📚 IGCSE CIE Science: Worked Examples Explained | IGCSE CIE 科学：典型例题详解
Typical worked examples are an essential part of revision for IGCSE CIE Co-ordinated Sciences or Combined Science. This article covers representative problems from Physics, Chemistry and Biology, giving step-by-step solutions and the key conceptual explanations behind each one. Mastering these examples will help you tackle similar questions confidently in your examination.
典型例题是 IGCSE CIE 协调科学或综合科学复习的重要组成部分。本文涵盖物理、化学和生物中的代表性题目，提供逐步解答以及每道题背后的关键概念解释。掌握这些例题，将帮助你在考试中自信应对类似问题。
1. How to Use Worked Examples in CIE Science Revision | 如何在 CIE 科学复习中使用典型例题
A worked example is not a story to read - it is a drill to repeat. The correct routine is simple: cover the solution, attempt the question with your own pen, then compare your working line by line with the model answer. Every difference you find is a gap in your method rather than a gap in your knowledge, and those are the gaps that cost marks.
典型例题不是用来"读"的故事，而是用来"练"的动作。正确流程很简单：遮住答案，自己动笔做一遍，然后逐行与标准答案对照。你发现的每一处差异，都是解题方法上的漏洞，而不是知识上的漏洞，而这类漏洞恰恰是丢分的主因。
Cambridge IGCSE Combined Science (0653) and Co-ordinated Sciences (0654) assess you through multiple-choice questions, structured theory papers and a practical or alternative-to-practical paper. Each format rewards a slightly different skill, but all three reward the same underlying habit of recognising a familiar problem type and applying a standard method quickly and accurately.
剑桥 IGCSE 综合科学（0653）与协调科学（0654）通过选择题、结构化理论卷，以及实验操作卷或实验替代卷来考查你。每种形式侧重的能力略有不同，但三者都奖励同一个底层习惯：识别熟悉题型，并快速、准确地套用标准解法。
Use this checklist for every worked example in this article:
- Write down the formula before substituting numbers. | 先写出公式，再代入数字。
- Show every stage of working, even when the arithmetic is easy. | 写出每一步过程，哪怕算术很简单。
- Attach the correct unit to the final answer. | 最终答案必须带上正确单位。
- Round only at the end, and state the number of significant figures. | 只在最后一步取整，并说明有效数字位数。
- Write a one-line conclusion in words. | 用一句话文字作答，点明结论。
2. Physics - Interpreting Distance-Time Graphs | 物理 - 解读距离-时间图
Problem: A student walks from home to school. The distance-time graph shows three sections: from 0 to 300 s the distance increases from 0 to 400 m; from 300 to 420 s the distance remains constant at 400 m; finally, from 420 to 600 s the distance increases to 600 m. Calculate (a) the speed during the first 300 s, and (b) the average speed for the whole journey.
题目： 一位学生从家步行去学校。距离-时间图显示三段：0 至 300 秒，距离由 0 增加到 400 米；300 至 420 秒，距离保持 400 米不变；最后 420 至 600 秒，距离继续增加到 600 米。计算 (a) 前 300 秒内的速度；(b) 整段行程的平均速度。
Step 1 - Identify the three sections. distance 400 → 600 m. Reading the shape of the graph correctly is worth marks on its own.
第一步 - 划分三段。 A 段（0-300 秒）：离开出发点，距离由 0 到 400 米。B 段（300-420 秒）：距离保持 400 米不变。C 段（420-600 秒）：距离由 400 米到 600 米。正确读出图形形状本身就有分值。
Step 2 - Recall the key rule. On a distance-time graph, the gradient equals the speed. A horizontal line means the gradient is zero, so the object is stationary. A straight sloping line means constant speed; a curve means changing speed.
第二步 - 回忆核心规则。 在距离-时间图上，斜率就是速度。水平线斜率为零，表示物体静止。倾斜的直线表示匀速运动；曲线表示速度在变化。
speed = gradient = Δdistance ÷ Δtime
average speed = total distance ÷ total time
Step 3 - Part (a), speed in the first 300 s. m/s. The value is written to three significant figures, and the unit is metres per second.
第三步 - (a) 前 300 秒的速度。 代入距离变化量与时间变化量：400 米 ÷ 300 秒 = 1.33 m/s。结果保留三位有效数字，单位为米每秒。
Step 4 - Part (b), average speed. is is lower than the speed in section A, because the student was stationary for 120 s and walked more slowly in section C.
第四步 - (b) 平均速度。 要用整段行程，而不是某一段：总距离 600 米，总时间 600 秒，所以 600 ÷ 600 = 1.00 m/s。注意这个值低于 A 段速度，因为学生在 B 段静止了 120 秒，且在 C 段走得更慢。
| Section | Time / s | Distance / m | Gradient | Motion |
| A | 0-300 | 0 → 400 | 400 ÷ 300 = 1.33 m/s | constant speed |
| B | 300-420 | 400 → 400 | 0 | stationary |
| C | 420-600 | 400 → 600 | 200 ÷ 180 = 1.11 m/s | slower constant speed |
3. Physics - Velocity-Time Graphs and Acceleration | 物理 - 速度-时间图与加速度
Problem: A car accelerates uniformly from rest to 20 m/s in 5 s, travels at a constant 20 m/s for the next 10 s, then decelerates uniformly to rest in a further 4 s. Find the acceleration in each stage and the total distance travelled.
题目： 一辆汽车从静止匀加速到 20 m/s 用时 5 秒，随后以 20 m/s 匀速行驶 10 秒，再用 4 秒匀减速到静止。求各阶段加速度及总行驶距离。
Working. eration is 0 m/s². Stage 3: (0 - 20) ÷ 4 = -5.0 m/s², which is a deceleration of 5.0 m/s².
