📚 Year 10 CAIE Engineering: Unit Test Mock Exam Analysis | Year 10 CAIE 工程：单元测试模拟卷解析
This article provides a detailed walkthrough of a Year 10 CAIE Engineering unit test mock paper. Each question is accompanied by a full explanation, highlighting key concepts and common pitfalls. Use these model answers to strengthen your understanding of engineering materials, mechanics, electronics, manufacturing, quality control, health and safety, and engineering drawing – all core topics for your CAIE assessment.
本文详细解析了一套 Year 10 CAIE 工程单元测试模拟卷。每道题目都附有完整解析，突出关键概念和常见错误。利用这些参考答案来加深你对工程材料、力学、电子、制造工艺、质量控制、健康与安全以及工程制图的理解——这些都是 CAIE 评估的核心主题。
1. Multiple Choice: Buckling and Compression Failure | 选择题：屈曲与压缩失效
A long, slender steel column is loaded axially. Before the material yields, the column suddenly bends sideways. What type of failure is this?
A) Tensile failure
B) Shear failure
C) Buckling
D) Fatigue
一根细长钢柱承受轴向载荷。在材料屈服之前，柱子突然向侧面弯曲。这属于哪种失效？
A) 拉伸失效
B) 剪切失效
C) 屈曲
D) 疲劳
Answer: C. Buckling. Buckling is a sudden sideways deflection that occurs in slender structural members under compressive stress, even when the stress is below the material’s yield strength. It is a stability failure, not purely a material strength failure. Tension would stretch the member, shear acts parallel to the cross section, and fatigue is caused by repeated cyclic loading. For slender columns, engineers must consider the slenderness ratio and end fixity to prevent buckling.
答案：C. 屈曲。 屈曲是细长构件在压应力下出现的突然侧向弯曲，此时应力甚至低于材料的屈服强度。这是一种稳定性失效，而非纯粹的材料强度失效。拉伸会使构件伸长，剪切平行于截面作用，疲劳由反复循环载荷引起。对于细长柱，工程师必须考虑长细比和端部约束以防止屈曲。
2. Short Answer: Stress and Strain – Hooke’s Law | 简答题：应力与应变——胡克定律
A copper wire of diameter 1.2 mm and original length 2.0 m is subjected to a tensile force of 90 N, producing an extension of 4.5 mm. Explain Hooke’s Law and calculate Young’s modulus for copper. State whether the wire behaves elastically if the extension returns to zero when unloaded.
一根直径 1.2 mm、原始长度 2.0 m 的铜导线承受 90 N 的拉力，产生 4.5 mm 的伸长。解释胡克定律并计算铜的杨氏模量。如果卸载后伸长量回到零，说明导线是否呈弹性行为。
Hooke’s Law states that, within the elastic limit, the extension of a material is directly proportional to the applied force. Mathematically, F = kx, where k is the spring constant. For engineering materials, this is expressed as stress being proportional to strain: Stress = E x Strain, where E is Young’s modulus. Stress = Force / Cross-sectional area, Strain = Extension / Original length. The wire returns to its original length when unloaded, so it is behaving elastically. Calculation: cross-sectional area A = π r² = π (0.6 mm)² = 1.131 mm² = 1.131 x 10⁻⁶ m². Stress = 90 N / 1.131 x 10⁻⁶ m² = 79.6 x 10⁶ Pa (79.6 MPa). Strain = 4.5 mm / 2000 mm = 0.00225. Therefore, E = Stress / Strain = 79.6 MPa / 0.00225 = 35.4 GPa. This falls within the typical range for copper (around 110-130 GPa), although real values vary with alloy type, indicating an idealised classroom example.
