📚 Interdisciplinary Physics Questions Training | 跨学科物理综合题型训练
In Year 7 Cambridge Physics, many problems go beyond a single subject. You will often need to combine physics with mathematics, chemistry, biology, geography or even history to find the answer. This article provides an integrated training programme that helps you practise these skills through real mixed-style questions. You will learn how to apply formulas, interpret data, link concepts across subjects and develop a scientific way of thinking that prepares you for the Checkpoint exam and beyond.
在剑桥 Year 7 物理中，许多题目并不局限于单一学科。你常常需要将物理与数学、化学、生物、地理甚至历史相结合才能得出答案。本文提供了一套综合训练方案，通过真实的混合题型帮助你锻炼这些技能。你将学会如何运用公式、解读数据、跨学科联系概念，并培养科学的思维方式，为 Checkpoint 考试及后续学习做好准备。
1. Physics Meets Mathematics | 物理遇上数学
Many physics calculations rely on basic arithmetic, rearranging simple formulas and converting units. You should be comfortable using the equation for average speed, density, pressure or moment. Before solving, always check that units are consistent – for example, distance in metres and time in seconds give speed in m/s. When a question gives mixed units, convert everything to a single system first.
很多物理计算依赖基本的算术、简单的公式变形和单位换算。你必须能熟练使用平均速度、密度、压强或力矩的公式。解题前务必核对单位是否一致——例如，距离用米、时间用秒，速度才是 m/s。如果题目给出的单位混合，先把所有量统一成一种单位制。
Example problem:
例题：
A cyclist travels 15 km in 30 minutes and then another 9 km in 20 minutes. Calculate the average speed in m/s.
一名自行车手先骑行 15 km，用时 30 分钟，然后又骑行了 9 km，用时 20 分钟。计算平均速度，单位用 m/s。
Solution steps (pairing):
解题步骤（配对）：
Total distance = (15 + 9) km = 24 km = 24 000 m. Total time = (30 + 20) min = 50 min = 50 × 60 s = 3000 s.
总距离 = (15 + 9) 千米 = 24 千米 = 24 000 米。总时间 = (30 + 20) 分钟 = 50 分钟 = 50 × 60 秒 = 3000 秒。
Average speed = total distance ÷ total time = 24 000 m ÷ 3000 s = 8 m/s.
平均速度 = 总距离 ÷ 总时间 = 24 000 米 ÷ 3000 秒 = 8 米/秒。
Common trap: Many students forget to convert minutes to seconds, or they average the two speeds wrongly. Speed is not simply the average of two separate speeds; it is total distance divided by total time.
常见陷阱：很多学生忘记把分钟换成秒，或者错误地直接把两个速度平均。速度并不是两个速度的简单平均，而是总距离除以总时间。
2. Density, Mass and Volume – Linking Maths, Physics and Chemistry | 密度、质量与体积——连接数学、物理和化学
The density equation ρ = m / V is used across physics and chemistry. When you identify an unknown substance, you often measure its mass and volume, calculate density and compare with reference tables. In the lab, you may use a displacement can to find the volume of an irregular solid – this is an interdisciplinary skill combining physics (density concept) with practical chemistry / general science techniques.
密度公式 ρ = m / V 在物理和化学中都会用到。当你鉴定一种未知物质时，通常要测量它的质量和体积，计算密度并对照参考表。实验中，你可能用排水法测量不规则固体的体积——这是一项跨学科技能，融合了物理（密度概念）与化学/通用实验方法。
Interdisciplinary question style:
跨学科题型：
A student drops a rock into a measuring cylinder containing 50 cm³ of water. The water level rises to 80 cm³. The mass of the rock is 78 grams. Calculate the density of the rock in g/cm³. Suggest what material the rock might be, using the table below.
