📚 Common Mistakes in IGCSE CIE Physics: Detailed Solutions | IGCSE CIE 物理易错题精讲
Many IGCSE CIE Physics students lose marks not because they do not know the content, but because they fall into predictable traps set by exam questions. Misreading graphs, confusing similar concepts, or applying formulas mechanically can cost precious marks. This article dissects eight of the most common error types with example questions, wrong answers, and step-by-step corrections. Read on to sharpen your exam technique and avoid repeating the same mistakes.
许多 IGCSE CIE 物理学生丢分并非因为不懂知识，而是落入了考题中常见的陷阱。读错图像、混淆相似概念、或机械套用公式，都可能白白失分。本文剖析八个最常见的错误类型，每个都以例题、错误答案和逐步修正的方式呈现。仔细阅读，提升你的应考技巧，避免重蹈覆辙。
1. Misunderstanding Speed-Time Graphs | 误解速度-时间图
Error: Students often think that a sloping line going downwards on a speed-time graph means the object is moving backwards. When asked whether the object changes direction, they answer ‘yes’ because the graph line goes down. They also mistakenly treat the area as speed multiplied by time without accounting for the triangle.
错误：学生常以为速度-时间图上向下的斜线表示物体在向后运动。当被问到物体是否改变方向时，他们回答“是”，因为图线向下。他们还错误地将面积直接当作速度乘以时间，忽略了三角形部分。
Problem: A car travels along a straight road. The speed-time graph shows it slowing down uniformly from 20 m/s to rest in 5 seconds. Does the car change direction? Calculate the distance travelled.
问题：一辆汽车在直路上行驶。速度-时间图显示它在5秒内从 20 m/s 均匀减速到静止。汽车是否改变了方向？计算经过的距离。
Common wrong answer: “Yes, it goes backwards. Distance = 20 × 5 = 100 m.”
常见错误答案：“是的，它向后行驶。距离 = 20 × 5 = 100 米。”
Correct analysis: The velocity remains positive (above the time axis), so the car continues moving forward the whole time. It does not change direction. Distance = area under graph = ½ × base × height = ½ × 5 s × 20 m/s = 50 m.
正确分析：速度始终为正值（位于时间轴上方），因此汽车全程向前运动，没有改变方向。距离 = 图线下方面积 = ½ × 底 × 高 = ½ × 5 s × 20 m/s = 50 米。
Key takeaway: On a speed-time graph, the sign of the velocity (above or below the axis) tells you direction; the slope tells you acceleration; and the area represents distance. Never assume that a downward trend implies reversing.
关键点：在速度-时间图上，速度的正负（在轴上方还是下方）表示运动方向；斜率表示加速度；面积代表距离。决不要认为下降趋势就意味着反向运动。
2. Confusing Mass and Weight | 混淆质量和重量
Error: Many candidates use mass and weight interchangeably, saying “the mass of the astronaut is 800 N” or thinking that mass changes when going to the Moon. They also incorrectly apply W = m × g.
错误：许多考生混用质量和重量，说出“宇航员的质量是800牛顿”，或者认为到了月球质量会改变。他们还错误地运用 W = m × g。
Problem: An astronaut has a weight of 800 N on Earth where g = 10 N/kg. Calculate his mass. On the Moon, the gravitational field strength is 1/6 of Earth’s. What is his weight on the Moon?
问题：一名宇航员在地球上的重量为 800 N，取 g = 10 N/kg。计算他的质量。在月球上，引力场强度是地球的 1/6，他在月球上的重量是多少？
Common wrong answer: “Mass = 80 N. On the Moon, mass becomes 80 ÷ 6 = 13.3 kg, weight = 13.3 N.”
常见错误答案：“质量 = 80 N。在月球上，质量变为 80 ÷ 6 = 13.3 kg，重量 = 13.3 N。”
Correct working: m = WEarth / gEarth = 800 N ÷ 10 N/kg = 80 kg. Mass is constant everywhere. On the Moon, gMoon = (1/6) × 10 ≈ 1.67 N/kg. So WMoon = m × gMoon = 80 × 1.67 ≈ 133 N.
正确计算：m = W地球 / g地球 = 800 N ÷ 10 N/kg = 80 kg。质量处处相同。在月球上，g月球 = (1/6) × 10 ≈ 1.67 N/kg，因此 W月球 = m × g月球 = 80 × 1.67 ≈ 133 N。
Remember: Mass (scalar, kg) measures the amount of matter and never changes; weight (vector, N) is the gravitational force and depends on g. Always include correct units.
