📚 In-Depth Analysis of CAIE IGCSE Physics Past Paper Questions | CAIE IGCSE 物理历年真题深度解析
CAIE IGCSE Physics (0625) past papers are an essential revision resource, revealing recurring question patterns, common pitfalls, and the depth of understanding expected by examiners. This article provides a step-by-step dissection of selected past paper questions covering core topics such as motion, forces, energy, electricity, waves, and radioactivity. Each section breaks down a typical exam question, explains the underlying principles, outlines a clear solution path, and highlights the most frequent mistakes made by students. By working through these worked examples, you will learn how to apply your knowledge under timed conditions and avoid losing marks to careless errors. Bookmark this guide and use it alongside your active past paper practice.
CAIE IGCSE 物理（0625）的历年真题是极其重要的复习资源，它们能揭示反复出现的命题规律、常见失分点以及考官所期望的知识深度。本文将精选力、运动、能量、电学、波和放射性等核心主题的典型真题，进行逐步拆解分析。每个小节都会拆解一道典型试题，解释背后的物理原理，给出清晰的解题路径，并重点指出学生最常犯的错误。通过练习这些范例，你将学会如何在限时条件下应用知识，并避免因粗心而丢分。请收藏本指南，并结合主动的真题训练一起使用。
1. Interpreting a Velocity-Time Graph | 解读速度-时间图像
A past paper question often presents a velocity–time graph for a sprinter and asks for the description of motion, calculation of acceleration, and total distance travelled. A typical graph might show a straight rising line from rest to 10 m/s in 4 s, then a flat line at 10 m/s for the next 6 s.
真题中经常给出短跑运动员的速度-时间图像，要求描述运动状态、计算加速度以及总移动距离。典型的图像可能是：从静止开始，一条倾斜直线在 4 秒内升至 10 m/s，之后以 10 m/s 的恒定速度水平延伸 6 秒。
- Description: From 0 to 4 s the object accelerates uniformly; from 4 s to 10 s it moves at constant velocity.
- 描述：0 到 4 秒物体匀加速运动；4 秒到 10 秒以恒定速度运动。
- Acceleration: Use a = (v − u) / t = 2.5 m/s². Many students forget to write the correct unit or use the wrong time interval.
- 加速度：用公式 a = (v − u) / t。u = 0, v = 10 m/s, t = 4 s, 得 a = 10 / 4 = 2.5 m/s²。许多学生忘记写单位或选错时间区间。
- Distance: istance with displacement when the motion is in a straight line without change of direction; here they are equal.
- 距离：图像下方的面积表示距离。分割为三角形（面积 = ½ × 4 × 10 = 20 m）和矩形（6 × 10 = 60 m），总距离 = 80 m。当运动沿直线且不改变方向时，不要混淆距离与位移；本题两者相等。
2. Calculating Moments and the Principle of Equilibrium | 计算力矩与平衡原理
A classic past paper scenario shows a uniform beam pivoted at its centre, with two weights hung on one side and one weight on the other. The question asks to find an unknown weight or distance. For example: 'A 2 N weight is hung 30 cm to the right of the pivot. Where must a 3 N weight be placed on the left to balance the beam?'
经典的真题场景是：一根均匀横梁在中点支起，一侧挂两个重物，另一侧挂一个重物，要求求出未知重量或距离。例如："一个 2 N 的重物挂在支点右侧 30 cm 处。一个 3 N 的重物应挂在左侧何处，才能使横梁平衡？"
- Principle of moments: ent = total anticlockwise moment about the pivot. Moment = force × perpendicular distance.
- 力矩原理：平衡时，绕支点的顺时针力矩之和等于逆时针力矩之和。力矩 = 力 × 垂直距离。
- Calculation: distance on the left; anticlockwise moment = 3 N × d. Set 3d = 0.60 → d = 0.20 m = 20 cm.
