📚 AS OCR Engineering: Unit Test Mock Paper Analysis | AS OCR 工程：单元测试模拟卷解析
This article provides a detailed analysis of a mock unit test for AS Level OCR Engineering, covering core principles such as mechanics, materials, and electrical systems. By working through each question, you will reinforce key calculation methods, understand common pitfalls, and improve exam technique.
本文对 AS 级别 OCR 工程的一份单元测试模拟卷进行了详细解析，涵盖力学、材料和电气系统等核心原理。通过逐步解答每个问题，你将巩固关键计算方法、了解常见错误并提高应试技巧。
1. Stress and Strain in a Tensile Member | 受拉构件中的应力和应变
A copper tie rod of diameter 8 mm carries an axial load of 5 kN. Its original gauge length is 50 mm and it extends by 0.12 mm under load. We need to calculate the stress, strain, and Young’s modulus.
一根直径为 8 mm 的铜拉杆承受 5 kN 的轴向载荷。其原始标距为 50 mm，在载荷作用下伸长了 0.12 mm。我们需要计算应力、应变和杨氏模量。
First, cross-sectional area: A = π d² / 4 = π × (8 mm)² / 4 = 50.27 mm². Convert to square metres: A = 50.27 × 10⁻⁶ m².
首先计算横截面积：A = π d² / 4 = π × (8 mm)² / 4 = 50.27 mm²。换算为平方米：A = 50.27 × 10⁻⁶ m²。
σ = F / A = 5000 N / (50.27 × 10⁻⁶ m²) ≈ 99.5 × 10⁶ Pa = 99.5 MPa
Engineering strain is the ratio of extension to original length.
工程应变是伸长量与原始长度的比值。
ε = ΔL / L₀ = 0.12 mm / 50 mm = 0.0024
Young’s modulus E links stress and strain in the elastic region.
杨氏模量 E 将弹性区域的应力和应变联系起来。
E = σ / ε = 99.5 MPa / 0.0024 ≈ 41.5 GPa
Always check unit conversions; using millimetres consistently for area gives MPa if force is in newtons and area in mm², but converting to base SI units is safer.
务必检查单位换算；一致使用毫米时，若力以牛顿为单位、面积以 mm² 为单位可得 MPa，但转换为基本国际单位更为稳妥。
2. Principle of Moments for Equilibrium | 力矩平衡原理
A uniform beam 4 m long weighs 200 N and is pivoted at its centre. A 300 N load is placed 0.8 m from the left end. Determine where a 150 N weight must be placed on the right side to balance the beam.
一根长 4 m 的均匀梁重 200 N，在其中点设支点。一个 300 N 的载荷放置在距离左端 0.8 m 处。求需要在右侧何处放置一个 150 N 的重物才能使梁平衡。
Because the beam is uniform and pivoted at the centre, its weight acts through the pivot and does not produce a moment about it.
由于梁是均匀的并且支点在中心，其重量通过支点作用，不会对支点产生力矩。
Measure distances from the pivot. The left load is 300 N at a distance of (2.0 m – 0.8 m) = 1.2 m from the pivot, creating an anticlockwise moment.
从支点测量距离。左侧 300 N 载荷距离支点 (2.0 m – 0.8 m) = 1.2 m，产生逆时针力矩。
Σ M_acw = 300 N × 1.2 m = 360 Nm
For equilibrium, the clockwise moment must equal 360 Nm. The 150 N weight must be placed at a distance d from the pivot on the right.
为达到平衡，顺时针力矩必须等于 360 Nm。150 N 重物需放置在支点右侧距离 d 处。
150 N × d = 360 Nm ⇒ d = 2.4 m
Thus, the weight should be positioned 2.4 m from the pivot on the right side, which is 4.4 m from the left end.
因此，重物应放置在支点右侧 2.4 m 处，即距离左端 4.4 m 的位置。
3. Ohm’s Law and Resistor Calculations | 欧姆定律与电阻计算
A circuit consists of a 9 V battery and a 220 Ω resistor. Find the current flowing through the resistor and the power dissipated.
一个电路由一个 9 V 电池和一个 220 Ω 电阻组成。求流过电阻的电流以及耗散的功率。
Ohm’s law directly relates voltage, current, and resistance.
欧姆定律直接关联电压、电流和电阻。
V = I × R ⇒ I = V / R = 9 V / 220 Ω ≈ 0.0409 A = 40.9 mA
Electrical power can be expressed in several forms; using P = I² R is convenient when current is known.
