An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V2) Cambridge IGCSE Physics (0625) paper. Not affiliated with or reproduced from Cambridge.
部分 Structured Extended Theory
Answer all 11 structured questions. Show all working clearly. Use standard values: g = 9.8 N/kg.
11 题目 · 79.15999999999998 分
题目 1 · structured
7.27 分
A research remotely operated vehicle (ROV) of mass \(150\text{ kg}\) is descending vertically through seawater. At \(t = 0\), it has a downward speed of \(3.5\text{ m/s}\). It decelerates uniformly to a speed of \(1.1\text{ m/s}\) over a time interval of \(6.0\text{ s}\), and then continues descending at this constant speed.
(a) Define acceleration.
(b) Calculate the deceleration of the ROV during the first \(6.0\text{ s}\).
(c) Calculate the total distance descended by the ROV between \(t = 0\) and \(t = 10.0\text{ s}\).
查看答案详解收起答案详解
解题
(a) Acceleration is defined as the rate of change of velocity, or change in velocity per unit time.
(b) Using the acceleration formula: \(a = \frac{v - u}{t} = \frac{1.1\text{ m/s} - 3.5\text{ m/s}}{6.0\text{ s}} = -0.40\text{ m/s}^2\) Thus, the deceleration is \(0.40\text{ m/s}^2\).
(c) The total distance descended is the area under the speed-time graph from \(t = 0\) to \(t = 10.0\text{ s}\): - For the first \(6.0\text{ s}\) (trapezium area): \(s_1 = \frac{3.5 + 1.1}{2} \times 6.0 = 13.8\text{ m}\) - For the remaining \(4.0\text{ s}\) (constant speed): \(s_2 = 1.1 \times 4.0 = 4.4\text{ m}\) Total distance descended = \(13.8\text{ m} + 4.4\text{ m} = 18.2\text{ m}\).
评分标准
(a) B1: Rate of change of velocity OR change in velocity per unit time. [1.0 mark] B1: Mention of direction/vector nature or correct unit context if fully defined. [1.0 mark]
(c) C1: Calculation of distance in first phase: \(13.8\text{ m}\) (area of trapezium/integration). [1.0 mark] C1: Calculation of distance in second phase: \(1.1 \times 4.0 = 4.4\text{ m}\). [1.0 mark] A1.27: Total distance of \(18.2\text{ m}\) with correct unit. [1.27 marks]
题目 2 · structured
7.27 分
An offshore wind turbine generator produces an alternating voltage of \(690\text{ V}\) a.c. This is connected to a step-up transformer with \(120\text{ turns}\) on the primary coil to raise the voltage to \(33\text{ kV}\) (\(33\,000\text{ V}\)) for marine cable transmission.
(a) Calculate the number of turns on the secondary coil of this transformer.
(b) State the material used for the core of the transformer and explain why this material is chosen.
(c) The power input to the transformer is \(2.4\text{ MW}\). Calculate the primary current, assuming the transformer is \(100\%\) efficient.
查看答案详解收起答案详解
解题
(a) Using the transformer equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\) \(\frac{690}{33\,000} = \frac{120}{N_s}\) \(N_s = \frac{120 \times 33\,000}{690} \approx 5739\text{ turns}\)
(b) The material used for the core is soft iron. Soft iron is chosen because it is easily magnetised and demagnetised, which minimises energy losses due to alternating magnetic fields.
(c) C1: Recall of \(P = V \times I\) or substitution of \(2.4 \times 10^6\text{ W}\). [1.0 mark] A1.27: \(3478\text{ A}\) (accept \(3500\text{ A}\) or \(3.5\text{ kA}\)) with correct unit. [1.27 marks]
题目 3 · structured
7.27 分
A ray of light is directed from a container of liquid paraffin towards the boundary with air. The refractive index of the liquid paraffin is \(1.48\).
(a) Calculate the critical angle \(c\) for the liquid paraffin-air boundary.
(b) Explain what happens to the ray of light if it strikes the boundary at an angle of incidence of \(40^\circ\).
(c) State the two conditions required for total internal reflection to occur.
查看答案详解收起答案详解
解题
(a) Using the formula for the critical angle: \(\sin c = \frac{1}{n}\) \(\sin c = \frac{1}{1.48} \approx 0.6757\) \(c = \arcsin(0.6757) \approx 42.5^\circ\)
(b) Since the angle of incidence (\(40^\circ\)) is less than the critical angle (\(42.5^\circ\)), the light ray will be refracted out into the air, bending away from the normal. A faint partially reflected ray will also be observed inside the liquid.
