Cambridge IGCSE · thinka-original Practice Paper

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

Thinka Jun 2026 (V3) 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 (V3) Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.

Section 1: Biology

Answer all four structured questions covering Biology topics.

4 Question · 40 marks

Question 1 · structured

10 marks

1 (a) Define the term transpiration. [2]

(b) A student investigated the rate of transpiration in a leafy shoot under different environmental conditions.
Explain how an increase in wind speed affects the rate of transpiration. [2]

(c) Water is transported from the roots to the leaves in xylem vessels.
(i) State two structural features of xylem vessels that adapt them for the transport of water. [2]
(ii) Describe how root hair cells absorb water from the soil by osmosis. [2]

(d) Translocation is the transport of sucrose and amino acids through phloem vessels.
(i) Define the term translocation. [1]
(ii) Name one plant organ that can act as a source and one plant organ that can act as a sink for sucrose. [1]

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Worked solution

(a) Transpiration is the loss of water vapour from the aerial parts of a plant (specifically leaves) caused by the evaporation of water at the surfaces of the mesophyll cells followed by diffusion of water vapour through the stomata.

(b) Increased wind speed moves humid air away from the leaf surface. This maintains a steep concentration gradient of water vapour between the inside of the leaf and the external air, increasing the rate of diffusion.

(c)(i) Xylem vessels are adapted by being hollow continuous tubes (no end walls / dead cells) and having walls reinforced with lignin for structural support.
(ii) The soil water has a higher water potential than the cytoplasm of the root hair cell. Water enters the cell across the partially permeable cell membrane down a water potential gradient by osmosis.

(d)(i) Translocation is the movement of sucrose and amino acids in phloem from a source to a sink.
(ii) Source: mature leaf / photosynthetic tissue. Sink: root / growing shoot / fruit / flower.

Marking scheme

(a) loss of water vapour from (plant) leaves / aerial parts; [1]
evaporation (at mesophyll surface) AND diffusion through stomata; [1]

(b) (wind) removes water vapour / humid air (from outside leaf / around stomata); [1]
maintains / increases concentration gradient (of water vapour); [1]

(c)(i) any two from:
hollow / no cytoplasm / no cell contents / no end walls; [1]
lignified / thick walls (for strength / to prevent collapse); [1]
continuous tube / uninterrupted column; [1]

(ii) soil has a higher water potential than root hair cell (cytoplasm) / ORA; [1]
water moves (in) by osmosis across a partially permeable membrane / down a water potential gradient; [1]

(d)(i) movement of sucrose / amino acids from source to sink; [1]

(ii) source: leaf / photosynthesising tissue AND sink: root / tuber / seed / fruit / growing shoot; [1]

Question 2 · structured

10 marks

2 (a) Mammals have a double circulatory system.
Explain what is meant by a double circulatory system. [2]

(b) (i) Name the chamber of the human heart that pumps oxygenated blood into the aorta. [1]
(ii) Explain why the muscular wall of the left ventricle is thicker than the muscular wall of the right ventricle. [2]

(c) Describe two structural differences between an artery and a vein. [2]

(d) Blood consists of several components.
(i) State the function of white blood cells (phagocytes and lymphocytes). [1]
(ii) State the function of platelets. [1]
(iii) State one substance transported by blood plasma, other than water. [1]

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Worked solution

(a) In a double circulatory system, blood passes through the heart twice during one complete cycle: once to the lungs for gas exchange (pulmonary circulation) and once to the rest of the body (systemic circulation).

(b)(i) The left ventricle pumps blood into the aorta.
(ii) The left ventricle must pump blood under high pressure around the entire body, whereas the right ventricle only pumps blood a short distance to the lungs under lower pressure.

(c) Arteries have a thick muscular wall and narrow lumen to withstand high blood pressure, while veins have a thinner muscular wall, wider lumen, and valves to prevent the backflow of blood under low pressure.

(d)(i) Phagocytes engulf and digest pathogens; lymphocytes produce specific antibodies.
(ii) Platelets are fragments of cells involved in blood clotting and preventing blood loss / entry of pathogens.
(iii) Blood plasma transports dissolved nutrients such as glucose, amino acids, mineral ions, hormones, carbon dioxide, or urea.

