📚 Edexcel A-Level Engineering: Mock Unit Test Walkthrough | Edexcel A-Level 工程：单元测试模拟卷解析
This article provides a full walkthrough of a mock unit test for Edexcel A-Level Engineering, covering key areas of the Engineering Principles paper. Each question is carefully broken down with step-by-step solutions, and all explanations are given in both English and Chinese to support bilingual learning and revision.
本文提供了 Edexcel A-Level 工程科目单元测试模拟卷的完整讲解，涵盖工程原理试卷的核心领域。每一道题目都通过逐步解析进行拆解，并以中英双语提供所有解释，以支持双语学习和复习。
1. Mock Test Structure and Key Topics | 模拟卷结构及核心考点
This mock test includes seven compulsory questions drawn from mechanics, electronics, materials, fluid dynamics, digital systems, and machine efficiency. The structure mirrors a typical Edexcel unit test, with a mix of calculations and principle applications. Understanding these topics is essential for success in the externally assessed Engineering Principles unit.
本模拟卷包含七道必答题，涵盖力学、电子学、材料、流体力学、数字系统以及机械效率。其结构模拟了典型的 Edexcel 单元测试，包含计算题和原理应用题。掌握这些主题对于顺利通过工程原理外部评估单元至关重要。
2. Question 1: Beam Reaction Forces | 问题1：梁的支座反力
Problem: A uniform beam of length 4 m and weight 200 N rests horizontally on two supports at its ends. A concentrated load of 300 N is placed 1 m from the left end. Calculate the reaction forces at the left support (RA) and the right support (RB).
题目：一根均匀梁长4 m，重200 N，两端由支座水平支撑。在距左端1 m处施加一个300 N的集中荷载。求左支座反力 RA 和右支座反力 RB。
Step 1: Draw the free body diagram, placing the beam’s weight (200 N) at its centre (2 m from either end). Label the unknown reactions RA and RB pointing upward.
步骤1：绘制隔离体图，将梁自重（200 N）置于梁中点（距两端2 m处）。标出向上的未知反力 RA 和 RB。
Step 2: Apply the equilibrium of vertical forces: ΣFy = 0 → RA + RB – 200 N – 300 N = 0.
步骤2：应用竖直力平衡：ΣFy = 0 → RA + RB – 200 N – 300 N = 0。
Step 3: Take moments about point A (left support). The clockwise moments are (300 N × 1 m) and (200 N × 2 m). The anticlockwise moment is (RB × 4 m). Set ΣMA = 0: 300 + 400 – 4RB = 0 → RB = 175 N.
步骤3：对 A 点（左支座）取矩。顺时针力矩为 (300 N × 1 m) 和 (200 N × 2 m)，逆时针力矩为 (RB × 4 m)。令 ΣMA = 0：300 + 400 – 4RB = 0 → RB = 175 N。
Step 4: Substitute RB back into the force equation: RA + 175 = 500 → RA = 325 N.
步骤4：将 RB 代入力平衡方程：RA + 175 = 500 → RA = 325 N。
RA = 325 N, RB = 175 N
3. Question 2: Series Circuit Calculation | 问题2：串联电路计算
Problem: A 12 V DC supply is connected in series with a 10 Ω resistor and a 15 Ω resistor. Determine the current flowing in the circuit and the power dissipated by the 10 Ω resistor.
题目：一台12 V直流电源与一只10 Ω电阻和一只15 Ω电阻串联。求电路中的电流以及10 Ω电阻上消耗的功率。
Step 1: For a series circuit, the total resistance Rtotal = R1 + R2 = 10 Ω + 15 Ω = 25 Ω.
步骤1：对于串联电路，总电阻 Rtotal = R1 + R2 = 10 Ω + 15 Ω = 25 Ω。
Step 2: Apply Ohm’s law to find current: I = V / Rtotal = 12 V / 25 Ω = 0.48 A.
步骤2：应用欧姆定律求电流：I = V / Rtotal = 12 V / 25 Ω = 0.48 A。
Step 3: Power dissipated by the 10 Ω resistor is given by P = I²R. P = (0.48 A)² × 10 Ω = 0.2304 × 10 = 2.304 W.
步骤3：10 Ω电阻消耗的功率由 P = I²R 求得。P = (0.48 A)² × 10 Ω = 0.2304 × 10 = 2.304 W。
I = 0.48 A, P = 2.30 W (approx.)
4. Question 3: Stress, Strain and Young’s Modulus | 问题3：应力、应变与杨氏模量
Problem: A steel rod has an original length L = 2 m and a cross-sectional area A = 0.001 m². It is subjected to a tensile force F = 50 kN. The Young’s modulus of steel is 200 GPa. Calculate the stress, strain, and elongation of the rod.
题目：一根钢杆原长 L = 2 m，横截面积 A = 0.001 m²，承受50 kN的拉力。钢的杨氏模量为200 GPa。计算应力、应变与杆的伸长量。
Step 1: Convert force to newtons: F = 50 kN = 50,000 N. Stress σ = F / A = 50,000 N / 0.001 m² = 50 × 10⁶ Pa = 50 MPa.
步骤1：将力换算为牛顿：F = 50 kN = 50,000 N。应力 σ = F / A = 50,000 N / 0.001 m² = 50 × 10⁶ Pa = 50 MPa。
Step 2: Young’s modulus E = 200 GPa = 200 × 10⁹ Pa. Strain ε = σ / E = 50 × 10⁶ / (200 × 10⁹) = 2.5 × 10⁻⁴.
步骤2：杨氏模量 E = 200 GPa = 200 × 10⁹ Pa。应变 ε = σ / E = 50 × 10⁶ / (200 × 10⁹) = 2.5 × 10⁻⁴。
Step 3: Elongation ΔL = ε × L = (2.5 × 10⁻⁴) × 2 m = 5.0 × 10⁻⁴ m = 0.5 mm.