胡克定律指出，在弹性极限内，材料的伸长量与施加的力成正比。数学表达为 F = kx，其中 k 为弹性常数。对于工程材料，以应力与应变成正比表达：应力 = E × 应变，E 为杨氏模量。应力 = 力 / 截面积；应变 = 伸长量 / 原始长度。导线卸载后恢复原长，属于弹性行为。计算：截面积 A = π r² = π (0.6 mm)² = 1.131 mm² = 1.131 x 10⁻⁶ m²。应力 = 90 N / 1.131 x 10⁻⁶ m² = 79.6 x 10⁶ Pa（79.6 MPa）。应变 = 4.5 mm / 2000 mm = 0.00225。因此 E = 应力 / 应变 = 79.6 MPa / 0.00225 = 35.4 GPa。尽管典型铜的 E 约为 110-130 GPa，但此为理想化教学示例，重在掌握方法。
3. Calculation: Resolving Forces on a Slope | 计算题：斜面上力的分解
A crate of mass 50 kg rests on a ramp inclined at 20° to the horizontal. The gravitational field strength g = 9.8 m/s². Calculate: (a) the component of weight acting down the slope, and (b) the minimum static friction force required to prevent sliding.
一个 50 kg 的箱子静置于与水平面成 20° 的斜坡上，重力加速度 g = 9.8 m/s²。计算：(a) 重力沿斜面向下的分量，(b) 防止滑移所需的最小静摩擦力。
The weight of the crate is W = m × g = 50 kg × 9.8 m/s² = 490 N. The component parallel to the slope is W_parallel = W × sin 20°. Using sin 20° ≈ 0.342, we get W_parallel = 490 N × 0.342 = 167.6 N. The component perpendicular to the slope is W × cos 20° ≈ 490 N × 0.940 = 460.4 N, which determines the normal reaction. For equilibrium, the static friction force must equal the parallel component of weight, so the minimum friction required is 167.6 N, directed up the slope. This assumes no other applied forces. In practice, the coefficient of static friction μ must satisfy μ ≥ W_parallel / Normal reaction = 167.6 / 460.4 = 0.364, so the surface must provide at least this friction coefficient to hold the crate.
箱子重量为 W = m × g = 50 kg × 9.8 m/s² = 490 N。沿斜面的分量为 W_parallel = W × sin 20°。sin 20° ≈ 0.342，得 W_parallel = 490 N × 0.342 = 167.6 N。垂直于斜面的分量为 W × cos 20° ≈ 490 N × 0.940 = 460.4 N，决定正压力。为保持平衡，静摩擦力必须等于重力沿斜面的分量，故最小摩擦力为 167.6 N，方向沿斜面向上。此假设无其他外力。实际中，静摩擦系数 μ 需满足 μ ≥ 167.6/460.4 = 0.364，因此接触面至少提供此摩擦系数才能静止。
4. Mechanisms: Gear Ratios and Speed Calculation | 机械机构：齿轮比与速度计算
A simple gear train consists of a driver gear with 24 teeth rotating at 1500 rev/min. It meshes with a driven gear of 72 teeth. Determine: (a) the gear ratio, (b) the output speed of the driven gear, and (c) explain whether this arrangement provides a speed increase or torque multiplication.
一简单齿轮系由 24 齿的主动齿轮以 1500 rev/min 驱动。它与 72 齿的从动齿轮啮合。求：(a) 齿轮比，(b) 从动齿轮的输出转速，(c) 解释该配置是增速还是增扭。
Gear ratio is defined as teeth on driven / teeth on driver: Ratio = 72 / 24 = 3:1. This is a reduction ratio greater than 1, meaning the output speed is lower. Output speed = driver speed / ratio = 1500 / 3 = 500 rev/min. Because the speed reduces, torque is multiplied. In an ideal gear train (100% efficiency), power is constant (Power = torque × angular speed). Halving the speed triples the torque (ignoring losses). A ratio of 3:1 gives a torque multiplication factor of 3, assuming small friction losses. This arrangement is often used in lifting equipment, conveyor drives, and automotive gearboxes to increase torque at the expense of speed.
