📚 Year 9 CCEA Physics: Case Study Practice | CCEA 九年级物理：案例分析实战演练
In Year 9 CCEA Physics, being able to apply your knowledge to real‑world scenarios is just as important as learning the theory. This article walks you through a series of case studies designed to strengthen your analytical skills. Each case breaks down a problem step by step, showing you how to interpret data, use equations, and explain what is happening in scientific terms. The examples cover forces and motion, energy, electricity, waves, and matter – all core topics for your course.
在 CCEA 九年级物理课程中，将所学知识应用于真实情境与理论学习同等重要。本文通过一系列案例分析，帮助你提升分析能力。每个案例都会逐步拆解问题，演示如何解读数据、运用公式并用科学语言解释现象。这些例子涵盖力与运动、能量、电学、波以及物质等课程核心主题。
1. Case Study: Stopping Distances on Wet Roads | 案例分析：湿滑路面的刹车距离
A driver is travelling at 13 m/s (about 30 mph) when a child runs into the road. The driver has a reaction time of 0.7 s. On a dry road the braking distance is 14 m; on a wet road it increases to 22 m. Calculate the thinking distance, then find the total stopping distance for both conditions. Explain why the braking distance is greater when the road is wet.
一位司机正以 13 m/s（约 30 mph）的速度行驶，突然一名儿童冲上马路。司机反应时间为 0.7 s。在干燥路面上制动距离为 14 m，在湿滑路面上增加到 22 m。请计算思考距离，再分别计算两种路面情况下的总刹车距离，并解释为何湿滑路面的制动距离更大。
= 9.1 m. This is the same on any surface because it depends only on the driver, not the road.
思考距离是车辆在司机反应时间内行驶的距离。根据 距离 = 速度 × 时间，思考距离 = 13 m/s × 0.7 s = 9.1 m。这在任何路面上都是一样的，因为它只取决于驾驶员，与路面无关。
= 31.1 m. The extra 8 m could be the difference between stopping in time and a collision.
总刹车距离 = 思考距离 + 制动距离。干燥路面：9.1 m + 14 m = 23.1 m。湿滑路面：9.1 m + 22 m = 31.1 m。多出的 8 m 可能就是能否及时停下的关键。
e work to remove the car’s kinetic energy. Since work done = braking force × distance, for the same braking force a smaller friction force leads to a longer distance to stop.
湿滑路面的制动距离更大，因为轮胎与路面之间的摩擦力减小了。水起到了润滑作用，这意味着刹车需要做更多的功来消除汽车的动能。根据 做功 = 制动力 × 距离，在相同制动力下，较小的摩擦力会导致更长的停车距离。
2. Case Study: Choosing the Right Fuse | 案例分析：选择合适的保险丝
A table lamp operates on the UK mains supply (230 V) and has a power rating of 60 W. The lamp’s plug must be fitted with either a 3 A, 5 A or 13 A fuse. Determine the correct fuse and explain why it is safer than using a 13 A fuse.
一盏台灯使用英国市电（230 V），额定功率为 60 W。台灯插头需要安装 3 A、5 A 或 13 A 保险丝。请确定正确的保险丝，并解释为什么它比使用 13 A 保险丝更安全。
ower ÷ voltage = 60 W ÷ 230 V ≈ 0.26 A. This is the current when the lamp works correctly.
首先计算台灯的正常工作电流：功率 = 电压 × 电流，所以 电流 = 功率 ÷ 电压 = 60 W ÷ 230 V ≈ 0.26 A。这是台灯正常工作时的电流。
The fuse rating should be slightly higher than the normal operating current. A 3 A fuse is the smallest available standard rating; 3 A is just above 0.26 A, so it is the correct choice. A 5 A fuse would also work but gives less precise protection, while a 13 A fuse is far too high.
保险丝的额定电流应略高于正常工作电流。3 A 保险丝是最小的标准额定值；3 A 刚好高于 0.26 A，因此是正确选择。5 A 保险丝也可行，但保护不够精细；而 13 A 保险丝则过高了。
If a fault develops and the current rises to, say, 8 A, a 13 A fuse might not blow because it is designed to withstand up to 13 A. The wiring could overheat and cause a fire. A 3 A fuse will melt quickly at 8 A, breaking the circuit and keeping the appliance safe. This demonstrates why the fuse rating must match the appliance’s power.
如果发生故障，电流升高到 8 A，13 A 保险丝可能不会熔断，因为它设计承受高达 13 A 的电流。导线可能过热并引发火灾。而 3 A 保险丝在 8 A 时会迅速熔断，切断电路并保护电器。这表明保险丝额定值必须与电器的功率相匹配。
3. Case Study: Keeping Drinks Hot – the Vacuum Flask | 案例分析：保温瓶如何保温
A vacuum flask is designed to keep liquids hot for several hours. Analyse how its structure reduces heat transfer by conduction, convection and radiation. Identify which parts of the flask tackle each process.
