Year 10 Eduqas Engineering: Unit Test Mock Paper Walkthrough | 10年级Eduqas工程：单元测试模拟卷解析

📚 Year 10 Eduqas Engineering: Unit Test Mock Paper Walkthrough | 10年级Eduqas工程：单元测试模拟卷解析

This walkthrough is designed to help Year 10 students tackle a typical Eduqas Engineering unit test with confidence. The mock paper covers core topics from the specification, including engineering materials, manufacturing processes, electronic circuits, mechanisms, and design communication. Each section models a question type you are likely to encounter, then unpacks the answer step by step, highlighting key command words and mark schemes.

本文旨在帮助10年级学生自信应对典型的Eduqas工程单元测试。模拟试卷涵盖了考纲的核心主题，包括工程材料、制造工艺、电子电路、机械机构以及设计沟通。每个部分模拟你可能会遇到的题型，然后逐步解析答案，突出关键词题和评分标准。

1. Multiple-Choice: Material Properties | 材料性质选择题

Question: An engineer selects a material for a bike frame that must be light, strong in tension, and resistant to fatigue. Which combination of properties is most critical?

题目：一位工程师为自行车车架选材，要求轻质、抗拉强度高且耐疲劳。哪组性质最为关键？

Correct Answer: High specific strength and high endurance limit. Specific strength is the ratio of tensile strength to density, so a light frame with high strength requires a material like titanium alloy or carbon-fibre composite. The endurance limit determines how well the material withstands repeated loading without cracking.

正确答案：高比强度和高疲劳极限。比强度是抗拉强度与密度的比值，轻质高强车架需要钛合金或碳纤维复合材料这类材料。疲劳极限决定了材料在反复加载下不开裂的能力。

Common mistake: choosing a material with high toughness. While toughness absorbs energy, it does not guarantee fatigue resistance under cyclic loads. Always link the property to the functional requirement.

常见错误：选择高韧性的材料。虽然韧性吸收能量，但不能保证循环载荷下的抗疲劳性。始终要把性质和功能要求联系起来。

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2. Short Answer: Ferrous vs Non-Ferrous Metals | 简答题：黑色金属与有色金属

Question: Explain the difference between ferrous and non-ferrous metals, giving one example of each and stating an engineering application. (4 marks)

题目：解释黑色金属与有色金属的区别，各举一例并说明一个工程应用。(4分)

Answer: Ferrous metals contain iron as the main element; most are magnetic and prone to rusting. Example: mild steel (low-carbon steel) – used for structural beams in buildings because of its high tensile strength and low cost. Non-ferrous metals do not contain significant iron; they are typically non-magnetic and more corrosion-resistant. Example: aluminium alloy – used in aircraft skins due to its low density and good strength-to-weight ratio.

答案：黑色金属以铁为主要元素；大多数具有磁性且易生锈。例如：低碳钢（软钢）——用于建筑结构梁，因其抗拉强度高且成本低。有色金属不含明显铁成分；通常无磁性且更耐腐蚀。例如：铝合金——用于飞机蒙皮，因其低密度和良好的比强度。

Marking tip: define the difference, give a correctly matched example and a justified application. Do not simply name the metal – describe a real use that shows you understand why the property matters.

评分提示：要给出定义区别，搭配正确的例子和合理的应用。不要只说出金属名称——描述一个实际用途，表明你理解性质的重要性。

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3. Diagram Analysis: Engineering Drawing Symbols | 工程图符号分析

Question: The drawing shows a steel bracket dimensioned with Ø20, M12, and a surface finish symbol resembling a ‘lay’ indicator. Interpret the Ø20 dimension, the M12 thread, and explain what the surface finish symbol conveys about the manufacturing process.

题目：图纸显示一个钢支架，标注有Ø20、M12以及一个类似纹理指示符的表面光洁度符号。解读Ø20尺寸和M12螺纹，并解释表面光洁度符号传达了怎样的制造工艺要求。

Answer: Ø20 indicates a through hole or cylindrical feature with a diameter of 20 mm. M12 specifies an ISO metric coarse thread with a nominal diameter of 12 mm. The surface finish symbol with a lay indicator (e.g., a horizontal bar with a tick) tells the machinist the required surface roughness (e.g., Ra 3.2 µm) and the direction of the machining marks, which could be parallel, perpendicular, or concentric to the surface. This affects how the part fits, wears, and seals.

