Cambridge IGCSE · thinka-original Practice Paper

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

Thinka Jun 2026 (V2) Cambridge IGCSE-Style Mock — Sciences - Co-ordinated (Double) (0654)

120 marks120 mins2026

An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V2) Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.

Section A: Biology

Answer all questions. Show working for calculations.

4 Question · 40 marks

Question 1 · Structured Theory & Data Analysis

10 marks

(a) State the balanced chemical symbol equation for photosynthesis.

[2]

(b) A student investigated the effect of light intensity on the rate of photosynthesis in an aquatic plant, Cabomba caroliniana. The plant was placed in a beaker containing dilute sodium hydrogencarbonate solution at $20^{\circ}\text{C}$. A lamp was positioned at varying distances from the plant, and the volume of oxygen gas produced in $5\text{ minutes}$ was recorded.

The results are shown in Table 1.1.

Table 1.1

| Distance of lamp / $\text{cm}$ | Volume of oxygen collected in $5\text{ min}$ / $\text{cm}^3$ | Rate of oxygen production / $\text{cm}^3\,\text{min}^{-1}$ |
| :--- | :--- | :--- |
| 10 | 4.5 | 0.90 |
| 20 | 3.2 | 0.64 |
| 30 | 1.8 | 0.36 |
| 40 | 0.9 | 0.18 |
| 50 | 0.4 | 0.08 |

(i) Describe the relationship between the distance of the lamp and the rate of oxygen production.

[1]

(ii) Explain why the rate of oxygen production decreases as the distance of the lamp increases.

[2]

(c) The experiment was repeated at $35^{\circ}\text{C}$ with the lamp at a fixed distance of $10\text{ cm}$. The rate of oxygen production increased from $0.90\text{ cm}^3\,\text{min}^{-1}$ to $1.45\text{ cm}^3\,\text{min}^{-1}$.

Explain, in terms of enzymes and kinetic energy, why this increase in temperature caused an increase in the rate of photosynthesis.

[3]

(d) Plants also require mineral ions absorbed from the soil for healthy growth.

(i) Name the mineral ion required to make chlorophyll.

[1]

(ii) State one symptom shown by a plant grown in soil deficient in the mineral ion named in (d)(i).

[1]

Show answer & marking scheme

Worked solution

(a) The balanced chemical equation for photosynthesis is $6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$.

(b)(i) As the distance of the lamp increases, the rate of oxygen production decreases (inverse relationship).

(b)(ii) Light is an essential energy source for photosynthesis. As the lamp is moved further away, light intensity decreases, meaning light becomes a limiting factor, leading to a reduced rate of photosynthesis and less oxygen released.

(c) Raising the temperature from $20^{\circ}\text{C}$ to $35^{\circ}\text{C}$ gives photosynthetic enzymes (and substrate molecules) more kinetic energy. They move faster, leading to a higher frequency of successful collisions between enzyme active sites and substrates, forming more enzyme-substrate complexes per unit time.

(d)(i) Magnesium ions ($\text{Mg}^{2+}$).

(d)(ii) Yellowing of leaves (chlorosis) or poor growth due to reduced chlorophyll.

Marking scheme

(a) $6\text{CO}_2 + 6\text{H}_2\text{O}$ [1] ;
$\rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$ [1] ;

(b)(i) as distance increases, rate of oxygen production / photosynthesis decreases / inversely proportional / AW [1] ;

(b)(ii) increasing distance decreases light intensity [1] ;
light is a limiting factor / less light energy available (for photosynthesis) [1] ;

(c) (higher temperature) increases kinetic energy of molecules / enzymes / substrates [1] ;
molecules move faster / more frequent collisions [1] ;
more successful collisions / more enzyme-substrate complexes formed (per unit time) [1] ;

(d)(i) magnesium / $\text{Mg}^{2+}$ [1] ;
(d)(ii) yellow leaves / yellowing between veins / chlorosis [1] ; (I: stunted growth alone unless linked to yellowing)

Question 2 · Structured Theory & Data Analysis

10 marks

(a) Fig. 2.1 shows the pathway involved in a simple withdrawal reflex when a person touches a hot object.

(i) State the type of neurone that transmits impulses from the receptor to the central nervous system (CNS).

[1]

(ii) State the role of the effector in this reflex action and identify one specific example of an effector involved in withdrawing the hand.

[2]

(iii) Describe how an electrical impulse travels across a synapse between two neurones.

[2]

(b) The skin plays an important role in maintaining a constant body temperature.

(i) Explain how arterioles in the skin and sweat glands respond to a rise in body temperature to prevent overheating.