保温瓶的设计可以让液体保温数小时。请分析其结构如何减少热传导、热对流和热辐射，并指出瓶子的各个部分分别应对哪种传热方式。
The flask consists of a double‑walled glass container with a vacuum between the walls. The vacuum contains no particles, so conduction and convection cannot occur through the walls. This stops the main escape routes for thermal energy from the liquid to the outside.
保温瓶的玻璃内胆为双层结构，两层之间为真空。真空中没有微粒，因此热传导和对流无法通过瓶壁发生。这就切断了液体热量向外散失的主要途径。
The inner glass surfaces are silvered. Shiny surfaces are poor emitters and poor absorbers of infrared radiation. The inner silvering reflects radiated heat back into the liquid, minimising heat loss by radiation.
玻璃内壁上镀有银层。光亮表面是红外辐射的不良发射体和吸收体。内壁的银层将辐射热反射回液体，从而最大限度地减少辐射热损失。
The stopper at the top is made of plastic or cork, which are good thermal insulators. It prevents convection currents from carrying hot air out and cold air in. Without the stopper, convection would quickly cool the drink. Together, these design features make the flask very efficient at keeping heat in.
瓶口的塞子由塑料或软木制成，这些都是良好的绝热体。它可以阻止对流将热空气带出、冷空气带入。如果没有塞子，对流会迅速使饮料冷却。这些设计特点共同作用，使保温瓶能高效地保持热量。
4. Case Study: Using a Lever to Lift a Rock | 案例分析：用杠杆撬动石块
A gardener uses a rigid steel bar as a lever to lift a 400 N rock. The pivot is placed 0.3 m from the rock, and the gardener applies an effort 1.2 m from the pivot. Calculate the effort needed to just lift the rock. State what happens to the required effort if the effort distance is increased.
一位园丁用一根硬质钢棒作杠杆来抬起一块重 400 N 的石头。支点距离石头 0.3 m，园丁在距离支点 1.2 m 处施加力。计算刚好能抬起石头所需的力，并说明如果施力距离增大，所需力会如何变化。
distance from pivot. The rock’s weight acts as the load: load × load distance = effort × effort distance.
这是一个第一类杠杆。力矩原理告诉我们，平衡时，顺时针力矩 = 逆时针力矩。力矩 = 力 × 到支点的垂直距离。石头的重力即为负载：负载 × 负载距离 = 施力 × 施力距离。
100 N. The gardener only needs to push down with 100 N, even though the rock weighs 400 N.
负载 = 400 N，负载距离 = 0.3 m，施力距离 = 1.2 m。因此 400 N × 0.3 m = 施力 × 1.2 m → 120 = 施力 × 1.2 → 施力 = 120 ÷ 1.2 = 100 N。即便石头重 400 N，园丁只需施加 100 N 的力。
0 N. Increasing the distance from the pivot reduces the force needed, which is why a longer crowbar makes the job easier. This is an example of a force multiplier.
如果施力距离增大到 2.0 m，那么施力 = 120 ÷ 2.0 = 60 N。增大与支点的距离可以减小所需力，这就是为什么长撬棍更省力。这就是一个“力放大器”的实例。
5. Case Study: Density and Floating in the Dead Sea | 案例分析：死海的密度与漂浮
The Dead Sea has a very high salt concentration, giving it a density of about 1240 kg/m³. A person with mass 70 kg and volume 0.065 m³ enters the water. Determine whether the person will float, and explain your reasoning using the concepts of density and upthrust.
死海的盐度极高，其密度约为 1240 kg/m³。一个质量为 70 kg、体积为 0.065 m³ 的人进入水中。判断此人是否会浮起，并运用密度和浮力的概念解释你的理由。
kg ÷ 0.065 m³ ≈ 1077 kg/m³. This is less than the density of Dead Sea water (1240 kg/m³).
先计算此人的平均密度：密度 = 质量 ÷ 体积 = 70 kg ÷ 0.065 m³ ≈ 1077 kg/m³。这小于死海水的密度（1240 kg/m³）。
An object floats when its density is less than the density of the fluid it is placed in. A less dense object displaces a weight of fluid equal to its own weight before it is fully submerged. Because the person’s density is lower, they will float more easily than in fresh water.
