📚 Year 7 AQA Engineering: In-depth Analysis of Past Exam Questions | 7年级AQA工程：历年真题深度解析
Year 7 Engineering exams under the AQA framework focus on core principles: forces, materials, structures, mechanical systems and the design process. By examining real past‑paper questions, students can identify recurring themes, understand what examiners expect and build confidence. This article presents a curated selection of typical questions with detailed answers and mark‑scheme commentary, providing a comprehensive revision tool for young engineers.
AQA 7年级工程考试围绕核心原理展开：力、材料、结构、机械系统与设计流程。通过分析真实的历年真题，学生能够发现常考主题，理解评分要求并建立自信。本文精选了典型题目，附有详细答案与评分标准解析，为小工程师们提供一份全面的复习工具。
1. Forces and Levers | 力与杠杆
Exam Question: A first‑class lever has a fulcrum placed 0.3 m from a load of 150 N. The effort is applied 0.9 m from the fulcrum. State the class of lever, calculate the mechanical advantage and explain how the lever makes work easier. (5 marks)
考试题目：一个第一类杠杆的支点距离 150 N 的负载 0.3 m，施力点距离支点 0.9 m。指出杠杆类别，计算机械效益，并解释该杠杆如何使工作更省力。（5分）
This is a first‑class lever because the fulcrum lies between the load and the effort. First‑class levers can multiply force or distance, and here the effort arm is longer, so the lever acts as a force multiplier.
这是一个第一类杠杆，因为支点位于负载与施力之间。第一类杠杆可以增大力或增大距离，此处施力臂较长，因此该杠杆起着增力器的作用。
Mechanical Advantage = Load ÷ Effort
机械效益 = 负载 ÷ 施力
N ÷ 45 N = 3.33. The lever reduces the required effort by a factor of about 3.3, meaning the user applies only 45 N to lift 150 N.
先用力矩原理求施力：负载 × 负载臂 = 施力 × 施力臂。150 N × 0.3 m = 施力 × 0.9 m → 施力 = 45 N。然后 MA = 150 N ÷ 45 N ≈ 3.33。杠杆将所需施力减小了约 3.3 倍，即使用者只需施加 45 N 即可举起 150 N。
Mark scheme insight: 1 mark for identifying the class, 1 for stating the moment equation, 1 for correct effort, 1 for correct MA, 1 for explaining force multiplication. Many students lose marks by forgetting to convert the MA concept into a practical explanation.
评分标准解读：正确指出类别得1分，写出力矩关系式得1分，施力计算正确得1分，MA正确得1分，解释力放大作用得1分。许多学生因未能将机械效益的概念转化为实际解释而失分。
2. Mechanical Advantage Calculations | 机械效益计算
Exam Question: A wheelbarrow carries a load of 400 N. The wheel acts as the fulcrum, the load is 0.4 m from the wheel, and the handles are 1.2 m from the wheel. Calculate the effort required and state why the wheelbarrow is classified as a class‑2 lever. (4 marks)
考试题目：一辆手推车承载 400 N 的负载。车轮作为支点，负载距离车轮 0.4 m，手柄距离车轮 1.2 m。计算所需施力，并说明为何手推车属于第二类杠杆。（4分）
The wheelbarrow is a class‑2 lever because the load sits between the fulcrum (wheel) and the effort (handles). In class‑2 levers, the effort arm is always longer than the load arm, yielding a mechanical advantage greater than 1.
手推车属于第二类杠杆，因为负载在支点（车轮）与施力（手柄）之间。第二类杠杆的施力臂始终长于负载臂，因此机械效益总大于1。
Load × Load arm = Effort × Effort arm
负载 × 负载臂 = 施力 × 施力臂
400 N × 0.4 m = Effort × 1.2 m → Effort = 160 N ÷ 1.2 ≈ 133.3 N. The effort is one‑third of the load, illustrating the advantage of placing the load near the fulcrum.
400 N × 0.4 m = 施力 × 1.2 m → 施力 = 160 N ÷ 1.2 ≈ 133.3 N。施力仅为负载的三分之一，体现了将负载靠近支点放置的优势。
Examiner’s comment: Students often struggle to distinguish between the effort arm and the distance from fulcrum to handle. Remember that the arm length is measured perpendicularly from the fulcrum to the line of action of the force.