Marking scheme

(a) blood passes through heart twice; [1]
for each complete circuit / cycle (pulmonary and systemic); [1]

(b)(i) left ventricle; [1]

(ii) left ventricle pumps blood to the rest of the body / further distance (than right ventricle); [1]
needs to generate higher pressure / overcome higher resistance; [1]

(c) any two from:
artery has thicker (muscular / elastic) wall (vein has thinner wall); [1]
artery has narrower lumen (vein has wider lumen); [1]
vein has valves (artery does not have valves); [1]

(d)(i) ingest / engulf pathogens / produce antibodies / fight infection / defence against disease; [1]

(ii) blood clotting / prevent blood loss / scab formation; [1]

(iii) any one from: glucose / amino acids / carbon dioxide / urea / hormones / mineral ions; [1]
IGNORE oxygen / red blood cells

Question 3 · structured

10 marks

3 (a) State the balanced chemical symbol equation for aerobic respiration. [2]

(b) During intense exercise, muscles may respire anaerobically.
(i) State the word equation for anaerobic respiration in human muscle cells. [1]
(ii) Explain why anaerobic respiration releases much less energy per glucose molecule than aerobic respiration. [2]
(iii) After vigorous exercise, an oxygen debt is created. Describe what happens to the lactic acid produced and explain why breathing rate remains high during recovery. [3]

(c) State two differences between aerobic respiration and breathing (gas exchange). [2]

Show answer & marking scheme

Worked solution

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

(b)(i) Glucose $\rightarrow$ lactic acid.
(ii) Glucose is not completely broken down because oxygen is absent; much of the chemical energy remains stored in the bonds of lactic acid molecules.
(iii) Lactic acid builds up in muscle cells and enters the blood, which transports it to the liver. There, it is oxidised aerobically using oxygen into carbon dioxide and water (or converted back into glucose/glycogen). Breathing rate remains high to take in the extra oxygen needed to pay back the oxygen debt.

(c) Respiration is a chemical reaction that occurs inside mitochondria/cells to release energy from glucose, whereas breathing is a mechanical process involving the movement of ribs, diaphragm, and lungs to ventilate gas exchange surfaces.

Marking scheme

(a) C6H12O6 + 6O2; [1]
-> 6CO2 + 6H2O; [1]

(b)(i) glucose -> lactic acid; [1]

(ii) incomplete breakdown of glucose / glucose is not fully oxidised; [1]
energy remains stored / trapped in bonds of lactic acid; [1]

(iii) lactic acid transported in blood / to the liver; [1]
lactic acid is broken down / oxidised / converted to glucose; [1]
extra oxygen required / repaid so breathing rate remains elevated; [1]

(c) any two from:
respiration is a chemical / metabolic reaction (breathing is physical / mechanical / ventilation); [1]
respiration occurs in all living cells / mitochondria (breathing occurs in lungs / respiratory system); [1]
respiration releases energy / ATP (breathing moves gases into/out of body); [1]

Question 4 · structured

10 marks

4 (a) A food chain in a temperate woodland is shown:
$$\text{Oak tree} \longrightarrow \text{Aphid} \longrightarrow \text{Ladybird} \longrightarrow \text{Robin} \longrightarrow \text{Hawk}$$
(i) Identify the producer and the tertiary consumer in this food chain. [2]
(ii) State the principal source of energy for this food chain. [1]
(iii) Explain why food chains rarely have more than five trophic levels. Use ideas about energy transfer. [2]

(b) Carbon atoms cycle through woodland organisms and the atmosphere.
(i) Name the process by which carbon dioxide is removed from the atmosphere by the oak tree. [1]
(ii) State two processes in the woodland ecosystem that release carbon dioxide into the atmosphere. [2]
(iii) Describe the role of decomposers in recycling carbon. [2]

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Worked solution

(a)(i) Producer = Oak tree (trophic level 1). Tertiary consumer = Robin (trophic level 4).
(ii) The Sun provides light energy for photosynthesis.
(iii) Energy is lost between trophic levels (approximately 90% lost) through metabolic heat, respiration, movement, excretion, and unconsumed parts. After four to five trophic levels, the amount of energy remaining is insufficient to support another viable population.

(b)(i) Photosynthesis removes carbon dioxide from the atmosphere.
(ii) Respiration by organisms and combustion of organic material (e.g. wood) release carbon dioxide.
(iii) Decomposers (bacteria and fungi) secrete enzymes to digest dead organic matter and waste products. They absorb nutrients and carry out cellular respiration, which converts organic carbon compounds into carbon dioxide that is released into the atmosphere.

Marking scheme

(a)(i) producer: oak tree; [1]
tertiary consumer: robin; [1]

(ii) the Sun / sunlight / light; [1]

(iii) energy is lost at / between each trophic level; [1]
any valid reason for energy loss (e.g. respiration / heat loss / excretion / uneaten parts) OR not enough energy left to support another trophic level; [1]

(b)(i) photosynthesis; [1]

(ii) any two from:
respiration (by animals / plants / decomposers); [1]
combustion / burning (of wood / fossil fuels); [1]

(iii) break down / digest dead organic matter / waste / feces; [1]
respire to release carbon dioxide (into atmosphere); [1]

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Practise This Topic

Section 2: Chemistry

Answer all four structured questions covering Chemistry topics. Use the Periodic Table provided.

