📚 CCEA Year 10 Physics: In-Depth Analysis of Past Paper Questions | CCEA 十年级物理历年真题深度解析
Welcome to our comprehensive analysis of CCEA Year 10 Physics past paper questions. This article breaks down the most common question types, provides model answers, and highlights key marking points. By understanding exam trends and practising targeted techniques, you can boost your confidence and performance in both structured questions and required practicals.
欢迎阅读我们对 CCEA 十年级物理历年真题的全面解析。本文分解最常见的题型，提供模范答案，并强调关键得分点。通过了解考试趋势并练习有针对性的技巧，你可以在结构化题目和必做实验题中提升信心与成绩。
1. Kinematics and Motion Graphs | 运动学与运动图像
A typical past paper question gives a velocity-time graph for a cyclist. The graph shows a straight line sloping upwards for 4 seconds, then a horizontal line for 6 seconds. The question asks for the acceleration during the first stage and the total distance travelled.
一道典型的真题给出一位骑行者的速度-时间图像。图像显示一条直线在 4 秒内向上倾斜，然后是一条水平线持续 6 秒。题目要求计算第一阶段的加速度和总行驶距离。
Acceleration is found using a = (v − u) / t. If the cyclist reaches 8 m/s from rest in 4 s, then a = (8 − 0) / 4 = 2 m/s². Always substitute the correct values and include the unit.
加速度用 a = (v − u) / t 计算。如果骑行者 4 秒内从静止达到 8 m/s，则 a = (8 − 0) / 4 = 2 m/s²。务必代入正确数值并写明单位。
The total distance equals the area under the velocity-time graph. Split the shape into a triangle (½ × base × height = ½ × 4 s × 8 m/s = 16 m) and a rectangle (6 s × 8 m/s = 48 m), giving a total of 64 m. A common mistake is to forget the ½ factor for the triangle.
总距离等于速度-时间图像下的面积。将形状分解为一个三角形（½ × 底 × 高 = ½ × 4 s × 8 m/s = 16 m）和一个矩形（6 s × 8 m/s = 48 m），总计 64 m。常见错误是忘记三角形的 ½ 因子。
Examiners frequently see students confuse velocity-time and distance-time graphs. Check the label on the y-axis: the slope on a velocity-time graph gives acceleration, not speed.
考官经常发现学生混淆速度-时间图和距离-时间图。检查 y 轴标签：速度-时间图的斜率表示加速度，而非路程。
2. Forces and Newton’s Laws | 力与牛顿定律
A past paper might describe a 5 kg box pulled along a rough surface with a force of 20 N. A frictional force of 5 N acts in the opposite direction. Students must calculate the resultant force and the acceleration.
真题可能会描述一个 5 kg 的箱子被 20 N 的力拉着在粗糙表面上运动。一个 5 N 的摩擦力反方向作用。学生需要计算合力和加速度。
Resultant force = forward force − friction = 20 N − 5 N = 15 N. Using F = m × a, the acceleration a = F / m = 15 N / 5 kg = 3 m/s². Always draw a free-body diagram to visualise the forces.
合力 = 前向力 − 摩擦力 = 20 N − 5 N = 15 N。使用 F = m × a，加速度 a = F/m = 15 N / 5 kg = 3 m/s²。务必画出受力分析图来直观显示各个力。
Many candidates lose marks by using the applied force instead of the resultant force. Only the net force causes acceleration. Also, ensure the mass is in kg, not grams.
许多考生因使用施加力而非合力而失分。只有净力产生加速度。此外，质量必须用 kg，而非 g。
A follow-up question may ask for the distance travelled in 4 seconds if starting from rest: s = ut + ½at² = 0 + ½ × 3 × 4² = 24 m. Remember to square the time.
后续问题可能要求计算静止启动 4 秒内的行驶距离：s = ut + ½at² = 0 + ½ × 3 × 4² = 24 m。记得将时间平方。
3. Energy Efficiency and Sankey Diagrams | 能源效率与桑基图
An electric motor does 500 J of useful work while consuming 800 J of electrical energy. A typical exam question asks: ‘Calculate the efficiency and draw a Sankey diagram for this motor.’
一台电动机做了 500 J 的有用功，同时消耗了 800 J 的电能。典型考题问：“计算该电动机的效率并画出其桑基图。”
Efficiency = (useful output energy / total input energy) × 100% = (500 J / 800 J) × 100% = 62.5%. The wasted energy is 300 J, usually dissipated as heat and sound.
效率 = (有用输出能 / 总输入能) × 100% = (500 J / 800 J) × 100% = 62.5%。浪费的能量为 300 J，通常以热和声的形式耗散。
In the Sankey diagram, the input arrow splits into a wide useful arrow and a narrower wasted arrow. The width of each arrow is proportional to the energy it represents. Examiners look for accurate widths and labelled amounts.
在桑基图中，输入箭头分成一个宽的有用箭头和一个窄的浪费箭头。每个箭头的宽度与其代表的能量成正比。考官关注准确的宽度和标注的数值。
A common error is to swap the useful and wasted arrows or to forget the percentage conversion. Always show the formula, substitution and final percentage with the correct unit (% not J).
