📚 Year 10 Cambridge Physics: Interdisciplinary Practice | 剑桥10年级物理：跨学科综合题型训练
In the Cambridge IGCSE Physics syllabus, you are often expected to apply physical principles to problems that draw on knowledge from mathematics, chemistry, biology, geography, and even music. These interdisciplinary practice questions not only deepen your understanding of physics but also train you to think like a scientist who connects multiple fields. The ability to move confidently between subjects is a core skill assessed through scenario-based and data-analysis questions. This article presents a selection of such cross-curricular problem types with worked examples, explanations, and strategies.
在剑桥 IGCSE 物理大纲中，你经常需要将物理原理应用于涉及数学、化学、生物、地理甚至音乐等多学科知识的题目中。这些跨学科综合题型不仅能加深你对物理的理解，还能训练你像一名会连接多个领域的科学家一样思考。自信地在学科之间迁移是一个核心技能，常通过情景题和数据分析题来考查。本文精选了这类跨学科题型并附有示例、讲解与解题策略。
1. Motion and Graphical Analysis (Maths) | 运动与图形分析（数学）
A velocity–time graph provides direct links to the mathematics of gradients and areas under curves. The slope gives the acceleration, and the area under the graph represents displacement. These ideas are identical to finding the derivative and the definite integral in calculus, making this a natural interdisciplinary bridge.
速度–时间图直接联系了关于斜率和曲线下面积的数学知识。斜率给出加速度，图线下的面积表示位移。这些概念与微积分中的导数和定积分完全一致，因此这门学科自然成了跨学科的桥梁。
Example problem: m s⁻¹; t = 15 s, v = 5 m s⁻¹; t = 20 s, v = 0 m s⁻¹. The cyclist moves in a straight line.
典型例题：一名自行车运动员的速度数据如下：在 t = 0 s 时 v = 0 m s⁻¹；t = 5 s 时 v = 10 m s⁻¹；t = 10 s 时 v = 10 m s⁻¹；t = 15 s 时 v = 5 m s⁻¹；t = 20 s 时 v = 0 m s⁻¹。运动员沿直线运动。
We plot the velocity–time graph. From 0–5 s the graph is a straight line; the acceleration is uniform. The slope is (10 – 0) m s⁻¹ / (5 – rapezium under this segment gives displacement for the first 5 s: (1/2)×(0 + 10)×5 = 25 m.
我们画出速度–时间图。0–5 s 内图像是一条直线，加速度均匀。斜率 = (10 – 0) m s⁻¹ / (5 – 0) s = 2 m s⁻²。此段图线下的梯形面积为前 5 s 的位移：(1/2)×(0 + 10)×5 = 25 m。
From 5– . The displacement for this interval is the rectangle area: 10 m s⁻¹ × 5 s = 50 m. From 10–15 s the velocity decreases linearly; acceleration = (5 – 10) / (15 – r this triangle and rectangle gives displacement = (10+5)/2 × 5 = 37.5 m. Finally, from 15– is the same value you would obtain by applying the geometric formula for the area of a trapezium for the entire graph.
5–10 s 内速度不变，加速度为零。该段时间位移为矩形面积：10 m s⁻¹ × 5 s = 50 m。10–15 s 内速度均匀减小；加速度 = (5 – 10) / (15 – 10) = −1 m s⁻²。此段包含三角形与矩形，位移 = (10+5)/2 × 5 = 37.5 m。最后 15–20 s 速度从 5 降至 0，加速度 = −1 m s⁻²，位移 = (1/2)×5×5 = 12.5 m。总位移 = 25 + 50 + 37.5 + 12.5 = 125 m。该值与用整个图形梯形面积公式求得的结果完全一致。
This combination of graphical skills and algebraic calculation highlights the close synergy between physics and mathematics.
这种图形技巧与代数计算的结合，凸显了物理与数学之间的紧密协同。
2. Energy Changes and Chemical Reactions (Chemistry) | 能量变化与化学反应（化学）
Thermochemistry links the concepts of internal energy, enthalpy, and the law of conservation of energy. When a fuel burns, chemical potential energy is converted to thermal energy, and we can calculate efficiency using the standard formula from physics.
热化学将内能、焓变和能量守恒定律联系起来。燃料燃烧时，化学势能转化为热能，我们可以利用物理学中的标准公式计算效率。
Example: he mass of ethanol burned is 0.92 g. The specific heat capacity of water is 4.2 J g⁻¹ °C⁻¹. The enthalpy of combustion of ethanol is −1367 kJ mol⁻¹. (Mᵣ of ethanol = 46).
例题：一个装有乙醇（C₂H₅OH）的酒精灯被用来加热 200 g 水。水的初始温度为 22 °C，最终温度为 58 °C。燃烧掉的乙醇质量为 0.92 g。水的比热容为 4.2 J g⁻¹ °C⁻¹。乙醇的燃烧焓为 −1367 kJ mol⁻¹。（乙醇相对分子质量 = 46）
First, calculate the thermal energy gained by the water using Q = m c Δθ. Δθ = 58 – 22 = 36 °C. Qwater = 200 g × 4.2 J g⁻¹ °C⁻¹ × 36 °C = 30 240 J = 30.24 kJ.
首先，利用 Q = m c Δθ 计算水获得的热能。Δθ = 58 – 22 = 36 °C。Q_水 = 200 g × 4.2 J g⁻¹ °C⁻¹ × 36 °C = 30 240 J = 30.24 kJ。
eoretical energy released = 0.020 mol × 1367 kJ mol⁻¹ = 27.34 kJ. (We take the magnitude).
接着，计算乙醇释放的能量。乙醇物质的量 = 质量 / 相对分子质量 = 0.92 g / 46 g mol⁻¹ = 0.020 mol。理论上释放的能量 = 0.020 mol × 1367 kJ mol⁻¹ = 27.34 kJ（取绝对值）。
Efficiency = (useful energy output / total energy input) × 100%
η = (Qwater / Qfuel) × 100% = (30.24 / 27.34) × 100% ≈ 110.6%
This result appears to exceed 100%, which suggests systematic errors such as an underestimated fuel mass or the spirit burner absorbed additional water vapour, illustrating how physics and chemistry together demand careful evaluation of experimental procedures.
结果似乎超过 100%，这表明存在系统误差，例如燃料质量称量偏小，或酒精灯吸收了额外的水蒸气。这正好说明物理与化学共同要求我们仔细评估实验过程。
3. Optics and the Human Eye (Biology) | 光学与人眼（生物）
The eye is a natural optical instrument. Light is refracted by the cornea and the lens to form a focused image on the retina. Long-sightedness (hyperopia) and short-sightedness (myopia) are corrected using convex and concave lenses, directly applying the lens formula and the concept of focal length from physics.
眼睛是一个天然的光学仪器。光线通过角膜和晶状体折射后在视网膜上形成清晰的像。远视和近视分别用凸透镜和凹透镜来矫正，直接应用了物理中的透镜公式和焦距概念。
Problem: A person cannot see objects closer than 80 cm clearly. Calculate the power and focal length of the lens required to correct her near point to 25 cm. Assume the lens is placed 2 cm from the eye.
问题：某人无法看清 80 cm 以内的物体。要将其近点矫正到 25 cm，求所需透镜的焦距和光焦度。假设镜片距眼 2 cm。
ich is 80 cm from the eye (78 cm from the lens). Using the lens equation 1/f = 1/v + 1/u, with sign convention: u = object distance = +23 cm, v = image distance = –78 cm (virtual image on same side as object).
物理告诉我们，透镜应使放在眼前 25 cm（距镜片 23 cm）的物体在她的非矫正近点（距眼 80 cm，即距镜片 78 cm）处成一个虚像。使用透镜公式 1/f = 1/v + 1/u，按符号规则：u = 物距 = +23 cm，v = 像距 = –78 cm（虚像与物同侧）。
1/f = 1/(+23) + 1/(–78) = 1/23 – 1/78
= +1 / 0.326 ≈ +3.07 D. The positive sign indicates a converging (convex) lens, which is used to correct hyperopia.
计算：通分 23×78 = 1794。1/23 ≈ 0.04348, 1/78 ≈ 0.01282, 差 ≈ 0.03066 cm⁻¹。f ≈ +32.6 cm = +0.326 m。光焦度 P = 1/f（以米为单位）= +1/0.326 ≈ +3.07 D。正号表示会聚（凸透镜），正用于矫正远视。
This integrated problem uses biological knowledge of the eye and the physics of lenses. Understanding such crossovers is essential for medical physics questions in Cambridge IGCSE.
这道综合题用到了眼睛的生物学知识与透镜的物理原理。理解这种交叉对于剑桥 IGCSE 中的医学物理题至关重要。
4. Thermal Conduction and Earth's Interior (Geography) | 热传导与地球内部（地理）
The Earth's internal heat is transferred through the mantle mainly by conduction and convection. Physics explains how the rate of heat transfer depends on the thermal conductivity of rocks, while geography describes tectonic activity driven by these heat flows.
地球内部热量主要通过传导和对流在地幔中传递。物理知识解释了热传递速率如何取决于岩石的导热系数，而地理则描述了由这些热流驱动的板块构造活动。
Data analysis: A borehole in a continental crust region shows a temperature increase from 15 °C at the surface to 150 °C at a depth of 3 km. The average thermal conductivity of granite is 2.5 W m⁻¹ K⁻¹. Estimate the geothermal heat flux and comment on its significance for geothermal energy.
数据分析：在大陆地壳某地区的一个钻孔显示，从地表 15 °C 到 3 km 深处温度升至 150 °C。花岗岩的平均导热系数为 2.5 W m⁻¹ K⁻¹。估算地热热流密度，并讨论其对地热能的意义。
The temperature gradient = ΔT / Δz = (150 – 15) K / 3000 m = 135 / 3000 = 0.045 K m⁻¹. Fourier's law of heat conduction gives heat flux q = – hy, such heat flow is linked to the thickness of the lithosphere and can indicate areas suitable for low-enthalpy geothermal energy extraction. Physics provides the quantitative tool to assess renewable energy potential.
地温梯度 = ΔT / Δz = (150 – 15) K / 3000 m = 135 / 3000 = 0.045 K m⁻¹。傅里叶导热定律给出热流密度 q = –k (dT/dz)。大小为 q = k × 温度梯度 = 2.5 × 0.045 = 0.1125 W m⁻²。这是稳定大陆区域的典型值。在地理中，这种热流与岩石圈厚度有关，并可指示适合低焓地热能开发的地区。物理则为评估可再生能源潜力提供了定量工具。
5. Electrical Circuits and Algebraic Equations (Maths) | 电路与代数方程（数学）
Solving series and parallel circuits requires systematic use of algebraic equations such as Ohm's law and Kirchhoff's rules. This is a direct application of simultaneous linear equations from mathematics.
求解串并联电路需要系统运用欧姆定律和基尔霍夫定律等代数方程。这正是数学中联立一次方程的直接应用。
Example: attery is connected across the whole network. Determine the current through each resistor.
例题：两个电阻 R₁ = 6 Ω 和 R₂ = 3 Ω 并联后，再与另一个电阻 R₃ = 4 Ω 串联。整个网络两端接有 12 V 电池。求通过每个电阻的电流。
1/R_parallel = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so R_parallel = 2 Ω. Total resistance Rtotal = R_parallel + R₃ = 2 + 4 = 6 Ω. Total current Itotal = V / Rtotal 2 A. This current passes through R₃, so I₃ = 2 A. The voltage across the parallel pair = Itotal h the total current. Algebraic manipulation of fractions is essential here, linking mathematical confidence with circuit analysis.
并联部分：1/R_并 = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2，所以 R_并 = 2 Ω。总电阻 R_总 = R_并 + R₃ = 2 + 4 = 6 Ω。总电流 I_总 = V / R_总 = 12 V / 6 Ω = 2 A。该电流通过 R₃，故 I₃ = 2 A。并联部分两端电压 = I_总 × R_并 = 2 × 2 = 4 V。因此 R₁ 和 R₂ 两端电压均为 4 V。I₁ = 4 V / 6 Ω = 2/3 A ≈ 0.667 A；I₂ = 4 V / 3 Ω = 4/3 A ≈ 1.333 A。检验：I₁ + I₂ = 2 A，与总电流一致。此处必须熟练处理分数，这恰好将数学自信心与电路分析联系了起来。
6. Levers and the Musculoskeletal System (Biology/PE) | 杠杆与肌肉骨骼系统（生物/体育）
Biomechanics applies the principle of moments to the human body. Bones act as levers, joints as fulcrums, and muscles provide effort forces. This is a perfect integration of physics and biological movement.
生物力学将力矩原理应用于人体。骨骼充当杠杆，关节为支点，肌肉提供动力。这是物理与生物运动学的完美结合。
Scenario: the forearm plus hand (mass 2 kg) is 15 cm from the elbow. The dumbbell is 35 cm from the elbow. Calculate the force exerted by the biceps muscle to maintain equilibrium. (g = 10 N kg⁻¹)
情景：某人手持 5 kg 哑铃，前臂保持水平。肱二头肌附着在距离肘关节 4 cm 处，前臂和手的质心（质量 2 kg）距肘关节 15 cm。哑铃距肘关节 35 cm。计算肱二头肌维持平衡所需的作用力。（g = 10 N kg⁻¹）
nt = 20 N × 0.15 m = 3.0 N m. Dumbbell weight = 5 kg × 10 = 50 N, moment = 50 × 0.35 = 17.5 N m. Total clockwise moment = 20.5 N m. The biceps exerts an upward force F at distance 0.04 m, producing an anticlockwise moment F × 0.04. For equilibrium: F × 0.04 = 20.5, so F = 20.5 / 0.04 = 512.5 N. This force is much larger than the total weight supported (70 N), demonstrating the mechanical disadvantage of a third-class lever but allowing a large range of motion—a key biological adaptation.
以肘关节为支点取矩。顺时针力矩由前臂重力和哑铃重力产生。前臂重力 = 2 kg × 10 = 20 N，力矩 = 20 N × 0.15 m = 3.0 N m。哑铃重力 = 50 N，力矩 = 50 × 0.35 = 17.5 N m。总顺时针力矩 = 20.5 N m。肱二头肌在 0.04 m 处向上施力 F，产生逆时针力矩 F × 0.04。平衡时：F × 0.04 = 20.5，得 F = 20.5 / 0.04 = 512.5 N。该力远大于所支撑的总重量（70 N），这体现了第三类杠杆的机械劣势，但它换来了大的运动范围——一项关键的生物适应性。
7. Sound Waves and Musical Instruments (Music) | 声波与乐器（音乐）
The physics of standing waves in strings and air columns determines the pitch of musical notes. The frequency depends on the length, tension, and mass per unit length of a string, or on the length of an air column. Music theory uses these physical relationships.
弦和空气柱中驻波的物理原理决定了乐音的音高。频率取决于弦的长度、张力和线密度，或者空气柱的长度。音乐理论正是建立在这些物理关系之上。
Problem: n in the string. The formula for the fundamental frequency of a fixed string is f = (1/2L) √T/μ, where T is tension and μ is linear density.
问题：一根吉他弦长 0.65 m，线密度为 0.75 g m⁻¹。其基频为 196 Hz（音符 G）。求弦中的张力。固定弦的基频公式为 f = (1/2L) √T/μ，其中 T 为张力，μ 为线密度。
e L² = 0.4225, f² = 38416. T = 4 × 0.4225 × 38416 × 0.00075 = 4 × 0.4225 × 28.812 ≈ 4 × 12.176 ≈ 48.7 N. This mathematical manipulation shows the direct link between a physical variable (tension) and the musical output. A musician tuning the guitar adjusts tension to change frequency, a perfect real-world example of physics in the arts.
先将 μ 转换为 kg m⁻¹：0.75 g m⁻¹ = 0.00075 kg m⁻¹。变换公式得 T = 4 L² f² μ。代入：T = 4 × (0.65 m)² × (196 Hz)² × 0.00075 kg m⁻¹。计算 L² = 0.4225，f² = 38416。T = 4 × 0.4225 × 38416 × 0.00075 = 4 × 0.4225 × 28.812 ≈ 4 × 12.176 ≈ 48.7 N。该数学运算展示了物理量（张力）与音乐输出之间的直接联系。音乐家调音时通过调整张力来改变频率，这正是艺术中物理原理的完美实例。
8. Energy Resources and Carbon Footprint (Environmental Science) | 能源资源与碳足迹（环境科学）
Evaluating energy resources involves physics (energy calculations, efficiency) and environmental science (greenhouse gas emissions, sustainability). Students must interpret numerical data to make informed choices about wind, solar, fossil, and nuclear options.
评估能源资源需要物理（能量计算、效率）和环境科学（温室气体排放、可持续性）的知识。学生必须解读数值数据，从而对风能、太阳能、化石燃料与核能做出明智的选择。
Table of data:
| Energy Source | Power Output (MW) | Capacity Factor (%) | CO₂ Emissions (g kWh⁻¹) |
| Coal | 3000 | 85 | 900 |
| Onshore Wind | 500 | 30 | 10 |
| Solar PV | 150 | 15 | 40 |
Task: Calculate the annual energy generated by each installation in GWh, and the annual CO₂ avoided if wind displaces an equivalent energy from coal.
任务： Published by TutorHao | Year 10 Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图