一位学生把一块石块放入盛有 50 cm³ 水的量筒中，水面上升到 80 cm³。石块质量为 78 克。计算石块的密度，单位 g/cm³，并利用下表推测石块可能是什么物质。
| Material | Density (g/cm³) |
|---|---|
| Aluminium | 2.7 |
| Iron | 7.9 |
| Granite | 2.6 – 2.8 |
| Glass | 2.4 – 2.8 |
Volume of rock = 80 – 50 = 30 cm³. Density = mass ÷ volume = 78 ÷ 30 = 2.6 g/cm³. The rock is likely granite or glass, but granite is more common as a rock.
石块体积 = 80 – 50 = 30 立方厘米。密度 = 质量 ÷ 体积 = 78 ÷ 30 = 2.6 克/立方厘米。石块可能是花岗岩或玻璃，但作为岩石更可能是花岗岩。
This problem tests your understanding of volume by displacement (chemistry / practical skills), density calculation (physics + maths) and data interpretation.
这道题考察你对排水法测体积（化学/实验技能）、密度计算（物理+数学）以及数据解读的理解。
3. Energy and Living Things – Physics Meets Biology | 能量与生物——物理遇见生物学
Energy is a key theme in both physics and biology. When you study food labels in biology, you see energy content in kilojoules (kJ). The physics concept of energy transfer helps explain how our bodies convert chemical energy stored in food into kinetic energy for movement and thermal energy to keep warm.
能量是物理和生物的共同核心主题。在生物课上学习食品标签时，你会看到以千焦（kJ）为单位的能量值。物理中的能量传递概念有助于解释人体如何将食物中储存的化学能转化为运动的动能和保暖的热能。
Typical question:
常见题型：
A cereal bar provides 450 kJ of energy. A student of mass 40 kg climbs stairs that are 5 m high. The gravitational field strength g = 10 N/kg. Calculate the gravitational potential energy gained. How many cereal bars would the student need to eat to have enough energy for 20 such climbs, assuming 100% efficiency?
一块谷物棒提供 450 kJ 能量。一名质量为 40 kg 的学生爬上 5 m 高的楼梯。重力场强度 g = 10 N/kg。计算获得的重力势能。假设效率为 100%，该学生需要吃多少块谷物棒才能有足够能量完成 20 次这样的爬楼？
GPE = m × g × h = 40 kg × 10 N/kg × 5 m = 2000 J = 2 kJ per climb. For 20 climbs: 20 × 2 kJ = 40 kJ. Number of bars = 40 kJ ÷ 450 kJ ≈ 0.089 bars. Wait – that seems small… Actually recalculate: 2000 J = 2 kJ; 450 kJ is a lot. 40 kJ is less than one bar. The question highlights that our body does not use energy at 100% efficiency, and 450 kJ is far more than needed for this activity. The bio-physics twist: real muscles are only about 25% efficient, so you would need more food. This builds interdisciplinary thinking.
重力势能 = m × g × h = 40 千克 × 10 牛/千克 × 5 米 = 2000 焦耳 = 2 千焦（一次）。20 次攀爬：20 × 2 kJ = 40 kJ。所需谷物棒数量 = 40 kJ ÷ 450 kJ ≈ 0.089 块。等等——看起来很少。重新检查：2000 J = 2 kJ；450 kJ 相当多。40 kJ 还不到一块。题目凸显了人体并非 100% 效率利用能量，而且 450 kJ 远超所需。生物-物理的结合点在于：真实肌肉效率仅约 25%，因此需要更多食物。这培养了跨学科思维。
4. Forces, Motion and the Human Body – Physics and Sport Science | 力、运动与人体——物理与运动科学
Understanding forces helps explain sports performance. The idea of friction between sports shoes and the ground, air resistance on a cyclist, and the moment applied by a tennis racket all link physics with physical education. Students can design simple experiments to measure grip or reaction time.
理解力有助于解释运动表现。运动鞋与地面的摩擦力、自行车手受到的空气阻力、网球拍施加的力矩等概念，都将物理与体育关联起来。学生可以设计简单实验测量抓地力或反应时间。
Interdisciplinary task:
跨学科任务：
Design an investigation to find out how the type of shoe sole affects the force needed to pull a shoe across a surface. You have a spring balance, different shoes and a wooden plank. Explain how you would ensure a fair test and what variables you would control. Link your findings to the choice of shoes for a basketball player.
