📚 Year 10 WJEC Engineering: In-Depth Past Paper Analysis | Year 10 WJEC 工程：历年真题深度解析
Past papers are the most effective way for Year 10 students to master the WJEC Engineering specification. By working through real exam questions, you not only become familiar with the command words and mark schemes but also learn to apply engineering principles to unfamiliar contexts. This article provides a detailed breakdown of typical past-paper questions, covering the core topics assessed in Year 10, with step-by-step bilingual explanations to boost your confidence and performance.
历年真题是 Year 10 学生掌握 WJEC 工程课程最有效的途径。通过练习真实的考题，你不仅能熟悉指令词和评分方案，还能学会将工程原理应用于陌生的情境。本文将深入解析典型的真题，覆盖 Year 10 的核心考查主题，并提供分步骤的中英双语解释，帮助你增强信心、提高成绩。
1. Interpreting Engineering Drawings and Symbols | 理解工程图与符号
Typical Question: " on an engineering drawing. State what it represents and explain its importance. (3 marks)" [A surface roughness symbol √ Ra 1.6 is provided.]
典型真题：“下图所示符号出现在工程图纸上。说明它代表什么并解释其重要性。（3分）” [提供一个表面粗糙度符号 √ Ra 1.6。]
This surface texture symbol indicates that the machined surface must have a roughness average (Ra) of 1.6 μm. The check mark shape tells the machinist which surface is controlled, while the number specifies the maximum allowable roughness.
这个表面纹理符号表示加工表面的平均粗糙度 (Ra) 必须达到 1.6 μm。勾号形状告诉机械师哪个表面受控，而数字规定了最大允许粗糙度。
In your answer, always mention the unit (micrometres, μm) and link roughness to function – for example, a sealing surface requires a finer finish to prevent leaks. The symbol is defined by ISO 1302, and citing the standard can demonstrate deeper understanding.
在作答时，一定要提及单位（微米，μm）并将粗糙度与功能联系起来——例如，密封表面需要更精细的光洁度以防止泄漏。该符号由 ISO 1302 定义，引用该标准可以展示更深入的理解。
Mark schemes reward precise language: 'Ra' stands for arithmetic mean roughness, and a lower value means a smoother surface. Avoid vague terms like 'smooth' – use the given value.
评分方案奖励精确的术语：'Ra' 代表算术平均粗糙度，数值越低表面越光滑。避免使用“光滑”之类的模糊词语——必须使用给定的数值。
2. Material Properties and Selection | 材料性能与选择
Typical Question: "Mild steel is often used for car body panels. Explain how two of its properties make it suitable for this application. (4 marks)"
典型真题：“低碳钢常用于汽车车身板。解释它的两个性能如何使其适合这一应用。（4分）”
First, mild steel has excellent ductility, which means it can be pressed into complex curved shapes without cracking during cold forming. This is essential for manufacturing aerodynamic body panels.
首先，低碳钢具有出色的延展性，这意味着它可以在冷成型过程中被冲压成复杂的曲面而不会开裂。这对于制造流线型的车身板至关重要。
Second, its moderate tensile strength (around 400–550 MPa) provides sufficient stiffness and crash resistance while keeping the material cost low. The steel can also be easily spot-welded, enabling rapid assembly on the production line.
其次，它适中的抗拉强度（约 400–550 MPa）提供了足够的刚度和耐撞性，同时保持了较低的材料成本。这种钢还可以轻松进行点焊，从而在装配线上实现快速组装。
When comparing materials, use a table of properties if a question provides values. For instance:
当问题提供数值时，可使用性能表格进行比较。例如：
| Property | Mild Steel | Aluminium Alloy |
|---|---|---|
| Ductility (%) | 20–25 | 12–18 |
| Tensile Strength (MPa) | 400–550 | 200–500 (heat-treated) |
| Density (g/cm³) | 7.85 | 2.70 |
| Cost | Low | Medium–High |
Always explain why the chosen property matters for the specific product – don't just list properties in isolation.
务必解释为什么所选性能对该特定产品很重要——不要只是孤立地罗列性能。
3. Manufacturing Processes: Casting and Welding | 制造工艺：铸造与焊接
Typical Question: "Explain why sand casting is chosen for producing a large cast iron engine block. Refer to the process and material in your answer. (5 marks)"
典型真题：“解释为什么砂型铸造被选择用于生产大型铸铁发动机缸体。回答时涉及工艺和材料。（5分）”
Sand casting uses a disposable mould made of sand mixed with a binder. This process can accommodate very large, complex geometries because the mould is not limited by tooling cost for one-off or low-volume production – a key advantage for engine blocks.
