IGCSE AQA Physics: Common Exam Mistakes & Model Solutions | IGCSE AQA 物理：易错题精讲

📚 IGCSE AQA Physics: Common Exam Mistakes & Model Solutions | IGCSE AQA 物理：易错题精讲

In IGCSE AQA Physics, many marks are lost not because of a lack of knowledge, but because of common misunderstandings that resurface in exams year after year. This article unpacks ten of the most frequent mistakes seen on real papers, explains why they happen, and shows you how to avoid them with clear reasoning and worked solutions.

在 IGCSE AQA 物理考试中，许多失分并非因为知识欠缺，而是由于那些年复一年出现的常见误解。本文精选了真题中最为常见的十个易错点，分析错误根源，并通过清晰的推理和解题过程，帮助你彻底避开这些陷阱。

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1. Misconceptions in Ohm's Law | 欧姆定律的误解

Question: A circuit contains a fixed resistor and a variable power supply. The voltage is doubled. What happens to the current? Many students answer that the current halves because they mistakenly recall "resistance opposes current".

题目：一个电路包含一个固定电阻和一个可调电源。当电压加倍时，电流如何变化？许多学生回答电流减半，因为他们错误地记住了“电阻阻碍电流”。

Common mistake: Thinking that a larger voltage somehow increases the resistance, or that current is inversely proportional to voltage. This often stems from confusing the roles of voltage and resistance in Ohm’s law.

常见错误：认为更高的电压会以某种方式增大电阻，或者电流与电压成反比。这往往源于混淆了欧姆定律中电压和电阻的角色。

Correct approach: Ohm' , since I = V / R. Always identify which variables are constant before predicting changes.

正确解法：欧姆定律指出 V = I × R。对于固定电阻，R 不变。若 V 加倍，I 也必须加倍才能维持等式平衡，因为 I = V / R。在预测变化前，务必先确定哪些量是恒定的。

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2. Errors in Resultant Force Diagrams | 合力示意图的错误

Question: Two forces of 5 N and 12 N act at right angles on an object. Find the magnitude of the resultant force. A common wrong answer is 13 N, but the reasoning behind it is often flawed or absent.

题目：两个力 5 N 和 12 N 成直角作用在一个物体上。求合力的大小。常见的错误答案是 13 N，但其推理过程往往有缺陷或缺失。

use it is a Pythagorean triple, without drawing a proper vector diagram. Some also get the direction wrong.

常见错误：学生直接将力相加（5 + 12 = 17），或仅凭 13 是勾股数便猜测答案，却没有绘制正确的矢量图。部分学生还弄错了方向。

Correct approach: Draw the forces tip-to-tail to form a right-angled triangle. The resultant is the hypotenuse: √5² + 12² = √25 + 144 the angle using tan θ = 5/12, giving θ ≈ 22.6° from the 12 N force. Always show a sketch.

正确解法：将两个力首尾相接，构成一个直角三角形。合力为斜边：√5² + 12² = √25 + 144 = √169 = 13 N。再用 tan θ = 5/12 算出方向，θ ≈ 22.6°（相对于 12 N 力）。务必画出草图。

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3. Efficiency Calculations Without Units | 缺失单位的效率计算

Question: A motor lifts a weight, doing 200 J of useful work while drawing 250 J of electrical energy. Calculate its efficiency. Answers like '0.8' or '80' lose marks.

题目：一台电动机提起重物，做了 200 J 有用功，同时消耗了 250 J 电能。计算其效率。像“0.8”或“80”这样的答案会丢分。

Common mistake: Giving a decimal or a number without the percentage sign. Efficiency can be expressed as a decimal or a percentage, but the question usually demands a percentage, and the '%' sign is essential.

常见错误：给出小数或没有百分号的数字。效率可以用小数或百分数表示，但题目通常要求百分数，且“%”符号必不可少。

Correct approach: Efficiency = (useful output energy / total input energy) × 100%. So (200 / 250) × 100% = 0.8 × 100% = 80% the full calculation with units, and circle your final answer including the percent sign.

正确解法：效率 =（有用输出能量 / 总输入能量）× 100%。即 (200 / 250) × 100% = 0.8 × 100% = 80%。写出完整的计算过程并带单位，最后圈出含百分号的答案。

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4. Wave Reflection and Echo Time | 波的反射与回声时间

Question: A sound wave is sent from a boat to the seabed and back in 0.8 s. The speed of sound in water is 1500 m/s. Find the depth of the sea. Students often forget to halve the distance.

