An original Thinka practice paper modelled on the structure and difficulty of the Jun 2022 AQA GCSE Physics 8463 paper. Not affiliated with or reproduced from AQA.
卷一 Foundation
Answer all questions. You must have a scientific calculator and the Physics Equations Sheet.
11 題目 · 99.99000000000002 分
題目 1 · Structured
9.09 分
A student investigates the energy of a toy car rolling down a ramp. The toy car has a mass of 0.20 kg. The top of the ramp is 0.25 m above the table. (a) Name the energy store of the car that decreases as it rolls down the ramp. [1 mark] (b) Write down the equation that links gravitational potential energy, mass, gravitational field strength, and height. [1 mark] (c) Calculate the gravitational potential energy of the car at the top of the ramp. Gravitational field strength, \(g = 9.8\text{ N/kg}\). [3 marks] (d) At the bottom of the ramp, the kinetic energy of the car is only 0.41 J. Describe what has happened to the remaining 0.08 J of energy. [2 marks]
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解題
a) As the car is high up, it has gravitational potential energy which decreases as its height decreases. b) The formula is \(E_p = m g h\). c) Using the formula: \(E_p = 0.20 \times 9.8 \times 0.25 = 0.49\text{ J}\). d) The energy was transferred to the thermal energy store of the car's wheels, the ramp, and the surrounding air due to work done against friction and air resistance.
評分準則
a) Gravitational potential (energy) store [1 mark] b) \(E_p = m g h\) [1 mark] c) Substitution: \(0.20 \times 9.8 \times 0.25\) [1 mark], calculation: 0.49 [1 mark], unit: Joules / J [1 mark] d) Work is done against friction / air resistance [1 mark], transferring energy to the thermal store of the surroundings / ramp [1 mark]
題目 2 · Structured
9.09 分
A student is investigating how the current in a filament lamp varies with potential difference. (a) Describe the circuit symbol for a filament lamp. [1 mark] (b) The potential difference across the lamp is 6.0 V. The current through the lamp is 1.5 A. Calculate the resistance of the lamp. Use the equation: potential difference = current \times resistance. [3 marks] (c) As the potential difference increases, the temperature of the filament lamp increases. State what happens to the resistance of the filament lamp. [1 mark] (d) Explain why the resistance of a filament increases as the temperature increases. [2 marks]
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解題
a) The symbol is a circle with a cross (X) inside. b) Rearranging the equation: \(R = V / I\). Substituting the values: \(R = 6.0 / 1.5 = 4.0\text{ }\Omega\). c) The resistance of the filament lamp increases. d) As temperature increases, the positive metal ions in the filament vibrate more rapidly. This increases the frequency of collisions between the vibrating ions and the conduction electrons flowing through the filament, resisting the flow of charge.
評分準則
a) Circle with a cross (X) inside [1 mark] b) Rearrangement: \(R = V / I\) [1 mark], substitution: \(6.0 / 1.5\) [1 mark], answer: 4.0 (\(\Omega\)) [1 mark] c) Resistance increases [1 mark] d) Metal ions/atoms vibrate more [1 mark], increasing collisions with flowing electrons [1 mark]
題目 3 · Structured
9.09 分
An experiment is carried out to measure the specific heat capacity of an aluminum block of mass 1.0 kg. (a) State what is meant by 'specific heat capacity'. [2 marks] (b) An electric heater transfers 18,000 J of energy to the 1.0 kg aluminum block. The temperature of the block rises from 20 °C to 40 °C. Calculate the specific heat capacity of aluminum. Use the Physics Equations Sheet. [3 marks] (c) The calculated value of specific heat capacity in this experiment is often higher than the actual value. Explain how heat loss to the surroundings causes this error. [2 marks]
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解題
a) Specific heat capacity is the amount of energy required to raise the temperature of 1 kg of a substance by 1 °C. b) Temperature change \(\Delta\theta = 40 - 20 = 20\text{ }^\circ\text{C}\). Equation: \(\Delta E = m c \Delta\theta\). Rearranging for \(c\): \(c = \Delta E / (m \Delta\theta)\). Substituting: \(c = 18000 / (1.0 \times 20) = 900\text{ J/kg}^\circ\text{C}\). c) Some thermal energy escapes to the surroundings instead of heating the block. Therefore, the heater must supply more energy than necessary to achieve the temperature rise, leading to a calculated value of \(c\) that is too high.
