📚 Year 12 Edexcel Engineering Unit Test Mock Paper Walkthrough | 爱德思工程单元测试模拟卷解析
This article provides a detailed walkthrough of a mock examination paper designed for Year 12 Edexcel Engineering, covering core topics such as statics, dynamics, electrical principles, materials, and energy systems. Each question is explained step by step to consolidate understanding and improve exam technique.
本文为爱德思工程学科 12 年级单元测试模拟卷提供详细解析，涵盖静力学、动力学、电学原理、工程材料与能量系统等核心模块。每道题均配有分步讲解，帮助学生巩固知识并提升应试技巧。
1. Question 1: Reaction Forces in a Simply Supported Beam | 第 1 题：简支梁的支反力计算
A uniform beam of length 6.0 m and weight 200 N rests on two supports at its ends. A concentrated load of 500 N is placed 2.0 m from the left support. The task is to determine the vertical reaction forces at the left and right supports, RA and RB.
一根长 6.0 m、重 200 N 的均质梁两端简支，在距左支点 2.0 m 处作用一个 500 N 的集中载荷，需求左右支座的垂直反力 RA 和 RB。
Apply the equilibrium condition for moments about point A. The clockwise moments are caused by the weight of the beam (acting at the centre, 3.0 m from A) and the load. The anticlockwise moment is due to RB. Sum of moments about A: RB × 6.0 = (200 × 3.0) + (500 × 2.0) → RB = (600 + 1000) / 6.0 = 266.7 N. Then, using vertical equilibrium, RA + RB = 200 + 500 = 700 N, so RA = 700 − 266.7 = 433.3 N.
对点 A 取矩，顺时针力矩由梁的自重（作用于中点，距 A 3.0 m）和集中载荷产生，逆时针力矩由 RB 提供。力矩平衡：RB × 6.0 = (200 × 3.0) + (500 × 2.0) → RB = 1600 / 6.0 = 266.7 N。竖直方向力平衡：RA + RB = 700 N，得 RA = 433.3 N。
Key principle: the principle of moments states that for an object in rotational equilibrium, the sum of clockwise moments about any pivot equals the sum of anticlockwise moments. Always remember to include the self-weight of the beam acting at its centre of gravity.
关键原理：力矩平衡原理指出，对于处于转动平衡的物体，对任意支点取矩，顺时针力矩之和等于逆时针力矩之和。务必考虑梁的自重并作用于其重心位置。
2. Question 2: Conservation of Energy – Roller Coaster | 第 2 题：能量守恒 – 过山车
A roller coaster car of mass 500 kg is released from rest at point A, 30 m above the ground. It travels down a frictionless track to point B, which is 10 m above the ground. Calculate the speed of the car at point B. Take g = 9.81 m/s².
一辆质量 500 kg 的过山车从离地 30 m 高的 A 点静止释放，沿无摩擦轨道滑行至离地 10 m 高的 B 点。求小车在 B 点的速度，取 g = 9.81 m/s²。
Since the track is frictionless, mechanical energy is conserved. Loss in gravitational potential energy = gain in kinetic energy: mg(hA − hB) = ½ mv². Cancel m: 9.81 × (30 − 10) = ½ v² → v² = 2 × 9.81 × 20 = 392.4 → v = √392.4 ≈ 19.8 m/s.
由于轨道无摩擦，机械能守恒。重力势能的减少量等于动能的增加量：mg(hA − hB) = ½ mv²。消去 m，得 v² = 2 × 9.81 × 20 = 392.4，v ≈ 19.8 m/s。
The question tests understanding of energy conversion and the conservation principle. In real engineering applications, friction would dissipate some energy, and calculations would involve work done against resistive forces.
本题考察能量转换与守恒定律的理解。在实际工程应用中，摩擦力会耗散部分能量，计算时需考虑克服阻力所做的功。
3. Question 3: Ohm’s Law and Series-Parallel Combination | 第 3 题：欧姆定律与串并联组合
A 12 V battery is connected to a network consisting of a 4 Ω resistor in series with a parallel combination of a 6 Ω and a 3 Ω resistor. Find the total current drawn from the battery and the current through the 6 Ω resistor.
