📚 Pre-U CAIE Physics Unit Test Mock Exam Analysis | Pre-U CAIE 物理单元测试模拟卷解析
Mock exams are an essential part of preparation for Pre-U CAIE Physics. They do more than test memory: they train you to read a question precisely, choose the right equation, keep units consistent, and present a multi-step solution that an examiner can follow mark by mark. This article provides a detailed breakdown of a unit test covering mechanics, waves, electricity and quantum physics. We will work through selected questions, share key formulas, and discuss common errors and marking pitfalls so that you can master the concepts and improve your exam technique.
模拟考试是 Pre-U CAIE 物理备考的重要环节。它不仅考查记忆，更训练你精确读题、选对公式、保持单位一致，并写出阅卷老师能逐步给分的多步解答。本文详细解析一份涵盖力学、波、电学和量子物理的单元测试卷。我们将逐题讲解，分享关键公式，并讨论常见错误与评分陷阱，帮助你掌握概念、提升应试技巧。
1. Structure of the Mock Test | 模拟卷结构
The mock test consists of two sections: Section A has 10 multiple-choice questions (20 marks), and Section B contains 4 structured questions (30 marks). The total is therefore 50 marks, and the paper is designed to be completed in about 75 minutes, giving roughly 1.5 minutes per mark. The topic coverage is deliberately broad: kinematics, dynamics, circular motion, simple harmonic motion, wave interference, DC circuits with internal resistance, and the photoelectric effect. One question involves data analysis requiring logarithmic plotting, which is the part students most often rush and lose marks on.
模拟卷包含两部分：Section A 为 10 道选择题（20 分），Section B 为 4 道结构化题目（30 分）。全卷共 50 分，建议用时约 75 分钟，平均每分约 1.5 分钟。考点覆盖较广：运动学、动力学、圆周运动、简谐运动、波的干涉、含内阻的直流电路和光电效应。还有一道需要对数作图的数据分析题，这正是学生最容易赶时间、最容易丢分的部分。
Before you attempt any calculation, look at the mark allocation. A 2-mark calculation normally expects a substituted equation plus a final answer with the correct unit; a 4-mark 'show that' question expects intermediate reasoning steps that are visible even if you already know the answer. Structured questions in Pre-U style usually build up: part (a) may ask for a definition, part (b) for a derived quantity, and part (c) for an interpretation or an evaluation of the model used.
动笔计算之前，先看分值分配。2 分的计算题通常要求写出代入的公式，再加带正确单位的最终答案；4 分的 "证明" 题则要求中间推理步骤可见，即便你已知答案。Pre-U 风格的结构化题目往往层层递进：小问 (a) 可能要求下定义，小问 (b) 求某个量，小问 (c) 要求解释含义或评价所用模型。
| Part | Questions | Marks | Main skills tested |
| Section A | 10 MCQ | 20 | Recall, unit reasoning, graph reading |
| Section B | 4 structured | 30 | Multi-step calculation, explanation, data analysis |
2. Question 1: Projectile Motion | 问题1：抛体运动
A ball is projected from ground level with an initial speed of 20 m s⁻¹ at an angle of 30° above the horizontal. Air resistance is negligible. Calculate (a) the time of flight, (b) the maximum height reached, and (c) the horizontal range. Take g = 9.81 m s⁻².
一球从地面以 20 m s⁻¹ 的初速度、与水平方向成 30° 角抛出，空气阻力可忽略。计算 (a) 飞行时间，(b) 最大高度，(c) 水平射程。取 g = 9.81 m s⁻²。
Resolve the initial velocity into components: uₓ = u cos θ and u_y = u sin θ. Horizontal motion has zero acceleration; vertical motion has constant acceleration −g (taking upward as positive). Treating the two directions independently is the single most important idea in projectile problems.
