A-Level OCR Science: Worked Example Breakdown | A-Level OCR 科学：典型例题详解

📚 A-Level OCR Science: Worked Example Breakdown | A-Level OCR 科学：典型例题详解

Mastering A-Level OCR Science requires more than just knowing facts — it demands the ability to apply concepts to unfamiliar problems. This article walks through a carefully selected set of worked examples spanning Physics, Chemistry, and Biology. Each example is broken down step by step, highlighting exam technique, common pitfalls, and the precise OCR mark scheme requirements.

掌握 A-Level OCR 科学不仅需要记住知识点，更要求能将概念应用到陌生的题目中。本文精选了一系列涵盖物理、化学和生物的典型例题，逐一拆解步骤，突出考试技巧、常见陷阱以及 OCR 评分方案的具体要求。

1. Physics: Resolving Vectors in Equilibrium | 物理：平衡状态下力的分解

A sign of mass 12 kg is hung from two cables attached to a horizontal beam. The left cable makes an angle of 30° to the horizontal, the right cable makes an angle of 45° to the horizontal. Determine the tension in each cable. (g = 9.81 m s⁻²)

一个质量为 12 kg 的标牌由两根连接至水平横梁的缆绳悬挂。左缆绳与水平方向成 30° 角，右缆绳与水平方向成 45° 角。求每根缆绳中的张力。（g = 9.81 m s⁻²）

Draw the free-body diagram. Weight W = mg = 12 × 9.81 = 117.72 N acting vertically downwards. Let tensions be T₁ (left) and T₂ (right). Resolve forces horizontally: T₁ cos30° = T₂ cos45°. Vertically: T₁ sin30° + T₂ sin45° = W. Substitute T₂ from the horizontal equation into the vertical equation, solve for T₁ then T₂.

画出受力分析图。重力 W = mg = 12 × 9.81 = 117.72 N，方向竖直向下。设两根缆绳张力分别为 T₁（左）和 T₂（右）。水平方向分解：T₁ cos30° = T₂ cos45°。竖直方向：T₁ sin30° + T₂ sin45° = W。将水平关系代入竖直方程，解得 T₁，再求 T₂。

Using cos30° = √3/2, sin30° = 1/2, cos45° = sin45° = 1/√2. From horizontal: T₂ = T₁ (cos30°/cos45°) = T₁ (√3/√2). Vertical: T₁(1/2) + T₂(1/√2) = 117.72. Substitute T₂: T₁/2 + (T₁ √3/√2)(1/√2) = 117.72 → T₁/2 + T₁ (√3/2) = 117.72 → T₁ (1 + √3)/2 = 117.72 → T₁ = 2 × 117.72 / (1 + √3) ≈ 235.44 / 2.732 ≈ 86.2 N. Then T₂ ≈ 86.2 × (√3/√2) ≈ 105.6 N.

利用 cos30° = √3/2，sin30° = 1/2，cos45° = sin45° = 1/√2。水平分解得 T₂ = T₁ (cos30°/cos45°) = T₁ (√3/√2)。竖直：T₁/2 + T₂/√2 = 117.72。代入 T₂ 得 T₁/2 + T₁(√3/2) = 117.72 → T₁(1+√3)/2 = 117.72 → T₁ ≈ 86.2 N，T₂ ≈ 105.6 N。

OCR examiners expect clear resolution statements and final answers to an appropriate number of significant figures (here 3 s.f.).

OCR 考官期望清晰的分解陈述，最终结果保留合适的有效数字（此处为三位有效数字）。

2. Physics: Capacitor Discharge | 物理：电容器放电

A 470 µF capacitor is charged to 6.0 V and then discharged through a 22 kΩ resistor. Calculate the time taken for the voltage to fall to 1.5 V.

一个 470 µF 的电容器充电至 6.0 V，然后通过 22 kΩ 电阻放电。计算电压降至 1.5 V 所需的时间。

Use V = V₀ e^(−t/RC). Time constant RC = 22×10³ × 470×10⁻⁶ = 10.34 s. Rearranged: t = −RC ln(V/V₀) = −10.34 × ln(1.5/6.0) = −10.34 × ln(0.25) = −10.34 × (−1.386) ≈ 14.3 s.

