📚 Case Study Practice for Year 11 Eduqas Physics | 案例分析实战演练
In the Eduqas GCSE Physics examination, case study questions are designed to assess your ability to apply scientific knowledge to real-world scenarios. These questions often present a passage of text, data tables, graphs, or diagrams, and require you to extract information, perform calculations, and evaluate implications. Mastering case studies is essential for achieving top grades. This article provides a practical guide with worked examples covering key topics in the Year 11 syllabus.
在 Eduqas GCSE 物理考试中，案例分析题旨在考查你将科学知识应用于真实情境的能力。这类题目通常给出一段文字、数据表、图表或示意图，要求你提取信息、进行计算并评估影响。掌握案例分析对于取得高分至关重要。本文将提供实战指南，通过覆盖 Year 11 考纲核心主题的例题进行详细讲解。
1. Understanding Case Study Questions | 理解案例分析题
Case study questions integrate multiple concepts from different areas of physics. You need to read the provided context carefully, identify key variables and keywords, and decide which equations or principles to apply. The Eduqas board frequently uses contexts such as energy transfers, transport safety, medical physics, and electricity generation. Always show your working clearly, state correct units, and refer explicitly to the data or information given in the stem. Marks are awarded for using the data, not just for memorised facts.
案例分析题融合了来自不同物理领域的多个概念。你需要仔细阅读所给情景，识别关键变量和关键词，并决定应用哪些方程或原理。Eduqas 考试局经常使用能量转移、交通安全、医学物理和发电等情景。你务必清晰展示解题过程，标注正确的单位，并明确引用题干中给出的数据或信息。评分时会看重你是否使用了所给数据，而不仅仅是记忆性的事实。
2. Key Skills: Extracting Data from Text and Diagrams | 关键技能：从文本和图表中提取数据
Keywords such as 'efficiency', 'power', 'terminal velocity', 'half-life', or 'potential difference' indicate which area of physics is relevant. Look for numerical values accompanied by units, for example 'the panel generates 200 W per square metre' or 'the car stops in 0.12 s'. Diagrams and graphs require careful interpretation: you may need to find gradients, areas under curves, or intercepts. Always check whether the units need conversion (e.g. cm² to m², minutes to seconds) before substituting into equations.
像“效率”、“功率”、“终端速度”、“半衰期”或“电势差”这样的关键词会提示你相关的物理领域。寻找带单位的数值，例如“该电池板每平方米发电200瓦”或“汽车在0.12秒内停下”。图表需要仔细解读：你可能会需要求斜率、曲线下的面积或截距。在代入方程之前，务必检查单位是否需要换算（例如将平方厘米换算为平方米，分钟换算为秒）。
3. Example 1: Solar Panels and Energy Efficiency | 实例一：太阳能电池板与能效
A homeowner installs solar panels covering an area of 20 m². The solar radiation intensity at the location is 500 W/m². The panels have an efficiency of 15%. Calculate the electrical power output of the array. If the panels deliver a total of 4.5 kWh of electrical energy on a particular day, calculate the total solar energy incident on the panels that day.
一位房主安装了覆盖20平方米的太阳能电池板。当地的太阳辐射强度为 500 W/m²。电池板的效率是15%。计算该阵列的电功率输出。如果在某一天电池板总共输出了4.5千瓦时的电能，计算当天入射到电池板上的总太阳能。
Solution: 5 × 10 000 W = 1500 W. For the energy calculation, rearrange the efficiency formula: η = Euseful / Etotal, so Etotal = Euseful / η. Using the electrical energy output as the useful energy: Etotal o 108 MJ. When discussing environmental benefits, you could mention that this reduces reliance on fossil fuels and lowers CO₂ emissions.
解答：总入射功率 = 辐照度 × 面积 = 500 W/m² × 20 m² = 10 000 W。电功率输出 = 效率 × 入射功率 = 0.15 × 10 000 W = 1500 W。关于能量的计算，将效率公式变形：η = E_有用 / E_总，因此 E_总 = E_有用 / η。将电能输出视为有用能量：E_总 = 4.5 kWh / 0.15 = 30 kWh。注意：1 kWh = 3.6 × 10⁶ J，因此入射能量也是108 MJ。在讨论环境效益时，你可以提及这减少了对化石燃料的依赖并降低了二氧化碳排放。
Pout = η × I × A
4. Example 2: Car Safety and Momentum | 实例二：汽车安全与动量
A car of mass 1200 kg is travelling at 18 m/s when it crashes into a wall and comes to rest. In the first scenario, the front of the car crumples by 0.15 m, bringing the car to a stop in 0.05 s. In the second scenario, improved crumple zones increase the stopping time to 0.25 s. Calculate the average force exerted on the car in each case and explain how the crumple zone protects the occupants.
