📚 Case Study Practical Exercises | 案例分析实战演练
Physics is full of everyday mysteries waiting to be solved. This article presents a series of real-world scenarios for Year 8 Cambridge learners, encouraging you to apply scientific knowledge, spot patterns, and justify your reasoning. Each case study is followed by a step-by-step analysis, helping you build the skills of observation, measurement, and logical explanation that are essential for Cambridge Lower Secondary Checkpoint and beyond.
物理世界中充满了有待破解的日常谜题。本文为 Year 8 剑桥学生准备了一系列真实情境案例，鼓励你运用科学知识、发现规律并论证自己的推理。每个案例分析后都配有逐步解析，帮助你养成观察、测量和逻辑解释的能力，这些能力对于剑桥初中 Checkpoint 考试以及后续学习都至关重要。
1. The Curious Metre Rule | 尺子弯曲之谜
A student measures the length of a classroom table using a wooden metre rule. The rule is slightly warped, causing a gap between the rule and the table surface. The student reads 120.3 cm from the warped rule. Another student uses a steel tape measure and records 119.8 cm. Explain why the readings differ and suggest which measurement is more reliable.
一位学生用木制米尺测量教室课桌的长度。该尺子略微弯曲，导致尺身与桌面之间存在缝隙。该学生从弯曲的尺子上读得 120.3 cm。另一位学生使用钢卷尺测量，记录为 119.8 cm。请解释为何两次读数不同，并指出哪次测量更为可靠。
The warped ruler introduces a systematic error. Because it does not lie flat, the graduations are effectively stretched along the measured length; the curved path is longer than the straight distance. The steel tape measure is flexible and hugs the surface, giving a truer linear measurement. Therefore the 119.8 cm value is more reliable. This reminds us that instruments must be checked for zero error and physical distortion before use.
弯曲的尺子引入了系统误差。由于尺子未能平贴，其刻度沿测量路径被实际拉长；弯曲的路径比直线距离更长。而钢卷尺柔韧且能贴紧表面，给出的直线测量值更真实。因此 119.8 cm 的数值更为可靠。这提醒我们，在使用仪器前必须检查其零误差与物理变形。
2. The Mystery of the Floating Egg | 鸡蛋浮沉探秘
A raw egg is placed in a beaker of fresh water. It sinks to the bottom. Salt is gradually stirred into the water, and after several spoonfuls the egg begins to rise and eventually floats. Use the concepts of density and upthrust to explain this observation. What would happen if the egg were transferred back into pure water?
将一枚生鸡蛋放入盛有清水的烧杯中，鸡蛋沉入杯底。向水中逐步加入食盐并搅拌，几勺之后鸡蛋开始上浮并最终漂浮。请用密度和上浮力的概念解释这一现象。若将鸡蛋重新移入纯水中会发生什么？
Initially, the egg sinks because its density is greater than that of fresh water. Adding salt dissolves into the water, increasing the mass per unit volume — the liquid’s density rises. As the saltwater’s density becomes greater than the egg’s average density, the upthrust (weight of displaced fluid) becomes equal to the egg’s weight, making it float. If transferred back to pure water, the egg’s density would again exceed that of the liquid, so it would sink once more.
起初鸡蛋下沉是因为其密度大于纯水的密度。加入的盐溶于水后，增加了单位体积的质量——液体密度上升。当盐水的密度大于鸡蛋的平均密度时，上浮力（排开液体的重量）便与鸡蛋的重量相等，鸡蛋因此漂浮。若移回纯水中，鸡蛋的密度再次大于液体密度，因此会再次下沉。
3. Heat Race: Metal vs. Wood | 导热竞速：金属与木块
On a cold morning, a student touches a metal door handle and a wooden tabletop. The metal feels much colder, even though both objects have been in the same room all night. Is the metal actually at a lower temperature? Explain using thermal conductivity. The student then wraps a piece of ice cube in aluminium foil and another in a woollen cloth. Which ice cube melts first?
