📚 Year 13 Edexcel Engineering Unit Test: Mock Paper Analysis | Edexcel Year 13 工程单元测试模拟卷解析
This mock paper has been designed to mirror the style and demand of a typical Year 13 Edexcel Engineering unit test, covering key topics from the Engineering Principles specification. Working through these questions will help you reinforce essential calculation methods, apply physical relationships confidently, and identify common pitfalls under timed conditions. Take each question as a revision opportunity – method marks are as valuable as final answers.
本模拟卷旨在还原 Edexcel 工程 Year 13 单元测试的题型与难度，覆盖工程原理核心知识点。通过逐题解析，你既能巩固关键计算方法，又能提升在限时条件下灵活运用物理关系的能力。请把每道题当作一次梳理机会 —— 解题步骤的分值与最终答案同样重要。
1. Equilibrium and Resolving Forces | 力的平衡与分解
Mock question: A lifting eye experiences three coplanar forces: 120 N acting east, 90 N acting north-west, and a third force P keeping the system in equilibrium. Determine the magnitude of P and its direction measured anticlockwise from the positive x-axis.
Draw all forces on a free-body diagram and resolve the north-west force into horizontal and vertical components: Fₓ = –90 cos 45° = –63.6 N, Fᵧ = 90 sin 45° = 63.6 N. For equilibrium, the sum of horizontal components must be zero: 120 + (–63.6) + Pₓ = 0 → Pₓ = –56.4 N. Vertically: 0 + 63.6 + Pᵧ = 0 → Pᵧ = –63.6 N. So P = √(56.4² + 63.6²) = 85.0 N. The direction θ = tan⁻¹(63.6 / 56.4) = 48.4°, but since both components are negative, the angle measured anticlockwise from the positive x-axis is 180 + 48.4 = 228.4°.
模拟题目： 一个吊环受到三个共面力：120 N 向东，90 N 向西北，以及保持平衡的第三个力 P。求 P 的大小和从正 x 轴逆时针测量的方向角。
首先画出受力图，将西北方向的力分解：水平分量 Fₓ = –90 cos 45° = –63.6 N，竖直分量 Fᵧ = 90 sin 45° = 63.6 N。根据平衡条件，水平合力为零：120 + (–63.6) + Pₓ = 0 → Pₓ = –56.4 N。竖直合力为零：0 + 63.6 + Pᵧ = 0 → Pᵧ = –63.6 N。因此 P = √(56.4² + 63.6²) = 85.0 N。方向角 θ = tan⁻¹(63.6 / 56.4) = 48.4°，因两个分量均为负，实际从正 x 轴逆时针测得的角度为 228.4°。
2. Moments and Cantilever Beams | 力矩与悬臂梁
A uniform cantilever beam of length 2.5 m is fixed at one end and carries a point load of 400 N at its free end. The beam itself has a weight of 150 N. Calculate the bending moment at the fixed support.
Bending moment at the fixed end = sum of moments due to the point load and the beam’s own weight. The point load creates a moment of 400 N × 2.5 m = 1000 N m. The weight acts through the centre of gravity, which is at the midpoint (1.25 m from the support), giving a moment of 150 N × 1.25 m = 187.5 N m. Total bending moment = 1000 + 187.5 = 1187.5 N m.
一根长 2.5 m 的均匀悬臂梁一端固定，自由端承受 400 N 的集中载荷，梁的自重为 150 N。求固定端处的弯矩。
固定端弯矩等于集中载荷和梁自重引起的力矩之和。集中载荷产生的力矩 = 400 N × 2.5 m = 1000 N·m。自重视为作用在跨中（距固定端 1.25 m），产生的力矩 = 150 N × 1.25 m = 187.5 N·m。总弯矩 = 1000 + 187.5 = 1187.5 N·m。
3. Stress, Strain and Young Modulus | 应力、应变与杨氏模量
A steel tie rod of diameter 12 mm and original length 2.0 m extends by 1.8 mm under a tensile load of 24 kN. Determine the tensile stress, tensile strain, and Young modulus of the material. Also calculate the strain energy stored assuming elastic behaviour.
Cross-sectional area A = π(0.012/2)² = 1.131 × 10⁻⁴ m². Stress σ = Force / Area = 24000 N / 1.131 × 10⁻⁴ = 2.12 × 10⁸ Pa. Strain ε = extension / original length = 1.8 × 10⁻³ m / 2.0 m = 9.0 × 10⁻⁴. Young modulus E = σ / ε = 2.12 × 10⁸ / 9.0 × 10⁻⁴ = 2.36 × 10¹¹ Pa. Strain energy U = ½ × stress × strain × volume. Volume = A × L = 1.131 × 10⁻⁴ × 2.0 = 2.262 × 10⁻⁴ m³. U = 0.5 × 2.12 × 10⁸ × 9.0 × 10⁻⁴ × 2.262 × 10⁻⁴ ≈ 21.6 J.