答案：Ø20表示直径为20 mm的通孔或圆柱特征。M12指定了公称直径为12 mm的ISO米制粗牙螺纹。带有纹理指示符（例如一条横线和勾号）的表面光洁度符号告知机加工人员所需的表面粗糙度（例如Ra 3.2 µm）以及加工纹理的方向，如平行、垂直或同心于表面。这会影响零件的配合、磨损和密封。

Look for the tick symbol in BS 8888 drawings. The number gives the roughness average; the lay symbol tells whether the tool marks run in a circular, cross-hatched, or radial pattern.

观察BS 8888图纸中的勾号符号。数字给出粗糙度平均值；纹理符号告知刀痕是以圆形、交叉还是放射状排列。

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4. Calculation: Gear Ratio and Output Speed | 计算题：齿轮比与输出转速

Question: A motor drives a gear train. The driver gear has 20 teeth and rotates at 1500 rpm. It meshes with a driven gear of 60 teeth. Calculate the gear ratio and the output speed. State whether this arrangement increases torque or speed.

题目：电机驱动齿轮系。主动轮有20齿，转速为1500 rpm。它与一个60齿的从动轮啮合。计算齿轮比和输出转速。说明该配置是增矩还是增速。

Gear ratio = driven teeth / driver teeth = 60 / 20 = 3:1. Output speed = driver speed / gear ratio = 1500 / 3 = 500 rpm. Since the driver is smaller, the driven gear turns slower, so torque is increased (multiplied by a factor of 3, ignoring efficiency losses). This is a reduction gear train.

齿轮比 = 从动轮齿数 / 主动轮齿数 = 60 / 20 = 3:1。输出转速 = 输入转速 / 齿轮比 = 1500 / 3 = 500 rpm。由于主动轮较小，从动轮转动较慢，因此扭矩增大（增为原来的3倍，忽略效率损失）。这是一个减速齿轮系。

Speed ₒᵤₜ = Speed ᵢₙ × (Tₐ / Tₒ) = 1500 × (20/60) = 500 rpm

Always include units and clearly state the ratio in the order driver:driven or as a numerical ratio.

始终写明单位，并清楚地按驱动:从动的顺序或数值比表示齿轮比。

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5. Extended Response: The Iterative Design Process | 扩展回答：迭代设计过程

Question: A team is tasked with designing a portable phone-charging power bank. Describe the steps of the iterative design process they should follow, and explain why testing and evaluation lead to design improvements. (6 marks)

题目：团队受命设计一款便携式手机充电宝。描述他们应遵循的迭代设计过程步骤，并解释测试与评估如何促使设计改进。(6分)

The iterative design cycle includes: 1) defining the problem and setting a specification (capacity, size, weight); 2) researching existing products and technologies; 3) generating a range of initial concepts through sketching and CAD; 4) developing a prototype using suitable materials and processes (e.g., 3D-printed casing, basic circuitry); 5) testing the prototype against the specification – checking charge capacity, heat dissipation, drop resistance; 6) evaluating test data to identify weaknesses (e.g., overheating at 2.1 A output) and refining the design; then repeating the cycle until the product meets requirements. Testing provides objective evidence, while evaluation converts data into actionable design changes, reducing the risk of failure in production.

迭代设计循环包括：1) 定义问题并制定规格（容量、尺寸、重量）；2) 调研现有产品和技术；3) 通过草图和CAD生成一系列初始概念；4) 使用适当的材料和工艺（如3D打印外壳、基础电路板）制作原型；5) 根据规格测试原型——检查充电容量、散热、抗跌落性能；6) 评估测试数据以识别弱点（例如在2.1 A输出时过热），并改进设计；然后重复循环直至产品满足要求。测试提供客观证据，而评估将数据转化为可操作的设计变更，降低生产中的失效风险。

For top marks, use an example from the context, such as discovering that the LED indicator drains too much standby current, leading to a lower-power microcontroller in the next iteration.

想获得高分，需结合情境举例，比如发现LED指示灯在待机时耗电过多，导致下一轮迭代采用更低功耗的微控制器。

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6. Calculations with Ohm’s Law and Power | 欧姆定律与功率计算

Question: An LED requires 20 mA at 2.0 V to operate safely from a 9 V battery. Calculate the required series resistor value and its minimum power rating. Show your working.