(c) The two conditions required for total internal reflection are: 1. The light must be travelling from an optically denser medium to an optically less dense medium. 2. The angle of incidence must be greater than the critical angle.
评分标准
(a) C1: Recall of \(\sin c = \frac{1}{n}\). [1.0 mark] C1: Correct substitution: \(\sin c = \frac{1}{1.48}\). [1.0 mark] A1: \(42.5^\circ\) (accept \(43^\circ\)). [1.0 mark]
(b) B1: Light is refracted out into the air. [1.0 mark] B1: Bends away from the normal (since \(i < c\)). [1.0 mark]
(c) B1: Light travels from denser to less dense medium. [1.0 mark] B1.27: Angle of incidence is greater than critical angle. [1.27 marks]
题目 4 · structured
7.27 分
Light from a distant galaxy is observed on Earth. A specific spectral line of hydrogen, which has a wavelength of \(656.3\text{ nm}\) in a laboratory on Earth, is observed at a wavelength of \(678.5\text{ nm}\) in the light from this galaxy.
(a) State the name given to this shift in wavelength and explain what it indicates about the motion of the galaxy.
(b) State the name of the theory of the Universe supported by this observation.
(c) Explain what is meant by Cosmic Microwave Background Radiation (CMBR) and how its existence supports this theory of the Universe.
查看答案详解收起答案详解
解题
(a) This shift to a longer wavelength is called redshift. It indicates that the galaxy is moving away from the observer (Earth).
(b) This supports the Big Bang theory.
(c) CMBR is electromagnetic radiation in the microwave region received from all parts of the sky. It represents the remnant heat/radiation left over from the Big Bang. As the Universe expanded, this high-energy radiation from the early Universe was stretched to longer wavelengths (microwaves). Its highly uniform presence across space supports the Big Bang theory by showing that the Universe expanded from a single extremely hot and dense initial state.
评分标准
(a) B1: Redshift. [1.0 mark] B1: The galaxy is moving away (from Earth). [1.0 mark]
(b) B1: Big Bang theory. [1.0 mark]
(c) B1: Definition of CMBR as microwave radiation from all directions. [1.0 mark] B1: Remnant radiation/heat from the early hot phase of the Big Bang. [1.0 mark] B1: Stretched wavelengths due to the expansion of space. [1.0 mark] B1.27: Highly uniform distribution supports expansion from a single point. [1.27 marks]
题目 5 · structured
7.27 分
An engineer uses an ultrasound transducer to detect internal cracks in a vertical concrete pillar of thickness \(0.80\text{ m}\). A pulse of ultrasound is emitted into the concrete. The speed of sound in concrete is \(3200\text{ m/s}\).
(a) Calculate the time taken for the pulse to travel through the pillar to the opposite side and return to the transducer if there are no cracks.
(b) Define the term ultrasound.
(c) An echo is detected after \(0.35\text{ ms}\) (\(3.5 \times 10^{-4}\text{ s}\)). Explain whether this indicates a crack, and if so, calculate its depth from the surface where the transducer is placed.
查看答案详解收起答案详解
解题
(a) Total distance travelled through the pillar and back = \(2 \times 0.80\text{ m} = 1.60\text{ m}\). Using \(t = \frac{d}{v}\): \(t = \frac{1.60\text{ m}}{3200\text{ m/s}} = 5.0 \times 10^{-4}\text{ s}\) (or \(0.50\text{ ms}\)).
(b) Ultrasound is sound waves with a frequency higher than \(20\,000\text{ Hz}\) (or \(20\text{ kHz}\)), which is above the limit of human hearing.
(c) Since \(0.35\text{ ms}\) is less than the expected \(0.50\text{ ms}\) return time, this indicates a crack is present. Distance to the crack and back: \(d = v \times t = 3200 \times 3.5 \times 10^{-4} = 1.12\text{ m}\). Depth of crack = \(\frac{1.12}{2} = 0.56\text{ m}\).
(b) B1: Frequency greater than \(20\,000\text{ Hz}\) / \(20\text{ kHz}\). [1.0 mark] B1: Beyond/above human hearing limit. [1.0 mark]
(c) B1: Explanation that \(0.35\text{ ms} < 0.50\text{ ms}\) confirms a crack. [1.0 mark] A1.27: Correct calculation of crack depth = \(0.56\text{ m}\) with correct unit. [1.27 marks]
题目 6 · structured
7.27 分
An electric winch of power rating \(1.5\text{ kW}\) is used to lift a cargo container of mass \(450\text{ kg}\) vertically upwards from a ship's hold to a height of \(8.0\text{ m}\). The process takes \(30\text{ s}\).
