📚 Worked Examples for GCSE Edexcel Physics | GCSE Edexcel 物理典型例题详解
Mastering GCSE Edexcel Physics requires not only memorising formulas but also applying them confidently to unfamiliar scenarios. This article walks through ten carefully selected worked examples, each targeting a core topic from the Edexcel specification. Step-by-step reasoning is provided in both English and Chinese to help you develop a clear problem-solving method and avoid common pitfalls.
攻克 GCSE Edexcel 物理，不仅要熟记公式，更要在陌生情境中从容应用。本文精选十个典型例题，覆盖 Edexcel 考纲核心主题。每道题都用中英双语逐步拆解，帮助你形成清晰的解题思路，避开常见失分点。
1. Motion and SUVAT | 运动学与匀加速公式
A car accelerates uniformly from rest at 3 m/s² for 8 seconds. Calculate its final velocity and the distance travelled during this time.
一辆汽车从静止开始以 3 m/s² 的加速度匀加速运动 8 秒。求末速度以及这段时间内行驶的距离。
Identify the knowns: u = 0 m/s, a = 3 m/s², t = 8 s. Use v = u + at: v = 0 + (3 × 8) = 24 m/s. Then use s = ut + ½at²: s = 0 + ½ × 3 × (8)² = ½ × 3 × 64 = 96 m.
写出已知量：初速 u = 0 m/s，加速度 a = 3 m/s²，时间 t = 8 s。先用 v = u + at 得 v = 0 + (3 × 8) = 24 m/s。再用 s = ut + ½at² 得 s = 0 + ½ × 3 × 64 = 96 m。
Always write down the SUVAT variables and choose the equation that relates the knowns to the unknown. Remember that if an object starts from rest, u = 0.
解题时务必列出 SUVAT 五个量，选择包含已知量和所求量的公式。物体从静止开始运动时，初速 u = 0。
2. Resultant Force and Newton’s Second Law | 合力与牛顿第二定律
A block of mass 12 kg is pulled along a rough horizontal surface. The pulling force is 50 N, and the frictional force opposing motion is 14 N. Determine the acceleration of the block.
一个质量为 12 kg 的物块在粗糙水平面上被拉动。拉力为 50 N，阻碍运动的摩擦力为 14 N。求物块的加速度。
Resultant force F = pulling force − friction = 50 N − 14 N = 36 N. From F = ma, a = F / m = 36 / 12 = 3 m/s².
合力 = 拉力 − 摩擦力 = 50 N − 14 N = 36 N。由 F = ma 得 a = F / m = 36 / 12 = 3 m/s²。
Always determine the net force in the direction of motion before using F = ma. When forces act in opposite directions, subtract the opposing force.
使用 F = ma 前，必须先求出运动方向上的合力。当力方向相反时，用拉力减去阻力。
3. Energy Transfers and Efficiency | 能量转移与效率
An electric motor lifts a 5 kg mass through a vertical height of 2.5 m in 4.0 seconds. The motor has an input power of 40 W. Calculate the useful power output and the efficiency of the motor. (g = 10 m/s²)
一台电动机在 4.0 秒内将 5 kg 的重物竖直提升 2.5 m。电动机的输入功率为 40 W。求有用输出功率及电动机的效率。（取 g = 10 m/s²）
Work done on mass = mgh = 5 × 10 × 2.5 = 125 J. Useful power output = work done / time = 125 / 4.0 = 31.25 W. Efficiency = (useful power output / input power) × 100% = (31.25 / 40) × 100% = 78.125% ≈ 78%.
对重物做的功 = mgh = 5 × 10 × 2.5 = 125 J。有用功率输出 = 做功 / 时间 = 125 / 4.0 = 31.25 W。效率 = (有用功率输出 / 输入功率) × 100% = (31.25 / 40) × 100% = 78.125% ≈ 78%。
Efficiency can be calculated using either the energy or the power ratio. Both approaches give the same result because time cancels out.
效率可以用能量比或功率比计算，两者结果相同，因为时间可以在分子分母中约掉。
4. Ohm’s Law and Series Circuits | 欧姆定律与串联电路
A 6 Ω resistor and a 12 Ω resistor are connected in series across a 9 V battery. Calculate the current in the circuit and the potential difference across the 12 Ω resistor.