题目：一只LED在20 mA和2.0 V下安全运行，由9 V电池供电。计算所需的串联电阻值及其最小额定功率。写出计算过程。

Voltage across resistor = supply voltage – LED voltage = 9 V – 2.0 V = 7.0 V. Current through resistor = 20 mA = 0.020 A. Resistance R = V / I = 7.0 / 0.020 = 350 Ω. Power dissipated P = V × I = 7.0 × 0.020 = 0.14 W. For reliability, choose a resistor with at least double the power rating, so a 0.25 W or 0.5 W resistor is suitable.

电阻两端电压 = 电源电压 – LED电压 = 9 V – 2.0 V = 7.0 V。通过电阻的电流 = 20 mA = 0.020 A。电阻值 R = V / I = 7.0 / 0.020 = 350 Ω。功率消耗 P = V × I = 7.0 × 0.020 = 0.14 W。为可靠起见，选择至少两倍额定功率的电阻，因此0.25 W或0.5 W电阻合适。

R = (Vₛ – Vₗₑₔ) / Iₗₑₔ = (9 – 2) / 0.02 = 350 Ω

Always convert mA to A before calculation. The power rating calculation is crucial because using an undersized resistor would overheat and fail.

计算前务必将mA转换为A。功率额定值的计算至关重要，因为使用额定功率过小的电阻会导致过热失效。

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7. Manufacturing Process: Injection Moulding vs Vacuum Forming | 制造工艺：注塑成型与真空成型

Question: Compare injection moulding and vacuum forming for producing a batch of 5000 plastic enclosures. Consider tooling cost, cycle time, material waste, and surface finish.

题目：为生产5000个塑料外壳，比较注塑成型与真空成型。考虑模具成本、周期时间、材料浪费和表面光洁度。

Injection moulding uses a two-part steel mould; polymer granules are melted and injected under high pressure. Tooling costs are high (thousands of pounds), but cycle time is short (seconds per part), material waste is low because runners can be recycled, and surface finish is excellent with detailed features. Vacuum forming uses a single-sided mould and heated sheet; tooling is cheaper (often wood or aluminium), but cycle time is longer, there is significant web waste around the part, and surface detail is limited to one side. For 5000 units, injection moulding is more economical per part despite high initial cost, because the unit cost becomes low. Vacuum forming would be more suitable for smaller runs or larger shallow parts.

注塑成型使用两瓣钢制模具；聚合物颗粒被熔化并在高压下注射。模具成本高（数千英镑），但周期时间短（每件数秒），材料浪费低，因为流道可回收，且表面光洁度极好，可实现精细细节。真空成型使用单面模具和加热片材；模具成本较低（常为木材或铝），但周期时间较长，零件周围有显著夹持余料浪费，且表面细节限于单面。对于5000件，尽管初始成本高，注塑成型的单件成本更低，更具经济性。真空成型更适合小批量或大型浅层零件。

When answering, link the process choice to production volume. Use terms like ‘initial outlay’, ‘unit cost’, and ‘thermoplastic’ to show technical understanding.

作答时，要将工艺选择与生产量联系起来。使用“初始投资”“单位成本”和“热塑性塑料”等术语以体现技术理解。

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8. Load and Stress Analysis: Cantilever Beam | 载荷与应力分析：悬臂梁

Question: A cantilever beam of length 0.5 m supports a 200 N load at its free end. The beam has a rectangular cross-section of 40 mm × 20 mm, with the 40 mm side vertical. Calculate the maximum bending moment and the bending stress at the fixed end. (Use σ = M y / I)

题目：长0.5 m的悬臂梁在自由端支持200 N载荷。梁截面为40 mm × 20 mm矩形，40 mm边竖直。计算固定端最大弯矩和弯曲应力。(使用σ = M y / I)

Maximum bending moment M = force × perpendicular distance = 200 N × 0.5 m = 100 Nm. Second moment of area I for rectangle about horizontal neutral axis: I = b h³ / 12 = (0.020 m) × (0.040 m)³ / 12 = 1.067 × 10⁻⁷ m⁴. Distance from neutral axis to extreme fibre y = 0.020 m (half of 40 mm). Bending stress σ = M y / I = (100 Nm × 0.020 m) / (1.067 × 10⁻⁷ m⁴) = 18.75 × 10⁶ Pa = 18.75 MPa.