在一个寒冷的早晨，一位学生触摸金属门把手和木质桌面。金属感觉冷得多，尽管两个物体整晚都处于同一房间内。金属的实际温度是否真的更低？请用热传导率解释。随后该学生用铝箔包裹一块冰块，用羊毛布包裹另一块。哪一块冰块会先融？
Both objects are at the same room temperature, but metal is a good thermal conductor. When you touch it, heat flows rapidly from your hand into the metal, making it feel cold. Wood is an insulator; it draws heat away much more slowly. For the ice cubes, aluminium foil is a good conductor and allows room heat to reach the ice quickly, so that cube melts first. The woollen cloth is a poor conductor and traps air, providing insulation and delaying melting.
两个物体实际处在于相同的室温下，但金属是热的良导体。当触摸时，热量快速从手传向金属，因而感觉冰冷。木材是绝缘体，夺走热量的速度要慢得多。对于冰块，铝箔是良导体，能让室温热量迅速到达冰块，因此那块先融。羊毛布导热性差且能锁住空气，起到保温作用，从而延缓融解。
4. The Seesaw Balancing Act | 跷跷板平衡难题
A 30 kg child sits 2.0 m from the pivot of a seesaw. Where must a 40 kg child sit to balance the seesaw horizontally? What would happen if the 40 kg child sat at 1.2 m from the pivot? Draw the turning effects and calculate the missing distance using the principle of moments.
一名 30 kg 的儿童坐在离跷跷板支点 2.0 m 处。要使跷跷板水平平衡，一名 40 kg 的儿童应坐在何处？如果 40 kg 的儿童坐在离支点 1.2 m 处，会发生什么？请画出转动效果并利用力矩原理计算未知距离。
iclockwise moment: 30 kg × 10 N/kg = 300 N, moment = 300 N × 2.0 m = 600 N m. For the 40 kg child (weight 400 N), distance d from pivot must satisfy 400 N × d = 600 N m, so d = 1.5 m. If the 40 kg child sits at 1.2 m, the clockwise moment is 400 N × 1.2 m = 480 N m, which is less than 600 N m. The seesaw would tip down on the 30 kg child’s side.
为达到平衡，顺时针力矩必须等于逆时针力矩。力矩 = 力 × 距支点的垂直距离。假设 30 kg 儿童产生逆时针力矩：体重为 30 kg × 10 N/kg = 300 N，力矩 = 300 N × 2.0 m = 600 N m。对于 40 kg 儿童（体重 400 N），至支点距离 d 需满足 400 N × d = 600 N m，解得 d = 1.5 m。若 40 kg 儿童坐在 1.2 m 处，其顺时针力矩为 400 N × 1.2 m = 480 N m，小于 600 N m，跷跷板将向 30 kg 儿童一侧倾斜。
5. Interpreting the Journey Graph | 解读行程图
Below is a distance–time graph for a cyclist travelling from home to a park and back. The graph shows a steep straight line upwards for the first 10 minutes, a flat horizontal line from 10 to 15 minutes, and a less steep straight line downwards from 15 to 25 minutes. Describe the motion in each section and calculate the cyclist’s speed on the outward journey and the return journey. (Assume the outward distance covered is 4 km.)
下图是一位骑行者从家前往公园并折返的距离–时间关系图。前 10 分钟线段陡直上升，10 至 15 分钟为水平线段，15 至 25 分钟为一条较缓的下降线段。请描述每段运动状态，并计算去程和回程的速度。（假设去程行进距离为 4 km。）
gh speed away from the starting point. The horizontal line shows the cyclist is stationary — lower than the outward trip. Speed on outward journey: distance 4 km, time 10 min = 1/6 hour, speed = 4 km ÷ (1/6 h) = 24 km/h. Return journey: 4 km in 10 min (from 15 to 25 min) = 1/6 h, speed = 24 km/h as well. However, if the line is less steep, the time might be different; our assumption gives same speed, but in a real case a gentler slope would indicate a longer time and thus lower speed.
