📚 Unit Test Mock Paper Walkthrough | 单元测试模拟卷解析
Welcome to an in-depth walkthrough of a Unit Test mock paper designed for Year 13 CIE Engineering. This session aims to strengthen your grasp of core A2 topics such as stress-strain analysis, thermal effects, circuit theory, op-amps, and control systems. Each question is dissected step-by-step, revealing common pitfalls and sharpening your problem-solving skills for the final examination.
欢迎深入解析这份为Year 13 CIE工程学设计的单元测试模拟卷。本课程旨在巩固你对A2核心主题的掌握，包括应力—应变分析、热效应、电路理论、运算放大器和控制系统。我们将逐步剖析每一个问题，揭示常见陷阱，并提高你应对期末考试的解题能力。
1. Overview of the Mock Paper | 模拟卷概览
This mock unit test contains eight structured questions that together cover the Mechanics of Materials and Electrical & Electronic Principles modules. The questions progress from straightforward calculations to applied reasoning, including material selection justification and error analysis. You should aim to complete the paper in 75 minutes, allocating roughly 9 minutes per question. A solid performance relies on precise unit conversions, clear formula recall, and the ability to explain physical behaviour in engineering terms.
这份模拟单元测试包含八道结构题，涵盖了材料力学和电气与电子原理两个模块。题目从直接的计算逐步过渡到应用推理，包括材料选择的合理性和误差分析。你应在75分钟内完成试卷，每道题大约分配9分钟。扎实的答题表现离不开精确的单位换算、清晰的公式记忆以及用工程术语解释物理行为的能力。
2. Question 1: Stress and Strain Calculation | 问题1：应力和应变计算
A steel rod of diameter 10 mm and original length 2.0 m is subjected to a tensile force of 15 kN. During loading, the rod extends by 1.8 mm. Calculate the engineering stress, the engineering strain, and Young’s modulus of the steel. Comment on whether the result agrees with typical structural steel data.
一根直径为10 mm、原长为2.0 m的钢杆承受15 kN的拉伸载荷。在加载过程中，杆伸长了1.8 mm。请计算工程应力、工程应变和钢材的杨氏模量，并评述计算结果是否与典型结构钢的数据相符。
First, determine the cross-sectional area A. For a circular section, A = πd²/4. Convert diameter to metres: d = 10 mm = 0.010 m. Hence A = π × (0.010)² / 4 = 7.854×10⁻⁵ m².
首先计算横截面积 A。对于圆形截面，A = πd²/4。将直径转换为米：d = 10 mm = 0.010 m。因此 A = π × (0.010)² / 4 = 7.854×10⁻⁵ m²。
The tensile force is 15 kN = 15 000 N. Engineering stress σ is defined as force per unit original area:
拉伸载荷为15 kN = 15 000 N。工程应力 σ 定义为力除以原始截面积：
σ = F / A = 15 000 N / 7.854×10⁻⁵ m² = 1.91×10⁸ Pa = 191 MPa
Engineering strain ε is the extension per unit original length. Original length L₀ = 2.0 m and extension ΔL = 1.8 mm = 1.8×10⁻³ m:
工程应变 ε 为伸长量除以原长。原长 L₀ = 2.0 m，伸长量 ΔL = 1.8 mm = 1.8×10⁻³ m：
ε = ΔL / L₀ = 1.8×10⁻³ m / 2.0 m = 9.0×10⁻⁴ (or 0.0009)
Young’s modulus E is the ratio of stress to strain within the elastic region. Using the calculated values:
杨氏模量 E 是弹性区域内应力与应变之比。代入计算值：
E = σ / ε = 191×10⁶ Pa / 9.0×10⁻⁴ = 2.12×10¹¹ Pa = 212 GPa
This value is very close to the typical Young’s modulus for carbon steel (around 210 GPa). The slight difference can be attributed to rounding of measurements, so the result is physically consistent and suggests the rod is made from a standard structural steel.