一辆质量为1200千克的汽车以18米/秒的速度行驶，撞上墙壁后停下。在第一种情景中，汽车前部溃缩0.15米，使汽车在0.05秒内停下。在第二种情景中，改进的溃缩区将停车时间延长至0.25秒。计算每种情况下施加在汽车上的平均力，并解释溃缩区如何保护乘员。
Solution: m/s. For the first crash: F = 21 600 / 0.05 = 432 000 N. For the second crash: F = 21 600 / 0.25 = 86 400 N. The force is five times smaller when the stopping time is increased. Crumple zones absorb energy by deforming, which increases the time over which the momentum changes. According to the equation F = Δp / Δt, a longer time of impact leads to a smaller average force, reducing the risk of injury to passengers. Seatbelts and airbags further increase stopping time and spread the force over a larger area.
解答：使用冲量-动量关系：F = Δp / Δt，其中 Δp = m × Δv。初始动量 = 1200 kg × 18 m/s = 21 600 kg m/s。末动量为0，因此 Δp = 21 600 kg m/s。对于第一次碰撞：F = 21 600 / 0.05 = 432 000 N。对于第二次碰撞：F = 21 600 / 0.25 = 86 400 N。当停车时间增大时，力减小到原来的五分之一。溃缩区通过变形吸收能量，从而增加了动量变化的时间。根据方程 F = Δp / Δt，更长的碰撞时间导致更小的平均力，降低了乘员受伤的风险。安全带和安全气囊进一步延长了停止时间并将力分散到更大的面积上。
F = Δp / Δt = (mv - mu) / t
5. Example 3: National Grid and Transformer Calculations | 实例三：国家电网与变压器计算
A power station generates electricity at 25 kV. A step-up transformer increases the voltage to 400 kV for transmission through the National Grid. The output power from the power station is 80 MW. Calculate the turn ratio of the transformer (Ns / Np) and the current in the primary and secondary coils. Explain why a high transmission voltage reduces power losses in the cables.
一座发电站以25千伏的电压发电。一台升压变压器将电压升高至400千伏，以便通过国家电网输送。发电站的输出功率为80兆瓦。计算变压器的匝数比 (Ns / Np) 以及一次线圈和二次线圈中的电流。解释为什么高输电电压能减少电缆中的功率损耗。
Solution: For an ideal transformer, Vs / Vp = Ns / Np. Turn ratio = 400 kV / 25 kV = 16. So Ns : Np = 16 : 1. Primary current: Ip = P / Vp = 80 × 10⁶ W / 25 × 10³ V = 3200 A. Secondary current: Is = P / Vs 00 × 10³ V = 200 A. The cables have resistance R, and the power lost as heat is given by Ploss current is reduced for the same transmitted power. Since the loss depends on I², a 16-fold reduction in current results in a 256-fold reduction in power loss. This efficiency makes the National Grid feasible. In reality, transformers are not 100% efficient, but modern units can exceed 99% efficiency.
解答：对于理想变压器，Vs / Vp = Ns / Np。匝数比 = 400 kV / 25 kV = 16。因此 Ns : Np = 16 : 1。一次电流：Ip = P / Vp = 80 × 10⁶ W / 25 × 10³ V = 3200 A。二次电流：Is = P / Vs = 80 × 10⁶ W / 400 × 10³ V = 200 A。电缆具有电阻R，以热量形式损耗的功率由 Ploss = I² R 给出。通过升高电压，在输送相同功率的情况下电流减小了。由于损耗取决于 I²，电流减小16倍会使功率损耗减少256倍。这种效率使国家电网切实可行。实际上，变压器并非100%效率，但现代设备可以超过99%。
Ploss = I² R
6. Example 4: Radioactive Tracers in Medicine | 实例四：医学中的放射性示踪剂
Iodine-131 is a radioactive isotope used in treating thyroid conditions. It has a half-life of 8 days. A patient is given an initial activity of 640 MBq. Calculate the activity remaining after 32 days. Explain why a source with a half-life of a few days is chosen for medical tracers, and discuss the precautions medical staff must take when handling radioactive materials.
碘-131是一种用于治疗甲状腺疾病的放射性同位素。它的半衰期为8天。一位病人被给予初始活度为640 MBq的剂量。计算32天后剩余的活度。解释为什么医学示踪剂会选择半衰期为几天的放射源，并讨论医护人员在处理放射性物质时必须采取的预防措施。
Solution: l activity × (½)⁴ = 640 MBq × (1/16) = 40 MBq. A half-life of a few days is suitable because the source remains active long enough to allow diagnosis or treatment but decays quickly enough to limit long-term radiation exposure to the patient. Handling precautions include using lead shielding, keeping distance from the source with tongs, limiting exposure time, and wearing protective clothing. Radioactive waste must be disposed of safely according to regulations.