电功率有多种表达形式；已知电流时用 P = I² R 较方便。
P = I² R = (0.0409 A)² × 220 Ω ≈ 0.368 W
Alternatively, P = V × I = 9 V × 0.0409 A ≈ 0.368 W gives the same result. In a simple resistive load, all electrical energy is converted to heat.
或者，P = V × I = 9 V × 0.0409 A ≈ 0.368 W 得到相同结果。在纯电阻负载中，所有电能都转化为热能。
4. Kirchhoff’s Current and Voltage Laws | 基尔霍夫电流与电压定律
In a parallel circuit, a 12 V supply feeds two branches. Branch 1 has a 100 Ω resistor; branch 2 has an unknown resistor R. The total current from the supply is 0.3 A. Find R and the branch currents.
在一个并联电路中，12 V 电源为两条支路供电。支路 1 有一个 100 Ω 电阻；支路 2 有一个未知电阻 R。电源总电流为 0.3 A。求 R 和各支路电流。
In a parallel circuit, the voltage across each branch is the supply voltage, 12 V.
在并联电路中，每条支路的电压等于电源电压 12 V。
Current in branch 1: I₁ = V / R₁ = 12 V / 100 Ω = 0.12 A.
支路 1 的电流：I₁ = V / R₁ = 12 V / 100 Ω = 0.12 A。
By Kirchhoff’s current law:
根据基尔霍夫电流定律：
Σ I_in = Σ I_out ⇒ I_total = I₁ + I₂
Thus, I₂ = 0.3 A – 0.12 A = 0.18 A.
因此，I₂ = 0.3 A – 0.12 A = 0.18 A。
Now apply Ohm’s law to branch 2: R = V / I₂ = 12 V / 0.18 A ≈ 66.7 Ω.
现在对支路 2 应用欧姆定律：R = V / I₂ = 12 V / 0.18 A ≈ 66.7 Ω。
Kirchhoff’s voltage law also holds: the sum of EMFs equals the sum of pds in any loop.
基尔霍夫电压定律同样成立：在任何回路中，电动势之和等于电压降之和。
5. Tensile Test Data and Young’s Modulus | 拉伸试验数据与杨氏模量
A mild steel specimen has a gauge length of 100 mm and diameter 10 mm. A tensile test gives the following data: at a load of 20 kN, the extension is 0.1 mm. Calculate the Young’s modulus.
一根低碳钢试样的标距为 100 mm，直径为 10 mm。拉伸试验数据如下：在 20 kN 载荷下，伸长量为 0.1 mm。计算杨氏模量。
The cross-sectional area A₀ = π d² / 4 = π × (10 mm)² / 4 = 78.54 mm² = 78.54 × 10⁻⁶ m².
横截面积 A₀ = π d² / 4 = π × (10 mm)² / 4 = 78.54 mm² = 78.54 × 10⁻⁶ m²。
σ = F / A₀ = 20,000 N / 78.54 × 10⁻⁶ m² = 254.6 MPa
ε = ΔL / L₀ = 0.1 mm / 100 mm = 0.001
E = σ / ε = 254.6 MPa / 0.001 = 204.6 GPa
This value is close to the typical Young’s modulus for steel (~210 GPa), confirming the measurement is in the linear elastic region.
此数值接近钢的典型杨氏模量（约 210 GPa），证实测量位于线弹性区域。
6. Kinetic Energy and Gravitational Potential Energy | 动能与重力势能
A mass of 5 kg is dropped from rest at a height of 20 m. Ignoring air resistance, calculate its velocity just before impact. Use g = 9.81 m/s².
一个 5 kg 的物体从 20 m 高处静止落下。忽略空气阻力，计算其撞击地面前的速度。取 g = 9.81 m/s²。
As it falls, gravitational potential energy is converted into kinetic energy.
下落过程中，重力势能转化为动能。
GPE_initial = m g h = 5 kg × 9.81 m/s² × 20 m = 981 J
KE_final = ½ m v²
By conservation of mechanical energy, KE_final = GPE_initial.
根据机械能守恒，KE_final = GPE_initial。
½ × 5 kg × v² = 981 J ⇒ v² = (981 J × 2) / 5 kg = 392.4 ⇒ v ≈ 19.8 m/s
Kinematics also gives the same result: v² = u² + 2gh = 0 + 2 × 9.81 × 20 = 392.4, so v = 19.8 m/s.
