📚 GCSE AQA Physics: Worked Examples Explained | GCSE AQA 物理：典型例题详解
Mastering GCSE AQA Physics means going far beyond memorising formulas — it demands the confidence to apply them in unfamiliar contexts. This article walks through authentic worked examples, each broken into clear steps with paired English–Chinese explanations. Use these as a model for your own revision, and learn to show full working in every exam question.
掌握 GCSE AQA 物理，绝不仅仅是死记公式——它要求你具备将公式应用到陌生物理情境中的信心。本文精选若干典型例题，步步拆解，并配有中英对照解释。以此作为你复习的范本，学会在每道考题中展示完整的推导过程。
1. Calculating Speed from a Distance–Time Graph | 通过距离–时间图求速度
A cyclist travels in a straight line. The distance–time graph shows a steady rise from 0 m at 0 s to 150 m at 30 s. Determine the speed.
骑自行车的人沿直线行驶。距离–时间图显示从 0 秒的 0 米匀速上升到 30 秒的 150 米。请计算其速度。
Step 1: Identify the change in distance Δd = 150 m – 0 m = 150 m, and change in time Δt = 30 s – 0 s = 30 s. Speed v = Δd / Δt = 150 m / 30 s = 5.0 m/s.
步骤一：确定距离变化量 Δd = 150 m – 0 m = 150 m，时间变化量 Δt = 30 s – 0 s = 30 s。速度 v = Δd / Δt = 150 m / 30 s = 5.0 m/s。
| Quantity | Value |
| Initial distance | 0 m |
| Final distance | 150 m |
| Time taken | 30 s |
The gradient of a distance–time graph gives speed. A constant gradient means constant speed.
距离–时间图的斜率代表速度。恒定的斜率意味着匀速运动。
2. Using F = m × a (Newton’s Second Law) | 牛顿第二定律 F = m × a 的应用
A 2000 kg car accelerates at 3.0 m/s². Find the resultant force acting on it.
一辆 2000 kg 的汽车以 3.0 m/s² 的加速度行驶。求作用在它上面的合力。
Resultant force F = m × a = 2000 kg × 3.0 m/s² = 6000 N (or 6.0 kN).
合力 F = m × a = 2000 kg × 3.0 m/s² = 6000 N（即 6.0 kN）。
In an exam, always state the equation, substitute numbers with units, and give the final answer to at least 2 significant figures.
在考试中，务必先写出公式，代入含单位的数值，最后结果至少保留两位有效数字。
3. Gravitational Potential Energy to Kinetic Energy | 重力势能与动能的转换
nore air resistance. Use g = 9.8 N/kg. Calculate the speed just before it hits the ground.
一块质量为 0.50 kg 的石头从 10 m 高处落下。忽略空气阻力，取 g = 9.8 N/kg。求它着地前瞬间的速度。
: Eₖ = ½ × m × v². So 49 = ½ × 0.50 × v² → v² = (49 × 2) / 0.50 = 196 → v = √196 = 14 m/s.
重力势能减少量：ΔEₚ = m × g × h = 0.50 × 9.8 × 10 = 49 J。该能量转化为动能：Eₖ = ½ × m × v²。因此 49 = ½ × 0.50 × v² → v² = (49 × 2) / 0.50 = 196 → v = √196 = 14 m/s。
ΔEₚ = m g h Eₖ = ½ m v²
ss cancels if you equate mgh = ½mv², but always substitute values to avoid algebra errors.
注意，若将 mgh = ½mv² 联立，质量会约掉，但为减少代数错误建议代入数值计算。
4. Specific Heat Capacity Calculation | 比热容计算
An electric kettle supplies 84 000 J of energy to 0.80 kg of water. The water temperature rises from 20 °C to 45 °C. Calculate the specific heat capacity of water using these data.
电热水壶向 0.80 kg 的水提供了 84 000 J 的热量，水温从 20 °C 升至 45 °C。请利用这些数据计算水的比热容。
Energy supplied E = m c Δθ, where Δθ = 45 – Rearranging: c = E / (m Δθ) = 84 000 J / (0.80 kg × 25 °C) = 84 000 / 20 = 4200 J/(kg °C).
提供的能量 E = m c Δθ，其中 Δθ = 45 – 20 = 25 °C。变形得：c = E / (m Δθ) = 84 000 J / (0.80 kg × 25 °C) = 84 000 / 20 = 4200 J/(kg °C)。
In practice, the accepted value for water is 4180–4200 J/(kg °C). Always show full working out.
实际应用中，水的比热容公认值在 4180–4200 J/(kg °C) 之间。解答时需展示完整推导过程。
5. Ohm’s Law and Series Circuit | 欧姆定律与串联电路
A 6.0 Ω resistor and a 12 Ω resistor are connected in series to a 9.0 V battery. Calculate the current in the circuit and the potential difference across the 12 Ω resistor.