题目：从船上向海底发射声波，0.8 秒后收到回波。水中声速为 1500 m/s。求海水深度。学生常忘记将距离除以二。

that is the depth. This gives the distance travelled by the sound wave, which is twice the depth (down and back up).

常见错误：直接用 距离 = 速度 × 时间 = 1500 × 0.8 = 1200 m，并认为这就是深度。这实际上是声波走过的路程，是深度的两倍（往返）。

is half of this, so depth = 1200 / 2 = 600 m. Always draw a clear ray diagram showing the 'there and back' path.

正确解法：总路程 = 速度 × 时间 = 1500 × 0.8 = 1200 m。深度为此值的一半，因此深度 = 1200 / 2 = 600 m。务必画出清晰的光线示意图，标明“去程”和“回程”。

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5. Density and Irregular Object Volume | 密度与不规则物体体积

Question: A stone is placed into a measuring cylinder containing 50 cm³ of water. The water level rises to 72 cm³. The mass of the stone is 66 g. Calculate the density. Many students subtract incorrectly.

题目：将一块石头放入装有 50 cm³ 水的量筒中，水面升至 72 cm³。石头的质量为 66 g。计算密度。许多学生减法出错。

Common mistake: Using 72 cm³ as the volume of the stone, or subtracting mass values instead of volume readings. Some also give the density in wrong units, such as g/cm².

常见错误：把 72 cm³ 当作石头的体积，或误用质量值相减。部分学生还给出了错误的密度单位，如 g/cm²。

Correct approach: Volume of stone = final water level – initial water level = 72 cm³ – /cm³. Always write the formula first, show the substitution, and include the correct unit.

正确解法：石头体积 = 最终水位 − 初始水位 = 72 cm³ − 50 cm³ = 22 cm³。密度 = 质量 / 体积 = 66 g / 22 cm³ = 3.0 g/cm³。始终先写公式，再代入数字，并带上正确单位。

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6. Series Circuit Voltage Division | 串联电路电压分配

Question: Two resistors, 2 Ω and 4 Ω, are connected in series across a 12 V battery. What is the voltage across the 4 Ω resistor? A typical wrong answer is 6 V.

题目：两个电阻 2 Ω 和 4 Ω 串联接在 12 V 电池上。4 Ω 电阻两端的电压是多少？典型的错误答案是 6 V。

Common mistake: Students often assume the voltage splits equally because there are two resistors, or they divide by resistance values without using ratios correctly. Another error is using the formula for parallel circuits accidentally.

常见错误：学生常因为有两个电阻就认为电压平分，或者在未正确使用比例的情况下直接用电阻值除。另一种错误是误用了并联电路公式。

₄ = (4 / (2+4)) × 12 = (4/6) × 12 = 8 V. Voltage is shared in proportion to resistance in series.

正确解法：总电阻 Rₜ = 2 + 4 = 6 Ω。电路电流 I = V / Rₜ = 12 / 6 = 2 A。4 Ω 两端电压 = I × 4 = 2 × 4 = 8 V。或用比例法：V₄ = (4 / (2+4)) × 12 = (4/6) × 12 = 8 V。串联电路中电压按电阻比例分配。

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7. Interpreting Velocity-Time Graphs | 速度-时间图的解读

Question: A velocity-time graph shows a straight horizontal line at 5 m/s for 10 s, then a straight diagonal line down to 0 m/s over the next 5 s. Find the total distance travelled. Many students find only the first section's distance.

题目：一个速度-时间图显示，0–10 秒为 5 m/s 的水平直线，然后 10–15 秒为一条斜向下的直线降至 0 m/s。求总路程。许多学生只求了第一段的路程。

Common mistake: Confusing distance with area under the graph, or calculating only the area of the rectangle and forgetting the triangle. Some also incorrectly use final velocity as average velocity.

常见错误：混淆路程与图下的面积，或只计算矩形面积而忽略了三角形。部分学生还错误地将末速度当作平均速度。

.5 m. Total distance = 50 + 12.5 = 62.5 m. Always shade the area you are calculating to avoid missing a section.