4 Question · 40 marks

Question 1 · theory

10 marks

(a) A student is provided with a green solid, X, which contains one cation and one anion.

(i) Solid X is dissolved in distilled water. Aqueous sodium hydroxide is added dropwise until in excess. A green precipitate is formed which remains insoluble in excess sodium hydroxide.

Identify the cation present in solid X.

........................................................................................................................ [1]

(ii) To a separate portion of the solution of solid X, dilute nitric acid followed by aqueous barium nitrate is added. A white precipitate forms.

Identify the anion present in solid X.

........................................................................................................................ [1]

(iii) Deduce the chemical formula of solid X.

........................................................................................................................ [1]

(b) Insoluble barium sulfate can be prepared by a precipitation reaction.

(i) Name two soluble salts that can be mixed together in aqueous solution to produce barium sulfate.

Salt 1: ................................................................................................................

Salt 2: ................................................................................................................ [2]

(ii) Describe how a pure, dry sample of barium sulfate is obtained from the reaction mixture.

........................................................................................................................

........................................................................................................................

........................................................................................................................ [3]

(c) Paper chromatography is used to separate and identify coloured substances in a dye mixture.

In a chromatography experiment, a spot of a red dye travels a distance of $3.6\text{ cm}$ from the baseline. The solvent front travels $6.0\text{ cm}$ from the baseline.

Calculate the $R_\text{f}$ value of the red dye.

$R_\text{f}$ value = .......................................................................................... [2]

Show answer & marking scheme

Worked solution

(a)(i) The green precipitate with aqueous sodium hydroxide that is insoluble in excess indicates the presence of iron(II) ions, $\text{Fe}^{2+}$.

(a)(ii) The addition of dilute nitric acid followed by aqueous barium nitrate yielding a white precipitate confirms the presence of sulfate ions, $\text{SO}_4^{2-}$.

(a)(iii) Combining $\text{Fe}^{2+}$ and $\text{SO}_4^{2-}$ gives $\text{FeSO}_4$.

(b)(i) Any pair of soluble barium and sulfate salts, e.g. barium chloride ($\text{BaCl}_2$) or barium nitrate ($\text{Ba(NO}_3)_2$) and sodium sulfate ($\text{Na}_2\text{SO}_4$) or potassium sulfate ($\text{K}_2\text{SO}_4$).

(b)(ii)
1. Filter the mixture to collect the solid precipitate (residue).
2. Wash the solid residue with distilled / deionised water to remove soluble impurities.
3. Dry the residue on filter paper / in a low-temperature oven.

(c) $R_\text{f} = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent front}} = \frac{3.6\text{ cm}}{6.0\text{ cm}} = 0.60$.

Marking scheme

(a)(i) iron(II) / $\text{Fe}^{2+}$; [1]
DO NOT ALLOW iron(III) / $\text{Fe}^{3+}$

(a)(ii) sulfate / $\text{SO}_4^{2-}$; [1]
DO NOT ALLOW sulfite / sulfide

(a)(iii) $\text{FeSO}_4$; [1]
ALLOW ecf from (a)(i) and (a)(ii)

(b)(i) barium chloride / barium nitrate; [1]
sodium sulfate / potassium sulfate / magnesium sulfate / sulfuric acid; [1]

(b)(ii) filter / filtration (to collect the precipitate); [1]
wash the residue / precipitate with (distilled / deionised) water; [1]
dry with filter paper / in an oven / in a desiccator; [1]

(c) $\frac{3.6}{6.0}$; [1]
$0.60$ / $0.6$; [1]

Question 2 · theory

10 marks

(a) Propane, $\text{C}_3\text{H}_8$, burns completely in oxygen according to the equation:

\[\text{C}_3\text{H}_8(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l})\]

(i) Calculate the relative molecular mass, $M_\text{r}$, of propane, $\text{C}_3\text{H}_8$.

[$A_\text{r}$: $\text{C} = 12$, $\text{H} = 1$]

$M_\text{r}$ = .......................................................................................... [1]

(ii) A sample containing $0.20\text{ mol}$ of propane is burned completely in excess oxygen.

Calculate the mass of carbon dioxide, $\text{CO}_2$, produced.

[$A_\text{r}$: $\text{C} = 12$, $\text{O} = 16$]

mass of $\text{CO}_2$ = ..................................................................... $\text{g}$ [2]

(iii) Calculate the volume of oxygen gas, $\text{O}_2$, measured at room temperature and pressure (r.t.p.), required to react completely with $0.20\text{ mol}$ of propane.