陡峭的上升线段表示骑行者以恒定高速远离起点。水平线段显示骑行者静止不动——可能在公园休息。下降线段表示返程，速度恒定，但较缓的斜率说明速度比去程慢。去程速度计算：距离 4 km，时间 10 min = 1/6 h，速度 = 4 km ÷ (1/6 h) = 24 km/h。返程若从 15 至 25 分钟也是 10 分钟，距离相同，速度同样是 24 km/h。但若图表中下降线段斜率较小，实际用时会更长，速度则更低。绘图时需始终依据斜率判断速度快慢。
6. The Faulty Fairy Lights | 彩灯电路故障排查
A set of 20 fairy lights is connected in series. One bulb blows and all the lights go out. Explain why. A student uses a multimeter to measure voltage across each bulb holder. At 19 holders the voltage is nearly zero, but across one holder it reads the full supply voltage of 240 V. What does this tell you about the fault? How would the situation differ if the lights were wired in parallel?
一串由 20 个小彩灯串联而成。其中一个灯泡烧坏，所有灯全部熄灭。请解释原因。一位学生用万用表测量每个灯座两端的电压。在 19 个灯座上电压几乎为零，但在一个灯座上读到了 240 V 的全幅电源电压。这说明了什么故障？如果这些灯是并联连接的，情况会有什么不同？
In a series circuit, all components share the same current. A blown bulb creates an open circuit, stopping current everywhere. The voltage across a healthy bulb in a complete series circuit is a fraction of the supply; across an open fault (infinite resistance) virtually all the supply voltage appears. Thus the holder showing 240 V indicates the location of the broken bulb. If wired in parallel, each bulb receives the full supply voltage independently. One blown bulb would not affect the others — they would remain lit, and the faulty bulb would simply be dark.
在串联电路中，所有元件共享同一电流。烧坏的灯泡造成断路，使各处电流中断。正常工作的串联灯泡两端的电压只是电源电压的一部分；而断路故障处（电阻无穷大）会承受几乎全部电源电压。因此显示 240 V 的那个灯座就是损坏灯泡的位置。若采用并联连接，每个灯泡独立获得全部电源电压。一个灯泡烧坏不会影响其他灯泡——其余灯泡仍旧发光，坏的那个灯泡则会熄灭。
7. Reflection Rescue: Mirror Placement | 反射救援：镜面定位
A rescue team uses a signalling mirror to reflect sunlight towards a distant helicopter. The sun is behind the rescuers, and the helicopter is ahead and slightly to the left. Draw a ray diagram to show how a flat mirror should be angled. State the law of reflection and explain why a plane mirror produces a sharp beam only when precisely aligned.
一支救援队使用信号镜将阳光反射给远处的直升机。太阳在救援人员身后，直升机在前方偏左位置。画出光路图说明平面镜应如何倾斜。陈述反射定律并解释为何平面镜只有在精确对准时才能产生锐利光束。
The law of reflection states: the angle of incidence equals the angle of reflection, both measured from the normal (a line perpendicular to the mirror). The mirror must be rotated until the reflected ray travels directly to the helicopter. Because the sun’s rays are nearly parallel, a slight misalignment causes the reflected beam to miss the target. A flat mirror redirects light without focusing it, so the beam remains narrow and directional. Rescuers adjust the mirror until they see the reflected sunlight on the target — a technique often practised with small sighting devices.
反射定律指出：入射角等于反射角，两者均从法线（垂直于镜面的线）开始测量。必须旋转镜子直到反射光线恰好射向直升机。由于太阳光线近乎平行，即使极微小的角度误差也会导致反射光束偏离目标。平面镜只是改变光的方向而不聚焦，因此光束保持狭窄且定向。救援人员调整镜子直到看到反射阳光照在目标上——通常会借助小型瞄准装置进行练习。
8. Pitch and the Vibrating Ruler | 音调与振动的尺子
A wooden ruler is clamped to a desk with 20 cm overhanging. When the free end is plucked, a low-pitched twang is heard. The overhang is shortened to 10 cm and plucked again. The sound is noticeably higher in pitch. Explain this change in terms of frequency and amplitude. What variable is being altered, and how does it affect the vibration?