设计一个探究实验，确定鞋底类型如何影响拉着鞋子在表面上移动所需的力。你有一个弹簧秤、不同的鞋子和一块木板。说明如何确保公平测试以及控制哪些变量。将你的发现与篮球运动员选鞋联系起来。
This combines experimental design (science enquiry), friction (physics) and practical application to sport. A good answer would mention keeping the mass inside the shoe constant, pulling at a steady speed, measuring the force when the shoe just starts to move, repeating and averaging, and comparing the coefficient of friction. Then discuss that basketball players need high friction for quick stops and turns, so soles with a high traction pattern are preferred.
这融合了实验设计（科学探究）、摩擦力（物理）以及在运动中的实际应用。好的答案会提到：保持鞋内质量不变、匀速拉动、测量鞋刚刚开始移动时的力、重复并求平均值、比较摩擦系数，然后讨论篮球运动员需要高摩擦力以急停和转向，因此偏好高抓地花纹的鞋底。
5. Heat, Temperature and Weather – Physics Meets Geography | 热、温度与天气——物理遇见地理
In geography, you study weather and climate. The physics of heat transfer – conduction, convection and radiation – explains many atmospheric phenomena. Sea breezes, for example, occur because land heats up and cools down faster than water. This is a direct application of specific heat capacity and convection currents.
在地理课上，你会学习天气与气候。热传递的物理原理——传导、对流和辐射——解释了许多大气现象。例如，海陆风的形成是因为陆地比水升温快、降温也快。这是比热容和对流气流原理的直接应用。
Cross-curricular question:
跨学科题目：
Explain why during a hot summer day, the wind often blows from the sea towards the land (sea breeze) in the afternoon. Use the ideas of heat absorption, expansion and density in your answer.
解释为什么在炎热的夏天，下午经常吹从海到陆的海风。在答案中运用热吸收、膨胀和密度的概念。
During the day, the sun heats both land and sea. Land has a lower specific heat capacity than water, so it warms up more quickly. The air above the land becomes hotter, expands and becomes less dense. This warm, less dense air rises, creating a low-pressure area. Cooler, denser air from over the sea then moves in to replace it, causing a sea breeze. This type of question explicitly tests whether you can link physics (heat, density, pressure) to geographical patterns.
白天，太阳同时加热陆地和海洋。陆地的比热容比水小，所以升温更快。陆地上方的空气变得更热，膨胀，密度变小。这些暖而轻的空气上升，形成低压区。海洋上方较冷、密度较大的空气随之吹来补充，形成海风。这类题目直接测试你是否能将物理（热、密度、压强）与地理规律联系起来。
6. Sound Waves and Music – Physics Embraces Art and Biology | 声波与音乐——物理拥抱艺术与生物
Sound is a topic that naturally crosses into music (pitch and loudness) and biology (the structure of the ear). In Year 7, you learn how vibrations create sound, how frequency relates to pitch, and amplitude to loudness. You can build simple musical instruments to demonstrate these concepts.
声音是一个天然跨入音乐（音高和响度）和生物（耳朵结构）的主题。在 Year 7，你会学到振动如何产生声音、频率如何决定音高、振幅如何决定响度。你可以制作简单乐器来演示这些概念。
Integrated question:
综合题：
A guitarist tightens a string. Explain what happens to the pitch produced and why. Then, describe how the sound travels from the guitar to the listener’s brain, naming the parts of the ear in order.
一位吉他手拧紧琴弦。解释音高会发生什么变化以及原因。然后描述声音如何从吉他传到听者的大脑，按顺序列出耳朵的各个部分。
Physics part: Tightening the string increases the tension, which increases the frequency of vibration, producing a higher pitch.
物理部分：拧紧琴弦增加了张力，从而提高了振动频率，音高变高。
Biology part: Sound waves travel through the air into the outer ear (pinna, ear canal), hit the eardrum causing it to vibrate, pass through the ossicles (hammer, anvil, stirrup) in the middle ear, enter the cochlea in the inner ear where hair cells convert vibrations into electrical signals, which then travel along the auditory nerve to the brain.
