An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V1) Cambridge IGCSE Physics (0625) paper. Not affiliated with or reproduced from Cambridge.
部分 Theory Questions
Answer all questions. Show your working clearly. Use a calculator if necessary. Take the weight of 1.0 kg to be 9.8 N.
10 题目 · 80 分
题目 1 · subjective
8 分
(a) A toy car accelerates from rest at a constant rate of 1.5 m/s^2 for 4.0 s. Calculate the speed of the toy car at 4.0 s. [2 marks] (b) It then travels at this constant speed for a further 6.0 s. Describe the shape of the distance-time graph for the car during this 6.0 s interval. [2 marks] (c) Finally, the car decelerates uniformly to a standstill in 5.0 s. (i) Calculate the total distance travelled by the car from the start of its motion. [3 marks] (ii) State the velocity of the car once it has completely stopped. [1 mark]
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解题
(a) speed = acceleration * time = 1.5 * 4.0 = 6.0 m/s. (b) Since the speed is constant, the distance increases at a constant rate, which is represented by a straight diagonal line with a constant positive gradient. (c)(i) Distance during acceleration = area under graph = 0.5 * base * height = 0.5 * 4.0 * 6.0 = 12 m. Distance during constant speed = speed * time = 6.0 * 6.0 = 36 m. Distance during deceleration = area under graph = 0.5 * base * height = 0.5 * 5.0 * 6.0 = 15 m. Total distance = 12 + 36 + 15 = 57 m. (ii) The velocity is 0 m/s because the car has stopped.
评分标准
(a) Recall of v = u + at or v = at (1 mark), Correct calculation to give 6.0 m/s (1 mark). (b) Statement that it is a straight line (1 mark), Statement that it has a constant positive gradient (1 mark). (c)(i) Calculation of distance in any one phase (1 mark), Calculation of distance in all three phases (12 m, 36 m, 15 m) (1 mark), Correct addition to give total distance of 57 m (1 mark). (ii) Correctly states 0 m/s (1 mark).
题目 2 · subjective
8 分
(a) State the two conditions required for an object to be in equilibrium. [2 marks] (b) A uniform wooden beam of length 2.4 m and weight 80 N is pivoted at its midpoint. A block of weight W is placed 0.90 m to the left of the pivot. To balance the beam horizontally, a vertical downward force of 30 N is applied at the right-hand end of the beam (1.2 m from the pivot). Calculate the weight W of the block. [3 marks] (c) An additional force of 50 N is applied downwards directly at the pivot. Explain what effect, if any, this has on the equilibrium of the beam. [3 marks]
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解题
(a) The two conditions are: 1. No resultant force acting on the object. 2. No resultant moment acting on the object. (b) Taking moments about the pivot: Anticlockwise moment = W * 0.90 m. Clockwise moment = 30 N * 1.2 m = 36 N m. For equilibrium, W * 0.90 = 36, which gives W = 40 N. (c) A force applied directly at the pivot has a perpendicular distance of zero from the pivot. Therefore, it exerts zero moment about the pivot. It increases the downward force on the pivot (which is balanced by an increased upward reaction force from the pivot), so the beam remains in both translational and rotational equilibrium.
评分标准
(a) No resultant force (1 mark) and no resultant moment (1 mark). (b) Use of moment = force * perpendicular distance (1 mark), Setting clockwise moments equal to anticlockwise moments: W * 0.90 = 30 * 1.2 (1 mark), Correct calculation to give W = 40 N (1 mark). (c) Statement that it has no effect on rotational equilibrium / the beam remains balanced (1 mark), Explanation that the perpendicular distance from the pivot is zero (1 mark), hence it creates no moment about the pivot (1 mark).
题目 3 · subjective
8 分
(a) A heavy cylindrical metal block has a mass of 45 kg. Calculate its weight. (g = 9.8 m/s^2) [2 marks] (b) The block has a flat circular base of radius 0.15 m. Calculate the pressure exerted by the block on a flat horizontal floor when standing upright. [3 marks] (c) The block is now lowered into a tank of oil of density 900 kg/m^3. Calculate the depth in the oil at which the pressure due to the oil alone is 1.8 * 10^4 Pa. (g = 9.8 m/s^2) [3 marks]
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解题
(a) Weight = mass * g = 45 * 9.8 = 441 N. (b) Base area = pi * r^2 = pi * 0.15^2 = 0.0707 m^2. Pressure = Force / Area = 441 / 0.0707 = 6238 Pa. (c) Pressure in liquid = depth * density * g, which rearranged gives depth = P / (density * g) = 1.8 * 10^4 / (900 * 9.8) = 18000 / 8820 = 2.04 m.