解答：确定半衰期个数：32天 / 8天 = 4个半衰期。每经过一个半衰期，活度减半。4个半衰期后的活度 = 初始活度 × (½)⁴ = 640 MBq × (1/16) = 40 MBq。几天的半衰期是合适的，因为放射源保持足够长时间的活度以进行诊断或治疗，但又迅速衰变，以限制患者受到的长期辐射。处理预防措施包括使用铅屏蔽、用钳子保持距离、限制接触时间以及穿着防护服。放射性废物必须根据规定安全处置。
A = A₀ (½)t / T₁/₂
7. Example 5: Terminal Velocity of a Skydiver | 实例五：跳伞者的终端速度
A skydiver of mass 75 kg jumps from a plane. After a few seconds, she reaches a terminal velocity of 55 m/s. At this point, the forces are balanced. Calculate the magnitude of the air resistance acting on her. Draw a labelled free-body diagram to show the forces, and sketch a velocity-time graph from the moment she leaves the plane until she opens her parachute. Explain the shape of the graph in terms of resultant force and acceleration.
一位质量为75千克的跳伞者从飞机上跳下。几秒钟后，她达到55米/秒的终端速度。此时，力达到平衡。计算作用在她身上的空气阻力大小。画出标有力名称的自由体图来表示这些力，并画出从她离开飞机到打开降落伞期间的速度-时间图。结合合力与加速度解释该图的形状。
Solution: s. The velocity-time graph: initially, the velocity increases rapidly (steep gradient) because weight is much greater than air resistance, producing a large downward acceleration. As speed increases, air resistance grows, reducing the resultant force and the acceleration. The gradient of the velocity-time graph decreases until it becomes horizontal at terminal velocity, where acceleration is zero. When the parachute opens, the area increases dramatically, air resistance briefly becomes much larger than weight, causing a sharp deceleration (negative gradient) until a new, lower terminal velocity is reached.
解答：在终端速度时，净力为零，因此重力 = 空气阻力。重量 W = m × g = 75 kg × 9.8 N/kg = 735 N。因此，空气阻力为向上的735 N。速度-时间图：最初，速度快速增加（陡峭的斜率），因为重力远大于空气阻力，产生一个很大的向下加速度。随着速度增加，空气阻力增长，减小了合力和加速度。速度-时间图的斜率逐渐减小，直到在终端速度时变为水平，此时加速度为零。当降落伞打开时，面积急剧增加，空气阻力瞬间远大于重力，导致急剧减速（负斜率），直到达到一个新的、更低的终端速度。
8. Example 6: Resistance in a Wire and Temperature | 实例六：导线电阻与温度
A metal wire has a resistance of 3.5 Ω at 20 °C. The temperature coefficient of resistance for the metal is 0.0045 per °C. Calculate its resistance at 120 °C. Explain, in terms of electrons and ions, why the resistance increases with temperature.
一根金属导线在20 °C时的电阻为3.5 Ω。该金属的电阻温度系数为0.0045每°C。计算它在120 °C时的电阻。从电子和离子的角度解释为什么电阻随温度升高而增大。
Solution: 0] = 3.5 × [1 + 0.45] = 3.5 × 1.45 = 5.075 Ω (≈ 5.1 Ω to 2 significant figures). As temperature increases, the metal lattice ions vibrate more vigorously. These increased vibrations impede the flow of free electrons, effectively increasing the collision rate between electrons and the lattice. Therefore, the resistance rises. In a filament lamp, this effect causes the resistance to be much higher at operating temperature than when cold, which can be demonstrated by the non-linear I-V characteristic.
解答：使用公式 R = R₀ [1 + α (θ - θ₀)]，其中 R₀ 为参考温度 θ₀ 下的电阻。这里 θ₀ = 20 °C，Δθ = 120 - 20 = 100 °C。因此 R = 3.5 × [1 + 0.0045 × 100] = 3.5 × [1 + 0.45] = 3.5 × 1.45 = 5.075 Ω（保留两位有效数字约为5.1 Ω）。随着温度升高，金属晶格中的离子振动得更剧烈。这些增强的振动阻碍了自由电子的流动，有效地增加了电子与晶格之间的碰撞率。因此，电阻升高。在灯丝灯泡中，这种效应导致工作温度下的电阻比冷态时高得多，这可以通过非线性的I-V特性来证明。
R = R₀ (1 + α Δθ)
9. Common Pitfalls and Exam Tips | 常见错误与应试技巧
Many students lose marks by not using data from the source material. Always make explicit references: "According to the graph...", "The passage states that...". Another common error is unit mismatches: converting cm² to m² requires dividing by 10 000, not 100. Also, check if the question expects an explanation in terms of particle behaviour or energy transfers; simply stating an equation is sometimes insufficient. When evaluating, consider both advantages and disadvantages, and support your arguments with figures or scientific reasoning.
许多学生因为没有使用源材料中的数据而丢分。一定要明确引用：“根据图表……”、“文章中指出……”。另一个常见错误
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