考官点评：学生经常混淆施力臂与支点到手柄的距离。记住，力臂是从支点到力的作用线的垂直距离。
3. Gears and Motion | 齿轮与运动
Exam Question: A driver gear with 20 teeth turns a driven gear with 60 teeth. If the driver rotates at 90 revolutions per minute (rpm), calculate the speed of the driven gear and describe the direction of rotation. (3 marks)
考试题目：一个主动齿轮有 20 齿，带动一个 60 齿的从动齿轮。如果主动轮转速为 90 转/分 (rpm)，计算从动轮的转速，并说明旋转方向。（3分）
tates more slowly. Speed of driven gear = driver speed ÷ gear ratio = 90 rpm ÷ 3 = 30 rpm.
齿轮比为从动齿数 ÷ 主动齿数 = 60 ÷ 20 = 3。因齿比大于1，此为减速传动：从动轮转速较慢。从动轮转速 = 主动轮转速 ÷ 齿比 = 90 rpm ÷ 3 = 30 rpm。
Direction: when two external gears mesh, the driven gear rotates in the opposite direction to the driver. So if the driver turns clockwise, the driven turns anticlockwise.
方向：两外啮合齿轮传动时，从动轮旋转方向与主动轮相反。若主动轮顺时针旋转，则从动轮逆时针旋转。
Common mistake: Forgetting that an idler gear between them would make both driver and final gear rotate in the same direction. Examiners often award a mark for stating “opposite direction” explicitly.
常见错误：忘记中间加入惰轮会使主动轮与最终从动轮同向旋转。考官通常对明确说出 “方向相反” 给一分。
4. Material Properties | 材料特性
Exam Question: A manufacturer is choosing a material for a bicycle frame. The frame must be strong, lightweight and resist corrosion. Evaluate the suitability of aluminium and steel by referring to at least two properties of each. (6 marks)
考试题目：某制造商为自行车车架选材。车架需强度高、重量轻且耐腐蚀。通过至少两种特性，评价铝和钢的适用性。（6分）
| Property | Aluminium | Steel |
| Density | Low (~2.7 g/cm³) | High (~7.8 g/cm³) |
| Tensile strength | Moderate | High |
| Corrosion resistance | Excellent (forms oxide layer) | Poor (rusts without coating) |
| Cost | Higher | Lower |
中文对照表：
| 特性 | 铝 | 钢 |
| 密度 | 低 (约2.7 g/cm³) | 高 (约7.8 g/cm³) |
| 抗拉强度 | 中等 | 高 |
| 耐腐蚀性 | 优良（形成氧化膜） | 差（无涂层易生锈） |
| 成本 | 较高 | 较低 |
Aluminium scores highly on weight and corrosion resistance, making it ideal for lightweight performance bikes. However, its moderate strength may require thicker tubes, slightly offsetting the weight saving. Steel provides exceptional strength and toughness, and it is cheaper, but the density makes the frame heavier and it must be painted or galvanised to prevent rust.
铝在重量轻和耐腐蚀性方面优势显著，适合高性能轻型自行车。但强度中等可能需要更厚的管材，稍微抵消重量优势。钢具有优异的强度和韧性，且成本较低，但密度大导致车架较重，并且必须涂漆或电镀以防生锈。
Examiner tip: To reach high marks, compare the trade‑offs rather than just listing properties. Use phrases like “although aluminium is lighter, steel offers higher strength per cost.”
考官提示：欲得高分，应比较权衡关系而非简单罗列特性。使用诸如“虽然铝更轻，但钢在单位成本上提供更高强度”的表述。
5. Structures and Triangular Bracing | 结构与三角支撑
Exam Question: A square frame made of four bars pin‑jointed at the corners easily deforms when pushed sideways. Explain why adding a diagonal brace prevents this deformation. Use the terms tension, compression and triangle in your answer. (4 marks)
考试题目：一个由四根杆件铰接而成的方形框架，在侧推时容易变形。解释为何增加一根斜撑能够防止变形。回答中请使用拉力、压力和三角形等术语。（4分）
A square can change shape because its angles can vary while the side lengths stay the same, making it a non‑rigid mechanism. When a diagonal brace is added, the square is divided into two triangles. Triangles are inherently stable shapes whose angles cannot alter without changing side lengths. Under lateral load, one diagonal of the frame experiences tension (being pulled) while the other would be in compression (being pushed); by installing a brace, that tension/compression is resisted by the material’s strength, locking the geometry.