評分準則
a) Energy required to change the temperature of 1 kg of a substance [1 mark] by 1 degree Celsius [1 mark] b) Calculating temperature change of 20 [1 mark], substitution: \(18000 / (1.0 \times 20)\) [1 mark], answer: 900 (J/kg°C) [1 mark] c) Thermal energy is lost to the surroundings [1 mark], so more energy is supplied than is actually absorbed by the block [1 mark]
題目 4 · Structured
9.09 分
Radioactive isotopes emit different types of nuclear radiation to become more stable. (a) Name the three types of nuclear radiation. [3 marks] (b) State which of these three types of nuclear radiation is: (i) the most ionizing [1 mark] (ii) the most penetrating [1 mark] (c) A radioactive source has a half-life of 20 minutes. If the initial activity of the source is 800 Bq, calculate the activity of the source after 60 minutes. [2 marks]
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解題
a) The three types of radiation are alpha, beta, and gamma. b) (i) Alpha is the most ionizing because of its relatively large mass and +2 charge. (ii) Gamma is the most penetrating as it has no mass or charge and can pass through many materials. c) The number of half-lives in 60 minutes is \(60 / 20 = 3\) half-lives. After 1 half-life: \(800 / 2 = 400\text{ Bq}\). After 2 half-lives: \(400 / 2 = 200\text{ Bq}\). After 3 half-lives: \(200 / 2 = 100\text{ Bq}\).
評分準則
a) Alpha, beta, gamma [3 marks, 1 for each] b)(i) Alpha [1 mark] b)(ii) Gamma [1 mark] c) Determining 3 half-lives have occurred [1 mark], correct final activity of 100 (Bq) [1 mark]
題目 5 · Structured
9.09 分
Electricity can be generated from both renewable and non-renewable energy resources. (a) Define the term 'renewable energy resource'. [1 mark] (b) State one environmental advantage of using wind power instead of burning coal. [1 mark] (c) State one disadvantage of using wind power compared to burning coal. [1 mark] (d) A wind turbine has an input kinetic energy of 500,000 J from the wind. It produces 150,000 J of useful electrical energy. Calculate the efficiency of the wind turbine. Use the equation: efficiency = useful output energy transfer / total input energy transfer. [3 marks]
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解題
a) A renewable resource is one that is being replenished at the same rate as, or faster than, it is being consumed. b) Wind power does not release greenhouse gases like carbon dioxide, which contribute to global warming. c) Wind is unreliable; electricity is only generated when the wind blows. d) Using the efficiency formula: \(\text{efficiency} = 150000 / 500000 = 0.3\). This can also be written as 30%.
評分準則
a) Resource that is replenished / will not run out [1 mark] b) Does not emit carbon dioxide / greenhouse gases [1 mark] c) Unreliable / wind does not always blow [1 mark] d) Substitution: \(150000 / 500000\) [1 mark], calculation: 0.3 [1 mark], percentage conversion: 30% (accept either 0.3 or 30%) [1 mark]
題目 6 · Structured
9.09 分
A student sets up a series circuit containing a 6.0 V battery, an ammeter, and two resistors, \(R_1\) and \(R_2\). (a) Resistor \(R_1\) has a resistance of 4.0 \(\Omega\). Resistor \(R_2\) has a resistance of 8.0 \(\Omega\). Calculate the total resistance of the series circuit. [1 mark] (b) Calculate the current in the circuit. Use the equation: potential difference = current \times resistance. [3 marks] (c) An identical third resistor of 4.0 \(\Omega\) is added in series to the circuit. State what happens to the total current in the circuit and explain why. [3 marks]
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解題
a) For series resistors, total resistance \(R_T = R_1 + R_2 = 4.0 + 8.0 = 12.0\text{ }\Omega\). b) Current \(I = V / R_T = 6.0 / 12.0 = 0.5\text{ A}\). c) The total current in the circuit will decrease. This is because adding another resistor in series increases the total resistance of the circuit to 16.0 \(\Omega\). Since the potential difference remains constant at 6.0 V, a larger resistance results in a smaller current.
評分準則
a) 12.0 (\(\Omega\)) [1 mark] b) Rearrangement: \(I = V / R\) [1 mark], substitution: \(6.0 / 12.0\) [1 mark], calculation: 0.5 (A) [1 mark] c) Current decreases [1 mark], because total resistance increases [1 mark], and voltage remains constant [1 mark]
題目 7 · Structured
9.09 分
A sample of ice at -10 °C is heated until it melts to form liquid water at room temperature. (a) Describe the arrangement and motion of particles in a solid. [2 marks] (b) During the melting process, the temperature of the ice remains constant at 0 °C even though it is still being heated. Explain why the temperature does not increase during melting. [2 marks] (c) State the name given to the energy required for a substance to change state without a change in temperature. [1 mark] (d) Describe the difference between evaporation and boiling. [2 marks]
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解題
a) In a solid, particles are closely packed in a regular arrangement, and they vibrate about fixed positions. b) During melting, the thermal energy supplied is used to overcome and break the attractive forces (bonds) between the particles of ice, rather than increasing their kinetic energy. Since temperature is a measure of average kinetic energy, the temperature remains constant. c) This energy is called latent heat (specifically, latent heat of fusion). d) Evaporation occurs at any temperature and only at the surface of the liquid, whereas boiling occurs throughout the liquid at a specific temperature (the boiling point).