[3]

(ii) Name the process of maintaining a constant internal environment in the body.

[1]

(c) Adrenaline is a hormone released during stressful situations.

State one effect of adrenaline on the human body.

[1]

Show answer & marking scheme

Worked solution

(a)(i) The sensory neurone carries electrical impulses from receptors to the central nervous system.
(a)(ii) The effector carries out the action/response. In this case, it is a muscle (e.g. biceps/arm muscle) that contracts to pull the hand away.
(a)(iii) When an electrical impulse reaches the end of a neurone (presynaptic knob), chemical neurotransmitters are released from vesicles. They diffuse across the narrow synaptic gap and bind to specific receptor molecules on the membrane of the postsynaptic neurone, initiating a new impulse.
(b)(i) In response to high body temperature:
1. Arterioles supplying skin surface capillaries dilate (vasodilation), allowing more blood to flow close to the skin surface, increasing heat loss by radiation.
2. Sweat glands produce more sweat onto the skin surface; the evaporation of water in sweat removes heat energy from the skin.
(b)(ii) Homeostasis.
(c) Increased heart rate / dilation of pupils / increased rate of breathing / increased blood glucose concentration.

Marking scheme

(a)(i) sensory (neurone) [1] ;

(a)(ii) carries out the response / converts impulse into action [1] ;
muscle / named muscle (e.g. biceps) / gland [1] ;

(a)(iii) neurotransmitter / chemical released (from presynaptic neurone) [1] ;
diffuses across the synaptic gap / cleft [1] ;
binds to receptors on the next neurone [1] ; (max 2)

(b)(i) arterioles dilate / widen / vasodilation [1] ;
more blood flows to (surface) capillaries / near skin surface (so more heat lost by radiation) [1] ;
sweat produced / secreted AND evaporates from skin removing heat [1] ;

(b)(ii) homeostasis [1] ;

(c) any one from: increases heart rate / increases breathing rate / dilates pupils / increases blood glucose / stimulates breakdown of glycogen / diversion of blood to muscles [1] ;

Question 3 · Structured Theory & Data Analysis

10 marks

(a) Flowering plants can reproduce sexually through pollination.

Table 3.1 compares the features of insect-pollinated flowers and wind-pollinated flowers.

Complete Table 3.1 by describing the differences in the structures listed.

Table 3.1

| Feature | Insect-pollinated flower | Wind-pollinated flower |
| :--- | :--- | :--- |
| Petals | Large, brightly coloured | ............................................................ |
| Pollen grains | ............................................................ | Light, smooth, produced in large numbers |
| Stigma | Small, sticky, enclosed inside petals | ............................................................ |

[3]

(b) State three environmental conditions required for the successful germination of seeds.

1. .........................................................................................................

2. .........................................................................................................

3. .........................................................................................................

[3]

(c) In human reproduction, gametes are specialised for their functions.

(i) State whether a human sperm cell is haploid or diploid.

[1]

(ii) Describe two structural adaptations of a human sperm cell and explain how each adaptation helps it to carry out its function.

[2]

(iii) State the name of the organ in the female reproductive system where fertilisation typically takes place.

[1]

Show answer & marking scheme

Worked solution

(a) In wind-pollinated flowers: petals are small, dull (often green) or absent; pollen grains of insect-pollinated flowers are sticky, spiky, and relatively large to adhere to insects; stigmas of wind-pollinated flowers are feathery/branched and protrude outside the flower to catch airborne pollen.
(b) The three conditions needed for germination are: water (to activate enzymes and swell the seed), oxygen (for aerobic respiration to produce ATP for growth), and a suitable temperature (warmth, to ensure enzyme activity at optimum rates).
(c)(i) A sperm cell contains half the set of chromosomes (23 in humans), so it is haploid.
(c)(ii) Adaptations of a sperm cell:
- Flagellum (tail): beats/moves to swim towards the egg.
- Acrosome at the tip of the head: contains digestive enzymes to digest the outer layer (zona pellucida) of the egg.
- High density of mitochondria in the midpiece: releases energy from aerobic respiration for movement.
(c)(iii) Fertilisation takes place in the oviduct (Fallopian tube).