把初速度分解为分量：uₓ = u cos θ，u_y = u sin θ。水平方向加速度为零；竖直方向加速度恒为 −g（取向上为正）。把两个方向独立处理，是抛体问题中最重要的思想。
uₓ = 20 cos 30° = 17.32 m s⁻¹ u_y = 20 sin 30° = 10 m s⁻¹
For part (a), use the vertical motion and set the displacement equal to zero when the ball returns to the ground. Using s = u_y t + ½ a t² with s = 0, u_y = 10 m s⁻¹ and a = −9.81 m s⁻²:
(a) 用竖直方向运动，令球落回地面时位移为零。使用 s = u_y t + ½ a t²，其中 s = 0、u_y = 10 m s⁻¹、a = −9.81 m s⁻²：
0 = 10 t − ½ (9.81) t² ⇒ t (10 − 4.905 t) = 0
Discarding the trivial root t = 0, we obtain t ight time, from launch to landing at the same height. Note that the answer is independent of the horizontal component.
舍去无意义的根 t = 0，得 t = 10 / 4.905 ≈ 2.04 s。这就是从抛出到落回同一高度的完整飞行时间。注意该结果与水平分量无关。
For part (b), at the maximum height the vertical velocity component is momentarily zero, so use v_y² = u_y² + 2as with v_y = 0, u_y = 10 m s⁻¹ and a = −9.81 m s⁻², giving s = h:
(b) 在最大高度处，竖直速度分量瞬时为零，使用 v_y² = u_y² + 2as，其中 v_y = 0、u_y = 10 m s⁻¹、a = −9.81 m s⁻²，位移 s 即 h：
0 = 10² − 2 × 9.81 × h ⇒ h = 100 / (2 × 9.81) = 5.10 m
For part (c), the horizontal range follows either from R = uₓ × t or from R = u² sin 2θ / g. Using the first route avoids algebraic slips and links directly to part (a):
(c) 水平射程可由 R = uₓ × t 得到，也可用 R = u² sin 2θ / g。用第一种方法可避免代数错误，并与 (a) 小问的结果直接衔接：
R = 17.32 × 2.04 = 35.3 m
Check with the second formula: R 5.3 m. The agreement confirms both the time and the range. A useful examination habit is to quote the unrounded value (2.0387 s here) when you carry it forward, then round only at the end to three significant figures.
用第二个公式验证：R = (400 × sin 60°) / 9.81 = 346.4 / 9.81 = 35.3 m。两者一致，说明时间和射程都算对了。一个实用的考试习惯是：中间步骤保留未取整值（此处为 2.0387 s）继续计算，只在最后按三位有效数字取整。
3. Question 2: Circular Motion and Centripetal Force | 问题2：圆周运动与向心力
A ball of mass 0.25 kg is attached to a string of length 0.80 m and swung in a horizontal circle at a constant speed of 3.0 m s⁻¹. Calculate the centripetal acceleration, the tension in the string, the angular velocity and the period of rotation.
质量为 0.25 kg 的小球系在长 0.80 m 的细绳上，以 3.0 m s⁻¹ 的恒定速率在水平圆周上运动。计算向心加速度、绳中张力、角速度与转动周期。
Centripetal acceleration depends on speed and radius only: a = v² / r. Substituting gives a ected towards the centre of the circle at every instant. The direction changes continuously even though the magnitude is constant, which is why the velocity vector is always changing and the motion is accelerating.
向心加速度只取决于速率和半径：a = v² / r。代入得 a = (3.0)² / 0.80 = 11.25 ≈ 11.3 m s⁻²，方向在每一瞬间都指向圆心。其大小不变但方向不断改变，这正是速度矢量始终在变化、因而存在加速度的原因。
The resultant force must be centripetal: F = mv² / r tal-circle arrangement, the string is not perfectly horizontal in practice; the tension has a vertical component that balances weight, so the true tension is larger than the value quoted unless the string is treated as ideal and the plane is assumed flat. Examiners accept the simplified model but reward candidates who state the assumption.
