-product gas of this fusion process is helium, an inert, non-toxic gas.
Final answer: (a) proton +1/1, neutron 0/1, electron −1/negligible; (b) 2 protons, 2 neutrons, 2 electrons; (c) small nuclei fuse together under extreme heat and pressure, releasing energy, producing helium gas.

(a) 1 mark for proton correct (charge +1, mass 1); 1 mark for neutron correct (charge 0, mass 1); 1 mark for electron correct (charge −1, mass very small/negligible) — max 3. (b) 1 mark for 2 protons AND 2 electrons; 1 mark for 2 neutrons — max 2. (c) 1 mark for describing small/hydrogen nuclei joining/fusing together under extreme heat and pressure, releasing energy; 1 mark for 'helium' as the gas produced — max 2.

(a) The half-life of a radioactive isotope is the average time it takes for the activity (or count rate) of a sample — equivalently, the number of undecayed radioactive nuclei remaining — to fall to half of its original (starting) value. (b) The activity falls from 640 counts/min to 80 counts/min. Each half-life halves the activity: \( 640 \rightarrow 320 \rightarrow 160 \rightarrow 80 \), which is three successive halvings, so 90 minutes represents 3 half-lives. The half-life is therefore \( \dfrac{90 \text{ minutes}}{3} = 30 \text{ minutes} \). (c) Precautions that should be taken include using tongs to hold or handle a radioactive source (rather than the fingers), keeping the source as far away from the body as possible, being exposed to the source for as short a time as possible, and storing sources in lead-lined containers when not in use. Using tongs and keeping the source at a distance both work by increasing the distance between the source and the user's body; since the intensity of radiation reaching a person decreases as the distance from the source increases, this reduces the dose of radiation the user receives, lowering the risk of harmful ionisation damaging their cells.
Final answer: (a) the average time for activity/count rate to fall to half its original value; (b) 30 minutes; (c) using tongs / keeping the source at a distance reduces the radiation dose received because intensity falls with distance from the source.

(a) 1 mark for 'time taken' language; 1 mark for '(activity/count rate) to fall to half its original/starting value' — max 2. (b) 1 mark for correctly identifying that 640 → 80 represents 3 half-lives (e.g. showing 640→320→160→80); 1 mark for the correct method (90 ÷ 3); 1 mark for the correct final answer, 30 minutes — max 3. (c) 1 mark for a valid, named precaution (e.g. tongs, distance, shielding, minimal exposure time); 1 mark for linking the precaution to increasing distance from / reducing time near / reducing exposure to the source; 1 mark for explaining this reduces the dose of radiation received by the user — max 3. Accept any one valid precaution, fully explained.

(a) Weight is calculated using \( W = mg \): \( W = 45 \times 10 = 450 \text{ N} \). (b) The base is square with side 0.6 m, so its area is \( A = 0.6 \times 0.6 = 0.36 \text{ m}^2 \). (c) Pressure is calculated using \( P = \dfrac{F}{A} \), where F is the force (here, the crate's weight) acting perpendicular to the surface: \( P = \dfrac{450}{0.36} = 1250 \text{ Pa} \) (since \( 1 \text{ Pa} = 1 \text{ N/m}^2 \)). (d) Since the crate's weight (the force, F) stays the same but it now rests on a smaller face (a smaller area, A), and \( P = \dfrac{F}{A} \), reducing A while keeping F constant increases the pressure exerted on the ground.
Final answer: (a) 450 N; (b) 0.36 m²; (c) 1250 Pa; (d) pressure increases, because the same force now acts over a smaller area.

(a) 1 mark for using \( W = mg \); 1 mark for correct answer 450 N — max 2. (b) 1 mark for method (0.6 × 0.6); 1 mark for correct answer 0.36 m² — max 2. (c) 1 mark for using \( P = F/A \) with F = 450 N (or their (a)) and A = 0.36 m² (or their (b)); 1 mark for correct numerical answer 1250; 1 mark for correct unit (Pa or N/m²) — max 3 (allow error carried forward from (a)/(b)). (d) 1 mark for 'increases' with a valid reason referencing the same force over a smaller area (both parts needed for the mark).

