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

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

This walkthrough takes you through a representative CAIE Engineering unit test paper, explaining each question, the working required, and the examiner’s key marking points. It is designed to help you move from knowing formulas to applying them under timed conditions.

本解析带你逐题完成一份具有代表性的 CAIE 工程单元测试模拟卷，讲解每道题的解答步骤、所需计算和考官关键评分点，帮助你从记住公式提升到在限时条件下正确应用。

1. Mock Paper Overview | 模拟卷概览

This mock unit test contains ten compulsory questions, worth a total of 60 raw marks. You should allow about 75 minutes. The topics cover mechanics of materials, statics, dynamics, electrical and electronic principles, thermofluids, and engineering design.

本模拟单元测试包含 10 道必答题，满分 60 分，建议用时 75 分钟。考查范围包括材料力学、静力学、动力学、电工电子原理、热流体以及工程设计。

• Answer all questions. 回答所有题目。
• Show all working; marks are awarded for method, not just the final answer. 写出完整步骤；分数不仅给最终答案，也给方法分。
• Use SI units unless stated otherwise. 除非另有说明，使用国际单位制。
• Give final answers to three significant figures. 最终答案保留三位有效数字。

---

2. Q1: Tensile Stress and Unit Conversion | 第1题：拉伸应力与单位换算

Question: A steel rod has a diameter of 12 mm and carries an axial load of 15 kN. Calculate the cross-sectional area in m² and the tensile stress in MPa. State one assumption.

题目： 一根钢杆直径为 12 mm，承受 15 kN 的轴向载荷。计算横截面积（m²）和拉伸应力（MPa），并说明一个假设。

Answer: Cross-sectional area A = πd²/4 = π(0.012 m)²/4 = 1.131 × 10⁻⁴ m². Tensile stress σ = F/A = 15000 N / 1.131 × 10⁻⁴ m² = 1.326 × 10⁸ Pa = 132.6 MPa. Assumption: the load is applied axially and is uniformly distributed across the section.

答案： 横截面积 A = πd²/4 = π(0.012 m)²/4 = 1.131 × 10⁻⁴ m²。拉伸应力 σ = F/A = 15000 N / 1.131 × 10⁻⁴ m² = 1.326 × 10⁸ Pa = 132.6 MPa。假设：载荷沿轴向施加，并且均匀分布在截面上。

Examiner note: The most common error is using millimetres directly in the area formula. Always convert diameter to metres before calculating stress in pascals.

考官提示： 最常见的错误是在面积公式中直接使用毫米。计算应力（Pa）前，必须先将直径转换为米。

---

3. Q2: Young’s Modulus, Strain and Factor of Safety | 第2题：杨氏模量、应变与安全系数

Question: A 2.0 m wire stretches 1.4 mm under a stress of 180 MPa. Calculate the strain and Young’s modulus. If the wire’s ultimate tensile stress is 450 MPa, determine the factor of safety.

题目： 一段 2.0 m 的金属丝在 180 MPa 应力下伸长 1.4 mm。计算应变和杨氏模量。如果该金属丝的抗拉极限强度为 450 MPa，求安全系数。

Answer: Strain ε = ΔL/L₀ = 1.4 × 10⁻³ m / 2.0 m = 7.0 × 10⁻⁴. Young’s modulus E = σ/ε = 180 × 10⁶ Pa / 7.0 × 10⁻⁴ = 2.571 × 10¹¹ Pa ≈ 257 GPa. Factor of safety = ultimate stress / working stress = 450 MPa / 180 MPa = 2.5.

答案： 应变 ε = ΔL/L₀ = 1.4 × 10⁻³ m / 2.0 m = 7.0 × 10⁻⁴。杨氏模量 E = σ/ε = 180 × 10⁶ Pa / 7.0 × 10⁻⁴ = 2.571 × 10¹¹ Pa ≈ 257 GPa。安全系数 = 极限应力 / 工作应力 = 450 MPa / 180 MPa = 2.5。

Examiner note: Strain is dimensionless, and Young’s modulus must carry a unit, usually Pa or GPa. The factor of safety has no unit and must always be greater than 1.