[The molar gas volume at r.t.p. is $24\text{ dm}^3$]

volume of $\text{O}_2$ = .................................................................... $\text{dm}^3$ [2]

(b) The combustion of propane is an exothermic reaction.

(i) State what is meant by an exothermic reaction in terms of thermal energy transfer.

........................................................................................................................ [1]

(ii) Explain, in terms of bond breaking and bond forming, why the combustion of propane is exothermic.

........................................................................................................................

........................................................................................................................ [2]

(iii) State the sign of the enthalpy change, $\Delta H$, for this reaction.

........................................................................................................................ [1]

(c) Name one environmental problem caused by increased levels of carbon dioxide in the atmosphere.

........................................................................................................................ [1]

Show answer & marking scheme

Worked solution

(a)(i) $M_\text{r}(\text{C}_3\text{H}_8) = (3 \times 12) + (8 \times 1) = 36 + 8 = 44$.

(a)(ii) From the balanced equation, $1\text{ mol}$ of $\text{C}_3\text{H}_8$ produces $3\text{ mol}$ of $\text{CO}_2$.
$\text{Moles of } \text{CO}_2 = 0.20 \times 3 = 0.60\text{ mol}$.
$M_\text{r}(\text{CO}_2) = 12 + (2 \times 16) = 44\text{ g/mol}$.
$\text{Mass of } \text{CO}_2 = 0.60\text{ mol} \times 44\text{ g/mol} = 26.4\text{ g}$.

(a)(iii) From the equation, $1\text{ mol}$ of $\text{C}_3\text{H}_8$ reacts with $5\text{ mol}$ of $\text{O}_2$.
$\text{Moles of } \text{O}_2 = 0.20 \times 5 = 1.0\text{ mol}$.
$\text{Volume of } \text{O}_2 = 1.0\text{ mol} \times 24\text{ dm}^3\text{/mol} = 24\text{ dm}^3$.

(b)(i) An exothermic reaction transfers thermal energy to the surroundings (causing the temperature of the surroundings to increase).

(b)(ii) Bond breaking is endothermic (requires energy) and bond forming is exothermic (releases energy). More energy is released during the formation of new bonds in the products than is absorbed in breaking bonds in the reactants.

(b)(iii) Negative ($-$).

(c) Enhanced greenhouse effect / global warming / climate change / sea level rise.

Marking scheme

(a)(i) $44$; [1]

(a)(ii) $\text{moles of } \text{CO}_2 = 0.60\text{ (mol)}$ OR $M_\text{r}(\text{CO}_2) = 44$; [1]
$26.4\text{ (g)}$; [1]
ALLOW ecf from incorrect mole value

(a)(iii) $\text{moles of } \text{O}_2 = 1.0\text{ (mol)}$; [1]
$24\text{ (dm}^3\text{)}$; [1]
ALLOW ecf from incorrect moles of $\text{O}_2$

(b)(i) releases / gives out heat / thermal energy (to the surroundings); [1]
ALLOW temperature of the surroundings increases

(b)(ii) energy needed to break bonds is LESS than the energy released when bonds are formed / more energy is released making bonds than taken in breaking bonds; [2]
(Award [1] for stating bond breaking is endothermic / takes in energy AND bond making is exothermic / gives out energy)

(b)(iii) negative / $-$; [1]

(c) enhanced greenhouse effect / global warming / climate change / rising sea levels / ocean acidification; [1]
IGNORE pollution / acid rain

Question 3 · theory

10 marks

(a) A student investigates the rate of reaction between calcium carbonate and excess dilute hydrochloric acid.

\[\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\]

(i) State two different experimental methods by which the rate of this reaction can be monitored over time.

Method 1: .............................................................................................................

Method 2: ............................................................................................................. [2]

(ii) The student repeats the experiment using the same mass of calcium carbonate, but as a fine powder instead of large chips. All other variables are kept constant.

State the effect of using powdered calcium carbonate on the initial rate of reaction.

........................................................................................................................ [1]

(iii) Explain your answer to (a)(ii) using collision theory.

........................................................................................................................

........................................................................................................................ [2]

(b) Explain, using collision theory, why increasing the temperature increases the rate of reaction.

........................................................................................................................

........................................................................................................................

........................................................................................................................ [3]

(c) In a blast furnace, iron(III) oxide reacts with carbon monoxide as shown:

\[\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\]

Identify the substance that is oxidised and explain your answer in terms of oxygen transfer.

Substance oxidised: .........................................................................................

Explanation: .................................................................................................... [2]

Show answer & marking scheme

Worked solution

(a)(i) Two suitable methods are: 1. Measuring the volume of $\text{CO}_2$ gas evolved at regular time intervals using a gas syringe; 2. Measuring the decrease in mass of the flask and reaction mixture over time using an electronic balance.

