📚 Year 10 CAIE Engineering: Unit Test Mock Paper Breakdown | CAIE 工程 Year 10 单元测试模拟卷解析
This article provides a full breakdown of a mock unit test tailored for Year 10 CAIE Engineering. Each section presents a typical exam-style question, followed by a clear explanation of the correct answer and the underlying engineering principles. The topics range from materials testing and manufacturing processes to mechanical systems and electrical components, covering the knowledge you need for a unit test.
本文详细解析了一套针对 Year 10 CAIE 工程课程的单元测试模拟卷。每节呈现一道典型试题，然后给出正确答案和清晰的原理解析，涵盖材料测试、制造工艺、机械系统与电子元件等主题，帮助你全面复习单元测试的核心知识点。
1. Multiple-Choice: Hardness Testing | 选择题：硬度测试
Question: A technician presses a hardened steel ball into a metal surface under a set load and measures the diameter of the indentation. Which hardness test is being performed? A Rockwell B Brinell C Vickers D Knoop
问题：技术人员使用硬化钢球在一定载荷下压入金属表面，并测量压痕直径。正在执行的是哪种硬度测试？A 洛氏 B 布氏 C 维氏 D 努氏
Answer and Explanation: The correct choice is B, the Brinell test. The Brinell hardness test uses a tungsten carbide or hardened steel ball indenter, typically 10 mm in diameter, applied with a standard load. The indentation diameter is measured optically and converted to a Brinell Hardness Number (BHN).
答案与解析：正确答案是 B，布氏硬度试验。布氏试验使用碳化钨或硬化钢球压头，直径通常为 10 mm，施加标准载荷。压痕直径通过光学测量并转换为布氏硬度值（BHN）。
In contrast, the Rockwell test measures depth of penetration with a diamond cone or steel ball under preliminary and major loads; it is faster but uses different scales. The Vickers test employs a diamond pyramid indenter and is suitable for very thin materials, while the Knoop test is similar but produces an elongated indentation.
相比之下，洛氏试验通过预载荷和主载荷测量金刚石圆锥或钢球的压入深度，速度更快但尺度不同。维氏试验使用金刚石四棱锥压头，适用于极薄材料；努氏试验类似但压痕为长菱形。
For a ductile metal like mild steel, Brinell testing provides reliable bulk hardness values and is often used in quality control of castings and forgings.
对于低碳钢等韧性金属，布氏测试能提供可靠的整体硬度值，常用于铸件和锻件的质量控制。
2. Crystal Structure and Formability | 晶体结构与成形性
Question: Aluminium sheet can be deep-drawn into complex shapes without cracking. This is mainly because aluminium has a face-centred cubic (FCC) structure. Which of the following metals also exists in an FCC structure at room temperature? A Iron (alpha phase) B Titanium C Copper D Zinc
问题：铝板可深冲拉深成复杂形状而不开裂，主要是因为铝具有面心立方（FCC）结构。以下哪种金属在室温下也是 FCC 结构？A 铁（α 相） B 钛 C 铜 D 锌
The correct answer is C, copper. FCC metals such as aluminium, copper, gold and nickel have 12 slip systems, which allow extensive plastic deformation before fracture. This makes them extremely ductile and suitable for processes like deep drawing, wire drawing and bending.
正确答案是 C，铜。FCC 金属如铝、铜、金和镍拥有 12 个滑移系，断裂前可发生大量塑性变形，因此具有极高延展性，适用于深冲、拔丝和弯曲等工艺。
Iron at room temperature (alpha iron) has a body-centred cubic (BCC) structure, which provides only 48 slip systems? Actually BCC has 48 possible slip systems, but less densely packed, making it less ductile than FCC. Titanium is hexagonal close-packed (HCP) and zinc is also HCP, both having limited slip and being relatively brittle at room temperature.
室温下的铁（α 铁）为体心立方（BCC），滑移系虽多但原子堆积不紧密，延展性不如 FCC。钛和锌均为密排六方（HCP）结构，室温下滑移系少，相对较脆。
Understanding crystal structure helps engineers select materials for manufacturing: FCC metals are preferred when severe forming is required, while BCC metals like low-carbon steel are often hot-worked to increase ductility.
理解晶体结构有助于工程师选择制造材料：需要剧烈成形时首选 FCC 金属，而 BCC 金属如低碳钢常采用热加工来提升延展性。
3. Sand Casting Process Steps | 砂型铸造工艺步骤
Question: Describe the basic sequence of operations involved in sand casting, from pattern to finished casting. Your answer should name at least five key stages.
问题：描述砂型铸造从模型到成品铸件的基本操作顺序，至少列出五个关键阶段。
Model answer: The first stage is pattern making. A pattern, usually made of wood, metal or plastic, is a replica of the final component with shrinkage allowance. The pattern is used to form the mould cavity in two halves (cope and drag).
