📚 Pre-U CAIE Engineering: Unit Test Mock Paper Analysis | Pre-U CAIE 工程：单元测试模拟卷解析
This mock paper analysis is designed to help Pre-U CAIE Engineering students consolidate core concepts and examination techniques. Each question is broken down with clear reasoning, common pitfalls, and step‑by‑step solutions. The topics cover mechanics, materials, thermodynamics, fluid statics, and electrical principles — all within the scope of a typical unit test.
本模拟卷解析旨在帮助 Pre‑U CAIE 工程学子巩固核心概念与应试技巧。每道题均配有清晰的解题思路、常见误区及分步解答。所涉知识点涵盖力学、材料、热力学、流体静力学与电路原理，均属于典型单元测试范围。
1. Resolving Forces and Moment Equilibrium | 力的分解与力矩平衡
A beam of length 4 m is supported at A and B, 1 m apart from the ends respectively. A vertical downward force of 500 N acts at the midpoint. Calculate the reactions at supports A and B, taking moments about A.
一根长 4 m 的梁在距离两端各 1 m 处由 A、B 两点支撑。梁的中点作用有垂直向下的 500 N 力。以 A 点为矩心计算支座 A 和 B 的反力。
Considering moment equilibrium about A: the 500 N force is 1 m to the right of mid‑span, so its lever arm from A is 2 m (midpoint) minus 1 m (overhang) = 1 m. Thus, clockwise moment = 500 N × 1 m. Reaction at B acts upward at 2 m from A, providing anticlockwise moment. Hence, 500 × 1 = RB × 2 → RB = 250 N. Vertically, RA + RB = 500 → RA = 250 N.
对 A 点取矩：500 N 力位于中点偏右 1 m 处，距 A 的力臂为 2 m – 1 m = 1 m。顺时针力矩 = 500 N × 1 m。B 点反力向上，力臂为 2 m，产生逆时针力矩。故 500 × 1 = RB × 2，得 RB = 250 N。竖直方向合力为零，RA + RB = 500，故 RA = 250 N。
2. Truss Analysis by Method of Joints | 节点法分析桁架
A simple triangular truss supports a load of 2 kN at the apex. The base length is 3 m, height 2 m. Determine the force in the top chord members, stating whether they are in tension or compression.
一简单三角形桁架在顶点承受 2 kN 荷载。底边长 3 m，高 2 m。求上弦杆的内力，并注明受拉或受压。
At the apex joint, two inclined members meet at an angle θ to the horizontal where tan θ = 2/1.5 → θ = 53.13°. Resolving vertically: 2F sin 53.13° = 2 kN → F = 1.25 kN. Since the force arrows point away from the joint, both members are in compression. (Alternatively, the equilibrium of the joint shows the members push towards the joint, confirming compression.)
在顶点节点，两斜杆与水平方向夹角 θ 满足 tan θ = 2/1.5 → θ = 53.13°。竖向分解：2F sin 53.13° = 2 kN → F = 1.25 kN。由于力箭头背离节点，两杆均受压。（亦可通过节点平衡判断杆件推向节点，确认受压。）
3. Stress, Strain and Young’s Modulus | 应力、应变与杨氏模量
A steel wire of diameter 2 mm and original length 1.5 m stretches by 0.8 mm under a load of 150 N. Calculate the tensile stress, strain, and Young’s modulus of the material.
一直径 2 mm、原长 1.5 m 的钢丝在 150 N 荷载下伸长 0.8 mm。计算材料的拉应力、拉应变及杨氏模量。
Cross‑sectional area A = π × (1×10⁻³)² = 3.142×10⁻⁶ m². Stress σ = F/A = 150 / 3.142×10⁻⁶ = 47.75 MPa. Strain ε = ΔL/L₀ = 0.8×10⁻³ / 1.5 = 5.333×10⁻⁴. Young’s modulus E = σ/ε = 47.75×10⁶ / 5.333×10⁻⁴ = 89.5 GPa.
横截面积 A = π × (1×10⁻³)² = 3.142×10⁻⁶ m²。应力 σ = F/A = 150 / 3.142×10⁻⁶ ≈ 47.75 MPa。应变 ε = ΔL/L₀ = 0.8×10⁻³ / 1.5 = 5.333×10⁻⁴。杨氏模量 E = σ/ε = 47.75×10⁶ / 5.333×10⁻⁴ ≈ 89.5 GPa。
4. Torsion of Circular Shafts | 圆轴扭转
A solid circular shaft of radius 25 mm transmits a torque of 800 N·m. Determine the maximum shear stress in the shaft and the angle of twist per metre length if the shear modulus G = 80 GPa.
