Year 9 SQA Physics: Case Study Practical Drills | SQA 物理案例分析实战演练

📚 Year 9 SQA Physics: Case Study Practical Drills | SQA 物理案例分析实战演练

In the SQA Physics National 4 course (typically studied in Year 9/S3), the ability to apply principles to real-world scenarios is crucial. Examination papers and unit assessments often present case studies – everyday situations where you must identify variables, select correct equations, and calculate results. This article provides a series of worked case studies covering motion, forces, energy, electricity, and waves. Each case is broken down into logical steps, helping you build confidence and accuracy. Follow the analysis drills, and you will develop the problem-solving mindset needed for success.

在 SQA 物理 National 4 课程（通常在九年级/S3 学习）中，把原理应用到真实世界情景中的能力至关重要。考试卷和单元评估常常给出案例分析——即你必须识别变量、选择正确方程并计算结果的日常情景。本文提供一系列已完成的案例分析，涵盖运动、力、能量、电学和波。每个案例都按逻辑步骤拆解，帮助你建立信心和准确度。跟随分析演练，你将培养成功所需的解题思维。

1. Understanding the Case Study Approach | 理解案例分析方法

A typical SQA Physics case study presents a paragraph of information describing an activity or device, together with a table of data. You are asked to calculate quantities such as speed, acceleration, energy transferred, power, or resistance. The key is to extract the known values, match them to the correct formula, substitute carefully, and check the units. Always start by highlighting the data given and the quantity to find.

典型的 SQA 物理案例研究会给出一段描述某项活动或设备的信息，并附有数据表格。要求你计算速度、加速度、能量转移、功率或电阻等物理量。关键是从中提取已知值，将它们与正确的公式匹配，小心代入，并检查单位。始终先从标记给出的数据和要计算的量开始。

Step 1: Read the text and underline the numbers with units. / 第一步：阅读文字，在带有单位的数字下划线。
Step 2: Write down what you know (d, t, m, F, etc.). / 第二步：写下你已知的量（d, t, m, F 等）。
Step 3: Convert any units to SI base units (metres, seconds, kilograms, etc.) if needed. / 第三步：如果需要，将单位转换为国际基本单位（米、秒、千克等）。
Step 4: Select the appropriate relationship from the SQA data sheet. / 第四步：从 SQA 数据表中选取合适的关系式。
Step 5: Substitute and calculate, showing all working. / 第五步：代入并计算，写出全部过程。

2. Case Study 1: The Sprinting Athlete | 案例分析一：短跑运动员

A student runs 100 m in 14.2 seconds during a school sports day. Calculate her average speed in metres per second and also express it in kilometres per hour. Then determine how far she would travel at this average speed in 25 seconds. This case study allows us to practise speed = distance/time and unit conversions.

一名学生在学校运动会上用 14.2 秒跑完 100 米。计算她的平均速度，以米/秒表示，并转换为千米/小时。然后计算以此平均速度 25 秒内能跑多远。这个案例让我们练习速度 = 距离/时间以及单位换算。

Known: d = 100 m, t = 14.2 s. / 已知：d = 100 m，t = 14.2 s。

Average speed v = d / t = 100 / 14.2 ≈ 7.04 m/s

To convert m/s to km/h: multiply by 3.6. / 转换为 km/h：乘以 3.6。

7.04 × 3.6 ≈ 25.3 km/h

Distance in 25 s: d = v × t = 7.04 × 25 = 176 m. / 25 秒内的距离：d = v × t = 7.04 × 25 = 176 m。

Always remember: if t is in seconds and d in metres, the unit of speed is m/s. For km/h, divide by 1000 then multiply by 3600, which is the same as multiplying by 3.6. / 始终记住：如果 t 以秒为单位、d 以米为单位，速度单位是 m/s。要得到 km/h，先除以 1000 再乘以 3600，相当于乘以 3.6。

3. Case Study 2: Forces on a Cyclist | 案例分析二：自行车手的受力

A cyclist of total mass 75 kg experiences a forward push from the road of 150 N when pedalling. Friction and air resistance total 60 N opposing motion. Calculate the resultant force and the acceleration of the cyclist. This illustrates Newton’s second law (F = m a) and vector addition of forces.