该值与碳钢的典型杨氏模量（约210 GPa）非常接近。微小差异可归因于测量值的舍入，因此结果在物理上是一致的，表明杆材为普通结构钢。
3. Question 2: Young’s Modulus from Experimental Data | 问题2：从实验数据求杨氏模量
A tensile test specimen has an initial cross-sectional area of 50 mm² and a gauge length of 80 mm. When a tensile force of 4.5 kN is applied, the recorded extension is 0.12 mm. Determine the Young’s modulus of the material. Explain why using a specimen with a longer gauge length can reduce the percentage uncertainty in the strain measurement.
一个拉伸试样的原始截面积为50 mm²，标距长度为80 mm。当施加4.5 kN的拉伸力时，记录的伸长量为0.12 mm。求材料的杨氏模量，并解释为何采用更长的标距长度可减小应变测量中的百分比不确定度。
The direct formula that links force, geometry and modulus is E = (F L₀) / (A ΔL). Start by converting all given values to consistent SI units: A = 50 mm² = 50×10⁻⁶ m², L₀ = 80 mm = 0.080 m, ΔL = 0.12 mm = 0.12×10⁻³ m, F = 4500 N.
直接联系力、几何尺寸和模量的公式为 E = (F L₀) / (A ΔL)。首先将所有给定值转换为一致的SI单位：A = 50 mm² = 50×10⁻⁶ m²，L₀ = 80 mm = 0.080 m，ΔL = 0.12 mm = 0.12×10⁻³ m，F = 4500 N。
E = (4500 N × 0.080 m) / (50×10⁻⁶ m² × 0.12×10⁻³ m) = 360 / (6.0×10⁻⁹) = 6.0×10¹⁰ Pa = 60 GPa
This modulus is typical for aluminium alloys. Now, regarding uncertainty: strain ε = ΔL / L₀. If the absolute uncertainty in the measured extension is δ(ΔL), the percentage uncertainty in strain is (δ(ΔL)/ΔL)×100%. Increasing the gauge length L₀ (for the same material and force) produces a proportionally larger extension ΔL, which makes the ratio δ(ΔL)/ΔL smaller, thus reducing the percentage uncertainty. This is a standard technique to improve accuracy in materials testing.
该模量是铝合金的典型值。现在来看不确定度：应变 ε = ΔL / L₀。若测量伸长量的绝对不确定度为 δ(ΔL)，则应变的百分比不确定度为 (δ(ΔL)/ΔL)×100%。增加标距长度 L₀（在相同材料和力的条件下）会按比例产生更大的伸长量 ΔL，从而使比值 δ(ΔL)/ΔL 变小，因此可减小百分比不确定度。这是材料试验中提高精度的常用方法。
4. Question 3: Thermal Expansion in a Bimetallic Strip | 问题3：双金属片的热膨胀
A bimetallic strip is made by bonding a brass strip (coefficient of linear expansion α = 19×10⁻⁶ /°C) and a steel strip (α = 11×10⁻⁶ /°C), each exactly 200 mm long at 20°C. The strip is uniformly heated to 100°C. Calculate the difference in free expansion of the two metals and state which metal lies on the outer (convex) side when the strip bends. Illustrate why the strip curves.
由黄铜片（线膨胀系数 α = 19×10⁻⁶ /°C）和钢片（α = 11×10⁻⁶ /°C）粘合而成的双金属片，在20°C时长均为200 mm。将该双金属片均匀加热至100°C。计算两种金属自由膨胀量的差值，并说明当金属片弯曲时，哪种金属位于外侧（凸面）。阐述为何金属片会产生弯曲。
Free thermal expansion ΔL = L₀ × α × ΔT. Temperature change ΔT = 100 – 20 = 80°C. For brass: ΔL_brass = 200 mm × 19×10⁻⁶ × 80 = 0.304 mm. For steel: ΔL_steel = 200 mm × 11×10⁻⁶ × 80 = 0.176 mm. The difference in expansion is 0.304 – 0.176 = 0.128 mm.
