A-Level Eduqas Engineering: Mock Unit Test Paper Walkthrough | A-Level Eduqas 工程：单元测试模拟卷解析

📚 A-Level Eduqas Engineering: Mock Unit Test Paper Walkthrough | A-Level Eduqas 工程：单元测试模拟卷解析

Mock unit test papers are an essential revision tool for A-Level Engineering students following the Eduqas specification. They mirror the style and difficulty of real examinations, allowing you to practise time management, apply theoretical knowledge, and identify weak spots before the final assessment. This walkthrough dissects eight representative problems from a typical mock paper, covering statics, materials, kinematics, circuits, thermodynamics, fluids, digital electronics, and manufacturing. Each section presents a problem statement, a step-by-step solution, and a clear answer, pairing English and Chinese explanations to support bilingual learners.

模拟单元测试卷是 A-Level 工程学生备考 Eduqas 考试局的重要复习工具。它们模仿真实考试的题型和难度，帮助你练习时间管理、运用理论知识并在最终评估前发现薄弱环节。本文解析一份典型模拟卷中的八道代表性题目，涵盖静力学、材料、运动学、电路、热力学、流体、数字电子和制造工程。每个部分给出问题陈述、逐步解答和明确答案，并以中英对照的方式帮助双语学习者深入理解。

1. Statics – Force Equilibrium in a Truss Joint | 静力学 – 桁架节点的力平衡

A truss joint at point A is loaded by a horizontal force of 5 kN acting to the right. Three members meet at A: member AB is inclined at 30° above the horizontal, member AC is vertical, and the third member (not shown) balances the joint. Assuming all members are pin-connected and the joint is in static equilibrium, determine the internal forces in members AB and AC. Indicate whether each member is in tension or compression.

点 A 处的桁架节点受到一个水平向右的 5 kN 外力作用。节点 A 连接三根杆件：杆件 AB 与水平线成 30° 夹角，杆件 AC 为竖直方向，第三根杆件（未画出）用以平衡节点。假设所有杆件均为铰接且节点处于静力平衡状态，求杆件 AB 和 AC 的内力大小，并指明各杆件受拉还是受压。

Resolve the forces horizontally and vertically, taking the direction to the right and upward as positive. Let F_AB be the force in AB (tension positive) and F_AC be the force in AC. The horizontal component of F_AB is F_AB cos30°, and the vertical component is F_AB sin30°. The applied force is 5 kN to the right.

分解水平和竖直方向上的力，以向右和向上为正方向。设杆件 AB 的内力为 F_AB（拉为正），杆件 AC 的内力为 F_AC。F_AB 的水平分量为 F_AB cos30°，竖直分量为 F_AB sin30°。外加力为向右的 5 kN。

Write the equilibrium equations. Horizontally: ∑F_x = 0 → F_AB cos30° − 5 kN = 0. Vertically: ∑F_y = 0 → F_AB sin30° + F_AC = 0. From the first equation, F_AB = 5 / cos30° = 5 / (√3/2) = 10/√3 ≈ 5.77 kN. The positive value indicates tension in AB. Substituting into the second equation: F_AC = −F_AB sin30° ≈ −5.77 × 0.5 = −2.89 kN. The negative sign shows that AC is in compression.

列出平衡方程。水平方向：∑F_x = 0 → F_AB cos30° − 5 kN = 0。竖直方向：∑F_y = 0 → F_AB sin30° + F_AC = 0。由第一个方程得 F_AB = 5 / cos30° = 5 / (√3/2) = 10/√3 ≈ 5.77 kN。正号表示 AB 受拉。代入第二个方程：F_AC = −F_AB sin30° ≈ −5.77 × 0.5 = −2.89 kN。负号表示 AC 受压。

F_AB ≈ 5.77 kN (tension), F_AC ≈ 2.89 kN (compression)

2. Stress, Strain and Young’s Modulus | 应力、应变与杨氏模量

A steel wire with a diameter of 2 mm and an original length of 1.5 m is subjected to an axial tensile load of 300 N. The wire extends by 0.75 mm. Calculate the tensile stress, the tensile strain, and Young’s modulus of the material. Give your answers in appropriate units.

一根直径为 2 mm、原长为 1.5 m 的钢丝，受到 300 N 的轴向拉伸载荷，伸长量为 0.75 mm。计算钢丝的拉伸应力、拉伸应变以及材料的杨氏模量，并使用合适的单位表示结果。

First, find the cross-sectional area A of the wire. A = πd²/4. Substituting d = 2 mm = 2 × 10⁻³ m: A = π(2 × 10⁻³)² / 4 = π × 4 × 10⁻⁶ / 4 = π × 10⁻⁶ m² ≈ 3.1416 × 10⁻⁶ m². Stress σ is defined as force per unit area: σ = F / A = 300 N / (π × 10⁻⁶) ≈ 9.549 × 10⁷ Pa = 95.5 MPa.

