📚 Year 9 Edexcel Science: Case Study Practice | Year 9 Edexcel 科学：案例分析实战演练
Case study questions in Year 9 Edexcel Science challenge you to apply your knowledge to real-world scenarios. You will be given data, graphs, or descriptions of experiments and then asked to analyse, calculate, and explain trends. This kind of practice builds the skills you need for higher-level science and exams.
在 Year 9 Edexcel 科学中，案例分析题要求你把知识运用到真实情境里。题目会给出数据、图表或实验描述，然后让你分析、计算并解释趋势。这类练习能帮你培养高阶科学思维和考试所需的能力。
1. Introduction to Case Study Questions | 案例分析题型简介
Case studies test how well you can combine scientific ideas with practical interpretation. A typical question provides background context, a table of results, and a series of sub-questions. You might need to identify variables, draw a conclusion, or perform a simple calculation.
案例分析考查你能否将科学概念与实际解读结合起来。典型题目会提供背景信息、一张结果表格以及若干小问。你可能需要识别变量、得出结论或进行简单计算。
In Edexcel Year 9 Science, the contexts often come from familiar topics: forces and motion, rates of reaction, ecosystems, and environmental chemistry. The key is to read the information carefully, use data to support your answers, and write in full, logical sentences.
在 Edexcel Year 9 科学中，情境常来自熟悉的主题：力与运动、反应速率、生态系统和环境化学。关键是要仔细阅读信息，用数据支撑答案，并用完整、有逻辑的句子作答。
2. Case Study 1: The Stopping Distance of a Car | 案例一：汽车的刹车距离
A student investigated how speed affects the braking distance of a model car. The car was released from different heights on a ramp to vary its speed. The braking distance was measured from the point where the brakes were applied. The results are shown below.
一位学生探究了速度对模型汽车刹车距离的影响。通过从坡道不同高度释放小车来改变它的速度，刹车距离从刹车点开始测量。结果如下所示。
| Speed v (m/s) | Braking distance d (m) |
|---|---|
| 1.0 | 0.25 |
| 2.0 | 1.00 |
| 3.0 | 2.25 |
| 4.0 | 4.00 |
The thinking distance (reaction time) was assumed to be constant at 0.75 s. The total stopping distance is the sum of thinking distance and braking distance.
假设反应距离（反应时间恒定）为 0.75 秒。总刹车距离等于反应距离与制动距离之和。
3. Data Analysis: Plotting and Interpreting Graphs | 数据分析：作图与解读
Plot the braking distance against speed on a graph. The pattern shows that braking distance is not proportional to speed. In fact, doubling the speed from 2.0 m/s to 4.0 m/s quadruples the braking distance from 1.00 m to 4.00 m.
在图上标出刹车距离与速度的关系。可以看出刹车距离与速度不成正比。事实上，速度从 2.0 m/s 加倍到 4.0 m/s，刹车距离从 1.00 m 增加到了 4.00 m，变成了四倍。
This indicates a squared relationship: d ∝ v². You can test this by calculating d/v² for each pair. All values give d/v² ≈ 0.25, confirming the proportionality.
这表明二者存在平方关系：d ∝ v²。你可以通过计算每一组的 d/v² 来验证，所有数值都约为 0.25，证实了这一比例关系。
The gradient of a d vs. v² graph would then be the constant 0.25 m⁻¹s². Always label axes clearly and include units when plotting graphs in case study questions.
因此 d 对 v² 图的斜率就是常数 0.25 m⁻¹s²。在案例分析题中绘图时，务必清晰标注坐标轴并注明单位。
4. Applying Physics: Forces and Motion Equations | 物理应用：力与运动方程
To find the thinking distance, use the equation: distance = speed × time. At 3.0 m/s, thinking distance = 3.0 m/s × 0.75 s = 2.25 m. The total stopping distance at this speed is 2.25 m (thinking) + 2.25 m (braking) = 4.50 m.
