IB AQA Physics: Common Mistakes and Detailed Solutions | IB AQA 物理：易错题精讲

📚 IB AQA Physics: Common Mistakes and Detailed Solutions | IB AQA 物理：易错题精讲

Students often lose marks not because a topic is inherently impossible, but because a subtle misconception has gone unchallenged. This article unpacks the most persistent pitfalls in IB and AQA physics—from Newton’s third law through to nuclear reactions—with worked examples and clear reasoning that build true understanding.

很多同学丢分并非因为知识点本身有多难，而是因为某个微小的错误理解从未被纠正。本文拆解 IB 与 AQA 物理中最顽固的易错点——从牛顿第三定律一直到核反应——配合典型例题和清晰的推理，帮助你真正抓住原理。

1. Newton’s Third Law Pairs vs Balanced Forces | 牛顿第三定律力对与平衡力的区分

A classic error is to claim that the weight of a book on a table and the normal force from the table form an action–reaction pair. They are equal, opposite, and act on the same body—so they are a balanced force pair, not a third-law pair. Newton’s third law requires the two forces to act on different bodies.

一个经典错误是认为放在桌上的书所受的重力与桌面的支持力构成作用力与反作用力。它们大小相等、方向相反，却作用在同一物体上，因此是平衡力，不是第三定律力对。牛顿第三定律要求两个力分别作用在不同物体上。

Example: A book of mass 2 kg rests on a table. Many students identify the third-law pair as ‘weight’ and ‘normal force’. The true third-law pair for weight is the gravitational pull of the Earth on the book and the gravitational pull of the book on the Earth. For the normal force, it is the push of the table on the book and the push of the book on the table.

例题：一本质量为 2 kg 的书静止在桌上。很多学生认为“重力”和“支持力”是第三定律力对。重力的真正第三定律伙伴是地球对书的引力和书对地球的引力。对于支持力，第三定律伙伴是桌面对书向上的推力和书对桌面向下的压力。

Weight: F_{Earth on book} = –F_{book on Earth}; Normal: F_{table on book} = –F_{book on table}

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2. Projectile Motion: Velocity at the Top | 抛体运动：最高点的速度与加速度

Many believe that because the vertical velocity is zero at the peak, the acceleration must also be zero. Acceleration due to gravity is always 9.81 m s⁻² downwards, even when the object is momentarily at rest in the vertical direction. The horizontal component of velocity remains unchanged throughout (ignoring air resistance).

很多同学误以为最高点竖直速度为零，所以加速度也为零。重力加速度始终为 9.81 m s⁻² 向下，即便物体在竖直方向瞬时静止。水平方向的速度在整个过程中保持不变（忽略空气阻力）。

Common mistake: using v_y = 0 ⇒ a_y = 0. Correct: At the peak, v_y = 0 but a_y = –g. If a ball is kicked at 20 m s⁻¹ at 30° to the horizontal, its vertical acceleration is always –9.81 m s⁻².

常见错误：由 v_y = 0 推出 a_y = 0。正确：最高点 v_y = 0，但 a_y = –g。若球以 20 m s⁻¹、仰角 30° 踢出，其竖直加速度始终为 –9.81 m s⁻²。

v_x = u cosθ, v_y = u sinθ – gt

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3. Ohm’s Law and the I–V Graph for a Filament Lamp | 欧姆定律与灯丝的 I–V 特性曲线

Students often misread the shape of the filament lamp’s I–V curve. The resistance increases with current because the temperature rises, so the graph bends towards the voltage axis. A common slip is to think resistance stays constant or that the lamp obeys Ohm’s law. Ohm’s law only applies when resistance is constant, i.e., for ohmic conductors at constant temperature.

学生经常误读灯丝的 I–V 曲线。电流增大、温度升高导致电阻增大，因此曲线向电压轴弯曲。常见错误是认为电阻不变或灯丝满足欧姆定律。欧姆定律仅在电阻恒定（即恒温下的欧姆导体）时成立。

For a filament lamp, as V increases, current increases less steeply. Resistance R = V/I is not constant; it increases. The gradient of the I–V graph is 1/R, so the gradient decreases with voltage. In contrast, a fixed resistor (ohmic) shows a straight line through the origin.