(a)(ii) Powdering the solid increases the initial rate of reaction.

(a)(iii) Powder has a larger surface area (for the same mass). This increases the area exposed to collisions, resulting in a higher frequency of collisions (more collisions per second) between the reactant particles.

(b) At higher temperatures, particles have greater kinetic energy and move faster. This causes an increase in collision frequency. More importantly, a significantly larger proportion of colliding particles possess energy equal to or greater than the activation energy ($E \ge E_\text{a}$), leading to a higher frequency of successful / effective collisions.

(c) Carbon monoxide ($\text{CO}$) is oxidised because it gains oxygen to become carbon dioxide ($\text{CO}_2$).

Marking scheme

(a)(i) ANY TWO FROM:
- measure the volume of gas / $\text{CO}_2$ produced over time / per unit time; [1]
- measure the loss in mass / decrease in mass over time / per unit time; [1]
- measure change in pH over time; [1]

(a)(ii) increases / reaction is faster; [1]

(a)(iii) (powder has a) larger / greater surface area; [1]
higher frequency of collisions / more collisions per second / more frequent collisions (between particles); [1]
DO NOT ALLOW "more collisions" without reference to time/rate/frequency

(b) particles have more kinetic energy / move faster; [1]
more frequent collisions / higher frequency of collisions; [1]
greater proportion / fraction / percentage of particles have energy $\ge$ activation energy / more collisions are successful / effective; [1]

(c) carbon monoxide / $\text{CO}$; [1]
gains oxygen / receives oxygen; [1]
ALLOW oxidation number of carbon increases from +2 to +4

Question 4 · theory

10 marks

(a) Molten lead(II) bromide, $\text{PbBr}_2$, is electrolysed using inert carbon (graphite) electrodes.

(i) Name the product formed at the cathode (negative electrode) and at the anode (positive electrode).

Cathode: ........................................................................................................

Anode: ............................................................................................................ [2]

(ii) Write an ionic half-equation for the reaction occurring at the cathode.

........................................................................................................................ [1]

(b) Aqueous copper(II) sulfate is electrolysed using inert platinum electrodes.

(i) Describe what is observed at the cathode.

........................................................................................................................ [1]

(ii) State the name of the gas evolved at the anode.

........................................................................................................................ [1]

(c) Steel is an alloy containing iron. When steel is exposed to air and water, it undergoes corrosion (rusting).

(i) State the two substances that must be present for iron to rust.

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

2. ..................................................................................................................... [2]

(ii) Steel sheets can be protected from rusting by coating them with a thin layer of zinc (galvanising).

Explain why zinc continues to protect the steel from rusting even if the zinc layer is scratched and the steel is exposed.

........................................................................................................................

........................................................................................................................ [2]

(iii) State one difference in physical properties between pure iron and steel.

........................................................................................................................ [1]

Show answer & marking scheme

Worked solution

(a)(i) In molten $\text{PbBr}_2$, $\text{Pb}^{2+}$ ions move to the cathode and are reduced to lead. $\text{Br}^-$\ ions move to the anode and are oxidised to bromine.

(a)(ii) At the cathode: $\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}$.

(b)(i) At the cathode, $\text{Cu}^{2+}$ ions are discharged to form metallic copper, which appears as a pink / brown / salmon-pink solid deposit.

(b)(ii) Hydroxide ions ($\text{OH}^-$) from water are discharged at the anode to produce oxygen gas ($\text{O}_2$).

(c)(i) Rusting requires both water (or moisture) and oxygen (or air).

(c)(ii) Zinc is higher in the reactivity series than iron (more reactive). Therefore, zinc acts as a sacrificial metal and oxidises / loses electrons in preference to iron ($\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-$), preventing the iron from rusting even when exposed.

(c)(iii) Steel is harder / stronger / less malleable than pure iron because the different sized carbon atoms disrupt the regular lattice layers, preventing them from sliding easily.

Marking scheme

(a)(i) cathode: lead / $\text{Pb}$; [1]
anode: bromine / $\text{Br}_2$; [1]
DO NOT ALLOW bromide

(a)(ii) $\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}$; [1]
ALLOW multiples

(b)(i) pink / brown / reddish-brown solid / coating / deposit (on electrode); [1]
IGNORE copper formed

(b)(ii) oxygen / $\text{O}_2$; [1]

(c)(i) oxygen / air; [1]
water / moisture; [1]

(c)(ii) zinc is more reactive than iron; [1]
zinc corrodes / oxidises / loses electrons / reacts in preference to iron / sacrificial protection; [1]

(c)(iii) steel is harder / stronger / less malleable / has higher tensile strength (than pure iron); [1]
ALLOW ORA for pure iron

Section 3: Physics

Answer all four structured questions covering Physics topics. Show all working for calculations.