一名总质量为 75 kg 的自行车手在蹬踏时受到来自路面的 150 N 向前推力。摩擦力和空气阻力总共为 60 N，方向与运动相反。计算合力与骑手的加速度。这展示了牛顿第二定律（F = m a）和力的矢量叠加。

Resultant force F_net = forward force – backward force = 150 N – 60 N = 90 N forward. / 合力 F_net = 向前力 – 向后力 = 150 N – 60 N = 90 N 向前。

Acceleration a = F_net / m = 90 N / 75 kg = 1.2 m/s². / 加速度 a = F_net / m = 90 N / 75 kg = 1.2 m/s²。

Key point: acceleration is always in the direction of the resultant force. In SQA questions, draw a simple diagram showing arrows to help visualise forces. / 关键点：加速度始终与合力方向相同。在 SQA 题目中，画一个简单示意图，用箭头表示力，以帮助形象化理解。

4. Case Study 3: The Electric Kettle’s Efficiency | 案例分析三：电热水壶的效率

An electric kettle has a power rating of 2200 W. In a test, it heats 0.50 kg of water from 20 °C to 80 °C in 70 seconds. The specific heat capacity of water is 4200 J/kg°C. Calculate (a) the electrical energy input, (b) the useful heat energy gained by the water, and (c) the efficiency of the kettle. This case links power, energy, and thermal physics.

一个电热水壶的额定功率为 2200 W。在一次测试中，它在 70 秒内将 0.50 kg 水从 20 °C 加热到 80 °C。水的比热容为 4200 J/kg°C。计算 (a) 输入的电能，(b) 水获得的有用热能，以及 (c) 水壶的效率。此案例将功率、能量和热物理联系起来。

(a) Electrical energy E_input = P × t = 2200 W × 70 s = 154 000 J (or 154 kJ). / 输入电能：E_input = P × t = 2200 W × 70 s = 154 000 J（或 154 kJ）。

(b) Temperature change Δθ = 80 °C – 20 °C = 60 °C. / 温度变化 Δθ = 80 °C – 20 °C = 60 °C。

Heat energy gained E_output = c × m × Δθ = 4200 × 0.50 × 60 = 126 000 J. / 获得的热能 E_output = c × m × Δθ = 4200 × 0.50 × 60 = 126 000 J。

(c) Efficiency = (useful energy output / total energy input) × 100% = (126 000 / 154 000) × 100% ≈ 81.8%. / 效率 = (有用能量输出 / 总能量输入) × 100% = (126 000 / 154 000) × 100% ≈ 81.8%。

Always express efficiency as a percentage and round to a suitable number of significant figures. Note that some energy is always lost to heating the kettle body and surroundings. / 始终以百分数表示效率，并四舍五入到合适的有效数字位数。注意，总有一部分能量损失在加热壶体和周围环境上。

5. Case Study 4: Ohm’s Law and Series Circuits | 案例分析四：欧姆定律与串联电路

A student builds a series circuit with a 12 V battery, a 4 Ω fixed resistor, and a variable resistor. When the variable resistor is set to 8 Ω, calculate the total resistance, the current in the circuit, and the voltage across each resistor. This case reinforces Ohm’s law (V = I R) and series circuit rules.

一名学生用 12 V 电池、一个 4 Ω 固定电阻和一个可变电阻搭建串联电路。当可变电阻调至 8 Ω 时，计算总电阻、电路中的电流和每个电阻两端的电压。此案例巩固欧姆定律（V = I R）和串联电路规则。

Total resistance R_total = 4 Ω + 8 Ω = 12 Ω. / 总电阻 R_total = 4 Ω + 8 Ω = 12 Ω。

Current I = V_battery / R_total = 12 V / 12 Ω = 1.0 A. / 电流 I = V_battery / R_total = 12 V / 12 Ω = 1.0 A。

