📚 Year 7 CAIE Physics Case Study Workout | 七年级CAIE物理案例分析实战演练
This workbook-style case study workout is specifically designed for Year 7 CAIE Physics learners who want to practise applying physics ideas to short, realistic situations. It keeps the original cycling case and extends the same method to ten more cases covering speed, weight, density, electric circuits, energy transformations, sound, light, heat transfer, springs and friction.
本实战演练面向七年级CAIE物理学生，帮助你把物理概念应用到简短而真实的情境中。文章保留原有的骑行者案例，并用同样的方法扩展到另外十个案例，涵盖速度、重量、密度、电路、能量转换、声音、光、热传递、弹簧和摩擦。
Each case follows a simple pattern: read the problem, try it yourself, then compare your working with the bilingual analysis. The focus is not on memorising answers, but on choosing the right equation, substituting correctly, and writing units clearly.
每个案例都遵循简单的模式：先读题，自己尝试解答，再与双语解析对照。重点不是背答案，而是学会选择正确公式、正确代入数据，并清楚写出单位。
1. How to Use This Case Study Workout | 如何使用本案例分析演练
Start by reading only the problem statement and covering the solution. Write down the quantities you are given, the quantity you need, and the equation that links them. This three-step habit prevents most careless mistakes in Year 7 physics.
先只读问题陈述，遮住解答部分。写下已知量、待求量和联系它们的公式。这个三步习惯可以避免七年级物理中大多数粗心错误。
Use the bilingual pairs to check understanding. If the English sentence is clear but the Chinese sentence is not, you may have missed a physics term rather than a language point. Key terms in this article include average speed, weight, density, current, voltage, efficiency, echo, reflection, conduction, extension and friction.
用中英配对来检查理解。如果英文清楚但中文不清楚，可能你漏掉的是物理术语，而不是语言问题。本文关键词包括平均速度、重量、密度、电流、电压、效率、回声、反射、传导、伸长量和摩擦。
A useful formula summary for the whole workout is shown below. Learn the quantity, the symbol and the unit together, because examiners award marks for correct units as well as correct numbers.
下面给出整套演练中有用的公式汇总。要把物理量、符号和单位一起记，因为考官不仅看数字，也看单位是否正确。
| Quantity | 物理量 | Equation | 公式 | Unit | 单位 |
|---|---|---|
| Average speed | 平均速度 | speed = distance ÷ time | m/s |
| Weight | 重量 | weight = mass × g | N |
| Density | 密度 | density = mass ÷ volume | g/cm³ or kg/m³ |
| Resistance | 电阻 | resistance = voltage ÷ current | Ω |
| Efficiency | 效率 | efficiency = useful energy ÷ total energy × 100% | % |
| Hooke's law | 胡克定律 | force = spring constant × extension | N = N/m × m |
2. Case 1: The Speeding Cyclist | 案例1：飞驰的骑行者
A cyclist travels along a straight road for 150 metres in 30 seconds. Calculate her average speed. Then determine how far she travels in 10 seconds if she maintains this speed.
一位骑行者沿直路行驶150米，用时30秒。计算她的平均速度。然后，如果她保持该速度，求她在10秒内通过的距离。
ven values: speed = 150 m ÷ 30 s = 5 m/s. The unit m/s means metres travelled each second.
步骤1：平均速度公式为 速度 = 距离 ÷ 时间。代入已知数值：速度 = 150米 ÷ 30秒 = 5米/秒。单位 m/s 表示每秒通过的米数。
speed = distance ÷ time = 150 m ÷ 30 s = 5 m/s
earrange the same relationship as distance = speed × time. Distance = 5 m/s × 10 s = 50 m.
步骤2：求10秒内通过的距离，把同一关系式变形为 距离 = 速度 × 时间。距离 = 5米/秒 × 10秒 = 50米。
distance = speed × time = 5 m/s × 10 s = 50 m
Step 3: This calculation assumes constant speed. In practice, cyclists may change speed up, slow down or stop, so 5 m/s is an average speed, not necessarily the instantaneous speed at every moment.
步骤3：该计算假定速度恒定。实际骑行中，骑行者可能加速、减速或停下，因此 5米/秒 是平均速度，不一定是每一时刻的瞬时速度。
Exam tip: always write the equation first, then substitute, then give the unit. A final answer of "50" without "m" may lose a mark, while "50 m" shows that you know distance is measured in metres.
考试提示：先写公式，再代入，最后写单位。只写 "50" 而没有 "m" 可能丢分，而 "50 m" 表明你知道距离用米测量。
3. Case 2: Weight on Earth and the Moon | 案例2：地球与月球上的重量
re g ≈ 10 N/kg, and on the Moon, where g ≈ 1.6 N/kg. Explain why her mass does not change.