- 计算：顺时针力矩 = 2 N × 0.30 m = 0.60 N m。设左侧距离为 d，逆时针力矩 = 3 N × d。令 3d = 0.60 → d = 0.20 m = 20 cm。
- Common mistakes: Forgetting to convert cm to metres when using N m, or including the beam's weight (ignore if it says 'uniform beam pivoted at centre'). Also, ensure distances are measured from the pivot, not from the ends.
- 常见错误：当使用 N m 时忘记将厘米换算为米，或错误计入横梁自重（若题目说"均匀横梁支于中心"可忽略自重）。另外，距离必须从支点量起，而非从端点量。
3. Series and Parallel Bulb Brightness | 串联与并联灯泡的亮度
Past papers frequently ask: 'Two identical lamps are connected first in series with a battery, then in parallel. Compare the brightness of the lamps in each circuit.' This tests understanding of potential difference and current division.
真题常问："将两个相同的灯泡先与电池串联，再并联。比较两种电路中灯泡的亮度。" 这考察了对电位差和电流分配的理解。
- Series: ttery voltage. Since power P = V² / R, reduced voltage means lower power and dimmer light.
- 串联：通过两灯的电流相同，但总电压被分担，每灯得到电池电压的一半。由 P = V² / R 可知，电压降低导致功率减小，灯泡变暗。
- Parallel: Each lamp is connected directly across the battery, receiving the full battery voltage. The full voltage gives maximum power, so both lamps shine at normal brightness. Current from the battery doubles but each branch only takes its rated current.
- 并联：每灯直接连在电池两端，获得全部电池电压。满电压产生最大功率，两灯均正常发光。干路电流加倍，但每条支路只流过额定电流。
- Key trap: Some students think parallel circuits always make bulbs brighter 'because there are two paths'; they must link this to full voltage across each component. Also, remember that lamp brightness is determined by power, not just current.
- 关键陷阱：有学生认为并联总能使灯泡更亮"因为有两条路径"；必须将其归因于每个元件均获得满电压。此外要牢记，灯泡亮度取决于功率，而非仅仅电流。
4. Energy Efficiency in a Sankey Diagram | 桑基图与能量效率
A past paper might show a Sankey diagram for an electric motor lifting a weight, with input energy 500 J, useful gravitational potential energy 300 J, and waste thermal/sound energy 200 J. It asks: 'Calculate the efficiency and suggest one way to reduce waste energy.'
真题可能给出电动机提升重物的桑基图：输入能量 500 J，有用的重力势能 300 J，浪费的热能/声能 200 J。问题："计算效率，并提出一种减少浪费能量的方法。"
- Efficiency formula: η = (useful output energy / total input energy) × 100%. Here, η = (300 / 500) × 100% = 60%.
- 效率公式：η = （有用的输出能量 / 总输入能量）× 100%。此处 η = (300 / 500) × 100% = 60%。
- Improving efficiency: Lubricate moving parts to reduce friction, or use more streamlined shapes to reduce air resistance. Always link the suggestion to the specific system.
- 提高效率的方法：润滑运动部件以减少摩擦，或采用更流线型的外形以减少空气阻力。建议必须与具体系统相关联。
- Common pitfall: Writing the efficiency as a fraction without converting to a percentage, or forgetting that the Sankey diagram width is proportional to energy; the waste arrows should be drawn narrower for higher efficiency.
- 常见错误：只用分数表示效率而未转换为百分数，或忘记桑基图的宽度与能量成正比；效率更高时，浪费能量的箭头应画得更窄。
5. Specific Heat Capacity – Heating a Metal Block | 比热容 – 加热金属块
A typical practical-based question: 'A 2 kg aluminium block is heated by a 50 W heater for 10 minutes. Its temperature rises from 20 °C to 45 °C. Calculate the specific heat capacity of aluminium.'
典型的实验题："一个 2 kg 的铝块被 50 W 的加热器加热 10 分钟。其温度从 20 °C 升至 45 °C。计算铝的比热容。"
- Energy supplied: E = P × t = 50 W × (10 × 60 s) = 50 × 600 = 30 000 J.
- 提供的能量：E = P × t = 50 W × (10 × 60 s) = 50 × 600 = 30 000 J。
- Temperature change: ecific heat capacity, a change in °C equals a change in K, so conversion is not necessary.
