📚 A-Level Science: Worked Example Problems Explained | A-Level 科学：典型例题详解
Mastering A-Level Science requires not only understanding theoretical concepts but also applying them to solve problems. Worked examples bridge the gap between theory and exam success. This article walks through typical A-level questions from Physics, Chemistry, and Biology, showing step-by-step solutions and highlighting common mistakes.
掌握 A-Level 科学不仅需要理解理论概念，还需要将它们应用于解决问题。典型例题详解能够弥合理论与考试成功之间的鸿沟。本文精选物理、化学、生物的典型 A-level 题目，逐步展示解题过程，并指出常见错误。
1. Physics: SUVAT Equations – Finding Displacement | 物理：匀加速运动方程 – 求位移
A car accelerates uniformly from rest at 2.0 m/s² for 5.0 seconds. Calculate the displacement during this time.
一辆汽车从静止开始以 2.0 m/s² 的加速度匀加速行驶 5.0 秒。求这段时间内的位移。
Step 1 – Identify known quantities: u = 0 m/s, a = 2.0 m/s², t = 5.0 s. The unknown is s.
步骤 1 – 确定已知量：u = 0 m/s，a = 2.0 m/s²，t = 5.0 s。未知量为 s。
Step 2 – Select the appropriate SUVAT equation: s = ut + ½at².
步骤 2 – 选择恰当的匀加速运动方程：s = ut + ½at²。
Step 3 – Substitute and calculate: s = (0)(5.0) + ½(2.0)(5.0)² = 0 + 1.0 × 25 = 25 m.
步骤 3 – 代入并计算：s = (0)(5.0) + ½(2.0)(5.0)² = 0 + 1.0 × 25 = 25 m。
Common mistake: forgetting to square the time or using v = u + at instead of the displacement equation.
常见错误：忘记将时间平方，或者错误地使用 v = u + at 而未能求出位移。
2. Physics: Resolving Forces on an Inclined Plane | 物理：斜面受力分解
A 5.0 kg block rests on a frictionless incline of 30° to the horizontal. Calculate the component of weight acting parallel to the incline and the acceleration of the block down the incline. (g = 9.8 m/s²)
一个 5.0 kg 的物块静置于与水平面成 30° 的光滑斜面上。求重力沿斜面方向的分力和物块沿斜面下滑的加速度。（g = 9.8 m/s²）
Step 1 – Find the weight: W = mg = 5.0 × 9.8 = 49 N.
步骤 1 – 计算重力：W = mg = 5.0 × 9.8 = 49 N。
Step 2 – Resolve the weight parallel to the slope: F_parallel = mg sin θ = 49 × sin 30° = 49 × 0.5 = 24.5 N.
步骤 2 – 分解重力沿斜面方向的分力：F_平行 = mg sin θ = 49 × sin 30° = 49 × 0.5 = 24.5 N。
Step 3 – Apply Newton’s second law: a = F_parallel / m = 24.5 / 5.0 = 4.9 m/s².
步骤 3 – 应用牛顿第二定律：a = F_平行 / m = 24.5 / 5.0 = 4.9 m/s²。
Pitfall: confusing sin and cos when resolving components. Always check whether you need the side opposite (sin) or adjacent (cos) to the angle.
易错点：分解时混淆正弦和余弦。务必确认使用的是角的对边（sin）还是邻边（cos）。
3. Physics: Ohm’s Law and Kirchhoff’s Voltage Law | 物理：欧姆定律与基尔霍夫电压定律
A series circuit contains a 12 V battery and two resistors R₁ = 4 Ω and R₂ = 6 Ω. Find the current and the voltage across each resistor.
一个串联电路包含一个 12 V 电池和两个电阻 R₁ = 4 Ω、R₂ = 6 Ω。求电流和每个电阻两端的电压。
Step 1 – Total resistance in series: R_total = R₁ + R₂ = 4 + 6 = 10 Ω.
步骤 1 – 串联总电阻：R_total = R₁ + R₂ = 4 + 6 = 10 Ω。
Step 2 – Current using Ohm’s law: I = V / R_total = 12 / 10 = 1.2 A.
步骤 2 – 利用欧姆定律求电流：I = V / R_total = 12 / 10 = 1.2 A。
Step 3 – Voltage across each resistor: V₁ = I × R₁ = 1.2 × 4 = 4.8 V; V₂ = I × R₂ = 1.2 × 6 = 7.2 V. Check: V₁ + V₂ = 4.8 + 7.2 = 12 V = battery voltage.
步骤 3 – 各电阻电压：V₁ = I × R₁ = 1.2 × 4 = 4.8 V；V₂ = I × R₂ = 1.2 × 6 = 7.2 V。验证：V₁ + V₂ = 4.8 + 7.2 = 12 V，等于电池电压。
Key insight: In a series circuit, current is the same everywhere; voltages add up to the supply voltage.
关键点：在串联电路中，各处电流相同；各电阻电压之和等于电源电压。
4. Chemistry: Mole Calculations with Gas Volumes | 化学：摩尔计算与气体体积
What volume of hydrogen gas (at RTP, 24 dm³/mol) is produced when 0.54 g of aluminium reacts with excess hydrochloric acid? 2Al + 6HCl → 2AlCl₃ + 3H₂.
0.54 g 铝与过量盐酸反应，在室温常压下（24 dm³/mol）产生多少体积的氢气？反应方程式：2Al + 6HCl → 2AlCl₃ + 3H₂。
Step 1 – Moles of Al: n(Al) = mass / Aᵣ = 0.54 g / 27 g/mol = 0.020 mol.