一名学生的质量为30千克。计算她在地球上的重量（取 g ≈ 10牛/千克）和在月球上的重量（取 g ≈ 1.6牛/千克）。并解释为什么她的质量不变。
nal field strength, or W = m × g. Weight is measured in newtons, N, because it is a force.
步骤1：重量是重力作用在质量上的力。公式为 重量 = 质量 × 重力场强度，即 W = m × g。重量用牛顿（N）测量，因为它是力。
W = m × g
g × 10 N/kg = 300 N. The kilogram unit cancels with the kilogram in N/kg, leaving newtons.
步骤2：在地球上，W = 30千克 × 10牛/千克 = 300牛。千克单位与 N/kg 中的千克相消，剩下牛顿。
Step 3: On the Moon, W = 30 kg × 1.6 N/kg = 48 N. The Moon's gravitational field is weaker, so the same mass has a smaller weight.
步骤3：在月球上，W = 30千克 × 1.6牛/千克 = 48牛。月球的重力场更弱，所以同一质量具有更小的重量。
Mass is the amount of matter in an object and is measured in kilograms. It does not depend on location. Weight depends on the gravitational field strength, so the student's mass stays 30 kg on Earth, on the Moon or in space.
质量是物体所含物质的多少，用千克测量，不随位置改变。重量取决于重力场强度，因此该学生在地球、月球或太空中质量始终是30千克。
4. Case 3: Identifying an Unknown Metal | 案例3：鉴定未知金属
A small metal block has a mass of 270 g and a volume of 100 cm³. Calculate its density and suggest what the metal might be. Then express the density in kg/m³.
一块小金属块质量为270克，体积为100立方厘米。计算它的密度，并推测它可能是什么金属。然后把密度换算成千克每立方米。
ures how much mass is packed into a given volume. The equation is density = mass ÷ volume.
步骤1：密度衡量单位体积内包含多少质量。公式为 密度 = 质量 ÷ 体积。
density = mass ÷ volume = 270 g ÷ 100 cm³ = 2.7 g/cm³
Step 2: A density of 2.7 g/cm³ is close to the accepted value for aluminium. This is why the block could be aluminium rather than iron, which has a density of about 7.9 g/cm³.
步骤2：2.7克/立方厘米接近铝的公认密度值。因此该金属块可能是铝，而不是密度约为7.9克/立方厘米的铁。
, remember that 1 g/cm³ = 1000 kg/m³. Therefore 2.7 g/cm³ = 2.7 × 1000 kg/m³ = 2700 kg/m³.
步骤3：换算成千克每立方米时，记住 1克/立方厘米 = 1000千克/立方米。因此 2.7克/立方厘米 = 2.7 × 1000千克/立方米 = 2700千克/立方米。
Step 4: Density also explains floating and sinking. An object floats in water if its average density is less than the density of water, which is about 1 g/cm³. Aluminium is denser than water, so a solid aluminium block sinks.
步骤4：密度也能解释漂浮和下沉。如果物体的平均密度小于水的密度（约1克/立方厘米），它就会浮在水面上。铝比水密度大，所以实心铝块会下沉。
5. Case 4: A Simple Series Circuit | 案例4：简单串联电路
A series circuit contains two 1.5 V cells, a lamp and an ammeter. The current is 0.2 A. Calculate the total voltage and the total resistance. State what happens if the lamp breaks.
一个串联电路包含两节1.5伏电池、一只灯泡和一个电流表。电流为0.2安。计算总电压和总电阻。说明如果灯泡损坏会发生什么。
V = 3.0 V. The cells push charge around the circuit, and the ammeter measures the current.
步骤1：串联电池的电压相加。总电压 = 1.5伏 + 1.5伏 = 3.0伏。电池推动电荷在电路中流动，电流表测量电流。
Vtotal = 1.5 V + 1.5 V = 3.0 V
ce = voltage ÷ current. R = 3.0 V ÷ 0.2 A = 15 Ω. The ohm symbol Ω is used for resistance.
步骤2：使用电阻公式 电阻 = 电压 ÷ 电流。R = 3.0伏 ÷ 0.2安 = 15欧姆。电阻的单位符号是 Ω。
R = V ÷ I = 3.0 V ÷ 0.2 A = 15 Ω
Step 3: In a series circuit, the current is the same at every point. If the lamp breaks, the circuit becomes an open circuit, so the current stops and all components in that loop stop working.
步骤3：在串联电路中，电流在任意位置都相同。如果灯泡损坏，电路变成断路，电流停止，该回路中的所有元件都停止工作。
Step 4: A voltmeter is connected in parallel across a component to measure voltage, while an ammeter is connected in series to measure current. Reversing the cells reverses the current direction but does not change the size of the current in a simple circuit.
步骤4：电压表并联在元件两端测量电压，而电流表串联在电路中测量电流。把电池反接会改变电流方向，但在简单电路中不会改变电流大小。
6. Case 5: Energy Transformations in a Torch | 案例5：手电筒中的能量转换
A torch uses a battery to light a bulb. The battery transfers 10 J of chemical energy, and the bulb transfers 4 J as useful light energy. Calculate the efficiency and describe the wasted energy.