方形能够改变形状，因为其角度可变而边长不变，故为非刚性机构。添加一根斜撑后，方形被分为两个三角形。三角形本身是稳定形状，其角度在不改变边长时无法变动。在侧向载荷下，框架的一条对角线承受拉力（被拉），另一条承受压力（被推）；安装斜撑后，材料强度抵抗该拉力/压力，从而锁定几何形状。
Mark allocation: 1 mark for stating that a square is not rigid, 1 mark for explaining that a triangle is stable, 1 mark for correctly using tension/compression, 1 mark for linking the brace to resisting these forces.
分数分配：指出方形非刚性得1分，解释三角形稳定得1分，正确使用拉力/压力得1分，联系斜撑抵抗这些力得1分。
6. Basic Electrical Circuits | 基础电路
Exam Question: Draw a series circuit containing a battery, a switch, a lamp and a fixed resistor. Then explain what happens to the brightness of the lamp if a second identical lamp is added in series. (5 marks)
考试题目：绘制一个包含电池、开关、一个灯泡和一个固定电阻的串联电路。然后解释如果再串联一个相同的灯泡，前一个灯泡的亮度会发生什么变化。（5分）
Adding a second lamp in series increases the total resistance of the circuit. In a series circuit, the current is the same everywhere and is given by I = V ÷ Rtotal. Since the resistance doubles (each lamp has the same resistance R, so Rtotal = R + Rfixed + Rlamp2 ent drops, power decreases, so both lamps glow dimmer than the single lamp did originally.
串联增加第二个灯泡会使电路总电阻增大。串联电路中电流处处相等，并由 I = V ÷ R总 决定。由于电阻加倍（假设每个灯泡电阻相同为R，则 R总 = R + R固定 + R灯泡2），电流减小。灯泡亮度取决于其消耗的功率 P = I²R。电流下降，功率减小，因此两个灯泡都比原先单独一个灯泡更暗。
Drawing expectations: The circuit diagram must use correct symbols: battery (two cells, long line positive), switch (break in a line), lamp (circle with a cross) and resistor (rectangle). Neatness and straight lines are expected; labels are not mandatory but can help.
绘图要求：电路图须使用正确符号：电池（两个单元，长线为正极）、开关（线段开口）、灯泡（带叉的圆）和电阻（矩形）。要求整洁、直线连接；标签非强制但有助于表达。
7. The Engineering Design Process | 工程设计流程
Exam Question: A toy company wants to develop a new pull‑along wooden duck. Outline the key stages of the design process they should follow, from initial idea to final product. Explain why testing and evaluation are critical. (6 marks)
考试题目：一家玩具公司欲开发一款新的牵拉式木鸭。概述从最初构想到最终产品应遵循的关键设计阶段，并解释为何测试与评估至关重要。（6分）
The design process typically follows these iterative stages: 1. Research – studying existing products, user needs and materials. 2. Design brief & specification – a clear statement of what the product must do, e.g. safe, durable, appealing to toddlers. 3. Idea generation – sketching multiple concepts. 4. Development – choosing the best idea and refining details like dimensions, axle placement. 5. Modelling/prototyping – building a working model (card, 3D print). 6. Testing & evaluation – checking against the specification: Does it roll smoothly? Is the cord child‑safe? Weaknesses are identified and the design is improved. 7. Manufacturing planning – preparing for mass production.
设计流程通常包括以下迭代阶段：1. 调研 – 研究现有产品、用户需求与材料。2. 设计简介与规格 – 明确产品须满足的要求，如安全、耐用、吸引幼儿。3. 创意构思 – 绘制多个方案草图。4. 方案发展 – 选定最佳方案并细化尺寸、轮轴位置等细节。5. 建模/原型制作 – 制作可动作模型（纸板、3D打印）。6. 测试与评估 – 对照规格检验：滚动是否顺畅？牵引绳是否对儿童安全？找出弱点并改进设计。7. 生产规划 – 准备批量生产。
Testing and evaluation are critical because they reveal flaws not obvious on paper, such as stability or choking hazards. Without this stage, a product might fail in the market or cause harm. The process is cyclical – evaluation leads back to development until the specification is met.
