Pre-U CAIE Physics Unit Test Mock Exam Analysis | Pre-U CAIE 物理单元测试模拟卷解析

📚 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. 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 to help you master the concepts and improve exam technique.

模拟考试是Pre-U CAIE物理备考的重要环节。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，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). Topics include 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.

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）。涉及运动学、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。

2. Question 1: Projectile Motion | 问题1：抛体运动

A ball is projected from ground level with speed 20 m s⁻¹ at 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: uₓ = u cosθ, uᵧ = u sinθ.

分解初速度：uₓ = u cosθ, uᵧ = u sinθ。

uₓ = 20 cos30° = 17.32 m s⁻¹, uᵧ = 20 sin30° = 10 m s⁻¹

For time of flight, consider vertical motion. The displacement is zero when it returns to the ground. Using s = uᵧ t + ½ a t² with s=0, uᵧ=10 m s⁻¹, a = -9.81 m s⁻².

计算飞行时间，考虑竖直方向运动。落回地面时位移为零。使用 s = uᵧ t + ½ a t²，其中 s=0, uᵧ=10 m s⁻¹, a = -9.81 m s⁻²。

0 = 10 t – ½ (9.81) t² ⇒ t (10 – 4.905 t) = 0. Discarding t=0, t = 10 / 4.905 ≈ 2.04 s.

0 = 10 t – ½ (9.81) t² ⇒ t (10 – 4.905 t) = 0。舍去 t=0，得 t = 10 / 4.905 ≈ 2.04 s。

At maximum height, vertical velocity is zero. vᵧ² = uᵧ² + 2a s ⇒ 0 = 10² – 2 × 9.81 × h ⇒ h = 100 / (2 × 9.81) = 5.10 m.

最大高度时竖直速度为零。vᵧ² = uᵧ² + 2a s ⇒ 0 = 10² – 2 × 9.81 × h ⇒ h = 100 / (2 × 9.81) = 5.10 m

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