4 Question · 40 marks

Question 1 · structured

10 marks

A robotic cart of mass $45\text{ kg}$ is used to transport materials up a ramp in an automated warehouse.

(a) The cart accelerates uniformly from rest to a speed of $3.6\text{ m/s}$ in a time of $4.0\text{ s}$.

(i) Calculate the acceleration of the cart. [1]

(ii) Calculate the resultant force acting on the cart during this acceleration. [2]

(b) The cart then moves at a constant speed of $3.6\text{ m/s}$ up the ramp for a distance of $18\text{ m}$, reaching a vertical height of $5.4\text{ m}$ above the ground.

(i) Calculate the gain in gravitational potential energy ($\Delta E_p$) of the cart. (gravitational field strength $g = 9.8\text{ N/kg}$) [2]

(ii) During this motion, the electric motor consumes $3200\text{ J}$ of electrical energy. Calculate the efficiency of the cart system in transferring electrical energy to gravitational potential energy. [2]

(c) The cart is fitted with four identical wide rubber wheels. Each wheel has a contact area of $0.0075\text{ m}^2$ with the floor.

(i) Calculate the total weight of the $45\text{ kg}$ cart. [1]

(ii) Calculate the pressure exerted by the cart on the ground when stationary on a horizontal floor. State the unit of your answer. [2]

Show answer & marking scheme

Worked solution

(a)(i) $a = \frac{v - u}{t} = \frac{3.6 - 0}{4.0} = 0.90\text{ m/s}^2$

(a)(ii) $F = m a = 45 \times 0.90 = 40.5\text{ N}$

(b)(i) $\Delta E_p = m g \Delta h = 45 \times 9.8 \times 5.4 = 2381.4\text{ J}$ (accept $2380\text{ J}$ or $2400\text{ J}$)

(b)(ii) $\text{Efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\% = \frac{2381.4}{3200} \times 100\% = 74.4\%$

(c)(i) $W = m g = 45 \times 9.8 = 441\text{ N}$

(c)(ii) Total contact area $A = 4 \times 0.0075 = 0.030\text{ m}^2$.
$p = \frac{F}{A} = \frac{441}{0.030} = 14\,700\text{ N/m}^2$ (or $\text{Pa}$).

Marking scheme

(a)(i) $0.90\text{ m/s}^2$ [1]

(a)(ii) $F = ma$ OR $45 \times 0.90$ [1];
$40.5\text{ N}$ / $41\text{ N}$ [1]

(b)(i) $\Delta E_p = mg\Delta h$ OR $45 \times 9.8 \times 5.4$ [1];
$2380\text{ J}$ / $2381.4\text{ J}$ / $2400\text{ J}$ [1]

(b)(ii) $\frac{2381.4}{3200} \times 100$ (allow ecf from (b)(i)) [1];
$74.4\%$ / $74\%$ (or $0.744$) [1]

(c)(i) $441\text{ N}$ [1]

(c)(ii) Total area $= 4 \times 0.0075 = 0.030\text{ m}^2$ AND $p = \frac{F}{A}$ [1];
$14\,700$ AND unit $\text{N/m}^2$ / $\text{Pa}$ [1]

Question 2 · structured

10 marks

Waves are used extensively in imaging and communication.

(a) A ray of monochromatic light enters a rectangular glass block from air at an angle of incidence of $38^\circ$. The refractive index of the glass is $1.52$.

(i) State the equation linking refractive index $n$, angle of incidence $i$, and angle of refraction $r$. [1]

(ii) Calculate the angle of refraction $r$ inside the glass block. [2]

(b) Fig. 10.1 shows three sections of the electromagnetic spectrum in order of increasing frequency:

$\text{Radio waves} \longrightarrow \mathbf{X} \longrightarrow \text{Infrared} \longrightarrow \text{Visible light} \longrightarrow \text{Ultraviolet} \longrightarrow \mathbf{Y} \longrightarrow \text{Gamma rays}$

(i) Identify electromagnetic radiation $\mathbf{X}$ and electromagnetic radiation $\mathbf{Y}$. [2]

(ii) State one harmful effect of excessive human exposure to ultraviolet radiation. [1]

(iii) State the speed at which all electromagnetic waves travel in a vacuum. [1]

(c) An ultrasound scanner is used to measure the depth of a crack inside a solid aluminium block. A pulse of ultrasound with a frequency of $2.5\text{ MHz}$ ($2.5 \times 10^6\text{ Hz}$) is emitted into the metal.

