An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V2) Cambridge IGCSE Science - Combined (0653) paper. Not affiliated with or reproduced from Cambridge.
部分 Structured Written Questions
Answer all structured questions in the spaces provided. Show working for calculations and include appropriate units.
9 题目 · 81 分
题目 1 · Structured
9 分
Answer all structured questions in the spaces provided. Show working for calculations and include appropriate units.
A cyclist of mass \(65\text{ kg}\) accelerates from rest along a straight horizontal path. The journey is described as follows: - From \(t = 0\) to \(t = 5.0\text{ s}\), the speed increases uniformly from \(0\) to \(8.0\text{ m/s}\). - From \(t = 5.0\text{ s}\) to \(t = 20.0\text{ s}\), the speed remains constant at \(8.0\text{ m/s}\). - From \(t = 20.0\text{ s}\) to \(t = 25.0\text{ s}\), the cyclist slows down uniformly to a stop.
(a) (i) Calculate the acceleration of the cyclist during the first \(5.0\) seconds. Show your working. [2]
(ii) Calculate the total distance travelled by the cyclist during the \(25.0\) seconds of the journey. Show your working. [3]
(b) (i) The total mass of the cyclist and bicycle is \(80\text{ kg}\). Calculate the kinetic energy of the cyclist and bicycle when travelling at \(8.0\text{ m/s}\). Show your working and state the unit. [3]
(ii) State the form of energy stored in the cyclist's muscles that is transferred to kinetic energy. [1]
(a) (i) - Correct formula or substitution: \(\frac{8.0}{5.0}\) [1] - Correct value with unit: \(1.6\text{ m/s}^2\) [1]
(ii) - Correct distance in any section (e.g., constant speed section \(120\text{ m}\) or either triangle \(20\text{ m}\)) [1] - Correct sum of areas: \(20 + 120 + 20\) [1] - Correct final answer: \(160\text{ m}\) [1]
(b) (i) - Correct formula or substitution: \(\frac{1}{2} \times 80 \times 8.0^2\) [1] - Correct calculation: \(2560\) [1] - Correct unit: \(\text{J}\) or Joules [1]
(ii) - Chemical (potential) energy [1]
题目 2 · Structured
9 分
A circuit contains a \(12\text{ V}\) battery, a switch, an ammeter, and two resistors connected in parallel with each other. One resistor has a resistance of \(4.0\ \Omega\) and the other has a resistance of \(6.0\ \Omega\).
(a) (i) State the reading on the ammeter when the switch is open. [1]
(ii) When the switch is closed, calculate the current in each of the two resistors. - Current in the \(4.0\ \Omega\) resistor. Show your working. [2] - Current in the \(6.0\ \Omega\) resistor. Show your working. [2]
(iii) Calculate the total resistance of the parallel combination of resistors. Show your working. [2]
(b) Explain, in terms of current and resistance, why connecting resistors in parallel decreases the total resistance of the circuit compared to a single resistor. [2]
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解题
(a) (i) \(0\text{ A}\) (no current flows when circuit is incomplete).
(ii) - Current in \(4.0\ \Omega\) resistor: \(I = \frac{V}{R} = \frac{12}{4.0} = 3.0\text{ A}\). - Current in \(6.0\ \Omega\) resistor: \(I = \frac{V}{R} = \frac{12}{6.0} = 2.0\text{ A}\).
(b) In parallel, there are more paths for the electric current to flow through. The total current from the battery increases for the same voltage, which means the overall resistance is reduced according to \(R = \frac{V}{I}\).
评分标准
(a) (i) \(0\) / zero (accept \(0\text{ A}\)) [1]
(ii) - Correct working for \(4.0\ \Omega\) resistor: \(\frac{12}{4}\) [1] - Correct current: \(3.0\text{ A}\) [1] - Correct working for \(6.0\ \Omega\) resistor: \(\frac{12}{6}\) [1] - Correct current: \(2.0\text{ A}\) [1]
(iii) - Correct formula or substitution: \(\frac{1}{R} = \frac{1}{4} + \frac{1}{6}\) or \(\frac{4 \times 6}{4 + 6}\) [1] - Correct resistance value: \(2.4\ \Omega\) [1]
(b) - Mention of more paths/routes for the current (which increases total current) [1] - Relationship described: increased total current for same potential difference reduces total resistance / according to \(R = V/I\) [1]
题目 3 · Structured
9 分
A student investigates the reaction between dilute sulfuric acid and solid copper(II) carbonate.
