IB WJEC Physics: Unit Test Paper | IB WJEC 物理：单元测试卷

📚 IB WJEC Physics: Unit Test Paper | IB WJEC 物理：单元测试卷

Unit tests in IB WJEC Physics serve as essential milestones that check your understanding of key concepts before you move on to more advanced topics. They mirror the style and rigour of final examinations, combining multiple-choice questions, short structured problems, and longer data-analysis tasks. A well-prepared student not only revises theory but also develops the skill to apply formulas under timed conditions, interpret graphs, and explain physical phenomena using precise scientific language. This article unpacks what makes a unit test paper challenging, offers sample questions with full walkthroughs, and equips you with strategies to boost your score.

IB WJEC 物理的单元测试是检验关键概念掌握程度的重要节点，帮助你在进入高阶内容前查漏补缺。这些测试模拟最终大考的题型和难度，涵盖选择题、简答题和数据分析题。充分备考不仅需要复习理论，还要练就在限时条件下运用公式、解读图表以及用精准科学语言解释现象的能力。本文拆解单元测试卷的难点，提供样题并完整解析，助你掌握提分技巧。

1. Understanding the Unit Test Format | 了解单元测试格式

A typical IB WJEC unit test paper lasts 45–60 minutes and carries 40–50 marks. It begins with several multiple-choice questions that target breadth of knowledge, followed by structured questions where you must show working, draw diagrams, or analyse data. The final section often presents an unfamiliar context—such as a sports scenario or an industrial application—to test transferable skills. Marks are allocated not only for the correct answer but also for clear logical steps, correct units, and appropriate significant figures.

一份典型的 IB WJEC 单元测试卷时长 45–60 分钟，总分 40–50 分。卷首是几道考查知识广度的选择题，然后是需写出解题步骤、作图或分析数据的结构化问题。最后一题常设置陌生情境——如体育运动或工业应用——考察知识迁移能力。得分点不仅在于正确答案，清晰的逻辑步骤、正确的单位和恰当的有效数字同样计分。

2. Key Topics in Mechanics | 力学关键主题

Mechanics is the backbone of any physics course. For the IB WJEC syllabus, you must be confident with kinematics equations, vector resolution, Newton’s laws, equilibrium, work–energy principle, conservation of momentum, and circular motion. Graphs—especially displacement–time, velocity–time, and acceleration–time—are examined intensively. Recognising gradients, areas under curves, and intercepts is essential.

力学是物理课程的基石。在 IB WJEC 大纲中，你必须熟练掌握运动学方程、矢量分解、牛顿定律、平衡条件、功能原理、动量守恒以及圆周运动。图像——尤其是位移–时间、速度–时间和加速度–时间图——是考查重点，读懂斜率、面积和截距至关重要。

3. Sample Multiple-Choice Questions | 选择题样题

Question 1: A car accelerates uniformly from rest at 3.0 m s⁻² for 8.0 s. What is its final velocity?
A. 11 m s⁻¹ B. 24 m s⁻¹ C. 48 m s⁻¹ D. 64 m s⁻¹

问题 1：一辆汽车从静止开始以 3.0 m s⁻² 的加速度匀加速行驶 8.0 s，末速度是多少？
A. 11 m s⁻¹ B. 24 m s⁻¹ C. 48 m s⁻¹ D. 64 m s⁻¹

Solution: Using v = u + at with u = 0, a = 3.0 m s⁻², t = 8.0 s gives v = 24 m s⁻¹. Answer: B. This tests basic substitution into the first suvat equation.

解析：用 v = u + at，代入 u = 0, a = 3.0 m s⁻², t = 8.0 s 得 v = 24 m s⁻¹。答案 B。考查基础 suvat 方程的直接代入。

Question 2: Two forces act on a block: 10 N east and 10 N north. What is the magnitude of the resultant force?
A. 10 N B. 14 N C. 20 N D. 0 N

问题 2：物体受两个力：10 N 向东和 10 N 向北。合力的大小是多少？
A. 10 N B. 14 N C. 20 N D. 0 N

Solution: The forces are perpendicular, so resultant = √(10² + 10²) = √200 ≈ 14.1 N. Answer: B. Always draw a vector triangle to avoid confusion.

解析：两力垂直，合力 = √(10² + 10²) = √200 ≈ 14.1 N。答案 B。务必画矢量三角形避免混淆。

4. Worked Example: Kinematics | 运动学示例解析

A ball is thrown vertically upwards with an initial speed of 20 m s⁻¹. Ignore air resistance and take g = 9.8 m s⁻².
(a) Calculate the maximum height reached.
(b) Determine the total time the ball is in the air before returning to the thrower’s hand.

一球以 20 m s⁻¹ 的初速度竖直上抛。忽略空气阻力，重力加速度 g = 9.8 m s⁻²。
(a) 计算到达的最大高度。
(b) 求球在空中到落回手中的总时间。

Part (a): At maximum height, velocity v = 0. Using v² = u² – 2g s (upward positive), 0 = (20)² – 2 × 9.8 × h_max → h_max = 400 / 19.6 = 20.4 m. Always write the equation and show substitution clearly.

(a) 部分：最高点速度 v = 0。用 v² = u² – 2gs（向上为正），0 = 20² – 2 × 9.8 × h_max → h_max = 400 / 19.6 = 20.4 m。一定要写出方程并清楚代入数据。

Part (b): Time to reach top: v = u – g t_top → 0 = 20 – 9.8 t_top → t_top = 2.04 s. Total time = 2 × t_top = 4.08 s. Alternatively, use s = 0 in s = u t – ½ g t² and solve the quadratic, but symmetry is faster.

(b) 部分：到最高点时间：v = u – g t_top → 0 = 20 – 9.8 t_top → t_top = 2.04 s。总时间 = 2 × 2.04 = 4.08 s。也可设 s = 0 代入 s = u t – ½ g t² 解二次方程，但对称法更快。

5. Worked Example: Forces and Newton’s Laws | 力与牛顿定律示例

A 5.0 kg block rests on a rough horizontal surface. The coefficient of static friction μₛ = 0.40, and kinetic friction μₖ = 0.25. A horizontal force F is gradually increased from zero. (a) What is the minimum force needed to start the block moving? (b) If the force is maintained at 20 N after motion starts, find the acceleration.

一 5.0 kg 的物块放在粗糙水平面上。静摩擦因数 μₛ = 0.40，动摩擦因数 μₖ = 0.25。水平力 F 从零逐渐增大。(a) 使物块开始运动的最小力是多少？(b) 若物块开始运动后保持 20 N 的力，求加速度。

Part (a): The limiting static friction F_f(max) = μₛ × normal reaction = 0.40 × (5.0 × 9.8) = 19.6 N. So a force just exceeding 19.6 N initiates motion. In practical exam answers, state 19.6 N as the threshold.

(a) 部分：最大静摩擦力 F_f(max) = μₛ × 法向反作用力 = 0.40 × (5.0 × 9.8) = 19.6 N。因此推力略大于 19.6 N 即可启动。考试中回答阈值 19.6 N 即可。

Part (b): Once moving, friction becomes kinetic: F_k = μₖ × 5.0 × 9.8 = 12.25 N. Net force = 20 – 12.25 = 7.75 N. a = F_net / m = 7.75 / 5.0 = 1.55 m s⁻². The direction is the same as applied force.

(b) 部分：运动后摩擦力变为动摩擦：F_k = 0.25 × 5.0 × 9.8 = 12.25 N。净力 = 20 – 12.25 = 7.75 N。加速度 a = 7.75 / 5.0 = 1.55 m s⁻²，方向与推力相同。

6. Worked Example: Energy and Momentum | 能量与动量示例

A 0.50 kg trolley moving at 3.0 m s⁻¹ collides with a stationary 1.0 kg trolley on a frictionless track. They stick together. (a) Find the speed after collision. (b) Calculate the kinetic energy lost in the collision and suggest where the energy goes.