生物部分：声波通过空气进入外耳（耳廓、耳道），撞击鼓膜使之振动，经中耳的听小骨（锤骨、砧骨、镫骨）传入内耳的耳蜗，耳蜗中的毛细胞将振动转换为电信号，信号沿听神经传到大脑。
This question clearly asks for two different disciplinary knowledge sets to be combined in one coherent answer.
这道题明确要求将两套不同学科的知识融合在一个连贯的答案中。
7. Light, Colour and Biology – The Eye and Beyond | 光、颜色与生物——眼睛及其他
The physics of light (reflection, refraction, dispersion) directly underpins our understanding of how the eye forms images. When white light passes through a prism, it disperses into the colours of the rainbow. The human eye contains a lens that refracts light to focus an image on the retina, where rods and cones detect brightness and colour.
光的物理特性（反射、折射、色散）直接支撑了我们对眼睛如何成像的理解。当白光通过棱镜时，会色散成彩虹的颜色。人眼包含一个晶状体，能折射光线，将图像聚焦在视网膜上，视网膜上的视杆细胞和视锥细胞则感知亮度和颜色。
Interdisciplinary question:
跨学科题目：
Compare how a camera and the human eye both use a converging lens to form an image. In your answer, refer to the terms ‘refraction’, ‘focal point’ and ‘real image’. Also, explain why we can see a green leaf as green, linking the concept of selective reflection of light.
比较照相机与人眼如何都使用会聚透镜成像。在答案中提到“折射”、“焦点”和“实像”。并解释为什么我们看到绿叶是绿色的，结合光的选择性反射概念。
The eye and camera both use a convex lens to refract light rays and bring them to a focus on a light-sensitive surface (retina / film or sensor). A real, inverted image is formed. The green leaf absorbs all colours of white light except green; it reflects green light into our eyes, stimulating the cone cells most sensitive to green light. This answer draws on physics (optics) and biology (eye structure, colour perception).
眼睛和相机都使用凸透镜折射光线，将其汇聚到感光面（视网膜/胶片或传感器）上，形成一个倒立的实像。绿叶吸收白光中除绿色以外的所有颜色，将绿光反射进我们的眼睛，刺激了对绿光最敏感的视锥细胞。这个答案同时运用了物理（光学）和生物（眼睛结构、色觉）。
8. Simple Machines and History – Physics and the Story of Technology | 简单机械与历史——物理与技术的故事
Levers, pulleys and inclined planes are not just physics topics; they have shaped human civilisation. When you calculate the mechanical advantage of a lever, you can link it to how ancient Egyptians may have used ramps to build pyramids, or how Archimedes understood the power of levers.
杠杆、滑轮和斜面不仅仅是物理话题，它们还塑造了人类文明。当你计算杠杆的机械效益时，可以将其与古埃及人可能如何使用斜坡建造金字塔，或阿基米德如何理解杠杆的力量联系起来。
Integrative task:
综合性任务：
Write a short paragraph that explains how a class–1 lever works, using the terms ‘effort’, ‘load’ and ‘fulcrum’. Then calculate the mechanical advantage if the effort arm is 2 m and the load arm is 0.5 m. Finally, comment on why this type of lever has been important in historical construction.
写一段简短的文字，解释一类杠杆的工作原理，使用“动力”、“阻力”和“支点”三个术语。然后计算当动力臂为 2 m、阻力臂为 0.5 m 时的机械效益。最后，评论为什么这类杠杆在历史建筑中很重要。
Physics: Mechanical advantage = effort arm ÷ load arm = 2 m / 0.5 m = 4. This means the lever multiplies the effort by 4. History: A class–1 lever can help lift heavy stones with less force, making it possible to move large blocks for temples or fortifications. This simple machine reduces the effort needed, which was crucial before modern engines.
