An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V3) Cambridge IGCSE Physics (0625) paper. Not affiliated with or reproduced from Cambridge.
Paper 4 Mock Structure
Answer all questions. Write your answers in the spaces provided on the question paper. Show all your working and use appropriate units.
10 题目 · 80 分
题目 1 · structured
8 分
An irregularly shaped metal alloy object is investigated. (a) Describe an experimental method to determine the volume of this irregularly shaped metal object using laboratory apparatus. [3] (b) The mass of the metal object is measured as \(340\text{ g}\). The volume is determined to be \(40\text{ cm}^3\). (i) Calculate the density of the metal alloy in \(\text{g/cm}^3\). [2] (ii) Calculate the density of the metal alloy in \(\text{kg/m}^3\). [2] (iii) State one precaution that should be taken when measuring the volume to ensure accuracy. [1]
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解题
(a) Pour a known volume of water into a measuring cylinder and record the initial volume \(V_1\). Carefully submerge the metal object fully in the water. Record the new total volume \(V_2\). The volume of the object is \(V = V_2 - V_1\). (b)(i) \(\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{340\text{ g}}{40\text{ cm}^3} = 8.5\text{ g/cm}^3\). (b)(ii) To convert \(\text{g/cm}^3\) to \(\text{kg/m}^3\), multiply by 1000: \(8.5 \times 1000 = 8500\text{ kg/m}^3\). (b)(iii) Any one of: view the scale perpendicularly to avoid parallax error / measure from the bottom of the meniscus / ensure the object does not splash water out of the cylinder / ensure no air bubbles are trapped on the submerged object.
评分标准
(a) [3 marks]: 1 mark for measuring initial volume of water in a measuring cylinder; 1 mark for fully submerging the object and measuring the new volume; 1 mark for calculating the volume by subtraction. (b)(i) [2 marks]: 1 mark for formula \(\rho = \frac{m}{V}\) or correct substitution; 1 mark for correct value \(8.5\) with unit \(\text{g/cm}^3\). (b)(ii) [2 marks]: 1 mark for multiplying by 1000 or converting units systematically; 1 mark for \(8500\text{ kg/m}^3\). (b)(iii) [1 mark]: Any valid experimental precaution to ensure accuracy (e.g., perpendicular line of sight to read meniscus).
题目 2 · structured
8 分
A student carries out an experiment to investigate a vertical spring. (a) State Hooke's Law. [1] (b) A load of \(6.0\text{ N}\) is hung from the spring. The unstretched length of the spring is \(12.0\text{ cm}\) and its stretched length becomes \(15.0\text{ cm}\). (i) Calculate the spring constant \(k\) of the spring, including its unit. [3] (ii) An additional load of \(4.0\text{ N}\) is added to the spring. Assuming the spring does not exceed its limit of proportionality, calculate the new total length of the spring. [3] (c) Define the term limit of proportionality. [1]
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解题
(a) Hooke's Law states that the extension of a spring is directly proportional to the applied load, provided the limit of proportionality is not exceeded. (b)(i) \(\text{Extension } x = 15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\). Using \(F = kx\), \(k = \frac{F}{x} = \frac{6.0\text{ N}}{3.0\text{ cm}} = 2.0\text{ N/cm}\). (b)(ii) Total load \(F = 6.0\text{ N} + 4.0\text{ N} = 10.0\text{ N}\). Using \(F = kx\), total extension \(x = \frac{F}{k} = \frac{10.0\text{ N}}{2.0\text{ N/cm}} = 5.0\text{ cm}\). New total length \(= \text{original length} + \text{extension} = 12.0\text{ cm} + 5.0\text{ cm} = 17.0\text{ cm}\). (c) The limit of proportionality is the point beyond which the extension of a material is no longer directly proportional to the applied force.