对于灯丝，随电压增大，电流的增长变缓。电阻 R = V/I 不是常数，而是增大。I–V 图的斜率为 1/R，因此斜率随电压增大而减小。相比之下，定值电阻（欧姆导体）显示一条过原点的直线。

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4. Series and Parallel Resistor Confusion | 串联与并联电阻的混淆

A very frequent mistake is to apply the formula 1/R_total = 1/R₁ + 1/R₂ to series circuits, or R_total = R₁ + R₂ to parallel circuits. In series, resistances simply add. In parallel, the reciprocals add.

一个非常常见的错误是在串联电路中套用公式 1/R_总 = 1/R₁ + 1/R₂，或在并联电路中使用 R_总 = R₁ + R₂。串联电路电阻直接相加，并联电路倒数相加。

Series: R_total = R₁ + R₂ + … ; current same through all. Parallel: 1/R_total = 1/R₁ + 1/R₂ + … ; potential difference same across each branch. Always check which quantity is shared before selecting the formula.

串联：R_总 = R₁ + R₂ + …；电流处处相等。并联：1/R_总 = 1/R₁ + 1/R₂ + …；各支路两端电压相等。选择公式之前，先确认哪个物理量是共享的。

Worked example: A 4 Ω and a 6 Ω resistor in parallel give total resistance: 1/R = 1/4 + 1/6 = 5/12 ⇒ R = 12/5 = 2.4 Ω. Common mistake: directly adding to get 10 Ω.

例题：一个 4 Ω 和一个 6 Ω 的电阻并联，总电阻为 1/R = 1/4 + 1/6 = 5/12 ⇒ R = 12/5 = 2.4 Ω。常见错误：直接相加得 10 Ω。

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5. Work Done and Displacement Direction | 做功与位移方向

Work done is calculated by W = F s cosθ, where s is the displacement in the direction of the force. A common error is to use the total distance travelled regardless of direction, or to ignore the angle when the force is not parallel to the displacement. If a crate is dragged horizontally by a rope at an angle, only the horizontal component of the force does work.

功的计算公式为 W = F s cosθ，其中 s 是物体在力的方向上发生的位移。常见错误是使用运动的总路程而不考虑方向，或当力与位移不平行时忽略夹角。如果用一根与水平方向成角度的绳子拖拉箱子，只有拉力的水平分力做功。

Another pitfall: lifting an object and carrying it horizontally. During the horizontal movement, the lifting force is vertical and displacement is horizontal ⇒ cos90° = 0 ⇒ no work is done by the lifting force on the object, despite the person feeling tired.

另一个陷阱：提起一个物体然后水平搬运。在水平移动过程中，提力竖直向上而位移水平 ⇒ cos90° = 0 ⇒ 提力对物体不做功，尽管人感到疲劳。

W = Fs cosθ; if θ = 90°, W = 0

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6. Conservation of Momentum: External Forces Veto | 动量守恒：外力使条件失效

Momentum is conserved only when the net external force on the system is zero. Students frequently apply conservation to situations where friction or gravity provides an external impulse—e.g., a collision on a rough surface or a rocket taking off. Always define the system and verify ΣF_ext = 0 for the time interval considered.

动量守恒仅在系统所受合外力为零时成立。学生常常在有摩擦或重力提供外冲量的情况下使用守恒——例如粗糙表面上的碰撞或火箭起飞。必须明确系统，并确认在所考虑的时间间隔内 ΣF_外 = 0。

Also, momentum is a vector; its direction matters. In a two-dimensional collision, students sometimes add magnitudes without considering vector components. Use p_x and p_y conservation separately.

此外，动量是矢量，方向很重要。在二维碰撞中，有时学生会直接相加大小而不考虑矢量分量。应分别对 p_x 和 p_y 使用守恒。

Example: A ball of mass 0.5 kg moving at 4 m s⁻¹ strikes a stationary ball of mass 0.3 kg on a frictionless table. After collision, the first ball moves at 1 m s⁻¹ in the original direction. Find the second ball’s velocity. Solution: 0.5×4 + 0 = 0.5×1 + 0.3×v ⇒ v = 5 m s⁻¹. This is valid because no external horizontal forces.