一把木尺被夹在课桌上，伸出桌面 20 cm。拨动自由端时，听到低沉的嗡嗡声。将伸出长度缩短至 10 cm 并再次拨动，声音明显升高。从频率和振幅的角度解释这一变化。哪个变量被改变了？它如何影响振动？
The pitch of a sound depends on the frequency of vibration. A shorter overhang has a stiffer effective spring and lower mass in motion, so it vibrates more quickly — higher frequency — and produces a higher pitch. The amplitude (size of vibration) may be similar if plucked with the same force, but the ear perceives the frequency change. The variable being altered is the length of the vibrating part; shorter length leads to faster oscillations and thus a higher-pitched sound. This principle is used in many musical instruments, such as the strings of a guitar.
声音的音调取决于振动频率。伸出部分较短时，等效弹性更强且运动质量更小，因而振动得更快——频率更高——产生更高的音调。若以相同力度拨动，振幅（振动幅度）可能相似，但人耳能分辨出频率的变化。被改变的变量是振动部分的长度；长度越短，振荡越快，音调也就越高。许多乐器中的原理与此无异，例如吉他弦的不同长度控制音高。
9. Energy Chain in a Wind-Up Toy | 发条玩具的能量链
A child winds the key of a toy car, then places it on the floor. The car accelerates, travels in a straight line, slows down, and stops. Identify the main energy stores and transfers at each stage: winding, moving, and stopping. Account for the energy dissipated.
一个孩子拧紧玩具小车的发条钥匙，然后把它放在地板上。小车加速，直线行驶，然后减速并停下。请指出在拧发条、运动以及停止这三个阶段中主要的能量储存与转移。解释能量是如何耗散的。
While winding, chemical energy from the child’s muscles is transferred mechanically to an elastic potential energy store in the spring. When released, the elastic potential energy is transferred to kinetic energy of the car. As the car moves, friction between the wheels and the floor, as well as air resistance, does work against motion, gradually transferring kinetic energy into thermal energy, raising the temperature of the surroundings slightly. The car stops when the initial elastic store is completely converted into thermal energy through mechanical work.
拧发条时，儿童肌肉的化学能通过机械做功转移为弹簧的弹性势能储存起来。释放后，弹性势能转化为小车的动能。行驶过程中，车轮与地面的摩擦以及空气阻力对运动做负功，逐步将动能转化为内能，使周围环境略微升温。当初始的弹性储能全部通过机械做功转变成热能时，小车便停下。
10. Electromagnet Crane’s Load Limit | 电磁铁吊臂的负载极限
A scrapyard crane uses an electromagnet to lift iron cars. One day the magnet struggles to lift a car that it used to manage easily. List three possible reasons for the reduced lifting force, explaining how each factor affects the magnetic field strength. The operator increases the current but notices the coil becomes very hot. Why is this a concern?
废品场吊臂使用电磁铁吸起铁质汽车。一天，磁铁对以前能轻松吸起的车辆显得有些吃力。列举导致吸力下降的三个可能原因，解释各因素如何影响磁场强度。操作员增大了电流，但发现线圈变得非常烫。为何这是一个隐患？
Possible reasons: (1) The number of turns in the coil may have reduced due to a broken wire — fewer turns weaken the magnetic field. (2) The current from the power supply could be lower than normal because of worn batteries or poor connections — less current gives a weaker field. (3) The iron core might have lost some magnetic permeability or be displaced — the core concentrates field lines, and a damaged core reduces flux density. Increasing current boosts the field but also raises resistive heating (I²R loss). Excessive heat can melt insulation, cause short circuits, and even start a fire, so the coil should be cooled or rated appropriately.
可能原因：(1) 线圈匝数可能因导线断裂而减少——匝数越少磁场越弱。(2) 电源供给的电流可能低于正常值，因电池老化或接触不良——电流减小导致磁场变弱。(3) 铁芯可能部分失磁或移位——铁芯能集中磁感线，受损的芯会削弱磁通密度。增大电流能增强磁场，但也会加剧电阻发热（I²R 损耗）。过热会熔化绝缘层，引发短路甚至火灾，因此线圈应适当冷却或按额定值使用。
Published by TutorHao | Physics Revision Series | aleveler.com
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