評分準則
a) Regular arrangement / closely packed [1 mark], vibrating about fixed positions [1 mark] b) Energy is used to break intermolecular forces / bonds [1 mark], so kinetic energy of particles does not increase [1 mark] c) Latent heat [1 mark] d) Evaporation only at surface / any temperature [1 mark], boiling throughout liquid / at boiling point [1 mark]
題目 8 · Structured
9.09 分
An electric kettle is designed to boil water quickly. (a) The kettle transfers 48,000 J of electrical energy in 20 seconds. Calculate the power of the kettle. Use the equation: power = energy transferred / time. [3 marks] (b) The kettle has a power rating of 2400 W and operates at a current of 10 A. Calculate the resistance of the heating element. Use the equation: power = (current)\(^2\) \times resistance. [3 marks] (c) The kettle is connected to the UK mains supply. Explain the difference between direct current (DC) and alternating current (AC). [2 marks]
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解題
a) Power \(P = E / t = 48000 / 20 = 2400\text{ W}\). b) Equation: \(P = I^2 \times R\). Rearranging for \(R\): \(R = P / I^2\). Substituting: \(R = 2400 / 10^2 = 2400 / 100 = 24\text{ }\Omega\). c) Direct current flows in only one direction around a circuit, whereas alternating current continuously reverses its direction of flow.
評分準則
a) Substitution: \(48000 / 20\) [1 mark], calculation: 2400 [1 mark], unit: Watts / W [1 mark] b) Rearrangement: \(R = P / I^2\) [1 mark], substitution: \(2400 / 10^2\) [1 mark], calculation: 24 (\(\Omega\)) [1 mark] c) DC flows in one direction only [1 mark], AC constantly changes/reverses direction [1 mark]
題目 9 · structured
9.09 分
A child of mass 25 kg slides down a playground slide from a vertical height of 3.2 m. The gravitational field strength, g, is 9.8 N/kg. Part A: Calculate the gravitational potential energy of the child at the top of the slide. Show your working and state the unit. Part B: State the maximum possible kinetic energy of the child at the bottom of the slide, assuming no energy is lost to the surroundings. Part C: During the slide, some energy is transferred to the surroundings. Explain what happens to this 'wasted' energy. Part D: At the bottom of the slide, the child's actual speed is less than the theoretical maximum speed. Explain why.
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解題
Part A: Using the formula Ep = m * g * h, we substitute the values: Ep = 25 kg * 9.8 N/kg * 3.2 m = 784. The unit for energy is Joules (J). Part B: According to the conservation of energy, if no energy is lost, all gravitational potential energy is converted to kinetic energy, so the maximum kinetic energy is 784 J. Part C: Frictional forces between the child and the slide, as well as air resistance, cause energy to be transferred to the thermal stores of the slide and air. This causes a temperature increase. Part D: Friction acts on the child, doing work. This means some of the initial potential energy is dissipated as heat rather than being converted into kinetic energy, resulting in a lower final speed.
評分準則
Part A [3 marks]: 1 mark for correct substitution: 25 * 9.8 * 3.2. 1 mark for the correct calculation: 784. 1 mark for the correct unit: J or Joules. Part B [1 mark]: 1 mark for 784 (accept value from Part A). Part C [3 marks]: 1 mark for stating that energy is transferred to the thermal energy store of the slide or air. 1 mark for identifying friction or air resistance as the cause. 1 mark for stating that the temperature of the slide or air increases. Part D [2 marks]: 1 mark for stating that work is done against friction or air resistance. 1 mark for identifying that this dissipates energy as heat or reduces the kinetic energy.
題目 10 · structured
9.09 分
A student sets up a circuit to investigate how the current through a filament lamp changes with the potential difference across it. Part A: State the name of the component that the student should use to vary the current in the circuit. Part B: Explain how the voltmeter and the ammeter should be connected in the circuit to measure the potential difference across the lamp and the current through it. Part C: The student measures a potential difference of 6.0 V and a current of 1.5 A. Calculate the resistance of the filament lamp. Use the equation: potential difference = current * resistance. Show your working and state the unit. Part D: Explain what happens to the resistance of the filament lamp as the potential difference across it increases.