Marking scheme

(a) (petals): small / dull / green / absent [1] ;
(pollen grains): large / sticky / spiky / heavy [1] ;
(stigma): feathery / branched / hangs outside (the flower) [1] ;

(b) water / moisture [1] ;
oxygen [1] ;
suitable temperature / warmth [1] ; (R: sunlight / light / soil)

(c)(i) haploid [1] ;

(c)(ii) any two pairs (adaptation + function) [2] :
- flagellum / tail [1] AND enables swimming / movement towards egg [1] ;
- acrosome / enzymes in head [1] AND digests outer coat of egg / allows entry into egg [1] ;
- many mitochondria / midpiece packed with mitochondria [1] AND provides energy / ATP (for swimming) [1] ;

(c)(iii) oviduct / Fallopian tube [1] ; (R: uterus / ovary)

Question 4 · Structured Theory & Data Analysis

10 marks

(a) Fig. 4.1 shows a food web in a marine ecosystem.

$$\text{Microscopic algae (phytoplankton)} \longrightarrow \text{Krill} \longrightarrow \text{Mackerel} \longrightarrow \text{Tuna} \longrightarrow \text{Shark}$$
$$\text{Microscopic algae (phytoplankton)} \longrightarrow \text{Krill} \longrightarrow \text{Baleen whale}$$

(i) Identify the producer in this food web.

[1]

(ii) State the trophic level of the mackerel in the food chain shown above.

[1]

(iii) Name the principal source of energy for this entire ecosystem.

[1]

(b) The energy content of the organisms at each trophic level was measured in a section of this marine ecosystem.

- Energy stored in phytoplankton = $120\,000\text{ kJ}$
- Energy transferred to krill = $14\,400\text{ kJ}$

(i) Calculate the percentage efficiency of energy transfer from the phytoplankton to the krill. Show your working.

Percentage efficiency = ........................................................ %

[2]

(ii) Explain why food chains rarely contain more than five trophic levels.

[3]

(c) Carbon is cycled continuously between living organisms and the environment.

(i) State the name of the biological process carried out by both marine animals and phytoplankton that releases carbon dioxide into the water.

[1]

(ii) State one way in which human activity disrupts the global carbon cycle, leading to an increased concentration of atmospheric carbon dioxide.

[1]

Show answer & marking scheme

Worked solution

(a)(i) The producer is microscopic algae (phytoplankton) as it synthesises organic molecules using light energy.
(a)(ii) Trophic level 1: Phytoplankton, Trophic level 2: Krill, Trophic level 3: Mackerel. Thus, mackerel is at the third trophic level (or is a secondary consumer).
(a)(iii) The Sun (light energy).

(b)(i) $\text{Efficiency} = \left(\frac{14\,400}{120\,000}\right) \times 100 = 12\%$.

(b)(ii) Energy is lost at each trophic level (around 90% is lost). Energy is lost via:
1. Metabolic heat released during respiration.
2. Excretion and egestion (faeces/urine).
3. Uneaten or indigestible body parts.
Because of these cumulative losses, there is insufficient energy remaining after 4–5 trophic levels to sustain a viable population of top predators.

(c)(i) Aerobic respiration.
(c)(ii) Burning/combustion of fossil fuels (coal, oil, natural gas) or deforestation (cutting down trees which reduces carbon dioxide absorption).

Marking scheme

(a)(i) microscopic algae / phytoplankton [1] ;

(a)(ii) third (trophic level) / 3 / secondary consumer [1] ;

(a)(iii) the Sun / sunlight / solar radiation / light [1] ;

(b)(i) $\frac{14\,400}{120\,000} \times 100$ [1] ;
$12$ (%) [1] ;

(b)(ii) any three from:
- energy is lost at each trophic level / only $\approx 10\%$ transferred [1] ;
- energy lost as heat (from respiration) [1] ;
- energy lost through excretion / egestion / urine / faeces [1] ;
- not all parts of organisms are eaten / digested [1] ;
- insufficient energy left to support a viable population at higher levels [1] ;

(c)(i) (aerobic) respiration [1] ;

(c)(ii) combustion / burning of fossil fuels OR deforestation / burning forests [1] ;

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Practise This Topic

Section B: Chemistry

Answer all questions. Periodic Table provided on page 32.

4 Question · 40 marks

Question 1 · Structured Chemical Reactions & Theory

10 marks

A student investigates the rate of reaction between excess zinc granules and dilute hydrochloric acid.