一根半径为 25 mm 的实心圆轴传递 800 N·m 的扭矩。求轴内最大剪应力及每米长度的扭转角，已知剪切模量 G = 80 GPa。
Polar second moment of area J = (π/2)R⁴ = (π/2)×(0.025)⁴ = 6.136×10⁻⁷ m⁴. Maximum shear stress τmax = TR/J = 800 × 0.025 / 6.136×10⁻⁷ = 32.6 MPa. Angle of twist per unit length θ = T/(GJ) = 800 / (80×10⁹ × 6.136×10⁻⁷) = 0.0163 rad/m = 0.934°/m.
极惯性矩 J = (π/2)R⁴ = (π/2)×(0.025)⁴ = 6.136×10⁻⁷ m⁴。最大剪应力 τmax = TR/J = 800 × 0.025 / 6.136×10⁻⁷ = 32.6 MPa。单位长度扭转角 θ = T/(GJ) = 800 / (80×10⁹ × 6.136×10⁻⁷) = 0.0163 rad/m = 0.934°/m。
5. Bending Stress in Beams | 梁的弯曲应力
A simply supported beam of rectangular cross‑section 120 mm wide × 200 mm deep carries a central point load of 30 kN over a span of 3 m. Find the maximum bending stress.
一简支梁截面为矩形，宽 120 mm、高 200 mm，跨中承受 30 kN 集中荷载，跨度 3 m。求最大弯曲应力。
Maximum bending moment M = PL/4 = 30×10³ × 3 / 4 = 22.5 kN·m. Second moment of area I = bd³/12 = 0.12 × (0.2)³ / 12 = 8×10⁻⁵ m⁴. Distance from neutral axis ymax = 0.1 m. Bending stress σ = M y / I = 22.5×10³ × 0.1 / 8×10⁻⁵ = 28.125 MPa.
最大弯矩 M = PL/4 = 30×10³ × 3 / 4 = 22.5 kN·m。截面惯性矩 I = bd³/12 = 0.12 × (0.2)³ / 12 = 8×10⁻⁵ m⁴。到中性轴的距离 ymax = 0.1 m。弯曲应力 σ = M y / I = 22.5×10³ × 0.1 / 8×10⁻⁵ = 28.125 MPa。
6. Kinematics of a Projectile | 抛体运动学
An object is launched from ground level with an initial speed of 25 m/s at 40° above the horizontal. Neglecting air resistance, calculate the time of flight, horizontal range, and maximum height reached. Take g = 9.81 m/s².
一物体从地面以初速 25 m/s、仰角 40° 发射，忽略空气阻力。计算飞行时间、水平射程及最大高度。取 g = 9.81 m/s²。
Initial vertical component uy = 25 sin 40° = 16.07 m/s. Time of flight t = 2uy/g = 2 × 16.07 / 9.81 = 3.276 s. Horizontal component ux = 25 cos 40° = 19.15 m/s, so range = uxt = 19.15 × 3.276 = 62.74 m. Maximum height H = uy²/(2g) = 16.07² / (2 × 9.81) = 13.16 m.
竖直初速 uy = 25 sin 40° = 16.07 m/s。飞行时间 t = 2uy/g = 2×16.07/9.81 = 3.276 s。水平初速 ux = 25 cos 40° = 19.15 m/s，水平射程 = uxt = 62.74 m。最大高度 H = uy²/(2g) = 16.07²/(2×9.81) = 13.16 m。
7. Newton’s Second Law and Friction | 牛顿第二定律与摩擦
A block of mass 15 kg rests on a rough horizontal surface with coefficient of static friction 0.4. A horizontal force of 50 N is applied. Determine whether the block moves and, if so, its acceleration. Take g = 9.81 m/s².
一质量 15 kg 的木块置于粗糙水平面上，静摩擦系数为 0.4。施加 50 N 水平力。判断木块是否运动；若运动，求其加速度。取 g = 9.81 m/s²。
Maximum static friction fmax = μs N = 0.4 × 15 × 9.81 = 58.86 N. Applied force (50 N) is less than 58.86 N, so the block remains stationary; acceleration = 0.
最大静摩擦力 fmax = μs N = 0.4 × 15 × 9.81 = 58.86 N。施加的 50 N 小于 58.86 N，故木块保持静止，加速度为 0。
8. Conservation of Energy in a Mechanical System | 机械能守恒
A pendulum bob of mass 0.5 kg is released from rest at a point where the string makes an angle of 30° with the vertical. The string length is 1.2 m. Find the speed of the bob as it passes through the lowest point. Ignore air resistance.