最大弯矩 M = 力 × 垂直距离 = 200 N × 0.5 m = 100 Nm。关于水平中性轴的矩形截面惯性矩：I = b h³ / 12 = (0.020 m) × (0.040 m)³ / 12 = 1.067 × 10⁻⁷ m⁴。中性轴到最外层纤维的距离 y = 0.020 m（40 mm的一半）。弯曲应力 σ = M y / I = (100 Nm × 0.020 m) / (1.067 × 10⁻⁷ m⁴) = 18.75 × 10⁶ Pa = 18.75 MPa。

σ = (M × y) / I = (200 × 0.5 × 0.02) / (0.02 × 0.04³ / 12) = 18.75 MPa

Always convert all lengths to metres for consistency in SI units. The bending stress is well below the yield strength of mild steel (~250 MPa), so the beam is safe under this load.

始终将所有长度单位转换为米，以保持SI单位一致。该弯曲应力远低于低碳钢的屈服强度（~250 MPa），因此梁在此载荷下安全。

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9. Circuit Interpretation: Potential Divider | 电路解读：电位分压器

Question: A potential divider consists of a 10 kΩ fixed resistor (R1) and a thermistor (R2) with a resistance of 5 kΩ at 25 °C, connected across a 6 V supply. Calculate the output voltage across the thermistor at 25 °C. Then explain what happens to the output voltage as temperature increases.

题目：分压电路由10 kΩ固定电阻（R1）和25 °C时阻值为5 kΩ的热敏电阻（R2）组成，接在6 V电源上。计算25 °C时热敏电阻两端的输出电压。然后解释温度升高时输出电压如何变化。

Output voltage Vₒᵤₜ = Vₛ × R2 / (R1 + R2) = 6 V × 5 kΩ / (10 kΩ + 5 kΩ) = 6 × 5/15 = 2 V. If the thermistor is NTC (negative temperature coefficient), its resistance drops as temperature rises. As R2 decreases, the fraction R2/(R1+R2) becomes smaller, so Vₒᵤₜ decreases – eventually approaching zero at very high temperatures. This circuit could be used as a temperature sensor feeding into a comparator to trigger a cooling fan.

输出电压 Vₒᵤₜ = Vₛ × R2 / (R1 + R2) = 6 V × 5 kΩ / (10 kΩ + 5 kΩ) = 6 × 5/15 = 2 V。若热敏电阻为NTC（负温度系数），其阻值随温度升高而减小。随着R2减小，R2/(R1+R2)的分数变小，因此Vₒᵤₜ下降——在极高温度时趋近于零。该电路可用作温度传感器，输入比较器以触发冷却风扇。

Vₒᵤₜ = Vₛ × (R₂ / (R₁ + R₂))

Always check whether the thermistor is NTC or PTC. NTC thermistors are more common for temperature sensing. The equation is the same regardless of which resistor is the sensor.

务必检查热敏电阻是NTC还是PTC。NTC热敏电阻在温度传感中更常见。无论哪个电阻作为传感器，公式相同。

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10. Evaluation: Design for Manufacture and Assembly (DFMA) | 评价题：面向制造与装配的设计

Question: A product originally used 12 separate metal parts bolted together. The redesigned version uses two injection-moulded plastic parts with snap-fit joints. Evaluate the impact of this change on manufacturing cost, assembly time, and environmental sustainability.

题目：某产品原用12个独立金属零件螺栓连接。重新设计的版本采用两个注塑塑料零件并带有卡扣配合。评价这一变更对制造成本、装配时间及环境可持续性的影响。

The reduction from 12 metal parts to two plastic parts drastically simplifies the bill of materials. Tooling cost for injection moulding may be high initially, but per-unit material and labour costs drop significantly. Assembly time is cut: snap-fits join instantly without tools, compared to aligning, inserting, and tightening 12 bolts. Worker fatigue and error rates reduce. For sustainability, the product is lighter, reducing transport emissions. Plastic parts can be designed for disassembly and recycling. However, end-of-life metal parts are easier to recycle infinitely, while plastic may degrade. A full life-cycle assessment (LCA) would weigh raw material extraction, manufacturing energy, use-phase efficiency, and end-of-life disposal.