自由热膨胀量 ΔL = L₀ × α × ΔT。温度变化 ΔT = 100 – 20 = 80°C。对于黄铜：ΔL_brass = 200 mm × 19×10⁻⁶ × 80 = 0.304 mm。对于钢：ΔL_steel = 200 mm × 11×10⁻⁶ × 80 = 0.176 mm。膨胀量之差为 0.304 – 0.176 = 0.128 mm。
When heated, brass expands more than steel. Since the two strips are rigidly bonded, the differential expansion forces the composite strip to bend. The brass, trying to become longer, is pushed to the outer circumference of the curve, forming the convex side, while the steel forms the concave inner side. This principle is exploited in thermostats.
加热时，黄铜的膨胀量大于钢。由于两片金属被牢固地粘合在一起，膨胀差异迫使复合金属片弯曲。试图变得更长的黄铜被推到曲线的外圆周，形成凸面，而钢则形成凹面内侧。这一原理被应用于温控器中。
5. Question 4: DC Circuit Analysis with Kirchhoff’s Laws | 问题4：运用基尔霍夫定律的直流电路分析
Consider a two-loop DC network. Loop 1 contains a 10 V battery, a 2 Ω resistor, and a 4 Ω resistor shared with Loop 2. Loop 2 contains a 6 V battery (opposing polarity), the same 4 Ω resistor, and a 3 Ω resistor. Using Kirchhoff’s Voltage Law, find the currents I₁ through the 2 Ω resistor and I₂ through the 3 Ω resistor. Assume the current through the 4 Ω shared resistor is I₁ – I₂ in the direction from left to right.
考虑一个双回路直流网络。回路1包含一个10 V电池、一个2 Ω电阻和一个与回路2共用的4 Ω电阻。回路2包含一个6 V电池（极性相反）、同一个4 Ω电阻和一个3 Ω电阻。利用基尔霍夫电压定律，求流过2 Ω电阻的电流 I₁ 和流过3 Ω电阻的电流 I₂。假设流过共用4 Ω电阻的电流为 I₁ – I₂，方向从左向右。
Apply KVL to Loop 1: start at the 10 V source and traverse clockwise. Voltage rises and drops sum to zero: 10 – 2I₁ – 4(I₁ – I₂) = 0. Simplify to 10 – 6I₁ + 4I₂ = 0 or 6I₁ – 4I₂ = 10. (Equation 1)
对回路1应用KVL：从10 V电源出发，顺时针绕行。电压升与电压降之和为零：10 – 2I₁ – 4(I₁ – I₂) = 0。化简得 10 – 6I₁ + 4I₂ = 0 或 6I₁ – 4I₂ = 10。（方程1）
For Loop 2, again traverse clockwise, noting the 6 V source opposes the loop direction: –6 – 4(I₂ – I₁) – 3I₂ = 0 → –6 + 4I₁ – 4I₂ – 3I₂ = 0 → 4I₁ – 7I₂ = 6. (Equation 2)
对于回路2，同样顺时针绕行，注意6 V电池的极性与回路方向相反：–6 – 4(I₂ – I₁) – 3I₂ = 0 → –6 + 4I₁ – 4I₂ – 3I₂ = 0 → 4I₁ – 7I₂ = 6。（方程2）
Solve simultaneously. Multiply Equation 2 by 1.5 to align coefficients: 6I₁ – 10.5I₂ = 9. Subtract from Equation 1: (6I₁ – 4I₂) – (6I₁ – 10.5I₂) = 10 – 9 → 6.5I₂ = 1 → I₂ = 0.154 A. Substitute back into Equation 1: 6I₁ – 4(0.154) = 10 → 6I₁ = 10.616 → I₁ = 1.769 A. The current through the shared branch is I₁ – I₂ = 1.615 A.