一个 6.0 Ω 的电阻与一个 12 Ω 的电阻串联后接到 9.0 V 的电池上。求电路中的电流以及 12 Ω 电阻两端的电压。
Total resistance Rtotal = 6.0 Ω + 12 Ω = 18 Ω. Current I = V / Rtotal .0 V / 18 Ω = 0.50 A. Potential difference across 12 Ω: V = I × R = 0.50 A × 12 Ω = 6.0 V.
总电阻 Rtotal = 6.0 Ω + 12 Ω = 18 Ω。电流 I = V / Rtotal = 9.0 V / 18 Ω = 0.50 A。12 Ω 电阻两端的电压：V = I × R = 0.50 A × 12 Ω = 6.0 V。
ets 0.50 A × 6.0 Ω = 3.0 V, and 6.0 V + 3.0 V = 9.0 V, satisfying Kirchhoff’s voltage law.
快速检验：另一电阻分得 0.50 A × 6.0 Ω = 3.0 V，6.0 V + 3.0 V = 9.0 V，符合基尔霍夫电压定律。
6. Power and Efficiency | 功率与效率
An electric motor lifts a 150 N weight through a vertical distance of 4.0 m in 2.0 s. The motor draws 50 W of electrical power. Calculate the useful output power and the efficiency.
一台电动机在 2.0 秒内将 150 N 的重物竖直提升 4.0 m。电动机消耗的电功率为 50 W。求输出的有用功率及效率。
one / time = 600 J / 2.0 s = 300 W. Efficiency = (useful power out / total power in) × 100% = (300 W / 50 W? motor draws 50 W, so total input power is 50 W. But output power 300 W would exceed input—impossible. Let’s adjust values: maybe input power is 150 W? The given 50 W is too low. We must model reasonable numbers. Let's modify: input power 150 W. Then efficiency = (300 / 150) × 100% = 200%? No, must be ≤100% ted 2.5 m in 5.0 s, work = 500 J, output power = 100 W; input power 125 W, efficiency = 80%. Let's redo the problem properly.
= 360 J. Useful power = 360 J / 2.0 s = 180 W. Input power = 540 J / 2.0 s = 270 W. Efficiency = (180/270)×100% = 66.7%. So we will present that.
工作完成量 = 力 × 距离 = 120 N × 3.0 m = 360 J。有用功率 = 360 J / 2.0 s = 180 W。输入功率 = 540 J / 2.0 s = 270 W。效率 = (180/270)×100% = 66.7%。
6. Power and Efficiency (Corrected) | 功率与效率（修正版例题）
A motor lifts a 120 N weight through 3.0 m vertically in 2.0 s. The total electrical energy supplied is 540 J. Calculate the useful mechanical power and the efficiency.
电动机在 2.0 s 内将 120 N 的重物竖直提升 3.0 m。提供的电能为 540 J。计算有用的机械功率和效率。
2.0 s = 180 W. Total power in = 540 J / 2.0 s = 270 W. Efficiency = (180 W / 270 W) × 100% ≈ 66.7%.
对重物做的功 = 力 × 距离 = 120 N × 3.0 m = 360 J。输出有用功率 = 功 / 时间 = 360 J / 2.0 s = 180 W。总输入功率 = 540 J / 2.0 s = 270 W。效率 = (180 W / 270 W) × 100% ≈ 66.7%。
Always express efficiency either as a decimal between 0 and 1 or as a percentage. In AQA GCSE, percentage is usually preferred.
效率通常表示为 0 到 1 之间的小数，或百分数。在 AQA GCSE 考试中常用百分数表示。
7. Radioactive Half-life | 放射性半衰期
A sample of a radioactive isotope has an initial activity of 1600 Bq. Its half-life is 8.0 days. Calculate the activity after 24 days.
某放射性同位素样品的初始活度为 1600 Bq，其半衰期为 8.0 天。求 24 天后的活度。
-life, activity halves. So after 3 half-lives, activity = 1600 × (½)³ = 1600 × ⅛ = 200 Bq.
半衰期个数：n = 总时间 / 半衰期 = 24 天 / 8.0 天 = 3。每经过一个半衰期，活度减半。因此经过 3 个半衰期后，活度 = 1600 × (½)³ = 1600 × ⅛ = 200 Bq。
uild a decay table: 0 days → 1600 Bq; 8 days → 800 Bq; 16 days → 400 Bq; 24 days → 200 Bq.
也可以列衰变表格：0 天 → 1600 Bq；8 天 → 800 Bq；16 天 → 400 Bq；24 天 → 200 Bq。
8. Wave Speed Equation | 波速公式计算
A water wave has a frequency of 5.0 Hz and a wavelength of 2.0 m. Calculate its speed.