使用 V = V₀ e^(−t/RC)。时间常数 RC = 22×10³ × 470×10⁻⁶ = 10.34 s。整理得 t = −RC ln(V/V₀) = −10.34 × ln(1.5/6.0) = −10.34 × ln(0.25) = −10.34 × (−1.386) ≈ 14.3 s。

If the question asks for time to halve, use half-life t₁/₂ = RC ln2 ≈ 0.693 RC. Show formula and substitution; always include units.

若题目问半衰期，可用 t₁/₂ = RC ln2 ≈ 0.693 RC。须展示公式和代入过程，始终包含单位。

3. Physics: Photoelectric Effect Calculation | 物理：光电效应计算

Light of wavelength 240 nm is incident on a metal surface with work function 3.2 eV. Determine the maximum kinetic energy of emitted electrons in eV and their stopping potential. (h = 6.63×10⁻³⁴ J s, c = 3.00×10⁸ m s⁻¹, e = 1.60×10⁻¹⁹ C)

波长为 240 nm 的光照射到功函数为 3.2 eV 的金属表面。求发射电子的最大动能（以 eV 表示）和遏止电势。（h = 6.63×10⁻³⁴ J s，c = 3.00×10⁸ m s⁻¹，e = 1.60×10⁻¹⁹ C）

Photon energy E = hf = hc/λ = (6.63×10⁻³⁴ × 3.00×10⁸) / (240×10⁻⁹) = 8.29×10⁻¹⁹ J. Convert to eV: 8.29×10⁻¹⁹ / 1.60×10⁻¹⁹ = 5.18 eV. KEmax = E − φ = 5.18 − 3.2 = 1.98 eV. Stopping potential Vs = KEmax / e = 1.98 V.

光子能量 E = hf = hc/λ = (6.63×10⁻³⁴ × 3.00×10⁸) / (240×10⁻⁹) = 8.29×10⁻¹⁹ J。转换为 eV：8.29×10⁻¹⁹ / 1.60×10⁻¹⁹ = 5.18 eV。最大动能 KEmax = E − φ = 5.18 − 3.2 = 1.98 eV。遏止电势 Vs = KEmax / e = 1.98 V。

Always check unit conversions carefully. OCR mark schemes reward clear steps showing energy in joules before converting to eV.

务必仔细检查单位换算。OCR 评分方案鼓励先以焦耳表示能量，再转换为电子伏特的清晰步骤。

4. Chemistry: Born–Haber Cycle for NaCl | 化学：NaCl 的波恩–哈伯循环

Construct a Born–Haber cycle for sodium chloride and use the data below to calculate the lattice enthalpy of NaCl(s). Data (kJ mol⁻¹): ΔHf°(NaCl) = −411, Na(s) → Na(g) = +108, Na(g) → Na⁺(g) + e⁻ = +496, ½Cl₂(g) → Cl(g) = +121, Cl(g) + e⁻ → Cl⁻(g) = −349.

构建氯化钠的波恩–哈伯循环，并利用下列数据计算 NaCl(s) 的晶格焓。数据（kJ mol⁻¹）：ΔHf°(NaCl) = −411，Na(s) → Na(g) = +108，Na(g) → Na⁺(g) + e⁻ = +496，½Cl₂(g) → Cl(g) = +121，Cl(g) + e⁻ → Cl⁻(g) = −349。

Cycle: Start with elements in standard states Na(s) + ½Cl₂(g). Route 1: direct formation ΔHf = −411. Route 2: atomisation of Na (+108), atomisation of Cl (+121), ionisation of Na (+496), electron affinity of Cl (−349), and lattice enthalpy (ΔH L). By Hess’s law: −411 = +108 + 121 + 496 − 349 + ΔH L. ΔH L = −411 − (108+121+496−349) = −411 − 376 = −787 kJ mol⁻¹.

循环：从标准状态下的元素 Na(s) + ½Cl₂(g) 出发。路径 1：直接生成焓 ΔHf = −411。路径 2：Na 的原子化（+108），Cl 的原子化（+121），Na 的电离（+496），Cl 的电子亲和（−349），晶格焓（ΔH L）。根据赫斯定律：−411 = +108 + 121 + 496 − 349 + ΔH L，得 ΔH L = −411 − (108+121+496−349) = −411 − 376 = −787 kJ mol⁻¹。

Always define lattice enthalpy as the exothermic process when gaseous ions form solid. Negative sign indicates energy released.