评分标准
(a) Recall of W = mg (1 mark), Correct calculation to give 441 N (1 mark). (b) Calculation of area using pi * r^2 (1 mark), Recall of P = F/A (1 mark), Correct calculation to give 6200 Pa (allow 6200 to 6300 Pa) (1 mark). (c) Recall of P = h * rho * g (1 mark), Rearrangement to h = P / (rho * g) (1 mark), Correct calculation to give 2.0 m (allow 2.0 to 2.04 m) (1 mark).
题目 4 · subjective
8 分
(a) Define the term specific heat capacity. [2 marks] (b) An electric heater rated at 48 W is used to heat a 1.2 kg block of metal. The heater is switched on for 5.0 minutes. (i) Calculate the thermal energy supplied by the heater in this time. [2 marks] (ii) The temperature of the block rises from 20 degrees Celsius to 45 degrees Celsius. Calculate the specific heat capacity of the metal. [3 marks] (c) Suggest one reason why this calculated value might be higher than the actual specific heat capacity of the metal. [1 mark]
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解题
(a) Specific heat capacity is the thermal energy required per unit mass to raise the temperature of a substance by one degree Celsius. (b)(i) Time = 5.0 minutes = 300 s. Energy = Power * time = 48 * 300 = 14400 J. (ii) Temperature rise = 45 - 20 = 25 degrees Celsius. c = Q / (m * delta_T) = 14400 / (1.2 * 25) = 480 J/(kg degrees Celsius). (c) Some thermal energy from the heater is lost to the surroundings or the heater itself, so the energy actually absorbed by the block is less than the calculated 14400 J, resulting in an overestimate of the specific heat capacity.
评分标准
(a) Energy per unit mass (1 mark), to raise temperature by one degree (1 mark). (b)(i) Conversion of minutes to seconds (1 mark), Correct calculation of energy to give 14400 J (1 mark). (ii) Calculation of temperature change as 25 degrees Celsius (1 mark), Recall of E = mc * delta_T (1 mark), Correct calculation to give 480 J/(kg degrees Celsius) (1 mark). (c) Mention of heat loss to surroundings / air / heater (1 mark).
题目 5 · subjective
8 分
(a) The critical angle for a semi-circular glass block is 41 degrees. Calculate the refractive index of the glass. [2 marks] (b) A ray of light enters the curved boundary of the glass block along a normal. State and explain what happens to: (i) the direction of the ray as it enters the block, [2 marks] (ii) the speed of the light as it enters the block. [2 marks] (c) The light ray now strikes the flat boundary inside the glass at an angle of incidence of 45 degrees. State and explain the path of the light ray at this boundary. [2 marks]
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解题
(a) n = 1 / sin(c) = 1 / sin(41) = 1 / 0.656 = 1.52 (accept 1.5). (b)(i) The direction does not change because the ray enters along the normal (angle of incidence is 0 degrees, so angle of refraction is also 0 degrees). (ii) The speed decreases because glass has a higher refractive index than air. (c) Total internal reflection occurs. This is because the light is travelling in the denser medium and the angle of incidence (45 degrees) is greater than the critical angle (41 degrees).
评分标准
(a) Recall of n = 1 / sin(c) (1 mark), Correct calculation to give 1.5 (allow 1.5 to 1.52) (1 mark). (b)(i) States direction does not change (1 mark), explains that it enters along the normal / angle of incidence is zero (1 mark). (ii) States speed decreases (1 mark), explains because glass is more dense than air / higher refractive index (1 mark). (c) States total internal reflection occurs (1 mark), explains that angle of incidence (45 degrees) is greater than the critical angle (41 degrees) (1 mark).
题目 6 · subjective
8 分
(a) Draw a circuit diagram showing a battery connected in series with a switch, a variable resistor, and a fixed resistor R. Include a voltmeter connected to measure the potential difference across resistor R. [3 marks] (b) The fixed resistor R has a resistance of 12 ohms. When the switch is closed, the reading on the voltmeter is 6.0 V. Calculate the current in resistor R. [2 marks] (c) The variable resistor is now adjusted so that its resistance increases. State and explain the effect of this change on: (i) the current in the circuit, [2 marks] (ii) the reading on the voltmeter. [1 mark]
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解题
(a) The diagram must show the battery, switch, variable resistor, and fixed resistor in a single loop (series). The voltmeter must be connected in parallel across the fixed resistor R. (b) I = V / R = 6.0 / 12 = 0.50 A. (c)(i) The current in the circuit decreases because the total resistance of the circuit increases when the resistance of the variable resistor increases, and current is inversely proportional to resistance. (ii) The voltmeter reading decreases because the current through R decreases, and V = I * R.