砂型铸造使用由砂和粘结剂混合制成的一次性模具。该工艺能够适应极为庞大、复杂的几何形状，因为对于单件或小批量生产，模具不受工装成本的限制——这对发动机缸体是一个关键优势。
Cast iron has excellent fluidity when molten, allowing it to flow into intricate sand mould cavities and reproduce fine details such as cylinder bores and water jackets. It also exhibits good vibration damping, which reduces engine noise.
铸铁熔融时流动性极佳，能够流入复杂的砂型型腔并复制出气缸筒和水套等精细细节。它还具有良好的减振性能，可降低发动机噪声。
However, the surface finish from sand casting is relatively rough and requires machining on critical surfaces. Machining marks are common on cylinder bores; mention the need for subsequent operations to achieve tolerance.
然而，砂型铸造的表面光洁度相对粗糙，关键表面需要进行机加工。气缸筒上常见加工痕迹；需要提及后续工序以达到公差要求。
Contrast with welding: fusion welding is better for joining pre-formed steel sections, but cannot easily create the internal hollow channels of an engine block. Understanding the limitations of each process is essential for high marks.
与焊接对比：熔焊更适合连接预成型的钢制构件，但难以制造出发动机缸体内部的空心流道。理解每种工艺的局限性是获得高分的关键。
4. Analysing Electronic Circuits | 电子电路分析
Typical Question: " a potential divider with R1 = 10 kΩ and R2 = 5 kΩ connected to a 9 V battery. Calculate Vout across R2. Show your working. (3 marks)"
典型真题：“图中显示一个分压器，R1 = 10 kΩ，R2 = 5 kΩ，连接至 9 V 电池。计算 R2 两端的 Vout。请写出计算过程。（3分）”
The potential divider formula is:
分压器公式为：
Vout = Vin × (R₂ / (R₁ + R₂))
Substitute the values: R₁ = 10 kΩ, R₂ = 5 kΩ, Vin = 9 V.
代入数值：R₁ = 10 kΩ，R₂ = 5 kΩ，Vin = 9 V。
Vout = 9 V × (5 kΩ / (10 kΩ + 5 kΩ)) = 9 V × (5 / 15) = 9 V × ⅓ = 3 V
Always include the unit (V) and show cancellation of kΩ. The answer is 3 V. Students often lose marks by forgetting to state the final unit or by incorrectly adding resistances in parallel when the formula requires series addition.
必须始终包括单位 (V)，并体现 kΩ 的约分。答案为 3 V。学生常因忘记标明最终单位或将电阻用并联方式相加而丢分，而该公式需要串联相加。
When explaining the circuit's purpose, mention that a potential divider scales down a voltage and can be used with a sensor (e.g. LDR, thermistor) to create a varying output for an alarm or control circuit.
在解释电路用途时，要提到分压器可用于降低电压，并能与传感器（如光敏电阻、热敏电阻）配合使用，为报警或控制电路提供变化的输出。
5. Mechanical Systems: Gears and Levers | 机械系统：齿轮与杠杆
Typical Question: "A simple gear train consists of a driver gear with 20 teeth and a driven gear with 60 teeth. The driver rotates at 180 rpm. Calculate the output speed and state whether the system increases torque or speed. (4 marks)"
典型真题：“一个简单的齿轮系由一个 20 齿的主动齿轮和一个 60 齿的从动齿轮组成。主动轮转速为 180 rpm。计算输出转速，并说明该系统是增加扭矩还是增加速度。（4分）”
Gear ratio (velocity ratio) = number of teeth on driven / number of teeth on driver.
齿轮比（速度比）= 从动轮齿数 / 主动轮齿数。
Gear Ratio = Ndriven / Ndriver = 60 / 20 = 3 : 1
Output speed = driver speed / gear ratio = 180 rpm / 3 = 60 rpm.
输出转速 = 主动轮转速 / 齿轮比 = 180 rpm / 3 = 60 rpm。
A gear ratio greater than 1 means the driven gear turns more slowly, so the system acts as a torque multiplier; torque is increased roughly three times (neglecting friction). So the output speed decreases to 60 rpm, and torque is increased.
齿轮比大于 1 意味着从动轮转速更慢，因此该系统充当扭矩倍增器；扭矩大约增加三倍（忽略摩擦）。因此输出转速降低至 60 rpm，扭矩增大。
r principles. The mechanical advantage (MA) of a lever = effort arm / load arm. If a lever' pplied effort is multiplied by 5. Always draw a free-body sketch to visualise pivot points – a common exam mistake is confusing the arm lengths.
