Year 13 CIE Physics: Unit Test Mock Exam Analysis | Year 13 CIE 物理：单元测试模拟卷解析

📚 Year 13 CIE Physics: Unit Test Mock Exam Analysis | Year 13 CIE 物理：单元测试模拟卷解析

This article provides a complete walkthrough of a mock unit test designed for Year 13 CIE Physics students. Every question is carefully analysed with step-by-step working, key formulas and conceptual explanations, helping you reinforce essential topics such as circular motion, gravitational fields, thermodynamics, electric and magnetic fields, capacitance and nuclear physics. Use this analysis to identify common pitfalls and strengthen your problem-solving skills ahead of the real examination.

本文为 CIE Year 13 物理学生提供一份完整的单元测试模拟卷解析。每道题都经过逐步演算，并给出关键公式与概念解释，帮助你巩固圆周运动、引力场、热力学、电场与磁场、电容以及原子核物理等核心内容。请利用这份解析查漏补缺，提升解题能力，为实考做好充分准备。

1. Overview of the Mock Paper | 模拟卷概述

The mock paper consists of eight compulsory questions covering the main topics of the Year 13 CIE Physics syllabus. It is designed to be completed in 60 minutes and carries a total of 60 marks. The questions range from straightforward calculations to multi-step problems that require a clear understanding of underlying principles. The following breakdown shows the topic distribution and mark allocation.

这份模拟卷包含八道必答题，覆盖 Year 13 CIE 物理教学大纲的主要课题。考试时间设计为 60 分钟，满分 60 分。题目从直接计算到多步推理均有涉及，要求学生透彻理解基本原理。下表给出了各题的知识点分布及分值。

Each question is presented with its complete worked solution. Pay attention to the way units are handled and how final answers are rounded appropriately. Let’s dive into the detailed solutions.

每道题均给出完整解答过程。请注意单位的处理方式以及最终答案的恰当舍入。现在让我们进入详细解析。

2. Question 1 – Circular Motion | 第 1 题 – 圆周运动

Question: A car of mass 1200 kg travels around a roundabout of radius 15 m at a constant speed of 12 m/s. Calculate the centripetal force acting on the car and determine the minimum coefficient of static friction between the tyres and the road needed to prevent the car from skidding.

题目：一辆质量为 1200 kg 的汽车以 12 m/s 的恒定速率在半径为 15 m 的环岛行驶。计算作用在汽车上的向心力，并求出防止汽车打滑所需轮胎与路面之间的最小静摩擦系数。

The centripetal force is given by F_c = m v² / r. Substituting the values: F_c = 1200 × (12)² / 15 = 1200 × 144 / 15 = 11520 N.

向心力公式为 F_c = m v² / r。代入数值：F_c = 1200 × (12)² / 15 = 1200 × 144 / 15 = 11520 N。

The frictional force supplies this centripetal force, and the maximum static friction is f_max = μ_s N = μ_s m g. Setting f_max = F_c gives μ_s = F_c / (m g). Using g = 9.81 m/s²: μ_s = 11520 / (1200 × 9.81) = 11520 / 11772 ≈ 0.98. Therefore, the minimum coefficient of friction required is about 0.98.

摩擦力提供该向心力，而最大静摩擦为 f_max = μ_s N = μ_s m g。令 f_max = F_c，得 μ_s = F_c / (m g)。取 g = 9.81 m/s²：μ_s = 11520 / (1200 × 9.81) = 11520 / 11772 ≈ 0.98。因此所需最小摩擦系数约为 0.98。

3. Question 2 – Gravitational Fields | 第 2 题 – 引力场

Question: A planet has mass 5.0 × 10²³ kg and radius 3.0 × 10⁶ m. Calculate the escape velocity from the surface of the planet. (Gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻²)

题目：某行星质量为 5.0 × 10²³ kg，半径为 3.0 × 10⁶ m。计算该行星表面的逃逸速度。（引力常量 G = 6.67 × 10⁻¹¹ N m² kg⁻²）

Escape velocity is given by v_esc = √(2 G M / R). Inserting the data: v_esc = √( 2 × 6.67 × 10⁻¹¹ × 5.0 × 10²³ / (3.0 × 10⁶) ).

逃逸速度公式为 v_esc = √(2 G M / R)。代入数据：v_esc = √( 2 × 6.67 × 10⁻¹¹ × 5.0 × 10²³ / (3.0 × 10⁶) )。

Calculate the numerator: 2 × 6.67 × 10⁻¹¹ × 5.0 × 10²³ = 6.67 × 10¹³. Divide by 3.0 × 10⁶ gives 2.223 × 10⁷. Taking the square root: v_esc = √(2.223 × 10⁷) ≈ 4.71 × 10³ m/s. So the escape velocity is approximately 4.7 km/s.

计算分子部分：2 × 6.67 × 10⁻¹¹ × 5.0 × 10²³ = 6.67 × 10¹³。除以 3.0 × 10⁶ 得 2.223 × 10⁷。开平方：v_esc = √(2.223 × 10⁷) ≈ 4.71 × 10³ m/s。因此逃逸速度约为 4.7 km/s。

4. Question 3 – Simple Harmonic Motion | 第 3 题 – 简谐运动

Question: A mass attached to a spring performs simple harmonic motion with amplitude 0.040 m and frequency 2.5 Hz. Determine the maximum speed and the maximum acceleration of the mass.

题目：一个固定在弹簧上的物体做简谐运动，振幅为 0.040 m，频率为 2.5 Hz。求物体的最大速率和最大加速度。

First find the angular frequency: ω = 2π f = 2π × 2.5 = 5π rad/s. Maximum speed is v_max = ω A = (5π) × 0.040 ≈ 0.628 m/s.

首先求角频率：ω = 2π f = 2π × 2.5 = 5π rad/s。最大速率 v_max = ω A = (5π) × 0.040 ≈ 0.628 m/s。

Maximum acceleration is a_max = ω² A = (5π)² × 0.040 = 25π² × 0.040 ≈ 9.87 m/s². Both values are reached when the displacement is zero (for speed) and at maximum displacement (for acceleration).

最大加速度 a_max = ω² A = (5π)² × 0.040 = 25π² × 0.040 ≈ 9.87 m/s²。这两个极值分别出现在位移为零时（最大速率）和最大位移处（最大加速度）。

5. Question 4 – Thermodynamics | 第 4 题 – 热力学

Question: A fixed mass of an ideal gas at 2.0 × 10⁵ Pa and 300 K occupies a volume of 0.020 m³. The gas is heated at constant volume until its temperature reaches 450 K. Calculate the final pressure and the work done by the gas during this process.

题目：一定质量的理想气体在 2.0 × 10⁵ Pa 和 300 K 下体积为 0.020 m³。气体在体积不变的条件下被加热，直至温度达到 450 K。计算最终压强以及此过程中气体做的功。

For a constant volume process, P₁/T₁ = P₂/T₂. Hence P₂ = P₁ × (T₂ / T₁) = 2.0 × 10⁵ × (450 / 300) = 3.0 × 10⁵ Pa.

对于等容过程，P₁/T₁ = P₂/T₂。因此 P₂ = P₁ × (T₂ / T₁) =

Published by TutorHao | Year 13 Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）