2026 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) Practice Paper with Answers

Use the magnification formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\). Substitute the values: \(\text{Magnification} = \frac{48\text{ mm}}{0.08\text{ mm}} = 600\).

Correct formula or substitution \(\frac{48}{0.08}\) [0.5]; Correct value of 600 or \(\times 600\) [1.0]. A: \(\times 600\) / 600 / 600x. R: Units such as mm attached to the final answer.

At high temperatures, the excess kinetic energy disrupts the specific three-dimensional shape of the protein. This causes the active site to change shape (the enzyme is denatured), meaning substrate molecules can no longer bind to form enzyme-substrate complexes.

Mention of active site changing shape / enzyme denatures [1.0]; (Substrate) no longer fits / cannot form enzyme-substrate complexes [0.5]. R: 'Enzyme dies' (scores 0 for denaturation mark).

Magnesium is a central mineral component required for the synthesis of chlorophyll. Without sufficient magnesium, chlorophyll cannot be produced, causing leaves to turn yellow, a condition known as chlorosis.

Making / synthesis of chlorophyll [1.0]; Yellow leaves / yellowing between veins / chlorosis [0.5]. I: Poor growth unqualified.

Arteries carry blood under high pressure and thus have thick muscular/elastic walls and narrow lumens. Veins carry blood under low pressure, have thinner walls, wider lumens, and possess valves to prevent backflow.

Any two comparative points: (1) Arteries have thicker muscular/elastic walls / veins have thinner walls [0.75]; (2) Arteries have narrower lumen / veins have wider lumen OR veins have valves / arteries do not have valves [0.75]. A: Tabulated differences.

Antibodies are specific proteins produced by lymphocytes. When exposed to a pathogen, memory cells remain in the blood/lymph; upon re-infection by the same pathogen, memory cells recognise the antigens and produce large amounts of antibodies very quickly.

Protein [0.5]; Memory cells remain in the circulation / produce antibodies faster / in greater quantity on re-infection [1.0].

During vasoconstriction in response to cold, the muscular walls of arterioles constrict (narrow). This diverts blood away from surface capillary loops, decreasing heat loss by radiation and convection.

Arterioles constrict / narrow / lumen diameter decreases [1.0]; (Less blood flows to surface capillaries leading to) reduced heat loss [0.5]. R: 'Capillaries constrict'.

1. High thermal energy causes bonds maintaining the tertiary structure of the enzyme protein to break, altering the specific 3D conformation of the active site (denaturation).
2. The shape of the active site is no longer complementary to the substrate molecule.
3. The substrate can no longer bind to the enzyme, preventing the formation of enzyme-substrate complexes.

bonds break / active site changes shape / enzyme is denatured; [1 mark] [R: enzyme killed/died]
active site no longer complementary to substrate / substrate no longer fits; [1 mark]
no / fewer enzyme-substrate complexes can form; [0.5 marks]

1. The weakened pathogen carries antigens that are recognized as foreign by white blood cells.
2. Lymphocytes are stimulated to produce antibodies specific/complementary to these antigens.
3. Memory cells are generated and remain in circulation, enabling a much faster and greater antibody response upon subsequent exposure to the live pathogen.

vaccine contains / introduces antigens (from the pathogen); [0.5 marks]
stimulates lymphocytes to produce specific / complementary antibodies; [1 mark]
memory cells are produced / remain in the blood to provide a rapid secondary immune response; [1 mark]

1. Moving air blows away the humid layer of water vapour that collects immediately outside the stomatal pores.
2. This maintains a steep concentration (or water potential) gradient of water vapour from the inside of the leaf to the external air.
3. Water vapour diffuses out of the open stomata into the atmosphere at a faster rate.

moving air / wind removes water vapour from the leaf surface / outside stomata; [1 mark]
maintains a steep water vapour concentration gradient / diffusion gradient; [1 mark]
increases the rate of diffusion of water vapour out through stomata; [0.5 marks]

1. Thermoreceptors detect elevated core body temperature and signal the hypothalamus.
2. Muscles in the walls of arterioles supplying skin surface capillaries relax, causing the arterioles to widen (vasodilation).
3. Increased volume of warm blood flows through capillaries close to the skin surface, increasing heat loss to the surrounding environment by radiation and convection.