合力必须提供向心力：F = mv² / r = 0.25 × 11.25 = 2.81 N。在实际的水平圆周运动中，绳子并非完全水平；张力的竖直分量要平衡重力，因此若不作理想化处理，真实张力会大于上述数值。阅卷人接受简化模型，但会奖励主动说明假设的考生。
ω = v / r = 3.0 / 0.80 = 3.75 rad s⁻¹ T = 2π / ω = 1.68 s
For vertical circles, a standard extension is the minimum speed at the top of the loop, where the only force available is weight: mg = mv² / r, so v = √gr. For r = 0.80 m this gives v = √9.81 × 0.80 m s⁻¹. Below this speed, the object cannot complete the loop and leaves the circular path.
对于竖直圆周运动，常见的拓展是求最高点的最小速率，此时唯一可用的力是重力：mg = mv² / r，故 v = √gr。取 r = 0.80 m，得 v = √9.81 × 0.80 = 2.80 m s⁻¹。低于此速率，物体无法完成整圈，会脱离圆周轨道。
4. Question 3: Simple Harmonic Motion | 问题3：简谐运动
A 0.50 kg mass on a spring of stiffness 20 N m⁻¹ is displaced 0.060 m from equilibrium and released. Calculate the period, the maximum speed, the maximum acceleration and the total energy of the oscillation. Then state how the period would change if the mass were doubled.
劲度系数为 20 N m⁻¹ 的弹簧上挂着 0.50 kg 的物体，将其从平衡位置拉开 0.060 m 后释放。计算周期、最大速率、最大加速度与振动总能量，并说明若质量加倍周期如何变化。
For a mass–spring system the period is T = 2π√m / k, which is independent of amplitude. This independence is the defining feature of simple harmonic motion and is a frequent examination point.
质量–弹簧系统的周期为 T = 2π√m / k，与振幅无关。这一无关性正是简谐运动的标志性特征，也是考试高频考点。
T = 2π√0.50 / 20 = 2π√0.025 = 0.99 s
The angular frequency is ω = √k / m = √40 = 6.32 rad s⁻¹. Maximum speed occurs at the equilibrium position and equals A acceleration occurs at the extremes of the motion, where the restoring force is greatest: a = Aω² = 0.060 × 40 = 2.4 m s⁻².
角频率 ω = √k / m = √40 = 6.32 rad s⁻¹。最大速率出现在平衡位置，vmax = Aω = 0.060 × 6.32 = 0.38 m s⁻¹。最大加速度出现在运动的两端，此时回复力最大：a = Aω² = 0.060 × 40 = 2.4 m s⁻²。
E = ½ k A² = ½ × 20 × (0.060)² = 0.036 J
During the oscillation this energy converts continuously between elastic potential energy (maximum at the extremes) and kinetic energy (maximum at the centre); the total remains constant if damping is neglected. If the mass is doubled, the period increases by a factor of √2 to about 1.40 s, because T ∝ √m. Amplitude, however, has no effect on period.
振动过程中，该能量在弹性势能（两端最大）与动能（中心最大）之间不断转化；忽略阻尼时总量守恒。若质量加倍，周期增大 √2 倍，约为 1.40 s，因为 T ∝ √m。而振幅对周期没有影响。
A common extension adds the displacement equation x = A cos(ωt). When using it, ensure your calculator is in radian mode, since ω is measured in rad s⁻¹. Mixing degrees with radians is one of the most frequent silent errors in this topic.
常见拓展会给出位移方程 x = A cos(ωt)。使用它时务必把计算器设为弧度制，因为 ω 的单位是 rad s⁻¹。混用角度与弧度是本专题中最常见的隐性错误之一。
5. Question 4: Wave Interference and Superposition | 问题4：波的干涉与叠加
Two loudspeakers are driven in phase by the same signal generator at 680 Hz and are placed 0.60 m apart. The speed of sound in air is 340 m s⁻¹. A microphone is moved along a line 2.5 m from the speakers. Determine the wavelength and explain the pattern of loud and quiet regions observed.
两只扬声器由同一信号发生器同相驱动，频率为 680 Hz，相距 0.60 m。空气中声
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