一个 12 V 电池连接至由 4 Ω 电阻与 6 Ω 和 3 Ω 电阻并联组合串联而成的网络。求电池输出的总电流和流过 6 Ω 电阻的电流。
First, calculate the equivalent resistance of the parallel branch: 1/Rp = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = ½, so Rp = 2 Ω. Total circuit resistance Rtotal = 4 + 2 = 6 Ω. Total current I = V / Rtotal = 12 / 6 = 2 A. The voltage across the parallel branch is Vp = I × Rp = 2 × 2 = 4 V. Current through 6 Ω resistor: I6 = Vp / 6 = 4 / 6 = 0.667 A.
先计算并联支路等效电阻：1/Rp = 1/6 + 1/3 = ½，得 Rp = 2 Ω。总电阻 Rtotal = 4 + 2 = 6 Ω。总电流 I = 12 / 6 = 2 A。并联支路电压 Vp = 2 × 2 = 4 V。流过 6 Ω 电阻的电流 I6 = 4 / 6 ≈ 0.667 A。
Ohm’s law and rules for series and parallel circuits are fundamental in electrical engineering. A common mistake is forgetting that the voltage across parallel resistors is the same, while the current divides inversely with resistance.
欧姆定律及串并联电路规则是电气工程的基础。常见错误是忘记并联电阻两端电压相同，而电流按电阻反比分配。
4. Question 4: Stress, Strain and Young’s Modulus | 第 4 题：应力、应变与杨氏模量
A steel rod of diameter 10 mm and original length 2.0 m is subjected to a tensile force of 15 kN. The extension measured is 1.8 mm. Calculate (a) the stress in the rod, (b) the strain, and (c) Young’s modulus of the material. Cross-sectional area A = πd²/4.
一根直径 10 mm、原长 2.0 m 的钢杆承受 15 kN 的拉力，测得伸长量为 1.8 mm。计算 (a) 杆的应力，(b) 应变，(c) 材料的杨氏模量。截面积 A = πd²/4。
First, area A = π × (10 × 10⁻³)² / 4 = 7.854 × 10⁻⁵ m². Stress σ = Force / Area = 15 000 N / 7.854 × 10⁻⁵ m² = 1.91 × 10⁸ Pa (191 MPa). Strain ε = extension / original length = 1.8 × 10⁻³ m / 2.0 m = 9.0 × 10⁻⁴ (dimensionless). Young’s modulus E = σ / ε = 1.91 × 10⁸ / 9.0 × 10⁻⁴ = 2.12 × 10¹¹ Pa (212 GPa), which is typical for steel.
截面积 A = π × (10×10⁻³)² / 4 = 7.854×10⁻⁵ m²。应力 σ = 15000 / 7.854×10⁻⁵ ≈ 1.91×10⁸ Pa (191 MPa)。应变 ε = 1.8×10⁻³ / 2.0 = 9.0×10⁻⁴。杨氏模量 E = σ/ε ≈ 2.12×10¹¹ Pa (212 GPa)，符合钢材的典型值。
This question reinforces the definitions of stress and strain and the calculation of Young’s modulus within the elastic limit. Always convert units to SI (metres, pascals) to avoid errors.
本题强化了应力与应变的定义及在弹性范围内杨氏模量的计算。务必统一使用国际单位制（米、帕斯卡）以免出错。
5. Question 5: Vector Addition of Forces | 第 5 题：力的矢量加法
Two forces act on a point: 50 N at 30° to the horizontal, and 80 N at 120° to the horizontal. Determine the magnitude and direction of the resultant force using resolution of components.
两个力作用于同一点：50 N，与水平方向夹角 30°；80 N，与水平方向夹角 120°。请用分量法求合力的大小和方向。
Resolve each force: F1x = 50 cos30° = 43.30 N, F1y = 50 sin30° = 25.00 N. F2x = 80 cos120° = −40.00 N, F2y = 80 sin120° = 69.28 N. Resultant components: Rx = 43.30 + (−40.00) = 3.30 N, Ry = 25.00 + 69.28 = 94.28 N. Magnitude R = √(3.30² + 94.28²) ≈ 94.34 N. Direction θ = tan⁻¹(Ry / Rx) = tan⁻¹(94.28/3.30) ≈ 88.0° above the horizontal. Because Rx is very small, the resultant is nearly vertical.