评分标准
(a) [1 mark]: For stating force/load is proportional to extension (provided the limit of proportionality is not exceeded). (b)(i) [3 marks]: 1 mark for finding extension \(x = 3.0\text{ cm}\); 1 mark for formula \(F = kx\) or correct substitution; 1 mark for correct value and unit: \(2.0\text{ N/cm}\) (or \(200\text{ N/m}\)). (b)(ii) [3 marks]: 1 mark for finding new total load of \(10.0\text{ N}\) or additional extension of \(2.0\text{ cm}\); 1 mark for calculating new total extension as \(5.0\text{ cm}\); 1 mark for correct final length \(17.0\text{ cm}\). (c) [1 mark]: For defining it as the limit beyond which force and extension are no longer proportional.
题目 3 · structured
8 分
(a) Define momentum in terms of mass and velocity. [1] (b) Railway truck A of mass \(1200\text{ kg}\) is travelling at \(4.0\text{ m/s}\) along a straight horizontal track. It collides with truck B of mass \(1800\text{ kg}\) which is stationary. After the collision, the two trucks couple together and move off with a common velocity \(v\). (i) Show that the total momentum before the collision is \(4800\text{ kg m/s}\). [2] (ii) Calculate the common velocity \(v\) of the coupled trucks after the collision. [2] (iii) Determine the loss in total kinetic energy of the trucks during the collision. [3]
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解题
(a) Momentum is the product of mass and velocity (\(p = mv\)). (b)(i) \(\text{Total momentum before} = (m_A \times v_A) + (m_B \times v_B) = (1200\text{ kg} \times 4.0\text{ m/s}) + (1800\text{ kg} \times 0) = 4800\text{ kg m/s}\). (b)(ii) By conservation of momentum, \(\text{momentum before} = \text{momentum after}\). \(4800 = (m_A + m_B) \times v = (1200 + 1800) \times v = 3000 \times v\). Thus, \(v = \frac{4800}{3000} = 1.6\text{ m/s}\). (b)(iii) \(\text{Initial kinetic energy } E_k = \frac{1}{2} m_A v_A^2 = \frac{1}{2} \times 1200 \times 4.0^2 = 9600\text{ J}\). \(\text{Final kinetic energy } E_k = \frac{1}{2} (m_A + m_B) v^2 = \frac{1}{2} \times 3000 \times 1.6^2 = 3840\text{ J}\). \(\text{Loss in kinetic energy} = 9600\text{ J} - 3840\text{ J} = 5760\text{ J}\).
评分标准
(a) [1 mark]: For stating \(\text{momentum} = \text{mass} \times \text{velocity}\). (b)(i) [2 marks]: 1 mark for substituting values \(1200 \times 4.0\); 1 mark for showing it equals \(4800\text{ kg m/s}\). (b)(ii) [2 marks]: 1 mark for equating momentum before and after \(4800 = 3000 \times v\); 1 mark for correct value \(1.6\text{ m/s}\) with correct unit. (b)(iii) [3 marks]: 1 mark for calculating initial kinetic energy \(9600\text{ J}\); 1 mark for calculating final kinetic energy \(3840\text{ J}\); 1 mark for correct subtraction to get \(5760\text{ J}\) (or \(5800\text{ J}\)).
题目 4 · structured
8 分
(a) Explain, in terms of particles, how a gas exerts a pressure on the walls of its container. [3] (b) A cylinder contains a fixed mass of gas at a constant temperature. The initial pressure of the gas is \(1.5 \times 10^5\text{ Pa}\) and its volume is \(0.080\text{ m}^3\). The piston is pushed in so that the volume is reduced to \(0.030\text{ m}^3\). (i) Calculate the new pressure of the gas. [3] (ii) State and explain, in terms of gas particles, why the pressure increases when the volume is reduced at constant temperature. [2]
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解题
(a) Gas particles are in constant, random motion. They collide with the walls of the container. During these collisions, the particles undergo a change in momentum, which exerts a force on the walls. The total force per unit area equals the gas pressure. (b)(i) Since temperature is constant, \(p_1 V_1 = p_2 V_2\). \(1.5 \times 10^5 \times 0.080 = p_2 \times 0.030\). \(12000 = p_2 \times 0.030 \implies p_2 = \frac{12000}{0.030} = 4.0 \times 10^5\text{ Pa}\). (b)(ii) When volume is reduced, the particles are packed closer together (higher concentration). Consequently, they collide more frequently with the container walls, which increases the total force per unit area (pressure).