(a) (i) State three observations that can be made during this reaction. [3]
(ii) Write the word equation for this reaction. [2]
(b) The student wants to obtain pure, dry crystals of copper(II) sulfate from the mixture after the reaction has finished.
Describe the experimental steps required to do this. [4]
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解题
(a) (i) Observations: effervescence/fizzing/bubbles, the green solid dissolves/disappears, the solution turns blue, temperature increase. (ii) copper(II) carbonate + sulfuric acid \(\rightarrow\) copper(II) sulfate + carbon dioxide + water (b) 1. Filter the mixture to remove any unreacted copper(II) carbonate solid. 2. Heat the filtrate (copper(II) sulfate solution) to evaporate some water until the crystallization point is reached. 3. Leave the hot solution to cool slowly so that crystals form. 4. Filter the crystals to separate them from the remaining liquid, and dry them between filter papers.
评分标准
(a) (i) Any three from: [3] - Effervescence / bubbles / fizzing - Green solid dissolves / disappears - Solution turns blue / green - Temperature increases / mixture gets warm
(b) - Filtration to remove excess/unreacted solid copper(II) carbonate [1] - Heat / evaporate the solution to saturation / crystallization point [1] - Cool slowly to allow crystals to grow [1] - Filter crystals and dry using filter paper [1]
题目 4 · Structured
9 分
Nitrogen gas, \(\text{N}_2\), is the most abundant gas in Earth's atmosphere. At room temperature, it behaves as a typical gas.
(a) (i) Describe the arrangement and motion of molecules in nitrogen gas. [2]
(ii) Explain, in terms of particles, what happens to the pressure of nitrogen gas in a closed container of fixed volume when it is heated. [3]
(b) Liquid nitrogen boils at \(-196\ ^\circ\text{C}\).
(i) Describe what is meant by boiling in terms of particles and energy. [2]
(ii) State two differences between boiling and evaporation. [2]
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解题
(a) (i) Arrangement: random, widely spaced/far apart. Motion: rapid, random, in all directions. (ii) As temperature increases, the kinetic energy of the particles increases, so they move faster. This causes them to collide with the walls of the container more frequently and with greater force, increasing the pressure. (b) (i) Boiling is the transition from liquid to gas where particles gain enough thermal energy to overcome the attractive forces holding them together in the liquid state, occurring throughout the liquid at a constant temperature. (ii) 1. Boiling occurs at a fixed temperature (the boiling point), whereas evaporation occurs at any temperature. 2. Boiling occurs throughout the liquid, whereas evaporation only occurs at the surface.
评分标准
(a) (i) - Random arrangement AND widely spaced [1] - Rapid / constant / random motion in all directions [1]
(ii) - Particles gain kinetic energy / move faster [1] - More frequent collisions with container walls [1] - Collisions are with greater force, resulting in higher pressure [1]
(b) (i) - Heat energy is absorbed / particles gain energy [1] - Overcomes the intermolecular forces / attractive forces between particles [1]
(ii) Any two differences: [2] - Boiling occurs at a specific/fixed temperature, evaporation at all temperatures - Boiling occurs throughout the liquid, evaporation only at the surface - Boiling is rapid, evaporation is slow
题目 5 · Structured
9 分
An investigation is carried out on the effect of temperature on the rate of an amylase-catalysed reaction.