首先求钢丝的横截面积 A。A = πd²/4。代入 d = 2 mm = 2 × 10⁻³ m：A = π(2 × 10⁻³)² / 4 = π × 4 × 10⁻⁶ / 4 = π × 10⁻⁶ m² ≈ 3.1416 × 10⁻⁶ m²。应力 σ 定义为单位面积上的力：σ = F / A = 300 N / (π × 10⁻⁶) ≈ 9.549 × 10⁷ Pa = 95.5 MPa。

Strain ε is the extension per unit original length. The change in length ΔL = 0.75 mm = 0.75 × 10⁻³ m, and L₀ = 1.5 m. Thus ε = ΔL / L₀ = 0.75 × 10⁻³ / 1.5 = 0.5 × 10⁻³ = 0.0005 (dimensionless). Young’s modulus E = σ / ε = 95.5 × 10⁶ Pa / 0.0005 = 1.91 × 10¹¹ Pa = 191 GPa. This value is typical for steel.

应变 ε 是单位原长的伸长量。长度变化 ΔL = 0.75 mm = 0.75 × 10⁻³ m，原长 L₀ = 1.5 m。因此 ε = ΔL / L₀ = 0.75 × 10⁻³ / 1.5 = 0.5 × 10⁻³ = 0.0005（无量纲）。杨氏模量 E = σ / ε = 95.5 × 10⁶ Pa / 0.0005 = 1.91 × 10¹¹ Pa = 191 GPa。该数值符合钢材的典型特征。

σ ≈ 95.5 MPa, ε = 0.0005, E ≈ 191 GPa

3. Kinematics – Uniformly Accelerated Motion | 运动学 – 匀加速运动

A car moves along a straight road with an initial velocity of 15 m/s. It accelerates uniformly and reaches a velocity of 25 m/s after 10 seconds. Calculate the acceleration and the displacement of the car during this 10‑second interval.

一辆汽车沿直线道路行驶，初速度为 15 m/s。它做匀加速运动，10 秒后速度达到 25 m/s。计算汽车在这 10 秒内的加速度和位移。

Use the standard equations of motion for constant acceleration: v = u + a t, and s = u t + ½ a t², where u = 15 m/s, v = 25 m/s, and t = 10 s. Rearranging the first equation gives a = (v − u) / t = (25 − 15) / 10 = 10/10 = 1.0 m/s². The acceleration is therefore 1 m/s² in the direction of motion.

运用匀加速直线运动的标准公式：v = u + a t 和 s = u t + ½ a t²，其中 u = 15 m/s，v = 25 m/s，t = 10 s。由第一式得 a = (v − u) / t = (25 − 15) / 10 = 10/10 = 1.0 m/s²。因此加速度大小为 1 m/s²，方向与运动方向相同。

Now substitute into the displacement equation: s = 15 × 10 + ½ × 1 × (10)² = 150 + 0.5 × 100 = 150 + 50 = 200 m. The car travels 200 metres while accelerating.

将加速度代入位移公式：s = 15 × 10 + ½ × 1 × (10)² = 150 + 0.5 × 100 = 150 + 50 = 200 m。汽车在加速过程中行驶了 200 米。

a = 1 m/s², s = 200 m

4. DC Circuit Analysis with Kirchhoff’s Voltage Law | 基尔霍夫电压定律的直流电路分析

A simple series circuit consists of a 12 V DC battery and two resistors: R₁ = 4 Ω and R₂ = 6 Ω. Use Kirchhoff’s voltage law and Ohm’s law to determine the current flowing in the circuit and the voltage drop across each resistor.

一个简单的串联电路由 12 V 直流电源和两个电阻 R₁ = 4 Ω、R₂ = 6 Ω 组成。运用基尔霍夫电压定律和欧姆定律，求电路中的电流以及每个电阻上的电压降。

In a series circuit, the total resistance R_total = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω. According to Kirchhoff’s voltage law, the sum of voltage drops around a closed loop equals the supply voltage: V_s = V₁ + V₂ = 12 V. By Ohm’s law, the circuit current I = V_s / R_total = 12 V / 10 Ω = 1.2 A.

在串联电路中，总电阻 R_total = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω。根据基尔霍夫电压定律，闭合回路中各段电压降之和等于电源电压：V_s = V₁ + V₂ = 12 V。由欧姆定律，电路中的电流 I = V_s / R_total = 12 V / 10 Ω = 1.2 A。

The voltage drop across R₁ is V₁ = I × R₁ = 1.2 A × 4 Ω = 4.8 V. The voltage drop across R₂ is V₂ = I × R₂ = 1.2 A × 6 Ω = 7.2 V. As a check, 4.8 V + 7.2 V = 12.0 V, satisfying KVL.

R₁ 上的电压降 V₁ = I × R₁ = 1.2 A × 4 Ω = 4.8 V。R₂ 上的电压降 V₂ = I × R₂ = 1.2 A × 6 Ω = 7.2 V。检验：4.8 V + 7.2 V = 12.0 V，符合 KVL。

I = 1.2 A, V₁ = 4.8 V, V₂ = 7.2 V

5. First Law of Thermod

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