An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V3) Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.
部分 Theory (Extended)
Answer all questions. Use a black or dark blue pen. You may use a calculator. Show all your working and use appropriate units.
12 题目 · 120 分
题目 1 · structured
10 分
A cargo drone delivers a parcel. Its flight is divided into three sections: vertical take-off with constant acceleration, horizontal flight at constant speed, and deceleration to land.
(a) The drone starts from rest and accelerates vertically upwards. After 5.0 s, it reaches a speed of 12 m/s. (i) Calculate the acceleration of the drone during this vertical ascent. State the unit. [2] (ii) Calculate the vertical distance travelled by the drone in the first 5.0 s. [2]
(b) The drone then flies horizontally. The drone has a mass of 15 kg. It travels at a constant horizontal speed of 12 m/s. (i) State the value of the resultant force acting on the drone in the horizontal direction. Explain your answer. [2] (ii) Calculate the kinetic energy of the drone during this horizontal flight. [2]
(c) While hovering, the drone releases a parcel of mass 2.0 kg. The gravitational field strength \(g = 10\text{ N/kg}\). Calculate the weight of the parcel. [2]
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解题
(a)(i) \(a = \frac{v - u}{t} = \frac{12\text{ m/s} - 0}{5.0\text{ s}} = 2.4\text{ m/s}^2\) (ii) \(d = \frac{1}{2} \times t \times v = \frac{1}{2} \times 5.0\text{ s} \times 12\text{ m/s} = 30\text{ m}\) (b)(i) Resultant force = 0 N because the speed is constant / acceleration is zero. (ii) \(KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 15\text{ kg} \times (12\text{ m/s})^2 = 1080\text{ J}\) (c) \(W = m \times g = 2.0\text{ kg} \times 10\text{ N/kg} = 20\text{ N}\)
(a) An atom of chlorine is represented by \({}_{17}^{37}\text{Cl}\). (i) State the number of protons, neutrons, and electrons in this atom. [2] (ii) Define the term *isotopes*. [2]
(b) Carbon reacts with sulfur to form carbon disulfide, \(\text{CS}_2\). (i) Describe the bonding in a molecule of carbon disulfide, \(\text{CS}_2\). State whether it is ionic or covalent and explain your answer in terms of electrons. [2] (ii) Describe how outer-shell electrons are shared in a molecule of carbon disulfide, \(\text{CS}_2\). [2]
(c) State one difference in the physical properties of covalent compounds, like carbon disulfide, compared to ionic compounds, like sodium chloride. [2]
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解题
(a)(i) Protons: 17, Neutrons: 20 (37 - 17), Electrons: 17 (ii) Atoms of the same element with the same number of protons but different numbers of neutrons. (b)(i) Covalent bonding; electrons are shared between non-metal atoms. (ii) The carbon atom shares two pairs of electrons with each of the two sulfur atoms, forming two double covalent bonds. This allows all three atoms to achieve a stable outer shell of 8 electrons. (c) Covalent compounds have lower melting/boiling points than ionic compounds (or do not conduct electricity in molten/liquid states).
评分标准
(a)(i) - 2 marks for all three correct (Protons = 17, Neutrons = 20, Electrons = 17) - 1 mark if two of these are correct (a)(ii) - Atoms of same element / same proton number [1] - Different number of neutrons [1] (b)(i) - Covalent [1] - Electrons are shared [1] (b)(ii) - Two double bonds / four shared electrons between C and each S atom [1] - Total of four shared pairs / eight bonding electrons in the molecule [1] (c) - Lower melting/boiling point [1] OR poor/no electrical conductivity when liquid [1] - Accept any correct comparison [1]
题目 3 · structured
10 分
(a) Describe how water moves from the soil into the root hair cells of a plant. Use the term *water potential* in your answer. [3]
(b) Xylem vessels are adapted for the transport of water and mineral ions up the stem. (i) State two structural features of xylem vessels that adapt them for this function. [2] (ii) Explain the mechanism of transpiration pull that draws water up the xylem vessels. [3]
(c) State the name of the tissue that transports sucrose and amino acids in a plant, and state the direction of transport of these substances. [2]
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解题
(a) Water moves from the soil into the root hair cells by osmosis, down a water potential gradient (soil has higher water potential than root hair cell cytoplasm), across a partially permeable membrane. (b)(i) Hollow tubes with no end walls, and walls thickened with lignin. (ii) Water evaporates from the mesophyll cell surfaces into the air spaces, and water vapour diffuses out of the leaves through stomata. This loss of water creates a tension/pull (transpiration pull) that draws up a continuous column of water molecules, held together by cohesion. (c) Phloem tissue; transports substances both upwards and downwards (from source to sink).