(a) (i) Describe the effect of temperature on the rate of this reaction. Use values (e.g. rate increases from \(10\ ^\circ\text{C}\) to \(40\ ^\circ\text{C}\), peaks at \(40\ ^\circ\text{C}\), then falls to \(0\) at \(60\ ^\circ\text{C}\)) to support your description. [3]
(ii) State the term used to describe the temperature at which the rate of an enzyme-catalysed reaction is greatest. [1]
(b) Explain, in terms of kinetic energy, collision theory, and active sites, why the rate of reaction:
(i) increases as the temperature is raised from \(10\ ^\circ\text{C}\) to \(40\ ^\circ\text{C}\). [2]
(ii) decreases rapidly above \(45\ ^\circ\text{C}\) and stops completely at \(60\ ^\circ\text{C}\). [3]
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解题
(a) (i) As temperature increases from \(10\ ^\circ\text{C}\) to \(40\ ^\circ\text{C}\), the rate of reaction increases. The rate reaches a maximum at \(40\ ^\circ\text{C}\). Above \(40\ ^\circ\text{C}\), the rate of reaction decreases rapidly and becomes zero at \(60\ ^\circ\text{C}\). (ii) Optimum temperature. (b) (i) Increasing temperature increases the kinetic energy of both enzyme and substrate molecules, so they move faster. This leads to more frequent successful collisions between the substrate and the active site of the enzyme, increasing the rate of reaction. (ii) At high temperatures (above \(45\ ^\circ\text{C}\)), the shape of the enzyme's active site is permanently altered because the protein denatures. The substrate can no longer fit into the active site, so no reaction can occur, stopping the reaction completely by \(60\ ^\circ\text{C}\).
评分标准
(a) (i) - Rate increases with temperature up to a peak/optimum [1] - Peak rate occurs at \(40\ ^\circ\text{C}\) [1] - Rate decreases rapidly above \(40\ ^\circ\text{C}\) and becomes zero at \(60\ ^\circ\text{C}\) [1]
(ii) Optimum (temperature) [1]
(b) (i) - Molecules gain kinetic energy / move faster [1] - More frequent successful collisions between enzyme (active site) and substrate [1]
(ii) - At high temperatures, the enzyme is denatured [1] - Shape of the active site changes / is destroyed [1] - Substrate can no longer fit / bind to the active site [1]
题目 6 · Structured
9 分
The human gas exchange system is adapted for the efficient exchange of gases between the air and the blood.
(a) (i) Name the tiny air sacs in the lungs where gas exchange takes place. [1]
(ii) List three features of these air sacs that adapt them for rapid and efficient diffusion of gases. [3]
(b) (i) State two differences between the composition of inspired (inhaled) air and expired (exhaled) air. [2]
(ii) Explain the biological reasons for each of the two differences stated in (b)(i). [3]
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解题
(a) (i) Alveoli (singular: alveolus). (ii) 1. Thin walls (only one cell thick) providing a short diffusion distance. 2. Large surface area for a high rate of diffusion. 3. Surrounded by an extensive network of capillaries (good blood supply) to maintain a steep concentration gradient. (b) (i) 1. Inspired air has more oxygen (\(\sim 21\%\)) than expired air (\(\sim 16\%\)). 2. Expired air has more carbon dioxide (\(\sim 4\%\)) than inspired air (\(\sim 0.04\%\)). (ii) Oxygen is absorbed by the blood and used in aerobic respiration by body cells to produce energy. Carbon dioxide is produced by aerobic respiration as a waste product, carried by the blood to the lungs, and excreted into the alveoli to be exhaled.
评分标准
(a) (i) Alveoli / alveolus [1]
(ii) Any three from: [3] - Large surface area - Very thin walls / one cell thick / short diffusion path - Good blood supply / dense capillary network (maintains gradient) - Moist lining (helps gases dissolve) - Good ventilation / airflow (maintains gradient)
(b) (i) Any two differences: [2] - Inspired air contains more oxygen than expired air (or vice versa) - Expired air contains more carbon dioxide than inspired air (or vice versa) - Expired air contains more water vapour (is saturated) - Expired air is warmer than inspired air
(ii) - Oxygen is used in (aerobic) respiration [1] - Carbon dioxide is produced as a waste product of respiration [1] - Water vapour is produced in respiration / evaporates from moist lining of alveoli [1]
题目 7 · Structured
9 分
The human circulatory system transports blood containing oxygen, nutrients, and waste products to and from body cells.
(a) (i) Compare the structures of an artery and a vein by completing the statements below. - Compare the thickness of their walls. [1] - Compare the size of their lumens. [1] - State which vessel contains valves and explain their function. [1]
(ii) Explain how the structure of a capillary is adapted to its function of exchanging substances with body cells. [2]
(b) (i) Name the chamber of the heart that pumps oxygenated blood into the aorta. [1]
(ii) Explain why the muscular wall of this chamber is much thicker than the wall of the right ventricle. [3]
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解题
(a) (i) - Arteries have thick, muscular, and elastic walls, while veins have thin walls with less muscle and elastic tissue. - Arteries have a narrow/small lumen, while veins have a wide/large lumen. - Veins contain valves to prevent the backflow of blood, ensuring blood flows towards the heart under low pressure.