始终将晶格焓定义为气态离子形成固态时的放热过程。负号表示释放能量。

5. Chemistry: Equilibrium Constant Kc | 化学：平衡常数 Kc

0.20 mol of N₂O₄ is allowed to reach equilibrium in a vessel of volume 2.0 dm³ at 298 K. N₂O₄(g) ⇌ 2NO₂(g). At equilibrium, 0.16 mol of N₂O₄ remains. Calculate Kc.

0.20 mol N₂O₄ 在容积为 2.0 dm³ 的容器中于 298 K 达到平衡。N₂O₄(g) ⇌ 2NO₂(g)。平衡时剩余 0.16 mol N₂O₄。计算 Kc。

Initial N₂O₄ = 0.20, change: −x, equilibrium = 0.16 mol, so x = 0.04 mol. NO₂ produced = 2x = 0.08 mol. Concentrations: [N₂O₄] = 0.16 / 2.0 = 0.080 mol dm⁻³; [NO₂] = 0.08 / 2.0 = 0.040 mol dm⁻³. Kc = [NO₂]² / [N₂O₄] = (0.040)² / 0.080 = 0.0016 / 0.080 = 0.020 mol dm⁻³.

初始 N₂O₄ = 0.20，变化量：−x，平衡时剩余 0.16 mol，故 x = 0.04 mol。生成 NO₂ = 2x = 0.08 mol。浓度：[N₂O₄] = 0.16/2.0 = 0.080 mol dm⁻³；[NO₂] = 0.08/2.0 = 0.040 mol dm⁻³。Kc = [NO₂]² / [N₂O₄] = (0.040)² / 0.080 = 0.020 mol dm⁻³。

Include units for Kc as it is not dimensionless here; OCR expects units.

Kc 在此并非无量纲，OCR 要求写明单位。

6. Chemistry: Rate Equation from Initial Rates | 化学：根据初始速率确定速率方程

Determine the rate equation for the reaction A + B → products using the following data:

[A] / mol dm⁻³	[B] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
0.10	0.20	1.6 × 10⁻³
0.20	0.20	3.2 × 10⁻³
0.20	0.40	1.28 × 10⁻²

利用以下数据确定反应 A + B → 产物的速率方程：

[A] / mol dm⁻³	[B] / mol dm⁻³	初始速率 / mol dm⁻³ s⁻¹
0.10	0.20	1.6 × 10⁻³
0.20	0.20	3.2 × 10⁻³
0.20	0.40	1.28 × 10⁻²

Compare expt 1 and 2: [B] constant, [A] doubles → rate doubles, so first order in A. Compare expt 2 and 3: [A] constant, [B] doubles → rate ×4 (from 3.2×10⁻³ to 1.28×10⁻²), so second order in B. Rate = k[A][B]². Calculate k using any row: e.g. k = rate / ([A][B]²) = (1.6×10⁻³) / (0.10 × 0.20²) = 1.6×10⁻³ / 0.004 = 0.40 dm⁶ mol⁻² s⁻¹.

比较实验 1 和 2：[B] 恒定，[A] 加倍，速率加倍，因此对 A 为一级。比较实验 2 和 3：[A] 恒定，[B] 加倍，速率变为四倍（3.2×10⁻³ 到 1.28×10⁻²），因此对 B 为二级。速率方程：rate = k[A][B]²。用任一行计算 k：如 k = rate/([A][B]²) = (1.6×10⁻³)/(0.10 × 0.20²) = 0.40 dm⁶ mol⁻² s⁻¹。

OCR requires stating orders with justification and giving units of k.

OCR 要求说明反应级数并提供理由，同时给出 k 的单位。

7. Biology: Cardiac Output Calculation | 生物：心输出量计算

An athlete has a resting heart rate of 55 beats per minute and a stroke volume of 80 cm³. During exercise, heart rate increases to 160 bpm and stroke volume to 120 cm³. Calculate the resting cardiac output, exercise cardiac output, and the percentage increase.