计算反应距离的公式是：距离 = 速度 × 时间。当速度为 3.0 m/s 时，反应距离 = 3.0 m/s × 0.75 s = 2.25 m。该速度下的总刹车距离为 2.25 m（反应）+ 2.25 m（制动）= 4.50 m。
A second equation, v² = u² + 2as, can describe the braking phase. Here u = initial speed, v = 0, a is deceleration (negative), and s is braking distance. Rearranging gives a = -u²/(2s). For u = 4.0 m/s, s = 4.00 m, a = -16/(8) = -2.0 m/s².
另一个方程 v² = u² + 2as 可描述制动过程，其中 u 为初速度，v = 0，a 为减速度（负值），s 为制动距离。变形得 a = -u²/(2s)。代入 u = 4.0 m/s, s = 4.00 m，得 a = -16/(8) = -2.0 m/s²。
Understanding these relationships helps explain why driving at high speeds dramatically increases danger. Always include these algebraic steps in your written answers to gain full marks.
理解这些关系有助于解释为何高速驾驶会大幅增加危险。在书面答案中一定要写出这些代数步骤，才能拿到满分。
5. Case Study 2: Investigating Reaction Rate of Magnesium and Acid | 案例二：探究镁与酸的反应速率
A class wanted to find out how the concentration of hydrochloric acid affects the rate of its reaction with magnesium ribbon. They measured the volume of hydrogen gas produced every 10 seconds. The acid concentrations used were 0.5 mol/dm³ and 1.0 mol/dm³.
某班级想探究盐酸浓度如何影响其与镁带的反应速率。他们每 10 秒测量一次产生的氢气体积。使用的酸浓度分别为 0.5 mol/dm³ 和 1.0 mol/dm³。
The reaction is: Mg + 2HCl → MgCl₂ + H₂. The students kept the mass of magnesium (0.12 g) and the volume of acid (50 cm³) the same. They also maintained the temperature at 20°C.
反应方程式为：Mg + 2HCl → MgCl₂ + H₂。学生保持镁的质量（0.12 g）、酸的体积（50 cm³）不变，温度维持在 20°C。
The results for 1.0 mol/dm³ acid showed 40 cm³ of hydrogen after 30 s, while 0.5 mol/dm³ acid produced only 20 cm³ in the same time. The rate is clearly faster with higher concentration.
1.0 mol/dm³ 酸在 30 秒后产生 40 cm³ 氢气，而 0.5 mol/dm³ 在同样时间内只产生 20 cm³。浓度越高，反应速率越快，这一点很明显。
6. Control Variables and Fair Testing | 控制变量与公平测试
To make the investigation valid, students must identify independent, dependent, and control variables. Here, the independent variable is the acid concentration, the dependent variable is the volume of hydrogen produced (or the rate), and control variables include the mass of Mg, volume of acid, and temperature.
为了使探究有效，学生必须识别独立变量、因变量和控制变量。这里独立变量是酸的浓度，因变量是氢气的体积（或速率），控制变量包括镁的质量、酸体积以及温度。
Why must the magnesium ribbon have the same surface area? Because cutting it into smaller pieces increases the surface area, which would make the reaction faster. Using the same length and width of ribbon ensures a fair test.
为什么镁带的表面积必须相同？因为切成更小的碎片会增加表面积，使反应更快。使用相同长度和宽度的镁带能保证公平测试。
Another important control is stirring the mixture gently. Without stirring, the acid around the magnesium becomes diluted as it reacts, slowing the rate. The same stirring rate must be used for all trials.
另一个重要的控制是轻轻搅拌混合物。不搅拌的话，镁周围的酸会随着反应进行而被稀释，导致速率下降。所有实验必须采用相同的搅拌速率。
7. Chemical Equations and Calculations | 化学方程式与计算
Write a word equation for the reaction: magnesium + hydrochloric acid → magnesium chloride + hydrogen. The balanced symbol equation is Mg + 2HCl → MgCl₂ + H₂. The arrow (→) shows the direction of change.
写出反应的文字表达式：镁 + 盐酸 → 氯化镁 + 氢气。配平的符号方程式为 Mg + 2HCl → MgCl₂ + H₂。箭头（→）表示变化的方向。
The rate of reaction can be calculated as the change in volume of H₂ divided by the time taken. For the first 30 s with 1.0 mol/dm³ acid, rate = 40 cm³ / 30 s = 1.33 cm³/s. For 0.5 mol/dm³, rate = 20 cm³ / 30 s = 0.67 cm³/s.
