📚 Case Study Practice: Physics in Context | 案例分析实战演练：情境中的物理
Case study questions in Edexcel Year 10 Physics require you to apply your knowledge to unfamiliar, real‑life situations. They are not about memorising facts; they test your ability to recognise the relevant physics principles, extract data, perform calculations, and explain phenomena clearly. This article guides you through a variety of practice scenarios, highlighting effective strategies for each topic. Working through these examples systematically will build your confidence and sharpen your problem‑solving skills for both assessments and the wider world.
Edexcel Year 10 物理的案例分析题要求你将所学知识应用于陌生的真实生活情境。这些题目不在于死记硬背，而是考查你识别相关物理原理、提取数据、进行计算并清晰解释现象的能力。本文将通过多种实战场景，逐个讲解有效的解题策略。系统练习这些范例，能帮助你建立信心，提升应对考试和现实世界的物理解题能力。
1. Introduction to Case Studies | 案例分析简介
A case study in physics typically presents a short description of a situation – such as a car braking on a wet road, a person in a lift, or a roller coaster ride. Your task is to identify the physical quantities involved, select the correct equations, perform calculations where needed, and write concise explanations using scientific vocabulary. Marks are awarded both for correct numerical answers and for the quality of your reasoning. Always begin by underlining or noting the key information: numbers, units, direction of motion, energy transfers, and any assumptions made.
物理中的案例分析通常会给出一个简短的情境描述，例如汽车在潮湿路面上刹车、一个人在电梯里，或者过山车的运行。你的任务是识别所涉及的物理量，选择正确的方程，在需要时进行计算，并使用科学术语写出简洁的解释。正确数值答案和推理质量都会得分。开始时务必划出或记下关键信息：数字、单位、运动方向、能量转换以及所作的任何假设。
2. Identifying the Relevant Physics Concepts | 识别相关的物理概念
Before reaching for a formula, decide which area of physics the case study belongs to. Is it about forces and motion, energy conservation, electricity, waves, or thermal physics? Look for trigger words: 'accelerates', 'constant speed', 'brakes' suggest kinematics; 'height gained', 'speed lost', 'friction' suggest energy changes; 'current', 'fuse', 'power' point to electricity; 'refraction', 'echo', 'vibration' indicate waves. Once you have identified the domain, the appropriate equations and principles become much easier to recall.
在想到公式之前，先判断这个案例属于物理学的哪个领域。是关于力与运动、能量守恒、电学、波动还是热物理？寻找关键词：“加速”、“匀速”、“刹车”暗示运动学；“高度增加”、“速度减小”、“摩擦”暗示能量变化；“电流”、“保险丝”、“功率”指向电学；“折射”、“回声”、“振动”表明是波。一旦确定了领域，相应的方程和原理就更容易回想起来。
3. Case Study: Car Braking on a Wet Road | 案例：湿滑路面上的汽车刹车
A driver travelling at 25 m/s sees a hazard and applies the brakes. On a dry road, the braking distance is 30 m, but on a wet road the maximum deceleration is reduced to 5.5 m/s². The driver’s reaction time is 0.7 s. Determine the total stopping distance on the wet road.
一辆汽车以 25 m/s 的速度行驶，司机发现危险后踩下刹车。在干燥路面上，刹车距离为 30 m，但在湿滑路面上最大减速度降至 5.5 m/s²。司机的反应时间为 0.7 s。计算湿滑路面上的总停车距离。
t, find the braking distance using v² = u² + 2as. Final velocity v = 0, initial u = 25 m/s, a = –5.5 m/s² (negative because deceleration). Rearranging: 0 = 25² + 2 × (– ble the dry braking distance, illustrating why speed limits are reduced in wet conditions.