分解各力：F1x = 50 cos30° = 43.30 N，F1y = 50 sin30° = 25.00 N。F2x = 80 cos120° = −40.00 N，F2y = 80 sin120° = 69.28 N。合力分量：Rx = 3.30 N，Ry = 94.28 N。合力大小 R ≈ 94.34 N，方向 θ = tan⁻¹(94.28/3.30) ≈ 88.0°（相对于水平线），因 Rx 极小，合力接近竖直方向。
Resolving vectors into perpendicular components is a fundamental skill in engineering mechanics. Pay close attention to the signs of cosine and sine for angles in different quadrants.
将矢量分解为相互垂直的分量是工程力学的基本技能。需特别注意不同象限角度的正余弦符号。
6. Question 6: Power and Efficiency of a Motor | 第 6 题：电动机的功率与效率
An electric motor lifts a 200 kg mass vertically at a constant speed of 0.5 m/s. The motor is supplied with 1.2 kW of electrical power. Calculate the mechanical output power and the efficiency of the motor. g = 9.81 m/s².
一台电动机以 0.5 m/s 的恒定速度竖直提升 200 kg 的重物，供电功率为 1.2 kW。计算机械输出功率和电动机的效率。g = 9.81 m/s²。
The force required to lift the mass at constant speed equals its weight: F = mg = 200 × 9.81 = 1962 N. Output power Pout = force × velocity = 1962 × 0.5 = 981 W. The input power Pin is 1200 W. Efficiency η = (Pout / Pin) × 100% = (981 / 1200) × 100% ≈ 81.75%.
匀速提升所需力等于重力：F = 200×9.81 = 1962 N。输出功率 Pout = 1962×0.5 = 981 W。输入功率 Pin = 1200 W。效率 η = (981/1200)×100% ≈ 81.75%。
Power is the rate of doing work. In lifting operations, output power can be found from F × v. Efficiency is a dimensionless ratio; expressing it as a percentage is conventional. The remaining 18.25% is lost mainly as heat in the motor and friction.
功率是做功的速率。在提升作业中，输出功率可由 F×v 求得。效率是无量纲比值，通常以百分数表示。其余 18.25% 主要耗散为电机发热及摩擦损失。
7. Question 7: Logic Gates and Truth Table | 第 7 题：逻辑门与真值表
Construct the truth table for a digital circuit comprising an AND gate and a NOT gate connected to form a NAND gate, with inputs A and B. Then, show the Boolean expression and draw the logic symbol.
构建由与门和非门连接形成与非门的真值表，输入为 A 和 B。随后写出其布尔表达式并画出逻辑符号。
The NAND gate outputs 0 only when both inputs are 1. Truth table: A=0, B=0 → out=1; A=0, B=1 → 1; A=1, B=0 → 1; A=1, B=1 → 0. Boolean expression: Q = NOT (A AND B) = (A · B)’. The symbol is an AND gate with a small circle (inversion bubble) at the output.
与非门仅在两个输入均为 1 时输出 0。真值表：A=0,B=0 → 输出 1；A=0,B=1 → 1；A=1,B=0 → 1；A=1,B=1 → 0。布尔表达式：Q = (A·B)′。逻辑符号为与门输出端带有小圆圈（反相泡）。
Understanding basic logic gates is essential for digital electronics engineering. NAND gates are universal gates, meaning any other logic function can be constructed using only NAND gates.
掌握基本逻辑门是数字电子工程的基础。与非门是通用门，即仅用与非门即可构建任何其他逻辑功能。
8. Question 8: Principle of Moments – Crane Jib | 第 8 题：力矩原理 – 起重机臂
A crane jib of length 4.0 m and mass 150 kg is pivoted at one end and held at an angle of 40° to the horizontal by a horizontal cable attached 3.0 m from the pivot. A load of 500 kg is suspended from the free end. Determine the tension in the cable. g = 9.81 m/s².
一起重机臂长 4.0 m，质量 150 kg，一端铰接，在与水平成 40° 角的位置由距铰点 3.0 m 处的水平拉索固定。自由端悬挂 500 kg 载荷。求拉索拉力。g = 9.81 m/s²。
Take moments about the pivot. The weight of the jib (1471.5 N) acts at its centre, 2.0 m from the pivot along the boom, with a perpendicular distance of 2.0 cos40°. The load (4905 N) acts 4.0 m from the pivot, perpendicular distance 4.0 cos40°. The cable tension T acts 3.0 m from the pivot, with perpendicular distance 3.0 sin40°. Equilibrium: T × 3.0 sin40° = (1471.5 × 2.0 cos40°) + (4905 × 4.0 cos40°). Solve for T: T = [cos40° (2943 + 19620)] / (3.0 sin40°) = (22563 cos40°) / (3.0 sin40°) = 22563 / (3.0 tan40°) ≈ 22563 / (3.0 × 0.8391) ≈ 8960 N.