反应速率可以用氢气体积的变化量除以所用时间来计算。使用 1.0 mol/dm³ 酸的前 30 秒，速率 = 40 cm³ / 30 s = 1.33 cm³/s。用 0.5 mol/dm³ 酸时，速率 = 20 cm³ / 30 s = 0.67 cm³/s。
You can also determine the mass of magnesium that reacted. If 24 cm³ of H₂ is collected at room temperature, 24 cm³ is roughly 0.001 mol, so 0.001 mol Mg reacted, mass = 0.001 × 24 = 0.024 g. This shows limited reactant calculations.
你还可以计算反应的镁的质量。如果在室温下收集到 24 cm³ H₂，大约是 0.001 mol，那么反应的 Mg 也是 0.001 mol，质量 = 0.001 × 24 = 0.024 g。这体现了限量反应物的计算。
8. Case Study 3: Food Webs in a Woodland Ecosystem | 案例三：林地生态系统中的食物网
Look at the following woodland food web: oak tree → aphid → ladybird → robin; oak tree → caterpillar → blue tit; also grass → rabbit → fox. A pesticide is sprayed to kill aphids. Explain how this action changes the populations of other organisms.
观察以下林地食物网：橡树 → 蚜虫 → 瓢虫 → 知更鸟；橡树 → 毛虫 → 蓝山雀；同时还有草 → 兔 → 狐狸。喷洒农药杀死了蚜虫。请解释这一行为如何改变其他生物的数量。
With aphids removed, the ladybirds will lose their main food source, so their population decreases. Robins that eat ladybirds may then decline due to lack of food. Alternatively, robins might switch to eating more caterpillars, increasing competition with blue tits.
蚜虫被清除后，瓢虫失去了主要食物来源，数量将减少。以瓢虫为食的知更鸟可能因食物短缺而数量下降，或者转而捕食更多毛虫，这会加剧与蓝山雀的竞争。
The caterpillar population might initially rise because fewer blue tits can survive, but this depends on how many alternate prey the blue tits can find. Ecosystem models like this show the interdependence of species.
毛虫数量最初可能上升，因为蓝山雀的数量减少，但这取决于蓝山雀能找到多少替代猎物。这类生态模型展示了物种间的相互依存关系。
9. Energy Flow and Trophic Levels | 能量流动与营养级
In this woodland, the oak tree is a producer. Aphids and caterpillars are primary consumers. Ladybirds and blue tits are secondary consumers. Robins and foxes are tertiary consumers. Energy passes along these trophic levels, but only about 10% is transferred between each level.
在这个林地生态系统中，橡树是生产者，蚜虫和毛虫是初级消费者，瓢虫和蓝山雀是次级消费者，知更鸟和狐狸是三级消费者。能量沿着这些营养级传递，但每一级之间只有约 10% 被传递。
Draw a pyramid of biomass: the oak tree has the largest biomass, followed by the primary consumers, then secondary, and finally a small biomass of tertiary consumers. Explain why most of the energy is lost: it is used for respiration, movement, growth, and lost as heat or in waste.
画出生物量金字塔：橡树的生物量最大，其次是初级消费者，然后是次级消费者，三级消费者的生物量最小。解释能量为何大部分会损失：用于呼吸、运动、生长，并以热能或废物形式散失。
Because of this energy loss, food chains rarely have more than four or five trophic levels. In the case study, the robin is near the top of one chain, so its population is limited by the energy available from the oak tree.
由于这种能量损耗，食物链很少能超过四到五个营养级。在这个案例中，知更鸟处于某条食物链的顶端，因此它的数量受限于橡树所能提供的能量。
10. Predicting the Impact of Environmental Change | 预测环境变化的影响
A new road is built near the woodland, causing air pollution that weakens the oak trees. Predict how the food web will be affected. Weakened trees produce fewer leaves, so aphid and caterpillar populations decline. This reduces the carrying capacity for all consumers above them.