杠杆遵循类似的原理。杠杆的机械效益 (MA) = 施力臂 / 负载臂。如果某杠杆的施力臂为 0.5 m，负载臂为 0.1 m，则 MA = 5，意味着施加的力被放大 5 倍。一定要画自由体草图以明确支点位置——常见的考试错误是将力臂的长度混淆。
6. Calculations: Stress and Strain | 计算：应力与应变
Typical Question: "A solid cylindrical strut of diameter 12 mm supports a compressive load of 8 kN. Calculate the compressive stress in the strut in MPa. (3 marks)"
典型真题：“一个直径为 12 mm 的实心圆柱支杆承受 8 kN 的压缩载荷。计算支杆中的压应力，单位为 MPa。（3分）”
a circle, A = π × d² / 4. Convert diameter to metres for consistency: d = 12 mm = 0.012 m.
首先，计算横截面积。对于圆形，A = π × d² / 4。为保持一致，将直径转换为米：d = 12 mm = 0.012 m。
A = π × (0.012 m)² / 4 = π × 1.44×10⁻⁴ m² / 4 ≈ 1.131×10⁻⁴ m²
Now stress (σ) = Force (F) / Area (A). Force = 8 kN = 8 × 10³ N.
现在应力 (σ) = 力 (F) / 面积 (A)。力 = 8 kN = 8 × 10³ N。
σ = (8 × 10³ N) / (1.131×10⁻⁴ m²) ≈ 7.07×10⁷ Pa = 70.7 MPa
The conversion from Pa to MPa is dividing by 10⁶. Always check significant figures: the diameter is given to two significant figures (12 mm), so round the stress to 71 MPa. Show your unit conversions clearly – many marks are lost through incorrect unit handling.
从 Pa 转换为 MPa 需要除以 10⁶。务必检查有效数字：直径以两位有效数字给出 (12 mm)，因此将应力四舍五入为 71 MPa。要清晰展示单位换算——很多分数因单位处理不正确而丢失。
Strain questions may ask for extension. Hooke's Law applies within the elastic limit: σ = E × ε, where E is Young' en ε = 70.7×10⁶ Pa / 200×10⁹ Pa = 3.54×10⁻⁴. Know the difference between stress and strain – stress has units of pressure, strain is dimensionless.
应变题目可能会要求计算伸长量。在弹性极限范围内适用胡克定律：σ = E × ε，其中 E 为杨氏模量。若 E = 200 GPa，则 ε = 70.7×10⁶ Pa / 200×10⁹ Pa = 3.54×10⁻⁴。要懂得应力与应变的区别——应力具有压力量纲，而应变无量纲。
7. Product Life Cycle and Sustainability | 产品生命周期与可持续性
Typical Question: "Explain why a life cycle assessment (LCA) should be carried out before a new electric kettle is manufactured. (6 marks)"
典型真题：“解释为什么在制造新型电热水壶之前应进行生命周期评估 (LCA)。（6分）”
An LCA evaluates environmental impacts from raw material extraction, manufacturing, distribution, use, and end-of-life disposal or recycling. For a kettle, the main phases include mining of copper and plastic feedstock, energy-intensive injection moulding and assembly, electricity consumption during the use phase, and eventual recycling of metals.
LCA 评估从原材料获取、制造、分销、使用到寿命终结处置或回收的各个阶段对环境的影响。对于水壶，主要阶段包括铜矿和塑料原料开采、耗能的注塑成型和装配、使用阶段的耗电，以及金属的最终回收。
Integrating LCA early in the design process allows engineers to choose recyclable polymers, reduce packaging, and design for disassembly. This supports circular economy principles and can reduce carbon footprint – a key selling point under eco-labelling regulations.
在早期设计阶段进行 LCA 能让工程师选择可回收聚合物、减少包装并设计为可拆卸式。这有助于循环经济原则，并能降低碳足迹——这也成为生态标签法规下的关键卖点。
Marks are awarded for naming specific stages and suggesting concrete improvements. For instance, using a limescale filter that extends product life, or designing a quick-boil element that cuts energy use by 30%. Also mention the six Rs: reduce, reuse, recycle, repair, refuse, rethink – these appear in mark schemes.
