📚 Year 12 Edexcel Engineering: Mock Unit Test Analysis | Year 12 Edexcel 工程：单元测试模拟卷解析
This article provides a complete walkthrough of a typical Year 12 Edexcel Engineering unit test mock paper. Each question is broken down with clear, step-by-step reasoning to help you master the essential principles of mechanics, electronics, materials and fluid systems. Use this analysis to identify common pitfalls and strengthen your problem-solving techniques for the real examination.
本文完整解析了一套 Year 12 Edexcel 工程单元测试模拟卷。每个问题都配有清晰的逐步推理，帮助你掌握力学、电子学、材料学和流体系统的基本原理。通过这份解析，你可以发现常见错误，并为真正的考试强化解题技巧。
1. Overview of the Mock Paper | 模拟卷概述
The mock test covers the core topics from Unit 1: Engineering Principles. It includes questions on resolving forces, moments, stress-strain calculations, electrical circuits, fluid statics, thermal expansion and material selection. The paper is designed to mimic the style and difficulty of the official Edexcel assessment, combining both short calculations and applied theory.
模拟卷涵盖了单元1：工程原理的核心主题。题目包括力的分解、力矩、应力应变计算、电路、流体静力学、热膨胀和材料选择。试卷仿照官方Edexcel 评估的风格和难度，既有简短计算也有应用理论。
2. Question 1 – Resolving Forces | 问题1 – 力的分解
A force of 200 N is applied at an angle of 30° to the horizontal. Calculate the horizontal and vertical components of this force.
一个200 N的力与水平方向成30°角施加。计算该力的水平分力和垂直分力。
To resolve the force, use trigonometry: horizontal component Fx = F cosθ, vertical component Fy = F sinθ.
分解力时使用三角函数：水平分量 Fx = F cosθ，垂直分量 Fy = F sinθ。
Substituting the given values: Fx = 200 × cos30° = 200 × 0.8660 = 173.2 N. Fy = 200 × sin30° = 200 × 0.5 = 100 N.
代入数值：Fx = 200 × cos30° = 200 × 0.8660 = 173.2 N。Fy = 200 × sin30° = 200 × 0.5 = 100 N。
Always check that the vector triangle makes sense: the horizontal part should be longer for a shallow angle.
务必检查矢量三角形的合理性：在较小角度下水平分量应更大。
3. Question 2 – Moments and Equilibrium | 问题2 – 力矩与平衡
A uniform beam of length 4.0 m is pivoted at its centre. A 50 N weight is hung 1.5 m to the left of the pivot. Determine the force required at the right end, 2.0 m from the pivot, to keep the beam horizontal.
一根长4.0 m的均匀梁在其中点处支撑。一个50 N的重物挂在支点左侧1.5 m处。求在支点右侧2.0 m处施加多大的力可使梁保持水平。
The principle of moments states that for equilibrium, clockwise moments equal anticlockwise moments about the pivot.
力矩原理指出，对于平衡状态，顺时针力矩等于逆时针力矩。
Anticlockwise moment = 50 N × 1.5 m = 75 Nm. The required clockwise moment must also be 75 Nm.
逆时针力矩 = 50 N × 1.5 m = 75 Nm。所需的顺时针力矩也必须为75 Nm。
Therefore, F × 2.0 m = 75 Nm, giving F = 75 / 2.0 = 37.5 N. The uniform beam’s weight acts at the pivot and creates no moment.
因此，F × 2.0 m = 75 Nm，得 F = 75 / 2.0 = 37.5 N。均匀梁的重力作用在支点上，不产生力矩。
4. Question 3 – Stress and Strain | 问题3 – 应力与应变
A steel rod of diameter 10 mm carries a tensile load of 5 kN. The original gauge length is 200 mm and it extends by 0.15 mm. Compute the tensile stress, tensile strain and Young’s modulus.
一根直径10 mm的钢杆承受5 kN的拉伸载荷。原始标距长度为200 mm，伸长量为0.15 mm。计算拉伸应力、拉伸应变和杨氏模量。
First, find the cross-sectional area: A = πd²/4 = π × (10 × 10⁻³ m)²/4 = 7.854 × 10⁻⁵ m². Stress σ = F / A.
首先计算横截面积：A = πd²/4 = π × (10 × 10⁻³ m)²/4 = 7.854 × 10⁻⁵ m²。应力 σ = F / A。
σ = 5000 N / 7.854×10⁻⁵ m² = 63.66 × 10⁶ Pa = 63.66 MPa
Strain ε = ΔL / L₀ = 0.15 mm / 200 mm = 7.5 × 10⁻⁴ (dimensionless).
应变 ε = ΔL / L₀ = 0.15 mm / 200 mm = 7.5 × 10⁻⁴（无量纲）。
Young’s modulus E = σ / ε = 63.66 MPa / 0.00075 ≈ 84.88 GPa. This matches typical values for structural steel.
杨氏模量 E = σ / ε = 63.66 MPa / 0.00075 ≈ 84.88 GPa。这与结构钢的典型值相符。
5. Question 4 – Ohm’s Law and Power | 问题4 – 欧姆定律与功率
A resistor is connected across a 12 V supply and draws a current of 2.0 A. Determine the resistance and the power dissipated.
一个电阻器接在12 V电源上，电流为2.0 A。求电阻值和耗散功率。
Using Ohm’s law: R = V / I = 12 V / 2.0 A = 6.0 Ω.
使用欧姆定律：R = V / I = 12 V / 2.0 A = 6.0 Ω。
Power can be calculated using P = V I: P = 12 V × 2.0 A = 24 W. Alternatively, P = I²R or V²/R give the same result.
功率可用 P = V I 计算：P = 12 V × 2.0 A = 24 W。也可以用 P = I²R 或 V²/R，结果相同。
6. Question 5 – Series and Parallel Resistors | 问题5 – 串并联电阻
Two resistors, 4.0 Ω and 6.0 Ω, are connected in parallel. This combination is then connected in series with a 3.0 Ω resistor to a 9.0 V battery. Find the total circuit current and the voltage across the 4.0 Ω resistor.
两个电阻4.0 Ω和6.0 Ω并联，然后与一个3.0 Ω电阻串联接到9.0 V电池上。求电路总电流和4.0 Ω电阻两端的电压。
First, calculate the equivalent parallel resistance: 1/Rp = 1/4 + 1/6 = 3/12 + 2/12 = 5/12, so Rp = 12/5 = 2.4 Ω.
首先计算并联等效电阻：1/Rp = 1/4 + 1/6 = 3/12 + 2/12 = 5/12，因此 Rp = 12/5 = 2.4 Ω。
Total resistance Rtotal = Rp + 3.0 Ω = 2.4 Ω + 3.0 Ω = 5.4 Ω. Total current I = V / Rtotal = 9.0 V / 5.4 Ω = 1.67 A (approx).
总电阻 Rtotal = Rp + 3.0 Ω = 2.4 Ω + 3.0 Ω = 5.4 Ω。总电流 I = V / Rtotal = 9.0 V / 5.4 Ω ≈ 1.67 A。
The voltage across the parallel bank is Vp = I × Rp = 1.67 A × 2.4 Ω ≈ 4.0 V. Since the 4.0 Ω resistor is in parallel, it has this same voltage: 4.0 V.
并联部分的电压 Vp = I × Rp = 1.67 A × 2.4 Ω ≈ 4.0 V。因为4.0 Ω电阻是并联的，其两端电压相同，为 4.0 V。
7. Question 6 – Energy and Efficiency | 问题6 – 能量与效率
An electric motor rated at 200 W runs for 5 minutes to lift a load. If the load gains 48 kJ of gravitational potential energy, calculate the total electrical energy supplied and the efficiency of the motor.
一台额定功率200 W的电动机运行5分钟提升重物。若重物获得48 kJ的重力势能，计算电动机提供的总电能和效率。
Electrical energy supplied: Eelec = power × time = 200 W × (5 × 60 s) = 60,000 J = 60 kJ.
提供的电能：Eelec = 功率 × 时间 = 200 W × (5 × 60 s) = 60,000 J = 60 kJ。
Useful output energy = 48 kJ. Efficiency η = (useful output / total input) × 100% = (48 kJ / 60 kJ) × 100% = 80%.
有用输出能量 = 48 kJ。效率 η =（有用输出 / 总输入）× 100% = (48 kJ / 60 kJ) × 100% = 80%。
The remaining 20% is lost as heat, sound and friction. Always express efficiency as a percentage unless asked otherwise.
其余20%以热、声音和摩擦等形式损失。除非另有要求，效率通常用百分比表示。
8. Question 7 – Hydrostatic Pressure | 问题7 – 流体静压强
A water storage tank is filled to a depth of 2.5 m. The density of water is 1000 kg/m³ and gravitational field strength is 9.81 N/kg. Determine the pressure at the bottom of the tank due to the water, and state the total pressure if atmospheric pressure is 101 kPa.
一个储水箱水深2.5 m。水的密度为1000 kg/m³，重力场强度为9.81 N/kg。计算水箱底部由水产生的压强，并说明若大气压强为101 kPa时的总压强。
Hydrostatic pressure: p = ρgh = 1000 × 9.81 × 2.5.
流体静压强：p = ρgh = 1000 × 9.81 × 2.5。
p = 24,525 Pa ≈ 24.5 kPa
Total absolute pressure at the bottom = gauge pressure + atmospheric pressure = 24.5 kPa + 101 kPa = 125.5 kPa.
