📚 Year 11 OCR Engineering: Unit Test Mock Paper Analysis | Year 11 OCR 工程：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test designed for the Year 11 OCR Engineering course. Each question is broken down step by step to reinforce core principles in mechanical systems, materials, electronics, and manufacturing processes. Use this analysis to deepen your understanding and boost your exam technique.
本文详细解析了一份为 Year 11 OCR 工程课程设计的单元测试模拟卷。每道题目均分步剖析，以巩固机械系统、材料、电子技术和制造工艺等核心原理。借助此解析深化理解并提升应试技巧。
1. Forces and Loads in Structures | 结构中的力与载荷
A beam is supported at both ends and a concentrated load is applied at its centre. Name the three internal forces experienced by the beam and mark their regions of maximum effect on a simple diagram.
一根梁两端支撑，集中载荷作用于中点。请说出梁承受的三种内力，并在简图上标出它们作用最显著的区域。
The beam experiences tension along the bottom edge, compression along the top edge, and shear which is greatest close to the supports. In a simply supported beam with a central point load, bending moment is maximum at the centre, leading to extreme fibre stresses – tension on the convex side and compression on the concave side. Shear force is constant in magnitude between the support and the load point, then changes sign.
梁的底边承受拉力，顶边承受压力，而剪切力在靠近支座处最大。对于中点受集中载荷的简支梁，弯矩在中心最大，导致外缘纤维应力——凸面受拉，凹面受压。剪力在支座与载荷作用点之间大小恒定，随后变号。
2. Material Selection: Steel vs Aluminium | 材料选择：钢与铝
An automotive suspension component needs high fatigue strength and low weight. Compare the suitability of mild steel and aluminium alloy for this application using two distinct material properties.
某汽车悬架部件需要高疲劳强度和低重量。使用两项不同的材料属性比较低碳钢和铝合金对此应用的适用性。
Aluminium alloy offers a superior strength-to-weight ratio ring adequate tensile strength. However, mild steel (density ≈7850 kg m⁻³) displays better fatigue limit, meaning it can endure more stress cycles without failure. If weight saving is critical, aluminium is preferred; for long-term durability under fluctuating loads, steel may be more reliable.
铝合金因密度低（约2700 kg·m⁻³）而具有优异的比强度，同时拉伸强度足够。但低碳钢（密度约7850 kg·m⁻³）展现出更佳的疲劳极限，即能承受更多应力循环而不断裂。若减重是关键，应选铝；若在波动载荷下追求长期耐久，钢更可靠。
3. Ohm’s Law and Series Circuits | 欧姆定律与串联电路
A 12 V battery is connected in series with a 220 Ω resistor and a 680 Ω resistor. Calculate the total resistance, circuit current, and the voltage drop across the 220 Ω resistor. Show your working.
一个12 V电池与一个220 Ω电阻和一个680 Ω电阻串联。计算总电阻、电路电流以及220 Ω电阻两端的电压降。写出计算过程。
Total resistance Rₜ = 220 Ω + 680 Ω = 900 Ω. Current I = V / Rₜ = 12 V / 900 Ω = 0.01333 A (13.33 mA). Voltage across 220 Ω: V₂₂₀ = I × 220 Ω = 0.01333 A × 220 Ω = 2.933 V (≈2.9 V).
总电阻 Rₜ = 220 Ω + 680 Ω = 900 Ω。电流 I = V / Rₜ = 12 V / 900 Ω = 0.01333 A（13.33 mA）。220 Ω 两端电压：V₂₂₀ = I × 220 Ω = 0.01333 A × 220 Ω = 2.933 V（约2.9 V）。
4. Manufacturing Process: Injection Moulding | 制造工艺：注塑成型
Describe the four key stages of the injection moulding cycle and explain why thermoplastics are suitable for this process.
描述注塑成型周期的四个关键阶段，并解释为何热塑性塑料适合该工艺。
The cycle comprises: 1. Clamping – mould halves are held together under pressure. 2. Injection – molten polymer is forced into the cavity. 3. Cooling – the part solidifies as heat is removed. 4. Ejection – mould opens and the part is pushed out. Thermoplastics are ideal because they can be melted, reshaped, and cooled repeatedly without significant degradation, allowing efficient recycling of scrap.
周期包括：1. 合模 – 模具两半在压力下闭合。2. 注射 – 熔融聚合物被压入模腔。3. 冷却 – 散热使零件固化。4. 脱模 – 模具打开并推出成品。热塑性塑料适合是因为可反复熔化、重塑和冷却而无显著降解，便于高效回收废料。
5. Gear Ratios and Torque | 齿轮比与扭矩
A driver gear with 15 teeth meshes with a driven gear of 45 teeth. If the driver rotates at 1200 rpm and delivers a torque of 4 N m, determine the output speed and ideal output torque. Assume 100% efficiency.
主动齿轮有15齿，与45齿的从动齿轮啮合。若主动轮转速为1200 rpm，输出扭矩为4 N m，求输出转速与理想输出扭矩。假设效率100%。
Gear ratio = driven/driver = 45/15 = 3:1. Output speed = driver speed / ratio = 1200 rpm / 3 = 400 rpm r an . For an ideal system, torque is multiplied by the ratio: output torque = input torque × 3 = 4 N m × 3 = 12 N m.
齿轮比 = 从动/主动 = 45/15 = 3:1。输出转速 = 主动轮转速 / 比 = 1200 rpm / 3 = 400 rpm。理想系统中，扭矩按比例放大：输出扭矩 = 输入扭矩 × 3 = 4 N m × 3 = 12 N m。
6. Pugh Matrix for Concept Selection | 用于方案选择的普氏矩阵
Three design concepts for a handheld garden tool (A, B, C) are evaluated against criteria: ease of manufacture, weight, durability, and cost. Using the Pugh matrix method with concept B as the datum, score each criterion as +1 (better), 0 (same), or –1 (worse). Explain how the scores lead to a final selection.