一辆 0.50 kg 的小车以 3.0 m s⁻¹ 的速度与一辆静止的 1.0 kg 小车在无摩擦轨道上碰撞，两车粘在一起。(a) 求碰后共同速度。(b) 计算碰撞中损失的动能，并说明能量去向。

Part (a): Momentum before = 0.50 × 3.0 = 1.5 kg m s⁻¹. After, total mass = 1.5 kg. Momentum conserved → 1.5 = 1.5 × v → v = 1.0 m s⁻¹. This is a classic inelastic collision.

(a) 部分：碰前动量 = 0.50 × 3.0 = 1.5 kg m s⁻¹。碰后总质量 1.5 kg。动量守恒 → 1.5 = 1.5 × v → v = 1.0 m s⁻¹。典型的完全非弹性碰撞。

Part (b): KE before = ½ × 0.50 × (3.0)² = 2.25 J. KE after = ½ × 1.5 × (1.0)² = 0.75 J. Loss = 1.5 J. This energy is transformed into internal energy (heat and sound) and work done in permanent deformation of the coupling.

(b) 部分：碰前动能 = ½ × 0.50 × 3.0² = 2.25 J。碰后动能 = ½ × 1.5 × 1.0² = 0.75 J。损失 1.5 J。这些能量转化为内能（热与声）以及使连接处发生永久形变所做的功。

7. Common Pitfalls and How to Avoid Them | 常见陷阱与避免方法

Many marks are lost through sign errors when applying suvat equations. Always define a positive direction at the start and stick to it consistently. In projectile motion, remember that acceleration due to gravity is toward the ground; if upward is positive, a = –9.8 m s⁻². Another common mistake is confusing weight (mg) with normal reaction in slope problems: only the component mg cos θ balances the normal force, while mg sin θ drives acceleration down the incline if friction is absent.

许多失分源于 suvat 方程中的符号错误。解题前务必定义正方向并一以贯之。在抛体运动中，重力加速度指向地面；若向上为正，则 a = –9.8 m s⁻²。另一常见错误是在斜面问题中混淆重力与法向力：只有 mg cos θ 与法向力平衡，而 mg sin θ 才是无摩擦时使物体沿斜面加速的力。

Units are another hazard. Energy must be in joules, but sometimes students mix centimetres with metres, or grams with kilograms. Convert all quantities to SI base units before substituting into formulas. Finally, always check the number of significant figures requested; a final answer to 5 s.f. when data is given to 2 s.f. loses a mark.

单位也是陷阱。能量必须用焦耳，但常有学生混用厘米与米、克与千克。代入公式前应把所有量换算为 SI 基本单位。最后，务必核对有效数字的要求；若题目数据仅保留两位有效数字，你却给出五位有效数字的答案会丢分。

8. Graphing Skills in Unit Tests | 单元测试中的作图技能

You are frequently asked to sketch and interpret graphs. For a velocity–time graph, the gradient gives acceleration, and the area under the line represents displacement. A curved line means varying acceleration. If you are given a displacement–time graph and asked to find instantaneous velocity, draw a tangent at the required time and calculate its slope. In force–extension experiments, the gradient of the linear region gives the spring constant k, but you may also need to identify the limit of proportionality.

你常被要求绘制和解读图像。在速度–时间图中，斜率表示加速度，图线下方面积代表位移。曲线表示加速度在变化。如果给定位移–时间图并求瞬时速度，在指定时刻画切线并计算斜率。在力–伸长量实验中，线性区域的斜率给出劲度系数 k，但你也可能需要识别比例极限。

9. Structured Data-Analysis Question | 结构化数据分析题

An experiment measures the period T of a simple pendulum for different lengths L. The data are: L = 0.25 m, T = 1.00 s; L = 0.50 m, T = 1.41 s; L = 1.00 m, T = 2.00 s. (a) Show that T² is proportional to L. (b) Determine the acceleration of free fall g from the gradient, given T = 2π √(L/g).

实验测量不同摆长 L 下单摆的周期 T。数据：L = 0.25 m, T = 1.00 s; L = 0.50 m, T = 1.41 s; L = 1.00 m, T = 2.00 s。(a) 证明 T² 与 L 成正比。(b) 根据 T = 2π √(L/g)，由斜率求自由落体加速度 g。

Part (a): Calculate T² for each: 1.00² = 1.00 s² for L=0.25 m; 1.41² ≈ 2.00 s² for L=0.50 m; 2.00² = 4.00 s² for L=1.00 m. The ratio T²/L is constant = 4.0 s² m⁻¹ for all points, verifying proportionality.