一名运动员安静心率为 55 次/分，每搏输出量为 80 cm³。运动中，心率升至 160 bpm，每搏输出量升至 120 cm³。计算安静时心输出量、运动时心输出量及增加的百分比。

Cardiac output (CO) = heart rate × stroke volume. Resting CO = 55 × 80 = 4400 cm³ min⁻¹ = 4.4 dm³ min⁻¹. Exercise CO = 160 × 120 = 19 200 cm³ min⁻¹ = 19.2 dm³ min⁻¹. Percentage increase = ((19.2 − 4.4) / 4.4) × 100% = (14.8 / 4.4) × 100% ≈ 336%.

心输出量 CO = 心率 × 每搏输出量。安静时 CO = 55 × 80 = 4400 cm³ min⁻¹ = 4.4 dm³ min⁻¹。运动时 CO = 160 × 120 = 19 200 cm³ min⁻¹ = 19.2 dm³ min⁻¹。增加百分比 = ((19.2 − 4.4) / 4.4) × 100% ≈ 336%。

Always show unit conversions. An answer in dm³ min⁻¹ is often preferred to cm³ min⁻¹ for large numbers.

始终展示单位换算。大数值下通常优先使用 dm³ min⁻¹。

8. Biology: Hardy–Weinberg Principle | 生物：哈迪–温伯格原理

In a population of 500 individuals, 20 show a recessive phenotype. Assuming Hardy–Weinberg equilibrium, calculate the frequency of heterozygous individuals.

在一个 500 个个体的种群中，有 20 个表现出隐性性状。假设符合哈迪–温伯格平衡，计算杂合子的频率。

Frequency of recessive homozygotes q² = 20/500 = 0.04. Therefore q = √0.04 = 0.2. p = 1 − q = 0.8. Frequency of heterozygotes = 2pq = 2 × 0.8 × 0.2 = 0.32. Number of heterozygotes ≈ 0.32 × 500 = 160.

隐性纯合子频率 q² = 20/500 = 0.04。因此 q = √0.04 = 0.2。p = 1 − q = 0.8。杂合子频率 = 2pq = 2 × 0.8 × 0.2 = 0.32。杂合子个数 ≈ 0.32 × 500 = 160。

Remember that p² + 2pq + q² = 1. This is a common OCR question; show all steps and avoid common error of calculating 2q instead of 2pq.

记住 p² + 2pq + q² = 1。这是 OCR 常见题目，展示全部步骤，避免常见错误——误算为 2q 而非 2pq。

9. Biology: Temperature Coefficient Q₁₀ | 生物：温度系数 Q₁₀

The rate of an enzyme-controlled reaction is 2.5 µmol min⁻¹ at 25°C and 6.4 µmol min⁻¹ at 35°C. Calculate the Q₁₀ for this reaction.

某酶促反应在 25°C 时速率为 2.5 µmol min⁻¹，在 35°C 时速率为 6.4 µmol min⁻¹。计算该反应的 Q₁₀。

Q₁₀ = rate at (T+10)°C / rate at T°C = 6.4 / 2.5 = 2.56. This indicates the rate more than doubles for a 10°C rise, typical for enzyme reactions up to the optimum temperature.

Q₁₀ = (T+10)°C 下的速率 / T°C 下的速率 = 6.4 / 2.5 = 2.56。这表明温度上升 10°C 速率增加超过一倍，在达到最适温度前为酶促反应的典型特征。

Explain biological significance: Q₁₀ values around 2–3 show metabolic sensitivity to temperature. OCR may ask for evaluation of deviation above optimum due to denaturation.

解释生物学意义：Q₁₀ 约为 2–3 表示代谢对温度敏感。OCR 可能要求评估超出最适温度后因变性产生的偏差。

10. Physics: Radioactive Decay and Half-life | 物理：放射性衰变与半衰期

A radioactive source has a half-life of 4.5 days and initial activity 1200 Bq. Determine the activity after 18 days and the time for activity to drop to 75 Bq.

某放射源的半衰期为 4.5 天，初始活度为 1200 Bq。求 18 天后的活度以及活度降至 75 Bq 所需的时间。

Number of half-lives n = 18 / 4.5 = 4. Activity A = A₀ (½)^n = 1200 × (½)^4 = 1200 / 16 = 75 Bq. For 75 Bq, we need n such that 1200 × (½)^n = 75 → (½)^n = 75/1200 = 0.0625. Since 0.0625 = (½)^4, n=4. Time = 4 × 4.5 = 18 days.