物理：机械效益 = 动力臂 ÷ 阻力臂 = 2 米 / 0.5 米 = 4。这意味着杠杆将动力放大 4 倍。历史：一类杠杆能用较小的力抬起沉重的石块，使得搬运建造庙宇或堡垒的大石块成为可能。这种简单机械减少了所需的力，这在现代发动机出现之前至关重要。
9. Electricity and Everyday Technology – Physics + Design & Technology | 电与日常科技——物理+设计与技术
Understanding circuits is essential for designing simple electronic products in DT lessons. When you build a steady-hand game or a burglar alarm, you are applying the physics of conductors, insulators, switches and buzzers. This topic also involves safety – fuses and earthing protect users from electric shocks.
理解电路对于在设计技术课上设计简单电子产品至关重要。当你制作一个“稳定手”游戏或防盗报警器时，你正在应用导体、绝缘体、开关和蜂鸣器等物理知识。这一主题还涉及安全知识——保险丝和接地保护使用者免受电击。
Design-based physics question:
基于设计的物理题：
You are asked to build a circuit that lights a lamp when the classroom door is opened. Which type of switch would you use? Draw a circuit diagram and explain how it works. Mention the terms ‘closed circuit’ and ‘open circuit’. Also, state one safety precaution you would include.
你被要求制作一个电路，当教室门被打开时点亮一盏灯。你会使用哪种开关？画出电路图并解释其工作原理。提及“闭合电路”和“断路”两个术语。还要说明你会采取的一项安全预防措施。
You would use a push-to-make switch or a magnetic reed switch attached to the door frame. When the door is closed, the switch is held open (open circuit). When the door opens, the switch closes (closed circuit), current flows and the lamp lights. A safety precaution: include a fuse in the circuit to prevent overheating if a fault occurs. This blends physics, electronics and practical design safety.
你可以使用一个常开触点开关或一个固定在门框上的磁簧开关。门关闭时，开关保持断开（断路）；门打开时，开关闭合（闭合电路），电流流过，灯亮起。安全预防措施：在电路中加入保险丝，以防发生故障时过热。这结合了物理、电子学和实际设计安全。
10. The Water Cycle and Changes of State – Physics and Geography | 水循环与物态变化——物理与地理
The water cycle is a classic bridging topic. Evaporation and condensation are physical changes of state, driven by energy from the sun. Geography describes the movement of water between oceans, atmosphere and land, but physics explains the why: particles gain energy, overcome attractive forces and escape as gas, or lose energy and come closer to form liquid.
水循环是一个经典的联结主题。蒸发和凝结是物理的物态变化，由来自太阳的能量驱动。地理描述了水在海洋、大气和陆地之间的运动，而物理则解释了原因：粒子获得能量，克服吸引力而逸出成为气体；或失去能量，相互靠近形成液体。
Explaining the process:
解释过程：
Describe evaporation from a puddle on a warm day in terms of particle behaviour. Then link this to cloud formation: as warm, moist air rises, it cools. Explain in particle terms what happens next (condensation). Finally, state the type of precipitation that might fall if the cloud temperature is below 0°C.
用粒子行为描述温暖天气下一个小水坑的蒸发过程。然后将其与云的形成联系起来：暖而潮湿的空气上升，遇冷。用粒子的语言解释接下来发生的事（凝结）。最后，如果云的温度低于 0°C，可能出现哪种降水类型。
Physics answer: Particles in the puddle have a range of kinetic energies. The most energetic ones near the surface escape into the air – this is evaporation. As the parcel of air rises, its particles lose thermal energy to the surroundings. Water vapour particles slow down, and when they collide, the attractive forces become enough to hold them together as tiny liquid droplets, forming a cloud – this is condensation. Below 0°C, ice crystals may form, leading to snow or hail. This integrated response connects particle theory, energy transfer and geographical weather outcomes.