联立求解。将方程2乘以1.5使系数对齐：6I₁ – 10.5I₂ = 9。从方程1中减去该式：(6I₁ – 4I₂) – (6I₁ – 10.5I₂) = 10 – 9 → 6.5I₂ = 1 → I₂ = 0.154 A。代回方程1：6I₁ – 4(0.154) = 10 → 6I₁ = 10.616 → I₁ = 1.769 A。流过共用支路的电流为 I₁ – I₂ = 1.615 A。
6. Question 5: Operational Amplifier – Inverting Configuration | 问题5：运算放大器——反相配置
An inverting operational amplifier circuit has a feedback resistor Rf = 100 kΩ and an input resistor Rin = 10 kΩ. The non-inverting input is grounded, and a DC input voltage of +0.5 V is applied to Rin. Calculate the output voltage and explain the concept of the virtual earth at the inverting terminal.
一个反相运算放大器电路具有反馈电阻 Rf = 100 kΩ 和输入电阻 Rin = 10 kΩ。同相输入端接地，并在 Rin 上施加 +0.5 V 的直流输入电压。计算输出电压，并解释反相输入端的“虚拟地”概念。
For an ideal inverting amplifier, the gain is given by Vout / Vin = –Rf / Rin. Substituting the resistor values: gain = –100 kΩ / 10 kΩ = –10. Therefore, Vout = –10 × 0.5 V = –5.0 V. The negative sign indicates a 180° phase shift, meaning the output is inverted.
对于理想反相放大器，增益由 Vout / Vin = –Rf / Rin 给出。代入电阻值：增益 = –100 kΩ / 10 kΩ = –10。因此，Vout = –10 × 0.5 V = –5.0 V。负号表示180°相移，即输出反相。
The virtual earth concept describes the condition where the inverting (-) input is held close to 0 V due to negative feedback, even though it is not directly connected to ground. Because the op-amp has very high open-loop gain, the differential voltage between the two inputs is forced to nearly zero. Since the non-inverting input is at 0 V, the inverting input must also be at approximately 0 V. This allows the circuit to set current I = Vin / Rin, which flows entirely through Rf, establishing Vout = –I Rf.
虚拟地概念描述了这样一种状态：由于负反馈，反相（-）输入端被维持在接近0 V，即使它并未直接接地。由于运放具有极高的开环增益，两输入端之间的差分电压被迫接近于零。既然同相输入端为0 V，反相输入端也必须近似为0 V。这使得电路中电流 I = Vin / Rin，该电流全部流经 Rf，从而建立起 Vout = –I Rf。
7. Question 6: Control System Block Diagram Reduction | 问题6：控制系统框图化简
A negative feedback control system has a forward-path transfer function G(s) = K / (s + 2) and a feedback-path transfer function H(s) = 1/s. Derive the overall closed-loop transfer function T(s) = C(s) / R(s). Determine the characteristic equation and state its order.
某负反馈控制系统具有前向通路传递函数 G(s) = K / (s + 2) 和反馈通路传递函数 H(s) = 1/s。推导整体闭环传递函数 T(s) = C(s) / R(s)。确定特征方程并给出其阶数。
The standard formula for a negative-feedback system is T(s) = G(s) / [1 + G(s)H(s)]. Replace G(s) and H(s):
负反馈系统的标准公式为 T(s) = G(s) / [1 + G(s)H(s)]。代入 G(s) 和 H(s)：
T(s) = (K/(s+2)) / [1 + (K/(s+2))×(1/s)] = (K/(s+2)) / [(s(s+2) + K) / (s(s+2))] = K / [s(s+2) + K]
Simplify the denominator: s(s+2) + K = s² + 2s + K. Hence the closed-loop transfer function is T(s) = K / (s² + 2s + K). The characteristic equation is the denominator set equal to zero: s² + 2s + K = 0. This is a second-order system, so its order is 2. The response type (overdamped, critically damped, or underdamped) depends on the value of K relative to the discriminant.
化简分母：s(s+2) + K = s² + 2s + K。因此闭环传递函数为 T(s) = K / (s² + 2s + K)。特征方程为分母等于零：s² + 2s + K = 0。这是一个
Published by TutorHao | Year 13 工程 Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply