Year 12 CAIE Engineering: Unit Test Mock Paper Analysis | Year 12 CAIE 工程：单元测试模拟卷解析

📚 Year 12 CAIE Engineering: Unit Test Mock Paper Analysis | Year 12 CAIE 工程：单元测试模拟卷解析

Mock papers are one of the most effective revision tools for CAIE Engineering. By attempting questions under timed conditions and then working through detailed solutions, you can identify gaps in your knowledge and sharpen your problem-solving skills. This article walks you through a typical Year 12 unit test, breaking down key questions in forces, materials, energy, thermal physics, and electrical systems. Every answer is explained step by step, highlighting common pitfalls and reinforcing essential formulas.

模拟卷是 CAIE 工程课程最高效的复习工具之一。在限时条件下完成试题并仔细研究解析，能帮助你发现知识漏洞并提升解题能力。本文带你解析一份典型的 Year 12 单元测试卷，涵盖力、材料、能量、热物理和电路等核心题目。每道题的答案都会逐步拆解，突出常见错误，并巩固重要公式。

1. Forces and Equilibrium | 力与平衡

Question: A uniform beam AB of length 5.0 m and weight 400 N rests on two supports C and D. C is 1.0 m from A, D is 1.0 m from B. A vertical load of 600 N is applied at a point 2.0 m from A. Determine the vertical reaction forces at C and D.

题目：一根长 5.0 m、重 400 N 的均匀梁 AB 放在两个支座 C 和 D 上。C 距 A 端 1.0 m，D 距 B 端 1.0 m。在距 A 端 2.0 m 处施加一个 600 N 的竖向载荷。求支座 C 和 D 的竖向反力。

Solution: Draw the free-body diagram. The weight of the beam acts through its centre (2.5 m from A). Take moments about C to eliminate R_C. Clockwise moments = R_D × 3.0 (since CD = 3.0 m), anticlockwise moments = (600 N × 1.0 m) + (400 N × 1.5 m). Equating gives R_D = (600 + 600) / 3.0 = 400 N. Then use vertical equilibrium: R_C + R_D = 600 + 400, so R_C = 600 N. Many students forget the beam’s own weight when taking moments.

解析：先画受力图。梁的自重作用在中心（距 A 2.5 m）。对 C 取矩消去 R_C。顺时针力矩 = R_D × 3.0（CD 间距 3.0 m），逆时针力矩 = (600 N × 1.0 m) + (400 N × 1.5 m)。令两者相等得 R_D = (600 + 600) / 3.0 = 400 N。再用竖向平衡：R_C + R_D = 600 + 400，故 R_C = 600 N。许多学生在取矩时忘记计入梁的自重。

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2. Stress and Strain Calculations | 应力与应变计算

Question: A steel tie rod of diameter 20 mm carries a tensile load of 80 kN. Young’s modulus for steel is 210 GPa. Calculate (a) the tensile stress in the rod, (b) the tensile strain, and (c) the extension over a 1.5 m original length.

题目：一根直径 20 mm 的钢拉杆承受 80 kN 的拉伸载荷。钢的杨氏模量为 210 GPa。计算：(a) 拉杆的拉应力，(b) 拉应变，(c) 原长 1.5 m 上的伸长量。

Solution: (a) Cross-sectional area A = π × (0.010 m)² = 3.142 × 10⁻⁴ m². Stress σ = F / A = 80 × 10³ N / 3.142 × 10⁻⁴ m² = 2.546 × 10⁸ Pa = 255 MPa. (b) Strain ε = σ / E = 2.55 × 10⁸ / 210 × 10⁹ = 1.214 × 10⁻³ (no unit). (c) Extension ΔL = ε × L₀ = 1.214 × 10⁻³ × 1.5 m = 1.82 × 10⁻³ m = 1.82 mm. Remember to convert diameter to radius in metres and to use consistent SI units.

解析：(a) 横截面积 A = π × (0.010 m)² = 3.142 × 10⁻⁴ m²。应力 σ = F / A = 80 × 10³ N / 3.142 × 10⁻⁴ m² = 2.546 × 10⁸ Pa = 255 MPa。(b) 应变 ε = σ / E = 2.55 × 10⁸ / 210 × 10⁹ = 1.214 × 10⁻³（无单位）。(c) 伸长量 ΔL = ε × L₀ = 1.214 × 10⁻³ × 1.5 m = 1.82 × 10⁻³ m = 1.82 mm。注意将直径转换为半径（米），并统一使用 SI 单位。

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3. Linear Motion with Constant Acceleration | 匀加速直线运动

Question: A vehicle accelerates uniformly from rest to 25 m/s in 8.0 s. It then maintains this speed for 12 s before decelerating uniformly to rest in a further 6.0 s. Sketch the velocity–time graph and determine the total distance travelled.

题目：一辆车由静止匀加速，在 8.0 s 内达到 25 m/s，之后保持该速度 12 s，再匀减速至静止，用时 6.0 s。画出速度—时间图并计算总行驶距离。

Solution: The v–t graph consists of a triangle (0–8 s), a rectangle (8–20 s), and a second triangle (20–26 s). Area under the graph gives distance. Area₁ = ½ × 8.0 × 25 = 100 m; Area₂ = 25 × 12 = 300 m; Area₃ = ½ × 6.0 × 25 = 75 m. Total distance = 100 + 300 + 75 = 475 m. Alternatively, use s = ut + ½at² for each phase after finding acceleration. Common mistake: misreading the time intervals, especially the end of the constant-speed phase.

解析：v–t 图由一个三角形 (0–8 s)、一个矩形 (8–20 s) 和第二个三角形 (20–26 s) 组成。图线下面积即为距离。面积₁ = ½ × 8.0 × 25 = 100 m；面积₂ = 25 × 12 = 300 m；面积₃ = ½ × 6.0 × 25 = 75 m。总距离 = 100 + 300 + 75 = 475 m。也可在求出加速度后，用 s = ut + ½at² 分段计算。常见错误：混淆时间区间，特别是匀速段的结束时刻。

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4. Conservation of Energy and Work Done | 能量守恒与做功

Question: A crate of mass 50 kg is pulled up a smooth incline of 30° by a force parallel to the plane. The crate moves 4.0 m along the incline. Find (a) the work done by the pulling force if the crate gains 1.5 m in vertical height, (b) the work done against gravity, and (c) the power required to achieve this in 5.0 s.

题目：一个 50 kg 的板条箱沿 30° 光滑斜面被平行于斜面的力拉上。箱沿斜面移动 4.0 m。求：(a) 若箱垂直升高 1.5 m，拉力所做的功；(b) 克服重力做的功；(c) 若在 5.0 s 内完成，所需的功率。

Solution: (a) By conservation of energy, work done by pulling force = gain in gravitational potential energy + work against friction (0 here) + change in kinetic energy (assume steady, ΔKE = 0). Gain in GPE = mgh = 50 × 9.81 × 1.5 = 735.75 J ≈ 736 J. (b) Work against gravity equals GPE gain = 736 J. (c) Power = work done / time = 736 / 5.0 = 147.2 W. Note that the pulling force can also be found from F = mg sinθ, giving work F × 4.0 = 50 × 9.81 × sin30° × 4.0 = 981 J if the height is calculated from geometry (h = 4.0 × sin30° = 2.0 m, not 1.5 m! The question gives a specific height of 1.5 m, so use that for GPE, implying the incline is not exactly 30° or friction is present; here it’s a simplified scenario).

解析：(a) 由能量守恒，拉力做功 = 重力势能增量 + 克服摩擦做功（此处为零）+ 动能增量（假设匀速，ΔKE = 0）。GPE 增量 = mgh = 50 × 9.81 × 1.5 = 735.75 J ≈ 736 J。(b) 克服重力做功等于 GPE 增量，736 J。(c) 功率 = 做功 / 时间 = 736 / 5.0 = 147.2 W。注意也可通过斜面几何关系计算：若斜面严格 30°，升高应为 4.0 × sin30° = 2.0 m，但本题给定 1.5 m，故直接用给定值，这在说明非理想情况或刻意简化。始终使用题目数据。

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5. Thermal Expansion of Solids | 固体的热膨胀

Question: A steel pipeline is 200 m long at an installation temperature of 15 °C. The coefficient of linear expansion for steel is 12 × 10⁻⁶ K⁻¹. If the pipeline experiences a maximum temperature of 45 °C, calculate (a) the change in length, and (b) the stress induced if the expansion is completely restrained. (E = 210 GPa)

题目：一条钢制管道在安装温度 15 °C 时长 200 m。钢的线膨胀系数为 12 × 10⁻⁶ K⁻¹。若管道经受的最高温度为 45 °C，计算：(a) 长度变化量，(b) 若膨胀完全被约束，产生的应力。（E = 210 GPa）

Solution: (a) ΔT = 45 – 15 = 30 °C (or 30 K). ΔL = α L₀ ΔT = 12 × 10⁻⁶ × 200 × 30 = 0.072 m = 72 mm. (b) If restrained, thermal strain ε = α ΔT = 12 × 10⁻⁶ × 30 = 3.6 × 10⁻⁴. Stress σ = E ε = 210 × 10⁹ × 3.6 × 10⁻⁴ = 75.6 × 10⁶ Pa = 75.6 MPa. This stress can be dangerous if not accommodated by expansion loops. Students often confuse the difference in temperature with the absolute temperature; always use ΔT in either Celsius or Kelvin.

解析：(a) ΔT = 45 – 15 = 30 °C（即 30 K）。ΔL = α L₀ ΔT = 12 × 10⁻⁶ × 200 × 30 = 0.072 m = 72 mm。(b) 若被约束，热应变 ε = α ΔT = 12 × 10⁻⁶ × 30 = 3.6 × 10⁻⁴。应力 σ = E ε = 210 × 10⁹ × 3.6 × 10⁻⁴ = 75.6 × 10⁶ Pa = 75.6 MPa。若未用膨胀弯管释放，此应力会带来危险。学生常混淆温差与绝对温度；一律用 ΔT，摄氏或开尔文均可。

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6. Heat Transfer Through a Composite Wall | 通过复合壁的热传递

Question: A furnace wall consists of an inner layer of firebrick (k = 1.2 W m⁻¹ K⁻¹) 150 mm thick, and an outer layer of insulating brick (k = 0.15 W m⁻¹ K⁻¹) 100 mm thick. The inside surface temperature is 800 °C and the outside surface temperature is 50 °C. Assuming steady-state conduction through a wall of area 3.0 m², calculate the rate of heat loss.

题目：一个炉壁由内层耐火砖（k = 1.2 W m⁻¹ K⁻¹，厚 150 mm）和外层保温砖（k = 0.15 W m⁻¹ K⁻¹，厚 100 mm）组成。内表面温度为 800 °C，外表面温度为 50 °C。假设稳态导热，壁面积为 3.0 m²，计算散热速率。

Solution: For series conduction, the thermal resistance R_total = (L₁ / k₁A) + (L₂ / k₂A). Thicknesses in metres: L₁ = 0.150 m, L₂ = 0.100 m. R_total = (0.150 / (1.2 × 3.0)) + (0.100 / (0.15 × 3.0)) = 0.04167 + 0.2222 = 0.2639 K W⁻¹. Temperature difference ΔT = 800 – 50 = 750 K. Heat transfer rate Q = ΔT / R_total = 750 / 0.2639 ≈ 2842 W (or 2.84 kW). Always check unit consistency (mm to m) and use the total temperature drop across both layers.

解析：串联导热的热阻 R_total = (L₁ / k₁A) + (L₂ / k₂A)。厚度单位米：L₁ = 0.150 m，L₂ = 0.100 m。R_total = (0.150 / (1.2 × 3.0)) + (0.100 / (0.15 × 3.0)) = 0.04167 + 0.2222 = 0.2639 K W⁻¹。温差 ΔT = 800 – 50 = 750 K。传热速率 Q = ΔT / R_total = 750 / 0.2639 ≈ 2842 W（2.84 kW）。务必检查单位一致性（mm 转 m），并使用两层的总温差。

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7. Basic Electrical Quantities and Ohm’s Law | 基本电学量与欧姆定律

Question: A circuit consists of a 24 V battery connected in series with a resistor R₁ = 10 Ω and a parallel combination of R₂ = 15 Ω and R₃ = 30 Ω. Determine (a) the total circuit resistance, (b) the current drawn from the battery, and (c) the current through R₂.

题目：一个电路由 24 V 电池串联一个 R₁ = 10 Ω 的电阻，再串联 R₂ = 15 Ω 和 R₃ = 30 Ω 的并联组合。求：(a) 电路总电阻，(b) 电池电流，(c) 流过 R₂ 的电流。

Solution: (a) Parallel equivalent: 1/R_par = 1/15 + 1/30 = 2/30 + 1/30 = 3/30, so R_par = 10 Ω. Total resistance R_total = R₁ + R_par = 10 + 10 = 20 Ω. (b) Total current I = V / R_total = 24 / 20 = 1.2 A. (c) Voltage across parallel branch = I × R_par = 1.2 × 10 = 12 V. Current through R₂ = 12 / 15 = 0.8 A. Check: current through R₃ = 12 / 30 = 0.4 A; sum = 1.2 A. Common mistake: using the supply voltage directly across R₂ without considering the series resistor.

解析：(a) 并联等效电阻：1/R_par = 1/15 + 1/30 = 2/30 + 1/30 = 3/30，故 R_par = 10 Ω。总电阻 R_total = R₁ + R_par = 10 + 10 = 20 Ω。(b) 总电流 I = V / R_total = 24 / 20 = 1.2 A。(c) 并联支路电压 = I × R_par = 1.2 × 10 = 12 V。通过 R₂ 的电流 = 12 / 15 = 0.8 A。验证：通过 R₃ 电流 = 12 / 30 = 0.4 A；和为 1.2 A。常见错误：未考虑串联电阻，直接将电源电压加在 R₂ 上。

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8. Fluid Pressure and Manometers | 流体压强与压力计

Question: A U-tube manometer contains mercury (density 13 600 kg/m³). One side is connected to a gas supply, the other is open to atmospheric pressure 101 kPa. The mercury level on the gas side is 80 mm higher than on the open side. Calculate the gauge pressure and the absolute pressure of the gas.

题目：一个 U 形管压力计装有水银（密度 13 600 kg/m³）。一侧连接气源，另一侧通大气压 101 kPa。连接气源一侧的水银柱比开口侧高 80 mm。求气源的表压和绝对压强。

Solution: The height difference indicates the gas pressure is less than atmospheric because the gas-side column is higher. Gauge pressure is negative. Δh = 80 mm = 0.080 m. Pressure difference ΔP = ρ g Δh = 13 600 × 9.81 × 0.080 = 10 673 Pa ≈ 10.7 kPa. So gauge pressure = –10.7 kPa (below atmospheric). Absolute pressure P_abs = P_atm + gauge pressure = 101 – 10.7 = 90.3 kPa. Always determine the direction of the pressure difference from the diagram.

解析：液柱高度差表明气源压力低于大气压，因为气侧液面更高。表压为负。Δh = 80 mm = 0.080 m。压差 ΔP = ρ g Δh = 13 600 × 9.81 × 0.080 = 10 673 Pa ≈ 10.7 kPa。故表压 = –10.7 kPa（低于大气压）。绝对压强 P_abs = P_atm + 表压 = 101 – 10.7 = 90.3 kPa。务必根据图示判断压差方向。

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9. Levers and Mechanical Advantage | 杠杆与机械效益

Question: A wheelbarrow is used to carry a load of 400 N. The load is located 0.40 m from the wheel axle. The effort is applied 1.2 m from the axle. Calculate (a) the mechanical advantage (MA), (b) the effort required to lift the load, assuming a vertical effort. Also state the velocity ratio (VR) if the effort moves 0.60 m while the load rises 0.20 m.