一根直径 12 mm、原长 2.0 m 的钢拉杆在 24 kN 拉伸载荷下伸长了 1.8 mm。求拉伸应力、拉伸应变以及材料的杨氏模量。假设为弹性行为，并计算储存的应变能。
截面积 A = π(0.012/2)² = 1.131 × 10⁻⁴ m²。应力 σ = 力/面积 = 24000 N / 1.131 × 10⁻⁴ = 2.12 × 10⁸ Pa。应变 ε = 伸长量/原长 = 1.8 × 10⁻³ m / 2.0 m = 9.0 × 10⁻⁴。杨氏模量 E = σ/ε = 2.12 × 10⁸ / 9.0 × 10⁻⁴ = 2.36 × 10¹¹ Pa。应变能 U = ½ × 应力 × 应变 × 体积。体积 = A × L = 1.131 × 10⁻⁴ × 2.0 = 2.262 × 10⁻⁴ m³。U = 0.5 × 2.12 × 10⁸ × 9.0 × 10⁻⁴ × 2.262 × 10⁻⁴ ≈ 21.6 J。
4. Thermal Expansion | 热膨胀
A stainless steel pipe (coefficient of linear expansion α = 1.6 × 10⁻⁵ K⁻¹) is 6.00 m long at 15 °C. It is installed in a process where it reaches 95 °C. Calculate the free expansion and the thermal stress that would arise if the pipe were completely restrained, given E = 210 GPa.
Change in temperature Δθ = 95 – 15 = 80 K. Free expansion ΔL = α L₀ Δθ = 1.6 × 10⁻⁵ × 6.00 × 80 = 7.68 × 10⁻³ m = 7.68 mm. If restrained, thermal strain ε = ΔL / L₀ = α Δθ = 1.6 × 10⁻⁵ × 80 = 1.28 × 10⁻³. Stress σ = E ε = 210 × 10⁹ × 1.28 × 10⁻³ = 2.688 × 10⁸ Pa = 269 MPa (compressive).
一根不锈钢管道（线膨胀系数 α = 1.6 × 10⁻⁵ K⁻¹）在 15 °C 时长 6.00 m，安装后在工艺中达到 95 °C。计算自由膨胀量和假设管道完全约束时产生的热应力，已知 E = 210 GPa。
温度变化 Δθ = 95 – 15 = 80 K。自由膨胀量 ΔL = α L₀ Δθ = 1.6 × 10⁻⁵ × 6.00 × 80 = 7.68 × 10⁻³ m = 7.68 mm。若被完全约束，热应变 ε = ΔL / L₀ = α Δθ = 1.6 × 10⁻⁵ × 80 = 1.28 × 10⁻³。应力 σ = E ε = 210 × 10⁹ × 1.28 × 10⁻³ = 2.688 × 10⁸ Pa = 269 MPa（压应力）。
5. Fluid Statics and Manometry | 流体静力学与压力计
A U-tube manometer contains oil of relative density 0.85 and is used to measure the pressure of water in a pipe. The water interface in the manometer limb connected to the pipe is 180 mm below the pipe centreline, while the oil level in the open limb is 320 mm above the interface. Determine the gauge pressure of the water in the pipe.
Use the hydrostatic equation. Gauge pressure at pipe centre p = ρₒₒ g h₁ + ρ_w g h₂, where h₁ = 320 mm oil column, h₂ = 180 mm water column. ρ_w = 1000 kg m⁻³, ρₒₒ = 850 kg m⁻³. p = (850 × 9.81 × 0.32) + (1000 × 9.81 × 0.18) = 2668 + 1766 = 4434 Pa, or approximately 4.43 kPa.