(i) The speed of ultrasound in aluminium is $6400\text{ m/s}$. Calculate the wavelength of this ultrasound in aluminium. [2]

(ii) The reflected pulse from the crack is detected $1.5 \times 10^{-5}\text{ s}$ after emission. Calculate the depth of the crack beneath the surface. [1]

Show answer & marking scheme

Worked solution

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

(a)(ii) $\sin r = \frac{\sin 38^\circ}{1.52} = \frac{0.61566}{1.52} = 0.4050$
$r = \sin^{-1}(0.4050) = 23.9^\circ$ (or $24^\circ$)

(b)(i) $\mathbf{X}$ = Microwaves; $\mathbf{Y}$ = X-rays

(b)(ii) Sunburn / skin cancer / premature aging of skin / cataracts (eye damage)

(b)(iii) $3.0 \times 10^8\text{ m/s}$

(c)(i) $v = f \lambda \implies \lambda = \frac{v}{f} = \frac{6400}{2.5 \times 10^6} = 2.56 \times 10^{-3}\text{ m}$ (or $2.6 \times 10^{-3}\text{ m}$)

(c)(ii) Total distance $= v \times t = 6400 \times 1.5 \times 10^{-5} = 0.096\text{ m}$
$\text{Depth} = \frac{0.096}{2} = 0.048\text{ m}$ (or $4.8\text{ cm}$)

Marking scheme

(a)(i) $n = \frac{\sin i}{\sin r}$ [1]

(a)(ii) $\sin r = \frac{\sin 38^\circ}{1.52}$ OR $0.405$ [1];
$23.9^\circ$ / $24^\circ$ [1]

(b)(i) $\mathbf{X}$: Microwaves [1];
$\mathbf{Y}$: X-rays [1]

(b)(ii) Skin cancer / sunburn / premature ageing of skin / eye damage / cataracts [1]
(DO NOT ALLOW cell mutation without named consequence / DO NOT ALLOW "danger")

(b)(iii) $3.0 \times 10^8\text{ m/s}$ (ALLOW $3 \times 10^8$) [1]

(c)(i) $\lambda = \frac{v}{f}$ OR $\frac{6400}{2.5 \times 10^6}$ [1];
$2.56 \times 10^{-3}\text{ m}$ / $0.00256\text{ m}$ / $2.6 \times 10^{-3}\text{ m}$ [1]

(c)(ii) $\text{Depth} = \frac{6400 \times 1.5 \times 10^{-5}}{2} = 0.048\text{ m}$ (or $4.8\text{ cm}$) [1]

Question 3 · structured

10 marks

An electric heater circuit and a mains transformer are investigated in a laboratory.

(a) A circuit contains a $12.0\text{ V}$ d.c. power supply connected to a $4.0\,\Omega$ resistor in series with a parallel combination of two identical $6.0\,\Omega$ resistors.

(i) Calculate the combined resistance of the two $6.0\,\Omega$ resistors connected in parallel. [1]

(ii) Show that the total resistance of the whole circuit is $7.0\,\Omega$. [1]

(iii) Calculate the current supplied by the $12.0\text{ V}$ power supply. [2]

(iv) Calculate the power dissipated in the $4.0\,\Omega$ series resistor. [2]

(b) The laboratory power supply uses a step-down transformer connected to a $230\text{ V}$ a.c. mains supply to provide an output voltage of $12.0\text{ V}$.

(i) The primary coil of the transformer has $1150\text{ turns}$. Calculate the number of turns on the secondary coil. [2]

(ii) Assuming the transformer is $100\%$ efficient, calculate the primary current when the secondary current is $3.5\text{ A}$. [2]

Show answer & marking scheme

Worked solution

(a)(i) $\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{6.0} = \frac{2}{6.0} = \frac{1}{3.0} \implies R_p = 3.0\,\Omega$

(a)(ii) $R_{\text{total}} = R_{\text{series}} + R_p = 4.0 + 3.0 = 7.0\,\Omega$

(a)(iii) $I = \frac{V}{R_{\text{total}}} = \frac{12.0}{7.0} = 1.71\text{ A}$ (or $1.7\text{ A}$)

(a)(iv) $P = I^2 R = (1.714)^2 \times 4.0 = 2.939 \times 4.0 = 11.76\text{ W}$ (or $11.8\text{ W}$ / $12\text{ W}$)
(Alternative: $V_4 = I R = 1.714 \times 4.0 = 6.86\text{ V}$, $P = V I = 6.86 \times 1.714 = 11.8\text{ W}$)

(b)(i) $\frac{V_p}{V_s} = \frac{N_p}{N_s} \implies N_s = N_p \times \frac{V_s}{V_p} = 1150 \times \frac{12.0}{230} = 60\text{ turns}$

(b)(ii) $I_p V_p = I_s V_s \implies I_p = \frac{I_s V_s}{V_p} = \frac{3.5 \times 12.0}{230} = \frac{42}{230} = 0.183\text{ A}$ (or $0.18\text{ A}$)