半衰期数 n = 18 / 4.5 = 4。活度 A = A₀ (½)^n = 1200 × (½)^4 = 1200/16 = 75 Bq。对于 75 Bq，需要 1200 × (½)^n = 75 → (½)^n = 75/1200 = 0.0625。由于 0.0625 = (½)^4，n=4。时间 = 4 × 4.5 = 18 天。

Alternatively use A = A₀ e^(−λt) where λ = ln2 / T₁/₂. Both methods accepted; ensure units match.

也可使用 A = A₀ e^(−λt)，其中 λ = ln2 / T₁/₂。两种方法均可，注意单位一致。

11. Chemistry: Titration Calculation | 化学：滴定计算

25.0 cm³ of 0.100 mol dm⁻³ NaOH is titrated against H₂SO₄ solution. The titre is 23.40 cm³. Calculate the concentration of H₂SO₄.

用 25.0 cm³ 浓度为 0.100 mol dm⁻³ 的 NaOH 滴定 H₂SO₄ 溶液，滴定体积为 23.40 cm³。计算 H₂SO₄ 的浓度。

Equation: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O. Moles NaOH = cV = 0.100 × (25.0/1000) = 0.00250 mol. From stoichiometry, moles H₂SO₄ = moles NaOH / 2 = 0.00125 mol. Concentration H₂SO₄ = moles / volume(dm³) = 0.00125 / (23.40/1000) = 0.0534 mol dm⁻³ (3 s.f.).

方程式：2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O。NaOH 物质的量 = cV = 0.100 × (25.0/1000) = 0.00250 mol。根据计量关系，H₂SO₄ 的量 = NaOH 的一半 = 0.00125 mol。浓度 = 物质的量 / 体积(dm³) = 0.00125 / (23.40/1000) = 0.0534 mol dm⁻³（三位有效数字）。

Consistent units are vital; always convert cm³ to dm³. OCR will penalise missing the 1:2 ratio.

单位统一至关重要，始终将 cm³ 转换为 dm³。OCR 会对遗漏 1:2 摩尔比扣分。

12. Integrated Question: Data Analysis in Biology | 综合题：生物数据分析

A student investigates osmosis in potato strips. They measure percentage mass change after placing strips in sucrose solutions of different concentrations (0.0, 0.2, 0.4, 0.6, 0.8, 1.0 mol dm⁻³). The percentage changes are: +15.2%, +8.3%, +1.1%, −6.0%, −11.4%, −16.8%. Plot a graph, determine the water potential of potato tissue, and explain the trend.

一个学生研究马铃薯条的渗透作用，测量马铃薯条在不同蔗糖浓度（0.0, 0.2, 0.4, 0.6, 0.8, 1.0 mol dm⁻³）中质量变化的百分比。百分比变化为：+15.2%，+8.3%，+1.1%，−6.0%，−11.4%，−16.8%。绘制图形，确定马铃薯组织的水势，并解释趋势。

Plot concentration on x-axis, % change on y-axis. Draw line of best fit; the point where line crosses zero % change corresponds to the sucrose concentration with equal water potential to potato cells – isotonic point. From graph, this occurs at approximately 0.42 mol dm⁻³. Positive change indicates water entering (hypotonic solution), negative change indicates water leaving (hypertonic).

绘制浓度–百分比变化图，画出最佳拟合线。线与零变化线交点的蔗糖浓度即为与马铃薯细胞水势相等的浓度——等渗点。据图，此浓度约为 0.42 mol dm⁻³。正变化表示水分进入（低渗溶液），负变化表示水分流出（高渗溶液）。

OCR practical-based questions demand accurate graph plotting, clear interpolation, and use of appropriate terminology. The water potential of potato tissue is equivalent to that of the sucrose solution at the isotonic point.

OCR 以实验为基础的题目要求准确绘图、清晰内插，并使用恰当术语。马铃薯组织的水势等同于等渗点处蔗糖溶液的水势。

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A-Level OCR Science: Worked Example Breakdown | A-Level OCR 科学：典型例题详解