评分标准
(a) - Osmosis [1] - Down a water potential gradient / soil has higher water potential than cytoplasm [1] - Across a partially permeable membrane [1] (b)(i) - No end walls / hollow tubes / dead cells [1] - Lignified/thickened walls [1] (b)(ii) - Water evaporates from mesophyll cells [1] - Water vapour diffuses out of stomata [1] - Creates tension/pull drawing up water column due to cohesion [1] (c) - Phloem [1] - Bidirectional / up and down / from source to sink [1]
题目 4 · structured
10 分
Chemical digestion of proteins starts in the stomach and is completed in the small intestine.
(a) (i) Name the enzyme that digests proteins in the stomach and state the pH at which it works most effectively. [2] (ii) State the product of protein digestion. [1]
(b) The teeth are involved in mechanical digestion. (i) Contrast the structure and function of incisors and molars. [3] (ii) Dental decay is caused by bacteria in the mouth. Describe how dental decay occurs. [3]
(c) Scurvy is a deficiency disease. Name the nutrient that is deficient in a person with scurvy. [1]
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解题
(a)(i) Pepsin / protease; pH 1.5 - 2.5 (acidic pH). (ii) Amino acids. (b)(i) Incisors have a chisel-shaped, sharp edge and are used for biting and cutting food, whereas molars have a broad, flat surface with ridges and are used for crushing and grinding food. (ii) Bacteria feed on sugar/food left on teeth, producing acid as they respire. This acid dissolves the calcium-rich enamel and dentine, leading to tooth decay. (c) Vitamin C (ascorbic acid).
(a) A student sets up a circuit with a 12 V battery connected in series to a variable resistor and a fixed resistor of resistance \(6.0\ \Omega\). (i) Draw a circuit diagram of this setup. [2] (ii) The variable resistor is adjusted so that the current in the circuit is 1.5 A. Calculate the potential difference across the \(6.0\ \Omega\) resistor. [2] (iii) Calculate the electrical power dissipated by the \(6.0\ \Omega\) resistor at this current. [2]
(b) An electromagnet is made by winding a coil of wire around a metal core. (i) State the name of the metal that should be used for the core of an electromagnet and explain why it is suitable. [2] (ii) State two ways to increase the strength of the magnetic field produced by the electromagnet. [2]
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解题
(a)(i) Circuit diagram must show: a battery symbol, a variable resistor symbol (rectangle with diagonal arrow), and a fixed resistor symbol (rectangle) all connected in a single loop (series). (ii) \(V = I \times R = 1.5\text{ A} \times 6.0\ \Omega = 9.0\text{ V}\) (iii) \(P = I^2 R = (1.5\text{ A})^2 \times 6.0\ \Omega = 2.25 \times 6.0 = 13.5\text{ W}\) (or \(P = V \times I = 9.0\text{ V} \times 1.5\text{ A} = 13.5\text{ W}\)) (b)(i) Iron; it is magnetically soft, meaning it is easily magnetised and demagnetised when the current is switched on and off. (ii) Increase the current flowing through the coil, and increase the number of turns in the coil.
评分标准
(a)(i) - Correct symbols for battery, variable resistor, and fixed resistor [1] - All components connected correctly in series [1] (a)(ii) - Formula/working: \(1.5 \times 6.0\) [1] - Value and unit: \(9.0\text{ V}\) [1] (a)(iii) - Formula/working: \(1.5^2 \times 6.0\) or \(9.0 \times 1.5\) [1] - Value and unit: \(13.5\text{ W}\) [1] (b)(i) - Iron [1] - Magnetically soft / easily demagnetised when current is off [1] (Do NOT accept steel) (b)(ii) - Increase the current [1] - Increase the number of turns/loops of the coil [1]
题目 6 · structured
10 分
(a) Lithium, sodium, and potassium are Group I alkali metals. (i) Describe the trend in the reactivity of alkali metals down the group. Explain this trend in terms of their electronic structure. [3] (ii) Sodium reacts vigorously with cold water. State two observations when a small piece of sodium is added to water. [2]
(b) Chlorine, bromine, and iodine are Group VII halogens. (i) State the physical state of bromine and iodine at room temperature and pressure. [2] (ii) Chlorine is bubbled into a solution of potassium iodide. State the colour change observed in the solution and write an ionic equation for the reaction. [3]
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解题
(a)(i) Reactivity increases down the group. Down the group, the atoms have more electron shells, so the outer electron is further from the positive nucleus. There is a weaker electrostatic attraction between the nucleus and the outer electron, making it easier to lose the single outer electron. (ii) Any two from: melts into a ball, floats/moves on the surface of the water, fizzes/bubbles/effervesces, dissolves/disappears. (b)(i) Bromine is a liquid; iodine is a solid. (ii) Colour change: colourless to brown (or orange/red-brown). Ionic equation: \(\text{Cl}_2(\text{g}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{I}_2(\text{aq})\)
评分标准
(a)(i) - Reactivity increases down the group [1] - Outer shell electron is further from the nucleus / atom gets larger [1] - Weaker attraction / electron lost more easily [1] (a)(ii) - Any two observations: melts into ball / floats / fizzes or bubbles / moves rapidly on surface / disappears [2] (b)(i) - Bromine is liquid [1] - Iodine is solid [1] (b)(ii) - Turns brown/orange/red-brown [1] - Correct ionic reactants and products: \(\text{Cl}_2 + 2\text{I}^- \rightarrow 2\text{Cl}^- + \text{I}_2\) [1] - Balanced equation [1]
题目 7 · structured
10 分
(a) Electromagnetic waves travel through a vacuum. (i) State the speed of all electromagnetic waves in a vacuum. [1] (ii) A microwave oven uses microwaves with a frequency of \(2.4 \times 10^9\text{ Hz}\). Calculate the wavelength of these microwaves. [2]
(b) Sound waves are longitudinal waves. (i) Describe the nature of a longitudinal wave, referring to the direction of vibration and the direction of energy transfer. [2] (ii) A sound wave travels from air into water. State what happens, if anything, to its speed and wavelength. [2]
(c) Light passes from air into a rectangular glass block. (i) State the name of the phenomenon that causes the light ray to bend as it enters the glass. [1] (ii) Describe how the path of the light ray changes as it enters the glass block and as it leaves the glass block. [2]
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解题
(a)(i) \(3.0 \times 10^8\text{ m/s}\) (ii) \(\lambda = \frac{v}{f} = \frac{3.0 \times 10^8\text{ m/s}}{2.4 \times 10^9\text{ Hz}} = 0.125\text{ m}\) (or \(1.3 \times 10^{-1}\text{ m}\) or \(12.5\text{ cm}\)) (b)(i) In a longitudinal wave, the vibrations of the particles are parallel to the direction of energy transfer. (ii) Speed increases and wavelength increases (frequency remains constant). (c)(i) Refraction (ii) As the ray enters the glass, it bends towards the normal. As it leaves the glass back into air, it bends away from the normal.
评分标准
(a)(i) - \(3.0 \times 10^8\text{ m/s}\) [1] (a)(ii) - Working: \(3.0 \times 10^8 / 2.4 \times 10^9\) [1] - Value and unit: \(0.125\text{ m}\) (or \(12.5\text{ cm}\)) [1] (b)(i) - Vibrations/oscillations [1] - Parallel to energy transfer direction [1] (b)(ii) - Speed increases [1] - Wavelength increases [1] (c)(i) - Refraction [1] (c)(ii) - Enters: bends towards normal [1] - Leaves: bends away from normal [1]
题目 8 · structured
10 分
Dilute hydrochloric acid reacts with calcium carbonate according to the equation: \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\)
(a) (i) Describe how the rate of this reaction can be measured experimentally. State the apparatus used. [3] (ii) The student repeats the experiment using the same mass of calcium carbonate but a higher concentration of hydrochloric acid. Explain, using collision theory, why the rate of reaction increases. [3]
(b) Catalysts are used to speed up chemical reactions. (i) Define the term *catalyst*. [2] (ii) Explain, in terms of activation energy, how a catalyst increases the rate of a chemical reaction. [2]
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解题
(a)(i) Connect the reaction flask to a gas syringe using a delivery tube. Measure the volume of carbon dioxide gas produced at regular time intervals using a stopwatch. (Alternatively, place the flask on a balance and record the decrease in mass over time.) (ii) A higher concentration means there are more acid particles per unit volume. This results in more frequent collisions (more collisions per second) between the reactant particles, which increases the rate of successful collisions and thus the rate of reaction. (b)(i) A substance that increases the rate of a chemical reaction, but remains chemically unchanged at the end of the reaction. (ii) A catalyst provides an alternative pathway for the reaction that has a lower activation energy, meaning a higher proportion of reacting particles have enough energy to react upon collision.
评分标准
(a)(i) - Measure volume of gas / decrease in mass [1] - Use gas syringe / balance [1] - Record time / use stopwatch [1] (a)(ii) - More particles per unit volume / particles closer [1] - More frequent collisions [1] - Increased rate of successful/effective collisions [1] (b)(i) - Speeds up reaction [1] - Chemically unchanged / mass remains constant at end [1] (b)(ii) - Alternative reaction pathway [1] - Lower activation energy [1]
题目 9 · structured
10 分
(a) A student investigates the rate of transpiration of a leafy shoot using a potometer.
(i) Describe how a potometer is used to measure the rate of transpiration. [2]
(ii) State the effect of increasing wind speed on the rate of transpiration and explain this effect in terms of water vapour concentration gradients. [3]
(b) (i) State two structural features of xylem vessels that adapt them for the transport of water. [2]
(ii) Describe how the mechanism of translocation of sucrose in phloem vessels differs from the transport of water in xylem vessels. [3]
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解题
(a) (i) The potometer is filled with water and a leafy shoot is inserted. The joint is sealed. An air bubble is introduced into the capillary tube. The distance moved by the bubble along a scale per unit time (e.g. per minute) is measured to estimate the rate of transpiration.
(ii) Increasing wind speed increases the rate of transpiration. Wind blows away the water vapour that accumulates around the surface of the leaf / stomata. This maintains a steep water vapour concentration gradient between the inside of the leaf and the outside air, causing faster diffusion of water vapour out of the stomata.
(b) (i) Any two from: - Hollow tubes / no cytoplasm / no living contents to allow unobstructed flow. - Lignified / thick walls to prevent the vessels from collapsing under tension. - End walls between cells are broken down / absent to form continuous tubes. - Pits in walls to allow lateral movement of water.
(ii) Translocation in phloem is an active process that requires energy / active transport (whereas xylem transport is passive / driven by transpiration pull). Translocation transports sucrose and amino acids (whereas xylem transports water and mineral ions). Translocation occurs in both upwards and downwards directions (whereas xylem transport is in the upwards direction only).
评分标准
(a)(i) Max 2 marks: - measure distance moved by bubble [1] - in a given/measured time / per unit time [1]
(a)(ii) Max 3 marks: - rate increases [1] - wind moves/removes water vapour from around the leaf / stomata [1] - maintains/increases the concentration gradient [1]
(b)(i) Max 2 marks: - hollow tubes / no cytoplasm / no living contents [1] - lignified/thick walls [1] - no end walls / continuous tubes [1]
(b)(ii) Max 3 marks: - active process / requires energy / active transport vs passive in xylem [1] - transport of sucrose/amino acids vs water/mineral ions in xylem [1] - transport is in both directions (up and down) vs upwards only in xylem [1]
题目 10 · structured
10 分
(a) Hydrocarbons can be cracked to produce smaller, more useful molecules.
(i) Write a balanced chemical equation for the cracking of decane, \(\text{C}_{10}\text{H}_{22}\), to form octane, \(\text{C}_8\text{H}_{18}\), and one other product. [2]
(ii) State the conditions required for catalytic cracking. [2]
(b) The other product formed in (a)(i) is ethene.
(i) Describe a chemical test to distinguish between octane and ethene. State the starting colour of the reagent and the final result with each hydrocarbon. [3]
(ii) Ethene undergoes addition polymerisation to form poly(ethene). Draw the structure of a section of poly(ethene) showing two repeat units. [3]
(ii) High temperature (approx. \(450^\circ\text{C}\) to \(800^\circ\text{C}\)) and a catalyst (silica / alumina / zeolite).
(b) (i) Test: Bromine water / aqueous bromine. Octane (alkane): Stays orange / brown / yellow / no change. Ethene (alkene): Decolourises / turns colourless.
(ii) Draw a chain of four carbon atoms linked by single bonds, with two hydrogen atoms attached to each carbon atom. Single bonds on the ends with extension lines to show the continuation of the polymer chain.
评分标准
(a)(i) Max 2 marks: - correct formula of ethene, \(\text{C}_2\text{H}_4\) [1] - fully balanced equation [1]
(a)(ii) Max 2 marks: - high temperature / temp in range \(450^\circ\text{C}\) to \(800^\circ\text{C}\) [1] - catalyst / named catalyst e.g. silica/alumina/zeolite [1]
(b)(i) Max 3 marks: - bromine water starting colour is orange/brown/yellow [1] - result with octane: stays orange/brown/yellow / no reaction [1] - result with ethene: decolourises / turns colourless [1]
(b)(ii) Max 3 marks: - only single bonds between carbon atoms in the chain [1] - four carbon atoms shown with extension bonds at the ends [1] - eight hydrogen atoms correctly attached to the carbon atoms [1]
题目 11 · structured
10 分
(a) A cyclist accelerates from rest to a speed of \(8.0\text{ m/s}\) in a time of \(12.0\text{ s}\).
