📚 Year 9 AQA Physics: Case Study Practical Drills | Year 9 AQA 物理：案例分析实战演练
Real-world physics is rarely a textbook diagram. In AQA Year 9, you are expected to apply concepts from energy, electricity, forces, waves and matter to everyday situations. Case study drills train you to spot the relevant physics hiding in a story, perform calculations with unfamiliar data and evaluate designs critically. This article walks you through ten carefully chosen scenarios, each paired with step-by-step analysis, typical of the problem-solving tasks you will meet in class and on assessments.
真实的物理很少像课本示意图那么简单。在 AQA Year 9 课程中，你需要把能量、电学、力学、波动物理和物质状态的概念应用到日常生活情境中。案例分析实战能训练你从一段故事中找出相关的物理原理、用不熟悉的数据进行计算并批判性地评估设计方案。本文为你准备了十个精心挑选的案例，每个都配有分步解析，正是你在课堂和测评中会遇到的典型问题。
1. Home Energy Bill Analysis | 家庭能源账单分析
The Jones family receives an electricity bill showing a previous meter reading of 34,560 kWh and a present reading of 35,100 kWh. The unit rate is £0.28 per kWh. Mrs Jones wants to know how much energy was used and how much it will cost. This case connects the energy unit kilowatt-hour with the power equation and everyday costs.
琼斯家收到的电费单显示上次电表读数为 34,560 kWh，当前读数为 35,100 kWh，单价为每千瓦时 0.28 英镑。琼斯太太想知道用了多少能量以及要花多少钱。这个案例将千瓦时这一能量单位与功率方程和日常开销联系起来。
The energy used is the difference in readings: 35,100 kWh – in kilowatts and time in hours. The cost is calculated by multiplying the energy used by the unit price: Cost = 540 kWh × £0.28/kWh = £151.20. This drill helps you practise unit conversion and recognise that 'kWh' on a bill is a measure of energy, not power.
已用电量是两次读数之差：35,100 kWh – 34,560 kWh = 540 kWh。回顾一下，1 kWh 是一台 1 kW 的电器运行 1 小时所传递的能量，因此它可以用 E = P × t 表示，其中功率以千瓦为单位，时间以小时为单位。成本是用电量乘以单价：成本 = 540 kWh × £0.28/kWh = 151.20 英镑。这个练习帮助你练习单位换算，并认识到账单上的 'kWh' 是能量而不是功率的度量。
Energy used = Final reading – Initial reading
已用电量 = 最终读数 – 初始读数
2. Insulated Cup Challenge | 保温杯挑战赛
A Year 9 class is asked to design a cup that keeps hot water warm for the longest time. Four prototypes are tested: a single-use paper cup, a polystyrene foam cup, a vacuum-insulated steel flask, and a steel can wrapped in aluminium foil. The temperature drop after 20 minutes is recorded. Which design works best and why? This case deepens understanding of thermal energy transfer by conduction, convection and radiation.
一个九年级班级要设计一个能让热水保温时间最长的杯子。他们测试了四种原型：一次性纸杯、聚苯乙烯泡沫杯、不锈钢真空保温杯和包裹铝箔的铁罐。记录 20 分钟后的温度降幅。哪种设计最好，为什么？这个案例加深对通过传导、对流和辐射传递热能的理解。
nd foil-wrapped can dropped to 56°C (ΔT = 24°C). The vacuum flask wins because the vacuum layer stops conduction and convection almost completely, and the reflective inner surface reduces radiation. The foam cup is a good insulator because trapped air is a poor conductor. The paper cup allows high conduction outward and evaporation, while the foil reflects some radiation but the metal can conducts heat quickly. Evaluating performance with data mirrors real engineering trade-offs between cost, mass and effectiveness.
实验结果表显示：纸杯从 80°C 降到 52°C (温差 28°C)，泡沫杯降到 64°C (温差 16°C)，真空保温杯降到 73°C (温差 7°C)，包裹铝箔的罐子降到 56°C (温差 24°C)。真空保温杯获胜，因为真空层几乎完全阻止了传导和对流，而反光内壁减少了辐射。泡沫杯是良好的隔热体，因为静止空气是热的不良导体。纸杯对外传导快速且蒸发散热，铝箔能反射部分辐射但金属罐本身导热快。用数据评估性能反映了实际工程中成本、重量和效果之间的权衡。
| Design | ΔT (°C) |
| Paper cup | 28 |
| Foam cup | 16 |
| Vacuum flask | 7 |
| Foil-wrapped can | 24 |
Design comparison table | 设计对比表
3. Christmas Light Fault Finding | 圣诞灯串故障排查
A string of 20 miniature filament bulbs in series stops glowing. One bulb has failed, breaking the circuit. You only have a single extra working bulb and a multimeter that can measure voltage and resistance. How would you identify the dead bulb efficiently? This exercise tests your understanding of series circuits and the effect of an open circuit on current and voltage distribution.
