📚 IGCSE CCEA Physics: Common Mistakes & Misconceptions – Worked Examples | IGCSE CCEA 物理：易错题精讲
This article highlights some of the most common mistakes students make in the IGCSE CCEA Physics examinations and provides step-by-step worked solutions to help you avoid similar pitfalls. Each section presents a typical error, explains the correct reasoning and gives you a confidence boost for your revision.
本文聚焦 IGCSE CCEA 物理考试中最常见的易错题型，通过逐步精讲帮助你避开这些陷阱。每节展示一个典型错误，解析正确思路，帮你巩固知识、提升备考信心。
1. Mass vs. Weight – The g Confusion | 质量与重量——g 的混淆
A classic blunder is using W = mg but forgetting that weight changes with gravitational field strength, while mass remains constant. Many candidates incorrectly state that a 60 kg astronaut weighs 60 kg on the Moon or use m = W/g without considering the correct g value.
经典错误是使用 W = mg 却忘记重量随引力场强度变化而质量不变。很多考生错误地认为一名 60 kg 的宇航员在月球上仍“重 60 kg”，或在使用 m = W/g 时未代入正确的 g 值。
Worked Example: An astronaut has a mass of 65 kg. The gravitational field strength on Earth is 10 N/kg and on the Moon is 1.6 N/kg. Find the astronaut’s weight on Earth and on the Moon, and their mass on the Moon.
精讲例题：一名宇航员质量为 65 kg。地球的引力场强度为 10 N/kg，月球为 1.6 N/kg。求宇航员在地球上与月球上的重量，以及在月球上的质量。
Correct solution: Weight on Earth = mg = 65 × 10 = 650 N. Weight on Moon = 65 × 1.6 = 104 N. Mass on Moon remains 65 kg. Many candidates wrongly write ‘650 kg’ for weight or claim mass decreases on the Moon. Remember: weight is a force, measured in newtons; mass is the amount of matter, measured in kilograms.
正确解法：地球上重量 = mg = 65 × 10 = 650 N。月球上重量 = 65 × 1.6 = 104 N。月球上的质量仍为 65 kg。许多考生错误地写出“重量为 650 kg”或声称质量在月球上减小。切记：重量是力，单位为牛顿；质量是物质的量，单位为千克。
2. Speed vs. Velocity – Direction Matters | 速率与速度——方向很重要
Students often treat speed and velocity as synonyms. In CCEA exams, if a question asks for velocity, you must give both magnitude and direction or a negative sign for opposite motion. Ignoring direction costs marks even if the number is correct.
学生常把速率和速度混为一谈。在 CCEA 考试中，若题目要求速度，必须给出大小和方向，或对反向运动使用负号。忽视方向即使数值正确也会丢分。
Example: A car travels 300 m north in 20 s, then 200 m south in 10 s. Calculate the average speed and average velocity for the whole journey.
例题：一辆汽车向北行驶 300 m 用时 20 s，然后向南行驶 200 m 用时 10 s。计算全程的平均速率和平均速度。
Common error: Using total distance for velocity. Correct approach: total distance = 500 m, total time = 30 s, average speed = 500/30 ≈ 16.7 m/s. For velocity, displacement = 300 m north − 200 m south = 100 m north (or +100 m if north is positive). Average velocity = 100/30 ≈ 3.3 m/s north. Always specify direction for velocity.
常见错误：用总路程计算速度。正确方法：总路程 = 500 m，总时间 = 30 s，平均速率 = 500/30 ≈ 16.7 m/s。对于速度，位移 = 300 m 北 − 200 m 南 = 100 m 北（或若北为正则为 +100 m）。平均速度 = 100/30 ≈ 3.3 m/s 北。速度必须指明方向。
3. Resultant Force and Acceleration – The F=ma Trap | 合力与加速度——F=ma 的陷阱
Many candidates apply F = ma directly without identifying all forces acting. A common mistake is using the driving force of a car as the resultant force, ignoring friction or air resistance. Another error is mixing up mass and weight in calculations.