参考答案：第一阶段为制作模型。模型通常由木材、金属或塑料制成，是最终零件的复制品并考虑了收缩余量。模型用于在上下型箱中形成型腔。
The second stage is mould preparation. The drag is filled with green sand (a mixture of silica sand, clay and water), the pattern is placed, and the sand is compacted. The cope is then assembled, incorporating a sprue and riser for pouring and venting.
第二阶段是制型。下型箱填入潮模砂（硅砂、黏土和水混合），放入模型并紧实。然后组装上型箱，设置浇口和冒口以用于浇注和排气。
Third, the mould is opened, the pattern is removed, and any cores are inserted to form internal hollows. The mould is then reassembled. Following this, the metal is melted in a furnace and poured through the sprue into the mould cavity.
第三，打开铸型、取出模型，并放入型芯以形成内部空腔。铸型重新合拢。随后在炉中熔化金属，经由浇口浇入型腔。
After solidification, the casting is allowed to cool. The mould is broken open, and the casting is removed. This is followed by fettling – removing the sprue, riser and any sand clinging to the surface. Finally, finishing operations such as shot blasting or machining are performed.
凝固后铸件冷却，打碎砂型取出铸件，随后进行清整——切除浇口、冒口并清除表面粘砂。最后进行喷丸或机加工等精整工序。
The process is versatile, capable of producing complex shapes at relatively low cost, but surface finish and dimensional accuracy are lower than machined parts.
该工艺适应性强，能以较低成本生产复杂形状，但表面光洁度和尺寸精度低于机加工零件。
4. Gear Ratio and Output Speed Calculation | 齿轮传动比与输出转速计算
Question: A driving pinion has 20 teeth and meshes with a driven gear of 60 teeth. The input shaft rotates at 300 revolutions per minute (rpm). Calculate (a) the gear ratio and (b) the output speed of the driven gear.
问题：主动小齿轮有 20 齿，与 60 齿的从动齿轮啮合。输入转速为 300 转/分。计算 (a) 传动比，(b) 从动齿轮输出转速。
Answer and method: The gear ratio (GR) is defined as the ratio of the number of teeth on the driven gear to the number of teeth on the driver.
答案与解法：传动比（GR）定义为从动轮齿数与主动轮齿数之比。
Gear Ratio = Ndriven / Ndriver
传动比 = 从动轮齿数 / 主动轮齿数
Substitute the given numbers: Ndriven = 60, Ndriver ratio is 3:1, which means the driver must rotate three times to turn the driven gear once.
代入已知数值：从动轮 60 齿，主动轮 20 齿。GR = 60 / 20 = 3。传动比为 3:1，即主动轮转三圈，从动轮转一圈。
d = 300 rpm / 3 = 100 rpm. The speed reduces while torque increases (neglecting friction).
求输出转速，使用关系式：输出转速 = 输入转速 / 传动比。因此输出 = 300 / 3 = 100 rpm。转速降低的同时扭矩增大（忽略摩擦）。
In a practical context, this gear set forms a simple reduction drive, often used in conveyor belts or winches to increase pulling force.
在实际应用中，这对齿轮构成简单减速驱动，常用于传送带或绞盘以增大牵引力。
5. Stress-Strain Terminology for Ductile Materials | 韧性材料的应力-应变术语
Question: Using a typical stress-strain graph for a low-carbon steel, explain the terms 'elastic limit' and 'yield point'. Why are these points important for engineering design?
问题：利用低碳钢的典型应力-应变图，解释术语“弹性极限”和“屈服点”。为什么这些点对工程设计很重要？
Explanation: The elastic limit is the maximum stress a material can withstand while still returning to its original shape when the load is removed. Up to this point, stress and strain are proportional (Hooke's law region), and the slope gives Young's Modulus.
解析：弹性极限是材料在卸载后仍能恢复原状所能承受的最大应力。在此之前，应力与应变成正比（胡克定律区），斜率即为杨氏模量。
The yield point is the stress at which plastic deformation starts. For low-carbon steel, this is often visible as a sudden elongation with no increase in load – the upper and lower yield points. Beyond the yield point, permanent strain occurs, and the material does not regain its original dimensions.
屈服点是塑性变形开始的应力。低碳钢常表现为载荷不增加而突然伸长——上屈服点和下屈服点。超过屈服点即发生永久变形，材料不再恢复原尺寸。
In engineering design, components must typically operate below the elastic limit to avoid permanent deformation. The yield strength is used to determine the safe load limit, with a factor of safety applied to account for uncertainties.
工程设计中，零件通常须在弹性极限以下工作以避免永久变形。屈服强度用于确定安全载荷极限，并引入安全系数应对不确定性。
If a part is loaded beyond its yield point, it may undergo necking and eventual fracture, leading to failure of the whole assembly.
