📚 A-Level CIE Science: Worked Examples Explained | A-Level CIE 科学：典型例题详解
This article presents a carefully selected collection of typical worked examples from the CIE A-Level Sciences – Physics, Chemistry, and Biology. Each example is broken down step by step to reinforce key concepts, common calculations, and exam techniques. By engaging with these bilingual explanations, you will deepen your understanding and build confidence for tackling CIE examination questions.
本文精心挑选了CIE A-Level科学（物理、化学、生物）中的典型例题，并逐题详解，旨在强化核心概念、常见计算与应试技巧。通过中英对照的讲解，帮助你深入理解并提升应对CIE考试题的信心。
1. Physics: Kinematics – Braking Distance | 物理：运动学 – 刹车距离
Question: A car travels at a constant speed of 25 m/s when the driver sees an obstacle. The reaction time before braking is 0.70 s. The car then decelerates uniformly at 6.0 m/s² until it stops. Calculate the total stopping distance.
问题：一辆汽车以25 m/s匀速行驶，司机发现障碍物。刹车前反应时间为0.70秒。之后汽车以6.0 m/s²匀减速至停止。计算总停车距离。
Solution – Thinking Distance: During reaction time, the car keeps moving at constant velocity. Distance = speed × time = 25 × 0.70 = 17.5 m.
解答 – 反应距离：在反应时间内，汽车保持匀速运动。距离 = 速度 × 时间 = 25 × 0.70 = 17.5 m。
Braking Distance: Use v² = u² + 2as, with final v = 0, u = 25 m/s, a = –6.0 m/s².
刹车距离：使用公式 v² = u² + 2as，末速度v = 0，初速u = 25 m/s，加速度a = –6.0 m/s²。
0 = (25)² + 2(–6.0)s → 0 = 625 – 12s → s = 52.08 m
Total stopping distance = thinking distance + braking distance = 17.5 + 52.08 ≈ 69.6 m.
总停车距离 = 反应距离 + 刹车距离 = 17.5 + 52.08 ≈ 69.6 m。
2. Physics: Newton’s Second Law – Connected Bodies | 物理：牛顿第二定律 – 连接体问题
Question: Two blocks of masses 5.0 kg and 3.0 kg are connected by a light inextensible string over a smooth pulley. The 5.0 kg mass rests on a frictionless horizontal table, while the 3.0 kg mass hangs vertically. Calculate the acceleration of the system and the tension in the string. (Take g = 9.8 m/s²)
问题：质量分别为5.0 kg和3.0 kg的两个物块用轻质不可伸长的绳子跨过光滑滑轮连接。5.0 kg滑块放在无摩擦的水平桌面上，3.0 kg物块竖直悬挂。计算系统的加速度和绳中张力。（取g = 9.8 m/s²）
Solution: For the hanging mass (3.0 kg): weight – tension = m₂a → 3.0g – T = 3.0a.
解答：对悬挂物块（3.0 kg）：重力 – 张力 = m₂a → 3.0g – T = 3.0a。
For the mass on the table (5.0 kg): T = m₁a → T = 5.0a.
对桌面上的物块（5.0 kg）：T = m₁a → T = 5.0a。
Solve simultaneously: 3.0g – 5.0a = 3.0a → 3.0g = 8.0a.
联立求解：3.0g – 5.0a = 3.0a → 3.0g = 8.0a。
a = (3.0 × 9.8) / 8.0 = 3.675 m/s² ≈ 3.7 m/s²
Tension T = 5.0a = 5.0 × 3.675 ≈ 18.4 N.
张力T = 5.0a = 5.0 × 3.675 ≈ 18.4 N。
3. Physics: Electricity – Series and Parallel Circuits | 物理：电学 – 串并联电路
Question: A 12 V battery is connected to a network: a 4.0 Ω resistor in series with a parallel combination of 6.0 Ω and 3.0 Ω resistors. Calculate (a) the total resistance, (b) the current supplied by the battery, and (c) the power dissipated in the 4.0 Ω resistor.
问题：一个12 V电池连接到网络：一个4.0 Ω电阻与一个由6.0 Ω和3.0 Ω并联组成的支路串联。计算：(a) 总电阻，(b) 电池提供的电流，(c) 4.0 Ω电阻耗散的功率。
(a) Parallel resistance: 1/R_par = 1/6.0 + 1/3.0 = 1/6.0 + 2/6.0 = 3/6.0 → R_par = 2.0 Ω.