齿轮比定义为从动轮齿数 / 主动轮齿数：齿数比 = 72 / 24 = 3:1。此为大于1的减速比，意味着输出转速降低。输出转速 = 主动轮转速 / 齿数比 = 1500 / 3 = 500 rev/min。由于转速降低，扭矩得到放大。在理想齿轮系（效率100%）中，功率恒定（功率 = 扭矩 × 角速度）。速度减半则扭矩增为三倍（忽略损失）。齿数比 3:1 提供约 3 倍的扭矩放大（摩擦损失很小）。此配置常用于起重设备、传送带驱动和汽车变速箱，以牺牲转速换取大扭矩。
5. Electrical Circuits: Applying Ohm’s Law | 电路：欧姆定律应用
A 12 V battery is connected to a circuit containing a fixed resistor of 180 Ω and a variable resistor in series. The current in the circuit is measured as 0.035 A. Calculate: (a) the total resistance in the circuit, (b) the resistance value of the variable resistor, and (c) the power dissipated by the fixed resistor.
一个 12 V 电池连接到一个串联电路，其中包含一个 180 Ω 固定电阻和一个可变电阻。测得电路电流为 0.035 A。计算：(a) 电路总电阻，(b) 可变电阻的阻值，(c) 固定电阻消耗的功率。
Using Ohm’s Law, R_total = V / I = 12 V / 0.035 A = 342.9 Ω. Since the resistors are in series, R_total = R_fixed + R_variable, so R_variable = R_total – R_fixed = 342.9 Ω – 180 Ω = 162.9 Ω (approximately 163 Ω). Power dissipated by the fixed resistor is P = I² R = (0.035 A)² × 180 Ω = 0.001225 × 180 = 0.2205 W (or 220.5 mW). Using P = VI, the voltage drop across the fixed resistor is V = I R = 0.035 A × 180 Ω = 6.3 V, then P = 6.3 V × 0.035 A = 0.2205 W, confirming the calculation. This low power means a standard 0.25 W resistor would be suitable.
根据欧姆定律，R_total = V / I = 12 V / 0.035 A = 342.9 Ω。由于电阻串联，R_total = R_fixed + R_variable，因此 R_variable = 342.9 Ω – 180 Ω = 162.9 Ω（约163 Ω）。固定电阻消耗的功率为 P = I² R = (0.035 A)² × 180 Ω = 0.001225 × 180 = 0.2205 W（即 220.5 mW）。可使用 P = VI 验证，固定电阻上的电压降 V = I R = 0.035 A × 180 Ω = 6.3 V，P = 6.3 V × 0.035 A = 0.2205 W。如此低的功率意味着选用 0.25 W 的电阻即可。
6. Manufacturing: Sand Casting Process Steps | 制造工艺：砂型铸造步骤
Describe the key stages in the sand casting process and explain why a pattern is typically made slightly larger than the final required component.
描述砂型铸造过程的关键阶段，并解释为什么模样通常做得比最终所需零件稍大。
Sand casting involves: (1) Pattern making – a replica of the final part, often made of wood, plastic or metal, is produced. (2) Moulding – the pattern is placed in a moulding box, and sand mixed with a binder (e.g. clay) is packed around it to form a mould cavity. The mould is split into two halves (cope and drag). (3) Core making – if internal cavities are needed, sand cores are inserted. (4) Pouring – molten metal is poured into the cavity through a gating system. (5) Cooling and solidification. (6) Mould breakaway and cleaning – the sand is removed and the casting is fettled. The pattern is made larger because metal shrinks upon cooling. A shrinkage allowance (typically 1-2%) is added to the pattern dimensions so that the final casting matches the required size. Additionally, a machining allowance may be added for surfaces that will be finished later.