评分标准
(a) [3 marks]: 1 mark for particles colliding with container walls; 1 mark for collisions causing force on the walls (due to change in momentum); 1 mark for pressure being force per unit area. (b)(i) [3 marks]: 1 mark for formula \(p_1 V_1 = p_2 V_2\); 1 mark for correct substitution \(1.5 \times 10^5 \times 0.080 = p_2 \times 0.030\); 1 mark for final correct value \(4.0 \times 10^5\text{ Pa}\) with unit. (b)(ii) [2 marks]: 1 mark for stating that particles collide more frequently with the walls (higher frequency of collision); 1 mark for relating this to the smaller volume/less space.
题目 5 · structured
8 分
A potential divider circuit is set up using a thermistor. (a) Describe how the electrical resistance of a thermistor changes as its temperature increases. [1] (b) A \(12\text{ V}\) battery of negligible internal resistance is connected in series with a fixed resistor of resistance \(400\ \Omega\) and the thermistor. At a temperature of \(20^\circ\text{C}\), the resistance of the thermistor is \(800\ \Omega\). (i) Calculate the current in the circuit. [2] (ii) Calculate the potential difference across the thermistor. [2] (c) The temperature of the thermistor is increased. (i) State and explain how this temperature increase affects the potential difference across the thermistor. [2] (ii) Suggest one practical application for this potential divider circuit. [1]
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解题
(a) The resistance of a thermistor decreases as its temperature increases. (b)(i) Total resistance \(R_{\text{total}} = R_{\text{fixed}} + R_{\text{thermistor}} = 400 + 800 = 1200\ \Omega\). Current \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{1200\ \Omega} = 0.010\text{ A}\) (or \(10\text{ mA}\)). (b)(ii) Potential difference across the thermistor \(V_{\text{thermistor}} = I \times R_{\text{thermistor}} = 0.010\text{ A} \times 800\ \Omega = 8.0\text{ V}\). (c)(i) As temperature increases, the resistance of the thermistor decreases. Since the resistance of the thermistor decreases relative to the fixed resistor, it takes a smaller fraction of the total battery voltage, so the potential difference across the thermistor decreases. (c)(ii) Temperature sensor / fire alarm / thermostat / cooling fan controller.
评分标准
(a) [1 mark]: For stating resistance decreases as temperature increases. (b)(i) [2 marks]: 1 mark for finding total resistance \(1200\ \Omega\); 1 mark for current calculation \(0.010\text{ A}\) (accept \(10\text{ mA}\)). (b)(ii) [2 marks]: 1 mark for formula \(V = IR\) or correct ratio method; 1 mark for correct value \(8.0\text{ V}\) with unit. (c)(i) [2 marks]: 1 mark for stating that potential difference across the thermistor decreases; 1 mark for explaining that its resistance decreases (reducing its share of the total voltage). (c)(ii) [1 mark]: Any valid temperature-related control application (e.g., thermostat/fire alarm).