arterioles supplying skin surface capillaries widen / dilate; [1 mark] [R: capillaries widen/move]
increased volume of blood flows through capillaries near the skin surface; [1 mark]
increases heat loss to the surroundings by radiation / convection; [0.5 marks]

1. Oxygen demand exceeds supply during intense exercise, causing muscle cells to perform anaerobic respiration which leads to the accumulation of lactic acid.
2. The oxygen debt represents the extra volume of oxygen needed to metabolise this lactic acid.
3. After the race, deep and fast breathing continues to supply extra oxygen to the liver, where lactic acid is aerobically oxidised to carbon dioxide and water (or converted to glucose).

insufficient oxygen supplied to muscles resulting in anaerobic respiration / buildup of lactic acid; [1 mark]
continued high breathing rate / elevated oxygen intake after exercise; [0.5 marks]
oxygen is used to aerobically break down / oxidise lactic acid in the liver; [1 mark]

1. Only approximately 10% of energy is passed from one trophic level to the next.
2. Most energy is lost through life processes such as aerobic respiration (heat loss), physical movement, egestion of undigested matter (faeces), excretion (urine), or unconsumed parts of organisms.
3. After 4 or 5 feeding levels, the total energy remaining is too small to support the metabolic and reproductive requirements of a further trophic level.

energy is lost between trophic levels / only ~10% of energy transferred; [1 mark]
credit named energy loss mechanism: respiration / heat / excretion / egestion / uneaten material / movement; [1 mark]
insufficient energy remains at higher levels to sustain another viable population; [0.5 marks]

1. Identify the genotypes of the parents: heterozygous purple stem is \(Aa\), green stem is \(aa\). 2. Construct the genetic cross / Punnett square: Gametes from \(Aa\) are \(A\) and \(a\); gametes from \(aa\) are \(a\). Offspring genotypes are \(Aa\) (purple stem) and \(aa\) (green stem) in a 1 : 1 ratio. 3. The probability of an offspring having green stems (\(aa\)) is \(0.5\) (or \(50\%\)). 4. Calculate expected number: \(480 \times 0.5 = 240\).

Award [1 mark] for correct parental genotypes (\(Aa\) and \(aa\)) and identifying offspring ratio of 1 : 1 (or probability of 0.5 / 50%); Award [1.5 marks] for correct calculation: \(480 \times 0.5 = 240\) (allow full 2.5 marks for correct final answer without working; allow ecf if wrong ratio correctly multiplied by 480).

1. Recall the magnification formula: \(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}}\). 2. Calculate actual size in millimetres: \(\text{Actual size} = \frac{48\text{ mm}}{600} = 0.08\text{ mm}\). 3. Convert millimetres to micrometres by multiplying by 1000: \(0.08\text{ mm} \times 1000 = 80\ \mu\text{m}\).

Award [1 mark] for correct substitution into formula: \(\frac{48}{600}\) (or \(0.08\text{ mm}\)); Award [0.5 mark] for unit conversion method (\(\times 1000\)); Award [1 mark] for correct final value with correct unit: \(80\ \mu\text{m}\) (allow full 2.5 marks for correct final answer with units).

1. Efficiency formula: \(\text{Efficiency} = \left( \frac{\text{Energy in secondary consumers}}{\text{Energy in primary consumers}} \right) \times 100\%\). 2. Substitute the values: \(\text{Efficiency} = \left( \frac{4.08 \times 10^3}{3.4 \times 10^4} \right) \times 100\% = \left( \frac{4080}{34000} \right) \times 100\%\). 3. Evaluate: \(0.12 \times 100 = 12\%\).

Award [1 mark] for correct formula or substitution of values (\(\frac{4.08 \times 10^3}{3.4 \times 10^4} \times 100\) or \(\frac{4080}{34000} \times 100\)); Award [1.5 marks] for correct evaluation of \(12\%\) (allow full 2.5 marks for correct final answer without working).

(a) Propene has three carbon atoms with one double bond between two carbon atoms and single bonds to six hydrogen atoms: \(\text{H}_2\text{C}=\text{CH}-\text{CH}_3\). Every C-H and C-C bond must be shown fully.
(b) Aqueous bromine is orange/brown. When it reacts across the double bond of an alkene via an addition reaction, the bromine is used up, turning the solution colourless.