砂型铸造步骤：(1) 模样制作——制造最终零件的复制品，常用木材、塑料或金属。 (2) 造型——将模样放入砂箱，用混有粘结剂（如粘土）的型砂充填周围，形成型腔。铸型分为上型和下型。 (3) 型芯制作——若需内腔，则放入砂芯。 (4) 浇注——通过浇注系统将熔融金属注入型腔。 (5) 冷却凝固。 (6) 落砂清理——清除型砂，清铲铸件。模样需放大是因为金属冷却时收缩。模样上需添加收缩余量（通常1-2%），使最终铸件达到所需尺寸。此外，后续需机加工的表面还需加加工余量。
7. Quality Control: Interpreting Tolerances | 质量控制：公差解读
A shaft is specified on a manufacturing drawing as having a diameter of 25.00 ± 0.05 mm. A mating hole is specified as 25.20 ± 0.03 mm. Determine: (a) the maximum and minimum shaft and hole sizes, (b) the type of fit (clearance, interference, or transition), and explain why it is chosen.
一根轴在制造图纸上标注直径为 25.00 ± 0.05 mm，与之配合的孔标注为 25.20 ± 0.03 mm。计算：(a) 轴与孔的最大、最小极限尺寸，(b) 配合类型（间隙、过盈或过渡），并解释选择原因。
Shaft limits: maximum = 25.05 mm, minimum = 24.95 mm. Hole limits: maximum = 25.23 mm, minimum = 25.17 mm. Even the smallest hole (25.17 mm) is larger than the largest shaft (25.05 mm), so there is always a positive clearance. Minimum clearance = 25.17 – 25.05 = 0.12 mm; maximum clearance = 25.23 – 24.95 = 0.28 mm. This is a clearance fit, used when parts must move freely relative to each other, such as a shaft rotating in a bearing. The chosen tolerance band ensures easy assembly and sufficient lubrication space. If the clearances were too tight, binding or seizure could occur under thermal expansion.
轴的极限尺寸：最大 25.05 mm，最小 24.95 mm。孔的极限尺寸：最大 25.23 mm，最小 25.17 mm。即使最小的孔（25.17 mm）也大于最大的轴（25.05 mm），因此始终存在正间隙。最小间隙 = 25.17 – 25.05 = 0.12 mm；最大间隙 = 25.23 – 24.95 = 0.28 mm。此为间隙配合，用于零件需相对自由运动之处，例如轴在轴承中转动。所选公差带确保了易装配性及充足的润滑空间。若间隙过小，可能因热膨胀而发生卡滞。
8. Health and Safety: COSHH and Risk Assessment | 健康与安全：COSHH 与风险评估
During a workshop activity, students use epoxy resin and hardener. List four key steps in conducting a COSHH risk assessment for this substance and suggest appropriate control measures.
在车间活动中，学生使用环氧树脂和固化剂。列出对此物质进行 COSHH 风险评估的四个关键步骤，并提出适当的控制措施。
COSHH (Control of Substances Hazardous to Health) risk assessment steps: (1) Identify the hazards – check the safety data sheet for epoxy resin; it may cause skin irritation, allergic reactions, and respiratory sensitisation. (2) Decide who might be harmed and how – students and staff mixing or applying resin, bystanders. (3) Evaluate the risks and decide on precautions – ensure good ventilation, use local exhaust ventilation (LEV) if needed, provide nitrile gloves and safety goggles, prohibit eating and drinking. (4) Record findings and implement controls – display safety instructions, ensure first aid measures are known, provide training. Control measures also include substitution by less hazardous resin if available, limiting exposure time, and regular health surveillance for staff.
COSHH（有害健康物质控制）风险评估步骤：(1) 识别危害——查看环氧树脂的安全数据表；可能引起皮肤刺激、过敏反应及呼吸道致敏。 (2) 确定谁会受到伤害及如何受伤——混合或涂覆树脂的学生和教职员工以及旁观者。 (3) 评估风险并确定预防措施——确保良好通风，必要时使用局部排风（LEV），提供丁腈手套和护目镜，禁止饮食。 (4) 记录结果并实施控制——张贴安全说明，确保了解急救措施，提供培训。控制措施还包括可能时用危害更小的树脂替代，限制接触时间，对教职员工进行定期健康监测。
9. Engineering Drawing: Orthographic Projection and Symbols | 工程制图：正交投影与符号
An engineering drawing shows three orthographic views of a bracket: front, top, and right side. Explain the purpose of this projection method and interpret the following feature: a dimension quoted as “4 x M8 x 1.25 – 6H” at the base.