题目 6 · structured
8 分
(a) Define electromotive force (e.m.f.). [2] (b) A charge of \(360\text{ C}\) passes through a resistor in a time of \(3.0\text{ minutes}\). (i) Calculate the current in the resistor. [2] (ii) The potential difference across the resistor is \(6.0\text{ V}\). Calculate the resistance of the resistor. [2] (iii) Calculate the electrical energy transferred in the resistor during this time. [2]
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解题
(a) Electromotive force (e.m.f.) is the energy supplied by a source in driving charge around a complete circuit per unit charge. (b)(i) Convert time to seconds: \(t = 3.0 \times 60 = 180\text{ s}\). Current \(I = \frac{Q}{t} = \frac{360\text{ C}}{180\text{ s}} = 2.0\text{ A}\). (b)(ii) Using Ohm's Law \(V = IR\), \(R = \frac{V}{I} = \frac{6.0\text{ V}}{2.0\text{ A}} = 3.0\ \Omega\). (b)(iii) Energy transferred \(E = VIt = VQ = 6.0\text{ V} \times 360\text{ C} = 2160\text{ J}\) (or \(2.16\text{ kJ}\)).
评分标准
(a) [2 marks]: 1 mark for energy supplied per unit charge (or work done per unit charge); 1 mark for mention of 'by a source' or 'around a complete circuit'. (b)(i) [2 marks]: 1 mark for converting time to \(180\text{ s}\); 1 mark for correct current \(2.0\text{ A}\) with unit. (b)(ii) [2 marks]: 1 mark for formula \(R = \frac{V}{I}\) or correct substitution; 1 mark for correct value \(3.0\ \Omega\) with unit. (b)(iii) [2 marks]: 1 mark for formula \(E = VQ\) or \(E = VIt\) or correct substitution; 1 mark for correct value \(2160\text{ J}\) (or \(2.2\text{ kJ}\)).
题目 7 · structured
8 分
(a) Carbon-14 (\(^{14}_{\phantom{0}6}\text{C}\)) decays by emitting a \(\beta\)-particle to form a stable isotope of nitrogen (\(\text{N}\)). (i) Determine the nucleon number and proton number of the nitrogen nucleus formed. [2] (ii) State the nucleon number and proton number of the \(\beta\)-particle. [1] (b) The half-life of Carbon-14 is \(5700\text{ years}\). A sample of wood from an ancient tree contains \(2.0\text{ mg}\) of Carbon-14 when first formed. (i) Determine the mass of Carbon-14 remaining in the sample after \(17\,100\text{ years}\). [2] (ii) State two different sources of background radiation. [2] (iii) State what is meant by a \(\beta\)-particle. [1]
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解题
(a)(i) During beta decay, a neutron decays into a proton and an electron. The nucleon number of nitrogen remains 14. The proton number of nitrogen increases by 1 to become 7. (a)(ii) A beta particle (electron) has a nucleon number of 0 and a proton number of -1. (b)(i) Number of half-lives \(n = \frac{17100\text{ years}}{5700\text{ years}} = 3\). Mass remaining \(= 2.0\text{ mg} \times \left(\frac{1}{2}\right)^3 = 2.0 \times \frac{1}{8} = 0.25\text{ mg}\). (b)(ii) Any two from: cosmic rays, radon gas, rocks/soil, medical sources (X-rays), food/drink, nuclear waste. (b)(iii) A beta particle is a high-speed electron emitted from the nucleus.
评分标准
(a)(i) [2 marks]: 1 mark for nucleon number = 14; 1 mark for proton number = 7. (a)(ii) [1 mark]: For nucleon number = 0 and proton number = -1. (b)(i) [2 marks]: 1 mark for identifying that 3 half-lives have elapsed; 1 mark for correct mass \(0.25\text{ mg}\) (accept \(2.5 \times 10^{-4}\text{ g}\)). (b)(ii) [2 marks]: 1 mark for each valid background source up to 2 (e.g., radon gas, cosmic rays). (b)(iii) [1 mark]: For defining beta particle as a (high-speed) electron.