(a) Calculate the work done in lifting the container.
(b) Calculate the useful power output of the winch.
(c) Calculate the efficiency of the winch system during this lift.
查看答案详解收起答案详解
解题
(a) The work done is equal to the gain in gravitational potential energy: \(W = m \times g \times h = 450\text{ kg} \times 9.8\text{ N/kg} \times 8.0\text{ m} = 35\,280\text{ J}\) (or \(35.3\text{ kJ}\)).
(c) C1: Correct matching of units (e.g. converting input power to \(1500\text{ W}\)). [1.0 mark] A1.27: \(78.4\%\) (accept \(78.7\%\) if using \(1180\text{ W}\)). [1.27 marks]
题目 7 · structured
7.27 分
A uniform diving board of length \(4.0\text{ m}\) and weight \(350\text{ N}\) is supported by a pivot at one end (A) and a vertical support pillar at a distance of \(1.2\text{ m}\) from A. A diver of weight \(680\text{ N}\) stands at the opposite end (B) of the board. The board is in equilibrium.
(a) State the two conditions required for an object to be in equilibrium.
(b) Calculate the moment of the diver's weight about the pivot A.
(c) Calculate the upward force exerted by the support pillar on the diving board.
查看答案详解收起答案详解
解题
(a) The two conditions for equilibrium are: 1. The resultant force acting on the object is zero. 2. The resultant moment acting on the object is zero.
(b) The diver's weight acts at end B, which is \(4.0\text{ m}\) from pivot A. \(\text{Moment} = \text{Force} \times \text{distance} = 680\text{ N} \times 4.0\text{ m} = 2720\text{ Nm}\) (clockwise).
(a) B1: Resultant/net force is zero. [1.0 mark] B1: Resultant/net moment is zero. [1.0 mark]
(b) C1: Moment = \(F \times d\) with \(d = 4.0\text{ m}\). [1.0 mark] A1: \(2720\text{ Nm}\). [1.0 mark]
(c) C1: Account for the uniform weight of the board acting at center \(2.0\text{ m}\). [1.0 mark] C1: Equation for sum of clockwise moments = \(3420\text{ Nm}\). [1.0 mark] A1.27: Correct value of \(2850\text{ N}\) upward with unit. [1.27 marks]
题目 8 · structured
7.27 分
Carbon-14 (\(^{14}_{6}\text{C}\)) is a radioactive isotope of carbon that decays by emitting a beta-particle (\(\beta^-\)) to form nitrogen-14 (\(^{14}_{7}\text{N}\)).
(a) Write a balanced nuclear equation for the decay of Carbon-14. Use the symbol \(\text{e}\) or \(\beta\) with appropriate proton and nucleon numbers for the beta-particle.
(b) Explain what happens to the proton number and neutron number of the nucleus during beta-minus (\(\beta^-\)) decay.
(c) A sample of organic material originally contained \(8.0\text{ \mu g}\) of Carbon-14. The half-life of Carbon-14 is \(5730\text{ years}\). Calculate the mass of Carbon-14 remaining in the sample after \(17\,190\text{ years}\).
查看答案详解收起答案详解
解题
(a) The decay equation is: \(^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + ^{0}_{-1}\text{e}\)
(b) During beta-minus decay, a neutron inside the nucleus decays into a proton and an electron. Therefore: - The proton number increases by 1. - The neutron number decreases by 1. (The mass number remains unchanged.)
(c) Number of half-lives elapsed: \(n = \frac{17\,190}{5730} = 3\text{ half-lives}\). Mass remaining: \(m = 8.0\text{ \mu g} \times \left(\frac{1}{2}\right)^3 = 8.0 \times \frac{1}{8} = 1.0\text{ \mu g}\).