The equation for the reaction is:
\[\text{Zn(s)} + 2\text{HCl(aq)} \rightarrow \text{ZnCl}_2\text{(aq)} + \text{H}_2\text{(g)}\]

(a) State the name of the salt formed in this reaction. [1]

(b) The student measures the volume of hydrogen gas collected every 30 seconds for 4 minutes.
(i) Name one piece of apparatus suitable for collecting and measuring the volume of hydrogen gas produced. [1]
(ii) State what happens to the rate of reaction as the reaction proceeds. [1]
(iii) Explain your answer to (b)(ii) in terms of collisions between reacting particles. [2]

(c) The experiment is repeated using the same mass of zinc powder instead of zinc granules. All other conditions are kept constant.
(i) State the effect of using zinc powder on the initial rate of reaction. [1]
(ii) Explain your answer to (c)(i) in terms of collision theory. [2]

(d) In a separate reaction, solid zinc reacts with copper(II) oxide at high temperature:
\[\text{Zn(s)} + \text{CuO(s)} \rightarrow \text{ZnO(s)} + \text{Cu(s)}\]
Identify the substance that is reduced in this reaction, and explain your answer in terms of electron transfer. [2]

Show answer & marking scheme

Worked solution

(a) The salt produced from zinc and hydrochloric acid is zinc chloride, $\text{ZnCl}_2$.

(b)(i) A gas syringe or an inverted measuring cylinder filled with water is used to collect and measure gas volume.
(ii) As reactants are used up, the rate decreases until it reaches zero when the reaction is complete.
(iii) Hydrochloric acid particles (or $\text{H}^+$ ions) are consumed during the reaction, so their concentration decreases. Consequently, there are fewer particles per unit volume, which reduces the frequency of collisions between reacting particles.

(c)(i) Using powder instead of large granules increases the initial rate of reaction.
(ii) Zinc powder has a much greater surface area exposed to the acid. This allows more reactant particles to collide at any instant, increasing the frequency of effective/successful collisions.

(d) $\text{CuO}$ (or $\text{Cu}^{2+}$) is reduced because each $\text{Cu}^{2+}$ ion gains two electrons to form a neutral copper atom ($\text{Cu}$). Reduction is the gain of electrons.

Marking scheme

(a) zinc chloride [1]
(b)(i) gas syringe / (graduated) measuring cylinder (inverted over water) / burette [1] (I: delivery tube alone)
(b)(ii) (rate) decreases / slows down / drops to zero [1]
(b)(iii) concentration of acid / $\text{H}^+$ ions decreases / fewer acid particles per unit volume [1]; lower collision frequency / fewer collisions per second / per unit time [1] (R: fewer collisions without time reference)
(c)(i) (rate) increases / faster [1]
(c)(ii) larger / greater surface area [1]; greater collision frequency / more collisions per second / per unit time [1]
(d) copper(II) oxide / $\text{CuO}$ / $\text{Cu}^{2+}$ [1]; gains electrons [1]

Question 2 · Structured Chemical Reactions & Theory

10 marks

Petroleum is a naturally occurring mixture of hydrocarbons.

(a) Fractional distillation is used to separate petroleum into useful fractions.
(i) State the physical property of hydrocarbons that allows fractional distillation to work. [1]
(ii) Name the fraction used as fuel for diesel engines. [1]

(b) Long-chain alkanes are cracked to produce shorter-chain molecules.
(i) State the conditions required for catalytic cracking. [2]
(ii) Complete the chemical equation for the cracking of decane, $\text{C}_{10}\text{H}_{22}$, to produce propene and one other hydrocarbon:
\[\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_3\text{H}_6 + \dots\dots\dots\dots\dots\] [1]

(c) Propene, $\text{C}_3\text{H}_6$, is an unsaturated hydrocarbon.
(i) Describe a chemical test to distinguish between propene and propane. State the test and the result for each compound. [2]
(ii) Propene undergoes polymerisation to form poly(propene).
- Name the type of polymerisation that occurs. [1]
- Draw the structure of the repeat unit of poly(propene). [2]

Show answer & marking scheme

Worked solution

(a)(i) Fractional distillation separates components based on differences in boiling points.
(ii) The fraction used in diesel engines is diesel oil (or gas oil).

(b)(i) Catalytic cracking requires high temperatures (typically $450\text{ }^\circ\text{C}$ to $750\text{ }^\circ\text{C}$) and a catalyst (such as silica / aluminium oxide / zeolite).
(ii) Subtracting $\text{C}_3\text{H}_6$ from $\text{C}_{10}\text{H}_{22}$:
Carbon atoms: $10 - 3 = 7$
Hydrogen atoms: $22 - 6 = 16$
The product is heptane, $\text{C}_7\text{H}_{16}$.

(c)(i) Aqueous bromine water is added. Propene reacts with bromine across the double bond, turning the solution from orange-brown to colourless. Propane does not react, so the solution remains orange-brown.
(ii) Propene undergoes addition polymerisation. In the repeat unit, the double bond becomes a single bond, with continuation bonds extending on both sides: $-\text{CH}(\text{CH}_3)-\text{CH}_2-$.