(a) 并联电阻：1/R_par = 1/6.0 + 1/3.0 = 1/6.0 + 2/6.0 = 3/6.0 → R_par = 2.0 Ω。
Total resistance R_total = 4.0 + 2.0 = 6.0 Ω.
总电阻R_total = 4.0 + 2.0 = 6.0 Ω。
(b) Current I = V / R_total = 12 / 6.0 = 2.0 A.
(b) 电流 I = V / R_total = 12 / 6.0 = 2.0 A。
(c) Power in 4.0 Ω: P = I²R = (2.0)² × 4.0 = 16 W.
(c) 4.0 Ω电阻的功率：P = I²R = (2.0)² × 4.0 = 16 W。
4. Physics: Waves – Double-Slit Interference | 物理：波动 – 双缝干涉
Question: In a Young’s double-slit experiment, light of wavelength 600 nm illuminates two slits separated by 0.50 mm. The interference pattern is observed on a screen 2.0 m away. Calculate the fringe spacing.
问题：在杨氏双缝实验中，波长为600 nm的光照射间距为0.50 mm的双缝。干涉图样在2.0 m远的屏幕上观察。计算条纹间距。
Solution: Fringe spacing Δx = λD / d, where λ = 600 × 10⁻⁹ m, D = 2.0 m, d = 0.50 × 10⁻³ m.
解答：条纹间距 Δx = λD / d，其中 λ = 600 × 10⁻⁹ m，D = 2.0 m，d = 0.50 × 10⁻³ m。
Δx = (600 × 10⁻⁹ × 2.0) / (0.50 × 10⁻³) = (1.2 × 10⁻⁶) / (5.0 × 10⁻⁴) = 2.4 × 10⁻³ m = 2.4 mm
Thus, the bright fringes are 2.4 mm apart.
因此，亮条纹间距为2.4 mm。
5. Chemistry: Stoichiometry – Moles and Yield | 化学：化学计量 – 摩尔与产率
Question: 5.00 g of calcium carbonate reacts with excess hydrochloric acid: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O. If 2.10 g of calcium chloride is obtained, calculate the percentage yield. (Molar masses: CaCO₃ = 100.1 g/mol, CaCl₂ = 111.0 g/mol)
问题：5.00 g碳酸钙与过量盐酸反应：CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O。如果得到2.10 g氯化钙，计算产率。（摩尔质量：CaCO₃ = 100.1 g/mol，CaCl₂ = 111.0 g/mol）
Moles of CaCO₃ used = mass / M = 5.00 / 100.1 = 0.04995 mol ≈ 0.0500 mol.
使用的CaCO₃物质的量 = 质量 / 摩尔质量 = 5.00 / 100.1 = 0.04995 mol ≈ 0.0500 mol。
From the 1:1 mole ratio, theoretical moles of CaCl₂ = 0.0500 mol.
根据1:1摩尔比，理论CaCl₂的物质的量 = 0.0500 mol。
Theoretical mass of CaCl₂ = 0.0500 × 111.0 = 5.55 g.
理论CaCl₂质量 = 0.0500 × 111.0 = 5.55 g。
Percentage yield = (actual / theoretical) × 100 = (2.10 / 5.55) × 100 ≈ 37.8%.
产率 = (实际产量 / 理论产量) × 100 = (2.10 / 5.55) × 100 ≈ 37.8%。
6. Chemistry: Organic – Nucleophilic Substitution | 化学：有机 – 亲核取代反应
Question: Bromoethane reacts with aqueous sodium hydroxide. Write the equation for the reaction and outline the mechanism using curly arrows. Identify the type of mechanism.
问题：溴乙烷与氢氧化钠水溶液反应。写出反应方程式并用弯箭头概述反应机理。指出机理类型。
Equation: C₂H₅Br + NaOH → C₂H₅OH + NaBr.
方程式：C₂H₅Br + NaOH → C₂H₅OH + NaBr。
Mechanism: The hydroxide ion, OH⁻, acts as a nucleophile and attacks the partially positive carbon atom bonded to bromine. The C–Br bond breaks heterolytically, with the electron pair moving to the bromine, which departs as a bromide ion.