Marking scheme

(a)(i) $3.0\,\Omega$ [1]

(a)(ii) $4.0 + 3.0 = 7.0\,\Omega$ shown clearly [1]

(a)(iii) $I = \frac{V}{R}$ OR $\frac{12.0}{7.0}$ [1];
$1.71\text{ A}$ / $1.7\text{ A}$ [1]

(a)(iv) $P = I^2 R$ OR $P = IV$ OR $P = \frac{V^2}{R}$ [1];
$11.7\text{ W}$ / $11.8\text{ W}$ / $12\text{ W}$ (allow ecf from (a)(iii)) [1]

(b)(i) $\frac{V_p}{V_s} = \frac{N_p}{N_s}$ OR $\frac{230}{12} = \frac{1150}{N_s}$ [1];
$60$ (turns) [1]

(b)(ii) $I_p V_p = I_s V_s$ OR $I_p \times 230 = 3.5 \times 12.0$ [1];
$0.183\text{ A}$ / $0.18\text{ A}$ [1]

Question 4 · structured

10 marks

This question is about nuclear physics and space physics.

(a) Radon-222 ($^{222}_{\phantom{0}86}\text{Rn}$) decays by alpha ($\alpha$) emission to form an isotope of polonium ($\text{Po}$).

(i) State the nature of an alpha particle. [1]

(ii) Complete the nuclear equation for this decay by determining the nucleon number and proton number of polonium and the alpha particle:

\[^{222}_{\phantom{0}86}\text{Rn} \longrightarrow \dots\dots\text{Po} + \dots\dots\alpha\] [2]

(b) A sample of radon-222 has an initial activity of $960\text{ Bq}$. The half-life of radon-222 is $3.8\text{ days}$.

(i) Define the term half-life. [1]

(ii) Calculate the activity of the sample after $11.4\text{ days}$. [2]

(c) Astronomers study planets and stars in our Solar System and beyond.

(i) Mars orbits the Sun in an approximately circular orbit of radius $2.28 \times 10^8\text{ km}$. The orbital period of Mars is $687\text{ days}$ ($5.94 \times 10^7\text{ s}$).
Calculate the orbital speed of Mars in $\text{km/s}$. [2]

(ii) State the name of the process that releases energy in the core of the Sun. [1]

(iii) Light observed from distant galaxies shows redshift. State what redshift tells astronomers about the motion of distant galaxies. [1]

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Worked solution

(a)(i) A helium nucleus OR 2 protons and 2 neutrons.

(a)(ii) Nucleon number of Po $= 222 - 4 = 218$; proton number of Po $= 86 - 2 = 84$.
Alpha particle is $^{4}_{2}\alpha$.
Equation: $^{222}_{\phantom{0}86}\text{Rn} \longrightarrow {^{218}_{\phantom{0}84}\text{Po}} + {^{4}_{2}\alpha}$

(b)(i) The time taken for half the radioactive nuclei (or count rate / activity) of an isotope to decay (or halve).

(b)(ii) Number of half-lives $= \frac{11.4}{3.8} = 3$.
Activity after 1 half-life $= 480\text{ Bq}$;
After 2 half-lives $= 240\text{ Bq}$;
After 3 half-lives $= 120\text{ Bq}$.

(c)(i) $v = \frac{2\pi r}{T} = \frac{2 \times \pi \times 2.28 \times 10^8}{5.94 \times 10^7} = \frac{1.4326 \times 10^9}{5.94 \times 10^7} = 24.1\text{ km/s}$

(c)(ii) (Nuclear) fusion

(c)(iii) The galaxies are moving away from us / receding (indicating expanding universe).

Marking scheme

(a)(i) Helium nucleus / 2 protons and 2 neutrons [1]
(DO NOT ALLOW helium atom)

(a)(ii) $^{4}_{2}\alpha$ [1];
$^{218}_{\phantom{0}84}\text{Po}$ [1]

(b)(i) Time taken for half the (radioactive) nuclei to decay / time for activity (or count rate) to halve [1]

(b)(ii) Evidence of 3 half-lives (e.g. $\frac{11.4}{3.8} = 3$ OR halving 3 times) [1];
$120\text{ Bq}$ [1]

(c)(i) $v = \frac{2\pi r}{T}$ OR $\frac{2 \times \pi \times 2.28 \times 10^8}{5.94 \times 10^7}$ [1];
$24.1\text{ km/s}$ (or $2.41 \times 10^4\text{ m/s}$) [1]

(c)(ii) (Nuclear) fusion [1]
(DO NOT ALLOW fission)

(c)(iii) Moving away (from Earth / observer) / receding [1]

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