当物体的密度小于其所处流体的密度时，物体会浮起。密度较小的物体会在完全沉没之前排开与自身重量相等的流体。由于此人的密度较低，他会比在淡水中更容易浮起来。
ght. Here, if fully submerged the upthrust = volume × fluid density × g = 0.065 m³ × 1240 kg/m³ × 10 N/kg = 806 N. The person’s weight = 70 kg × 10 N/kg = 700 N. Since 806 N > 700 N, the upthrust is greater than the weight, so the person will float easily.
浮力等于排开液体的重量。要使人浮起，完全浸没时所受的最大浮力必须至少等于其体重。此处若完全浸没，浮力 = 体积 × 流体密度 × g = 0.065 m³ × 1240 kg/m³ × 10 N/kg = 806 N。此人的体重 = 70 kg × 10 N/kg = 700 N。因为 806 N > 700 N，浮力大于体重，所以此人会轻松浮起。
6. Case Study: Series vs Parallel Circuits in a Model House | 案例分析：模型房屋中的串联与并联电路
A student builds a model with two lamps. In one circuit the lamps are connected in series to a 6 V battery; in another they are in parallel. Each lamp has a resistance of 12 Ω. For each arrangement, calculate the total resistance, the current from the battery, and explain which arrangement would be used for household lighting.
一名学生用两盏灯泡搭建模型。在一个电路中，灯泡串联连接在 6 V 电池上；在另一个电路中，灯泡并联。每盏灯泡的电阻均为 12 Ω。针对每种连接方式，计算总电阻和电池提供的电流，并解释家用照明应采用哪一种连接方式。
ame current flows through both lamps, so they share the voltage; each lamp gets 3 V, making them dim.
串联电路：总电阻 Rₜ = R₁ + R₂ = 12 Ω + 12 Ω = 24 Ω。电池提供的电流：I = V ÷ Rₜ = 6 V ÷ 24 Ω = 0.25 A。相同的电流流过两盏灯，因此它们分压，每盏灯只得到 3 V，灯泡较暗。
Each branch gets the full 6 V and draws 0.5 A (6 V ÷ 12 Ω), so both lamps shine brightly.
并联电路：对于两个相同电阻，总电阻 Rₜ = R / 2 = 12 Ω ÷ 2 = 6 Ω。电池电流：I = 6 V ÷ 6 Ω = 1 A。每条支路都得到完整的 6 V 电压，各自电流为 0.5 A，因此两盏灯都明亮发光。
Household lighting uses parallel circuits. If one lamp fails, the other remains on because each has its own separate loop. In a series circuit, a break anywhere stops all current. Also, parallel circuits provide full voltage to each appliance, so they work at the correct brightness.
家用照明使用并联电路。如果一盏灯熄灭，另一盏仍会亮着，因为每盏灯都有自己独立的回路。而在串联电路中，任何一处断路都会导致全部断电。此外，并联电路能为每个电器提供额定电压，确保它们以正确的亮度工作。
7. Case Study: Efficiency of a Motor Lifting Weights | 案例分析：电动机提升重物的效率
lculate the useful work done on the mass and the efficiency of the motor. Use g = 10 N/kg.
一台小型电动机将 2 kg 的重物垂直提升 1.5 m。提升过程中，电动机接收了 50 J 的电能。计算对重物所做的有用功以及电动机的效率。取 g = 10 N/kg。
g × 10 N/kg × 1.5 m = 30 J. This is the gravitational potential energy gained by the mass.
有用功 = 重力 × 高度 = 质量 × g × 高度 = 2 kg × 10 N/kg × 1.5 m = 30 J。这是重物增加的重力势能。
Efficiency = (useful output energy ÷ total input energy) × 100% = (30 J ÷ 50 J) × 100% = 60%. So 60% of the electrical energy is converted into lifting the mass; the remaining 40% is wasted, mostly as thermal energy in the motor and sound.
效率 = （有用输出能量 ÷ 总输入能量）× 100% = (30 J ÷ 50 J) × 100% = 60%。因此 60% 的电能转化为提升重物的功；其余 40% 被浪费，主要转化为电动机的热能和声能。
Energy cannot be created or destroyed; the 50 J input is entirely accounted for (30 J useful + 20 J wasted). Improving the motor’s design, e.g. by lubricating moving parts, can reduce friction and raise the efficiency.
能量既不会凭空产生也不会凭空消失；输入的 50 J 能量全部有去向（30 J 有用功 + 20 J 浪费）。改进电动机的设计，例如润滑运动部件，可以减少摩擦并提高效率。
8. Case Study: Echo Sounding to Measure Depth | 案例分析：用回声测深
A ship’s sonar sends a sound pulse straight down to the seabed. The pulse returns after 0.4 s. The speed of sound in seawater is 1500 m/s. Calculate the depth of the water. Explain why the time is halved in your calculation.