(a) Fully correct displayed formula showing \(\text{C}=\text{C}\) double bond, \(\text{C}-\text{C}\) single bond, and all 6 \(\text{C}-\text{H}\) single bonds clearly displayed [1]
[A: unambiguous structural connectivity]
[R: molecular or condensed formula]

(b) From orange / brown / red-brown / yellow to colourless / decolourised [0.5]
[R: clear instead of colourless]

From the balanced chemical equation, \(1\text{ mol}\) of \(\text{Mg}\) produces \(1\text{ mol}\) of \(\text{H}_2\).

Therefore, moles of \(\text{H}_2\text{ gas} = 0.050\text{ mol}\).

\(\text{Volume of gas} = \text{moles} \times 24\text{ dm}^3\text{/mol} = 0.050 \times 24 = 1.2\text{ dm}^3\).

Use of mole ratio to identify moles of \(\text{H}_2 = 0.050\text{ mol}\) [0.5]

Calculation: \(0.050 \times 24\) [0.5]

Correct evaluation: \(1.2\text{ dm}^3\) [0.5]

(a) During addition polymerisation, the carbon-carbon double bond (\(\text{C}=\text{C}\)) breaks to form single covalent bonds between monomer units. The repeat unit consists of two carbon atoms connected by a single bond, each bonded to two hydrogen atoms, with continuation open bonds on either side extending outside square brackets.
(b) The bond between the carbon atoms in the backbone of poly(ethene) is a single covalent bond.

(a) Correct repeat unit showing two carbon atoms bonded by a single covalent bond, 4 \(\text{C}-\text{H}\) bonds, and continuation bonds extending to the sides/brackets [1]
[R: presence of \(\text{C}=\text{C}\) double bond]

(b) Single covalent bond / single bond [0.5]

(a) Lead ions (\(\text{Pb}^{2+}\)) are positively charged cations attracted to the negative cathode, where they gain electrons to form lead metal.
(b) Bromide ions (\(\text{Br}^-\)) are negatively charged anions attracted to the positive anode, where they lose electrons (oxidation) to form bromine molecules: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\).

(a) Lead / \(\text{Pb}\) [0.5]

(b) \(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\) (or \(2\text{Br}^- - 2\text{e}^- \rightarrow \text{Br}_2\)) [1]
[1 mark for correct species and balance]

(a) Insoluble salts are formed by precipitation when two soluble solutions containing the required cation and anion are mixed. Barium chloride or barium nitrate provides soluble \(\text{Ba}^{2+}\) ions, while sodium sulfate, potassium sulfate, or sulfuric acid provides soluble \(\text{SO}_4^{2-}\) ions.
(b) Filtration is used to separate the solid precipitate (residue) from the aqueous solution (filtrate).

(a) Any soluble barium salt (e.g. barium nitrate / barium chloride) AND any soluble sulfate (e.g. sodium sulfate / potassium sulfate / dilute sulfuric acid) [1]
[R: insoluble compounds such as barium carbonate]

(b) Filtration / filter [0.5]

At higher temperatures, reacting particles gain thermal kinetic energy and move faster, leading to a higher collision frequency (more collisions per unit time). Furthermore, a much greater fraction of colliding particles have kinetic energy greater than or equal to the activation energy (\(E_a\)), resulting in a higher percentage of collisions being successful/effective.

Particles have more kinetic energy / move faster causing more frequent collisions (more collisions per unit time) [0.5]

More / greater proportion / greater fraction of colliding particles have energy greater than or equal to the activation energy [1]

(a) An oxygen atom has 6 valence electrons and each hydrogen atom has 1 valence electron. Oxygen forms two single covalent bonds by sharing one pair of electrons with each hydrogen atom (2 bonding pairs total). Oxygen retains 4 unshared outer-shell electrons (2 lone pairs).
(b) Simple covalent substances contain weak intermolecular forces between molecules, which require only a small amount of thermal energy to overcome, resulting in low boiling points.

(a) Two shared pairs of electrons (one pair between oxygen and each hydrogen) [0.5] AND two lone pairs / 4 non-bonding electrons on the oxygen atom [0.5]

(b) Weak intermolecular forces / weak forces between molecules (require little energy to break) [0.5]
[R: weak covalent bonds broken]

1. Calculate relative formula masses ($M_r$):
- $M_r(\text{CaCO}_3) = 40 + 12 + (3 \times 16) = 100$
- $M_r(\text{CaO}) = 40 + 16 = 56$

2. Calculate moles of $\text{CaCO}_3$ decomposed:
$$\text{moles of } \text{CaCO}_3 = \frac{25.0\text{ g}}{100\text{ g/mol}} = 0.250\text{ mol}$$

3. Determine moles of $\text{CaO}$ formed:
From the balanced equation, the mole ratio is $1 : 1$, so moles of $\text{CaO} = 0.250\text{ mol}$.