题目 8 · structured
8 分
(a) Light from distant galaxies is redshifted when observed on Earth. (i) State what redshift tells us about the motion of distant galaxies. [1] (ii) Explain how redshift provides evidence for the Big Bang theory. [2] (b) Hubble's constant \(H_0\) is used in space physics. (i) State the equation that defines \(H_0\) in terms of the recession speed \(v\) of a galaxy and its distance \(d\) from Earth. [1] (ii) A galaxy is at a distance of \(3.1 \times 10^{22}\text{ m}\) from Earth and has a recession speed of \(6.8 \times 10^4\text{ m/s}\). Use these values to calculate an estimate for Hubble's constant \(H_0\). Include its unit. [2] (iii) Using your calculated value of \(H_0\), estimate the age of the Universe in years. Note: \(1\text{ year} = 3.16 \times 10^7\text{ s}\). [2]
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解题
(a)(i) Redshift tells us that distant galaxies are moving away from Earth. (a)(ii) Redshift shows that further galaxies are receding faster, indicating that space is expanding. If we run this expansion backward in time, all matter must have originated from a single extremely hot, dense point (the Big Bang). (b)(i) The equation is \(H_0 = \frac{v}{d}\). (b)(ii) \(H_0 = \frac{6.8 \times 10^4\text{ m/s}}{3.1 \times 10^{22}\text{ m}} \approx 2.19 \times 10^{-18}\text{ s}^{-1}\). (b)(iii) The age of the Universe \(T \approx \frac{1}{H_0} = \frac{1}{2.19 \times 10^{-18}\text{ s}^{-1}} \approx 4.57 \times 10^{17}\text{ s}\). Converting to years: \(T = \frac{4.57 \times 10^{17}}{3.16 \times 10^7} \approx 1.44 \times 10^{10}\text{ years}\) (or \(14\text{ billion years}\)).
评分标准
(a)(i) [1 mark]: For stating that galaxies are moving away (receding) from Earth. (a)(ii) [2 marks]: 1 mark for stating that the expansion of space is shown; 1 mark for linking expansion to a single starting point in the past. (b)(i) [1 mark]: For formula \(H_0 = \frac{v}{d}\). (b)(ii) [2 marks]: 1 mark for correct substitution of values; 1 mark for correct value \(2.2 \times 10^{-18}\) with unit \(\text{s}^{-1}\) (or \(/\text{s}\)). (b)(iii) [2 marks]: 1 mark for relationship \(T = \frac{1}{H_0}\) or correct conversion method; 1 mark for correct value in years \(1.4 \times 10^{10}\text{ years}\) (allow range \(1.3 \times 10^{10}\) to \(1.5 \times 10^{10}\)).
题目 9 · structured
8 分
A toy cart of mass 0.40 kg travelling at a constant speed of 2.5 m/s along a straight track collides with a stationary toy cart of mass 0.60 kg. After the collision, the first cart bounces backward along the same track with a speed of 0.50 m/s.
(a) Calculate the momentum of the first cart before the collision.
(b) Calculate the velocity of the second cart after the collision.
(c) The collision lasts for a duration of 0.15 s. Calculate the average force exerted on the first cart during the collision.
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解题
(a) The momentum \(p\) of the first cart before the collision is calculated using: \(p = m_1 \times u_1\) \(p = 0.40 \text{ kg} \times 2.5 \text{ m/s} = 1.0 \text{ kg m/s}\) (or \(\text{N s}\)).
(b) Using the principle of conservation of momentum: \(\text{Total momentum before collision} = \text{Total momentum after collision}\) \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\) Taking the initial direction of motion as positive: \(1.0 + 0 = 0.40 \times (-0.50) + 0.60 \times v_2\) \(1.0 = -0.20 + 0.60 v_2\) \(1.2 = 0.60 v_2\) \(v_2 = 2.0 \text{ m/s}\) (in the original direction of the first cart).
(c) The average force \(F\) is given by the rate of change of momentum of the first cart: \(F = \frac{\Delta p}{t} = \frac{m_1 v_1 - m_1 u_1}{t}\) \(F = \frac{0.40 \times (-0.50) - 1.0}{0.15} = \frac{-0.20 - 1.0}{0.15} = \frac{-1.2}{0.15} = -8.0 \text{ N}\). The magnitude of the average force exerted on the cart is \(8.0 \text{ N}\).