(a) A vector quantity has both a magnitude (size) and a direction associated with it — for example, displacement (distance travelled in a specified direction) and velocity (speed in a specified direction) are vectors. A scalar quantity has only a magnitude, with no associated direction — for example, distance and speed are scalars. (b) Using \( a = \dfrac{v - u}{t} \), with \( u = 0 \), \( v = 12 \text{ m/s} \) and \( t = 8 \text{ s} \): \( a = \dfrac{12 - 0}{8} = 1.5 \text{ m/s}^2 \). (c) The velocity–time graph consists of three straight-line sections: it rises steadily from (0, 0) to (8, 12), reflecting the uniform acceleration phase; it is then a horizontal line at 12 m/s from t = 8 s to t = 28 s, reflecting constant velocity; finally it falls steadily from (28, 12) to (32, 0), reflecting uniform deceleration to rest. Since the area under a velocity–time graph gives the distance travelled, the total distance is the sum of the areas of the three sections: the first section is a triangle, area \( = \tfrac{1}{2} \times 8 \times 12 = 48 \text{ m} \); the second section is a rectangle, area \( = 20 \times 12 = 240 \text{ m} \); the third section is a triangle, area \( = \tfrac{1}{2} \times 4 \times 12 = 24 \text{ m} \). Total distance \( = 48 + 240 + 24 = 312 \text{ m} \).
Final answer: (a) vectors have direction (e.g. velocity), scalars do not (e.g. speed); (b) 1.5 m/s²; (c) rises, then flat, then falls; total distance = 312 m.

(a) 1 mark for correctly stating vectors have magnitude and direction, scalars have magnitude only; 1 mark for a correctly matched vector example (displacement or velocity); 1 mark for a correctly matched scalar example (distance or speed) — max 3. (b) 1 mark for correct method \( (12-0)/8 \); 1 mark for correct answer 1.5 m/s² — max 2. (c) 1 mark for correctly describing the three-section shape (rising, then horizontal, then falling); 1 mark for correct method — summing the three section areas (triangle + rectangle + triangle); 1 mark for correct final total, 312 m — max 3. Accept the three section distances (48 m, 240 m, 24 m) shown separately as valid working.

(a) Hooke's law states that the extension of a spring is directly proportional to the applied force, provided the limit of proportionality is not exceeded — this means the ratio \( F/e \) should be constant. Checking both readings: \( \dfrac{2}{0.04} = 50 \), and \( \dfrac{5}{0.10} = 50 \). Since both give the same value (50 N/m), force and extension are directly proportional for this spring over this range, so the data is consistent with Hooke's law. (b) Using \( F = ke \), rearranged to \( k = \dfrac{F}{e} \): \( k = \dfrac{2}{0.04} = 50 \text{ N/m} \) (or equivalently using the second data point, \( k = \dfrac{5}{0.10} = 50 \text{ N/m} \)); the unit of the spring constant is newtons per metre, N/m. (c) Using \( F = ke \), rearranged to \( e = \dfrac{F}{k} \): \( e = \dfrac{8}{50} = 0.16 \text{ m} \). (d) The limit of proportionality is the point up to which a spring's extension remains directly proportional to the force applied to it; if the spring is stretched beyond this point, the extension is no longer proportional to the force, and the spring may become permanently stretched (not return to its original length when the force is removed).
Final answer: (a) F/e = 50 N/m for both readings, confirming direct proportionality; (b) k = 50 N/m; (c) 0.16 m; (d) the point beyond which extension is no longer proportional to force.

(a) 1 mark for calculating F/e (or e/F) for both data points; 1 mark for obtaining the same value (50) for both; 1 mark for a valid conclusion that this confirms direct proportionality/Hooke's law — max 3. (b) 1 mark for correct numerical value 50; 1 mark for correct unit N/m — max 2 (award both marks for 50 N/m stated with correct working, allow error carried forward from (a)). (c) 1 mark for correct method \( e = F/k \); 1 mark for correct answer 0.16 m (allow error carried forward from their k in (b)) — max 2. (d) 1 mark for a correct description of the limit of proportionality.