一个水波的频率为 5.0 Hz，波长为 2.0 m。求其传播速度。
Wave speed v = f × λ = 5.0 Hz × 2.0 m = 10 m/s.
波速 v = f × λ = 5.0 Hz × 2.0 m = 10 m/s。
Remember that frequency is in hertz (Hz), which means per second (s⁻¹). Wavelength is in metres. The product gives m/s.
注意频率单位为赫兹（Hz），即每秒的次数（s⁻¹）；波长单位为米。相乘即得米/秒。
9. Density Calculation | 密度计算
A metal block has a mass of 540 g and a volume of 200 cm³. Determine its density in g/cm³ and in kg/m³.
一块金属的质量为 540 g，体积为 200 cm³。求其密度，分别以 g/cm³ 和 kg/m³ 表示。
: multiply by 1000, so 2.7 × 1000 = 2700 kg/m³. (Conversion factor: 1 g/cm³ = 1000 kg/m³.)
密度 ρ = m / V = 540 g / 200 cm³ = 2.7 g/cm³。换算成 kg/m³：乘以 1000，即 2.7 × 1000 = 2700 kg/m³。（换算因子：1 g/cm³ = 1000 kg/m³。）
This density matches aluminium, a common material in the AQA practical exam questions.
这一密度值与铝相符，铝是 AQA 实验题中常见的材料。
10. Work Done and Elastic Potential Energy | 做功与弹性势能
A spring has a spring constant of 300 N/m. It is stretched from its natural length by extending it 0.080 m. Calculate the elastic potential energy stored.
一根弹簧的劲度系数为 300 N/m。将其从原长拉伸 0.080 m。求储存的弹性势能。
= ½ k x², where x is extension. Eₑ = ½ × 300 N/m × (0.080 m)² = ½ × 300 × 0.0064 = 0.96 J.
弹性势能 Eₑ = ½ k x²，其中 x 为伸长量。Eₑ = ½ × 300 N/m × (0.080 m)² = ½ × 300 × 0.0064 = 0.96 J。
Make sure the extension is converted to metres if given in centimetres.
若伸长量以厘米给出，务必先转换为米。
11. Momentum and Impulse | 动量与冲量
A 0.15 kg cricket ball travels at 20 m/s before being caught. It is brought to rest in 0.10 s. Calculate the force exerted by the hands on the ball.
一个 0.15 kg 的板球以 20 m/s 的速度飞来后被接住，在 0.10 s 内停下。求手对球施加的力。
Initial momentum p_initial = m × u = 0.15 kg × 20 m/s = 3.0 kg m/s. Final momentum pfinal = 0. Change in momentum Δp = 0 – 3.0 = –3.0 kg m/s. Force F = Δp / Δt = –3.0 / 0.10 = –0 N. The negative sign indicates the force opposes the original motion. Magnitude is 30 N.
初动量 p_initial = m × u = 0.15 kg × 20 m/s = 3.0 kg m/s。末动量 pfinal = 0。动量变化 Δp = 0 – 3.0 = –3.0 kg m/s。力 F = Δp / Δt = –3.0 / 0.10 = –30 N。负号表示力的方向与原运动方向相反，其大小为 30 N。
of Newton’s second law in terms of momentum: resultant force = rate of change of momentum.
这是牛顿第二定律在动量形式下的应用：合力等于动量的变化率。
12. Series and Parallel Circuit Combination | 串并联电路组合分析
Two 10 Ω resistors are connected in parallel. This parallel combination is then connected in series with a 5.0 Ω resistor to a 12 V battery. Determine the total resistance and the current from the battery.
两个 10 Ω 的电阻并联，然后将该并联组合与一个 5.0 Ω 的电阻串联，接到 12 V 的电池上。求总电阻和电池提供的电流。
Resistance of parallel pair: 1/Rpara = 1/10 + 1/10 = 2/10, so Rpara = 5.0 Ω. Total resistance Rtotal = 5.0 Ω (parallel) + 5.0 Ω (series) = 10 Ω. Current I = V / Rtotal = 12 V / 10 Ω = 1.2 A.
并联部分的电阻：1/Rpara = 1/10 + 1/10 = 2/10，因此 Rpara = 5.0 Ω。总电阻 Rtotal = 5.0 Ω（并联部分） + 5.0 Ω（串联部分） = 10 Ω。电流 I = V / Rtotal = 12 V / 10 Ω = 1.2 A。
Always simplify stepwise: first combine parallel branches, then treat the circuit as a simple series loop.
始终分步简化：先合并并联分支，再将电路视为简单串联回路处理。
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