📚 Year 12 Cambridge Physics: Unit Test Mock Paper Analysis | 剑桥12年级物理：单元测试模拟卷解析
Mock unit tests are a vital part of Year 12 Cambridge Physics revision. They help you identify strengths and weaknesses in core AS-Level topics such as kinematics, dynamics, energy, materials, and electricity. This article walks you through a typical mock paper, providing full solutions, examiner‑style commentary, and key strategies. Each question is designed to mirror the Cambridge assessment style, reinforcing both conceptual understanding and numerical skills.
单元模拟测试是12年级物理复习的关键环节。它们能帮你发现在运动学、动力学、能量、材料和电学等AS级核心主题中的强项与弱项。本文逐题讲解一份典型模拟卷，提供完整解答、类似考官的点评和关键策略。每道题都仿照剑桥评估风格设计，以强化概念理解和计算能力。
1. Mock Paper Structure and Key Topics | 模拟试卷结构与核心主题
This mock unit test is built around six structured questions covering the first half of the AS syllabus. The topics include uniformly accelerated motion, Newton's laws with friction, conservation of momentum in collisions, work and energy with dissipative forces, the Young modulus of a wire, and DC circuit analysis. Each question contains two or three parts, with command words like 'calculate', 'draw', 'explain', and 'determine'. The total marks are typically 50, to be completed in 60 minutes.
这份模拟单元测试由六道结构化题目组成，覆盖了AS教学大纲的前半部分。主题包括匀加速运动、带摩擦的牛顿定律、碰撞中的动量守恒、含耗散力的功与能、金属丝的杨氏模量以及直流电路分析。每道题包含两到三个小题，使用的指令词有“计算”“画图”“解释”和“确定”等。卷面总分通常为50分，要求在60分钟内完成。
You should treat this mock as a real exam – no notes, a calculator, and strict timing. After attempting the paper, read through our solutions to understand common pitfalls and examiner expectations.
你应当把这次模拟当作真实考试——不允许查阅笔记，只使用计算器，并严格计时。尝试完成试卷后，仔细阅读我们的解答，以理解常见陷阱和考官的期望。
2. Kinematics: v-t Graph and Displacement | 运动学：速度–时间图与位移
A car accelerates uniformly from rest to 20 m s⁻¹ in 10 s, maintains this speed for 20 s, and then decelerates uniformly to rest in a further 30 s. (a) Sketch the velocity–time graph. (b) Calculate the total distance travelled. (c) Determine the deceleration in the final stage.
一辆汽车由静止开始匀加速，10秒内达到20 m s⁻¹，然后保持此速度20秒，最后在30秒内匀减速至静止。(a) 画出速度–时间图像。(b) 计算总行驶距离。(c) 计算最后阶段的减速度。
The v-t graph consists of three straight-line segments: a sloping line from (0,0) to (10,20), a horizontal line from (10,20) to (30,20), and a sloping line from (30,20) to (60,0). The area under the graph gives the displacement. The shape is a trapezium plus two triangles, which can be combined into a single trapezium of bases 20 s and 60 s and height 20 m s⁻¹.
速度–时间图由三条直线段构成：从(0,0)到(10,20)的倾斜线，从(10,20)到(30,20)的水平线，以及从(30,20)到(60,0)的倾斜线。图线下的面积表示位移。整个图形可以看作一个上底为20 s、下底为60 s、高为20 m s⁻¹的梯形。
ng phase (½ × 10 × 20 = 100 m), constant speed phase (20 × 20 = 400 m), and decelerating phase (½ × 30 × 20 = 300 m), giving a total of 800 m.
根据梯形面积公式：s = ½ × (两底之和) × 高 = ½ × (20 + 60) × 20 = 800 m。也可以将三个阶段的面积相加：加速段(½ × 10 × 20 = 100 m)、匀速段(20 × 20 = 400 m)和减速段(½ × 30 × 20 = 300 m)，总和为800 m。
667 m s⁻². The magnitude of the deceleration is 0.667 m s⁻². Remember to state the direction or indicate the negative sign, as Cambridge expects a clear sign convention.
最后阶段的减速度是恒定的：a = (v − u) / t = (0 − 20) / 30 = −0.667 m s⁻²。减速度的大小为0.667 m s⁻²。请务必注明方向或给出负号，因为剑桥试卷通常要求清晰的符号约定。
3. Dynamics with Friction | 有摩擦的动力学
block and the surface is 0.20. Take g = 9.8 m s⁻². Calculate the acceleration of the block.
一个质量为2.0 kg的物块放在粗糙水平面上。一个10 N的水平力拉动物块。物块与表面间的动摩擦系数为0.20。取g = 9.8 m s⁻²。计算物块的加速度。
92 N. Friction opposes motion, so the net horizontal force is F − fₖ = 10 − 3.92 = 6.08 N.
先进行竖直方向受力分析：法向反作用力R与重力平衡，故R = mg = 2.0 × 9.8 = 19.6 N。动摩擦力fₖ = μₖ R = 0.20 × 19.6 = 3.92 N。摩擦力阻碍运动，因此水平方向的合力为F − fₖ = 10 − 3.92 = 6.08 N。
Using Newton's second law: a = Fnet 2.0 = 3.04 m s⁻². Always give your answer to an appropriate number of significant figures – more precise. Show the full working to gain method marks even if a numerical slip occurs.