Voltage across 4 Ω resistor: V₁ = I × R₁ = 1.0 A × 4 Ω = 4 V. / 4 Ω 电阻两端电压：V₁ = I × R₁ = 1.0 A × 4 Ω = 4 V。

Voltage across 8 Ω resistor: V₂ = I × R₂ = 1.0 A × 8 Ω = 8 V. / 8 Ω 电阻两端电压：V₂ = I × R₂ = 1.0 A × 8 Ω = 8 V。

Check: 4 V + 8 V = 12 V, matching the battery. / 检查：4 V + 8 V = 12 V，与电池电压匹配。

Important: In a series circuit, current is the same everywhere, and voltages add up. Always verify your answers meet these rules. / 重要：在串联电路中，各处电流相同，电压相加。始终验证你的答案满足这些规则。

6. Case Study 5: Gravitational Potential Energy on a Slide | 案例分析五：滑梯上的重力势能

A child of mass 40 kg climbs to the top of a slide 2.5 m high. Calculate the gravitational potential energy gained (g = 10 N/kg). If 20% of this energy is lost due to friction as she slides down, what is her kinetic energy at the bottom? This case combines E_p = m g h and energy conservation.

一名 40 kg 的孩子爬上 2.5 m 高的滑梯顶端。计算获得的重力势能（g = 10 N/kg）。如果滑下时由于摩擦损失了 20% 的能量，她在底部时的动能是多少？此案例结合 E_p = m g h 和能量守恒。

GPE = m g h = 40 × 10 × 2.5 = 1000 J. / 重力势能 = 40 × 10 × 2.5 = 1000 J。

Energy lost to friction = 20% of 1000 J = 200 J. / 摩擦损失的能量 = 1000 J 的 20% = 200 J。

Kinetic energy at bottom E_k = 1000 J – 200 J = 800 J. / 底部动能 E_k = 1000 J – 200 J = 800 J。

You could also find her speed using E_k = ½ m v²: 800 = ½ × 40 × v², so v² = 40, v = √40 ≈ 6.32 m/s. / 你也可以利用 E_k = ½ m v² 求她的速度：800 = ½ × 40 × v²，所以 v² = 40，v = √40 ≈ 6.32 m/s。

In SQA problems, g is often given as 10 N/kg on Earth for simplicity. Use exactly the value provided in the question. / 在 SQA 题目中，为简便起见，地球上的 g 常取为 10 N/kg。要使用题目中明确给出的值。

7. Case Study 6: Wave Speed from Frequency and Wavelength | 案例分析六：由频率和波长求波速

A loudspeaker produces a sound wave of frequency 440 Hz. The wavelength of the sound in air is measured as 0.77 m. Calculate the speed of sound in air. Then explain what would happen to the wavelength if the frequency were increased to 880 Hz, assuming the speed remains constant. This practises the wave equation v = f λ.

一个扬声器发出 440 Hz 的声波。测得空气中声波波长为 0.77 m。计算空气中的声速。然后解释如果频率增加到 880 Hz，且声速保持不变，波长会发生什么变化。这练习波速公式 v = f λ。

v = f λ = 440 Hz × 0.77 m = 338.8 m/s, which approximates to 339 m/s. / v = f λ = 440 Hz × 0.77 m = 338.8 m/s，近似为 339 m/s。

If speed is constant, v fixed, then λ = v / f. Doubling f to 880 Hz will halve the wavelength to approximately 0.385 m. / 若速度恒定，v 不变，则 λ = v / f。f 加倍到 880 Hz 时，波长将减半，约为 0.385 m。

Remember: frequency and wavelength are inversely proportional when wave speed is constant. This is a common concept tested. / 记住：波速一定时，频率与波长成反比。这是常考概念。

8. Case Study 7: The Hot Water Tank – Power and Time | 案例分析七：热水箱——功率与时间

An immersion heater rated at 3.0 kW is used to heat a tank of water. The water requires 4.5 × 10⁶ J of energy to reach the desired temperature. Assuming no losses, calculate the time needed. Then discuss how real-world losses would affect the time. This ties E = P t to a practical context.