记住：质量（标量，kg）衡量物质的多少，从不改变；重量（矢量，N）是引力，取决于 g。务必使用正确的单位。
3. Misapplying Newton’s First Law | 误用牛顿第一定律
Error: Students often believe that a constant force is needed to keep an object moving at constant velocity. In equilibrium problems, they assume that if an object is moving there must be a resultant force in the direction of motion.
错误：学生常误认为需要恒定的力来维持物体匀速运动。在平衡问题中，他们以为只要物体在运动，就必然有与运动方向相同的合力。
Problem: A shopping trolley is pushed along a horizontal floor with a constant force of 30 N. It moves at a steady speed of 1.5 m/s. What is the size of the friction force acting on the trolley?
问题：一辆购物车在水平地板上被 30 N 的恒力推动，以 1.5 m/s 的稳定速度前进。作用在车上的摩擦力是多少？
Common wrong answer: “Less than 30 N, because it is moving slowly.” or “30 N, but only when it stops.”
常见错误答案：“小于 30 N，因为它移动得慢。”或“30 N，但只有它停下时才成立。”
Correct reasoning: Constant velocity means zero resultant force (Newton’s First Law). Therefore, friction must exactly balance the pushing force: friction = 30 N in the opposite direction.
正确推理：匀速直线运动意味着合力为零（牛顿第一定律）。因此，摩擦力必须恰好与推力平衡：摩擦力 = 30 N，方向相反。
Insight: If an object is moving steadily, all forces are balanced. The idea that a force is needed to “keep it going” is a pre-Newtonian misconception; an object continues with its velocity unless a resultant force acts on it.
洞见：若物体匀速运动，所有力相互平衡。需要力来“维持运动”的观念是前牛顿时代的误解；物体保持原有速度，除非有合力作用在其上。
4. Confusion Between Work Done and Energy Transferred | 混淆做功与能量转移
Error: Many candidates think that holding a heavy object or carrying it horizontally involves doing work on the object. They ignore the direction of force relative to displacement.
错误：许多考生认为拿着重物或水平搬运物体是在对物体做功。他们忽略了力相对于位移的方向。
Problem: A porter carries a 200 N suitcase and walks 10 m horizontally at constant speed. How much work does he do on the suitcase? A weightlifter lifts a 1000 N barbell vertically through 2 m. Calculate the work done.
问题：一位搬运工提着 200 N 的手提箱，匀速水平步行 10 米。他对箱子做了多少功？一位举重运动员将 1000 N 的杠铃垂直上举 2 米，计算他做的功。
Common wrong answer: “Porter: work = 200 × 10 = 2000 J. Weightlifter: work = 1000 × 2 = 2000 J (but maybe unsure).”
常见错误答案：“搬运工：功 = 200 × 10 = 2000 J。举重者：功 = 1000 × 2 = 2000 J （但可能不确定）。”
Correct analysis: Work done W = F × d × cos θ. For the porter, force (up) is perpendicular to displacement (horizontal), θ = 90°, cos 90° = 0, so work done = 0. For the weightlifter, force and displacement are in the same direction, cos0° = 1, so W = 1000 × 2 = 2000 J.
正确分析：做功 W = F × d × cos θ。对于搬运工，力（向上）与位移（水平）垂直，θ = 90°，cos 90° = 0，所以做功为零。对于举重者，力与位移同向，cos0° = 1，W = 1000 × 2 = 2000 J。
Key principle: Work is done only when a force moves its point of application in the direction of the force. Holding stationary objects or horizontal carrying does not transfer mechanical energy to the object.
关键原理：只有当力使其作用点沿力的方向移动时，才做功。静止提着物体或水平搬运并不向物体转移机械能。
5. Incorrect Use of Circuit Rules (Series vs Parallel) | 电路规则的错误使用（串联与并联）
Error: In series circuits, students often assume the voltage splits equally regardless of resistance. In parallel circuits, they mistakenly think current is the same in all branches or that total resistance is simply added.
错误：在串联电路中，学生常假设不论电阻大小电压都均分。在并联电路中，他们误以为各支路电流相同，或总电阻直接相加。
Problem: Two resistors R₁ = 2 Ω and R₂ = 4 Ω are connected in series to a 12 V battery. Calculate the voltage across R₁. In a second scenario, the same resistors are in parallel with the same battery; what is the current through R₁?