首先计算反应距离（反应时间内行驶的距离）：反应距离 = 速度 × 反应时间 = 25 × 0.7 = 17.5 m。接着利用 v² = u² + 2as 求出刹车距离。末速度 v = 0，初速度 u = 25 m/s，加速度 a = –5.5 m/s²（负号表示减速）。代入：0 = 25² + 2 × (–5.5) × s → s = 25² / (2 × 5.5) = 625 / 11 ≈ 56.8 m。总停车距离 = 反应距离 + 刹车距离 = 17.5 + 56.8 = 74.3 m。这比干燥路面刹车距离两倍还多，说明湿滑条件下为何要降低限速。
4. Case Study: Apparent Weight in a Lift | 案例：电梯中的视重
he lift accelerates upwards at 2.0 m/s², and when it travels upwards at a constant speed of 3.0 m/s. (Take g = 9.8 N/kg)
一个质量为 70 kg 的人站在电梯内的体重秤上。体重秤显示的读数是支持力（牛顿）。计算当电梯以 2.0 m/s² 向上加速时，以及当电梯以 3.0 m/s 匀速上升时，体重秤的读数。（g 取 9.8 N/kg）
When accelerating upwards, the net force equation is N – ht. When moving upwards at constant speed, acceleration is zero: N = mg = 70 × 9.8 = 686 N, equal to the person’s weight. This case study illustrates the principle behind 'weightlessness' and 'g‑forces' experienced in accelerating systems.
当电梯向上加速时，合力方程为 N – mg = ma。因此 N = m(g + a) = 70 × (9.8 + 2.0) = 70 × 11.8 = 826 N。人感到更重，是因为秤向上的推力大于自身重力。当电梯匀速上升时，加速度为零：N = mg = 70 × 9.8 = 686 N，等于人的体重。这个案例说明了在加速系统中“失重”和“超重”背后所依据的原理。
5. Case Study: Roller Coaster Energy Transfers | 案例：过山车中的能量转换
, estimate the speed at the bottom and at a point 10 m above the ground. Use g = 9.8 N/kg.
一辆质量为 500 kg 的过山车从 40 m 高处由静止释放。忽略摩擦，估算它在底部和距地面 10 m 处的速度。g 取 9.8 N/kg。
gy (KE). So ½mv² = 196 000 → v² = 2 × 196 000 / 500 = 784 → v = 28 m/s. At 10 m height, the GPE is mgh = 500 × 9.8 × 10 = 49 000 J. The remaining energy has become KE: KE = 196 000 – v ≈ 24.2 m/s. This demonstrates the conservation of mechanical energy in an ideal system.
在最高点，所有能量为重力势能（GPE）：GPE = mgh = 500 × 9.8 × 40 = 196 000 J。在底部，所有重力势能转化为动能（KE）。所以 ½mv² = 196 000 → v² = 2 × 196 000 / 500 = 784 → v = 28 m/s。在 10 m 高处，重力势能为 mgh = 500 × 9.8 × 10 = 49 000 J。剩余能量转为动能：KE = 196 000 – 49 000 = 147 000 J。于是 v² = 2 × 147 000 / 500 = 588 → v ≈ 24.2 m/s。这演示了理想系统中机械能守恒。
6. Case Study: Choosing a Fuse for a Home Appliance | 案例：为家用电器选择保险丝
A mains‑powered electric heater has a power rating of 2.5 kW and operates at 230 V in the UK. The available fuses are rated at 3 A, 5 A, and 13 A. Determine the most appropriate fuse and explain why it protects the appliance.
一台电暖器的额定功率为 2.5 kW，在英国使用 230 V 市电。可用的保险丝额定电流有 3 A、5 A 和 13 A。确定最合适的保险丝，并解释它如何保护电器。
ons, the current is about 10.9 A. A 3 A fuse would blow immediately as it is far below the operating current. A 5 A fuse would also blow. A 13 A fuse can safely carry this current but will blow if a fault causes the current to rise significantly above 13 A, protecting the wiring and appliance from overheating. Therefore the 13 A fuse is the correct choice. A fuse slightly above the operating current is chosen to prevent nuisance blowing.
首先用 P = IV 计算正常工作电流，I = P / V = 2500 W / 230 V ≈ 10.9 A。正常条件下电流约为 10.9 A。3 A 保险丝会立即熔断，因为它远低于工作电流。5 A 保险丝也不行。13 A 保险丝能安全承载该电流，并在故障电流显著超出 13 A 时熔断，保护线路和电器免于过热。因此选择 13 A 保险丝是合适的。通常会选择略高于工作电流的保险丝，以避免无故熔断。
7. Case Study: Water Waves in a Ripple Tank | 案例：水波槽中的水波
In a ripple tank, straight waves have a frequency of 12 Hz and a wavelength of 0.03 m in deep water. When they enter a shallower region, the wavelength decreases to 0.02 m. Determine the speed of the waves in both regions and explain the change.