从12个金属零件减少到两个塑料零件极大简化了物料清单。注塑模具的初始成本可能较高，但单件材料和人工成本大幅下降。装配时间缩短：卡扣瞬间接合，无需工具，而原来需要对齐、插入并拧紧12个螺栓。工人疲劳度和出错率降低。在可持续性方面，产品更轻，减少了运输排放。塑料零件可设计为易于拆解并回收。然而，金属零件在寿命终止时更易于无限次回收，而塑料可能降解。完整的生命周期评估需权衡原材料提取、制造能耗、使用阶段效率以及报废处置。

To score highly, compare specific figures if possible – e.g., ‘assembly time drops from 4 minutes to 30 seconds per unit’. Show you can think beyond cost alone.

要获得高分，尽可能对比具体数字——例如“装配时间从每件4分钟降至30秒”。展现你能超越成本进行思考。

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11. Graph Reading: Stress-Strain Curve | 图表阅读：应力-应变曲线

Question: The graph shows stress-strain curves for three materials: A (steep straight line, high peak), B (shallow straight line, low peak), and C (curved with large strain). Identify which material is stiff and brittle, which is ductile, and which is flexible but weak. Estimate the Young’s modulus for material A if its stress is 300 MPa at a strain of 0.002.

题目：图表显示三种材料的应力-应变曲线：A（陡峭直线，高峰值）、B（浅直线，低峰值）和C（曲线具大应变）。辨识哪种材料刚硬且脆，哪种延展性好，哪种柔韧但弱。若材料A在应变0.002时应力为300 MPa，估算其杨氏模量。

Material A with a steep linear region and abrupt failure is stiff and brittle (e.g., ceramic or high-carbon steel). Material B with a shallow linear slope and low failure stress is flexible but weak (e.g., low-density polyethylene). Material C with extensive plastic deformation is ductile (e.g., mild steel). Young’s modulus E = stress / strain in the linear region = 300 MPa / 0.002 = 150 000 MPa = 150 GPa, which is typical for steel.

材料A具有陡峭线性区并突然失效，属刚硬且脆（如陶瓷或高碳钢）。材料B具有浅线性斜率且失效应力低，柔韧但弱（如低密度聚乙烯）。材料C具有大量塑性变形，属延展性（如低碳钢）。杨氏模量 E = 应力 / 应变（线性区内）= 300 MPa / 0.002 = 150 000 MPa = 150 GPa，为钢材典型值。

E = σ / ε = 300 × 10⁶ Pa / 0.002 = 150 × 10⁹ Pa = 150 GPa

Remember: stiffness is slope, strength is the highest stress before failure, ductility is the amount of permanent strain before fracture.

记住：刚度是斜率，强度是失效前的最高应力，延展性是断裂前的永久应变量。

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12. Quality Control and Testing | 质量控制与测试

Question: A manufacturer tests a sample of 50 steel bolts for tensile strength. The specification states strength must be 800–850 MPa. Five bolts fail below 800 MPa. Calculate the failure rate and suggest two actions the company should take to improve quality.

题目：制造商对50个钢制螺栓样本进行抗拉强度测试。规范要求强度为800–850 MPa。有5个螺栓低于800 MPa。计算不合格率，并建议公司应采取的两项措施以提高质量。

Failure rate = (number of failures / total tested) × 100% = (5/50) × 100% = 10%. This is unacceptably high. Action 1: Review incoming raw material certification – switch to a supplier with tighter tolerances on steel composition. Action 2: Implement statistical process control (SPC) during heat treatment to monitor furnace temperature and time, ensuring consistent material properties. Additional actions could include increasing sample size for inspection or using non-destructive testing such as hardness testing to screen parts.

不合格率 = (不合格数量 / 总测试数量) × 100% = (5/50) × 100% = 10%。该比例高得无法接受。措施1：审查进厂原材料合格证——更换钢材成分公差更严格的供应商。措施2：在热处理过程中实施统计过程控制（SPC），监控炉温和时间，确保材料性能一致。其他措施可以包括增加检测样本量或采用无损检测（如硬度测试）来筛选零件。

Linking quality tools like SPC, cause-and-effect diagrams, and control charts demonstrates higher-level understanding. The goal is traceability and prevention, not just detection.

将SPC、因果图和控制图等质量工具联系起来，可展现出更高级的理解。目标是可追溯性和预防，而不仅仅是检测。

Published by TutorHao | Engineering Revision Series | aleveler.com

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