📚 Year 12 AQA Engineering: Unit Test Mock Paper Analysis | AQA工程：单元测试模拟卷解析
This article provides a detailed walkthrough of a typical Year 12 AQA Engineering unit test mock paper. Each section breaks down a key topic area, presenting a sample question, the thought process behind the solution, and the final answer. Use this analysis to reinforce your understanding of core principles in mechanics, materials, electronics, thermodynamics, and design methodology.
本文对一份典型的 AQA 工程 Year 12 单元测试模拟卷进行了详细解析。每个部分拆解一个核心知识领域，展示样题、解题思路和最终答案。通过这篇解析，你可以加深对力学、材料、电子、热力学和设计方法论等核心原理的理解。
1. Resolving Forces and Moments | 力与力矩的分解
Mock Question: A uniform horizontal beam AB of length 4.0 m and weight 200 N is supported by two vertical ropes at its ends. A load of 300 N is placed 1.2 m from A. Determine the tension in each rope.
模拟题：一根均匀水平横梁 AB 长 4.0 m、重 200 N，在两端由两根垂直绳索支撑。一个 300 N 的负载放在距 A 端 1.2 m 处。求每根绳索的拉力。
Begin by drawing a free-body diagram. Place the beam horizontally, mark the weight acting at the centre (2.0 m from either end), the load at 1.2 m from A, and the two tensions TA and TB acting upward at the ends.
首先画出受力图。将横梁水平放置，标记作用在中心点（距两端各 2.0 m）的重力、距 A 端 1.2 m 的负载，以及两端向上的拉力 TA 和 TB。
Apply the equilibrium condition for forces: upward forces = downward forces. Hence TA + TB = 200 N + 300 N = 500 N.
应用力的平衡条件：向上力 = 向下力。因此 TA + TB = 200 N + 300 N = 500 N。
Take moments about point A to eliminate TA. Clockwise moments = (200 N × 2.0 m) + (300 N × 1.2 m) = 400 N·m + 360 N·m = 760 N·m. This is balanced by the anticlockwise moment due to TB × 4.0 m.
对 A 点取矩以消去 TA。顺时针力矩 = (200 N × 2.0 m) + (300 N × 1.2 m) = 400 N·m + 360 N·m = 760 N·m。该值由 TB × 4.0 m 的逆时针力矩平衡。
Therefore TB = 760 N·m / 4.0 m = 190 N. Substitute back to get TA = 500 N – 190 N = 310 N.
因此 TB = 760 N·m / 4.0 m = 190 N。代回得 TA = 500 N – 190 N = 310 N。
The tension in the rope at A is 310 N and at B is 190 N. Always check that the sum equals 500 N to confirm the calculation.
A 端绳索拉力为 310 N，B 端为 190 N。务必检查两力之和等于 500 N，以验证计算正确。
2. Stress, Strain and Young’s Modulus | 应力、应变与杨氏模量
Mock Question: A metal wire of diameter 1.8 mm and original length 2.50 m stretches by 2.1 mm when a force of 85 N is applied. Calculate (a) stress, (b) strain, and (c) Young’s modulus for the wire material.
模拟题：一根直径 1.8 mm、原长 2.50 m 的金属丝在 85 N 的力作用下伸长 2.1 mm。计算 (a) 应力、(b) 应变以及 (c) 该金属丝的杨氏模量。
Convert all units to SI base units. Diameter d = 1.8 × 10⁻³ m, original length L₀ = 2.50 m, extension ΔL = 2.1 × 10⁻³ m, force F = 85 N.
将所有单位转换为国际单位制基本单位。直径 d = 1.8 × 10⁻³ m，原长 L₀ = 2.50 m，伸长量 ΔL = 2.1 × 10⁻³ m，力 F = 85 N。
Cross-sectional area A = πd²/4 = π (1.8 × 10⁻³)² / 4 = π × 3.24 × 10⁻⁶ / 4 ≈ 2.545 × 10⁻⁶ m².
横截面积 A = πd²/4 = π (1.8 × 10⁻³)² / 4 = π × 3.24 × 10⁻⁶ / 4 ≈ 2.545 × 10⁻⁶ m²。
Stress σ = F / A = 85 / 2.545 × 10⁻⁶ ≈ 3.34 × 10⁷ Pa (or 33.4 MPa).
应力 σ = F / A = 85 / 2.545 × 10⁻⁶ ≈ 3.34 × 10⁷ Pa（即 33.4 MPa）。
Strain ε = ΔL / L₀ = 2.1 × 10⁻³ / 2.50 = 8.4 × 10⁻⁴ (no units).
应变 ε = ΔL / L₀ = 2.1 × 10⁻³ / 2.50 = 8.4 × 10⁻⁴（无单位）。
Young’s modulus E = σ / ε = 3.34 × 10⁷ / 8.4 × 10⁻⁴ ≈ 3.98 × 10¹⁰ Pa (≈ 39.8 GPa). This value is typical for brass or mild steel.