一只手电筒用电池点亮灯泡。电池转移10焦耳的化学能，灯泡转移4焦耳的有用光能。计算效率并描述浪费的能量。
Step 1: Energy can be transferred from one store to another, but it cannot be created or destroyed. In a torch, chemical energy in the battery is transferred to electrical energy, then to light energy and thermal energy.
步骤1：能量可以从一种能量库转移到另一种能量库，但不能被创造或消灭。在手电筒中，电池里的化学能转化为电能，再转化为光能和热能。
Step 2: Efficiency = useful energy output ÷ total energy input × 100%. Here efficiency = 4 J ÷ 10 J × 100% = 40%.
步骤2：效率 = 有用能量输出 ÷ 总能量输入 × 100%。这里效率 = 4焦耳 ÷ 10焦耳 × 100% = 40%。
efficiency = 4 J ÷ 10 J × 100% = 40%
Step 3: The remaining 6 J is wasted, mainly as thermal energy in the bulb and wires. This is why a torch bulb becomes warm after being switched on for a while.
步骤3：剩余的6焦耳被浪费，主要以灯泡和导线中的热能形式散失。这就是手电筒灯泡点亮一段时间后会变热的原因。
Step 4: Useful energy transfers can be shown with a Sankey diagram. The input arrow is wider than the useful output arrow, and the wasted arrow branches away to show energy that spreads into the surroundings.
步骤4：有用的能量转移可以用桑基图表示。输入箭头比有用输出箭头宽，浪费的箭头分支出去，表示能量散失到周围环境中。
7. Case 6: Echo in a Canyon | 案例6：峡谷中的回声
A hiker shouts towards a canyon wall and hears an echo 2.0 seconds later. If the speed of sound in air is 340 m/s, how far away is the wall? Explain why sound cannot travel through a vacuum.
一名徒步者朝峡谷壁呼喊，2.0秒后听到回声。如果空气中的声速为340米/秒，峡谷壁有多远？并解释为什么声音不能在真空中传播。
Step 1: Sound is produced by vibrating objects and travels as a wave through a medium. The sound travels to the wall and back, so the total distance is twice the distance to the wall.
步骤1：声音由振动产生，并以波的形式通过介质传播。声音传到峡谷壁再返回，所以总距离是到峡谷壁距离的两倍。
me = 340 m/s × 2.0 s = 680 m. The distance to the wall is half of this: 680 m ÷ 2 = 340 m.
步骤2：总距离 = 速度 × 时间 = 340米/秒 × 2.0秒 = 680米。到峡谷壁的距离是它的一半：680米 ÷ 2 = 340米。
d = v × t ÷ 2 = 340 m/s × 2.0 s ÷ 2 = 340 m
Step 3: Sound needs particles to pass the vibration along. In a vacuum there are no particles, so sound cannot travel. Light can travel through a vacuum, which is why we can see the Sun but cannot hear explosions in space.
步骤3：声音需要粒子来传递振动。真空中没有粒子，所以声音无法传播。光可以在真空中传播，因此我们能看见太阳，却听不到太空中的爆炸声。
Step 4: Frequency affects pitch: a higher frequency gives a higher pitch. Amplitude affects loudness: a larger amplitude gives a louder sound. Echoes are reflected sound waves, and hard, smooth surfaces reflect sound well.
步骤4：频率影响音调：频率越高，音调越高。振幅影响响度：振幅越大，声音越响。回声是反射的声波，坚硬光滑的表面能很好地反射声音。
8. Case 7: Reflection in a Plane Mirror | 案例7：平面镜反射
A ray of light hits a plane mirror at an angle of incidence of 35°. State the angle of reflection and describe three properties of the image formed.
一束光以35°的入射角射到平面镜上。写出反射角，并描述所成像的三个性质。
Step 1: The law of reflection states that the angle of incidence equals the angle of reflection. Both angles are measured from the normal, which is an imaginary line perpendicular to the mirror surface.
步骤1：反射定律指出，入射角等于反射角。两个角都从法线量起，法线是垂直于镜面的假想线。
angle of incidence = angle of reflection = 35°
Step 2: The image in a plane mirror is virtual, because light rays only appear to come from behind the mirror. It is upright, and it is the same size as the object.
步骤2：平面镜中的像是虚像，因为光线看起来只是从镜后发出。像是正立的，且与物体大小相同。
Step 3: The image is laterally inverted, meaning left and right are swapped. It is also the same distance behind the mirror as the object is in front of the mirror.
步骤3：像是左右颠倒的，即左边和右边互换。像到镜后的距离等于物到镜前的距离。
diagrams use straight lines with arrows to show the direction of light, and the normal is
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