考官提示： 应变无量纲，杨氏模量必须带单位，通常为 Pa 或 GPa。安全系数无量纲，必须始终大于 1。

---

4. Q3: Resolving Forces and Equilibrium | 第3题：力的分解与平衡

Question: A 50 N force acts at 30° to the horizontal. Calculate the horizontal and vertical components. A block remains stationary on a rough horizontal surface under this force. State the friction force required for equilibrium.

题目： 一个 50 N 的力与水平方向成 30° 角。计算该力的水平分量和竖直分量。一个物块在该力作用下仍静止在粗糙水平面上，说明维持平衡所需的摩擦力。

Answer: Horizontal component Fₓ = 50 cos 30° = 43.3 N. Vertical component Fy = 50 sin 30° = 25.0 N. For equilibrium, the friction force must be 43.3 N opposite to the horizontal component. The upward vertical component reduces the normal reaction by 25 N.

答案： 水平分量 Fₓ = 50 cos 30° = 43.3 N。竖直分量 Fy = 50 sin 30° = 25.0 N。为保持平衡，摩擦力必须为 43.3 N，方向与水平分量相反。向上的竖直分量使法向反力减少 25 N。

Examiner note: Use the correct component: cosine gives the adjacent component to the angle. Always state the direction of friction because force is a vector.

考官提示： 正确使用分量：余弦给出与角度相邻的分量。力是矢量，因此必须说明摩擦力的方向。

---

5. Q4: Principle of Moments and Beam Reactions | 第4题：力矩原理与梁的支反力

Question: A uniform beam 6.0 m long weighs 200 N. It rests on supports at A on the left and B on the right. A 300 N load sits 2.0 m from A. Calculate the reaction at B by taking moments about A, then find the reaction at A.

题目： 一根长 6.0 m 的均匀梁重 200 N，左端支点 A、右端支点 B 支撑。一个 300 N 的载荷位于距 A 端 2.0 m 处。以 A 为支点取矩，计算 B 处反力，再求 A 处反力。

Answer: Clockwise moments about A = (300 N × 2.0 m) + (200 N × 3.0 m) = 600 Nm + 600 Nm = 1200 Nm. Anticlockwise moment = R_B × 6.0 m. Equating: R_B = 1200 Nm / 6.0 m = 200 N. Vertical equilibrium gives R_A = total downward force − R_B = 500 N − 200 N = 300 N.

答案： 绕 A 点的顺时针力矩 = (300 N × 2.0 m) + (200 N × 3.0 m) = 600 Nm + 600 Nm = 1200 Nm。逆时针力矩 = R_B × 6.0 m。力矩平衡：R_B = 1200 Nm / 6.0 m = 200 N。由竖直方向平衡得 R_A = 总向下力 − R_B = 500 N − 200 N = 300 N。

Examiner note: The beam’s own weight acts at its centre, which is 3.0 m from A because the beam is uniform. Always take moments about a point where an unknown reaction acts to eliminate it from the equation.

考官提示： 梁的自重作用在其中点，因为是均匀梁，该点距 A 端 3.0 m。通常对某一未知反力作用点取矩，可使该未知量不出现在力矩方程中。

---

6. Q5: Work, Energy and Power | 第5题：功、能量与功率

Question: A hoist lifts a 250 kg mass through a vertical height of 12 m in 20 s at constant speed. Calculate the work done, the useful power output, and state one source of energy loss.

题目： 一台提升机在 20 s 内将 250 kg 的质量匀速提升 12 m。计算所做的功、有用功率输出，并说明一种能量损失来源。

Answer: Weight = mg = 250 kg × 9.81 m/s² = 2452.5 N. Work done = force × distance = 2452.5 N × 12 m = 29,430 J

Published by TutorHao | A-Level 工程 Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）