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解題
Part A: A variable resistor (or rheostat) is used to continuously vary the resistance, which in turn varies the current in the circuit. Part B: The voltmeter must be connected in parallel (across) the filament lamp to measure its voltage. The ammeter must be connected in series with the lamp to measure the flow of current through it. Part C: Rearranging the formula V = I * R gives R = V / I. Substituting the values: R = 6.0 V / 1.5 A = 4.0. The unit of resistance is ohms. Part D: Increasing the potential difference increases the current. The increased current causes the temperature of the filament to rise. The increased vibration of ions in the metal lattice increases resistance to electron flow.
評分準則
Part A [1 mark]: 1 mark for variable resistor (allow rheostat). Part B [3 marks]: 1 mark for connecting the voltmeter in parallel across the lamp. 1 mark for connecting the ammeter in series with the lamp. 1 mark for indicating that both meters must be read. Part C [3 marks]: 1 mark for rearranging the formula: R = V / I (or substituting: 6.0 = 1.5 * R). 1 mark for the calculation: 4 (or 4.0). 1 mark for the correct unit: ohms or ̂. Part D [2 marks]: 1 mark for identifying that current increases the temperature of the filament. 1 mark for stating that this increased temperature increases the resistance of the lamp.
題目 11 · structured
9.09 分
This question is about atoms and isotopes. Carbon-12 has an atomic number of 6 and a mass number of 12. Part A: Describe the structure of a carbon-12 atom. State the number of protons, neutrons, and electrons it contains. Part B: Carbon-14 is an isotope of carbon with a mass number of 14. (i) Define the term 'isotope'. (ii) Determine the number of neutrons in a nucleus of carbon-14. Part C: Explain why a complete atom has no overall electrical charge.
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解題
Part A: The atomic number is 6, which is the number of protons. Since it is a neutral atom, there are also 6 electrons. The mass number is 12, so the number of neutrons is mass number minus atomic number: 12 - 6 = 6 neutrons. Part B: (i) An isotope is defined as atoms of the same element with the same atomic number (number of protons) but a different mass number (number of neutrons). (ii) Carbon-14 has mass number 14 and atomic number 6, so it has 14 - 6 = 8 neutrons. Part C: Protons carry a +1 relative charge, and electrons carry a -1 relative charge. Since there are equal numbers of protons and electrons in an atom, the total positive charge balances the total negative charge, leaving a net charge of zero.
評分準則
Part A [3 marks]: 1 mark for stating 6 protons. 1 mark for stating 6 neutrons. 1 mark for stating 6 electrons. Part B [3 marks]: (i) [2 marks]: 1 mark for stating same number of protons or same atomic number. 1 mark for stating different number of neutrons or different mass number. (ii) [1 mark]: 1 mark for 8. Part C [3 marks]: 1 mark for stating that protons are positive and electrons are negative. 1 mark for stating that there are equal numbers of protons and electrons. 1 mark for stating that the charges cancel each other out.
卷二 Foundation
Answer all questions. You must have a ruler, protractor, scientific calculator and the Physics Equations Sheet.
9 題目 · 99.99 分
題目 1 · Structured
11.11 分
A student is investigating water waves in a ripple tank. (a) What type of wave is a water wave? (b) Define the wavelength of a wave. (c) The student measures the frequency of the waves to be \(5\text{ Hz}\). The wavelength is \(0.04\text{ m}\). Calculate the wave speed. Use the equation: wave speed = frequency \(\times\) wavelength. (d) Describe how the student could measure the wavelength more accurately using a ruler and a photograph of the waves. (e) State what happens to the wavelength of the waves as they move into shallower water where they slow down, given that the frequency remains constant.
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解題
(a) Water waves are transverse waves because the oscillations are perpendicular to the direction of energy transfer. (b) Wavelength is defined as the distance between two consecutive identical points on a wave, such as from crest to crest. (c) Using \(v = f \lambda\), we substitute the values: \(v = 5\text{ Hz} \times 0.04\text{ m} = 0.2\text{ m/s}\). (d) To improve accuracy, the student should use a ruler to measure the total distance across a large number of wave crests (for example, 10 waves) from a clear photograph, and then divide this total length by the number of wave cycles to find the average wavelength. (e) Since wave speed is directly proportional to wavelength when frequency is constant (\(v = f \lambda\)), a decrease in speed causes a decrease in wavelength.
評分準則
(a) [1 mark] for stating 'transverse'. (b) [2 marks] for stating 'distance from a point on one wave to the equivalent point on the next wave' (1 mark for distance between crests/trough, 2 marks for complete equivalent point definition). (c) [3 marks total]: 1 mark for correct substitution: \(5 \times 0.04\), 1 mark for correct calculation: \(0.2\), 1 mark for correct unit: \(\text{m/s}\). (d) [3 marks total]: 1 mark for measuring across multiple waves, 1 mark for dividing by the number of waves, 1 mark for using a photograph to freeze the waves. (e) [2 marks total]: 1 mark for stating that the wavelength decreases, 1 mark for explaining that frequency remains constant so wavelength must decrease to match the lower speed.