(a) For the mass and volume of a substance to be directly proportional, the ratio mass ÷ volume must be constant for every reading. Checking the data: \( 27 \div 10 = 2.7 \), \( 54 \div 20 = 2.7 \), and \( 81 \div 30 = 2.7 \) — the ratio is the same (2.7) in each case, showing mass is directly proportional to volume for this metal; equivalently, a graph of mass (y-axis) against volume (x-axis) would give a straight line passing through the origin. (b) Since density is defined as \( D = \dfrac{\text{mass}}{\text{volume}} \), and this ratio is constant, the density can be found from any one data point (or as the gradient of a mass–volume graph): \( D = \dfrac{27}{10} = 2.7 \text{ g/cm}^3 \) (consistent with all three readings). (c) Because the sample is irregularly shaped, its volume cannot be measured directly with a ruler, so a displacement method is used. A displacement (eureka) can is filled with water exactly up to the level of its side spout, with an empty measuring cylinder placed underneath the spout to catch any overflow. The metal sample is then lowered gently into the can, fully submerged, using a thread — as it enters the water, it displaces a volume of water equal to its own volume, which flows out through the spout into the measuring cylinder below. The volume of water collected in the measuring cylinder is then read off and is equal to the volume of the metal sample.
Final answer: (a) mass/volume = 2.7 for every reading, showing direct proportionality; (b) 2.7 g/cm³; (c) submerge the sample in a full displacement can and measure the volume of water displaced/collected in a measuring cylinder.

(a) 1 mark for calculating the mass/volume ratio for at least two data points; 1 mark for a valid conclusion (ratio constant / graph would be a straight line through the origin) — max 2. (b) 1 mark for correct method (mass ÷ volume, using any consistent data point); 1 mark for correct numerical answer 2.7; 1 mark for correct unit g/cm³ — max 3. (c) 1 mark for using a displacement/eureka can filled to the spout; 1 mark for lowering the sample in (fully submerged) and collecting the displaced water in a measuring cylinder; 1 mark for correctly stating the volume of water collected equals the volume of the sample — max 3.

(a) Using \( a = \dfrac{v - u}{t} \), with \( u = 2 \text{ m/s} \), \( v = 8 \text{ m/s} \) and \( t = 6 \text{ s} \): \( a = \dfrac{8 - 2}{6} = \dfrac{6}{6} = 1 \text{ m/s}^2 \). (b) Using Newton's second law, \( F = m \times a \), with \( m = 75 \text{ kg} \) and \( a = 1 \text{ m/s}^2 \) from part (a): \( F = 75 \times 1 = 75 \text{ N} \). (c) Newton's second law states that a resultant (unbalanced) force acting on an object will cause it to accelerate, and that this acceleration is proportional to the size of the resultant force acting on it; for a given mass this is expressed by the equation \( F = m \times a \), where F is the resultant force, m is the mass and a is the acceleration produced. (d) As the cyclist moves forward, friction between the tyres and the road/moving parts and, especially at higher speeds, air resistance (drag) act on the cyclist and bicycle in the opposite direction to their motion, opposing it.
Final answer: (a) 1 m/s²; (b) 75 N; (c) resultant force causes acceleration proportional to it, F = ma; (d) friction or air resistance.

(a) 1 mark for correct method \( (8-2)/6 \); 1 mark for correct answer 1 m/s² — max 2. (b) 1 mark for using \( F = ma \) with their acceleration; 1 mark for correct answer 75 N (allow error carried forward from (a)) — max 2. (c) 1 mark for 'a resultant force causes an object to accelerate, proportional to the size of the force'; 1 mark for correctly stating/using \( F = m \times a \) — max 2. (d) 1 mark for 'friction' or 'air resistance/drag' (either accepted).