一个 6 Ω 电阻和一个 12 Ω 电阻串联后接在 9 V 电池两端。求电路中的电流以及 12 Ω 电阻两端的电压。
Total resistance R = 6 + 12 = 18 Ω. Current I = V / R = 9 / 18 = 0.5 A. The current is the same everywhere in a series circuit. Voltage across the 12 Ω resistor: V = IR = 0.5 × 12 = 6 V.
总电阻 R = 6 + 12 = 18 Ω。电流 I = V / R = 9 / 18 = 0.5 A。串联电路中各处电流相同。12 Ω 电阻两端的电压：V = IR = 0.5 × 12 = 6 V。
In series circuits, resistances add directly, and the current is constant. You can verify by checking that the sum of individual voltages equals the supply voltage: (0.5 × 6) + 6 = 3 + 6 = 9 V.
串联电路中，电阻直接相加，电流处处相等。验证：各电阻电压之和等于电源电压：(0.5 × 6) + 6 = 3 + 6 = 9 V。
5. Density and Material Properties | 密度与材料特性
A metal cylinder has a mass of 540 g and a volume of 200 cm³. Determine the density of the metal in g/cm³ and in kg/m³. State whether it will float in water (density of water = 1000 kg/m³).
一个金属圆柱体质量为 540 g，体积为 200 cm³。求该金属的密度（单位：g/cm³ 和 kg/m³），并判断它是否会浮在水面上（水的密度 = 1000 kg/m³）。
Density ρ = mass / volume = 540 / 200 = 2.7 g/cm³. To convert to kg/m³, multiply by 1000: 2.7 × 1000 = 2700 kg/m³. Since 2700 > 1000, the metal will sink in water.
密度 ρ = 质量 / 体积 = 540 / 200 = 2.7 g/cm³。换算为 kg/m³：乘以 1000，得 2700 kg/m³。因 2700 > 1000，该金属会沉入水中。
Remember the conversion: 1 g/cm³ = 1000 kg/m³. Objects with density greater than the fluid sink; those with lower density float.
记住换算关系：1 g/cm³ = 1000 kg/m³。物体密度大于液体密度时下沉，小于液体密度时上浮。
6. Hooke’s Law and Elastic Behaviour | 胡克定律与弹性行为
A spring stretches by 3.0 cm when a load of 6 N is hung from it. The spring obeys Hooke’s law up to a load of 15 N. Calculate the spring constant k and the extension produced by a 10 N load.
一根弹簧在 6 N 的负载下伸长 3.0 cm。该弹簧在 15 N 以下遵循胡克定律。求弹簧的劲度系数 k，以及 10 N 负载产生的伸长量。
First, convert extension to metres: 3.0 cm = 0.030 m. Using F = kx, k = F / x = 6 / 0.030 = 200 N/m. For a 10 N load, x = F / k = 10 / 200 = 0.050 m = 5.0 cm.
先将伸长量换算为米：3.0 cm = 0.030 m。由 F = kx 得 k = F / x = 6 / 0.030 = 200 N/m。对于 10 N 负载，x = F / k = 10 / 200 = 0.050 m = 5.0 cm。
Always use SI units when calculating spring constants: force in newtons, extension in metres. The extension is directly proportional to force only within the limit of proportionality.
计算劲度系数时务必使用国际单位：力用牛，伸长量用米。只有在弹性限度内，伸长才与力成正比。
7. Wave Speed, Frequency and Wavelength | 波速、频率与波长
A water wave has a wavelength of 0.80 m and a frequency of 5 Hz. Determine the wave speed. If the frequency is doubled while the speed remains constant, what happens to the wavelength?
一个水波的波长为 0.80 m，频率为 5 Hz。求波速。若频率加倍而波速不变，波长如何变化？
Wave speed v = f × λ = 5 × 0.80 = 4.0 m/s. If frequency doubles to 10 Hz and speed stays 4.0 m/s, then λ = v / f = 4.0 / 10 = 0.40 m. The wavelength halves.
波速 v = f × λ = 5 × 0.80 = 4.0 m/s。若频率加倍至 10 Hz 而波速仍为 4.0 m/s，则波长 λ = v / f = 4.0 / 10 = 0.40 m。波长减半。
The wave equation v = fλ is fundamental. For the same medium, wave speed is usually constant, so frequency and wavelength are inversely proportional.