许多考生不分析受力就直接套用 F = ma。常见错误是把汽车的驱动力当作合力，忽略摩擦或空气阻力。另一个错误是在计算中混淆质量与重量。
Worked Example: A 1200 kg car experiences a driving force of 2400 N and a total resistive force of 900 N. Calculate the acceleration.
精讲例题：一辆 1200 kg 的汽车受到 2400 N 的驱动力和 900 N 的总阻力。计算加速度。
Correct: Resultant force = 2400 − 900 = 1500 N. a = F/m = 1500/1200 = 1.25 m/s². Some students incorrectly use 2400 N as the net force, giving a = 2 m/s². Always subtract opposing forces to find the unbalanced force before using F = ma.
正确解法：合力 = 2400 − 900 = 1500 N。a = F/m = 1500/1200 = 1.25 m/s²。一些学生错误地把 2400 N 当作合力，得出 a = 2 m/s²。使用 F = ma 前必须先减去反向力求出合外力。
4. Momentum – Direction and Conservation | 动量——方向与守恒
Momentum calculations often go wrong when students forget that momentum is a vector. In collision or explosion problems, assigning positive and negative directions is essential. A frequent error is adding momenta without considering sign.
动量计算常因忘记矢量性而失分。在碰撞或爆炸问题中，必须设定正、负方向。常见错误是不考虑符号直接相加动量。
Example: A 3 kg trolley moving at 2 m/s to the right collides with a stationary 1 kg trolley. They stick together. Find the velocity after collision.
例题：一辆 3 kg 的小车以 2 m/s 向右运动，与静止的 1 kg 小车碰撞后粘在一起。求碰撞后的速度。
Wrong approach: (3×2 + 1×0) = (3+1)v ⇒ 6 = 4v ⇒ v = 1.5 m/s, but direction may be omitted. Correct: Take right as positive. Total momentum before = 6 + 0 = 6 kg m/s. After collision, momentum = 4v. So v = 1.5 m/s to the right. Always state direction. For explosions, be careful: total momentum before is zero, so momenta of fragments must be equal and opposite.
错误做法：(3×2 + 1×0) = (3+1)v ⇒ 6 = 4v ⇒ v = 1.5 m/s，但可能遗漏方向。正确做法：取向右为正。碰前总动量 = 6 + 0 = 6 kg m/s。碰后总动量 = 4v。因此 v = 1.5 m/s 向右。必须说明方向。对于爆炸问题注意：爆炸前总动量为零，碎片动量必须等大反向。
5. Energy Transfers – Kinetic vs. Potential Pitfalls | 能量转化——动能与势能的易错点
Candidates frequently misapply the formulas KE = ½mv² and GPE = mgh. A typical mistake is using velocity instead of speed squared, forgetting the ½ factor, or using mass in grams. Also, many believe that energy is ‘used up’ rather than transferred.
考生经常误用公式 KE = ½mv² 和 GPE = mgh。典型错误包括用速度代替速度的平方、遗漏 ½ 因子，或质量单位用克。许多人还误认为能量被“用尽”而非转化。
Sample question: A 0.5 kg ball is dropped from 8 m. Ignoring air resistance, find its speed just before hitting the ground. (g = 10 N/kg)
例题：一个 0.5 kg 的球从 8 m 高度落下。忽略空气阻力，求它撞击地面前的速率。(g = 10 N/kg)
Common mistake: Using KE = mgh directly without ½mv². Correct: loss of GPE = gain in KE ⇒ mgh = ½mv². Cancel m: 10×8 = ½ v² ⇒ 80 = ½ v² ⇒ v² = 160 ⇒ v = √160 ≈ 12.6 m/s. If you forget the ½, you’d get v² = 80 ⇒ v ≈ 8.94 m/s, which is incorrect. Always write the conservation equation clearly.