正确解法：路程等于速度-时间图下的面积。矩形面积（0–10 秒）：10 × 5 = 50 m。三角形面积（10–15 秒）：½ × 5 × 5 = 12.5 m。总路程 = 50 + 12.5 = 62.5 m。始终将你正在计算的区域涂上阴影，以免遗漏。

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8. Specific Heat Capacity Experiment Errors | 比热容实验的误差

Question: In a standard school experiment to measure the specific heat capacity of aluminium, a student uses an immersion heater and a joulemeter. They obtain a value much higher than the true value. Why? Common answer: 'heat loss to the surroundings' – but often poorly explained.

题目：在测量铝的比热容的标准学校实验中，学生使用浸没式加热器和焦耳计。他们测得的值比真实值大很多。为什么？常见回答是“热量散失到周围环境中”——但往往解释得不充分。

ature rise is smaller than expected, which makes c appear larger because c = E / (m × Δθ).

常见错误：只提热量散失，却未与公式 E = m × c × Δθ 联系起来。若测得的能量是输入能量，但其中一部分散失，则温升小于预期，从而导致计算出的 c 偏大，因为 c = E / (m × Δθ)。

ven Δθ, the measured E is too large, and when computing c = E / (m × Δθ), the numerator is too high, giving an overestimated c. To improve accuracy, lag the block and use a lid.

正确解法：焦耳计记录的是供给的总电能。其中一部分未被金属块吸收而散失。因此，对于给定的 Δθ，测得的 E 偏大，计算 c = E / (m × Δθ) 时分子偏大，导致 c 值偏高。提高精度的方法包括给金属块包裹隔热层并加盖。

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9. Electromagnetic Induction Confusion | 电磁感应的混淆

Question: A bar magnet is pushed into a coil connected to a sensitive ammeter. The needle flicks to the right. The magnet is then pulled out quickly. Which way does the needle deflect? A common wrong answer: 'to the right again because it's the same magnet'.

题目：将条形磁铁推入与灵敏电流计相连的线圈，指针向右偏转。随后快速将磁铁拉出。指针将如何偏转？常见错误回答：“再次向右，因为是同一块磁铁”。

Common mistake: Thinking the direction of deflection depends only on the magnet’s poles, not on the direction of motion. Students often fail to apply Lenz's law and the idea that the induced current opposes the change causing it.

常见错误：认为偏转方向只与磁铁极性有关，与运动方向无关。学生常未能运用楞次定律和感应电流阻碍变化的概念。

Correct approach: When the magnet is pulled out, the magnetic field through the coil decreases. By Lenz’s law, the induced current will try to keep the field by creating a field that attracts the magnet, i.e., a south pole at the near end if the magnet's north is moving away. This produces current in the opposite direction, so the needle flicks to the left. Direction depends on the change in flux, not just the poles.

正确解法：拉出磁铁时，穿过线圈的磁场减弱。根据楞次定律，感应电流将试图维持原磁场，产生一个吸引磁铁的磁场——若磁铁 N 极远离，线圈靠近端产生 S 极。这会产生反向电流，因而指针向左偏转。方向取决于磁通量的变化，而非仅仅是磁极。

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10. Half-Life and Activity Graphs | 半衰期与活度图线

Question: A radioactive sample has an initial activity of 800 Bq. After 15 hours, its activity is 100 Bq. What is its half-life? Many candidates answer 5 hours incorrectly because they misread the pattern.

题目：一个放射性样品的初始活度为 800 Bq。15 小时后，其活度变为 100 Bq。它的半衰期是多少？许多考生错误地给出 5 小时，因为他们误读了规律。

s, then dividing 15 hours by the number of steps guessed. In reality, activity halves each time.

常见错误：误以为活度每经过一个半衰期减少相等的数量，如 800 → 400 → 0 两步，然后将 15 小时除以猜测的步数。实际上，活度每次减少为原来的一半。

passed in 15 hours. Therefore, half-life = 15 / 3 = 5 hours. But note: the common mistake is reaching 5 hours without realising the need for three steps. Show the sequence clearly.

正确解法：从 800 Bq 开始。一个半衰期后：400 Bq。两个后：200 Bq。三个后：100 Bq。因此 15 小时内经过了 3 个半衰期，故半衰期 = 15 / 3 = 5 小时。但须注意：常见错误是未意识到需要三步便得出 5 小时。务必清晰展示衰减序列。

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