杨氏模量 E = σ / ε = 3.34 × 10⁷ / 8.4 × 10⁻⁴ ≈ 3.98 × 10¹⁰ Pa（约 39.8 GPa）。此数值是黄铜或低碳钢的典型值。
Always show your working step by step; examiners award marks for correct formulas, unit conversions, and final answers with proper units.
务必逐步展示计算过程；考官会根据正确的公式、单位换算以及带正确单位的最终答案给分。
3. Electrical Circuits and Kirchhoff’s Laws | 电路与基尔霍夫定律
Mock Question: In the circuit below, a 9 V battery is connected to a 10 Ω resistor in series with a parallel combination of a 30 Ω and a 15 Ω resistor. Find the current through each resistor.
模拟题：在以下电路中，9 V 电池与一个 10 Ω 电阻串联，再与一个 30 Ω 和一个 15 Ω 电阻的并联组合串联。求每个电阻中的电流。
First, calculate the equivalent resistance of the parallel branch: 1/Rp = 1/30 + 1/15 = 1/30 + 2/30 = 3/30, so Rp = 10 Ω.
首先，计算并联支路的等效电阻：1/Rp = 1/30 + 1/15 = 1/30 + 2/30 = 3/30，因此 Rp = 10 Ω。
Total circuit resistance RT = 10 Ω (series) + 10 Ω (parallel) = 20 Ω. Total current IT = V / RT = 9 V / 20 Ω = 0.45 A.
电路总电阻 RT = 10 Ω（串联） + 10 Ω（并联） = 20 Ω。总电流 IT = V / RT = 9 V / 20 Ω = 0.45 A。
This total current flows through the 10 Ω series resistor, so I10 = 0.45 A.
该总电流流过 10 Ω 串联电阻，因此 I10 = 0.45 A。
The voltage across the parallel branch is Vp = IT × Rp = 0.45 × 10 = 4.5 V. Using Ohm’s law, current through 30 Ω: I30 = 4.5 / 30 = 0.15 A; through 15 Ω: I15 = 4.5 / 15 = 0.30 A.
并联支路两端的电压 Vp = IT × Rp = 0.45 × 10 = 4.5 V。使用欧姆定律，通过 30 Ω 的电流：I30 = 4.5 / 30 = 0.15 A；通过 15 Ω：I15 = 4.5 / 15 = 0.30 A。
Check using Kirchhoff’s current law: I30 + I15 = 0.15 + 0.30 = 0.45 A = IT, which confirms the split. Practice drawing clear schematic diagrams to avoid confusion.
用基尔霍夫电流定律检验：I30 + I15 = 0.15 + 0.30 = 0.45 A = IT，验证了分流。练习绘制清晰的原理图，避免混淆。
4. Material Properties and Selection | 材料性能与选择
Mock Question: An aircraft wing spar must be lightweight yet stiff. Using the data table, explain why 7075 aluminium alloy is preferred over mild steel for this application.
模拟题：飞机翼梁必须轻质且高刚度。利用数据表，解释为什么 7075 铝合金在此应用中优于低碳钢。
| Property / 性能 | Mild Steel / 低碳钢 | 7075 Al Alloy / 7075 铝合金 |
|---|---|---|
| Density (kg/m³) / 密度 | 7850 | 2810 |
| Young’s Modulus (GPa) / 杨氏模量 | 210 | 72 |
| Specific Stiffness (E/ρ) / 比刚度 | 26.8 | 25.6 |
| Yield Strength (MPa) / 屈服强度 | 250 | 500 |
The specific stiffness (E/ρ) is a measure of stiffness per unit mass. Mild steel has a value of 26.8 × 10⁶ m²/s², while 7075 alloy has 25.6 × 10⁶ m²/s² — very similar, meaning for the same weight, the aluminium alloy offers almost equal structural stiffness.
比刚度（E/ρ）是单位质量的刚度度量。低碳钢的值为 26.8 × 10⁶ m²/s²，而 7075 合金为 25.6 × 10⁶ m²/s²——非常接近，意味着在同等重量下，铝合金可提供几乎相等的结构刚度。
However, the density of aluminium alloy is only about one-third that of steel (2810 vs 7850 kg/m³). This translates to a significant weight saving for the same volume, which is critical in aerospace design to improve fuel efficiency and payload capacity.
然而，铝合金的密度仅为钢的大约三分之一（2810 对比 7850 kg/m³）。这意味着在相同体积下可极大减轻重量，这在航空航天设计中对于提高燃油效率和有效载荷至关重要。
Moreover, the yield strength of 7075 alloy (500 MPa) is double that of mild steel, providing a better strength-to-weight ratio. Engineers use property charts and selection criteria to optimise for weight, strength, and cost.
此外，7075 合金的屈服强度（500 MPa）是低碳钢的两倍，具有更优的强度-重量比。工程师使用性能图表和选材准则来优化重量、强度和成本。
5. Thermodynamics: Heat Transfer and Efficiency | 热力学：传热与效率
Mock Question: An aluminium engine block of mass 15 kg is heated from 20 °C to 95 °C during operation. The specific heat capacity of aluminium is 900 J/(kg· °C). Calculate the heat energy supplied. If the engine burns 0.5 kg of fuel with a calorific value of 44 MJ/kg, what is the thermal efficiency of the heating process?