題目 2 · Structured
11.11 分
A student carries out an experiment to investigate the extension of a spring. (a) State the name of the law that describes the relationship between force and extension for a spring below its limit of proportionality. (b) A student hangs a weight of \(3\text{ N}\) from a spring. The spring extends by \(0.15\text{ m}\). Calculate the spring constant of the spring. Use the equation: force = spring constant \(\times\) extension. State the unit. (c) Explain what is meant by the 'limit of proportionality' of a spring. (d) Identify two safety precautions the student should take when carrying out this experiment. (e) When a very heavy weight is removed, the spring does not return to its original length. State the type of deformation that has occurred.
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解題
(a) Hooke's Law states that extension is directly proportional to the applied force up to the limit of proportionality. (b) Rearranging the formula \(F = k e\) gives \(k = F / e\). Substituting the values: \(k = 3\text{ N} / 0.15\text{ m} = 20\text{ N/m}\). (c) The limit of proportionality is the maximum force that can be applied to the spring where the relationship remains linear. Beyond this point, the extension is no longer proportional to the force. (d) Safety precautions include securing the base of the stand with a G-clamp to prevent tipping and wearing safety goggles to protect eyes in case the spring breaks or flies off. (e) Since the spring did not return to its original shape, it has undergone inelastic deformation.
評分準則
(a) [1 mark] for Hooke's Law. (b) [4 marks total]: 1 mark for rearranging the equation: \(k = F / e\), 1 mark for correct substitution: \(3 / 0.15\), 1 mark for correct value: \(20\), 1 mark for correct unit: \(\text{N/m}\) (or \(\text{N/cm}\) if adjusted, but here must match substitution). (c) [2 marks total]: 1 mark for mentioning 'limit beyond which force and extension are not proportional', 1 mark for indicating the relationship ceases to be linear. (d) [2 marks total]: 1 mark for each sensible safety measure (e.g., clamp stand to bench, wear goggles, place a soft pad below weights). (e) [2 marks total]: 1 mark for inelastic/plastic, 1 mark for explaining it is because it does not return to its original length.
題目 3 · Structured
11.11 分
A student is investigating the magnetic field around bar magnets. (a) Describe how a student could use a plotting compass to map the magnetic field pattern of a bar magnet. (b) State the direction of magnetic field lines. (c) Two North poles of bar magnets are brought close together. Describe the force acting between them. (d) Explain the difference between a permanent magnet and an induced magnet.
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解題
(a) To map the magnetic field: Place the bar magnet on paper. Position a plotting compass near the North pole. Draw a dot at each end of the compass needle. Move the compass so that the tail of the needle aligns with the dot just made. Mark the new tip position. Repeat this process until reaching the South pole, then join the dots to form a field line. Repeat for different starting points. (b) Magnetic field lines always run from the North pole to the South pole of a magnet. (c) Like poles repel, so two North poles brought together experience an electrostatic-like repulsion (magnetic repulsive force). (d) A permanent magnet has its own permanent magnetic field and retains its magnetism over time. An induced magnet is a material (like iron) that only gains magnetic properties when it is close to a permanent magnet, and loses its magnetism quickly when removed from the field.
評分準則
(a) [4 marks total]: 1 mark for placing compass near magnet and marking needle direction with dots, 1 mark for moving compass to the new dot, 1 mark for repeating this sequence to map a line, 1 mark for connecting dots with a continuous line. (b) [1 mark] for stating 'North to South'. (c) [2 marks total]: 1 mark for identifying the force as repulsive/repulsion, 1 mark for stating that like poles repel. (d) [4 marks total]: 2 marks for permanent magnet (produces its own field, does not lose magnetism), 2 marks for induced magnet (only magnetic when in a field, loses magnetism when removed).
題目 4 · Structured
11.11 分
A cyclist is travelling along a straight path. (a) The cyclist travels a distance of \(120\text{ m}\) in a time of \(15\text{ s}\). Calculate the average speed of the cyclist. Use the equation: speed = distance / time. (b) Explain what a horizontal flat line represents on a distance-time graph. (c) Explain the difference between a scalar quantity and a vector quantity. Give one example of each. (d) Write down the equation that links acceleration, change in velocity, and time taken.