一摆锤质量 0.5 kg，自绳与竖直线成 30° 处静止释放。绳长 1.2 m。忽略空气阻力，求摆锤经过最低点时的速率。
Height lost h = L(1 − cos 30°) = 1.2 × (1 − 0.8660) = 0.1608 m. By energy conservation, mgh = ½ mv² → v = √(2gh) = √(2 × 9.81 × 0.1608) = 1.776 m/s.
下落高度 h = L(1 − cos 30°) = 1.2 × (1 − 0.8660) = 0.1608 m。由机械能守恒：mgh = ½ mv² → v = √(2 × 9.81 × 0.1608) = 1.776 m/s。
9. First Law of Thermodynamics for a Closed System | 封闭系统热力学第一定律
A gas in a cylinder expands against a piston, doing 120 J of work while receiving 200 J of heat from a reservoir. Calculate the change in internal energy of the gas.
气缸内气体膨胀推动活塞对外做 120 J 功，同时从热源吸收 200 J 热量。求气体内能的变化量。
First law: ΔU = Q − W. Sign convention: work done by system is positive W. So ΔU = 200 − 120 = +80 J. Internal energy increases by 80 J.
第一定律：ΔU = Q − W。符号约定：系统对外做功取正。故 ΔU = 200 − 120 = +80 J。内能增加 80 J。
10. Hydrostatic Pressure and Manometry | 流体静压力与测压计
A U‑tube manometer contains mercury (density 13 600 kg/m³) to measure gas pressure. The difference in mercury levels is 150 mm, with the gas side lower. Atmospheric pressure is 101 kPa. Find the absolute pressure of the gas.
一 U 形管测压计中装有水银（密度 13 600 kg/m³）以测量气体压力。水银液面高度差为 150 mm，气体侧较低。大气压为 101 kPa。求气体的绝对压强。
Gauge pressure pgauge = ρgh = 13600 × 9.81 × 0.15 = 20.0 kPa. Since the gas side is lower, the gas pressure exceeds atmospheric, so absolute pressure pabs = patm + pgauge = 101 + 20 = 121 kPa.
表压 pgauge = ρgh = 13600 × 9.81 × 0.15 = 20.0 kPa。由于气体侧液面较低，说明气体压力大于大气压，因此绝对压强 pabs = patm + pgauge = 101 + 20 = 121 kPa。
11. DC Circuit Analysis with Series and Parallel Resistors | 串并联直流电路分析
A 12 V battery is connected across a network comprising a 10 Ω resistor in series with a parallel combination of 20 Ω and 30 Ω resistors. Find the total current drawn from the battery and the current through the 20 Ω resistor.
一只 12 V 电池接于一电阻网络，该网络由一个 10 Ω 电阻与 20 Ω 和 30 Ω 的并联组合串联而成。求电池提供的总电流及流经 20 Ω 电阻的电流。
Parallel equivalent: 1/Rp = 1/20 + 1/30 = 5/60 → Rp = 12 Ω. Total resistance RT = 10 + 12 = 22 Ω. Total current IT = V/RT = 12/22 = 0.5455 A. Voltage across parallel branch Vp = IT × Rp = 0.5455 × 12 = 6.545 V. Current through 20 Ω = Vp/20 = 6.545/20 = 0.3273 A.
并联等效电阻：1/Rp = 1/20 + 1/30 = 5/60 → Rp = 12 Ω。总电阻 RT = 10 + 12 = 22 Ω。总电流 IT = V/RT = 12/22 = 0.5455 A。并联支路电压 Vp = IT × Rp = 0.5455 × 12 = 6.545 V。流经 20 Ω 的电流 = 6.545/20 = 0.3273 A。
12. Measurement Uncertainty and Error Propagation | 测量不确定度与误差传递
The resistance R of a wire is obtained from R = V/I. Voltage V = 5.0 ± 0.1 V and current I = 2.0 ± 0.05 A. Calculate the nominal resistance and the maximum percentage uncertainty in R.
导线电阻 R = V/I。电压 V = 5.0 ± 0.1 V，电流 I = 2.0 ± 0.05 A。计算电阻标称值及 R 的最大百分比不确定度。
Nominal R = 5.0/2.0 = 2.5 Ω. Percentage uncertainty in V = (0.1/5.0)×100% = 2%. In I = (0.05/2.0)×100% = 2.5%. For division, add percentage uncertainties: total % uncertainty = 2% + 2.5% = 4.5%. Thus R = 2.5 Ω ± 4.5%.
标称值 R = 5.0/2.0 = 2.5 Ω。V 的百分比不确定度 = (0.1/5.0)×100% = 2%；I 的不确定度 = (0.05/2.0)×100% = 2.5%。乘除运算时，百分比不确定度相加，合计 4.5%。因此 R = 2.5 Ω ± 4.5%。
Published by TutorHao | Engineering 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