一个 U 形管压力计装有相对密度 0.85 的油，用于测量管道中的水压力。连接管道的支管中水界面位于管道中心线下方 180 mm，开敞支管中的油位比界面高 320 mm。求管道中水的表压。
根据流体静力学方程，管道中心处的表压 p = ρₒₒ g h₁ + ρ_w g h₂，其中 h₁ = 320 mm 油柱，h₂ = 180 mm 水柱。ρ_w = 1000 kg m⁻³，ρₒₒ = 850 kg m⁻³。p = (850 × 9.81 × 0.32) + (1000 × 9.81 × 0.18) = 2668 + 1766 = 4434 Pa，约 4.43 kPa。
6. Bernoulli’s Equation | 伯努利方程
Water flows steadily through a horizontal pipe that narrows from a diameter of 80 mm to 40 mm. The pressure in the wider section is 250 kPa, and the flow velocity there is 2.5 m s⁻¹. Assuming no energy losses, calculate the pressure in the narrower section.
Apply the continuity equation: A₁v₁ = A₂v₂. A₁ = π(0.04)² = 5.027 × 10⁻³ m²; A₂ = π(0.02)² = 1.257 × 10⁻³ m². v₂ = (A₁/A₂) v₁ = (5.027/1.257) × 2.5 = 10.0 m s⁻¹. Bernoulli’s equation for horizontal flow: p₁/ρg + v₁²/(2g) = p₂/ρg + v₂²/(2g). Rearranging: p₂ = p₁ + ½ρ(v₁² – v₂²). ρ = 1000 kg m⁻³. p₂ = 250 × 10³ + 0.5 × 1000 × (2.5² – 10.0²) = 250000 – 46875 = 203125 Pa ≈ 203 kPa.
水在水平管道中稳定流动，管道从直径 80 mm 缩至 40 mm。宽截面处的压力为 250 kPa，流速为 2.5 m s⁻¹。假设无能量损失，求窄截面处的压力。
由连续性方程：A₁v₁ = A₂v₂。A₁ = π(0.04)² = 5.027 × 10⁻³ m²；A₂ = π(0.02)² = 1.257 × 10⁻³ m²。v₂ = (A₁/A₂) v₁ = (5.027/1.257) × 2.5 = 10.0 m s⁻¹。水平流动的伯努利方程：p₁/ρg + v₁²/(2g) = p₂/ρg + v₂²/(2g)。整理得：p₂ = p₁ + ½ρ(v₁² – v₂²)。ρ = 1000 kg m⁻³。p₂ = 250 × 10³ + 0.5 × 1000 × (2.5² – 10.0²) = 250000 – 46875 = 203125 Pa ≈ 203 kPa。
7. Thermodynamics – First Law | 热力学第一定律
In a closed system, 500 J of heat energy is supplied to a gas. The gas expands and does 320 J of work on the surroundings. Calculate the change in internal energy of the gas and state whether it is an increase or decrease. If the same process took place in a cylinder with a piston of area 0.025 m² moving 30 mm against a constant external pressure, determine that pressure.
From the first law, ΔU = Q – W = 500 J – 320 J = 180 J. Positive ΔU indicates an increase in internal energy. Work done W = p ΔV. Here displacement Δx = 30 mm = 0.03 m, so ΔV = A × Δx = 0.025 × 0.03 = 7.5 × 10⁻⁴ m³. Then p = W / ΔV = 320 / (7.5 × 10⁻⁴) = 4.27 × 10⁵ Pa = 427 kPa.
在一个封闭系统中，向气体提供 500 J 的热量，气体膨胀并对周围做 320 J 的功。求气体内能的变化量，并说明是增加还是减少。如果该过程发生在气缸内，活塞面积为 0.025 m²，在恒定外压下移动了 30 mm，求该压力值。
根据热力学第一定律，ΔU = Q – W = 500 J – 320 J = 180 J。ΔU 为正，表示内能增加。做功 W = p ΔV，位移 Δx = 30 mm = 0.03 m，故 ΔV = A × Δx = 0.025 × 0.03 = 7.5 × 10⁻⁴ m³。则 p = W / ΔV = 320 / (7.5 × 10⁻⁴) = 4.27 × 10⁵ Pa = 427 kPa。
8. Electrical Circuits in Engineering | 工程电路分析
A sensor circuit consists of a 12 V battery and two resistors R₁ = 470 Ω and R₂ = 330 Ω connected in series. Determine the total resistance, circuit current, and the voltage across R₂. The R₂ resistor is then replaced by a parallel combination of 330 Ω and a thermistor that reads 200 Ω at room temperature; find the new current drawn from the battery.
Series resistance R_total = 470 + 330 = 800 Ω. Current I = V / R_total = 12 / 800 = 0.015 A = 15 mA. Voltage V_R₂ = I × R₂ = 0.015 × 330 = 4.95 V. For the new circuit, R₂ is replaced by parallel combination R_parallel = (330 × 200)/(330 + 200) = 66000/530 = 124.5 Ω. New total resistance = 470 + 124.5 = 594.5 Ω. New current = 12 / 594.5 = 0.0202 A = 20.2 mA.