一款手持园艺工具的三种设计方案（A、B、C）根据制造便利性、重量、耐久性和成本进行评估。使用普氏矩阵法，以方案B为基准，将每项评分设为+1（更好）、0（相同）或–1（更差）。阐述如何根据得分做出最终选择。
scoring, totals are summed. For example: A: manufacture +1, weight 0, durability +1, cost –1, total = +1. C: manufacture –1, weight +1, durability – promising on cost. The matrix makes trade-offs visible and supports data-driven decisions.
普氏矩阵有助于系统地比选方案。评分后进行总分加和。例如：A：制造+1、重量0、耐久+1、成本–1，总分=+1。C：制造–1、重量+1、耐久–1、成本+1，总分=0。方案A因在制造和耐久性上突出而仅微增成本，故最佳。矩阵使权衡清晰可见，支持数据驱动决策。
7. Tolerance and Quality Control | 公差与质量控制
A shaft is designed with a diameter of 20.00 ± 0.05 A sample of five shafts has the following measurements: 20.03, 20.06, 19.98, 20.04, 20.02 ulate the percentage of compliant parts. Identify whether any systematic error is present.
一根轴的设计直径为20.00 ± 0.05 mm。五根样本的测量值为：20.03、20.06、19.98、20.04、20.02 mm。判断哪些轴在公差内，计算合格率。分析是否存在系统误差。
Tolerance limits: 19.95 mm to 20.05 mm. Measurements: 20.03 ✓, 20.06 ✗ (over), 19.98 ✓, 20.04 ✓, 20.02 ✓. Compliant = 4 out of 5 = 80%. The readings are generally shifted upward, with a mean of 20.026 mm, suggesting a possible systematic error such as tool offset or calibration drift, which should be investigated.
公差范围：19.95 mm 至 20.05 mm。测量值：20.03 ✓、20.06 ✗（超差）、19.98 ✓、20.04 ✓、20.02 ✓。合格 = 4/5 = 80%。读数整体偏高，平均值20.026 mm，可能暗示存在系统误差，如刀具偏置或校准漂移，需予以排查。
8. Mechanical Power and Efficiency | 机械功率与效率
An electric motor lifts a mass of 50 kg through a vertical height of 8 m in 6 seconds. The motor draws a constant electrical power of 800 e the useful mechanical power output and the overall efficiency of the motor. Take g = 9.8 m s⁻².
一台电动机在6秒内将50 kg的重物垂直提升8 m。电动机输入恒定电功率800 W。计算输出的有用机械功率和电动机的总效率。取 g = 9.8 m·s⁻²。
Work done = m × g × h = 50 × 9.8 × 8 = 3920 J. Useful power output = work / time = 3920 / 6 = 653.33 W. Efficiency η = (useful output / electrical input) × 100% = (653.33 / 800) × 100% = 81.67%.
做功 = m × g × h = 50 × 9.8 × 8 = 3920 J。有用功率输出 = 功 / 时间 = 3920 / 6 = 653.33 W。效率 η =（有用输出 / 电功率输入）× 100% =（653.33 / 800）× 100% = 81.67%。
9. Pneumatic Circuit Symbols | 气动回路符号
Draw and label the standard ISO symbols for a single-acting cylinder (spring return) and a 3/2 way push-button valve. Describe the valve’s function in controlling the cylinder.
绘制并标注单作用气缸（弹簧复位）和3/2通按钮阀的标准ISO符号。描述该阀门在控制气缸时的功能。
The single-acting cylinder is represented by a rectangle with a spring symbol on one side and a single port for air supply. The 3/2 valve shows three ports and two positions; the push-button operator activates the spring-return mechanism. When pressed, compressed air flows from port 1 to port 2 extending the cylinder; when released, the valve blocks supply and exhausts port 2 to port 3, allowing the spring to retract the cylinder. This simple circuit is used in clamping devices.
单作用气缸由一个矩形表示，一侧画有弹簧符号，仅有一个供气口。3/2通阀有三个接口和两个工位；按钮操作器驱动弹簧复位机构。按下时，压缩空气从1口流向2口伸出气缸；松开时，阀门切断气源并将2口与3口相通排气，弹簧使气缸缩回。这一简单回路常用于夹紧装置中。
10. Sustainability and Life Cycle Assessment | 可持续性与生命周期评估
Discuss how an engineer can reduce the environmental impact of a product during the material extraction and end-of-life stages. Give one concrete strategy for each stage.
讨论工程师如何在材料提取和报废阶段降低产品的环境影响。每个阶段给出一个具体策略。
During material extraction, engineers can specify recycled or reclaimed materials – for instance, using recycled aluminium reduces the need for energy-intensive bauxite mining and refining by up to 95%. At the end-of-life stage, products should be designed for disassembly and material recovery: snap-fit joints instead of adhesives, clear labelling of polymers, and modular structures enable components to be reused or recycled, diverting waste from landfill.
在材料提取阶段，工程师可指定回收或再生材料——例如，使用再生铝可降低高能耗的铝土矿开采和精炼需求达95%。在报废阶段，产品应设计为易于拆解和材料回收：采用卡扣配合代替粘合剂、清楚标注聚合物类型、以及模块化构造，使零部件能被重复使用或回收，从而减少填埋废物。
Published by TutorHao | Engineering Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图