评分标准
(a) Battery, switch, variable resistor and fixed resistor R in a correct series loop (1 mark), Voltmeter connected in parallel across resistor R (1 mark), All correct standard symbols used (1 mark). (b) Recall of I = V / R (1 mark), Correct calculation to give 0.50 A (1 mark). (c)(i) States that current decreases (1 mark), explains that total resistance of circuit has increased (1 mark). (ii) States that the voltmeter reading decreases (1 mark).
题目 7 · subjective
8 分
(a) Explain the principle of operation of a simple iron-cored step-down transformer. [4 marks] (b) The primary coil of a transformer has 1200 turns and is connected to a 240 V a.c. supply. The secondary coil has 60 turns. Calculate the output voltage across the secondary coil. [2 marks] (c) The output power of the transformer is 24 W. Assuming 100% efficiency, calculate the current in the primary coil. [2 marks]
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解题
(a) An alternating current (a.c.) in the primary coil produces a changing magnetic field. The soft-iron core concentrates and transfers this changing magnetic field to the secondary coil. The changing magnetic field cuts the secondary coil, which induces an alternating electromotive force (e.m.f.) / voltage across the secondary coil. (b) V_s / V_p = N_s / N_p, which rearranged gives V_s = V_p * (N_s / N_p) = 240 * (60 / 1200) = 12 V. (c) Input power = Output power = 24 W. P_p = V_p * I_p, which rearranged gives I_p = P_p / V_p = 24 / 240 = 0.10 A.
评分标准
(a) Alternating current in primary coil creates a changing magnetic field (1 mark), Soft-iron core transfers magnetic field to secondary coil (1 mark), Changing magnetic field cuts secondary coil (1 mark), inducing an alternating voltage / e.m.f. in secondary coil (1 mark). (b) Recall of V_s/V_p = N_s/N_p (1 mark), Correct calculation to give 12 V (1 mark). (c) Statement that input power equals output power / P = V * I (1 mark), Correct calculation to give 0.10 A (1 mark).
题目 8 · subjective
8 分
(a) State what happens to the proton number and the nucleon number of a nucleus when it undergoes beta-minus (\beta^-) decay. [2 marks] (b) Iodine-131 is a beta-minus emitter with a half-life of 8.0 days. (i) Define the term half-life. [2 marks] (ii) A laboratory sample of Iodine-131 initially has an activity of 640 Bq. Calculate the activity of the sample after 24 days. [2 marks] (c) State two safety precautions that must be taken when handling radioactive sources in a school laboratory. [2 marks]
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解题
(a) In beta-minus decay, a neutron decays into a proton and an electron. Therefore, the proton number increases by 1, and the nucleon number remains the same. (b)(i) Half-life is the time taken for the number of radioactive nuclei (or the activity) in a sample to decrease to half of its initial value. (ii) Number of half-lives = 24 / 8.0 = 3. After 1 half-life: 320 Bq. After 2 half-lives: 160 Bq. After 3 half-lives: 80 Bq. (c) Safety precautions include using long-handled tongs to handle the source and storing the source in a lead-lined container when not in use.
评分标准
(a) Proton number increases by 1 (1 mark), Nucleon number remains the same (1 mark). (b)(i) Reference to time taken (1 mark), for number of radioactive nuclei / activity to halve (1 mark). (ii) Calculation of 3 half-lives (1 mark), Correct final activity of 80 Bq (1 mark). (c) Any two correct precautions: e.g., use tongs / keep distance, store in lead-lined container, wear protective clothing, do not point source at eyes (1 mark for each, max 2 marks).
题目 9 · Theory
8 分
A small motorized hoist is used on a construction site to lift a load of building materials.
(a) The hoist lifts a crate of mass 15 kg vertically through a height of 4.0 m. Calculate the work done in lifting the crate. [3]
(b) The lift takes a time of 5.0 s. Calculate the useful power output of the hoist's motor during this lift. [2]
(c) The motor of the hoist is supplied with 800 J of electrical energy to perform this lift.