机理：氢氧根离子 OH⁻ 作为亲核试剂进攻与溴相连的带部分正电荷的碳原子。C–Br键异裂，电子对移向溴原子，溴以溴离子形式离去。
Curly arrow from a lone pair on O in OH⁻ towards the carbon atom; a second curly arrow from the C–Br bond to the Br atom. Transition state involves a pentavalent carbon. The product is ethanol, and the mechanism is SN2 (bimolecular nucleophilic substitution).
用弯箭头表示：OH⁻中氧上的孤对电子进攻碳原子；第二个弯箭头从C–Br键指向Br原子。过渡态涉及五价碳。产物为乙醇，机理类型为SN2（双分子亲核取代）。
7. Chemistry: Equilibrium – Kc Calculation | 化学：平衡 – Kc计算
Question: For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), 0.80 mol of PCl₅ is placed in a 2.0 dm³ container and heated. At equilibrium, 0.20 mol of Cl₂ is present. Calculate Kc.
问题：对于反应 PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)，将0.80 mol PCl₅放入一个2.0 dm³容器中加热。达到平衡时，有0.20 mol Cl₂存在。计算Kc。
Initial moles: PCl₅ = 0.80, PCl₃ = 0, Cl₂ = 0.
初始物质的量：PCl₅ = 0.80，PCl₃ = 0，Cl₂ = 0。
Change: let x mol of PCl₅ dissociate. At equilibrium, Cl₂ = x = 0.20 mol. Thus PCl₃ = 0.20 mol, PCl₅ = 0.80 – 0.20 = 0.60 mol.
变化：设x mol PCl₅解离。平衡时Cl₂ = x = 0.20 mol。因此PCl₃ = 0.20 mol，PCl₅ = 0.80 – 0.20 = 0.60 mol。
Equilibrium concentrations (mol/dm³): [PCl₅] = 0.60/2.0 = 0.30, [PCl₃] = 0.20/2.0 = 0.10, [Cl₂] = 0.20/2.0 = 0.10.
平衡浓度（mol/dm³）：[PCl₅] = 0.60/2.0 = 0.30，[PCl₃] = 0.20/2.0 = 0.10，[Cl₂] = 0.20/2.0 = 0.10。
Kc = [PCl₃][Cl₂] / [PCl₅] = (0.10)(0.10) / 0.30 = 0.0333 mol/dm³
Thus Kc ≈ 3.33 × 10⁻² mol dm⁻³.
因此Kc ≈ 3.33 × 10⁻² mol dm⁻³。
8. Biology: Genetics – Monohybrid Cross Probability | 生物：遗传学 – 单基因杂交概率
Question: Cystic fibrosis is caused by a recessive allele ‘f’. Two carriers (Ff) have a child. (a) Calculate the probability that the child is affected. (b) What is the probability that a healthy child is a carrier?
问题：囊性纤维化由隐性等位基因’f’引起。两个携带者（Ff）生育一个孩子。(a) 计算孩子患病的概率。(b) 健康孩子是携带者的概率是多少？
(a) Using a Punnett square:
(a) 用庞纳特方格：
| Gametes | F | f |
| F | FF | Ff |
| f | Ff | ff |
Genotypic ratio: 1 FF : 2 Ff : 1 ff. Affected child = ff, probability = 1/4 or 25%.
基因型比例：1 FF : 2 Ff : 1 ff。患病孩子为ff，概率 = 1/4 或 25%。
(b) Healthy children are FF and Ff (3 out of 4). Among these healthy ones, 2 out of 3 are carriers. So probability = 2/3.
(b) 健康孩子为FF和Ff（4个中占3个）。在这些健康孩子中，3个中有2个是携带者。因此概率 = 2/3。
9. Biology: Respiration – Respiratory Quotient (RQ) | 生物：细胞呼吸 – 呼吸商
Question: A respirometer experiment shows that a germinating seed consumes 0.45 cm³ of oxygen and produces 0.40 cm³ of carbon dioxide in 30 minutes. Calculate the respiratory quotient (RQ) and suggest the likely respiratory substrate.