常见错误：直接使用 KE = mgh 而遗漏 ½mv²。正确方式：重力势能减少量 = 动能增加量 ⇒ mgh = ½mv²。消去 m：10×8 = ½ v² ⇒ 80 = ½ v² ⇒ v² = 160 ⇒ v = √160 ≈ 12.6 m/s。如果忘记乘 ½，会得到 v² = 80 ⇒ v ≈ 8.94 m/s，这个答案是错误的。务必清晰写出能量守恒方程。
6. Specific Heat Capacity vs. Specific Latent Heat – Mixing Up Formulas | 比热容与比潜热——公式混淆
A very common CCEA exam slip is using Q = mcΔθ when there is a change of state (temperature constant) or using Q = mL when the temperature is changing. Students also mix up the units of mass (g vs kg) and energy (J vs kJ).
CCEA 考试中极常见的失误是：在状态变化（温度不变）时使用 Q = mcΔθ，或在温度变化时使用 Q = mL。考生还常混淆质量单位（克与千克）和能量单位（焦耳与千焦）。
Worked example: How much energy is needed to melt 2.0 kg of ice at 0 °C? (Specific latent heat of fusion of ice = 334 000 J/kg)
精讲例题：熔化 2.0 kg 0 °C 的冰需要多少能量？（冰的熔化比潜热 = 334 000 J/kg）
Misconception: Some students multiply by specific heat capacity and a temperature change (Δθ). That is wrong because melting occurs at constant temperature. Correct: Q = mL = 2.0 × 334 000 = 668 000 J (or 668 kJ). Use Q = mcΔθ only when temperature changes without a change of state. When state changes, use Q = mL.
误解：有些学生乘上比热容和温度变化（Δθ）。这是错误的，因为熔化在恒定温度下发生。正确解法：Q = mL = 2.0 × 334 000 = 668 000 J（或 668 kJ）。只有在温度变化而无状态变化时使用 Q = mcΔθ；状态变化时使用 Q = mL。
7. Series and Parallel Circuits – Resistance and Current | 串联与并联电路——电阻与电流
Students often calculate total resistance incorrectly: adding reciprocals for series or simply adding resistances for parallel. Another error is assuming current remains constant across a parallel branch or voltage is the same in series.
学生常错误地计算总电阻：串联时用倒数相加，并联时直接相加电阻。另一个错误是认为并联支路中电流恒定，或串联中电压处处相等。
Example: Two resistors, 6 Ω and 3 Ω, are connected in parallel. Calculate the total resistance and the current through the 6 Ω resistor if the supply is 12 V.
例题：两个电阻 6 Ω 和 3 Ω 并联。计算总电阻，以及当电源电压为 12 V 时通过 6 Ω 电阻的电流。
Wrong: R_total = 6 + 3 = 9 Ω. Correct: 1/R_total = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 ⇒ R_total = 2 Ω. For current, voltage across each branch is 12 V. I_6Ω = V/R = 12/6 = 2 A. (Many try to split 12 V between resistors in parallel—voltage is the same in parallel.) In series, remember current is the same through all components, while voltage divides.
错误：R_total = 6 + 3 = 9 Ω。正确：1/R_total = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 ⇒ R_total = 2 Ω。对于电流，各支路电压均为 12 V。I_6Ω = V/R = 12/6 = 2 A。（许多人试图把 12 V “分给”并联电阻——并联电压相等。）在串联电路中，需注意电流处处相等，电压则按电阻分配。
8. Electromagnetic Induction – The Right-Hand Rule Slip | 电磁感应——右手定则失误
In explaining generators or dynamos, a common mistake is describing the induced current direction incorrectly. Students often confuse Fleming’s right-hand rule (for generators) with the left-hand rule (for motors), or they forget that an induced current is produced only when there is relative motion or changing magnetic field.
在解释发电机或直流发电机原理时，常见错误是搞错感应电流方向。学生经常混淆弗莱明右手定则（用于发电机）和左手定则（用于电动机），或者忘记只有存在相对运动或变化磁场时才会产生感应电流。
Typical CCEA question: A magnet is pushed into a coil connected to a sensitive ammeter. The needle deflects to the left. What happens when the magnet is pulled out faster?