常见错误是把有用和浪费箭头互换，或者忘记转换成百分比。一定要写出公式、代入过程以及带正确单位（% 而非 J）的最终百分比。
4. Electrical Circuits and Ohm’s Law | 电路与欧姆定律
A CCEA question often provides a 12 V battery connected to two resistors in series: 4 Ω and 6 Ω. Candidates must calculate total resistance, circuit current and the voltage across the 4 Ω resistor.
CCEA 常常出题：一个 12 V 电池与两个串联电阻连接：4 Ω 和 6 Ω。考生必须计算总电阻、电路电流以及 4 Ω 电阻两端的电压。
For series circuits, Rtotal = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω. Then current I = V / R = 12 V / 10 Ω = 1.2 A. Voltage across 4 Ω, V = I × R = 1.2 A × 4 Ω = 4.8 V.
对于串联电路，R总 = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω。然后电流 I = V / R = 12 V / 10 Ω = 1.2 A。4 Ω 两端的电压 V = I × R = 1.2 A × 4 Ω = 4.8 V。
When resistors are in parallel, students must use 1/Rtotal = 1/R₁ + 1/R₂. For a 4 Ω and 6 Ω parallel pair, 1/Rtotal = 1/4 + 1/6 = 5/12, so Rtotal = 12/5 = 2.4 Ω. A mistake is to treat parallel resistors as series and simply add them.
当电阻并联时，学生必须使用 1/R总 = 1/R₁ + 1/R₂。对于 4 Ω 和 6 Ω 并联，1/R总 = 1/4 + 1/6 = 5/12，因此 R总 = 12/5 = 2.4 Ω。一个错误是把并联电阻当成串联直接相加。
Exam technique: always write the formula, rearrange if needed, substitute values with units and state the final answer to two or three significant figures. Ohm’s law questions frequently test unit conversion, such as mA to A.
考试技巧：始终写出公式，需要时变形，代入带单位的数值，并以两或三位有效数字给出最终答案。欧姆定律题目经常考单位换算，如 mA 转换成 A。
5. Wave Properties and the Wave Equation | 波的性质与波动方程
Water waves with a frequency of 5 Hz and a wavelength of 0.2 m are a common starting point. The question asks for the wave speed and what happens to wavelength if frequency is doubled while the medium stays the same.
频率为 5 Hz、波长为 0.2 m 的水波是常见的起点。题目要求计算波速，并问如果频率加倍而介质不变，波长会怎样。
Wave speed v = f × λ = 5 Hz × 0.2 m = 1.0 m/s. Since wave speed depends on the medium, doubling the frequency to 10 Hz while speed remains 1.0 m/s forces the wavelength to halve: λ = v / f = 1.0 / 10 = 0.1 m.
波速 v = f × λ = 5 Hz × 0.2 m = 1.0 m/s。由于波速取决于介质，频率加倍到 10 Hz 而波速维持 1.0 m/s，则波长必须减半：λ = v / f = 1.0 / 10 = 0.1 m。
Examiners often include a ripple tank diagram and ask about refraction. When waves move from deep to shallow water, speed decreases, wavelength shortens, but frequency remains unchanged. Draw the wavefronts bending towards the normal.
考官常常配以波纹水槽图并问折射。当波从深水区进入浅水区时，波速减小，波长变短，但频率不变。画出波前向法线方向弯曲。
Many candidates incorrectly state that frequency changes when waves enter a new medium. Emphasise that the frequency is set by the source and never changes during refraction.
许多考生错误地声称波进入新介质时频率改变。强调频率由波源决定，折射过程中从不变。
6. Electromagnetic Spectrum and Calculations | 电磁波谱与计算
A typical CCEA question asks students to list the electromagnetic spectrum in order of increasing frequency. The correct order is: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays.
一道典型的 CCEA 题目要求学生按频率递增顺序列出电磁波谱。正确顺序是：无线电波、微波、红外线、可见光、紫外线、X 射线、伽马射线。
Given that microwaves have a wavelength of 0.1 m, calculate their frequency using f = c / λ, where c = 3 × 10⁸ m/s. f = 3 × 10⁸ / 0.1 = 3 × 10⁹ Hz. Always express large numbers in standard form.
已知微波波长为 0.1 m，使用 f = c / λ 计算频率，其中 c = 3 × 10⁸ m/s。f = 3 × 10⁸ / 0.1 = 3 × 10⁹ Hz。总是用科学记数法表达大数。
Applications questions test knowledge of uses: radio waves for TV and radio, microwaves for satellite communication, infrared for remote controls, X-rays for medical imaging. Link the penetrating ability to frequency.
应用题考查用途知识：无线电波用于电视和广播，微波用于卫星通信，红外线用于遥控器，X 射线用于医学成像。将穿透能力与频率关联起来。
Students sometimes confuse frequency and wavelength order. A helpful tip: as frequency increases, wavelength decreases. So gamma rays have the highest frequency but shortest wavelength.
学生有时混淆频率和波长顺序。一个有用的提示：频率升高时，波长减小。因此伽马射线频率最高但波长最短。
7. Radioactive Decay and Half-life | 放射性衰变与半衰期
A classic half-life question provides a decay graph or table for iodine-131 and asks how many nuclei remain after 24 days if the half-life is 8 days and the initial count is 1000.
经典的半衰期题给出碘-131 的衰变图或表格，问如果半衰期为 8 天、初始数量为 1000，24 天后还剩下多少原子核。
Number of half-lives = total time / half-life = 24
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