底部的绝对总压强 = 表压 + 大气压 = 24.5 kPa + 101 kPa = 125.5 kPa。
Remember that gauge pressure ignores atmospheric pressure and is the value read on many instruments.
注意，表压忽略大气压强，是许多仪器上读取的数值。
9. Question 8 – Thermal Expansion | 问题8 – 热膨胀
An aluminium rail has an original length of 1.500 m at 20 °C. It is heated to 100 °C. The linear expansion coefficient for aluminium is 23 × 10⁻⁶ /°C. Calculate the change in length and the new length.
一根铝轨在20 °C时的原始长度为1.500 m，被加热到100 °C。铝的线膨胀系数为23 × 10⁻⁶ /°C。计算长度变化和新长度。
Temperature change ΔT = 100 °C – 20 °C = 80 °C.
温度变化 ΔT = 100 °C – 20 °C = 80 °C。
Change in length ΔL = α L₀ ΔT = (23 × 10⁻⁶) × 1.500 × 80.
长度变化 ΔL = α L₀ ΔT = (23 × 10⁻⁶) × 1.500 × 80。
ΔL = 0.00276 m = 2.76 mm
New length L = L₀ + ΔL = 1.500 m + 0.00276 m = 1.50276 m. This small expansion must be accounted for in engineering design with expansion gaps.
新长度 L = L₀ + ΔL = 1.500 m + 0.00276 m = 1.50276 m。在工程设计中必须通过设置伸缩缝来容纳这种微小膨胀。
10. Question 9 – Engineering Materials Comparison | 问题9 – 工程材料对比
Compare the typical values of Young’s modulus and ultimate tensile strength for low-carbon steel and aluminium alloy. Explain why steel is preferred for structural frames despite its higher density.
比较低碳钢和铝合金的典型杨氏模量与抗拉强度。解释为什么钢尽管密度更高，但仍被首选用于结构框架。
| Property | Low-carbon steel | Aluminium alloy |
|---|---|---|
| Young’s modulus (GPa) | ~210 | ~70 |
| Tensile strength (MPa) | 400 – 800 | 200 – 500 |
| Density (kg/m³) | 7800 | 2700 |
Steel has a Young’s modulus almost three times that of aluminium alloy, meaning it deforms much less under the same tensile stress. Its higher tensile strength allows for thinner sections, partly offsetting the density disadvantage. Moreover, steel is cheaper and easier to weld, making it the backbone of building frames and bridges.
钢的杨氏模量几乎是铝合金的三倍，意味着在相同拉应力下变形小得多。其更高的抗拉强度允许使用更薄的截面，部分抵消了密度的劣势。此外，钢更便宜且易于焊接，使其成为建筑框架和桥梁的核心材料。
11. Question 10 – Failure Mode and Factor of Safety | 问题10 – 失效模式与安全系数
A component is designed with a yield strength of 250 MPa and must not exceed a working stress of 100 MPa. Name this design principle and calculate the factor of safety. State one possible failure mode if the load is repeatedly applied.
一个零件的屈服强度为250 MPa，工作应力不得超过100 MPa。说出这一设计原理并计算安全系数。如果载荷反复施加，陈述一种可能的失效模式。
The design principle is ‘factor of safety’. Factor of safety (FoS) = yield strength / allowable working stress = 250 MPa / 100 MPa = 2.5.
设计原理是“安全系数”。安全系数 (FoS) = 屈服强度 / 允许工作应力 = 250 MPa / 100 MPa = 2.5。
Under repeated (cyclic) loading, even if the stress is below yield strength, fatigue failure can occur. Cracks initiate and grow gradually, leading to sudden fracture. This is a time-dependent failure mode.
在反复（循环）载荷下，即使应力低于屈服强度，也可能发生疲劳失效。裂纹萌生并逐渐扩展，最终导致突然断裂。这是一种与时间相关的失效模式。
12. Revision Tips for the Unit Test | 单元测试备考建议
Always show your working clearly, including correct unit conversions. Practise drawing free-body diagrams for mechanics problems and redrawing circuits to simplify series/parallel combinations. Memorise the definitions of stress, strain, Young’s modulus, and be comfortable with the factor of safety concept. In the exam, manage your time by tackling the calculation questions you find easiest first.
务必清晰展示计算步骤，包括正确的单位换算。练习绘制力学问题的自由体受力图，并重画电路以简化串并联组合。牢记应力、应变、杨氏模量的定义，并熟练掌握安全系数概念。在考试中，先做你觉得最简单的计算题以合理管理时间。
Review past papers and unit test mocks regularly, as the same principles are often tested with different numbers. If you master these worked examples, you will be well prepared for the real assessment.
定期复习历年真题和单元模拟卷，因为相同的原理常会以不同数值出现。掌握这些解析例题后，你将为正式评估做好充分准备。
Published by TutorHao | Engineering Revision Series | aleveler.com
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