CCEA 典型题：一块磁铁推入与灵敏电流计相连的线圈，指针向左偏转。当磁铁更快地拉出时会发生什么？
Error: Some students say the needle deflects to the left again or there is no deflection. Correct reasoning: Pulling out reverses the direction of induced current (needle deflects to the right). Doing it faster increases the rate of change of magnetic flux, so the deflection is larger (but still to the right). Always link induced current direction to Lenz’s law – the induced field opposes the change causing it.
错误：部分学生会说指针再次向左偏转，或指针不偏转。正确推理：拉出磁铁使感应电流方向反转（指针向右偏转）。更快地拉出会增大磁通量变化率，因此偏转幅度更大（但仍向右）。始终将感应电流方向与楞次定律联系起来——感应磁场总是阻碍引起感应的变化。
9. Waves – Drawing Refraction and Diffraction Diagrams | 波——折射与衍射作图
In wave diagrams, pupils frequently forget to show wavelength change when waves enter a different medium. For refraction, the frequency remains constant but speed and wavelength change. A common error is drawing the refracted ray towards the normal when it should be away (or vice versa) or showing equal wavelengths on both sides.
在波动作图中，学生常忘记表现波进入不同介质时波长的变化。对于折射，频率不变，但波速与波长改变。常见错误是折射光线应远离法线时画成靠近（或反之），或在界面两侧画出相等的波长。
Example: Water waves travel from deep to shallow water at an angle. The speed decreases. Sketch the wavefronts.
例题：水波以一定角度从深水区传入浅水区，波速减小。画出波前示意图。
Correct: In shallow water, wavelength is shorter (since v = fλ, and f is constant). Wavefronts bend towards the normal. Many students draw the refracted wavefronts parallel to the original ones or keep the same spacing. Also, in diffraction diagrams, the amount of spreading increases as the gap size approaches the wavelength; candidates often draw slight spreading for a very small gap.
正确：浅水区波长变短（因为 v = fλ，且 f 恒定）。波前向法线弯折。许多学生把折射波前画得与原波前平行或保持相同间距。此外，在衍射作图中，当缝隙大小接近波长时，波的扩展程度增加；考生常常把极小缝隙画成只有微弱扩展。
10. Radioactive Decay – Half-life Calculations without Care | 放射性衰变——半衰期计算的粗心
Half-life problems cause trouble when students fail to convert time units or use the wrong number of half-lives. Some attempt to divide the total time by the half-life but then incorrectly apply the fraction left (e.g., using 1/3 instead of 1/2^n).
半衰期题目容易在时间单位换算或半衰期次数上出错。部分考生会把总时间除以半衰期，但应用剩余分数时出错（例如用 1/3 而非 1/2ⁿ）。
Worked example: A sample has a half-life of 6 hours. Its initial activity is 800 Bq. What is the activity after 18 hours?
精讲例题：某样本半衰期为 6 小时，初始活度为 800 Bq。问 18 小时后的活度是多少？
Common mistake: 18 ÷ 6 = 3, so activity = 800 ÷ 3 ≈ 267 Bq. Correct: Number of half-lives = 3. After each half-life, activity halves: 800 → 400 → 200 → 100 Bq. So activity is 800 × (½)³ = 100 Bq. Always use powers of 2, not division by the number of half-lives. Also watch for units: half-life may be given in days or minutes; ensure the time interval matches.
常见错误：18 ÷ 6 = 3，所以活度 = 800 ÷ 3 ≈ 267 Bq。正确：半衰期次数 = 3。每经过一个半衰期，活度减半：800 → 400 → 200 → 100 Bq。即活度 = 800 × (½)³ = 100 Bq。务必使用 2 的幂次，而非除以半衰期次数。还要注意单位：半衰期可能以天或分钟给出，保证时间间隔匹配。
Published by TutorHao | Physics Revision Series | aleveler.com
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