使用牛顿第二定律：a = Fnet / m = 6.08 / 2.0 = 3.04 m s⁻²。答案的有效数字位数要合理——这里可接受3.0 m s⁻²，但3.04 m s⁻²更为精确。要展示完整的计算过程，即使有计算错误也能得到步骤分。
4. Conservation of Momentum: Inelastic Collision | 动量守恒：完全非弹性碰撞
A trolley of mass 3.0 kg moves to the right with a speed of 4.0 m s⁻¹. It collides with a stationary trolley of mass 2.0 kg. After the collision the trolleys stick together. Find their common velocity just after the collision.
一辆质量为3.0 kg的实验小车以4.0 m s⁻¹的速度向右运动。它与一辆质量为2.0 kg的静止小车发生碰撞。碰撞后两车粘连在一起。求碰撞刚结束时它们的共同速度。
. Total momentum before = (3.0 × 4.0) + (2.0 × 0) = 12 kg m s⁻¹. After the collision, the combined mass is 5.0 kg, moving with velocity v. So 5.0v = 12, giving v = 12 / 5.0 = 2.4 m s⁻¹ to the right.
由于小车粘连，这是一次完全非弹性碰撞。在水平方向无外力作用下，动量守恒。碰撞前的总动量 = (3.0 × 4.0) + (2.0 × 0) = 12 kg m s⁻¹。碰撞后总质量为5.0 kg，以速度v运动。故有5.0v = 12，解得v = 12 / 5.0 = 2.4 m s⁻¹，方向向右。
s forget to specify the direction. In the Cambridge exam, velocity is a vector, so stating '2.4 m s⁻¹ to the right' or '2.4 m s⁻¹ in the original direction' re = ½ × 3 × 4² = 24 J, KE after = ½ × 5 × 2.4² = 14.4 J, confirming the inelastic nature.
很多学生忘记标注方向。在剑桥考试中，速度是矢量，因此写出“2.4 m s⁻¹向右”或“2.4 m s⁻¹沿原方向”才能拿到最后一分。同时，可验证动能减少了：碰前KE = ½ × 3 × 4² = 24 J，碰后KE = ½ × 5 × 2.4² = 14.4 J，证实了非弹性碰撞的特性。
5. Work, Energy, and Power with Dissipative Forces | 功、能和功率（含耗散力）
one. Using g = 9.8 m s⁻², calculate the speed of the stone just before it hits the ground.
一块质量为1.0 kg的石块从离地面5.0 m高处落下。下落过程中，空气阻力对石块做负功10 J。取g = 9.8 m s⁻²，计算石块即将撞击地面前的速度。
ical system, so the gain in kinetic energy is ΔKE = mgh − W_resistance = 49 − 10 = 39 J. Starting from rest, the final kinetic energy is ½ mv² = 39 J, so v = √2 × 39 / 1.0 = √78 ≈ 8.83 m s⁻¹.
初始重力势能为mgh = 1.0 × 9.8 × 5.0 = 49 J。克服空气阻力做的功是从机械系统中转移走的能量，因此动能增量为ΔKE = mgh − W_resistance = 49 − 10 = 39 J。由静止开始，末动能为½ mv² = 39 J，故v = √2 × 39 / 1.0 = √78 ≈ 8.83 m s⁻¹。
In questions involving non‑ mmon mistake of adding the resistance work to the potential energy. If the stone had an initial speed, you would need to include its initial kinetic energy as well.
在处理包含非保守力的题目时，务必清晰地写出能量等式：初始能量 − 克服摩擦/阻力所做的功 = 末能量。这能避免将阻力做功错加在势能上的常见错误。如果石块有初速度，还需计入其初始动能。
6. Material Properties: Young Modulus | 材料性质：杨氏模量
The wire extends by 4.0 mm. Using g = 9.8 m s⁻², calculate the Young modulus of the steel.
一根直径为0.50 mm、原长为2.0 m的钢丝在10 kg的负载下伸长4.0 mm。取g = 9.8 m s⁻²，计算钢的杨氏模量。
First, find the cross‑ ⁷) ≈ 5.00 × 10⁸ Pa. Strain ε = extension / original length = (4.0 × 10⁻³) / 2.0 = 2.0 × 10⁻³ (no units).
先求横截面积A。半径r = 0.25 mm = 2.5 × 10⁻⁴ m，因此A = πr² = π × (2.5 × 10⁻⁴)² ≈ 1.96 × 10⁻⁷ m²。拉力F = mg = 10 × 9.8 = 98 N。应力σ = F / A = 98 / (1.96 × 10⁻⁷) ≈ 5.00 × 10⁸ Pa。应变ε = 伸长量 / 原长 = (4.0 × 10⁻³) / 2.0 = 2.0 × 10⁻³（无量纲）。
nvert all lengths to metres before substituting. A common error is using the diameter instead of the radius, which overestimates E by a factor of four.
杨氏模量E = 应力 / 应变 = (5.00 × 10⁸) / (2.0 × 10⁻³) = 2.50 × 10¹¹ Pa。这与钢的公认值（约2.0 × 10¹¹ Pa）接近。代入数据前，务必将所有长度换算为米。常见错误是代入直径而非半径，这会使E值放大至四倍。
7. DC Circuits: Series‑Parallel Combinations | 直流电路：串并联组合
A 12 V battery is connected to a network consisting of a 4.0 Ω resistor in series with a parallel combination of 3.0 Ω and 6.0 Ω. Calculate (a) the total external resistance, (b) the current supplied by the battery, and (c) the power dissipated in the 4.0 Ω resistor.
一个12 V电池连接到一个电路网络：一个4.0 Ω电阻与一个由3.0 Ω和6.0 Ω并联组成的组合串联。(a) 计算总外电阻，(b) 求电池提供的电流，(c) 求4.0 Ω电阻上消耗的功率。
For the parallel branch: Rpar = (R₁ × R₂) / (R₁ + R₂) = (3.0 × 6.0
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