- 温度变化：Δθ = 45 − 20 = 25 °C。计算比热容时，摄氏温度的变化与开尔文变化相等，无需换算。
- Formula: = m c Δθ → c = Q / (m Δθ) = 30 000 / (2 × 25) = 30 000 / 50 = 600 J/(kg °C) [or J/(kg K)].
- 公式：Q = m c Δθ → c = Q / (m Δθ) = 30 000 / (2 × 25) = 30 000 / 50 = 600 J/(kg °C) [或 J/(kg K)]。
- Exam tip: o the surroundings. If asked why experimental value is higher than the true value, answer: ' rger temperature rise is recorded for the same energy input actually retained by the block? ise is smaller, giving a larger calculated c (since c = E / (mΔθ), smaller Δθ → larger c).' Be precise.
- 考试技巧：题目通常假设无热损失。若问为什么实验值比真实值大，应回答："因为有部分能量散失到环境中，导致实际被铝块吸收的能量小于电能输入，测得的温升偏小，由 c = E / (mΔθ) 可知，Δθ 偏小导致计算出的 c 偏大。" 表述要精确。
6. Wave Equation and Ripple Tank | 波动方程与波纹槽
Past paper diagram: 'A ripple tank produces waves with a frequency of 5 Hz. The distance between six successive crests is 20 cm. Calculate the wave speed.'
真题图示："波纹槽产生频率为 5 Hz 的波。六个连续波峰之间的距离为 20 cm。计算波速。"
- Wavelength determination: The distance between six crests covers five wavelengths (since 1–2, 2–3, ..., 5–6). So 5λ = 20 cm → λ = 4.0 cm = 0.04 m.
- 波长确定：六个波峰之间包含五个波长（第1到第2，第2到第3，…，第5到第6）。因此 5λ = 20 cm → λ = 4.0 cm = 0.04 m。
- Wave speed: v = f × λ = 5 Hz × 0.04 m = 0.20 m/s.
- 波速：v = f × λ = 5 Hz × 0.04 m = 0.20 m/s。
- Common error: (cm and m). Always convert to metres before substituting into v = fλ to get speed in m/s.
- 常见错误：数波峰时直接当作波长数，忘记减去1，或单位混用（cm 和 m）。应始终在代入 v = fλ 前换算为米，以得到以 m/s 为单位的波速。
- Refraction extension: If the wave moves into shallower water, speed decreases, wavelength decreases, but frequency remains constant. Many students incorrectly say frequency changes.
- 折射延伸：若波进入浅水区，波速减小，波长减小，但频率保持不变。许多学生错误地认为频率会改变。
7. Half-Life from a Decay Graph | 从衰变曲线求半衰期
A radioactive decay curve is plotted with activity (counts per minute) against time. The initial activity is 800 counts/min. After 10 days, activity falls to 200 counts/min. Find the half-life.
给出放射性衰变曲线，纵轴为活度（每分钟计数），横轴为时间。初始活度 800 counts/min，10 天后降至 200 counts/min。求半衰期。
- Method: 0 → 400 → 200. That is 2 half-lives. So time elapsed = 2 × t₁/₂ = 10 days → t₁/₂ = 5 days.
- 方法：观察从 800 → 400 → 200 经过的半衰期个数。此为 2 个半衰期。因此经过时间 = 2 × t₁/₂ = 10 天 → t₁/₂ = 5 天。
- Direct reading: Alternatively, find the time for the activity to halve from any value. From 800 to 400 takes 5 days, confirming half-life is 5 days.
- 直接读取：也可找出活度从任一值减半所需的时间。从 800 降至 400 用时 5 天，确认半衰期为 5 天。
- Clarification: es (20 days), activity would be 800 × (½)⁴ = 50 counts/min. Use this to check predictions.