步骤 1 – 铝的物质的量：n(Al) = 质量 / 相对原子质量 = 0.54 g / 27 g/mol = 0.020 mol。
Step 2 – Use molar ratio from balanced equation: 2 mol Al : 3 mol H₂, so n(H₂) = 0.020 × (3/2) = 0.030 mol.
步骤 2 – 运用配平方程式的摩尔比：2 mol Al : 3 mol H₂，因此 n(H₂) = 0.020 × (3/2) = 0.030 mol。
Step 3 – Convert moles to volume: Volume = n × V_m = 0.030 mol × 24 dm³/mol = 0.72 dm³ (or 720 cm³).
步骤 3 – 物质的量换算为体积：体积 = n × V_m = 0.030 mol × 24 dm³/mol = 0.72 dm³（或 720 cm³）。
Watch out: Many students invert the molar ratio; always write the ratio as (wanted substance / given substance).
注意：许多学生会将摩尔比弄反；始终将比例写成（目标物质 / 已知物质）。
5. Chemistry: Equilibrium Constant Kc Calculation | 化学：平衡常数 Kc 计算
For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), at equilibrium the concentrations are: [H₂] = 0.10 mol/dm³, [I₂] = 0.10 mol/dm³, [HI] = 0.50 mol/dm³. Calculate Kc.
对于反应 H₂(g) + I₂(g) ⇌ 2HI(g)，平衡时各物质浓度为：[H₂] = 0.10 mol/dm³，[I₂] = 0.10 mol/dm³，[HI] = 0.50 mol/dm³。计算 Kc。
Step 1 – Write the equilibrium expression: Kc = [HI]² / ([H₂][I₂]).
步骤 1 – 写出平衡常数表达式：Kc = [HI]² / ([H₂][I₂])。
Step 2 – Substitute the equilibrium concentrations: Kc = (0.50)² / (0.10 × 0.10) = 0.25 / 0.01 = 25.
步骤 2 – 代入平衡浓度：Kc = (0.50)² / (0.10 × 0.10) = 0.25 / 0.01 = 25。
Step 3 – Determine units: Since the number of moles of gaseous reactants equals the number of moles of products, the units cancel, so Kc has no units.
步骤 3 – 确定单位：由于反应物气体摩尔总数等于生成物气体摩尔总数，单位约去，因此 Kc 无量纲。
Exam tip: Always check if the equation is balanced before writing Kc; stoichiometric coefficients become exponents.
考试技巧：在写 Kc 表达式前务必检查方程式是否已配平；化学计量数作为指数使用。
6. Chemistry: Acid-Base Titration Problem | 化学：酸碱滴定例题
25.0 cm³ of sodium hydroxide solution is titrated with 0.100 mol/dm³ sulfuric acid. 23.20 cm³ of acid is needed to reach the endpoint. Calculate the concentration of the NaOH solution. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
用 0.100 mol/dm³ 的硫酸滴定 25.0 cm³ 的氢氧化钠溶液，到达终点时消耗硫酸 23.20 cm³。求 NaOH 溶液的浓度。反应：H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O。
Step 1 – Moles of H₂SO₄ used: n = C × V = 0.100 mol/dm³ × (23.20/1000) dm³ = 0.00232 mol.
步骤 1 – 所用 H₂SO₄ 的物质的量：n = C × V = 0.100 mol/dm³ × (23.20/1000) dm³ = 0.00232 mol。
Step 2 – Moles of NaOH reacting: From the equation, 1 mol H₂SO₄ reacts with 2 mol NaOH, so n(NaOH) = 2 × 0.00232 = 0.00464 mol.
步骤 2 – 参与反应的 NaOH 物质的量：根据反应式，1 mol H₂SO₄ 与 2 mol NaOH 反应，所以 n(NaOH) = 2 × 0.00232 = 0.00464 mol。
Step 3 – Concentration of NaOH: C = n / V = 0.00464 mol / (25.0/1000) dm³ = 0.1856 mol/dm³ ≈ 0.186 mol/dm³ (3 sf).
步骤 3 – NaOH 浓度：C = n / V = 0.00464 mol / (25.0/1000) dm³ = 0.1856 mol/dm³ ≈ 0.186 mol/dm³（三位有效数字）。
Common error: forgetting to convert cm³ to dm³ by dividing by 1000.
常见错误：忘记将 cm³ 除以 1000 转换为 dm³。
7. Biology: Enzyme Kinetics – Determining Vmax and Km | 生物：酶动力学 – 求解 Vmax 与 Km
An enzyme-catalysed reaction gave the following initial rates at two substrate concentrations: at [S] = 0.20 mM, v = 20 µmol/min; at [S] = 0.80 mM, v = 50 µmol/min. Use the Michaelis–Menten equation to calculate Vmax and Km.
某酶促反应在两个底物浓度下测得初始速率：[S] = 0.20 mM 时，v = 20 µmol/min；[S] = 0.80 mM 时，v = 50 µmol/min。利用米氏方程计算 Vmax 和 Km。
Step 1 – Write the Michaelis–Menten equation: v = (Vmax [S]) / (Km + [S]).
步骤 1 – 写出米氏方程：v = (Vmax [S]) / (Km + [S])。
Step 2 – Set up two equations: 20 = (Vmax × 0.20) / (Km + 0.20) and 50 = (Vmax × 0.80) / (Km + 0.80).
步骤 2 – 建立两个方程：20 = (Vmax × 0.20) / (Km + 0.20) 和 50 = (Vmax × 0.80) / (Km + 0.80)。
Step 3 – Solve simultaneously: From the first, Vmax = 100(Km + 0.20). Substitute into the second: 50 = [100(Km+0.20) × 0.80] / (Km + 0.80) → 50 = 80(Km+0.20) / (Km+0.80) → 5/
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