4. Calculate mass of $\text{CaO}$:
$$\text{mass} = 0.250\text{ mol} \times 56\text{ g/mol} = 14.0\text{ g}$$

- Correct calculation of $M_r(\text{CaCO}_3) = 100$ and $M_r(\text{CaO}) = 56$ [0.5];
- Calculation of moles of $\text{CaCO}_3 = 0.250\text{ mol}$ [1.0];
- Correct final mass of $\text{CaO} = 14.0\text{ g}$ (allow ecf from incorrect molar masses or moles) [1.0];

1. Calculate moles of magnesium reacted:
$$\text{moles of } \text{Mg} = \frac{3.60\text{ g}}{24\text{ g/mol}} = 0.150\text{ mol}$$

2. Use the molar ratio to determine moles of $\text{H}_2$ produced:
From the equation, $1\text{ mol Mg} : 1\text{ mol H}_2$, so:
$$\text{moles of } \text{H}_2 = 0.150\text{ mol}$$

3. Calculate the volume of $\text{H}_2$ gas at r.t.p.:
$$\text{volume} = \text{moles} \times \text{molar gas volume} = 0.150\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 3.60\text{ dm}^3$$

- Calculation of moles of $\text{Mg} = \frac{3.60}{24} = 0.150\text{ mol}$ [1.0];
- Recognition of $1:1$ molar ratio yielding $0.150\text{ mol}$ of $\text{H}_2$ [0.5];
- Evaluation of volume of $\text{H}_2 = 0.150 \times 24.0 = 3.60\text{ dm}^3$ [1.0];

(a) From the balanced equation, $1\text{ mol } \text{C}_3\text{H}_8$ requires $5\text{ mol } \text{O}_2$.
$$\text{Moles of } \text{O}_2 = 0.40 \times 5 = 2.0\text{ mol}$$

(b) From the equation, $1\text{ mol } \text{C}_3\text{H}_8$ produces $3\text{ mol } \text{CO}_2$.
$$\text{Moles of } \text{CO}_2 = 0.40 \times 3 = 1.20\text{ mol}$$
$$\text{Volume of } \text{CO}_2 = 1.20\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 28.8\text{ dm}^3$$

(a) $2.0\text{ mol}$ / $2\text{ mol}$ [0.5];
(b) Moles of $\text{CO}_2 = 0.40 \times 3 = 1.20\text{ mol}$ [1.0];
Volume of $\text{CO}_2 = 1.20 \times 24.0 = 28.8\text{ dm}^3$ [1.0];

1. Calculate moles of $\text{H}_2\text{SO}_4$ used:
$$\text{moles of } \text{H}_2\text{SO}_4 = \text{concentration} \times \text{volume in dm}^3 = 0.100\text{ mol/dm}^3 \times \frac{20.0}{1000}\text{ dm}^3 = 0.00200\text{ mol}$$

2. Use the stoichiometric ratio to find moles of $\text{NaOH}$:
$$\text{moles of } \text{NaOH} = 2 \times \text{moles of } \text{H}_2\text{SO}_4 = 2 \times 0.00200 = 0.00400\text{ mol}$$

3. Calculate the concentration of $\text{NaOH}$:
$$\text{concentration} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00400\text{ mol}}{25.0 / 1000\text{ dm}^3} = 0.160\text{ mol/dm}^3$$

- Calculation of moles of $\text{H}_2\text{SO}_4 = 0.00200\text{ mol}$ [0.5];
- Using $2:1$ ratio to deduce moles of $\text{NaOH} = 0.00400\text{ mol}$ [1.0];
- Concentration of $\text{NaOH} = \frac{0.00400}{0.0250} = 0.160\text{ mol/dm}^3$ [1.0];

To calculate the rate of reaction from a tangent to a volume-time graph:

1. Find the change in volume (\(\Delta V\)) and change in time (\(\Delta t\)) using the coordinates of the tangent:
$$\Delta V = 58\text{ cm}^3 - 10\text{ cm}^3 = 48\text{ cm}^3$$
$$\Delta t = 80\text{ s} - 0\text{ s} = 80\text{ s}$$

2. Calculate the gradient:
$$\text{Rate} = \frac{\Delta V}{\Delta t} = \frac{48}{80} = 0.60$$

3. Determine the units: \(\text{cm}^3/\text{s}\) (or \(\text{cm}^3\text{ s}^{-1}\)).

gradient expression / \(\frac{\Delta V}{\Delta t}\) seen using tangent coordinates (e.g. \(\frac{58 - 10}{80 - 0}\) or \(\frac{48}{80}\)) [1];
\(0.60\) (or \(0.6\)) [1];
\(\text{cm}^3/\text{s}\) or \(\text{cm}^3\text{ s}^{-1}\) [1]

1. Description of rate: The rate of reaction is fastest at the start (steepest gradient), slows down/decreases over time, and becomes zero (curve levels off / graph becomes horizontal).
2. Explanation using collision theory: As the reaction proceeds, reactant particles (acid/calcium carbonate) are used up, so their concentration/surface area decreases. This leads to a lower frequency of collisions (fewer collisions per unit time / per second) between reacting particles.
3. Completion: When one of the reactants is completely used up (the limiting reactant, \(\text{CaCO}_3\)), no further collisions can occur, so the reaction stops.

rate decreases / reaction slows down over time (or fastest at the start / stops when horizontal) [1];
concentration of acid / number of reactant particles (per unit volume) decreases [1];
frequency of collisions decreases / fewer collisions per second / per unit time [1];
(Note: Reject 'fewer successful collisions' without reference to rate/time/frequency for mark 3)

(a) At higher temperature (\(45^\circ\text{C}\)), the initial rate of reaction is higher, so the initial gradient of the graph is steeper.

(b) At \(45^\circ\text{C}\):
- Particles have higher kinetic energy and move faster, leading to a higher frequency of collisions (more collisions per second / unit time).
- A greater proportion / fraction of colliding particles have energy equal to or greater than the activation energy (\(E \ge E_a\)), resulting in a higher proportion of successful / effective collisions.

(a) \(45^\circ\text{C}\) [1];
(b) particles have more kinetic energy / move faster leading to higher collision frequency / more collisions per second [1];
greater proportion / fraction of particles have energy greater than or equal to activation energy (or more collisions are successful / effective) [1]

(a) When an object is placed at \(2f\) from a thin converging lens, the rays converge on the opposite side to form an image that is real and inverted (also same size as the object).
(b) The image is located at a distance of \(2f\) on the opposite side of the lens from the object.

(a) Real AND inverted; [1] (A: same size / unmagnified)
(b) At \(2f\) / at twice the focal length (on the opposite side of the lens); [0.5]

(a) For total internal reflection to take place, the angle of incidence \(i\) within the denser medium must be greater than the critical angle \(c\) (\(i > c\)).
(b) Using the relation \(\sin c = \frac{1}{n}\):
\(\sin c = \frac{1}{1.52} \approx 0.6579\)
\(c = \sin^{-1}(0.6579) = 41.14^\circ \approx 41.1^\circ\).

(a) Angle of incidence is greater than the critical angle / \(i > c\); [0.5]
(b) \(\sin c = \frac{1}{n}\) / \(\sin c = \frac{1}{1.52}\) AND \(c = 41.1^\circ\) (accept \(41^\circ\) or \(41.14^\circ\)); [1]

(a) An image formed by a plane mirror is virtual, upright, laterally inverted, same size as the object, and located at the same distance behind the mirror as the object is in front.
(b) A ray directed along the normal strikes the mirror at an angle of incidence of \(0^\circ\), so the angle of reflection is \(0^\circ\) (it reflects back along its original path).

(a) Any two from: virtual; upright; same size / same height (as object); laterally inverted; same distance behind the mirror (as the object is in front);; [1] (1 mark for any two correct points)
(b) \(0^\circ\) / zero degrees; [0.5]

(a) The principal focus of a converging lens is the point on the principal axis where incident rays running parallel to the principal axis converge (meet) after refraction through the lens.
(b) Any ray passing through the optical centre of a thin lens continues in a straight line without bending or deviation.