一艘船上的声纳垂直向海底发射声脉冲。脉冲在 0.4 s 后返回。声波在海水中的传播速度为 1500 m/s。计算水深，并解释为何计算中要将时间减半。
herefore, depth = (speed × total time) ÷ 2 = (1500 m/s × 0.4 s) ÷ 2 = (600 m) ÷ 2 = 300 m.
声波从船向下传播到海底再返回，因此它在 0.4 s 内经过的距离是水深的 2 倍。所以，水深 = （速度 × 总时间）÷ 2 = (1500 m/s × 0.4 s) ÷ 2 = (600 m) ÷ 2 = 300 m。
We divide by two because the measured time accounts for the two‑way trip. If we used the full 0.4 s directly, we would be calculating the round‑trip distance, not the depth. This technique is used in sonar and ultrasound imaging.
需要除以 2，因为测得的时间包含了往返路程。若直接使用 0.4 s，算出的将是往返距离而非深度。这项技术广泛应用于声纳和超声成像中。
The frequency of the sound used is very high (ultrasound) so it can be directed in a narrow beam and will not be heard by marine mammals. The same principle can be applied to radar using electromagnetic waves.
所用声波频率非常高（超声波），因此可以形成窄波束，且不会被海洋哺乳动物听到。同样的原理也可应用于使用电磁波的雷达。
9. Case Study: Light Reflection in a Periscope | 案例分析：潜望镜中的光反射
A simple periscope uses two plane mirrors placed at 45° to the horizontal, inside a tube. Draw and describe the path of a light ray from an object to the observer’s eye. State the law of reflection and explain why the image appears the right way up.
一个简单的潜望镜在管内装有两面与水平面成 45° 的平面镜。请画出并描述从物体到观察者眼睛的光线路径。陈述反射定律，并解释为何最终成像看起来是正立的。
The light ray from the object hits the top mirror at an angle of incidence of 45°. According to the law of reflection, the angle of incidence equals the angle of reflection, so the ray reflects vertically downwards along the tube.
来自物体的光线以 45° 的入射角射到顶部平面镜上。根据反射定律，入射角等于反射角，因此光线沿管内垂直向下反射。
At the bottom mirror, the ray strikes again at 45° and reflects horizontally into the observer’s eye. The two reflections make the light travel around the corner. The image is not inverted because each mirror reverses the direction in one plane, and the second mirror reverses it back.
在底部平面镜上，光线再次以 45° 角入射，并水平反射进入观察者的眼睛。两次反射使光线绕过拐角。成像没有倒立，因为每面镜子在一个平面上颠倒方向，而第二面镜子又将其颠倒回来。
Plane mirrors produce virtual images that are the same size as the object and laterally inverted. In a periscope, the lateral inversion from the first mirror is corrected by the second, so the final image has the correct orientation. This is useful in submarines to see above the water surface while remaining hidden.
平面镜产生等大的、左右颠倒的虚像。在潜望镜中，第一面镜子造成的左右颠倒被第二面镜子纠正过来，因此最终成像的方向是正确的。这有助于潜艇在水下观察水面上方情况而不被发现。
10. Case Study: Mass, Weight and a Spring Balance | 案例分析：质量、重量与弹簧秤
s has a weight of 50 N (using g = 10 N/kg). Calculate the weight of the same mass on the Moon. Explain why the spring balance gives a different reading, but a beam balance would still compare masses correctly.
一位宇航员将弹簧秤和一组已知质量带到月球上，月球的引力场强度约为 1.6 N/kg。在地球上，5 kg 的物体重 50 N（g = 10 N/kg）。计算同一物体在月球上的重量。解释为何弹簧秤的读数会不同，而天平仍能正确比较质量。
ins 5 kg everywhere, but the weight is much smaller on the Moon because gravity is weaker.
重量 = 质量 × 引力场强度。地球上：重量 = 5 kg × 10 N/kg = 50 N。月球上：重量 = 5 kg × 1.6 N/kg = 8 N。质量在任何地方都是 5 kg，但由于月球引力较弱，重量小得多。
A spring balance measures weight – the force stretching the spring. Because weight changes with location, the reading changes. A beam balance, however, compares the gravitational force on an unknown mass with that on known masses. Since gravity affects both sides equally, the comparison is independent of the local g. Therefore a beam balance still shows 5 kg on the Moon, while a spring balance would show about 8 N (or a mass scale calibrated for Earth would read incorrectly).