在水波槽中，深水区直波的频率为 12 Hz，波长为 0.03 m。当波进入浅水区时，波长减小至 0.02 m。计算两区域中的波速，并解释这一变化。
ns the same because the source (the dipper) does not change frequency. In shallow water: v' = fλ' behaviour at a boundary and explains why water waves bend (refract) when entering shallower water at an angle.
波速 v = fλ。深水区：v = 12 × 0.03 = 0.36 m/s。当波进入浅水区时，频率保持不变，因为振源没有改变频率。浅水区中：v' = fλ' = 12 × 0.02 = 0.24 m/s。波速减小，导致波长减小而频率恒定。这是波在界面处行为的很好实例，也解释了为什么水波以一定角度进入浅水区时会发生弯曲（折射）。
8. Case Study: Greenhouse Effect and Radiation | 案例：温室效应与辐射
Explain, using physics principles, why the Earth’s surface is warmer than expected without an atmosphere, and why carbon dioxide is considered a greenhouse gas. In your answer, refer to the types of electromagnetic radiation involved and the energy balance.
运用物理原理，解释为什么地球表面在没有大气层时会比预期更冷，以及为什么二氧化碳被视为温室气体。回答时请提及所涉及的电磁波类型和能量平衡。
The Sun emits mainly visible and ultraviolet radiation because it is very hot (around 5500 °C). This short‑wavelength radiation passes through the atmosphere and is absorbed by the Earth's surface, warming it. The Earth, at about 15 °C, emits infrared radiation of much longer wavelength. Greenhouse gases such as CO₂, methane, and water vapour absorb this outgoing infrared radiation and re‑emit it in all directions, including back towards the surface. This traps thermal energy, raising the average surface temperature above what it would be without an atmosphere. The process is natural and essential for life, but increased CO₂ from human activities enhances the effect, causing global warming.
太阳主要发出可见光和紫外线，因为它温度极高（约 5500 °C）。这些短波辐射穿过大气层被地表吸收，使地面升温。地球温度约为 15 °C，发射波长更长的红外辐射。二氧化碳、甲烷和水蒸气等温室气体吸收这些向外放射的红外线，并向各个方向重新辐射，包括返回地面。这就将热能“困住”，使地表平均温度高于没有大气层时的温度。这个自然过程对生命至关重要，但人类活动增加的二氧化碳增强了这一效应，导致全球变暖。
9. Case Study: Designing a Parachute | 案例：设计降落伞
A physics class is designing a parachute for a 50 g egg. They want the egg to hit the ground at a terminal velocity no faster than 3 m/s. With a suitable parachute, the drag force is given by Fdrag what happens to the velocity over time from the moment the parachute opens. (g = 9.8 N/kg)
物理课上要为一只 50 g 的鸡蛋设计降落伞。他们希望鸡蛋撞击地面的终极速度不超过 3 m/s。合适的降落伞产生的阻力由 Fdrag = 0.25 v² 给出（v 以 m/s 为单位时，力以 N 为单位）。判断此终极速度是否满足设计要求，并描绘自降落伞打开起速度随时间的变化。（g = 9.8 N/kg）
ight mg = 0.050 kg × 9.8 N/kg = 0.49 N. Set 0.25 v² = 0.49 → v² = 0.49 / 0.25 = 1.96 → v = √1.96 = 1.4 m/s. This is well below the 3 m/s limit, so the design is safe. On a velocity– velocity then asymptotically approaches the terminal velocity of 1.4 m/s without overshooting. The key shape is a rapid deceleration followed by a smooth levelling off.