(a) 部分：计算各点 T²：L=0.25 m 时 1.00² = 1.00 s²；L=0.50 m 时 1.41² ≈ 2.00 s²；L=1.00 m 时 2.00² = 4.00 s²。T²/L 恒为 4.0 s² m⁻¹，证明正比关系成立。

Part (b): From T = 2π √(L/g), square both sides: T² = (4π²/g) L. So gradient m = 4π²/g. Using any point, e.g., (0.25, 1.00): m = 1.00 / 0.25 = 4.0 s² m⁻¹. Then g = 4π² / m = 4π² / 4.0 = π² ≈ 9.87 m s⁻². This is close to the accepted value 9.81 m s⁻²; experimental errors explain the small difference.

(b) 部分：由 T = 2π √(L/g) 两边平方：T² = (4π²/g) L。故斜率 m = 4π²/g。任取一点如 (0.25, 1.00)：m = 1.00 / 0.25 = 4.0 s² m⁻¹。g = 4π² / 4.0 = π² ≈ 9.87 m s⁻²。这接近公认值 9.81 m s⁻²，微小差异源自实验误差。

10. Exam Techniques for Success | 成功答题技巧

First, scan the whole paper and allocate time roughly according to marks. For multiple-choice, eliminate obviously wrong options to increase your guessing odds. In structured questions, even if you cannot obtain a numerical answer, write down relevant formulas—the exam board awards method marks. Always box your final answer and include the unit. If a question says ‘hence or otherwise’, you may use a given result from an earlier part even if you haven’t proved it. Manage your stress by deep breathing and positive self-talk.

首先，快速浏览全卷并按分值大致分配时间。对于选择题，排除明显错误选项以提高猜题几率。在简答题中，即使算不出数值，也要写下相关公式——考试局会给方法分。记得用方框标出最终答案并写出单位。如果题目说“由此或用其他方法”，即使没证明前一小问的结果，也可直接使用。通过深呼吸和积极自我暗示管理紧张情绪。

11. Building a Revision Timetable | 制定复习计划

Break the mechanics unit into subtopics: kinematics, forces, energy, momentum, circular motion. Dedicate two 45-minute sessions per subtopic: one for concept revision and worked examples, one for past-paper questions under timed conditions. After each session, mark your work and log errors in a ‘mistake diary’. Review this diary weekly to prevent repeating the same mistakes. A week before the unit test, do a full mock paper in one sitting to build stamina.

将力学单元拆分为子主题：运动学、力、能量、动量、圆周运动。每个子主题安排两个 45 分钟的学习时段：一个用于概念复习与例题研究，一个用于限时刷真题。每次完成后自行批改，将错误记录在“错题本”上。每周复习一次错题本，避免重蹈覆辙。单元测试前一周，一次做完一整份模拟卷以培养答题耐力。

12. Final Tips Before the Test | 考前最后提示

On the day before, avoid cramming new material. Instead, review your formula sheet and the mistake diary. Ensure your calculator has fresh batteries and you have a clear ruler and protractor for diagrams. During the test, read each question twice—many errors stem from misreading ‘acceleration’ as ‘velocity’ or ‘maximum height’ as ‘range’. If stuck, move on and return later; a fresh perspective often helps. Trust your preparation and think like a physicist: break complex motions into simple parts, apply conservation laws, and always check dimensions.

考前一天不要硬塞新知识，转而复习公式表和错题本。确保计算器电量充足，带好干净直尺和量角器以作图。考试时每题读两遍——很多错误源于看错“加速度”与“速度”或“最大高度”与“射程”。卡壳时先跳过，稍后再回来看；换一个视角常有奇效。相信自己的准备，像物理学家一样思考：把复杂运动分解为简单部分，运用守恒定律，并始终检查量纲。

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IB WJEC Physics: Unit Test Paper | IB WJEC 物理：单元测试卷