若零件加载超过屈服点，可能发生颈缩并最终断裂，导致整个组件失效。
6. Resistor Colour Code Identification | 电阻色码识别
brown, black, red, gold. Determine its resistance value and tolerance. Options: A 100 Ω ±5% B 1 kΩ ±5% C 10 kΩ ±5% D 100 kΩ ±5%
问题：一只电阻有四道色环，顺序为棕、黑、红、金。确定其阻值与容差。选项：A 100 Ω ±5% B 1 kΩ ±5% C 10 kΩ ±5% D 100 kΩ ±5%
Correct answer: B, 1 kΩ ±5% x100; therefore 10 × 100 = 1000 Ω = 1 kΩ. The fourth band gold indicates a tolerance of ±5%.
正确答案：B，1 kΩ ±5%。色码解读：第一环棕色 = 1，第二环黑色 = 0，第三环红色 = 乘数 ×100，故 10 × 100 = 1000 Ω = 1 kΩ。第四环金色表示容差 ±5%。
The standard E12 resistor colour code is essential for identifying through-hole components. Remember: black 0, brown 1, red 2, orange 3, yellow 4, green 5, blue 6, violet 7, grey 8, white 9. The multiplier ring uses black x1, brown x10, red x100, etc.
标准 E12 电阻色码对于识别直插元件至关重要。记忆：黑 0 棕 1 红 2 橙 3 黄 4 绿 5 蓝 6 紫 7 灰 8 白 9。乘数环：黑 ×1，棕 ×10，红 ×100，依此类推。
In circuit analysis, using the incorrect resistor value can alter current and voltage division, potentially damaging other components. Always double-check colour bands before soldering.
在电路分析中，用错阻值会改变分压与分流，可能损坏其他元件。焊接前务必核对色环。
7. Orthographic Projection Basics | 正投影基础
Question: Sketch the third-angle orthographic views (front, top and right side) of a simple L-bracket. Explain the alignment principle between the views.
问题：绘制简易 L 形支架的第三角正投影视图（前视、俯视和右视）。解释视图之间的对齐原则。
Guidance: In third-angle projection, the object is placed in the third quadrant. The front view is drawn first, typically the most descriptive face. The top view is projected directly above the front view, and the right side view is projected to the right of the front view.
指导：第三角投影中，物体置于第三象限。通常先绘制最能体现特征的正面图。俯视图正投影于前视图上方，右视图投影于前视图右侧。
The views must remain strictly aligned: vertical lines on the front and top views align, and horizontal lines on the front and side views align. The top and side views both show the depth of the bracket, but the depth dimension is transferred using a 45° miter line or direct measurement.
视图必须严格对齐：前视图与俯视图的竖直轮廓对齐，前视图与侧视图的水平轮廓对齐。俯视图和侧视图均显示支架深度，深度尺寸通过 45° 辅助线或直接测量转换。
For an L-bracket, the front view would show the base and vertical leg with a hole. The top view looks down to display the width of the base and the thickness of the leg. The right view shows the height and the width of the leg.
以 L 形支架为例，前视图显示底板和竖直臂及孔。俯视图从上方显示底板宽度和臂的厚度。右视图显示臂的高度和宽度。
Accurate orthographic drawings are the universal language of engineering, enabling manufacture without ambiguity. Hidden lines represent edges not directly visible, and centerlines mark symmetrical features.
准确的正投影图纸是工程通用语言，确保制造无歧义。隐藏线表示不可见棱边，中心线标注对称要素。
8. Truss Load Distribution | 桁架载荷分布
Question: Explain how a simple triangular truss bridge transfers a central vertical load to the supports. Identify which members are in tension and which are in compression.
问题：解释简易三角形桁架桥如何将中心垂直载荷传递至支座。指出哪些杆件受拉、哪些受压。
Answer: A triangular truss consists of top chords, bottom chords and diagonal web members. When a vertical load is applied at the centre, the top chords experience compression, while the bottom chords are placed in tension. The diagonal members transfer the shear force; those sloping downward toward the centre are in compression, and those sloping upward toward the centre are in tension.
答案：三角形桁架由上弦杆、下弦杆和腹杆组成。当垂直载荷作用于中心时，上弦杆受压，下弦杆受拉。斜腹杆传递剪力：向中心下斜的杆件受压，向中心上斜的杆件受拉。
The triangle is the simplest stable geometric shape; it resists deformation by converting applied loads into pure axial forces along its members. This is why truss bridges are efficient and lightweight.
三角形是最简单的稳定几何形状，通过将外加载荷转化为杆件的纯轴向力来抵抗变形。因此桁架桥既高效又轻便。
For example, in a Warren truss with a point load at the top centre node, the two top chords forming the peak are in compression, the vertical or diagonal webs distribute forces, and the bottom chord carries tension across the span to the pin and roller supports.
例如，在顶点受集中载荷的沃伦桁架中，构成顶尖的两根上弦杆受压，竖直或斜腹杆分配载荷，下弦杆在跨度间承受拉力并将力传递至铰支座和滚动支座。
Understanding tension and compression helps in material selection: steel is strong in both, but concrete is weak in tension, so reinforced concrete uses steel rebars to carry tensile forces in structural members.
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