Marking scheme

(a)(i) (different) boiling point(s) [1]
(a)(ii) diesel (oil) / gas oil [1]
(b)(i) high temperature / $450\text{--}750\text{ }^\circ\text{C}$ [1]; catalyst / named catalyst (e.g. alumina / silica / zeolite) [1]
(b)(ii) $\text{C}_7\text{H}_{16}$ [1]
(c)(i) add (aqueous) bromine / bromine water [1]; propene: turns colourless / decolourises AND propane: remains orange / brown / no change [1]
(c)(ii) addition (polymerisation) [1];
correct repeat unit structure showing a two-carbon backbone with single $\text{C}-\text{C}$ bond, one $-\text{CH}_3$ group attached to one carbon, three $-\text{H}$ atoms, and open extension bonds extending beyond the brackets [2] (1 mark for correct atoms attached to a $\text{C}-\text{C}$ single bond, 1 mark for correct continuation bonds)

Question 3 · Structured Chemical Reactions & Theory

10 marks

Hydrated copper(II) sulfate crystals, $\text{CuSO}_4\cdot 5\text{H}_2\text{O}$, can be prepared by reacting dilute sulfuric acid with insoluble copper(II) oxide.

(a) Write the balanced chemical equation for the reaction between copper(II) oxide and dilute sulfuric acid. [1]

(b) State why excess copper(II) oxide is added to the acid during the preparation. [1]

(c) Describe the practical procedure for preparing pure, dry crystals of hydrated copper(II) sulfate starting from the mixture of dilute sulfuric acid and excess copper(II) oxide. [3]

(d) A student tests an aqueous solution of the prepared copper(II) sulfate.
(i) Describe the observations when aqueous sodium hydroxide is added dropwise and then in excess to the copper(II) sulfate solution. [2]
(ii) Describe the test to confirm the presence of sulfate ions in the solution, including the reagents used and the positive observation. [2]

(e) Copper(II) oxide is a basic oxide. State what is meant by the term basic oxide in terms of its reaction with acids. [1]

Show answer & marking scheme

Worked solution

(a) $\text{CuO(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{CuSO}_4\text{(aq)} + \text{H}_2\text{O(l)}$

(b) Excess insoluble base is used to ensure that all the acid has reacted, so the resulting solution contains only copper(II) sulfate and water.

(c) The procedure is:
1. Filter the mixture to remove unreacted solid copper(II) oxide.
2. Heat the filtrate in an evaporating basin until the crystallisation point / saturation point is reached.
3. Allow the hot saturated solution to cool so crystals form, filter off the crystals, and dry them between sheets of filter paper.

(d)(i) Adding aqueous $\text{NaOH}$ to $\text{Cu}^{2+}$ ions produces a light blue precipitate of copper(II) hydroxide, $\text{Cu(OH)}_2$. This precipitate does not dissolve when excess sodium hydroxide is added.
(ii) Acidify the sample with dilute nitric acid (to eliminate any carbonate ions), then add aqueous barium nitrate (or acidify with $\text{HCl}$ and add $\text{BaCl}_2$). A white precipitate of barium sulfate, $\text{BaSO}_4$, confirms the presence of sulfate ions.

(e) A basic oxide is a metallic oxide that reacts with an acid to form a salt and water only.

Marking scheme

(a) $\text{CuO} + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O}$ [1]
(b) to ensure all the acid is neutralised / used up / reacts [1] (I: to make sure reaction goes to completion)
(c) filter (to remove excess / unreacted $\text{CuO}$) [1]; heat / evaporate filtrate to crystallisation point / saturate solution [1]; cool and dry crystals on filter paper / in a desiccator / warm oven [1]
(d)(i) light blue precipitate [1]; insoluble in excess [1]
(d)(ii) add dilute nitric acid (or dilute hydrochloric acid) AND aqueous barium nitrate (or aqueous barium chloride) [1]; white precipitate [1]
(e) reacts with / neutralises acid to form a salt and water [1]

Question 4 · Structured Chemical Reactions & Theory

10 marks

Electrolysis is the breakdown of an ionic compound, molten or in aqueous solution, by the passage of electricity.