测试与评估至关重要，因为它们能揭示图纸上不明显的问题，如稳定性或窒息风险。缺少这一阶段，产品可能在市场上失败或造成伤害。该过程是循环的—评估结果反馈至发展阶段，直至满足规格。
8. Bridge Design – Tension and Compression | 桥梁设计：拉力与压力
Exam Question: A simple beam bridge sags in the middle when a heavy load is applied. Explain, using diagrams if necessary, how adding a truss structure beneath the beam changes the distribution of forces. Identify which members are in tension and which are in compression. (5 marks)
考试题目：一座简支梁桥在重载作用下中部下挠。请说明（必要时图示）在梁下方添加桁架结构如何改变力的分布，并指出哪些构件受拉、哪些受压。（5分）
In a simple beam, the top surface experiences compression while the bottom surface experiences tension. The middle region, however, has low material efficiency. Adding a truss — typically using triangular arrangements of struts and ties — transforms the beam into a deeper structure. The top chord (upper horizontal beam) undergoes compression, the bottom chord undergoes tension, and the diagonal members alternate between tension and compression depending on their orientation: diagonals sloping downward toward the centre are in compression, while those sloping upward toward the centre are in tension. This spreads the load efficiently to the supports.
在简支梁中，上表面受压而下表面受拉，但中部区域材料利用效率低。加设桁架—通常采用支柱和拉杆的三角形排列—将梁转变为更厚的结构。上弦杆（上部水平梁）受压，下弦杆受拉，斜杆则根据方向交替受拉和受压：向中心下倾的斜杆受压，向中心上倾的斜杆受拉。这样可将载荷高效传递至支座。
Key to full marks: Correctly labelling a sketched truss with “T” and “C” earns explicit marks. It is not enough to say “triangles make it stronger” — you must link the shape to tension/compression paths.
得满分关键：在草图桁架上正确标注“T”（拉力）和“C”（压力）可直接得分。仅说“三角形使其更坚固”是不够的—必须将形状与拉/压传力路径联系起来。
9. Workshop Safety Rules | 车间安全规则
Exam Question: List four safety rules that must be followed when using a hot‑glue gun in a school workshop. Explain the reason behind each rule. (4 marks)
考试题目：列出在学校车间使用热熔胶枪时必须遵守的四条安全规则，并解释每条规则背后的原因。（4分）
- Wear safety goggles – Hot glue can splatter and cause eye burns; goggles protect against liquid adhesive.
- Keep the gun in its stand when not in use – Prevents the hot nozzle from contacting the bench or skin, reducing fire and burn risks.
- Do not touch the nozzle or melted glue – The nozzle reaches ~180 °C and can cause severe burns; molten glue also sticks and burns.
- Work in a well‑ventilated area – Some glue sticks release fumes when heated; ventilation prevents inhalation of irritants.
- 佩戴护目镜 – 热熔胶可能飞溅并导致眼部灼伤；护目镜可阻挡液态粘合剂。
- 不用时将胶枪放在支架上 – 防止热喷嘴接触桌面或皮肤，减少火灾与烫伤风险。
- 切勿触碰喷嘴或熔化的胶 – 喷嘴温度可达 ~180 °C，会造成严重烫伤；熔胶也会粘住并烧伤。
- 在通风良好处操作 – 某些胶棒加热时会释放烟雾；通风可防止吸入刺激性气体。
Scoring note: A rule without a reason can only gain half a mark. Always pair each rule with a clear “because...” explanation, ideally connecting to a specific hazard.
评分说明：只列规则而无理由只能得半分。务必将每条规则与清晰的“因为……”解释配对，最好能联系具体危险。
10. Measurement and Drawing Standards | 测量与绘图标准
Exam Question: In a workshop, you measure the diameter of a wooden dowel as 12.3 mm, 12.4 mm and 12.2 mm in three trials. Calculate the mean diameter and state one reason why repeating measurements improves accuracy. (3 marks)
考试题目：在车间中，你测得一根木榫的直径分别为 12.3 mm、12.4 mm 和 12.2 mm。计算平均直径，并说明重复测量提高准确度的原因。（3分）
12.3 mm. Repeating measurements reduces the effect of random errors, such as misreading the scale or slight variations in dowel roundness. It allows an average to be taken, which is more reliable than a single reading.
平均值 = (12.3 + 12.4 + 12.2) ÷ 3 = 36.9 ÷ 3 = 12.3 mm。重复测量能减少随机误差的影响，例如读错刻度或木榫圆度微小差异。取平均值比单次读数更可靠。
Engineering drawing tip (often tested): When dimensioning a simple part, use millimetres, add dimension lines with arrows, and place the measurement outside the view. Never cross dimension lines. In AQA questions, a well‑laid‑out orthographic drawing can earn up to 4 additional marks.
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