一个传感器电路由 12 V 电池和两个电阻 R₁ = 470 Ω、R₂ = 330 Ω 串联组成。求总电阻、电路电流以及 R₂ 两端的电压。随后，R₂ 替换为一个 330 Ω 电阻与室温下读数为 200 Ω 的热敏电阻并联；求此时从电池获取的新电流。
串联总电阻 R_total = 470 + 330 = 800 Ω。电流 I = V / R_total = 12 / 800 = 0.015 A = 15 mA。R₂ 两端电压 V_R₂ = I × R₂ = 0.015 × 330 = 4.95 V。新电路中，R₂ 被并联组合代替，并联阻值 R_parallel = (330 × 200)/(330 + 200) = 66000/530 = 124.5 Ω。新总电阻 = 470 + 124.5 = 594.5 Ω。新电流 = 12 / 594.5 = 0.0202 A = 20.2 mA。
9. Material Properties and Safety Factor | 材料性能与安全系数
A steel bolt is manufactured from a material with an ultimate tensile strength (UTS) of 520 MPa. In service, it must withstand a maximum working tensile stress of 160 MPa. Calculate the factor of safety based on UTS. If the bolt has a diameter of 16 mm, determine the maximum allowable tensile load using the working stress. Finally, comment on why a factor of safety is necessary even when the material operates below the yield point.
Factor of safety = UTS / working stress = 520 MPa / 160 MPa = 3.25. Cross-sectional area A = π(0.016/2)² = 2.011 × 10⁻⁴ m². Maximum allowable load F = working stress × A = 160 × 10⁶ × 2.011 × 10⁻⁴ = 32176 N ≈ 32.2 kN. A safety factor accounts for material imperfections, unexpected loads, fatigue, and stress concentrations. It ensures that localised stresses or transient overloads do not cause failure, even if the nominal stress is below yield.
一钢制螺栓采用极限抗拉强度（UTS）为 520 MPa 的材料制造。使用中必须承受的最大工作拉伸应力为 160 MPa。计算基于 UTS 的安全系数。若螺栓直径为 16 mm，利用工作应力确定最大允许拉伸载荷。并说明即使材料应力在屈服点以下，为什么仍需要安全系数。
安全系数 = UTS / 工作应力 = 520 MPa / 160 MPa = 3.25。截面积 A = π(0.016/2)² = 2.011 × 10⁻⁴ m²。最大允许载荷 F = 工作应力 × A = 160 × 10⁶ × 2.011 × 10⁻⁴ = 32176 N ≈ 32.2 kN。安全系数用于应对材料缺陷、意外载荷、疲劳和应力集中。即使名义应力低于屈服点，局部应力或瞬时过载也可能导致失效，安全系数正是为了防范此类风险。
10. Energy Methods and Efficiency | 能量方法与效率
An electric motor with an efficiency of 85% lifts a load of 2000 N at a constant speed of 0.5 m s⁻¹ through a hoist. Determine the mechanical output power and the electrical input power required. If the motor runs for 120 seconds, calculate the useful work done and the energy lost as heat.
Mechanical output power P_out = force × velocity = 2000 N × 0.5 m s⁻¹ = 1000 W. Efficiency η = P_out / P_in → P_in = P_out / η = 1000 / 0.85 = 1176.5 W. Useful work done = P_out × time = 1000 × 120 = 120000 J = 120 kJ. Energy lost = P_in × time – useful work = (1176.5 × 120) – 120000 = 141180 – 120000 = 21180 J ≈ 21.2 kJ. The lost energy appears as heat due to resistive and frictional losses in the motor and gearbox.
一台效率为 85% 的电动机通过卷扬机以 0.5 m/s 的恒定速度提升 2000 N 的重物。求机械输出功率和所需的电输入功率。若电动机运行 120 秒，计算有用功和以热量形式损失的能量。
机械输出功率 P_out = 力 × 速度 = 2000 N × 0.5 m/s = 1000 W。效率 η = P_out / P_in → P_in = P_out / η = 1000 / 0.85 = 1176.5 W。有用功 = P_out × 时间 = 1000 × 120 = 120000 J = 120 kJ。损失能量 = 输入总能量 – 有用功 = (1176.5 × 120) – 120000 = 141180 – 120000 = 21180 J ≈ 21.2 kJ。这部分能量因电机和齿轮箱中的电阻耗散与摩擦转化为热量散失。
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