(a) Molten lead(II) bromide is electrolysed using inert carbon electrodes.
(i) State why solid lead(II) bromide cannot conduct electricity, whereas molten lead(II) bromide can. [2]
(ii) Name the product formed at the cathode (negative electrode) and the product formed at the anode (positive electrode). [2]
(iii) Write the ionic half-equation for the reaction that occurs at the cathode. [1]

(b) An aqueous solution of copper(II) sulfate is electrolysed using inert carbon electrodes.
(i) State the observation made at the anode during this electrolysis. [1]
(ii) Name the substance formed at the anode. [1]

(c) Hydrogen-oxygen fuel cells are used to produce electrical energy in some vehicles.
(i) State the chemical equation for the overall reaction occurring in a hydrogen-oxygen fuel cell. [1]
(ii) Give two advantages of using a hydrogen-oxygen fuel cell compared with an internal combustion engine using petrol. [2]

Show answer & marking scheme

Worked solution

(a)(i) Solid lead(II) bromide has a giant ionic lattice where ions are fixed in place by strong electrostatic forces and are not free to move. When molten, the lattice breaks down and the ions ($\text{Pb}^{2+}$ and $\text{Br}^-$) become free to move and carry electrical current.
(ii) At the cathode, $\text{Pb}^{2+}$ ions gain electrons to form lead metal. At the anode, $\text{Br}^-$ ions lose electrons to form bromine vapour/liquid.
(iii) Cathode half-equation: $\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}$.

(b)(i) At the anode, hydroxide ions from water are oxidised, producing oxygen gas, which is seen as bubbling/effervescence.
(ii) The gas produced at the anode is oxygen ($\text{O}_2$).

(c)(i) The overall reaction is the combination of hydrogen and oxygen to produce water: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$.
(ii) Key advantages include: only water is produced (no carbon dioxide or greenhouse gas emissions / no toxic pollutants such as $\text{CO}$ or $\text{NO}_x$), and fuel cells are more efficient with fewer moving parts.

Marking scheme

(a)(i) in solid: ions are in fixed positions / cannot move [1]; in liquid / molten: ions are free to move [1] (R: electrons move)
(a)(ii) cathode: lead [1]; anode: bromine [1]
(a)(iii) $\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}$ [1]
(b)(i) bubbles / effervescence / gas given off [1]
(b)(ii) oxygen / $\text{O}_2$ [1]
(c)(i) $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ [1]
(c)(ii) any two from:
- only produces water / no greenhouse gases / no $\text{CO}_2$ / no carbon emissions [1]
- no toxic pollutant gases / no $\text{CO}$ / no $\text{NO}_x$ [1]
- higher efficiency / less energy lost as heat [1]
- quiet operation / fewer moving parts [1]

Section C: Physics

Answer all questions. Show full working and state appropriate units.

4 Question · 40 marks

Question 1 · theory

10 marks

A motorized research cart of mass $450\text{ kg}$ is tested along a straight horizontal track.

(a) The cart accelerates uniformly from rest to a speed of $18\text{ m/s}$ in a time of $6.0\text{ s}$.
(i) Calculate the acceleration of the cart. [2]
(ii) Calculate the resultant force acting on the cart during this acceleration. [2]

(b) After $6.0\text{ s}$, the cart travels at a constant speed of $18\text{ m/s}$ for a further $14.0\text{ s}$.
(i) Calculate the kinetic energy of the cart while travelling at this constant speed. [2]
(ii) Calculate the total distance travelled by the cart during the entire $20.0\text{ s}$ journey. [2]

(c) The cart then drives up a smooth slope to a platform at a vertical height of $4.5\text{ m}$ above the track. Calculate the gain in gravitational potential energy of the cart. (Take $g = 9.8\text{ N/kg}$) [2]

Show answer & marking scheme

Worked solution

(a)(i) Acceleration is calculated using $a = \frac{\Delta v}{\Delta t} = \frac{18 - 0}{6.0} = 3.0\text{ m/s}^2$.

(a)(ii) Resultant force is found using Newton's second law: $F = ma = 450 \times 3.0 = 1350\text{ N}$.

(b)(i) Kinetic energy: $E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 450 \times (18)^2 = 0.5 \times 450 \times 324 = 72900\text{ J} = 72.9\text{ kJ}$.

(b)(ii) Distance during acceleration phase: $d_1 = \text{average speed} \times t = \frac{18}{2} \times 6.0 = 54\text{ m}$.
Distance during constant speed phase: $d_2 = 18 \times 14.0 = 252\text{ m}$.
Total distance: $d_{\text{total}} = 54 + 252 = 306\text{ m}$.

(c) Gravitational potential energy gain: $\Delta E_p = mgh = 450 \times 9.8 \times 4.5 = 19845\text{ J} \approx 19800\text{ J}$ (or $19.8\text{ kJ}$).