林地附近新建了一条公路，空气污染削弱了橡树的长势。预测食物网会受到怎样的影响。长势衰弱的橡树叶片减少，因此蚜虫和毛虫的数量下降，这会降低所有上层消费者的环境容纳量。
Long-term effects could include local extinction of ladybirds and robins if they cannot find alternative habitats. Blue tits might move to other wooded areas, while rabbits and foxes might be less affected because they rely on grass, which is more resilient.
长期影响可能包括瓢虫和知更鸟在本地的灭绝，如果它们无法找到替代栖息地的话。蓝山雀可能会迁移到其他林地，而兔子和狐狸受影响较小，因为它们依赖的草地更有恢复力。
Explain why this is a case of bioaccumulation: pollutants from cars can be absorbed by oak leaves, then concentrate in aphids, ladybirds, and robins. This puts top predators at highest risk.
解释这为什么会涉及生物累积：汽车污染物可被橡树叶片吸收，然后在蚜虫、瓢虫和知更鸟体内富集，使顶级捕食者面临最高风险。
11. Case Study 4: Acid Rain and its Effects | 案例四：酸雨及其影响
Acid rain is formed when sulfur dioxide (SO₂) and nitrogen oxides (NOₓ) from burning fossil fuels react with water vapour in the air. The main acids produced are sulfurous acid (H₂SO₃) and nitric acid (HNO₃). This rain damages buildings, forests, and aquatic life.
酸雨是由于燃烧化石燃料产生的二氧化硫（SO₂）和氮氧化物（NOₓ）与空气中的水蒸气反应而形成的。主要生成的酸是亚硫酸（H₂SO₃）和硝酸（HNO₃）。酸雨会损害建筑、森林和水生生物。
An investigation measured the pH of rainwater in three locations: a city centre (pH 4.2), a rural area (pH 5.6), and near a coal power station (pH 3.9). The lowest pH indicates the most acidic conditions and therefore the most severe environmental impact.
一项调查测量了三个地点的雨水 pH：市中心（pH 4.2）、乡村地区（pH 5.6）和煤电厂附近（pH 3.9）。pH 最低表明酸性最强，环境影响也最严重。
Limestone buildings are made of calcium carbonate (CaCO₃). Acid rain reacts with CaCO₃: CaCO₃ + 2H⁺ → Ca²⁺ + CO₂ + H₂O. This causes erosion and weathering. Testing the mass of a limestone chip before and after soaking in acid can quantify the effect.
石灰石建筑由碳酸钙（CaCO₃）构成。酸雨与 CaCO₃ 反应：CaCO₃ + 2H⁺ → Ca²⁺ + CO₂ + H₂O，造成侵蚀和风化。将一块石灰石在酸中浸泡前后称重，就可以量化这种影响。
12. Writing a Balanced Conclusion | 撰写均衡的结论
In case study answers, a balanced conclusion means you state what the data shows, explain the science, and mention any limitations or further experiments needed. For the acid rain study, conclude that locations closer to pollution sources have lower pH, and the chemical reaction with limestone proves the destructive potential.
在案例分析题的答案中，均衡的结论意味着你要陈述数据反映了什么，解释其中的科学原理，并提及任何局限性或需要进一步实验的内容。对于酸雨研究，可以得出结论：越靠近污染源的地区雨水 pH 越低，而与石灰石的化学反应证明了其破坏潜力。
Always use data from the table or graph, e.g. ‘the city centre pH was 4.2 compared to rural 5.6, a difference of 1.4 pH units, which represents a roughly 15-fold increase in hydrogen ion concentration.’ Linking numbers to scientific concepts secures high marks.
务必使用表格或图表中的数据，例如：“市中心 pH 为 4.2，乡村为 5.6，相差 1.4 个 pH 单位，这代表氢离子浓度大约增加了 15 倍。” 将数字与科学概念联系起来能确保拿到高分。
Finally, suggest how reliability could be improved: collect more samples, repeat the measurement over several months, or compare different seasons. This shows you understand the nature of scientific evidence.
最后，提出如何提高可靠性的建议：收集更多样本、连续数月重复测量或比较不同季节的数据。这展示了你对科学证据本质的理解。
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