- 辨析：给定同位素的半衰期是恒定的。不可线性外推。4 个半衰期（20 天）后，活度应为 800 × (½)⁴ = 50 counts/min。可用此验证预测。
8. Induced Voltage and Generator Effect | 感生电压与发电机效应
A magnet is pushed into a coil connected to a sensitive meter. The question asks: 'Explain why a reading is obtained only while the magnet is moving.' This tests electromagnetic induction.
将磁铁推入连接灵敏电流计的线圈中。问题："解释为什么仅当磁铁运动时表针才会偏转。" 考察电磁感应。
- Key concept: An induced e.m.f. is produced when there is a change in the magnetic flux linking the coil. Movement of the magnet changes the flux.
- 核心概念：当穿过线圈的磁通量发生变化时，就会产生感生电动势。磁铁的运动改变了磁通量。
- Explanation: When the magnet is stationary, flux is constant, so no induced voltage. While moving in, the flux increases, inducing a voltage and causing a current. Moving out induces a voltage in the opposite direction. Faster motion gives a larger rate of flux change, producing a larger induced e.m.f.
- 解释：磁铁静止时，磁通量恒定，不产生感生电压。推入时磁通量增加，感应出电压并引起电流。拉出时感生电压方向相反。运动越快，磁通量变化率越大，感生电动势也越大。
- Common misconception: Many write 'current is produced' without mentioning the need for a changing magnetic field. Always use 'induced e.m.f.' first, then 'current if circuit is complete'.
- 常见误解：许多学生直接写"产生电流"，却不提及需要变化的磁场。务必先写"感生电动势"，再写"若电路闭合则有电流"。
9. Pressure in Liquids – The Manometer Question | 液体压强 – 压强计问题
ter connected to a gas supply. The liquid column is higher on the open side. The question: ' fference is 0.15 m. Density of liquid = 13600 kg/m³ (mercury). Calculate the gas pressure.'
真题展示一种连接气源的 U 形管压强计，开口侧液柱更高。已知大气压为 1.0 × 10⁵ Pa，液面高度差 0.15 m，液体密度 13600 kg/m³（水银）。计算气体压强。
- Pressure difference: 15 m. Using g = 10 m/s² for IGCSE unless specified 9.8. Δp = 13600 × 10 × 0.15 = 20400 Pa.
- 压强差：Δp = ρ g h = 13600 kg/m³ × 10 m/s² × 0.15 m。IGCSE 中若无特别说明，取 g = 10 m/s²。Δp = 13600 × 10 × 0.15 = 20400 Pa。
- Determining gas pressure: ressure = atmospheric pressure + ρgh = 1.0 × 10⁵ + 20400 = 1.204 × 10⁵ Pa (or 120 400 Pa).
- 判断气体压强：若气体侧将液柱压低，则气体压强大于大气压。气体压强 = 大气压 + ρgh = 1.0 × 10⁵ + 20400 = 1.204 × 10⁵ Pa（或 120 400 Pa）。
- Check orientation: Always draw the diagram and note which side is higher. The higher column is on the lower pressure side. If the open side is higher, atmospheric pressure is smaller than gas pressure.
- 检查液面：务必画图并观察哪侧液面更高。较高液面那一侧压强较低。若开口侧液面更高，则大气压小于气体压强。
10. Hooke’s Law and the Limit of Proportionality | 胡克定律与比例极限
A force-extension graph from a past paper shows an initial straight line passing through the origin, then curving. The question: 'Determine the spring constant and state what the point P where the graph starts to curve represents.'
真题中经常给出力-伸长量图像，最初为一条过原点的直线，随后弯曲。问题："求弹簧的劲度系数，并说明图像开始弯曲的点 P 代表什么。"
- Hooke’s Law: . k = F / x from any point on the line, e.g., F = 6 N, extension x = 0.03 m → k = 200 N/m.
- 胡克定律：F = k x 仅适用于直线部分。取直线上任一点，例如 F = 6 N，伸长量 x = 0.03 m → k = 200 N/m。
- Point P: It is the limit of proportionality. Beyond P, force is no longer directly proportional to extension, and the spring will not obey Hooke’s Law.