Marking scheme

(a)(i)
- $a = \frac{\Delta v}{\Delta t}$ or $\frac{18}{6.0}$; [1]
- $3.0\text{ m/s}^2$; [1]

(a)(ii)
- $F = ma$ or $450 \times 3.0$ (ecf from (a)(i)); [1]
- $1350\text{ N}$ (or $1.35 \times 10^3\text{ N}$); [1]

(b)(i)
- $E_k = \frac{1}{2}mv^2$ or $0.5 \times 450 \times 18^2$; [1]
- $72900\text{ J}$ / $72.9\text{ kJ}$; [1]

(b)(ii)
- Distance in first stage $54\text{ m}$ OR distance in second stage $252\text{ m}$ (or correct area calculation from speed-time graph); [1]
- $306\text{ m}$; [1]

(c)
- $\Delta E_p = mgh$ or $450 \times 9.8 \times 4.5$; [1]
- $19800\text{ J}$ / $19.8\text{ kJ}$ (accept $19845\text{ J}$); [1]

Question 2 · theory

10 marks

This question is about waves and optics.

(a) A mechanical wave generator produces ripples on the surface of water in a ripple tank. The frequency of the wave source is $12\text{ Hz}$ and the wavelength of the ripples is $0.035\text{ m}$.
(i) Calculate the speed of the water ripples. [2]
(ii) State whether water waves are transverse or longitudinal. Explain your answer in terms of the direction of particle oscillation and wave travel. [2]

(b) A ray of monochromatic light travels through air and strikes the flat surface of a transparent acrylic block at an angle of incidence of $42^\circ$. The refractive index of acrylic is $1.48$.
(i) State the formula relating refractive index $n$, angle of incidence $i$, and angle of refraction $r$. [1]
(ii) Calculate the angle of refraction $r$ of the light ray inside the acrylic block. [2]
(iii) Describe what happens to the speed, frequency, and wavelength of the light wave as it passes from air into the acrylic block. [3]

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Worked solution

(a)(i) Using the wave equation $v = f \lambda = 12 \times 0.035 = 0.42\text{ m/s}$.

(a)(ii) Water surface waves are transverse because the particles of the medium oscillate up and down perpendicular (at right angles) to the direction of energy transfer / wave propagation.

(b)(i) $n = \frac{\sin i}{\sin r}$.

(b)(ii) $\sin r = \frac{\sin i}{n} = \frac{\sin 42^\circ}{1.48} = \frac{0.6691}{1.48} \approx 0.4521$.
$r = \sin^{-1}(0.4521) = 26.9^\circ$ (or $27^\circ$).

(b)(iii) Speed decreases because the medium is optically denser; frequency remains unchanged / constant; wavelength decreases proportionally with speed (since $v = f \lambda$).

Marking scheme

(a)(i)
- $v = f \lambda$ or $12 \times 0.035$; [1]
- $0.42\text{ m/s}$; [1]

(a)(ii)
- Transverse; [1]
- Particle oscillations / vibrations are perpendicular / at right angles to the direction of wave propagation / energy transfer; [1]

(b)(i)
- $n = \frac{\sin i}{\sin r}$; [1]

(b)(ii)
- $\sin r = \frac{\sin 42^\circ}{1.48}$ OR $\sin r = 0.452$; [1]
- $26.9^\circ$ (accept $27^\circ$); [1]

(b)(iii)
- Speed decreases; [1]
- Frequency remains constant / unchanged; [1]
- Wavelength decreases; [1]

Question 3 · theory

10 marks

An electric circuit consists of a $24\text{ V}$ d.c. power supply connected to two heating resistors, $R_1 = 12\,\Omega$ and $R_2 = 24\,\Omega$, arranged in parallel.

(a) Calculate the combined equivalent resistance of the two resistors connected in parallel. [2]

(b) Calculate the total electric current supplied by the $24\text{ V}$ source. [2]

(c) Calculate the electrical power dissipated in resistor $R_1$. [2]

(d) The circuit is operated continuously for $15\text{ minutes}$. Calculate the total electrical energy transferred by the power supply in joules. [2]

(e) A fuse is included in the circuit near the power supply.
(i) State the function of the fuse. [1]
(ii) Explain how the fuse protects the circuit when a fault causes an excessive current. [1]

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Worked solution

(a) For resistors in parallel: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8.0}$.
Therefore, $R_p = 8.0\,\Omega$.

(b) Total current $I_{\text{total}} = \frac{V}{R_p} = \frac{24}{8.0} = 3.0\text{ A}$.

(c) The voltage across $R_1$ is $24\text{ V}$.
$P_1 = \frac{V^2}{R_1} = \frac{24^2}{12} = \frac{576}{12} = 48\text{ W}$ (or $I_1 = \frac{24}{12} = 2.0\text{ A}$, $P = IV = 2.0 \times 24 = 48\text{ W}$).