问题：呼吸计实验显示，发芽种子在30分钟内消耗0.45 cm³氧气并产生0.40 cm³二氧化碳。计算呼吸商（RQ）并推测可能的呼吸底物。
RQ = volume of CO₂ produced / volume of O₂ consumed = 0.40 / 0.45 ≈ 0.89.
RQ = 产生的CO₂体积 / 消耗的O₂体积 = 0.40 / 0.45 ≈ 0.89。
RQ values: carbohydrate ~1.0; lipid ~0.7; protein ~0.9. An RQ of 0.89 is close to the value for proteins, but it could also indicate a mixed substrate with predominantly proteins or a combination of lipid and carbohydrate.
RQ值：碳水化合物约1.0；脂类约0.7；蛋白质约0.9。RQ为0.89接近蛋白质的数值，但也可能表明为混合底物，以蛋白质为主或脂类与碳水化合物的组合。
10. Biology: Ecology – Energy Flow Efficiency | 生物：生态学 – 能量流动效率
Question: In a grassland ecosystem, the net primary production (NPP) is 28,000 kJ m⁻² yr⁻¹. Primary consumers ingest 4,200 kJ m⁻² yr⁻¹, of which 2,800 kJ is assimilated and 1,600 kJ is used for net secondary production. Calculate (a) the assimilation efficiency and (b) the ecological efficiency from primary producer to secondary consumer.
问题：在一片草原生态系统中，净初级生产量为28,000 kJ m⁻² yr⁻¹。初级消费者摄取4,200 kJ m⁻² yr⁻¹，其中2,800 kJ被同化，1,600 kJ用于净次级生产。计算 (a) 同化效率和 (b) 由初级生产者到次级消费者的生态效率。
(a) Assimilation efficiency = (assimilated / ingested) × 100 = (2,800 / 4,200) × 100 ≈ 66.7%.
(a) 同化效率 = (同化量 / 摄取量) × 100 = (2,800 / 4,200) × 100 ≈ 66.7%。
(b) Ecological efficiency = (net secondary production / NPP) × 100 = (1,600 / 28,000) × 100 ≈ 5.7%. This is typical for the transfer between trophic levels.
(b) 生态效率 = (净次级生产量 / 净初级生产量) × 100 = (1,600 / 28,000) × 100 ≈ 5.7%。该数值符合营养级间的典型传递效率。
11. Physics: Momentum – Impulse and Force | 物理：动量 – 冲量与力
Question: A tennis ball of mass 0.058 kg strikes a racket horizontally at 30 m/s and rebounds at 25 m/s in the opposite direction. The contact time is 0.015 s. Calculate (a) the impulse exerted on the ball and (b) the average force exerted by the racket.
问题：一个质量为0.058 kg的网球以30 m/s水平撞击球拍，并以25 m/s沿相反方向反弹。接触时间为0.015 s。计算 (a) 球获得的冲量和 (b) 球拍施加的平均力。
(a) Impulse = change in momentum = m(v – u). Taking the initial direction as positive: u = +30 m/s, v = –25 m/s.
(a) 冲量 = 动量变化量 = m(v – u)。设初始方向为正：u = +30 m/s，v = –25 m/s。
Impulse = 0.058 × (–25 – 30) = 0.058 × (–55) = –3.19 kg m/s
The magnitude is 3.19 N s, direction opposite to initial motion.
冲量大小为3.19 N s，方向与初始运动方向相反。
(b) Average force = impulse / time = 3.19 / 0.015 ≈ 213 N in the direction of the rebound.
(b) 平均力 = 冲量 / 时间 = 3.19 / 0.015 ≈ 213 N，方向沿反弹方向。
12. General Scientific Skill: Graph Interpretation | 通用科学技能：图表解读
Question: A student investigates the effect of temperature on the rate of an enzyme-controlled reaction. The rate (arbitrary units) is measured every 5°C between 10°C and 60°C. The data are: 10°C – 0.5, 20°C – 1.2, 30°C – 2.4, 40°C – 3.6, 50°C – 2.0, 60°C – 0.3. Describe the trend and explain the shape of the curve.
问题：一名学生研究温度对酶促反应速率的影响。在10°C至60°C之间每隔5°C测量反应速率（任意单位）。数据为：10°C – 0.5, 20°C
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