一串由 20 个小灯丝灯泡串联的灯饰不亮了，其中一颗灯泡损坏导致断路。你只有一个备用的好灯泡和一个可以测量电压和电阻的万用表。如何高效地找出坏掉的灯泡？这个练习考验你对串联电路以及断路对电流和电压分布影响的理解。
In a series circuit, the same current would normally flow through every bulb, but an open circuit stops current everywhere. If you use the multimeter set to resistance mode, you can check each bulb individually: a blown bulb shows infinite resistance (OL), while a good filament shows a few ohms. A quicker method is to replace each bulb one by one with the spare until the string lights up – systematic substitution. You could also set the multimeter to voltage mode and check the potential difference across each bulb or across the supply, but without current flowing, the dead bulb will have the full supply voltage across its terminals if tested correctly. This fault-finding drill is identical to troubleshooting old-style fairy lights and introduces the logic behind continuity testing.
在串联电路中，电流通常流经每一颗灯泡，但断路会使电流处处为零。若用万用表电阻档，可以单独测量每一颗灯泡：坏掉的灯泡显示无穷大电阻，好灯泡的灯丝则显示几欧姆。一个更快的办法是用备用灯泡逐一替换，直到灯串亮起——这就是系统替换法。你也可以用电压档测量每颗灯泡两端或电源两端的电位差：在断路状态且正确测量时，损坏的灯泡两端将显示电源全电压。这种查找故障的练习和检修老式节日彩灯的方法相同，也能初步介绍通断测试的逻辑。
In series: I is constant; open circuit → I = 0
串联中：电流处处相等；断路 → 电流为零
4. Ski Jump Energy Transformation | 滑雪跳台的能量转换
A 55 kg skier starts from rest at the top of a jump ramp 8.0 m above the landing zone. The ramp is smooth enough that friction and air resistance can be ignored. What speed does the skier reach just before touching the snow, and why does the height matter? This case models the conservation of energy between gravitational potential energy and kinetic energy.
一名 55 kg 的滑雪者从高出着落区 8.0 m 的跳台顶端由静止出发。滑道足够光滑，摩擦力与空气阻力可以忽略。滑雪者即将接触雪面时的速度是多少？高度为什么重要？这个案例用重力势能和动能之间的能量守恒进行建模。
o losses), so mgh = ½mv². Mass cancels out, showing speed depends only on height and g: v = √2gh. Taking g = 9.8 m/s², h = 8.0 m, v = √2 × 9.8 × 8.0 ghts that the mass of the skier does not affect the final speed in ideal conditions, a key insight linking the equations of motion and energy stores.
储存的重力势能转化为动能。用公式 GPE = mgh 和 KE = ½mv²。在底部，所有 GPE 转换为 KE（假设无损耗），所以 mgh = ½mv²。质量被约掉，表明速度只取决于高度和重力加速度 g：v = √2gh。取 g = 9.8 m/s², h = 8.0 m，则 v = √2 × 9.8 × 8.0 = √156.8 ≈ 12.5 m/s。这个练习强调在理想情况下滑雪者质量不影响最终速度，这是联系运动方程和能量储存的重要见解。
v = √2gh = √2 × 9.8 m/s² × 8.0 m ≈ 12.5 m/s
v = √2gh ≈ 12.5 m/s
5. Echo Sounding the Seabed | 回声探测海深
A research ship sends a pulse of sound vertically downward into the ocean. The echo returns 2.4 seconds later. The speed of sound in seawater is taken as 1500 m/s. Calculate the depth of the seafloor. The challenge here is to remember the pulse travels to the bottom and back, so the distance must be halved.
一艘科考船向海中垂直发射一个声脉冲，2.4 秒后收到回声。声音在海水中的速度取为 1500 m/s。计算海底深度。这里的难点是要记住脉冲走了一个来回，因此算出的距离要除以二。
cent. Therefore, the depth (one-way) = 3600 m ÷ 2 = 1800 m. This simple case reinforces the relationship v = s / t rearranged to s = v × t, and demands careful interpretation of the word ‘ a common error. Practising with sonar problems also links waves to real marine technology.
声脉冲往返的总距离为 距离 = 速度 × 时间 = 1500 m/s × 2.4 s = 3600 m。这个距离包括去程和回程。因此，深度（单程）= 3600 m ÷ 2 = 1800 m。这个简单案例巩固了 v = s / t 及其变形 s = v × t 的关系，并要求仔细解读“回声”一词。在考试中，忘记除以二是常见错误。用声纳问题练习还能将波动与真实的海洋技术联系起来。
Depth = (v × t) ÷ 2 = (1500 m/s × 2.4 s) ÷ 2 = 1800 m
深度 = (v × t) ÷ 2 = 1800 m
6. Electromagnet Crane Lift-off | 电磁铁起重机起吊
A scrapyard crane uses an electromagnet to lift a 300 kg steel car frame. The electromagnet's strength depends on the current flowing through its coil and the number of turns. The current is initially set to 4.0 A with 500 turns, and the lifting force is just enough to hold the frame. An operator wants to lift a heavier 450 kg frame. Propose two ways to increase the magnetic force, and calculate the new current if only the current can be changed while keeping the same 500 turns. Assume force is proportional to current for the same number of turns.
废料场的起重机用电磁铁起吊一个 300 kg 的钢制车架。电磁铁的强度取决于通过线圈的电流和线圈匝数。初始电流设为 4.0 A、匝数为 500，刚好能够吸住车架。操作员想吊起更重的 450 kg 车架。请提出两种增强磁力的方法，并计算如果在保持 500 匝的情况下只能改变电流，需要多大的新电流（假设相同匝数下磁力与电流成正比）。
nserting a soft iron core is already present, but using a material with higher permeability could also help. Since weight ≈ magnetic force under static hold and force ∝ roportion and the design of electromagnets, reminding you that current is a controllable variable in practical engineering.
增强电磁力的方法有：(1) 增大线圈中的电流；(2) 增加绕在铁芯上的线圈匝数。使用软铁芯已经是前提，但换用导磁率更高的材料也能改善。因为静态吸持时重量 ≈ 磁力，且磁力与电流成正比（匝数不变），所需力之比为 450 kg / 300 kg = 1.5。因此，新电流 = 1.5 × 4.0 A = 6.0 A。这个练习结合了正比例关系和电磁铁的设计，提醒你电流在实际工程中是一个可控变量。
Force ∝ Current (turns constant) → Inew = (450÷300) × 4.0 A = 6.0 A
磁力 ∝ 电流（匝数不变）→ 新电流 = 6.0 A
7. Is It Real Gold? Density Test | 是真金吗？密度测试
A souvenir shop sells a small statue claimed to be pure gold. The statue has a mass of 96.5 g. When completely submerged in a measuring cylinder of water, the water level rises from 50.0 cm³ to 55.0 cm³. Using the density equation, determine if the statue could be pure gold. The density of pure gold is 19.3 g/cm³.
纪念品商店出售一座声称是纯金的小雕像。雕像质量为 96.5 g。把它完全浸入量筒的水中，水面从 50.0 cm³ 上升到 55.0 cm³。运用密度方程判断这座像是否为纯金。纯金的密度是 19.3 g/cm³。
The volume displaced is the difference: 55.0 cm³ – e density of pure gold. However, in reality a skilled forger could make a hollow object or use a tungsten core plated with gold giving the same average density. So while the numbers suggest purity, you must remain critical of a single test. This drill practises the density measurement technique and the use of the displacement method for irregular-shaped objects.
排开的水体积是差值：55.0 cm³ – 50.0 cm³ = 5.0 cm³。雕像的密度 = 质量 ÷ 体积 = 96.5 g ÷ 5.0 cm³ = 19.3 g/cm³。这与纯金密度完全吻合。但在现实中，技艺高超的伪造者可以制造空心物体或用钨芯镀金得到相同的平均密度。所以尽管数据符合纯金，你必须对单一的测试结果保持批判性。这个练习训练不规则物体密度的排水法测量。
Density = m ÷ V = 96.5 g ÷ 5.0 cm³ = 19.3 g/cm³
密度 = m ÷ V = 96.5 g ÷ 5.0 cm³ = 19.3 g/cm³
8. Hand-Lens Image Examination | 手持放大镜成像观察
A stamp collector uses a convex lens (magnifying glass) with a focal length of 10 cm to examine a rare stamp. The stamp is placed 6.0 cm from the lens. Describe the image formed and calculate the magnification. This real-life application delves into ray diagrams and the thin lens equation for a simple magnifier.
一位集邮者用一个焦距为 10 cm 的凸透镜（放大镜）观察一枚珍稀邮票。邮票距透镜 6.0 cm。描述所成的像并计算放大率。这个真实应用深入探讨了简单放大镜的光线图和薄透镜公式。
arged image on the same side of the lens as the object. Using the lens formula 1/f = 1/u + 1/v: 1/10 = 1/6 + 1/v, giving 1/v = 1/10 – 1/6 = (3 – 5) / 30 = –2/30, so v = –15 cm. The negative sign confirms the image is virtual. Magnification M = –v / u = –(– his case strengthens sign conventions and the distinction between real and virtual images.
由于物距 (u = 6.0 cm) 小于焦距 (f = 10 cm)，透镜相当于一个简单放大镜，产生一个正立、放大的虚像，像与物在透镜同侧。运用透镜公式 1/f = 1/u + 1/v：1/10 = 1/6 + 1/v，得 1/v = 1/10 – 1/6 = (3 – 5)/30 = –2/30，所以 v = –15 cm。负号确认像为虚像。放大率 M = –v / u = –(–15) / 6 = 2.5 倍。集邮者看到的是实际邮票高度 2.5 倍的像。这个案例强化了正负号规则以及实像与虚像的区别。
1/f = 1/u + 1/v → v = –15 cm; M = 2.5
1/f = 1/u + 1/v → v = –15 cm; 放大倍率 2.5
9. Solar-Powered Light System | 太阳能灯具系统
A garden light contains a small photovoltaic panel with an efficiency of 15%. The panel receives an average solar power of 60 W during a 5-hour charge period. It stores the electrical energy in a battery and later runs a 0.50 W LED for the rest of the day. Estimate how long the LED can operate from one full charge. This case blends energy, power and efficiency in
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