物理答案：水坑中的粒子具有不同的动能。表面附近能量最高的粒子挣脱到空气中——这就是蒸发。当一团空气上升时，其中的粒子向周围散失热能。水蒸气粒子减速，碰撞时吸引力足以将它们束缚成微小的液态水滴，形成云——这就是凝结。在零度以下，可能会形成冰晶，导致雪或冰雹。这一综合回答将粒子理论、能量传递和地理天气结果联系起来。
11. Practical Investigation: Combining Graphs, Tables and Conclusions | 实验探究：结合图表和结论
The Checkpoint exam often includes a practical investigation question where you must record data in a table, draw a graph and write a conclusion. This requires skills from maths (graph plotting, line of best fit) and science (identifying variables, making predictions). A common topic is Hooke’s law or the effect of temperature on dissolving rate, both of which involve collecting numerical data and looking for patterns.
Checkpoint 考试经常包含一道实验探究题，要求你将数据记录在表格中、绘制图表并写下结论。这需要数学（绘制图表、最佳拟合线）和科学（识别变量、做出预测）的技能。常见题目是胡克定律或温度对溶解速率的影响，两者都需要收集数值数据并寻找规律。
Exam–style scenario:
考试情境：
A student stretches a spring by adding different masses and measures the length. The results are: 0 g → 10.0 cm, 100 g → 12.5 cm, 200 g → 15.0 cm, 300 g → 17.5 cm, 400 g → 20.0 cm. Plot a graph of extension (y-axis) against mass (x-axis). Calculate the extension for each mass. What pattern do you notice? Predict the length for 250 g. This is pure graph and pattern work (maths) inside a physics context.
一名学生通过添加不同质量拉伸一根弹簧并测量长度。结果如下：0 克 → 10.0 厘米，100 克 → 12.5 厘米，200 克 → 15.0 厘米，300 克 → 17.5 厘米，400 克 → 20.0 厘米。以伸长量（y 轴）为纵轴，质量（x 轴）为横轴绘制图表。计算每个质量下的伸长量。你注意到什么规律？预测 250 克时的弹簧长度。这是在物理情境中纯粹的图表和规律（数学）工作。
Extensions: 0 g → 0 cm; 100 g → 2.5 cm; 200 g → 5.0 cm; 300 g → 7.5 cm; 400 g → 10.0 cm. The pattern: extension is directly proportional to mass. For 250 g, extension = 2.5 cm × 2.5 = 6.25 cm, so total length ≈ 10.0 cm + 6.25 cm = 16.25 cm. This interdisciplinary approach is assessed regularly.
伸长量：0 克 → 0 厘米；100 克 → 2.5 厘米；200 克 → 5.0 厘米；300 克 → 7.5 厘米；400 克 → 10.0 厘米。规律：伸长量与质量成正比。250 克时，伸长量 = 2.5 厘米 × 2.5 = 6.25 厘米，因此总长度 ≈ 10.0 厘米 + 6.25 厘米 = 16.25 厘米。这种跨学科技能在考试中经常评估。
12. Key Strategies for Interdisciplinary Success | 跨学科成功的关键策略
Here are five strategies to help you master interdisciplinary questions:
以下是帮助你掌握跨学科题目的五个策略：
a) Identify the disciplines involved – ask yourself: “Is this asking for physics plus something else?”
a) 识别涉及的学科——问问自己：“这道题是否要求物理加上别的什么？”
b) Break the question into small parts. Tackle the pure physics part first, then the maths or the biology part.
b) 把问题分解成小部分。先解决纯粹的物理部分，然后再处理数学或生物部分。
c) Use correct units and convert when necessary. Show all steps clearly.
c) 使用正确的单位，必要时进行换算。清晰写出所有步骤。
d) When writing explanations, include scientific keywords from both subjects (e.g., ‘convection current’, ‘habitat’ in a joint geography–physics answer).
d) 在写解释时，要包含来自两个学科的科学关键词（例如，在地理-物理联合题中用“对流气流”“栖息地”等）。
e) Practise with past Checkpoint papers and look out for questions that mix topics. The more you practise, the better you become at switching between different modes of thinking.
e) 用历届 Checkpoint 真题练习，留意那些混合主题的题目。练得越多，你就越能在不同思维模式之间灵活切换。
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