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解題
(a) Using the formula \(v = s / t\), we calculate the speed: \(v = 120\text{ m} / 15\text{ s} = 8\text{ m/s}\). (b) On a distance-time graph, the gradient represents the speed of the object. A horizontal flat line has a gradient of zero, which means the speed is zero, indicating the object is stationary. (c) A scalar quantity has size (magnitude) only (e.g., speed, distance). A vector quantity has both size (magnitude) and direction (e.g., velocity, force). (d) The equation is: \(a = \Delta v / t\) where \(a\) is acceleration, \(\Delta v\) is the change in velocity, and \(t\) is the time taken.
評分準則
(a) [3 marks total]: 1 mark for substitution: \(120 / 15\), 1 mark for correct value: \(8\), 1 mark for correct unit: \(\text{m/s}\). (b) [2 marks total]: 1 mark for stating 'stationary' or 'at rest', 1 mark for explaining that the distance is not changing as time passes. (c) [4 marks total]: 1 mark for scalar definition, 1 mark for vector definition, 1 mark for scalar example (e.g. distance, speed, mass), 1 mark for vector example (e.g. displacement, velocity, force). (d) [2 marks total]: 1 mark for correct word formula or symbols, 1 mark for correct arrangement.
題目 5 · Structured
11.11 分
Electromagnetic waves form a continuous spectrum of transverse waves. (a) Put the following electromagnetic waves in order of increasing wavelength: Gamma rays, Radio waves, Visible light, Ultraviolet. (b) State one everyday use of infrared radiation and one everyday use of microwaves. (c) Explain why ultraviolet waves can be hazardous to human skin, whereas radio waves are generally considered safe. (d) X-rays are used in medical imaging. Explain how the properties of X-rays make them suitable for taking images of bones.
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解題
(a) The electromagnetic spectrum in order of increasing wavelength (shortest to longest) is: Gamma rays, Ultraviolet, Visible light, and Radio waves. (b) Infrared is commonly used for cooking (grills/toasters) and electrical heaters. Microwaves are used for heating food in microwave ovens and satellite communication. (c) Ultraviolet radiation has short wavelengths and high frequencies, meaning it carries high energy. It is ionizing radiation, which can penetrate skin cells, damaging DNA and potentially causing mutations or skin cancer. Radio waves have extremely long wavelengths, low frequencies, and low energy. They are non-ionizing and cannot cause chemical/biological damage in cells. (d) X-rays have high energy and short wavelengths, allowing them to pass easily through soft tissues such as muscle and skin. However, they are absorbed by denser materials like bone. This creates a high-contrast image on a photographic film or digital detector placed behind the patient.
評分準則
(a) [2 marks total]: 1 mark if partly correct (e.g., Gamma first, Radio last), 2 marks for completely correct order: Gamma rays, Ultraviolet, Visible light, Radio waves. (b) [2 marks total]: 1 mark for a valid infrared use (e.g., heaters, TV remote), 1 mark for a valid microwave use (e.g., communication, cooking). (c) [3 marks total]: 1 mark for stating UV is ionising / has high energy, 1 mark for stating UV can damage DNA/skin cells, 1 mark for stating radio waves are non-ionising / have low energy. (d) [4 marks total]: 1 mark for stating X-rays can penetrate soft tissue, 1 mark for stating X-rays are absorbed by dense bones, 1 mark for explaining that a detector/film records the transmitted radiation, 1 mark for linking this to creating an image.
題目 6 · Structured
11.11 分
Our Sun is a star in the Milky Way galaxy. (a) Name the force that keeps planets in orbit around the Sun. (b) Our Sun is a stable main sequence star. Explain how the forces inside a main sequence star are balanced. (c) Describe the life cycle of a star with a mass much larger than our Sun, starting after the main sequence stage. (d) Explain the difference between a natural satellite and an artificial satellite, giving an example of each.
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解題
(a) The gravitational force (gravity) acts towards the center of the Sun, keeping planets in their stable orbital paths. (b) A main sequence star is stable because it is in hydrostatic equilibrium. The inward-acting force of gravity is perfectly balanced by the outward-acting radiation pressure produced by nuclear fusion reactions occurring in the star's core. (c) After the main sequence, a massive star expands to form a Red Super Giant. It eventually collapses and explodes in a massive event called a supernova. The remnants of the explosion collapse further to form either a very dense neutron star or, if the mass is extremely large, a black hole. (d) Satellites are bodies that orbit a larger body. A natural satellite occurs naturally in space, such as the Moon orbiting Earth. An artificial satellite is an engineered machine constructed by humans and launched into orbit, such as the GPS satellites.
評分準則
(a) [1 mark] for gravity / gravitational attraction. (b) [3 marks total]: 1 mark for identifying inward force of gravity, 1 mark for identifying outward pressure (due to nuclear fusion), 1 mark for stating these forces are balanced/equal. (c) [4 marks total]: 1 mark for Red Super Giant, 1 mark for Supernova, 1 mark for Neutron Star, 1 mark for Black Hole (last two linked to the mass remnant). (d) [3 marks total]: 1 mark for defining natural satellite with example, 1 mark for defining artificial satellite with example, 1 mark for clear contrast (man-made vs natural).