评分标准
(a) - C1: For formula \(p = mv\) or correct substitution \(0.40 \times 2.5\) - A1: Correct value with unit: \(1.0 \text{ kg m/s}\) or \(1.0 \text{ N s}\)
(b) - C1: For stating principle of conservation of momentum or writing equation \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\) - C1: For correct substitution showing negative sign for bounce velocity: \(1.0 = 0.40 \times (-0.50) + 0.60 \times v_2\) or \(1.2 = 0.60 v_2\) - A1: Correct final value: \(2.0 \text{ m/s}\)
(c) - C1: For formula \(F = \frac{\Delta p}{t}\) or \(\text{Impulse} = F \times t\) - C1: For calculation of change in momentum: \(1.2 \text{ kg m/s}\) (or \(\text{N s}\)) or \(-1.2 \text{ kg m/s}\) (or \(\text{N s}\)) - A1: Correct final value: \(8.0 \text{ N}\) (or \(-8.0 \text{ N}\))
题目 10 · structured
8 分
A student designs a temperature-sensing potential divider circuit to monitor the temperature of a cold room. The circuit is powered by a 12 V battery of negligible internal resistance and consists of a thermistor connected in series with a fixed resistor of resistance 1.5 k\(\Omega\). A voltmeter is connected across the fixed resistor.
(a) Draw a circuit diagram of this temperature-sensing potential divider circuit.
(b) At a room temperature of 20 °C, the resistance of the thermistor is 4.5 k\(\Omega\). (i) Show that the reading on the voltmeter at this temperature is 3.0 V. (ii) Calculate the current in the circuit at this temperature.
(c) The temperature of the cold room decreases. State and explain the effect of this temperature change on the reading on the voltmeter.
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解题
(a) The circuit diagram should display: - A series loop containing a 12 V DC source (battery symbol), a thermistor symbol (rectangle with a diagonal line ending in a horizontal flat section, or similar standard thermistor symbol), and a fixed resistor symbol (rectangle). - A voltmeter symbol (circle with 'V') connected in parallel across the fixed resistor.
(b)(ii) Current in the circuit can be calculated using Ohm's Law across the fixed resistor: \(I = \frac{V_{\text{fixed}}}{R_{\text{fixed}}}\) \(I = \frac{3.0 \text{ V}}{1500 \text{ }\Omega} = 0.0020 \text{ A}\) (or \(2.0 \text{ mA}\)). Alternatively, using total resistance: \(R_{\text{total}} = 1500 \text{ }\Omega + 4500 \text{ }\Omega = 6000 \text{ }\Omega\) \(I = \frac{12 \text{ V}}{6000 \text{ }\Omega} = 0.0020 \text{ A}\).
(c) When the temperature decreases: - The resistance of the thermistor increases. - Since it has a higher resistance, it takes a larger share of the 12 V potential difference. - Consequently, the potential difference across the fixed resistor decreases, causing the reading on the voltmeter to decrease.
评分标准
(a) - B1: Correct symbols for battery, thermistor, fixed resistor connected in series. - B1: Voltmeter connected in parallel across the fixed resistor only.
(b)(i) - C1: Correct formula for potential divider \(V_{\text{out}} = V_{\text{in}} \times \frac{R_1}{R_1 + R_2}\) OR calculation of total resistance \(R_{\text{total}} = 6.0 \text{ k}\Omega\) and current \(I = 2.0 \text{ mA}\). - A1: Correct substitution showing the output is \(3.0 \text{ V}\).
(b)(ii) - C1: For using \(I = \frac{V}{R}\) with correct values (e.g., \(\frac{3.0}{1500}\) or \(\frac{12}{6000}\)). - A1: Correct value with unit: \(0.0020 \text{ A}\) or \(2.0 \text{ mA}\).
(c) - B1: State that the voltmeter reading decreases. - B1: Explain that as temperature decreases, thermistor resistance increases, so it takes a larger proportion of the supply voltage.