达到终极速度时，重力等于阻力。重力 mg = 0.050 kg × 9.8 N/kg = 0.49 N。令 0.25 v² = 0.49 → v² = 0.49 / 0.25 = 1.96 → v = √1.96 = 1.4 m/s。这远低于 3 m/s 的限制，所以设计安全。速度–时间图上，一开始速度较高（降落伞刚打开时），由于阻力初始远大于重力，速度急剧下降。速度随后逐渐趋近于 1.4 m/s 的终极速度，不会超越。图形特征为快速减速后平滑趋于水平。
10. Case Study: Analysing Spring Data (Hooke’s Law) | 案例：分析弹簧数据（胡克定律）
A student hangs different masses on a spring and measures the extension. The results are shown in the table. Determine whether the spring obeys Hooke’s law and find the spring constant k.
| Mass (g) | Weight (N) | Extension (cm) |
|---|---|---|
| 100 | 0.98 | 4.0 |
| 200 | 1.96 | 8.1 |
| 300 | 2.94 | 12.0 |
| 400 | 3.92 | 15.8 |
| 500 | 4.90 | 20.0 |
一位学生将不同质量的砝码悬挂在弹簧上并测量伸长量。结果如上表所示。判断弹簧是否遵循胡克定律，并求出弹簧常数 k。
he force per unit extension for each pair. Convert extensions to metres: 0.040, 0.081, 0.120, 0.158, 0.200 m. k = F / x gives: 0.98 / 0.040 = 24.5 N/m; 1.96 / 0.081 ≈ 24.2 N/m; 2.94 / 0.120 = 24.5 N/m; 3.92 / 0.158 ≈ 24.8 N/m; 4.90 / 0.200 = 24.5 N/m. The values are nearly constant, confirming the spring follows Hooke’s law. Average k ≈ 24.5 N/m. Small variations are due to experimental uncertainties.
胡克定律指出，只要不超过弹性极限，伸长量与所施加的力成正比（F = kx）。要检验这一点，计算每组数据的力与伸长量之比。将伸长量转换为米：0.040、0.081、0.120、0.158、0.200 m。k = F / x 得：0.98 / 0.040 = 24.5 N/m；1.96 / 0.081 ≈ 24.2 N/m；2.94 / 0.120 = 24.5 N/m；3.92 / 0.158 ≈ 24.8 N/m；4.90 / 0.200 = 24.5 N/m。这些值几乎不变，证实弹簧遵循胡克定律。平均值 k ≈ 24.5 N/m。微小差异源于实验误差。
11. Step‑by‑Step Strategy for Any Case Study | 应对任何案例分析的逐步策略
Use this checklist when tackling unfamiliar scenarios:
- Read the information carefully and highlight given quantities with their units.
- Identify the physics topic(s) involved – force, energy, waves, electricity, etc.
- Note down any assumptions (e.g. 'ignore air resistance', 'friction negligible').
- Select the relevant equation(s) and check that the quantities match the units expected.
- Substitute values carefully, carry out calculations, and give answers to appropriate significant figures.
- Always include units in the final answer.
- For explanation questions, use precise scientific language and link cause and effect.
- Evaluate your answer: does it make sense in the context? For example, a car stopping distance of 5 m at motorway speed is obviously wrong – revisit your working.
在应对陌生情境时，请使用以下清单：
- 仔细阅读信息，圈出所给的物理量及其单位。
- 识别涉及的物理主题——力、能量、波动、电学等。
- 记下所有假设（例如“忽略空气阻力”、“摩擦不计”）。
- 选择相关的方程，并检查物理量与预期单位是否匹配。
- 仔细代入数值进行计算，并给出适当有效数字的答案。
- 最终答案务必包含单位。
- 对于解释性问题，使用精准的科学用语，将因果联系起来。
- 评估答案：在情境中合理吗？比如，高速公路上停车距离仅 5 m 显然是错的——回头检查计算过程。
12. Conclusion: Practice Builds Confidence | 结论：练习建立信心
Mastering case studies in physics comes from repeated exposure to a variety of contexts. The scenarios you have worked through here span motion, forces, energy, electricity, waves, and thermal physics – exactly the range you will meet in Edexcel Year 10 assessments. Remember that the physics principles are always the same; the skill lies in recognising them inside a fresh description. Keep a log of the case studies you attempt, noting any common pitfalls and the key equations you used. With consistent practice, you will approach any case study question with clarity and precision.
掌握物理案例分析需要反复接触各种不同的情境。本文所演练的场景涵盖了运动、力、能量、电学、波动和热物理——正是你在 Edexcel Year 10 评估中会遇到的范围。请记住，物理原理始终不变；真正的技巧在于从全新的描述中识别出它们。建立案例练习日志，记下常见错误和用到的关键方程。通过持续练习，你将能够清晰、准确地应对任何案例分析题目。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图