波速方程 v = fλ 是基础。同一介质中波速通常不变，因此频率和波长成反比。
8. Radioactive Decay and Half-Life | 放射性衰变与半衰期
A sample of a radioactive isotope has an initial activity of 800 Bq. Its half-life is 3 hours. Calculate the activity after 9 hours, and determine how long it takes for the activity to fall to 100 Bq.
某种放射性同位素样品的初始活度为 800 Bq，半衰期为 3 小时。求 9 小时后的活度，并计算活度降至 100 Bq 所需的时间。
After 3 h: halved to 400 Bq; after 6 h: 200 Bq; after 9 h: 100 Bq. So activity after 9 hours is 100 Bq. The activity falls from 800 to 100 Bq, which is three half-lives, taking 3 × 3 = 9 hours.
3 小时后：减半至 400 Bq；6 小时后：200 Bq；9 小时后：100 Bq。因此 9 小时后的活度为 100 Bq。活度从 800 降至 100 Bq，经历了三个半衰期，所需时间为 3 × 3 = 9 小时。
For half-life problems, you can either halve repeatedly or use the formula: activity after n half-lives = initial activity / 2ⁿ. Here n = 9/3 = 3, so 800 / 2³ = 800 / 8 = 100 Bq.
半衰期题目可逐次减半，或用公式：n 个半衰期后的活度 = 初始活度 / 2ⁿ。本题 n = 9/3 = 3，故 800 / 2³ = 800 / 8 = 100 Bq。
9. Specific Heat Capacity | 比热容
An electric heater supplies 36 000 J of energy to a 2.0 kg aluminium block, raising its temperature by 20 °C. Calculate the specific heat capacity of aluminium. If the same energy were supplied to the same mass of water (c = 4200 J/kg°C), what temperature rise would occur?
一个电加热器向 2.0 kg 的铝块提供 36 000 J 能量，使其温度升高 20 °C。求铝的比热容。若将相同能量供给相同质量的水（c = 4200 J/kg°C），水温会升高多少？
Using ΔE = m c Δθ: c = ΔE / (m Δθ) = 36 000 / (2.0 × 20) = 36 000 / 40 = 900 J/kg°C. For water: Δθ = ΔE / (m c) = 36 000 / (2.0 × 4200) = 36 000 / 8400 ≈ 4.3 °C.
利用 ΔE = m c Δθ：c = ΔE / (m Δθ) = 36 000 / (2.0 × 20) = 36 000 / 40 = 900 J/kg°C。对于水：Δθ = ΔE / (m c) = 36 000 / (2.0 × 4200) = 36 000 / 8400 ≈ 4.3 °C。
Aluminium has a lower specific heat capacity than water, so it heats up more for the same amount of energy. Always show unit conversion if necessary, though here all units are consistent.
铝的比热容比水小，因此吸收相同能量时升温更多。注意单位统一，本题中单位已一致，无需换算。
10. Transformer Equation (Ideal) | 理想变压器方程
A step-down transformer has 500 turns on its primary coil and 50 turns on its secondary. It is connected to a 230 V mains supply. Calculate the secondary output voltage. If the primary current is 0.40 A, and assuming 100% efficiency, what is the secondary current?
一个降压变压器初级线圈有 500 匝，次级线圈有 50 匝，接到 230 V 的市电上。求次级输出电压。若初级电流为 0.40 A，并假设效率为 100%，求次级电流。
Using Vp / Vs = Np / Ns: Vs = Vp × (Ns / Np) = 230 × (50 / 500) = 230 × 0.1 = 23 V. For 100% efficiency, input power = output power: Vp × Ip = Vs × Is, so Is = (Vp × Ip) / Vs = (230 × 0.40) / 23 = 92 / 23 = 4.0 A.
利用 Vp / Vs = Np / Ns：Vs = Vp × (Ns / Np) = 230 × (50 / 500) = 23 V。效率 100% 时，输入功率 = 输出功率：Vp × Ip = Vs × Is，故 Is = (Vp × Ip) / Vs = (230 × 0.40) / 23 = 4.0 A。
The turns ratio determines whether the transformer steps voltage up or down. For an ideal transformer, power is conserved, so a lower voltage is accompanied by a proportionally higher current.
匝数比决定变压器升压还是降压。理想变压器功率守恒，因此电压降低时电流会按比例升高。
Published by TutorHao | Physics Revision Series | aleveler.com
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