📚 Interdisciplinary Integrated Question Training for Year 7 CAIE Engineering | 七年级CAIE工程跨学科综合题型训练
Engineering is not just about building things; it requires drawing together knowledge from science, mathematics, and design technology. This article provides a set of integrated question workouts that help Year 7 CAIE learners see how these subjects connect in real-world engineering challenges. Each section presents a typical scenario, followed by step-by-step analysis that blends physics concepts, number work, material choice, and practical thinking.
工程学不仅仅是制造物品，它需要整合科学、数学与设计技术的知识。本文提供了一系列跨学科综合题型训练，帮助七年级 CAIE 学生理解这些学科在真实工程挑战中如何相互关联。每个小节都展示一个典型情境，并通过逐步分析融合物理概念、数字运算、材料选择与实践思维。
1. Forces and Motion: Combining Physics and Maths | 力与运动：结合物理与数学
A remote-controlled car of mass 0.8 kg is pushed for 2 seconds with a constant force of 4 N. First, find the acceleration. Then calculate how fast the car is moving after these 2 seconds. This task links Newton’s second law with basic kinematics, and it requires careful unit handling – a key numerical skill in engineering.
一辆质量为 0.8 kg 的遥控小车被 4 N 的恒力推动 2 秒。先求加速度，再计算 2 秒后小车的速度。这个任务将牛顿第二定律与基础运动学联系起来，并要求仔细处理单位——这是工程中关键的数值技能。
We start from the force formula: F = m × a. Rearranging gives a = F / m. Substituting the values, a = 4 N ÷ 0.8 kg = 5 m/s². The acceleration is 5 metres per second squared. Now we use the equation v = u + a × t, where u = 0 (starting from rest). So v = 0 + 5 m/s² × 2 s = 10 m/s.
我们从力的公式开始：F = m × a。整理后得到 a = F / m。代入数值，a = 4 N ÷ 0.8 kg = 5 m/s²。加速度为每秒每秒 5 米。接下来使用方程 v = u + a × t，其中 u = 0（从静止开始）。因此 v = 0 + 5 m/s² × 2 s = 10 m/s。
The answer shows how a pure physics law (force–mass–acceleration) turns into a numerical prediction about speed. Without mathematics, we could describe motion, but with it, we can design motor power and braking systems precisely.
这个答案展示了纯粹的物理定律（力–质量–加速度）如何转化为对速度的数值预测。没有数学，我们可以描述运动，但有了数学，我们就能精确设计马达功率和刹车系统。
2. Simple Machines: Lever Calculations and Mechanical Advantage | 简单机械：杠杆计算与机械效益
A lever is used to lift a 300 N rock. The rock is placed 0.3 m from the fulcrum, and the effort is applied 1.2 m from the fulcrum on the opposite side. Find the minimum effort needed to balance the lever, and determine the ideal mechanical advantage.
用一根杠杆抬起 300 N 的岩石。岩石离支点 0.3 m，力施加在支点另一侧 1.2 m 处。求平衡杠杆所需的最小力，并计算理想机械效益。
The principle of moments states that for equilibrium, load × load arm = effort × effort arm. Here, load arm = 0.3 m, effort arm = 1.2 m. So 300 N × 0.3 m = effort × 1.2 m. Solving gives effort = (300 × 0.3) ÷ 1.2 = 75 N. The mechanical advantage (MA) is load ÷ effort = 300 N ÷ 75 N = 4. It can also be found from arm lengths: effort arm ÷ load arm = 1.2 ÷ 0.3 = 4.
力矩原理指出，平衡时 载荷 × 载荷臂 = 力 × 力臂。这里载荷臂 = 0.3 m，力臂 = 1.2 m。因此 300 N × 0.3 m = 力 × 1.2 m。解得 力 = (300 × 0.3) ÷ 1.2 = 75 N。机械效益 (MA) 为载荷 ÷ 力 = 300 N ÷ 75 N = 4。也可以从力臂长度比求出：力臂 ÷ 载荷臂 = 1.2 ÷ 0.3 = 4。
This problem connects physics (moments) with ratio and proportion in mathematics. Real levers also have friction, so the actual effort would be slightly higher – an important engineering insight when selecting motors or human operators.
这个问题把物理（力矩）与数学中的比和比例联系起来。真实的杠杆还存在摩擦，因此实际需要的力会略大——这是在选择马达或人力操作时重要的工程洞见。
3. Electrical Circuits: Ohm’s Law and Current Flow | 电路：欧姆定律与电流
A simple series circuit consists of a 9 V battery and a resistor of 4.5 Ω. Calculate the current flowing in the circuit. Then predict what happens to the current if a second identical resistor is added in series.