題目 7 · Structured
11.11 分
Forces act on objects when they interact. (a) Classify the following forces as either contact or non-contact forces: Gravity (weight), Friction, Air resistance, Magnetic force. (b) A car travels at a constant speed in a straight line. Describe the relationship between the forward driving force and the backward resistive forces. (c) The mass of the car is \(1200\text{ kg}\). Calculate the weight of the car on Earth, where the gravitational field strength is \(9.8\text{ N/kg}\). Use the equation: weight = mass \(\times\) gravitational field strength. (d) State Newton's Third Law of motion.
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解題
(a) Friction and Air resistance require physical contact between interacting bodies to exist, so they are contact forces. Gravity and Magnetic forces can act across a distance without direct contact, making them non-contact forces. (b) According to Newton's First Law, if an object travels at a constant velocity, the resultant force must be zero. Therefore, the forward driving force must be equal in magnitude and opposite in direction to the total backward resistive forces (drag and friction). (c) Using the equation \(W = m g\): \(W = 1200\text{ kg} \times 9.8\text{ N/kg} = 11760\text{ N}\). (d) Newton's Third Law states that forces always occur in matched pairs: if body A exerts a force on body B, body B exerts an equal and opposite force on body A.
評分準則
(a) [4 marks total]: 1 mark for each correctly classified force (Friction = contact, Air resistance = contact, Gravity = non-contact, Magnetic = non-contact). (b) [2 marks total]: 1 mark for stating they are equal in magnitude, 1 mark for stating they are opposite in direction (or stating 'resultant force is zero'). (c) [3 marks total]: 1 mark for correct substitution: \(1200 \times 9.8\), 1 mark for correct calculation: \(11760\), 1 mark for correct unit: \(\text{N}\). (d) [2 marks total]: 1 mark for mentioning 'equal forces', 1 mark for mentioning 'opposite direction' between two interacting objects.
題目 8 · Structured
11.11 分
A current-carrying conductor placed inside a magnetic field experiences a force due to the motor effect. (a) State three ways to increase the size of the force acting on a current-carrying wire in a magnetic field. (b) A straight wire of length \(0.20\text{ m}\) carries a current of \(3.0\text{ A}\). It is placed at right angles to a uniform magnetic field with a magnetic flux density of \(0.40\text{ T}\). Calculate the magnetic force acting on the wire. Use the equation: force = magnetic flux density \(\times\) current \(\times\) length. (c) State what each of the following represents in Fleming's Left-Hand Rule: (i) First finger, (ii) Second finger, (iii) Thumb. (d) State the effect of reversing the direction of the current on the direction of the force.
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解題
(a) The magnitude of the electromagnetic force on a conductor is given by \(F = B I l\). To increase this force, one can: increase the magnetic flux density (use a stronger magnet), increase the electric current flowing through the wire, or increase the length of the wire inside the magnetic field. (b) Substituting the given values into the formula: \(F = 0.40\text{ T} \times 3.0\text{ A} \times 0.20\text{ m} = 0.24\text{ N}\). (c) Fleming's Left-Hand Rule helps determine the direction of the force: (i) the First finger points in the direction of the Magnetic Field (North to South), (ii) the Second finger points in the direction of the conventional electric Current (Positive to Negative), and (iii) the Thumb points in the direction of the resulting Motion or Force. (d) Reversing the direction of either the magnetic field or the current will reverse the direction of the force. Thus, reversing current reverses the force direction.
評分準則
(a) [3 marks total]: 1 mark for increasing current, 1 mark for stronger magnetic field, 1 mark for increasing length of wire in the field. (b) [4 marks total]: 1 mark for correct formula identification, 1 mark for substitution: \(0.40 \times 3.0 \times 0.20\), 1 mark for correct calculation: \(0.24\), 1 mark for correct unit: \(\text{N}\). (c) [3 marks total]: 1 mark for First finger = field, 1 mark for Second finger = current, 1 mark for Thumb = force/motion. (d) [1 mark] for stating the force reverses direction.
題目 9 · Structured
11.11 分
A student investigated the relationship between the force applied to a spring and its extension.
**Figure 1** shows the apparatus used.
$$\text{[Apparatus: A vertical stand holding a clamped spring next to a vertical metre ruler with a pointer pointing to the ruler scale.]}$$
### Part (a) State one precaution the student should take to make sure that the measurement of the initial length of the spring is accurate. **[1 mark]**
### Part (b) The student attached a horizontal pointer to the bottom of the spring. Explain how using a pointer improves the accuracy of the measurements. **[1 mark]**
### Part (c) The student added weights to the spring and recorded the extension for each weight. The table below shows some of the results.