(a) The point on the principal axis where rays initially parallel to the axis meet / converge after passing through the lens; [1]
(b) Passes straight through / does not deviate / continues undeviated; [0.5]

(a) Glass is optically denser than air. When light travels from air into glass, its speed decreases. Since frequency remains unchanged (\(v = f \lambda\)), its wavelength also decreases.
(b) Because the light slows down upon entering a medium of higher refractive index, the ray refracts (bends) towards the normal.

(a) Speed decreases AND wavelength decreases; [1]
(b) (Bends) towards the normal; [0.5]

Step 1: Calculate the useful gravitational potential energy gained by the crate:
\[\Delta E_p = mgh = 45\text{ kg} \times 9.8\text{ m/s}^2 \times 12\text{ m} = 5292\text{ J}\]

Step 2: Calculate the total electrical energy supplied to the winch:
\[E_{\text{in}} = P \times t = 480\text{ W} \times 15\text{ s} = 7200\text{ J}\]
(Alternatively, useful power output \(P_{\text{out}} = \frac{5292}{15} = 352.8\text{ W}\))

Step 3: Calculate the efficiency:
\[\text{Efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\%\]
\[\text{Efficiency} = \frac{5292}{7200} \times 100\% = 73.5\%\]

1 mark for calculating useful energy output: \(\Delta E_p = mgh = 45 \times 9.8 \times 12 = 5292\text{ J}\) (or useful power \(P_{\text{out}} = 352.8\text{ W}\));
1 mark for calculating total electrical energy input: \(E_{\text{in}} = 480 \times 15 = 7200\text{ J}\) (or power ratio setup);
1 mark for correct evaluation of efficiency: \(73.5\%\) [A: 0.735 / 74%];

Step 1: Calculate the current through each parallel branch (or the total combined resistance):
\[I_1 = \frac{V}{R_1} = \frac{230}{46} = 5.0\text{ A}\]
\[I_2 = \frac{V}{R_2} = \frac{230}{92} = 2.5\text{ A}\]
\[I_{\text{total}} = 5.0 + 2.5 = 7.5\text{ A}\]
(Alternatively: \(R_{\text{total}} = \frac{46 \times 92}{46 + 92} = 30.67\ \Omega\))

Step 2: Calculate the total power of the heater:
\[P = V \times I_{\text{total}} = 230\text{ V} \times 7.5\text{ A} = 1725\text{ W}\]
(or \(P = \frac{V^2}{R_{\text{total}}} = \frac{230^2}{30.67} = 1725\text{ W}\))

Step 3: Convert time to seconds and calculate the total energy transferred:
\[t = 20 \times 60 = 1200\text{ s}\]
\[E = P \times t = 1725\text{ W} \times 1200\text{ s} = 2.07 \times 10^6\text{ J}\ \text{(or } 2\,070\,000\text{ J)}\]

1 mark for finding the total current \(I_{\text{total}} = 7.5\text{ A}\) OR combined resistance \(R = 30.67\ \Omega\) OR individual power values (\(1150\text{ W}\) and \(575\text{ W}\));
1 mark for time conversion \(t = 1200\text{ s}\) and recall of \(E = P t\) (or \(E = V I t\));
1 mark for correct final energy value: \(2.07 \times 10^6\text{ J}\) or \(2\,070\,000\text{ J}\) [A: \(2.07\text{ MJ}\)];

Step 1: Calculate the refractive index \(n\) of the acrylic resin:
\[n = \frac{v_{\text{air}}}{v_{\text{acrylic}}} = \frac{3.0 \times 10^8}{2.0 \times 10^8} = 1.50\]

Step 2: Use Snell's Law to relate the angle of incidence and angle of refraction:
\[n = \frac{\sin i}{\sin r} \implies \sin r = \frac{\sin i}{n}\]
\[\sin r = \frac{\sin(42^\circ)}{1.50} = \frac{0.6691}{1.50} = 0.4461\]

Step 3: Calculate the angle of refraction \(r\):
\[r = \arcsin(0.4461) = 26.5^\circ\]

1 mark for calculating the refractive index \(n = 1.50\) from the ratio of speeds;
1 mark for substituting into Snell's law \(\sin r = \frac{\sin(42^\circ)}{1.50}\) (or \(0.446\));
1 mark for correct angle \(r = 26.5^\circ\) [A: \(26.49^\circ\) to \(26.5^\circ\), accept \(26^\circ\)];