一个额定功率为 3.0 kW 的浸入式加热器用于加热水箱中的水。水需要 4.5 × 10⁶ J 的能量才能达到目标温度。假设没有损失，计算所需时间。然后讨论实际中的损失会如何影响时间。这使 E = P t 与实际情境相结合。

Convert 3.0 kW to watts: 3.0 kW = 3000 W. / 将 3.0 kW 转换为瓦特：3.0 kW = 3000 W。

Time t = Energy E / Power P = (4.5 × 10⁶ J) / 3000 W = 1500 seconds. / 时间 t = 能量 E / 功率 P = (4.5 × 10⁶ J) / 3000 W = 1500 秒。

In minutes: 1500 / 60 = 25 minutes. / 以分钟计：1500 / 60 = 25 分钟。

In reality, heat losses to the surroundings mean more energy is needed, so the heater would need to run longer to supply the same useful energy to the water. This explains why real heating times are longer. / 实际上，向环境散热意味着需要更多能量，因此加热器需要运行更长时间才能为水提供相同的有用能量。这就解释了为什么实际加热时间更长。

9. Common Pitfalls and Tips | 常见陷阱与建议

After working through these case studies, students often make similar mistakes. Watch out for mixing units: always convert km to m, minutes to seconds, grams to kilograms before substituting into formulas. Also, when calculating efficiency, use the formula correct way – useful/total × 100%, not the other way round. Ensure you show all working clearly, as marks are awarded for substitution and unit handling.

在完成这些案例研究之后，学生们常常犯类似的错误。注意单位混用：在代入公式之前，始终将 km 转换为 m，分钟转换为秒，克转换为千克。此外，计算效率时，正确使用公式——有用/总 × 100%，而不是反过来。务必清晰展示所有过程，因为代入和单位处理都会得分。

Pitfall / 陷阱	How to avoid / 如何避免
Forgetting to square the velocity in E_k = ½ m v²	Always write v² = v × v and calculate carefully.
Using non-SI units (e.g., km for distance in speed formula)	Convert to m and s before substitution. Use 1 km = 1000 m, 1 min = 60 s.
Confusing mass and weight	Weight = m × g; mass is measured in kg, weight in N.
Reading a scale incorrectly	Check the unit on the axis or the increment carefully.

10. Practice Drills: Sampling Real SQA-Style Questions | 练习演练：真实 SQA 风格题目抽样

Now put your skills to work. Try these short drill questions based on the same case structure. Answer them independently before checking the solutions below.

现在把你的技能付诸实践。尝试这些基于相同案例结构的简短演练题。在查阅下方答案之前先独立作答。

Drill A: A car travels 150 km in 2 hours and 30 minutes. Calculate the average speed in km/h and in m/s. / 汽车在 2 小时 30 分钟内行驶 150 km。计算平均速度，分别以 km/h 和 m/s 表示。
Drill B: A 12 Ω lamp is connected to a 6 V battery. What current flows? / 一个 12 Ω 的灯泡连接到 6 V 电池上。流过的电流是多少？
Drill C: A crane lifts a 200 kg mass through a height of 8 m. Determine the work done (g = 10 N/kg). / 起重机将 200 kg 重物提升 8 m。求做功多少（g = 10 N/kg）。
Drill D: A radio wave has frequency 100 MHz. Using v = 3 × 10⁸ m/s, find its wavelength. / 一个无线电波频率为 100 MHz。使用 v = 3 × 10⁸ m/s，求其波长。

Solutions: A: 2.5 h → v = 150/2.5 = 60 km/h; 60 ÷ 3.6 ≈ 16.7 m/s. B: I = V/R = 6/12 = 0.5 A. C: W = mgh = 200×10×8 = 16 000 J. D: f = 100 MHz = 1 × 10⁸ Hz; λ = v/f = 3×10⁸ / 1×10⁸ = 3 m.