运动学公式也可得出相同结果：v² = u² + 2gh = 0 + 2 × 9.81 × 20 = 392.4，故 v = 19.8 m/s。
7. Electrical Power and Energy Dissipation | 电功率与能量耗散
A heating element has a resistance of 50 Ω and is connected to a 230 V mains supply for 5 minutes. Determine the power rating and the energy consumed in kilowatt-hours.
一个加热元件的电阻为 50 Ω，接入 230 V 市电供电 5 分钟。求功率额定值以及以千瓦时为单位的消耗能量。
Current I = V / R = 230 V / 50 Ω = 4.6 A.
电流 I = V / R = 230 V / 50 Ω = 4.6 A。
P = V × I = 230 V × 4.6 A = 1058 W ≈ 1.058 kW
Time in hours: t = 5 min / 60 = 0.08333 h.
时间以小时计：t = 5 min / 60 = 0.08333 h。
Energy (kWh) = Power (kW) × time (h) = 1.058 kW × 0.08333 h ≈ 0.0882 kWh
Alternatively, energy in joules: P × t (s) = 1058 W × 300 s = 317,400 J. Divide by 3.6×10⁶ to convert to kWh: 0.0882 kWh.
或者，以焦耳计的能量：P × t (s) = 1058 W × 300 s = 317,400 J。除以 3.6×10⁶ 即换算为 0.0882 kWh。
8. Orthographic Projection Reading | 正投影图读图
An engineering drawing shows a bracket in third-angle projection. The front view shows a rectangular base with a vertical rib and a circular hole. The top view reveals the hole as a dashed circle, and the rib appears as a rectangle. Interpret the features.
一张工程图以第三角投影展示了一个支架。主视图显示矩形底板，带有垂直肋板和一个圆孔。俯视图中孔为虚线圆，肋板显示为矩形。解读这些特征。
In third-angle projection, the top view is placed above the front view, and the right-side view is to the right. The dashed circle in the top view indicates a hidden hole through the base.
在第三角投影中，俯视图位于主视图之上，右视图位于右侧。俯视图中的虚线圆表示一个穿过底板的隐藏孔。
The rectangular shape of the rib in the top view confirms the rib extends vertically from the base, without any taper, visible also in the front view as a raised rectangle.
俯视图中肋板的矩形形状确认该肋板从底板垂直延伸，无锥度，在主视图中也可见为一个凸起矩形。
Proper interpretation of hidden details ensures correct manufacturing; the dashed lines must be read accurately to avoid missing features.
正确解读隐藏细节可确保制造无误；必须准确识读虚线，以免遗漏特征。
9. Casting Process and Defects | 铸造工艺与缺陷
Sand casting is widely used for producing complex metal parts. The pattern is placed in a mould box, sand mixed with a binder is packed around it, and then the pattern is removed to leave a cavity. Describe the steps and identify two common defects.
砂型铸造广泛用于制造复杂金属零件。将模型放入砂箱，用混有粘结剂的型砂填充紧实，然后取出模型留下型腔。描述其步骤并指出两种常见缺陷。
Steps: pattern making, moulding (cope and drag assembly), core placement if needed, pouring molten metal, cooling, shakeout, and fettling.
步骤包括：制模、造型（上箱与下箱组装）、必要时下芯、浇注熔融金属、冷却、落砂和清理打磨。
Two common defects are porosity (gas holes due to trapped air or moisture) and shrinkage cavities caused by inadequate feeding of molten metal during solidification.
两种常见缺陷是气孔（由于卷入空气或水分造成）和缩孔，后者由凝固过程中补缩不足引起。
Understanding these defects helps engineers design proper risers and venting systems to ensure casting quality.
了解这些缺陷有助于工程师设计合适的冒口和排气系统，以确保铸件质量。
10. Material Properties and Selection Charts | 材料性能与选择图表
An engineer must select a material for a lightweight tie rod that requires high stiffness and strength. The table below shows density, Young’s modulus, and yield strength for three candidate materials. Analyse the specific properties to recommend the best choice.
工程人员需为轻质拉杆选材，要求高刚度与高强度。下表列出了三种候选材料的密度、杨氏模量和屈服强度。分析比性能以推荐最佳选择。
| Material | Density ρ (kg/m³) | Young’s modulus E (GPa) | Yield strength σ_y (MPa) |
|---|---|---|---|
| Aluminium alloy | 2700 | 70 | 250 |
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