对铰点取矩。臂自重 1471.5 N 作用于中点，沿臂方向距铰点 2.0 m，力臂为 2.0 cos40°。载荷 4905 N 作用于末端，力臂为 4.0 cos40°。拉索拉力 T 距铰点 3.0 m，力臂为 3.0 sin40°。力矩平衡：T × 3.0 sin40° = (1471.5 × 2.0 cos40°) + (4905 × 4.0 cos40°)。解得 T = [cos40° × 22563] / (3 sin40°) ≈ 8960 N。
For inclined structures, ensure you use the perpendicular distance from the line of action to the pivot. Breaking the geometry into components is often simpler when the angle is given relative to the horizontal.
对于倾斜结构，务必使用力作用线到支点的垂直距离。当给定角度为相对水平线时，分解几何关系往往更加简便。
9. Question 9: Thermal Expansion of a Pipeline | 第 9 题：管道热膨胀
A steel pipe is 25.0 m long at 15 °C. It will be used to carry steam at 120 °C, and the coefficient of linear expansion for steel is 12 × 10⁻⁶ /°C. Calculate the increase in length and the final length of the pipe. Explain why expansion loops are installed in long steam pipes.
一根钢管在 15 °C 时长 25.0 m，用于输送 120 °C 的蒸汽，钢材线膨胀系数为 12×10⁻⁶ /°C。计算管的伸长量和最终长度，并解释为何长蒸汽管道需要安装膨胀弯。
Change in temperature ΔT = 120 − 15 = 105 °C. ΔL = α L₀ ΔT = 12×10⁻⁶ × 25.0 × 105 = 0.0315 m = 31.5 mm. Final length L = 25.0 + 0.0315 = 25.0315 m. Expansion loops or U-bends are used to absorb thermal expansion without causing excessive stress that could damage the pipe or supports.
温升 ΔT = 105 °C。伸长量 ΔL = α L₀ ΔT = 12×10⁻⁶ × 25.0 × 105 = 0.0315 m（31.5 mm）。最终长度 25.0315 m。膨胀弯或 U 形弯用于吸收热膨胀，避免产生过大应力损坏管道或支架。
Thermal expansion is a critical consideration in engineering design, particularly for long pipelines, bridges, and railway tracks. Allowing movement prevents buckling and structural failure.
热膨胀是工程设计中的关键因素，尤其对长管道、桥梁和铁轨。设置膨胀间隙可防止屈曲和结构破坏。
10. Question 10: AC Circuit Concepts – Peak and RMS Values | 第 10 题：交流电路概念 – 峰值与有效值
The UK mains supply is rated at 230 V RMS, with a frequency of 50 Hz. Determine the peak voltage and write an expression for the instantaneous voltage as a function of time t. Also calculate the instantaneous voltage when t = 8 ms.
英国市电额定值为 230 V 有效值，频率 50 Hz。求峰值电压，写出瞬时电压随时间 t 变化的表达式，并计算 t = 8 ms 时的瞬时电压。
Peak voltage V₀ = √2 × VRMS = 1.414 × 230 ≈ 325 V. Angular frequency ω = 2πf = 2π × 50 = 100π ≈ 314.16 rad/s. Instantaneous voltage v(t) = V₀ sin(ωt) = 325 sin(100π t). At t = 8 ms = 0.008 s, v = 325 sin(100π × 0.008) = 325 sin(0.8π) = 325 sin(144°) ≈ 325 × 0.5878 ≈ 191 V.
峰值电压 V₀ = √2 × 230 ≈ 325 V。角频率 ω = 2π × 50 = 100π rad/s。瞬时电压 v(t) = 325 sin(100π t)。当 t = 8 ms 时，v = 325 sin(0.8π) ≈ 325 × 0.5878 ≈ 191 V。
RMS (root mean square) value is the effective DC equivalent of an AC waveform. For sinusoidal signals, VRMS = V₀ / √2. This concept is fundamental for power calculations in AC circuits.
有效值（均方根值）是交流波形等效为直流的数值。对于正弦信号，VRMS = V₀ / √2。此概念是交流电路功率计算的基础。
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