📚 Year 7 CIE Engineering: In-depth Analysis of Past Paper Questions | CIE 工程七年级历年真题深度解析
Welcome to this comprehensive analysis of past paper questions for Year 7 CIE Engineering. Designed for students following the Cambridge Lower Secondary or introductory IGCSE Engineering syllabus, this article dissects typical exam questions, explains core concepts, and provides model answers to build confidence and understanding. Each section features a real-exam-style question, followed by a detailed solution and key learning points.
欢迎阅读本次CIE工程七年级历年真题深度解析。本文面向剑桥初中或IGCSE工程入门阶段的学生，精选历年高频考题，逐题剖析核心概念，提供标准答案与解题思路，帮助你建立信心、深化理解。每个小节均包含一道真题风格的题目、详细解答与关键要点。
1. Levers and Mechanical Advantage | 杠杆与机械优势
Question: A first-class lever has an effort arm of 40 cm and a load arm of 10 cm. Calculate the mechanical advantage (MA). If an effort of 25 N is applied, what load can be lifted? (Assume no friction.)
Answer: oad arm = 40 cm / 10 cm = 4. The load that can be lifted = MA × effort = 4 × 25 N = 100 N.
Lever problems test your understanding of moments and mechanical advantage. Always identify which type of lever is used; first-class levers have the fulcrum between the load and effort. The mechanical advantage can also be expressed as load/effort, but here we used arm lengths. No units for MA, it's a ratio. The answer shows that this lever multiplies force by 4.
题目：一个第一类杠杆的动力臂长40厘米，阻力臂长10厘米。计算其机械优势（MA）。如果施加25牛的动力，能举起多重的负载？（假设无摩擦。）
解答：MA = 动力臂 / 阻力臂 = 40 cm / 10 cm = 4。可举起的负载 = MA × 动力 = 4 × 25 N = 100 N。
杠杆问题测试你对力矩和机械优势的理解。首先判别杠杆类型：第一类杠杆支点在负载和动力之间。机械优势亦可用负载/动力计算，此处用臂长比。注意MA无单位。答案显示该杠杆可将力放大4倍。
2. Identifying Forces in Structures | 结构中的力识别
Question: A simple bridge truss is shown (not drawn). Identify the type of force (tension, compression, or torsion) experienced by the top chord when a load is applied at the centre. Explain your reasoning.
Answer: The top chord experiences compression. When a downward load is applied to the bridge, the top members are pushed together, trying to shorten. The bottom chord experiences tension as it is pulled. This is typical for simply supported beams. Torsion is twisting, not present here.
题目：一座简单桥梁桁架如图所示（未画出）。当中心施加载荷时，上弦杆承受哪种力（拉伸、压缩或扭转）？请解释理由。
解答：上弦杆承受压缩。当向下的载荷作用于桥面时，顶部构件被挤压趋向缩短，故为压缩。下弦杆被拉而承受拉伸。简支梁典型如此。扭转属扭曲力，此处不存在。
3. Electrical Circuits: Series vs Parallel | 电路：串联与并联
Question: Two identical bulbs are connected in series to a 6 V battery. Describe the brightness of each bulb compared to a single bulb connected to the same battery. What happens if one bulb blows? Now repeat for a parallel connection.
Answer: In series, current is the same through both bulbs but total resistance doubles, so current halves. Each bulb gets about 3 V (half voltage) and glows dimly. If one blows, the circuit breaks and both go out. In parallel, each bulb gets the full 6 V, so brightness is normal. If one blows, the other remains on. Series circuits have voltage division; parallel circuits have independent branches.
题目：两个相同灯泡串联接至6 V电池。描述与单个灯泡接同一电池时相比，各灯泡亮度如何？若一盏灯泡烧坏会怎样？再回答并联接法。
解答：串联时两灯泡电流相同，但总电阻翻倍，电流减半。每灯分得约3 V（一半电压），变暗。若一灯断路，电路断开，两灯均灭。并联时各灯均获6 V全压，亮度正常。一灯坏，另一灯仍亮。串联分压；并联支路独立。
4. Material Properties: Hardness and Toughness | 材料特性：硬度与韧性
Question: Explain the difference between hardness and toughness. Give an example of a material that is hard but not tough, and a material that is tough but not hard.
Answer: Hardness is resistance to surface indentation or scratching. Toughness is the ability to absorb energy before fracturing, i.e., resistance to crack propagation. Glass is hard (can scratch) but brittle (not tough); it shatters easily. Rubber is tough (can stretch and absorb energy) but not hard (easily scratched). Engineers select materials based on these properties for tools, safety equipment, etc.
题目：解释硬度与韧性的区别。给出一种硬而不韧的材料例子，以及一种韧而不硬的例子。
解答：硬度指抵抗表面压入或划伤的能力。韧性指断裂前吸收能量的能力，即抵抗裂纹扩展。玻璃硬（可划伤它物）但脆（不韧），易碎。橡胶韧（可拉伸吸能）但不硬（易被划伤）。工程师根据这些特性选用材料，如工具需硬，安全装置需韧。
5. Gear Ratios and Speed | 齿轮比与速度
Question: A driving gear with 20 teeth meshes with a driven gear of 60 teeth. If the driving gear rotates at 90 rpm, calculate the speed of the driven gear and the gear ratio. State whether this is a speed multiplier or torque multiplier.