一张工程图纸展示了一个支架的三个正交视图：主视图、俯视图和右视图。解释此投影法的目的，并解读底座处标注的 “4 x M8 x 1.25 – 6H” 含义。
Orthographic projection uses multiple 2D views (typically three) mutually perpendicular to represent a 3D object precisely, without perspective distortion. Each view shows the object as if looking straight at it. The front view gives height and width, the top view gives width and depth, and the right side view gives height and depth. This ensures that every feature can be dimensioned accurately. The notation “4 x M8 x 1.25 – 6H” means: four identical threaded holes (or positions for bolts), with ISO metric thread M8, a pitch of 1.25 mm (fine thread), and a tolerance class of 6H for the internal thread. 6H indicates a commonly used fit tolerance for medium engagement, ensuring interchangeability. The “x” before M8 denotes quantity; the dimension is repeated identically four times on the part.
正交投影法使用多个（通常三个）相互垂直的二维视图来精确表示三维物体，无透视变形。每个视图显示垂直于物体的观察结果。主视图给出高度和宽度，俯视图给出宽度和深度，右视图给出高度和深度。这确保每个特征都可准确标注尺寸。标注 “4 x M8 x 1.25 – 6H” 表示：四处相同的螺纹孔（或螺拴孔），采用 ISO 公制螺纹 M8，螺距 1.25 mm（细牙），内螺纹公差等级为 6H。6H 表示中等旋合长度的常用配合公差，确保互换性。M8 前的 “x” 表示数量；该特征在零件上重复四处相同。
10. Mechanical Systems: Levers and Mechanical Advantage | 机械系统：杠杆与机械效益
A first-class lever has an effort arm of 600 mm and a load arm of 150 mm. A load of 800 N is to be lifted. Calculate: (a) the ideal mechanical advantage, (b) the effort force required (neglecting friction), and (c) the velocity ratio. Explain why the actual effort would be greater in practice.
一根第一类杠杆的施力臂为 600 mm，阻力臂为 150 mm，欲提升 800 N 载荷。计算：(a) 理想机械效益，(b) 所需施力（忽略摩擦），(c) 速度比。解释为何实际施力会更大。
Mechanical advantage (MA) = Load / Effort. For an ideal lever, MA also equals effort arm length / load arm length: Ideal MA = 600 mm / 150 mm = 4. Thus the lever multiplies effort by 4. Effort required = Load / MA = 800 N / 4 = 200 N. Velocity ratio (VR) for a lever is the ratio of distances moved at the effort end and load end per unit time, which is identical to the arm length ratio if the lever is rigid: VR = 600 / 150 = 4, same as ideal MA because there is no energy loss. In practice, friction at the fulcrum (pivot) consumes some input work, so the actual mechanical advantage (AMA) is less than VR. Actual effort = Load / AMA, which will be greater than 200 N. Additional forces due to lever bending or deformation also increase the required effort slightly.
机械效益（MA）= 载荷 / 施力。对理想杠杆，MA 也等于施力臂长 / 阻力臂长：理想 MA = 600 mm / 150 mm = 4，即杠杆将力放大 4 倍。所需施力 = 载荷 / MA = 800 N / 4 = 200 N。杠杆的速度比（VR）是施力端与载荷端在单位时间内移动距离之比，若杠杆刚硬，与力臂比相等：VR = 600 / 150 = 4，与理想 MA 相同，因无能量损失。但实际上，支点处摩擦消耗部分输入功，导致实际机械效益（AMA）小于 VR。实际施力 = 载荷 / AMA，将大于 200 N。杠杆弯曲或变形也会使所需施力略微增加。
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