弹簧秤测量的是重量——即拉伸弹簧的力。由于重量随位置变化，读数也会变化。然而，天平是将未知质量所受的重力与已知质量所受的重力进行比较。因为重力对两边的影响相同，这种比较与当地 g 值无关。因此，天平在月球上仍显示 5 kg，而弹簧秤则会显示约 8 N（若按地球重力校准的质量刻度则会读数错误）。
11. Case Study: Energy Transfers in a Bungee Jump | 案例分析：蹦极跳中的能量转换
A bungee jumper of mass 60 kg jumps from a bridge. At the top, before jumping, she has 30,000 J of gravitational potential energy. When she reaches the lowest point, all this energy has been transferred to elastic potential energy in the stretched rope and some thermal energy due to air resistance. If the elastic energy stored is 27,000 J, calculate the thermal energy produced and the average speed at the point halfway down, assuming no energy is wasted halfway.
一名质量为 60 kg 的蹦极者从桥上跳下。跳前在最高点，她的重力势能为 30000 J。到达最低点时，所有能量已转化为绳索的弹性势能和由于空气阻力产生的热能。如果储存的弹性势能为 27000 J，计算产生的热能，并估算在不考虑能量损失的前提下，下落到一半高度时的平均速度。
energy + thermal energy. So 30000 J = 27000 J + thermal energy → thermal energy = 30000 J – issipated to the surroundings because of air resistance and internal friction in the rope.
能量守恒：初始重力势能 = 弹性势能 + 热能。所以 30000 J = 27000 J + 热能 → 热能 = 30000 J – 27000 J = 3000 J。这 3000 J 由于空气阻力和绳索内摩擦而散失到周围环境中。
ted to kinetic energy, assuming no large elastic stretch yet. Kinetic energy = ½ × mass × speed² → 15000 J = ½ × 60 kg × v² → v² = (15000 × 2) ÷ 60 = 500 → v ≈ 22.4 m/s. This is an estimate; in reality some energy is already going into the rope and air resistance.
下落到一半高度时，她失去了一半的初始高度，因此剩余重力势能 = 15000 J。已损失的重力势能（15000 J）转化为动能，假设此时绳索尚未明显拉伸。动能 = ½ × 质量 × 速度² → 15000 J = ½ × 60 kg × v² → v² = (15000 × 2) ÷ 60 = 500 → v ≈ 22.4 m/s。这是估算值；实际上部分能量已用于绳索拉伸和空气阻力。
12. Case Study: Selecting Insulation for a Hot Water Pipe | 案例分析：为热水管选择保温材料
A pipe carries water at 70 °C through a cold basement at 10 °C. Two insulation materials are tested: material A with a U‑value of 0.8 W/m²°C and material B with a U‑value of 0.3 W/m²°C. For a pipe surface area of 2 m², calculate the rate of heat loss for each material and advise which is better. Explain what U‑value represents.
一根管道输送 70 °C 的热水，通过温度为 10 °C 的寒冷地下室。测试了两种保温材料：材料 A 的 U 值为 0.8 W/m²°C，材料 B 的 U 值为 0.3 W/m²°C。若管道表面积为 2 m²，计算每种材料的热损失速率，并建议哪种更好。解释 U 值代表什么。
Heat loss rate = U‑value × area × temperature difference. Temperature difference = 70 °C – . For material B: rate = 0.3 W/m²°C × 2 m² × 60 °C = 36 W. Material B loses far less heat.
热损失速率 = U 值 × 面积 × 温差。温差 = 70 °C – 10 °C = 60 °C。材料 A：速率 = 0.8 W/m²°C × 2 m² × 60 °C = 96 W。材料 B：速率 = 0.3 W/m²°C × 2 m² × 60 °C = 36 W。材料 B 的热损失少得多。
The U‑value measures how well a material conducts heat – it is the amount of energy that passes through 1 m² of the material each second for a 1 °C temperature difference. A low U‑value means the material is a good insulator. Therefore, material B is the better choice because it will keep the water hotter and reduce energy bills.
U 值衡量材料的导热能力——它表示在 1 °C 温差下，每秒通过 1 m² 材料的能量。U 值越低，材料的保温性能越好。因此，材料 B 是更好的选择，因为它能让热水保持高温并降低能源费用。
In addition, using a thick layer of material B will further reduce heat loss. Practical insulation often involves foam or fibre wraps, which trap air and have very low U‑values. This same principle is used to insulate lofts and walls in homes.
此外，使用厚层材料 B 可以进一步减少热损失。实际保温常采用泡沫或纤维包裹层，它们能锁住空气并具有极低的 U 值。同样的原理也用于家庭阁楼和墙体的保温。
Published by TutorHao | CCEA Physics Revision Series | aleveler.com
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