- P 点：它是比例极限。超过 P 点，力不再与伸长量成正比，弹簧不再服从胡克定律。
- Elastic limit: Do not confuse with the elastic limit (point beyond which the material is permanently deformed). In many IGCSE questions, the limit of proportionality and elastic limit are distinct; the curving starts at limit of proportionality.
- 弹性极限：不要与弹性极限（超过此点材料发生永久变形）混淆。在 IGCSE 中，比例极限和弹性极限往往不同；曲线弯曲的起点是比例极限。
- Energy stored: The area under the graph up to a certain extension represents elastic potential energy (½ F x for the linear region). Be ready to calculate this for a triangle area.
- 储能：图像下方至某伸长量的面积代表弹性势能（在线性区为 ½ F x）。应准备好通过三角形面积计算。
11. Refractive Index and Total Internal Reflection | 折射率与全内反射
A ray diagram shows light travelling from glass into air, with an angle of incidence of 35° inside the glass. The critical angle for this glass is given as 42°. Question: 'Explain what happens to the ray at the boundary and calculate the refractive index of the glass.'
光路图显示光线从玻璃射向空气，玻璃内入射角为 35°。已知该玻璃的临界角为 42°。问题："解释光线在界面处会发生什么，并计算玻璃的折射率。"
- First, compare angles: Since 35° < 42°, the angle of incidence is less than the critical angle. Therefore, total internal reflection will NOT occur; most light will refract out into air, with a weak partial reflection.
- 首先比较角度：因为 35° < 42°，入射角小于临界角。因此不会发生全内反射；大部分光将折射进入空气，并伴有微弱的部反射。
- Refractive index: e C is critical angle. n = 1 / sin 42°. sin 42° ≈ 0.669, so n ≈ 1.49. Always show working.
- 折射率：n = 1 / sin C，其中 C 为临界角。n = 1 / sin 42°。sin 42° ≈ 0.669，因此 n ≈ 1.49。务必展示计算过程。
- Full marks tip: If asked to describe the path, state: 'The ray refracts away from the normal, because it enters a less dense medium. A weak ray is reflected back into the glass.' Some candidates forget the weak reflection.
- 满分技巧：若要求描述光路，应指出："光线偏离法线折射，因为它进入光疏介质。同时有一条弱的光线反射回玻璃中。" 许多考生会忽略弱反射光。
12. Orbital Speed and Gravitational Force | 轨道速度与引力
h at a constant speed of 7.6 km/s in a circular orbit of radius 7.0 × 10⁶ m. The question: ' ccelerating even though its speed is constant, and calculate the centripetal acceleration.'
一颗人造卫星以 7.6 km/s 的恒定速率在半径为 7.0 × 10⁶ m 的圆形轨道上绕地球运行。问题："解释为何卫星速率恒定却在加速，并计算其向心加速度。"
- Acceleration explanation: Velocity is a vector. Although speed is constant, the direction of velocity is continuously changing as the satellite moves around the circle. Any change in velocity (speed or direction) means acceleration. The acceleration is directed towards the centre of the circle – centripetal acceleration.
- 加速解释：速度是矢量。虽然速率恒定，但卫星绕圆运动时速度方向不断变化。速度的任何变化（大小或方向）都意味着有加速度。加速度指向圆心，即向心加速度。
- Calculation: 00 000 ≈ 8.25 m/s². Many forget to square and then divide, or fail to convert km/s to m/s.
- 计算：a = v² / r。将速率换算为 m/s：v = 7600 m/s，r = 7.0 × 10⁶ m。a = (7600)² / (7.0 × 10⁶) = 57 760 000 / 7 000 000 ≈ 8.25 m/s²。许多学生忘记先平方再除以半径，或忘记将 km/s 换算为 m/s。
- Gravity provies the force: The gravitational force from Earth acts as the centripetal force, keeping the satellite in orbit. If the gravitational force suddenly disappeared, the satellite would move in a straight line at a tangent to the orbit.
- 引力提供向心力：地球对卫星的引力提供向心力，使卫星保持在轨道上。若引力突然消失，卫星将沿轨道切线方向做匀速直线运动。
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