步骤3：伸长量 ΔL = ε × L = (2.5 × 10⁻⁴) × 2 m = 5.0 × 10⁻⁴ m = 0.5 mm。
σ = 50 MPa, ε = 2.5 × 10⁻⁴, ΔL = 0.5 mm
5. Question 4: Fluid Flow Rates | 问题4：流体流量计算
Problem: Water flows through a circular pipe of internal diameter 0.05 m at a velocity of 2 m/s. The density of water is 1000 kg/m³. Calculate the volumetric flow rate Q and the mass flow rate ṁ.
题目：水流经一根内径0.05 m的圆形管道，流速为2 m/s。水的密度为1000 kg/m³。计算体积流量 Q 和质量流量 ṁ。
Step 1: Cross-sectional area A = πd² / 4 = π × (0.05)² / 4 = π × 0.0025 / 4 ≈ 1.9635 × 10⁻³ m².
步骤1：横截面积 A = πd² / 4 = π × (0.05)² / 4 = π × 0.0025 / 4 ≈ 1.9635 × 10⁻³ m²。
Step 2: Volumetric flow rate Q = A × v = 1.9635 × 10⁻³ m² × 2 m/s = 3.927 × 10⁻³ m³/s.
步骤2：体积流量 Q = A × v = 1.9635 × 10⁻³ m² × 2 m/s = 3.927 × 10⁻³ m³/s。
Step 3: Mass flow rate ṁ = ρ × Q = 1000 kg/m³ × 3.927 × 10⁻³ m³/s = 3.927 kg/s.
步骤3：质量流量 ṁ = ρ × Q = 1000 kg/m³ × 3.927 × 10⁻³ m³/s = 3.927 kg/s。
Q ≈ 3.93 × 10⁻³ m³/s, ṁ ≈ 3.93 kg/s
6. Question 5: Binary to Denary and Hexadecimal | 问题5：二进制转十进制与十六进制
Problem: Convert the binary number 1101₂ into its denary (decimal) and hexadecimal equivalents. Show your working.
题目：将二进制数 1101₂ 转换为十进制和十六进制。请写出过程。
Step 1: Write the place values for binary: (1×2³) + (1×2²) + (0×2¹) + (1×2⁰) = 8 + 4 + 0 + 1 = 13 in denary.
步骤1：写出二进制的位权值：(1×2³) + (1×2²) + (0×2¹) + (1×2⁰) = 8 + 4 + 0 + 1 = 13（十进制）。
Step 2: Group the binary digits into nibbles (4 bits) for hexadecimal. 1101₂ is already one nibble. Its denary value is 13, which corresponds to the hex digit D.
步骤2：将二进制位分为四位一组以便转换。1101₂ 本身即一个半字节，其十进制值为13，对应的十六进制数字为 D。
1101₂ = 1310 = D16
7. Question 6: Pulley System Efficiency | 问题6：滑轮系统效率
Problem: A pulley system lifts a load of 500 N through a vertical height of 2 m. The effort rope moves 6 m. The system’s efficiency is 80%. Calculate the velocity ratio (VR), the mechanical advantage (MA), and the effort required.
题目：一组滑轮系统将500 N的重物竖直提升2 m，施力绳移动6 m。系统效率为80%。计算速度比（VR）、机械效益（MA）以及所需的施力。
Step 1: Velocity ratio VR = distance moved by effort / distance moved by load = 6 m / 2 m = 3.
步骤1：速度比 VR = 施力端移动距离 / 负载端移动距离 = 6 m / 2 m = 3。
Step 2: Efficiency η = MA / VR, so MA = η × VR = 0.80 × 3 = 2.4.
步骤2：效率 η = MA / VR，因此 MA = η × VR = 0.80 × 3 = 2.4。
Step 3: MA = Load / Effort, so Effort = Load / MA = 500 N / 2.4 ≈ 208.33 N.
步骤3：MA = 负载 / 施力，故施力 = 负载 / MA = 500 N / 2.4 ≈ 208.33 N。
VR = 3, MA = 2.4, Effort ≈ 208 N
8. Question 7: Linear Motion and Work Done | 问题7：直线运动与做功
Problem: A car of mass 1200 kg accelerates uniformly from 10 m/s to 20 m/s over a time of 5 seconds on a straight, level road. Determine the acceleration, the resultant force acting on the car, and the work done by the engine during this acceleration.
题目：一辆质量为1200 kg的汽车在平直公路上在5秒内从10 m/s均匀加速至20 m/s。求加速度、作用在汽车上的合力以及该加速过程中发动机所做的功。
Step 1: Acceleration a = (v – u) / t = (20 – 10) / 5 = 2 m/s².
步骤1：加速度 a = (v – u) / t = (20 – 10) / 5 = 2 m/s²。
Step 2: Resultant force F = m × a = 1200 kg × 2 m/s² = 2400 N.
步骤2：合力 F = m × a = 1200 kg × 2 m/s² = 2400 N。
Step 3: To find work done, first calculate the displacement s = u t + ½ a t² = (10 × 5) + ½ × 2 × 5² = 50 + 25 = 75 m.
步骤3：为求做功，先计算位移 s = u t + ½ a t² = (10 × 5) + ½ × 2 × 5² = 50 + 25 = 75 m。
Step 4: Work done W = F × s = 2400 N × 75 m = 180,000 J = 180 kJ. (Alternatively, using energy: ΔKE = ½ m (v² – u²) = 0.5 × 1200 × (400 – 100) = 600 × 300 = 180,000 J.)
步骤4：做功 W = F × s = 2400 N × 75 m = 180,000 J = 180 kJ。(也可用
Published by TutorHao | A-Level 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