(i) Calculate the average acceleration of the cyclist. State the unit. [3]
(ii) The total mass of the cyclist and bicycle is \(75\text{ kg}\). Calculate the kinetic energy of the cyclist and bicycle when travelling at \(8.0\text{ m/s}\). [2]
(b) The cyclist then climbs a hill of vertical height \(15\text{ m}\) at a constant speed.
(i) Calculate the work done against gravity by the cyclist in climbing the hill. (The gravitational field strength \(g = 10\text{ N/kg}\)). [2]
(ii) During the climb, the power input by the cyclist is \(300\text{ W}\) and the climb takes \(50\text{ s}\). Calculate the efficiency of the cyclist in climbing the hill. [3]
(ii) \(KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 75 \times 8.0^2 = 2400\text{ J}\)
(b) (i) \(W = m g h = 75 \times 10 \times 15 = 11250\text{ J}\)
(ii) Total energy input = \(P \times t = 300 \times 50 = 15000\text{ J}\). Efficiency = \(\frac{\text{useful energy output}}{\text{total energy input}} \times 100\% = \frac{11250}{15000} \times 100\% = 75\%\).
(a)(ii) Max 2 marks: - formula: \(KE = \frac{1}{2} m v^2\) or correct substitution [1] - calculation: \(2400\text{ J}\) [1]
(b)(i) Max 2 marks: - formula: \(W = mgh\) or correct substitution [1] - calculation: \(11250\text{ J}\) [1]
(b)(ii) Max 3 marks: - calculate total energy input: \(15000\text{ J}\) [1] - formula: \(\text{efficiency} = (\text{useful energy} / \text{total energy}) \times 100\%\) [1] - calculation: \(75\%\) [1]
题目 12 · structured
10 分
(a) A battery of electromotive force (e.m.f.) \(12\text{ V}\) is connected to a parallel combination of two resistors, \(R_1\) and \(R_2\), of resistance \(4.0\ \Omega\) and \(6.0\ \Omega\) respectively.
(i) Calculate the combined resistance of the two resistors in parallel. [2]
(ii) Calculate the total current drawn from the battery. [2]
(iii) Calculate the electrical power dissipated in resistor \(R_1\). [2]
(b) A straight wire carrying a current is placed in a magnetic field. Alternatively, when a straight wire is moved through a magnetic field, an electromotive force (e.m.f.) is produced across the ends of the wire.
(i) State the name of the effect that produces this e.m.f. [1]
(ii) Describe three ways to increase the magnitude of the induced e.m.f. in the wire. [3]
(ii) \(I = \frac{V}{R_p} = \frac{12}{2.4} = 5.0\text{ A}\)
(iii) The voltage across \(R_1\) is \(12\text{ V}\) (parallel circuit). \(P = \frac{V^2}{R_1} = \frac{12^2}{4.0} = \frac{144}{4.0} = 36\text{ W}\) (Alternatively: current through \(R_1\) is \(I_1 = \frac{12}{4.0} = 3.0\text{ A}\), Power \(P = I_1^2 \times R_1 = 3.0^2 \times 4.0 = 36\text{ W}\))
(b) (i) Electromagnetic induction.
(ii) Any three from: - Move the wire faster / increase the speed of movement. - Use a stronger magnet / increase the magnetic field strength. - Increase the length of wire in the magnetic field / use a coil with more turns. - Move the wire at a right angle (\(90^\circ\)) to the magnetic field lines.
评分标准
(a)(i) Max 2 marks: - formula: \(1/R_p = 1/R_1 + 1/R_2\) or \(R_p = (R_1 \times R_2)/(R_1 + R_2)\) or correct substitution [1] - calculation: \(2.4\ \Omega\) [1]
(a)(ii) Max 2 marks: - formula: \(I = V/R\) or correct substitution [1] - calculation: \(5.0\text{ A}\) [1]
(a)(iii) Max 2 marks: - calculate current in \(R_1\) as \(3.0\text{ A}\) OR state voltage as \(12\text{ V}\) [1] - calculation: \(36\text{ W}\) [1]
(b)(i) Max 1 mark: - electromagnetic induction [1]
(b)(ii) Max 3 marks (any three from): - move the wire faster [1] - stronger magnetic field / stronger magnet [1] - more turns of wire / longer wire [1] - move perpendicular to field lines [1]