(ii) Capillaries have walls that are only one cell thick, which minimizes the diffusion distance for oxygen, carbon dioxide, glucose, and other substances. They are also highly branched, providing a very large surface area.
(b) (i) Left ventricle. (ii) The left ventricle must pump blood to the entire body (systemic circulation), which requires a very high pressure to overcome the resistance of all the systemic blood vessels. The right ventricle only pumps blood a short distance to the lungs (pulmonary circulation), which is at a much lower pressure.
评分标准
(a) (i) - Wall thickness: artery has thicker walls / vein has thinner walls [1] - Lumen size: artery has narrower lumen / vein has wider lumen [1] - Valves: veins have valves AND function is to prevent backflow of blood [1]
(ii) - Wall is only one cell thick [1] - Short diffusion distance for rapid exchange / large surface area [1]
(b) (i) Left ventricle [1]
(ii) - Left ventricle pumps blood to the whole body / systemic circulation [1] - Right ventricle only pumps blood to the lungs / pulmonary circulation [1] - Thicker muscle generates the higher pressure needed to pump blood further [1]
题目 8 · Structured
9 分
Sound waves and light waves transfer energy from one place to another through different mechanisms.
(a) (i) Describe the difference between longitudinal waves and transverse waves in terms of particle vibration and direction of wave travel. [2]
(ii) State whether sound waves can travel through a vacuum, and explain your answer. [2]
(b) A sound wave travels through air. The frequency of the sound wave is \(850\text{ Hz}\) and its speed in air is \(340\text{ m/s}\).
(i) Calculate the wavelength of this sound wave. Show your working. [2]
(ii) Describe how the wavelength of this sound wave changes, if at all, when it enters water where the speed of sound is \(1500\text{ m/s}\). Explain your answer. [3]
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解题
(a) (i) In longitudinal waves, the particles of the medium vibrate parallel to the direction of wave travel (energy transfer). In transverse waves, the particles vibrate perpendicular to the direction of wave travel. (ii) No, sound waves cannot travel through a vacuum because they are mechanical waves that require a material medium (particles) to transmit the compressions and rarefactions. (b) (i) Speed \(v = f \lambda \implies \lambda = \frac{v}{f} = \frac{340}{850} = 0.40\text{ m}\). (ii) The frequency of the wave remains constant at \(850\text{ Hz}\) as it passes from air into water. Since speed increases from \(340\text{ m/s}\) to \(1500\text{ m/s}\) and \(v = f \lambda\), the wavelength must increase. New wavelength \(\lambda = \frac{1500}{850} \approx 1.76\text{ m}\).
评分标准
(a) (i) - Longitudinal: vibration/oscillation is parallel to the direction of wave travel / energy transfer [1] - Transverse: vibration/oscillation is perpendicular to the direction of wave travel [1]
(ii) - Sound waves cannot travel through a vacuum [1] - Sound is a mechanical wave / requires a medium/particles to propagate (compressions/rarefactions) [1]
(b) (i) - Correct formula or substitution: \(\lambda = \frac{340}{850}\) [1] - Correct calculation: \(0.40\text{ m}\) (accept with unit) [1]
(ii) - Frequency remains constant [1] - Wavelength increases [1] - Explanation: since speed is much higher in water and \(v = f\lambda\), wavelength must increase proportionally [1]
题目 9 · Structured
9 分
Answer all parts of the question in the spaces provided. Show working for calculations and include appropriate units.
A student uses a small electric motor to lift a block of mass \( 0.40\text{ kg} \) vertically upwards. The block is lifted through a height of \( 1.5\text{ m} \) in a time of \( 3.0\text{ s} \). The gravitational field strength \( g \) is \( 10\text{ N/kg} \).
**(a)** Calculate the weight of the block.
weight = ................................... [1]
**(b)** Calculate the useful work done in lifting the block.
work done = ................................... [2]
**(c)** Calculate the useful power output of the motor.
power = ................................... [2]
**(d)** The electrical power input to the motor is \( 5.0\text{ W} \). Calculate the efficiency of the motor.