一个简单串联电路由一个 9 V 电池和一个 4.5 Ω 的电阻组成。计算电路中的电流。然后预测如果再串联一个相同的电阻，电流会如何变化。
Ohm’s law links voltage (V), current (I), and resistance (R): I = V / R. With one resistor, I = 9 V ÷ 4.5 Ω = 2 A. When a second 4.5 Ω resistor is added in series, total resistance becomes 4.5 Ω + 4.5 Ω = 9 Ω. The new current is I = 9 V ÷ 9 Ω = 1 A. The current halves.
欧姆定律将电压 (V)、电流 (I) 和电阻 (R) 联系起来：I = V / R。接一个电阻时，I = 9 V ÷ 4.5 Ω = 2 A。当串联第二个 4.5 Ω 电阻时，总电阻变为 4.5 Ω + 4.5 Ω = 9 Ω。新的电流 I = 9 V ÷ 9 Ω = 1 A。电流减半。
This exercise uses algebraic rearrangement and simple division, then adds a design thinking twist: how to control current by choosing resistor values. Engineering circuits require such calculations to avoid overheating and to ensure components work within safe limits.
这个练习运用了代数变换和简单的除法，并加入了一个设计思维转折：如何通过选择电阻值控制电流。工程电路需要进行这样的计算，以避免过热并确保元器件在安全范围内工作。
4. Material Properties: Choosing the Right Material | 材料性质：选择合适的材料
You need to make a lightweight bridge beam that is 2 m long, must support at least 5000 N without breaking, and must not cost more than £10. The table below gives properties of three candidate materials. Which one would you choose, and why? Use calculations of mass and cost to justify your decision.
你需要制造一根 2 m 长的轻质桥梁梁，必须能承受至少 5000 N 而不断裂，并且成本不能超过 10 英镑。下表列出了三种候选材料的性质。你会选择哪一种？请通过质量和成本计算说明理由。
| Material | Density (g/cm³) | Strength (MPa) | Cost per cm³ (£) |
|---|---|---|---|
| Pine wood | 0.6 | 40 | 0.002 |
| Steel | 7.8 | 250 | 0.015 |
| Aluminium alloy | 2.7 | 200 | 0.010 |
Assume the beam has a volume of 800 cm³. First check strength: all three materials can carry 5000 N because their cross-sectional strength far exceeds the load (stress = force / area, but given MPa values are well above needed stress for typical cross-section). So the decision is driven by mass and cost. Mass = density × volume. Pine mass = 0.6 × 800 = 480 g = 0.48 kg; cost = 800 × 0.002 = £1.60. Steel mass = 7.8 × 800 = 6240 g = 6.24 kg; cost = 800 × 0.015 = £12.00 (over budget). Aluminium mass = 2.7 × 800 = 2160 g = 2.16 kg; cost = 800 × 0.010 = £8.00.
假设梁的体积为 800 cm³。首先检查强度：三种材料都能承载 5000 N，因为它们的截面强度远超所需（应力 = 力 / 面积，给定的 MPa 值对于典型截面来说远高于需要的应力）。因此决策由质量和成本主导。质量 = 密度 × 体积。松木质量 = 0.6 × 800 = 480 g = 0.48 kg；成本 = 800 × 0.002 = £1.60。钢质量 = 7.8 × 800 = 6240 g = 6.24 kg；成本 = 800 × 0.015 = £12.00（超预算）。铝合金质量 = 2.7 × 800 = 2160 g = 2.16 kg；成本 = 800 × 0.010 = £8.00。
Aluminium is the best choice: it is light enough and within budget, while offering high strength. This task blends science (density, strength), mathematics (multiplication, comparison), and design criteria – exactly how engineers select materials in product development.
铝合金是最佳选择：它足够轻且在预算内，同时提供高强度。这个任务融合了科学（密度、强度）、数学（乘法、比较）和设计标准——这正是工程师在产品开发中选择材料的方式。
5. Energy Conversion and Efficiency | 能量转换与效率
An electric motor is used to lift a weight. The electrical energy supplied to the motor is 150 J. The motor does 105 J of useful work lifting the weight. Calculate the efficiency of the motor as a percentage. Also, suggest where the ‘missing’ energy goes and how you might improve the system.