解答。 加速度是速度-时间图的斜率：a = (v - u) ÷ t。第一阶段：(20 - 0) ÷ 5 = 4.0 m/s²。第二阶段：斜率为零，加速度为 0 m/s²。第三阶段：(0 - 20) ÷ 4 = -5.0 m/s²，即减速度 5.0 m/s²。
distance travelled = area under the velocity-time graph
g gives 290 m. Always show the three separate areas - examiners award method marks for each one.
总距离通过把图形面积拆成三角形、矩形、三角形来计算：½ × 5 × 20 = 50 米，10 × 20 = 200 米，½ × 4 × 20 = 40 米。三者相加得 290 米。务必分别写出三个面积，因为评分标准对每一步都给出方法分。
Common trap. The negative sign in stage 3 is not an error; it tells you the direction of the acceleration is opposite to the motion. If the question asks for deceleration, give the magnitude as a positive number and say "deceleration".
常见陷阱。 第三阶段的负号不是算错，它表示加速度方向与运动方向相反。如果题目问的是减速度，就给出正的数值并写明"减速度"。
4. Physics - Moments, Density and Pressure | 物理 - 力矩、密度与压强
Problem A: A uniform beam is pivoted at its centre. A 200 N weight hangs 1.5 m to the left of the pivot. How far to the right of the pivot must a 600 N weight be placed to balance the beam?
题目 A： 一根均匀横梁支点在中央。支点左侧 1.5 米处挂一个 200 牛的砝码。要在支点右侧多远处放置 600 牛的砝码才能使横梁平衡？
Working. he sum of anticlockwise moments. So 200 × 1.5 = 600 × d, giving 300 = 600d and d = 0.50 m.
解答。 应用力矩原理：系统平衡时，顺时针力矩之和等于逆时针力矩之和。因此 200 × 1.5 = 600 × d，即 300 = 600d，得 d = 0.50 米。
moment = force × perpendicular distance from the pivot
Problem B: A metal block has a mass of 540 g and a volume of 200 cm³. Calculate its density and identify the likely metal.
题目 B： 一块金属质量 540 克，体积 200 立方厘米。计算其密度并判断可能是什么金属。
Working. vert to SI units, multiply by 1000, giving 2700 kg/m³. Remember that 1 g/cm³ = 1000 kg/m³.
解答。 密度 = 质量 ÷ 体积 = 540 ÷ 200 = 2.7 g/cm³，与铝相符。换算成国际单位需乘以 1000，得 2700 kg/m³。记住 1 g/cm³ = 1000 kg/m³。
Problem C: A student of weight 600 N stands on two feet, each with an area of 0.01 m². Calculate the pressure on the floor.
题目 C： 一名体重 600 牛的学生双脚站立，每只脚接触面积为 0.01 平方米。求对地面的压强。
Working. Pa. If the student stands on one foot, the area halves and the pressure doubles to 60 kPa.
解答。 总接触面积 = 2 × 0.01 = 0.02 平方米。压强 = 压力 ÷ 面积 = 600 ÷ 0.02 = 30 000 帕 = 30 千帕。若单脚站立，面积减半，压强加倍到 60 千帕。
5. Physics - Series and Parallel Circuits | 物理 - 串联与并联电路
Problem: A 6.0 V battery is connected to a 6.0 Ω resistor and a 3.0 Ω resistor. Calculate the total resistance, the current from the battery, and the current in each resistor when they are connected (a) in series and (b) in parallel.
题目： 一个 6.0 伏电源连接 6.0 欧和 3.0 欧两个电阻。求 (a) 串联、(b) 并联两种情况下，总电阻、电源电流以及每个电阻中的电流。
Part (a) - series. 7 × 6.0 = 4.0 V and across the 3.0 Ω resistor is 0.67 × 3.0 = 2.0 V, and 4.0 + 2.0 = 6.0 V, which checks the supply voltage.
(a) 串联。 串联时电阻相加：R = 6.0 + 3.0 = 9.0 欧。电流处处相同：I = V ÷ R = 6.0 ÷ 9.0 = 0.67 安。6.0 欧电阻两端电压为 0.67 × 6.0 = 4.0 伏，3.0 欧电阻两端为 0.67 × 3.0 = 2.0 伏，4.0 + 2.0 = 6.0 伏，与电源电压吻合。
Part (b) - parallel. e full 6.0 V across it, so the current in the 6.0 Ω resistor is 1.0 A and in the 3.0 Ω resistor is 2.0 A. Adding gives 3.0 A, which confirms the total.
(b) 并联。 用 1/R = 1/R₁ + 1/R₂：1/R = 1/6.0 + 1/3.0 = 0.5，所以 R = 2.0 欧。电源电流为 6.0 ÷ 2.0 = 3.0 安。每条支路两端都是 6.0 伏，所以 6.0 欧电阻中电流为 1.0 安，3.0 欧电阻中为 2.0 安。相加为 3.0 安，与总电流一致。
- In series, current is constant and p.d. is shared in the ratio of the resistances. | 串联中电流恒定，电压按电阻比例分配。
- In parallel, p.d. is constant and current is shared in inverse proportion to the resistance. | 并联中电压恒定，电流按电阻的反比分配。
- Adding a resistor in parallel always lowers the total resistance. | 并联接入电阻总会使总电阻降低。
- Always check that the branch currents add up to the supply current. | 始终检验各支路电流之和是否等于电源电流。
6. Chemistry - Moles and Reacting Masses | 化学 - 摩尔与反应物质量计算
Problem: 4.8 g of magnesium ribbon reacts completely with excess dilute hydrochloric acid. Calculate (a) the number of moles of magnesium used, (b) the volume of hydrogen produced at room temperature and pressure, and (c) the mass of magnesium chloride formed.
题目： 4
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