答案：A: 2.5 h → v = 150/2.5 = 60 km/h；60 ÷ 3.6 ≈ 16.7 m/s。B: I = V/R = 6/12 = 0.5 A。C: W = mgh = 200×10×8 = 16 000 J。D: f = 100 MHz = 1 × 10⁸ Hz；λ = v/f = 3×10⁸ / 1×10⁸ = 3 m。

11. Extending to Unfamiliar Contexts | 扩展到陌生情境

SQA exams often include an ‘unfamiliar situation’ question where you must apply known principles to a new device or experiment. For instance, a question might describe a solar charger converting light into electrical energy to charge a battery. You could be given the area of the panel, the solar intensity (in W/m²), and time. By calculating total energy input and stored energy, you can find efficiency. Approach such questions by identifying the core physics: energy, power, efficiency, and the relevant formulas. Break down the description phrase by phrase, translate each into a physics quantity, and build your solution step by step.

SQA 考试中经常包括一个“陌生情境”题目，要求你将已知原理应用于新的设备或实验。例如，题目可能描述太阳能充电器把光能转化为电能为电池充电。你可能会得到电池板面积、太阳光强（W/m²）和时间。通过计算总输入能量和储存能量，可以求出效率。处理此类问题时，要识别核心物理量：能量、功率、效率和相关公式。逐句分解描述，将每一短语转化为一个物理量，然后逐步构建你的解答。

Example scenario: A solar panel of 0.5 m² receives 800 W/m² for 2 hours. The battery stores 0.8 kWh. Calculate the efficiency. / 示例情景：一个 0.5 m² 的太阳能电池板接收到 800 W/m² 的光照，持续 2 小时。电池储存了 0.8 kWh。计算效率。

Power received = 800 W/m² × 0.5 m² = 400 W. Time = 2×3600 = 7200 s. Energy input = 400×7200 = 2.88×10⁶ J. Stored energy: 0.8 kWh = 0.8×1000×3600 = 2.88×10⁶ J. Efficiency = (2.88×10⁶ / 2.88×10⁶)×100% = 100% (ideal case; real panels have lower efficiency). / 接收功率 = 800 W/m² × 0.5 m² = 400 W。时间 = 2×3600 = 7200 s。输入能量 = 400×7200 = 2.88×10⁶ J。储存能量：0.8 kWh = 0.8×1000×3600 = 2.88×10⁶ J。效率 = (2.88×10⁶ / 2.88×10⁶)×100% = 100%（理想情况；真实电池板效率更低）。

Always check unit conversions: 1 kWh = 3.6×10⁶ J. This is a valuable conversion to know. / 始终检查单位转换：1 kWh = 3.6×10⁶ J。这是一个值得掌握的转换。

12. Final Advice and Revision Resources | 最后建议与复习资源

Regular case study practice sharpens your ability to apply physics rather than just recall facts. Keep a formula sheet handy and annotate it with typical unit conversions. In the exam, read the case study twice, underline key numbers, and label them with standard symbols (e.g., d for distance, m for mass). Never leave a calculation blank; even a correct substitution without the final answer can earn marks. For the SQA National 4 Physics assignment, you will carry out a practical investigation and write a report, so these analytical skills directly transfer. Use past papers and mark schemes to familiarise yourself with the expected language and format.

经常性的案例分析练习磨练你应用物理的能力，而不仅仅是回忆事实。手边放一张公式表，在上面标注典型的单位转换。在考试中，阅读案例研究两遍，在关键数字下画线，并用标准符号（例如 d 代表距离，m 代表质量）标注。绝不要让计算题空白；即使仅正确代入而没有算出最终答案也可能得分。对于 SQA National 4 物理作业，你将进行一次实际调查并撰写报告，因此这些分析技能可直接迁移。使用历年真题和评分方案，让自己熟悉期望的语言和格式。

By working systematically through these drills, you are already building a robust problem-solving toolkit. Physics is not about memorising isolated facts; it is about understanding how the world works and proving it with numbers. Keep practising, and you will see the patterns clearly.

通过系统地进行这些演练，你已经在构建一套强大的解题工具箱。物理不是要记住孤立的事实；而是要理解世界如何运作，并用数字来证明。持续练习，你将会清晰地看出其中的规律。

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Year 9 SQA Physics: Case Study Practical Drills | SQA 物理案例分析实战演练