(i) State what happens to the energy that is not used as useful work. [1]
(ii) Calculate the efficiency of the hoisting system. [2]
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解题
(a) The work done in lifting the mass vertically is given by: \(W = F \times d = m \times g \times d\) Using \(g = 9.8\text{ N/kg}\): \(W = 15\text{ kg} \times 9.8\text{ N/kg} \times 4.0\text{ m} = 588\text{ J}\)
(b) Power is the rate of doing work: \(P = \frac{W}{t}\) \(P = \frac{588\text{ J}}{5.0\text{ s}} = 117.6\text{ W}\) (or approximately \(118\text{ W}\) or \(120\text{ W}\) to 2 s.f.).
(c) (i) The remaining energy is dissipated as thermal energy to the surroundings (or hoist components) due to friction and electrical resistance.
(ii) Efficiency is calculated using: \(\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100\%\) \(\text{Efficiency} = \frac{588\text{ J}}{800\text{ J}} \times 100\% = 73.5\%\) (or \(0.735\)).
评分标准
(a) - C1: Recall of formula \(W = F \times d\) or \(W = m \times g \times d\) - C1: Substitution of correct values: \(15 \times 9.8 \times 4.0\) - A1: Correct final value \(588\text{ J}\) (or \(590\text{ J}\) if using rounded 2 s.f. weight of 150 N)
(b) - C1: Recall of power formula \(P = W/t\) or \(588 / 5.0\) (allow ecf from (a)) - A1: Correct calculation \(118\text{ W}\) or \(120\text{ W}\) or \(117.6\text{ W}\) with correct unit
(c)(i) - B1: Stating that energy is wasted/dissipated to the surroundings as thermal energy / heat
(c)(ii) - C1: Recall of efficiency formula \(\frac{\text{useful output}}{\text{total input}}\) - A1: Correct calculation \(73.5\%\) or \(0.735\) (allow \(74\%\) or \(0.74\))
题目 10 · Theory
8 分
A radioactive source contains Cobalt-60 (\(^{60}\text{Co}\)), which decays by emitting beta-particles (\(\beta^-\)) and gamma-rays (\(\gamma\)) to form a stable isotope of Nickel-60 (\(^{60}\text{Ni}\)).
(a) (i) Describe how the ionizing ability and the penetrating power of beta-particles compare with those of gamma-rays. [2]
(ii) State which of these two types of radiation (beta-particles or gamma-rays) is deflected by a magnetic field. [1]
(b) The half-life of Cobalt-60 is 5.3 years. Initially, a sample of this source has an activity of 480 Bq.
(i) Calculate the activity of the sample after a time period of 15.9 years. [3]
(ii) State and explain how the activity of the Cobalt-60 source is affected if the temperature of the sample is increased significantly. [2]
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解题
(a) (i) Beta-particles are more strongly ionizing than gamma-rays, but have a much lower penetrating power (they are stopped by a few millimeters of aluminum, whereas gamma-rays require several centimeters of lead to be significantly absorbed).
(ii) Beta-particles are deflected by a magnetic field because they carry a negative electrical charge, whereas gamma-rays are uncharged electromagnetic waves and are not deflected.
(b) (i) First, calculate the number of half-lives that have passed: \(n = \frac{15.9\text{ years}}{5.3\text{ years}} = 3\text{ half-lives}\) After 1 half-life: \(480 / 2 = 240\text{ Bq}\) After 2 half-lives: \(240 / 2 = 120\text{ Bq}\) After 3 half-lives: \(120 / 2 = 60\text{ Bq}\) So, the activity after 15.9 years is \(60\text{ Bq}\).
(ii) The activity remains completely unchanged. Radioactive decay is a spontaneous nuclear process and is entirely unaffected by external physical conditions such as changes in temperature or pressure.
评分标准
(a)(i) - B1: Correct comparison of ionizing power (beta-particles are more ionizing than gamma-rays / gamma-rays are less ionizing) - B1: Correct comparison of penetrating power (gamma-rays are more penetrating than beta-particles / beta-particles are less penetrating)
(a)(ii) - B1: Beta-particles (or \(\beta\))
(b)(i) - C1: Determine the number of half-lives to be 3 - C1: Show step-by-step halving of the activity (e.g., \(480 \rightarrow 240 \rightarrow 120 \rightarrow 60\)) - A1: Final answer of \(60\text{ Bq}\) (unit must be present for the mark)
(b)(ii) - B1: State that activity is unaffected / remains the same - B1: Explain that radioactive decay is a random/spontaneous/nuclear process (not affected by physical conditions)