Step 1: Convert orbital radius and orbital period to standard SI units:
\[r = 4.22 \times 10^4\text{ km} = 4.22 \times 10^7\text{ m}\]
\[T = 24.0\text{ hours} = 24.0 \times 3600\text{ s} = 86\,400\text{ s}\]

Step 2: Use the orbital speed formula:
\[v = \frac{2 \pi r}{T}\]
\[v = \frac{2 \times \pi \times 4.22 \times 10^7\text{ m}}{86\,400\text{ s}}\]

Step 3: Evaluate the speed:
\[v = \frac{2.6515 \times 10^8}{86\,400} \approx 3068.9\text{ m/s} \approx 3070\text{ m/s}\]

1 mark for converting radius to metres (\(4.22 \times 10^7\text{ m}\)) AND time to seconds (\(86\,400\text{ s}\));
1 mark for substitution into \(v = \frac{2 \pi r}{T}\);
1 mark for correct final value: \(3070\text{ m/s}\) or \(3069\text{ m/s}\) [A: \(3.07 \times 10^3\text{ m/s}\)];

Step 1: Determine the corrected initial and final count rates by subtracting the background count rate:
\[\text{Corrected initial count rate} = 1944 - 24 = 1920\text{ counts/min}\]
\[\text{Corrected final count rate} = 54 - 24 = 30\text{ counts/min}\]

Step 2: Determine the number of half-lives that have elapsed:
\[\frac{1920}{30} = 64 = 2^6\]
Therefore, 6 half-lives have passed.

Step 3: Calculate the duration of one half-life:
\[T_{1/2} = \frac{\text{Total time}}{\text{Number of half-lives}} = \frac{18\text{ hours}}{6} = 3.0\text{ hours}\]

1 mark for calculating both corrected count rates: \(1920\text{ counts/min}\) and \(30\text{ counts/min}\);
1 mark for determining that \(6\) half-lives have elapsed (e.g. \(1920 \rightarrow 960 \rightarrow 480 \rightarrow 240 \rightarrow 120 \rightarrow 60 \rightarrow 30\));
1 mark for correct half-life: \(3.0\text{ hours}\) [A: \(3\text{ h}\), \(180\text{ minutes}\)];

Method 1 (Work-Energy Theorem):
Step 1: Calculate the initial kinetic energy of the car:
\[E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 1200\text{ kg} \times (25\text{ m/s})^2 = 375\,000\text{ J}\]

Step 2: Relate kinetic energy lost to work done by the braking force:
\[W = F \times d = 375\,000\text{ J}\]

Step 3: Calculate the braking force \(F\):
\[F = \frac{W}{d} = \frac{375\,000\text{ J}}{60\text{ m}} = 6250\text{ N}\]

Method 2 (Kinematics and Newton's Second Law):
Step 1: Calculate acceleration \(a\) using \(v_{\text{final}}^2 = v_{\text{initial}}^2 + 2ad\):
\[0 = 25^2 + 2a(60) \implies a = -\frac{625}{120} = -5.208\text{ m/s}^2\]

Step 2: Calculate force using \(F = ma\):
\[F = 1200 \times 5.208 = 6250\text{ N}\]

1 mark for calculating the initial kinetic energy \(E_k = \frac{1}{2}(1200)(25^2) = 375\,000\text{ J}\) OR for calculating acceleration \(a = 5.21\text{ m/s}^2\);
1 mark for using \(W = F d\) (or \(F = \frac{E_k}{d}\)) OR \(F = ma\);
1 mark for correct final value: \(6250\text{ N}\) [A: \(6.25\text{ kN}\) / \(6.3\text{ kN}\)];

1. Particles in the liquid film have a distribution of different kinetic energies.
2. More energetic / faster-moving particles near the surface overcome intermolecular forces of attraction and escape into the air as gas/vapour.
3. The loss of these high-energy particles decreases the average kinetic energy of the remaining particles.
4. Since temperature is proportional to the average kinetic energy of the particles, the temperature of the water drops, causing thermal energy to conduct from the swimmer's skin into the water.

• Energetic / fast-moving particles / molecules overcome attractive forces and escape from the liquid surface [1];
• Average kinetic energy of the remaining particles / molecules in the water decreases [1];
• Temperature decreases OR thermal energy is transferred from the skin / body to the water [0.5];
[Total: 2.5 marks]