问题：两个电阻 R₁ = 2 Ω 和 R₂ = 4 Ω 串联后接到 12 V 电池上。计算 R₁ 两端的电压。另一种情况：两电阻并联后接同一电池，求通过 R₁ 的电流。
Common wrong answer (series): “Voltage divides equally: 12 V ÷ 2 = 6 V across each.” (parallel): “Total R = 2 + 4 = 6 Ω, so current in R₁ = 12 ÷ 6 = 2 A.”
常见错误答案（串联）：“电压均分：每个电阻 6 V。”（并联）：“总电阻 = 2 + 4 = 6 Ω，故 R₁ 中电流 = 12 ÷ 6 = 2 A。”
Correct working (series): V₁ = (R₁ / (R₁+R₂)) × Vtotal = (2/6)×12 = 4 V. (parallel): In parallel, voltage across each branch = 12 V. So I₁ = V / R₁ = 12 V / 2 Ω = 6 A.
正确计算（串联）：V₁ = (R₁ / (R₁+R₂)) × V总 = (2/6)×12 = 4 V。（并联）：并联时每条支路电压均为 12 V，因此 I₁ = V / R₁ = 12 V / 2 Ω = 6 A。
Remember: In series, current is constant, voltage splits in proportion to resistance. In parallel, voltage is the same across each branch, and current divides inversely with resistance. Always redraw the circuit if needed.
记住：串联中电流处处相等，电压按电阻比例分配；并联中电压相同，电流与电阻成反比分配。必要时重新绘制电路图。
6. Drawing Ray Diagrams for Refraction Incorrectly | 折射光路图绘制错误
Error: When drawing refraction, students frequently bend the ray the wrong way at the boundary. They may make the refracted ray go away from the normal when entering a denser medium, or show refraction instead of total internal reflection when the angle exceeds the critical angle.
错误：画折射时，学生常常在界面把光线弯错方向。他们可能让光线进入光密介质时偏离法线，或在入射角大于临界角时仍画折射而非全内反射。
Problem: Light travels from air into glass (refractive index n > 1). Sketch the refracted ray. In another question, light in glass strikes the glass-air boundary at an angle of 50°, given the critical angle is 42°. What happens?
问题：光从空气进入玻璃（折射率 n > 1）。画出折射光线。另一题中，光在玻璃中以 50° 角入射到玻璃-空气界面，已知临界角为 42°，会发生什么？
Common wrong answer: “The ray bends away from the normal in glass.” and “It is refracted out into the air at a smaller angle.”
常见错误答案：“光线在玻璃中偏离法线。”以及“它以较小的角度折射进入空气。”
Correct explanation: When light enters a denser medium (air to glass), it slows down and bends towards the normal. For the second case, 50° > critical angle 42°, so total internal reflection occurs; the ray reflects entirely back into the glass.
正确解释：光进入光密介质（空气到玻璃）时速度减小，向法线偏折。第二种情况，50° > 临界角 42°，发生全内反射；光线全部反射回玻璃中。
Tip: Use the phrase “denser towards the normal, rarer away from the normal” to remember the bending direction. For total internal reflection, two conditions must be met: light must be in the denser medium and angle of incidence > critical angle.
技巧：用“密向法线靠，疏离法线去”来记忆弯折方向。全内反射须满足两个条件：光在光密介质中，且入射角大于临界角。
7. Half-Life Calculation Pitfalls | 半衰期计算的陷阱
Error: Students often mishandle half-life problems by dividing the initial count rate by the total time or by the number of half-lives in the wrong way. They may also fail to recognize that background radiation must sometimes be subtracted, though here we focus on the basic decay pattern.
错误：学生在处理半衰期问题时，常错误地直接除以总时间或误算半衰期次数。他们有时也忘记需要扣除本底辐射，不过这里我们重点关注基本的衰变模式。
Problem: A radioactive sample has an initial activity of 640 counts per minute. Its half-life is 3 hours. What is the activity after 9 hours?
问题：一个放射性样品的初始活度为 640 次计数每分钟，半衰期为 3 小时。求 9 小时后的活度。
Common wrong answer: “640 ÷ 9 = 71.1 cpm” or “640 ÷ 3 = 213.3 cpm.”
常见错误答案：“640 ÷ 9 = 71.1 cpm”或“640 ÷ 3 = 213.3 cpm。”
Correct approach: Number of half-lives = total time / half-life = 9 h / 3 h = 3. After each half-life, activity halves: 640 → 320 → 160 → 80. So activity after 9 hours = 80 counts per minute.
正确方法：半衰
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