题目：一个手推车用于搬运 400 N 的重物。重物距轮轴 0.40 m，动力施力点距轮轴 1.2 m。计算：(a) 机械效益 (MA)，(b) 抬起重物所需的动力（假设动力竖直），并给出当动力移动 0.60 m 而重物升高 0.20 m 时的速度比 (VR)。

Solution: (a) By moments about the axle: Load × 0.40 = Effort × 1.2. MA = Load / Effort = 1.2 / 0.40 = 3. (b) Effort = 400 / 3 = 133.3 N. (c) VR = distance moved by effort / distance moved by load = 0.60 / 0.20 = 3. For an ideal machine, MA = VR; here they match, indicating no friction. Students often misidentify the pivot point in wheelbarrow problems.

解析：(a) 对轮轴取矩：Load × 0.40 = Effort × 1.2。MA = 负载/动力 = 1.2 / 0.40 = 3。(b) 动力 = 400 / 3 = 133.3 N。(c) VR = 动力移动距离 / 重物移动距离 = 0.60 / 0.20 = 3。对理想机械，MA = VR；此处相等说明无摩擦。学生在手推车问题中常误判支点。

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10. Power and Efficiency in Rotating Systems | 旋转系统的功率与效率

Question: An electric motor drives a winch that lifts an 80 kg mass vertically at a constant speed of 0.50 m/s. The motor operates at 240 V and draws 4.2 A. Calculate (a) the output mechanical power, (b) the electrical input power, and (c) the overall efficiency of the motor-winch system.

题目：一台电动机驱动绞车以 0.50 m/s 的恒定速度竖直提升 80 kg 的重物。电机工作电压 240 V，电流 4.2 A。计算：(a) 输出机械功率，(b) 输入电功率，(c) 电机—绞车系统的总效率。

Solution: (a) Output power P_out = force × velocity = mg × v = 80 × 9.81 × 0.50 = 392.4 W. (b) Input electrical power P_in = V × I = 240 × 4.2 = 1008 W. (c) Efficiency η = (P_out / P_in) × 100% = (392.4 / 1008) × 100% ≈ 38.9%. The low efficiency suggests significant losses in the motor and possibly in the winch gearing. Remember to use the net weight force acting at constant speed.

解析：(a) 输出功率 P_out = 力 × 速度 = mg × v = 80 × 9.81 × 0.50 = 392.4 W。(b) 输入电功率 P_in = V × I = 240 × 4.2 = 1008 W。(c) 效率 η = (P_out / P_in) × 100% = (392.4 / 1008) × 100% ≈ 38.9%。效率较低说明电机和绞车齿轮存在较大损耗。注意恒速提升时有效力就是重物的重力。

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11. Common Mistakes to Avoid in Unit Tests | 单元测试中要避免的常见错误

Many students lose marks not because they don’t know the content, but because of avoidable slip-ups. Always write down the formula before substituting numbers. Double-check unit conversions—metres, not millimetres, in stress and thermal calculations. In circuit problems, redraw the diagram and label known values. For equilibrium, take moments about a point that eliminates one unknown. In energy problems, define the system clearly so that work done by the pulling force equals the total energy change. Lastly, show all workings; even if the final answer is wrong, method marks can be earned.

许多学生丢分并非因为不懂，而是由于可以避免的小错。务必先写出公式再代入数值。反复检查单位换算——应力和热计算中要用米而非毫米。电路问题中重新画图并标出已知值。在平衡问题中，对着能消去一个未知量的点取矩。在能量问题中，明确系统，使拉力做功等于总能量变化。最后，展示所有计算步骤；即使最终答案错了，也能得到方法分。

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12. Final Review and Practice Suggestions | 最终复习与练习建议

After completing a mock paper, spend as much time reviewing your answers as you did writing them. Identify which topics gave you trouble and return to your textbook or notes for targeted study. Create a formula sheet with only the equations you tend to forget. Practice under timed conditions at least twice before the real test. For CAIE Engineering, focus on units, clear diagrams, and systematic working. Engineering is about solving problems methodically—the mock paper analysis is your training ground.

完成一套模拟卷后，请花和答题同样多的时间来回顾答案。找出困扰你的主题，回到教材或笔记进行针对性学习。自制一份只收录易忘公式的公式表。在真正考试前，至少进行两次限时模拟训练。对于 CAIE 工程，关注单位、清晰的示意图和系统的解题步骤。工程学就是有条理地解决问题——模拟卷解析就是你的训练场。

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