Answer: 0 rpm. Since the output speed is lower than input, this is a torque multiplier (increases torque, reduces speed). A gear ratio > 1 means torque increase. Note: if driver smaller, driven slower, torque greater.
题目：主动齿轮20齿，从动齿轮60齿。若主动轮转速90 rpm，计算从动轮转速及齿轮比。并说明系统是增速还是增扭。
解答：齿轮比 = 从动齿数/主动齿数 = 60/20 = 3:1。从动轮转速 = 主动转速/齿轮比 = 90 rpm / 3 = 30 rpm。输出转速低于输入，故为增扭器（扭矩增大，转速降低）。齿轮比>1表示扭矩放大。小齿轮带动大齿轮，减速增扭。
6. Design Process: CAD and CAM | 设计过程：CAD与CAM
Question: Define CAD and CAM and explain how they work together in modern manufacturing.
Answer: CAD (Computer-Aided Design) is software used to create precise 2D or 3D models. CAM (Computer-Aided Manufacturing) uses software to control machine tools, like CNC mills or 3D printers. The CAD model is exported to CAM software, which generates toolpaths and G-code. This integration reduces errors, speeds up prototyping, and allows complex geometries. Common in engineering industries.
题目：定义CAD与CAM，并解释二者如何协同用于现代制造。
解答：CAD（计算机辅助设计）是用软件创建精确2D或3D模型。CAM（计算机辅助制造）以软件控制机床，如CNC铣床或3D打印机。CAD模型导出至CAM软件，生成刀具路径和G代码。这种集成减少错误、加速原型制造、实现复杂形状，广泛用于工程领域。
7. Energy Forms and Conversions | 能量形式与转换
Question: A wind turbine converts wind energy into electrical energy. Describe the energy transformations involved from wind to electricity. Include the names of components.
Answer: enerator. The gearbox increases rotational speed for the generator. The generator uses electromagnetic induction to convert mechanical rotation into electricity. There are losses as heat and sound. Key components: blades, gearbox, generator, tower.
题目：风力发电机把风能转为电能。描述从风到电的能量转换过程，并提及主要部件。
解答：风动能 → 叶片旋转动能 → 转轴机械能 → 发电机内电能。齿轮箱提高转速传递给发电机。发电机利用电磁感应将机械转动转化为电能。部分能量损失为热与声。主要部件：叶片、齿轮箱、发电机、塔架。
8. Pulley Systems and Efficiency | 滑轮系统与效率
Question: ocity ratio (VR), mechanical advantage (MA), and efficiency. (Take g = 10 m/s² if needed.)
Answer: y load = 4 m / 1 m = 4. MA = load / effort = 90 N / 30 N = 3. Efficiency = (MA / VR) × 100% = (3/4) × 100% = 75%. The system loses energy due to friction. Always compare MA to VR for efficiency.
题目：一两轮滑轮组，动力绳移动4米使负载升高1米。动力30 N，负载90 N。计算速比（VR）、机械优势（MA）和效率。（若需g取10 m/s²）
解答：VR = 动力移动距离/负载移动距离 = 4 m / 1 m = 4。MA = 负载/动力 = 90 N / 30 N = 3。效率 = (MA/VR) × 100% = (3/4) × 100% = 75%。系统因摩擦损失能量。效率恒为MA与VR之比。
9. Measurement and Units in Engineering | 工程测量与单位
Question: Convert the following: (a) 2.5 km to metres, (b) 450 g to kg, (c) 12 cm² to m². State the SI base units for length, mass, and time.
Answer: e units: metre (m), kilogram (kg), second (s). Be careful with area conversions; square the conversion factor.
题目：单位换算：(a) 2.5 km为米，(b) 450 g为千克，(c) 12 cm²为平方米。并写出长度、质量和时间的国际基本单位。
解答：(a) 2.5 km = 2500 m。(b) 450 g = 0.45 kg。(c) 1 cm² = (0.01 m)² = 10⁻⁴ m²，所以12 cm² = 12×10⁻⁴ m² = 0.0012 m²。国际基本单位：米(m)、千克(kg)、秒(s)。面积换算需平方进率。
10. Safety in the Workshop | 车间安全
Question: List four personal protective equipment (PPE) items required when using a pillar drill and explain why each is important.
Answer: 1. Safety goggles – protect eyes from flying chips/swarf. 2. Ear defenders – reduce noise exposure. 3. Apron/overalls – protect clothing and skin from hot sharp debris. 4. Steel-toe boots – protect feet from falling heavy objects. Additionally, tie back long hair, remove loose jewelry. Always follow safety rules to prevent accidents.
题目：列出使用台式钻床时所需的四种个人防护装备（PPE），并解释各自的重要性。
解答：1. 护目镜 – 防止飞屑溅入眼睛。2. 耳罩 – 减少噪声伤害。3. 工作围裙/防护服 – 保护衣物与皮肤免被热、锋利碎屑伤及。4. 钢头安全鞋 – 防止重物坠落伤脚。此外应束发、摘除松垮首饰。时刻遵守安全规范，预防事故。
Published by TutorHao | Engineering Revision Series | aleveler.com
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