一个电动机用来提升重物。供给电动机的电能为 150 J。电动机做了 105 J 的有用功提升重物。计算电动机的效率（百分比）。同时，指出“消失的”能量去了哪里，以及如何改进系统。
Efficiency formula: η = (useful output energy / total input energy) × 100%. Here, η = (105 J / 150 J) × 100% = 70%. The remaining 45 J is mostly converted into heat due to friction and electrical resistance in the wires. This is an energy loss that engineers always try to minimise.
效率公式：η = (有用输出能量 / 总输入能量) × 100%。这里，η = (105 J / 150 J) × 100% = 70%。剩下的 45 J 大部分因摩擦和导线电阻转化为热能。这是工程师始终试图最小化的能量损失。
To improve efficiency, we could lubricate moving parts, use thicker wires to reduce resistance, or redesign the motor with better magnets. This combines physics (energy conservation) with practical design technology, and the percentage calculation is a straightforward numeracy skill.
要提高效率，我们可以润滑运动部件、使用更粗的电线以减小电阻，或重新设计具有更好磁体的电机。这结合了物理（能量守恒）和实际设计技术，而百分比计算是直接的算术技能。
6. Structural Stability: Shapes and Load Distribution | 结构稳定性：形状与载荷分布
A tower is made from straws and connectors. Two designs are proposed: one uses only square frames, the other uses triangular frames. Explain, using the concept of force distribution, why triangular frames make a stiffer structure. If a side load of 20 N is applied at the top, a triangular tower resists by distributing forces as tension and compression along the members without changing shape. How does this relate to geometry?
用吸管和连接件搭建一座塔。提出了两种设计：一种仅使用方形框架，另一种使用三角形框架。请用力分布的概念解释为什么三角形框架能构成更刚性的结构。如果在塔顶施加 20 N 的侧向载荷，三角形塔通过沿杆件以拉力和压力分布力而不改变形状来抵抗。这与几何学有何关系？
A square can easily deform into a parallelogram because its joints can rotate; changing the angles does not change the side lengths. A triangle, however, is a rigid shape: for a given set of side lengths, the angles are fixed. When a force pushes one vertex, the triangle stays in shape and distributes the load through tension and compression in its sides.
正方形很容易变形为平行四边形，因为其节点可以旋转；改变角度并不改变边长。然而三角形是一种刚性形状：对于一组给定的边长，角度是固定的。当力推动一个顶点时，三角形保持形状，并通过其边的拉力和压力分布载荷。
This concept is used in bridges, cranes, and roof trusses. The mathematical property of triangle rigidity (SSS congruence) directly supports engineering design. Here, learning geometric rules is not just for maths class – it is essential for building safe structures.
这一概念在桥梁、起重机和屋架中都有运用。三角形的几何刚性（边边边全等）直接支撑了工程设计。在这里，学习几何规则不仅仅是为了数学课——它对建造安全的结构至关重要。
7. The Design Process: Integrating Skills Step by Step | 设计过程：逐步整合技能
You are asked to design a mechanical grabber to pick up a tin can from a shelf 1.5 m above the ground. Describe the stages of the design process you would follow, and for each stage mention at least one science or maths skill you would use.
要求你设计一个机械抓手，用来从离地 1.5 m 的架子上抓取一个罐头。描述你将遵循的设计过程各阶段，并在每个阶段至少提及一项你会用到的科学或数学技能。
A typical engineering design cycle includes: 1) Research – measuring arm reach (maths: length estimation) and studying levers (physics: moments). 2) Generate ideas – sketching and applying geometry for pivot positions. 3) Modelling – building a prototype and testing grip force (science: friction needed to hold can). 4) Evaluation – calculating mechanical advantage and comparing with required force (maths: ratio and proportion). 5) Improvement – adjusting materials or pivot lengths based on data.
一个典型的工程设计循环包括：1) 研究——测量手臂伸展范围（数学：长度估计）并研究杠杆（物理：力矩）。2) 产生想法——绘制草图并应用几何学确定支点位置。3) 建模——制作原型并测试抓取力（科学：抓取罐头所需的摩擦力）。4) 评估——计算机械效益并与所需力比较（数学：比和比例）。5) 改进——根据数据调整材料或支点长度。
Throughout, you are not just ‘making’; you are applying Newton’s laws, measurement, and ratio analysis. This problem-based learning mirrors how real engineering teams work, blending disciplines continuously.
在整个过程中，你不仅仅是在“制作”，而是在应用牛顿定律、测量和比率分析。这种基于问题的学习反映了真实工程团队的运作方式，持续融合各学科。
8. Costing a Project: Maths in Engineering Budgets | 项目成本：工程中的预算数学
For a solar oven project, you need the following materials: a cardboard box (£1.20), aluminium foil (0.5 m² at £2.40 per m²), black paper (3 sheets at £1.50 per 10-sheet pack), plastic wrap (£0.80), and glue (£1.10). Calculate the total cost. If you have a budget of £6.00, can you afford it? Show your working.
一个太阳能烤箱项目需要以下材料：一个纸板箱（£1.20）、铝箔（0.5 m²，每平方米 £2.40）、黑纸（3 张，每 10 张一包 £1.50）、保鲜膜（£0.80）和胶水（£1.10）。计算总成本。如果你的预算是 £6.00，你负担得起吗？请展示演算过程。
Aluminium foil cost: 0.5 × £2.40 = £1.20. Black paper: since a pack of 10 costs £1.50, cost per sheet = £1.50 ÷ 10 = £0.15. For 3 sheets: 3 × £0.15 = £0.45. Now sum all costs: £1.20 (box) + £1.20 (foil) + £0.45 (paper) + £0.80 (wrap) + £1.10 (glue) = £4.75. This is less than £6.00, so the project is within budget with £1.25 to spare.
铝箔成本：0.5 × £2.40 = £1.20。黑纸：因为 10 张一包 £1.50，每张成本 = £1.50 ÷ 10 = £0.15。3 张：3 × £0.15 = £0.45。现在汇总所有成本：£1.20 (箱) + £1.20 (箔) + £0.45 (纸) + £0.80 (膜) + £1.10 (胶水) = £4.75。低于 £6.00，因此项目在预算之内，还剩下 £1.25。
Costing is a vital real-world engineering skill. It requires accurate decimal arithmetic and unit price calculations. Even the most brilliant design is useless if it cannot be manufactured within a budget.
成本计算是一项至关重要的实际工程技能。它需要精确的小数运算和单价计算。如果不能在预算内制造，即使最出色的设计也是无用的。
9. Safety and Risk Assessment: Technical and Social Thinking | 安全与风险评估：技术与社会思考
When using a hot glue gun to assemble a model bridge, identify at least three hazards. For each hazard, rate the risk (low/medium/high) and suggest a control measure. This exercise combines technical awareness with decision-making tables often used in engineering.
在使用热熔胶枪组装模型桥时，指出至少三种危害。对每种危害，评定风险级别（低/中/高）并提出一项控制措施。这个练习将技术意识与工程中常用的决策表结合起来。
| Hazard | Risk | Control Measure |
|---|---|---|
| Burns from hot nozzle | High | Use in a stand, keep hands away, wear heat-resistant gloves |
| Dripping hot glue | Medium | Place a protective mat, clear workspace |
| Electrical cord trip | Low | Tape cord to floor, keep walkway clear |
Performing a risk assessment before starting any building task is standard practice. It forces you to think scientifically about cause and effect (the burn is rapid heat transfer) and to apply logical ordering to prevent harm. It also addresses the social responsibility of keeping yourself and others safe.
在开始任何建造任务之前进行风险评估是标准做法。它促使你科学地思考因果关系（烫伤是快速热传递），并运用逻辑顺序来防止伤害。它还涉及对自己和他人安全负责的社会责任。
10. Integrated Challenge: Designing a Spring-Powered Catapult | 综合挑战：设计弹簧动力投石器
A small catapult uses a spring with stiffness k = 200 N/m. You compress the spring by 0.05 m and use it to launch a 20 g (0.02 kg) ball horizontally. Assume all elastic potential energy converts to kinetic energy. Calculate the launch speed of the ball. Then explain one design change that would increase the range without using a stiffer spring.
一个小型投石器使用劲度系数 k = 200 N/m 的弹簧。你将弹簧压缩 0.05 m，用它水平发射一个 20 g (0.02 kg) 的小球。假设所有弹性势能都转化为动能。计算小球的发射速度。然后解释一项在不用更硬弹簧的情况下增加射程的设计变更。
Elastic potential energy stored: Eₑ = ½ × k × x². Substitute: Eₑ = 0.5 × 200 × (0.05)² = 100 × 0.0025 = 0.25 J. This energy becomes kinetic energy of the ball: Eₖ = ½ × m × v². So 0.25 = 0.5 × 0.02 × v², giving v² = 0.25 / 0.01 = 25, hence v = √25 = 5 m/s. (We use ½ as the fraction and × for multiplication.)
储存的弹性势能：Eₑ = ½ × k × x²。代入：Eₑ = 0.5 × 200 × (0
Published by TutorHao | Year 7 工程 Revision Series | aleveler.com
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