Explain how the results in the table show that the spring has exceeded its limit of proportionality. **[2 marks]**
### Part (d) Calculate the spring constant of the spring in its linear region. Your answer should be in newtons per metre (\(\text{N/m}\)). Use the equation: $$\text{force applied to a spring} = \text{spring constant} \times \text{extension}$$ Show your working and state the unit. **[4 marks]**
### Part (e) The student replaced the spring with a different spring. This new spring has a spring constant of \(25\text{ N/m}\). Calculate the elastic potential energy stored in this spring when it has an extension of \(0.20\text{ m}\). Use the equation: $$\text{elastic potential energy} = 0.5 \times \text{spring constant} \times (\text{extension})^2$$ Give your answer to 2 significant figures. **[3 marks]**
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解題
### Part (a) To obtain an accurate reading of the initial length, the student must avoid parallax error. This is achieved by reading the scale on the metre ruler at eye level.
### Part (b) A horizontal pointer makes it easier to align the bottom of the spring with the markings on the ruler, reducing parallax error and making the readings more precise.
### Part (c) In the linear region (from \(0.0\text{ N}\) to \(3.0\text{ N}\)), each additional \(1.0\text{ N}\) of weight causes an extension increase of exactly \(2.0\text{ cm}\). However, when the weight increases from \(3.0\text{ N}\) to \(4.0\text{ N}\), the extension increases by \(3.0\text{ cm}\) (from \(6.0\text{ cm}\) to \(9.0\text{ cm}\)). This change in gradient shows that force and extension are no longer directly proportional, meaning the limit of proportionality has been exceeded between \(3.0\text{ N}\) and \(4.0\text{ N}\).
### Part (d) 1. Identify a pair of values in the linear region: e.g., \(F = 1.0\text{ N}\), \(e = 2.0\text{ cm}\). 2. Convert the extension into metres: $$e = 2.0\text{ cm} = 0.02\text{ m}$$ 3. Rearrange the formula to solve for spring constant (\(k\)): $$k = \frac{F}{e}$$ 4. Substitute the values: $$k = \frac{1.0\text{ N}}{0.02\text{ m}} = 50\text{ N/m}$$
### Part (e) 1. Use the elastic potential energy formula: $$E_e = 0.5 \times k \times e^2$$ 2. Substitute the given values (\(k = 25\text{ N/m}\) and \(e = 0.20\text{ m}\)): $$E_e = 0.5 \times 25 \times (0.20)^2$$ $$E_e = 0.5 \times 25 \times 0.04$$ $$E_e = 0.5\text{ J}$$ 3. Express the final answer to 2 significant figures: $$E_e = 0.50\text{ J}$$
評分準則
### Part (a) [1 mark] * **1 mark**: Read at eye level (to avoid parallax error) OR make sure the ruler is vertical (using a set square or plumb line) OR use a fiducial marker. * *Do not accept: "measure carefully".*
### Part (b) [1 mark] * **1 mark**: Reduces parallax error / makes it easier to read the scale / ensures alignment with ruler markings.
### Part (c) [2 marks] * **1 mark**: Up to \(3.0\text{ N}\), the extension increases by a constant amount (\(2.0\text{ cm}\)) for each additional \(1.0\text{ N}\) added (or force and extension are directly proportional). * **1 mark**: Above \(3.0\text{ N}\) (or at \(4.0\text{ N}\)), the extension increases by more (\(3.0\text{ cm}\)), so it is no longer directly proportional / the limit of proportionality has been exceeded.
### Part (d) [4 marks] * **1 mark**: Convert extension from \(\text{cm}\) to \(\text{m}\) (e.g. \(2.0\text{ cm} = 0.02\text{ m}\) or \(4.0\text{ cm} = 0.04\text{ m}\)). * **1 mark**: Correct rearrangement of equation: \(k = \frac{F}{e}\) or substitution of correct values into equation: \(1.0 = k \times 0.02\). * **1 mark**: Evaluation of spring constant: \(50\). * **1 mark**: Correct unit: \(\text{N/m}\) (or \(\text{N m}^{-1}\)). * *Note: If no unit conversion is done, a maximum of 3 marks can be awarded for an answer of \(0.5\) with the unit \(\text{N/cm}\).*
### Part (e) [3 marks] * **1 mark**: Correct substitution into formula: \(E_e = 0.5 \times 25 \times (0.20)^2\). * **1 mark**: Correct calculation of numerical value: \(0.5\). * **1 mark**: Final answer given to 2 significant figures: \(0.50\) (with or without unit \(\text{J}\)).
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