(d) Total power $P_{\text{total}} = V \times I_{\text{total}} = 24 \times 3.0 = 72\text{ W}$.
Time $t = 15 \times 60 = 900\text{ s}$.
Total energy $E = P_{\text{total}} \times t = 72 \times 900 = 64800\text{ J} = 64.8\text{ kJ}$.

(e)(i) The fuse prevents excessive current from flowing through the circuit to prevent overheating / fire.
(e)(ii) When the current exceeds the fuse's rated value, the thin fuse wire heats up, melts, and breaks the circuit, stopping the current.

Marking scheme

(a)
- $\frac{1}{R} = \frac{1}{12} + \frac{1}{24}$ or $R = \frac{12 \times 24}{12 + 24}$; [1]
- $8.0\,\Omega$; [1]

(b)
- $I = \frac{V}{R}$ or $\frac{24}{8.0}$ (ecf from (a)); [1]
- $3.0\text{ A}$; [1]

(c)
- $P = \frac{V^2}{R}$ OR $I_1 = \frac{24}{12} = 2.0\text{ A}$ and $P = IV$ / $I^2R$; [1]
- $48\text{ W}$; [1]

(d)
- Conversion of time to seconds ($15 \times 60 = 900\text{ s}$) AND $E = Pt$ / $E = IVt$; [1]
- $64800\text{ J}$ / $64.8\text{ kJ}$; [1]

(e)(i)
- Protects circuit / appliance from excessive current / prevents overheating / fire risk; [1]

(e)(ii)
- (Excessive current causes the) wire to melt, breaking the circuit; [1]

Question 4 · theory

10 marks

This question is about space physics and nuclear processes in stars.

(a) A planet orbits a star in an approximately circular path of radius $1.50 \times 10^{11}\text{ m}$. The orbital period of the planet is $3.15 \times 10^7\text{ s}$.
(i) Calculate the orbital speed of the planet. Give your answer to 3 significant figures. [2]
(ii) State the name of the force that keeps the planet in circular orbit around the star. [1]

(b) The star produces huge amounts of energy in its core through nuclear fusion.
(i) Describe what occurs during nuclear fusion in the core of a star. [2]
(ii) State the name of the galaxy that contains our Solar System. [1]

(c) Light from distant galaxies displays redshift.
(i) State what is meant by redshift. [2]
(ii) Explain how the observation of redshift from distant galaxies provides evidence for the Big Bang theory. [2]

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Worked solution

(a)(i) Orbital speed is calculated using $v = \frac{2 \pi r}{T} = \frac{2 \times \pi \times 1.50 \times 10^{11}}{3.15 \times 10^7} = \frac{9.4248 \times 10^{11}}{3.15 \times 10^7} \approx 2.99 \times 10^4\text{ m/s}$ (or $29900\text{ m/s}$).

(a)(ii) Gravitational force (or gravity) between the planet and the star.

(b)(i) During nuclear fusion, light atomic nuclei (such as hydrogen nuclei/protons) collide and join/fuse together under extreme temperature and pressure to form heavier nuclei (such as helium), releasing energy in the process.

(b)(ii) The Milky Way.

(c)(i) Redshift is the observed increase in the wavelength (shift towards the red end of the spectrum) of electromagnetic radiation / light emitted by a star or galaxy moving away from the observer.

(c)(ii) Observations show that more distant galaxies have greater redshift, indicating they are moving away faster. This demonstrates that space itself is expanding in all directions, supporting the model that the Universe expanded outward from a single extremely dense and hot initial point (the Big Bang).

Marking scheme

(a)(i)
- $v = \frac{2 \pi r}{T}$ or $\frac{2 \times \pi \times 1.50 \times 10^{11}}{3.15 \times 10^7}$; [1]
- $2.99 \times 10^4\text{ m/s}$ (or $29900\text{ m/s}$); [1]

(a)(ii)
- Gravitational force / gravity; [1]

(b)(i)
- Two (smaller / lighter) nuclei join / combine; [1]
- to form a single larger / heavier nucleus (with release of energy); [1]

(b)(ii)
- Milky Way; [1]

(c)(i)
- Increase in (observed) wavelength / decrease in frequency of light; [1]
- from objects / galaxies moving away from the observer (shifted towards red end of spectrum); [1]

(c)(ii)
- More distant galaxies show greater redshift / are moving away faster; [1]
- Shows the Universe is expanding (from a single hot dense origin); [1]

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