模拟题：一个质量 15 kg 的铝制发动机缸体在运行中从 20 °C 加热到 95 °C。铝的比热容为 900 J/(kg· °C)。计算供给的热能。如果发动机燃烧 0.5 kg 热值为 44 MJ/kg 的燃料，加热过程的热效率是多少？
Temperature change ΔT = 95 °C – 20 °C = 75 °C. Heat energy Q = m c ΔT = 15 × 900 × 75 = 1,012,500 J = 1.0125 MJ.
温度变化 ΔT = 95 °C – 20 °C = 75 °C。热能 Q = m c ΔT = 15 × 900 × 75 = 1,012,500 J = 1.0125 MJ。
Energy input from fuel Ein = 0.5 kg × 44 MJ/kg = 22 MJ. Efficiency η = (useful energy output / energy input) × 100% = (1.0125 / 22) × 100% ≈ 4.6%.
燃料输入能量 Ein = 0.5 kg × 44 MJ/kg = 22 MJ。效率 η =（有用能量输出 / 能量输入）× 100% = (1.0125 / 22) × 100% ≈ 4.6%。
This low efficiency is expected because most of the fuel energy is lost as exhaust heat, friction, and in moving parts, not just heating the block. In real engine analysis, students should consider energy balance and Sankey diagrams.
这一低效率在意料之中，因为大部分燃料能量以废气热、摩擦及运动部件等形式损失，而不仅仅用于加热缸体。在实际发动机分析中，学生应考虑能量平衡和桑基图。
6. Engineering Mathematics: Vectors and Trigonometry | 工程数学：矢量与三角学
Mock Question: Two forces act on a bracket: F₁ = 120 N at 30° above the positive x-axis, and F₂ = 80 N at 120° from the positive x-axis. Determine the resultant force magnitude and direction.
模拟题：两个力作用在一个支架上：F₁ = 120 N，与 x 轴正方向成 30° 夹角；F₂ = 80 N，与 x 轴正方向成 120° 夹角。求合力的大小和方向。
Resolve each force into horizontal and vertical components. F₁: F₁ₓ = 120 cos30° = 120 × 0.866 = 103.9 N; F₁ᵧ = 120 sin30° = 60.0 N.
将每个力分解为水平分量和竖直分量。F₁：F₁ₓ = 120 cos30° = 120 × 0.866 = 103.9 N；F₁ᵧ = 120 sin30° = 60.0 N。
F₂: F₂ₓ = 80 cos120° = 80 × (−0.5) = −40.0 N; F₂ᵧ = 80 sin120° = 80 × 0.866 = 69.3 N.
F₂：F₂ₓ = 80 cos120° = 80 × (−0.5) = −40.0 N；F₂ᵧ = 80 sin120° = 80 × 0.866 = 69.3 N。
Sum x-components: Rₓ = 103.9 – 40.0 = 63.9 N. Sum y-components: Rᵧ = 60.0 + 69.3 = 129.3 N. Resultant magnitude R = √(63.9² + 129.3²) ≈ √(4083 + 16718) = √20801 ≈ 144.2 N.
x 方向分量之和：Rₓ = 103.9 – 40.0 = 63.9 N。y 方向分量之和：Rᵧ = 60.0 + 69.3 = 129.3 N。合力大小 R = √(63.9² + 129.3²) ≈ √(4083 + 16718) = √20801 ≈ 144.2 N。
Direction θ = tan⁻¹(Rᵧ / Rₓ) = tan⁻¹(129.3 / 63.9) ≈ tan⁻¹(2.024) ≈ 63.7° above the positive x-axis. Drawing a vector triangle helps visualise the addition.
方向 θ = tan⁻¹(Rᵧ / Rₓ) = tan⁻¹(129.3 / 63.9) ≈ tan⁻¹(2.024) ≈ 63.7°，在 x 轴正方向上方。画出矢量三角形有助于直观理解加法。
Vector resolution is a fundamental skill; always use a consistent sign convention and ensure your calculator is in degree mode.
矢量分解是一项基本技能；始终坚持一致的符号约定，并确保计算器处于角度模式。
7. Control Systems: Open and Closed Loop | 控制系统：开环与闭环
Mock Question: Distinguish between open-loop and closed-loop control systems, giving one practical engineering example of each. Draw a block diagram for a closed-loop temperature control system.
模拟题：区分开环和闭环控制系统，各给出一个工程中的实际例子。画出闭环温度控制系统的框图。
In an open-loop system, the control action is independent of the output. There is no feedback; the system cannot correct errors. Example: an electric toaster with a timer – it runs for a set time regardless of whether the bread is perfectly toasted.
在开环系统中，控制动作与输出无关。没有反馈；系统无法纠正误差。例如：带定时器的电烤箱——它
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