评分时会奖励命名具体阶段并提出切实改进的做法。例如，使用可延长产品寿命的水垢过滤器，或设计节能 30% 的快沸元件。还要提及 6R 原则：减少、重复使用、回收、修复、拒绝、重新思考——这些频繁出现在评分方案中。
8. Health and Safety in Engineering | 工程中的健康与安全
Typical Question: "Describe two safety precautions that must be taken when using a centre lathe, and explain the hazard each precaution controls. (4 marks)"
典型真题：“描述在使用中心车床时必须采取的两项安全预防措施，并说明每项措施所控制的风险。（4分）”
First, a machine guard or chuck guard must be in place to prevent entanglement with rotating parts. Loose clothing, long hair, or jewellery can be caught by the chuck, causing severe injury. The guard physically isolates the operator from the danger zone.
首先，必须安装机床护罩或卡盘护罩，以防止与旋转部件缠绕。宽松的衣物、长发或首饰可能被卡盘卷入，造成严重伤害。护罩可将操作人员与危险区域进行物理隔离。
Second, the workpiece must be securely clamped using the appropriate chuck and tailstock support. An unsecured workpiece can fly out at high speed, posing a projectile hazard. Always check that the clamping is tight before starting the spindle.
其次，必须使用合适的卡盘和尾座支撑将工件牢固夹紧。未固定的工件可能高速飞出，构成抛射危险。启动主轴前务必检查夹紧是否牢固。
Additionally, wear personal protective equipment (PPE): safety goggles to shield against swarf and cutting fluid, and steel-toe boots. However, examiners want you to prioritise collective protection (guards) over individual PPE. Use technical vocabulary like 'risk assessment', 'COSHH' for coolants, and 'PUWER' regulations to access higher marks.
还需佩戴个人防护装备 (PPE)：护目镜以防切屑和切削液飞溅，以及钢头安全鞋。然而，考官希望你优先考虑集体防护（护罩）而非个人 PPE。使用诸如“风险评估”、冷却液相关的“COSHH”以及“PUWER”法规等技术词汇，以获得更高分数。
9. The Design Process and Evaluation | 设计过程与评估
Typical Question: "A team is developing a portable phone charger. Outline how they could use an iterative design approach. (5 marks)"
典型真题：“一个团队正在开发一款便携式手机充电器。概述他们可以如何采用迭代设计方法。（5分）”
Iterative design involves cycling through stages of research, ideation, prototyping, testing, and refinement multiple times, not just once. The team would start by researching user needs – battery capacity, size, charging speed – then generate initial concept sketches and CAD models.
迭代设计包括多次循环进行研究、构思、原型制作、测试和完善等阶段，而非仅此一次。团队可从研究用户需求入手——电池容量、尺寸、充电速度——然后生成初始概念草图与 CAD 模型。
A first prototype, perhaps 3D-printed, would be tested for ergonomics and portability. Feedback from users might reveal that the casing is too slippery, leading to a second iteration with a rubberised grip and a redesigned USB port position. Each loop improves the design based on objective data and user testing.
第一个原型可能是 3D 打印的，将测试其人机工程学和便携性。用户的反馈可能揭示外壳太滑，从而带来第二次迭代，增加橡胶握把并重新设计 USB 端口位置。每个循环都基于客观数据和用户测试来改进设计。
Marks are gained by emphasizing that failure is a learning tool – a design that fails a drop test isn't useless, it provides data. The term 'rapid prototyping' and reference to 'design for manufacture' aluation criteria should be measurable, such as charging time ≤ 3 hours or weight ≤ 200 g.
通过强调“失败是学习工具”可以获得分数——通不过跌落测试的设计并非毫无价值，它提供了数据。提及“快速原型制作”和“面向制造的设计” (DFM) 可展现更深入的理解。评估标准应可量化，如充电时间 ≤ 3 小时或重量 ≤ 200 g。
10. Common Pitfalls and High-Scoring Strategies | 常见错误与高分策略
Across WJEC past papers, a few mistakes consistently cost students marks. One is failing to read the command word: "state" requires a brief fact, while "explain" demands a reason linked to engineering principles. Write in full sentences and avoid bullet points unless asked for a list.
纵观 WJEC 历年真题，有几种错误一直导致学生丢分。其一是未能理解指令词：“state（陈述）”只需给出简短的事实，而“explain（解释）”则要求结合工程原理阐明原因。要用完整的句子作答，除非题目要求列表，否则避免使用项目符号。
Another is neglecting units and numerical format. If a calculation gives 0.0053 m, convert to 5.3 mm where sensible. Always check that the unit matches the quantity: using N for force, not kg.
Published by TutorHao | Year 10 工程 Revision Series | aleveler.com
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