评分标准
(a) B1: Correct symbols for reactants and products (C, N). [1.0 mark] B1: Correct symbol and numbers for beta-particle (\(^{0}_{-1}\text{e}\) or \(^{0}_{-1}\beta\)). [1.0 mark] B1: Equation balanced on both sides for nucleon and proton numbers. [1.0 mark]
(b) B1: Proton number increases by 1. [1.0 mark] B1: Neutron number decreases by 1. [1.0 mark]
(c) C1: Identification of 3 half-lives. [1.0 mark] A1.27: \(1.0\text{ \mu g}\) (or \(1.0 \times 10^{-6}\text{ g}\)) with correct unit. [1.27 marks]
题目 9 · Structured
7 分
A toy railway car A of mass 0.40 kg is moving at a speed of 2.5 m/s along a straight horizontal track. It collides with a stationary toy railway car B of mass 0.60 kg. The two cars couple together during the collision. (a) Define momentum. [1] (b) Calculate the velocity of the joined cars after the collision. [3] (c) The collision lasts for 0.15 s. Calculate the average force exerted by car A on car B during the collision. [3]
查看答案详解收起答案详解
解题
(a) Momentum is defined as the product of mass and velocity (p = mv). (b) Initial momentum = m_A * u_A + m_B * u_B = 0.40 kg * 2.5 m/s + 0 = 1.0 kg m/s. Since total momentum is conserved, final velocity v = (total initial momentum) / (m_A + m_B) = 1.0 kg m/s / (0.40 kg + 0.60 kg) = 1.0 m/s. (c) Average force F = change in momentum of car B / time = (m_B * v - 0) / t = (0.60 kg * 1.0 m/s) / 0.15 s = 4.0 N.
评分标准
(a) Product of mass and velocity [1] (b) Use of conservation of momentum: m_A * u_A = (m_A + m_B) * v [1]; Substitution: 0.40 * 2.5 = (0.40 + 0.60) * v [1]; v = 1.0 m/s (with unit) [1] (c) Use of F = change in momentum / time [1]; Change in momentum of B = 0.60 kg m/s [1]; F = 4.0 N (with unit) [1]
题目 10 · Structured
7 分
A ray of monochromatic light is incident on the flat face of a semi-circular glass block at an angle of incidence of 40 degrees. The refractive index of the glass is 1.52. (a)(i) Explain what is meant by monochromatic light. [1] (ii) Calculate the angle of refraction inside the glass block. [3] (b) Calculate the critical angle for this glass block. [3]
查看答案详解收起答案详解
解题
(a)(i) Monochromatic light consists of light of a single frequency or wavelength. (ii) Using Snell's Law: n = sin(i) / sin(r) -> sin(r) = sin(i) / n = sin(40) / 1.52 = 0.6428 / 1.52 = 0.4229. Thus, r = arcsin(0.4229) = 25.0 degrees. (b) Critical angle c is found using: sin(c) = 1 / n = 1 / 1.52 = 0.6579. Thus, c = arcsin(0.6579) = 41.1 degrees.
评分标准
(a)(i) Light of a single frequency / wavelength [1] (ii) Use of n = sin(i) / sin(r) [1]; Substitution: sin(r) = sin(40) / 1.52 [1]; r = 25 degrees (accept 25.0) [1] (b) Use of sin(c) = 1 / n [1]; Substitution: sin(c) = 1 / 1.52 [1]; c = 41 degrees (accept 41.1) [1]
题目 11 · Structured
7 分
A radioactive Carbon-14 nucleus (nucleons = 14, protons = 6) decays by emitting a beta-minus particle to form a stable Nitrogen-14 nucleus (nucleons = 14, protons = 7). (a) Describe the composition of a Carbon-14 nucleus in terms of protons and neutrons. [2] (b) Complete the nuclide equation for the beta-decay of Carbon-14 by writing down the nucleon number and proton number for Nitrogen, and the symbol for the beta-particle. [3] (c) A beam of beta-particles enters a region of space between two oppositely charged horizontal metal plates. State and explain the direction in which the beam of beta-particles is deflected. [2]
查看答案详解收起答案详解
解题
(a) Carbon-14 has a proton number (Z) of 6 and a nucleon number (A) of 14. Therefore, it has 6 protons, and 14 - 6 = 8 neutrons. (b) Nitrogen-14 has nucleon number 14 and proton number 7. The emitted beta-particle is represented as an electron with nucleon number 0 and proton number -1. (c) Since beta-particles are fast-moving electrons, they carry a negative charge. In an electric field between oppositely charged plates, negative charges are attracted towards the positive plate and repelled by the negative plate, so the beam deflects towards the positive plate.
评分标准
(a) 6 protons [1]; 8 neutrons [1] (b) Nitrogen nucleon number = 14 [1]; Nitrogen proton number = 7 [1]; Beta-particle symbol with nucleon number 0 and proton number -1 [1] (c) Deflected towards the positive plate [1]; Because beta-particles are negatively charged and opposite charges attract / like charges repel [1]