例题：质量为 0.5 kg 的小球以 4 m s⁻¹ 的速度在光滑桌面上碰撞静止的 0.3 kg 小球，碰后第一球以 1 m s⁻¹ 原方向运动。求第二球速度。解：0.5×4 + 0 = 0.5×1 + 0.3×v ⇒ v = 5 m s⁻¹。由于水平方向无外力，守恒成立。

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7. Interference: Path Difference and Phase Difference | 干涉：路程差与相位差

A persistent mistake is to use path difference = nλ for destructive interference instead of constructive. The correct conditions: constructive interference when path difference = nλ (n = 0, 1, 2…); destructive interference when path difference = (n + ½)λ. Confusing these flips the entire pattern.

一个持续性的错误是把路程差 = nλ 当成减弱条件而非加强条件。正确条件：路程差 = nλ（n = 0, 1, 2…）时干涉加强；路程差 = (n + ½)λ 时干涉减弱。混淆这两个条件会将整个图样反转。

Phase difference is related by Δφ = (2π/λ) × path difference. Students sometimes mistake degrees for radians: 180° = π rad. Ensure consistent units when calculating. Also, when a wave reflects off a boundary of higher refractive index, a phase change of π occurs, equivalent to an extra half-wavelength path difference.

相位差由 Δφ = (2π/λ) × 路程差决定。学生有时弄混角度与弧度：180° = π rad。计算时单位要统一。另外，当波从较高折射率的边界反射时，会发生 π 的相位突变，相当于额外增加了半个波长的路程差。

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8. Radioactive Decay: Constant Activity vs Constant Count Rate | 放射性衰变：活度与计数率的区别

Students often confuse the count rate measured by a GM tube with the true activity of the sample. Count rate includes background radiation and depends on the detector’s efficiency. Activity (measured in becquerels) is the number of decays per second. Common mistake: using raw count rate for half-life calculations without background subtraction.

学生经常把盖革计数器测得的计数率与样品的真实活度混淆。计数率包含本底辐射且取决于探测器的效率。活度（单位贝克勒尔）是每秒发生的衰变次数。常见错误：在半衰期计算中使用未经本底扣除的原始计数率。

The decay law N = N₀ e^–λt or A = A₀ e^–λt describes exponential decay. Half-life T_½ = ln2/λ. Many treat decay as linear, halving the activity every half-life by simply subtracting a constant amount, which is incorrect. The same fraction decays each half-life, not the same number.

衰变规律 N = N₀ e^–λt 或 A = A₀ e^–λt 描述了指数衰减。半衰期 T_½ = ln2/λ。许多人把它当作线性衰减处理，每个半衰期减掉一个常数，这是错误的。每经过一个半衰期，衰变的是相同的比例，而非相同的数目。

Example: A sample has initial activity 800 Bq. After 6 hours, activity is 100 Bq. Half-life? Not a simple division; 800 → 400 → 200 → 100 is three half-lives, so 3 T_½ = 6 h ⇒ T_½ = 2 h.

例题：某样品初始活度 800 Bq，6 小时后活度为 100 Bq。求半衰期？并非简单除法；800 → 400 → 200 → 100 经历了三个半衰期，因此 3 T_½ = 6 h ⇒ T_½ = 2 h。

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9. Specific Heat Capacity and Latent Heat: The Formula Trap | 比热容与潜热：公式陷阱

Mixing up Q = mcΔθ and Q = mL is a common error. Remember: use specific heat capacity when the substance changes temperature without changing state; use specific latent heat when there is a change of state at constant temperature. A classic mistake is to apply mcΔθ to melting ice or condensing steam.

混淆 Q = mcΔθ 和 Q = mL 是常见的错误。记住：当物质温度改变而状态不变时使用比热容；当温度不变而改变状态时使用比潜热。典型错误是将 mcΔθ 用于冰的熔化或水蒸气的凝结。

For a heating curve, the flat sections indicate phase changes where added energy goes into breaking bonds (latent heat), not raising temperature. Students often forget that during melting, the temperature stays at 0 °C for water, and all energy contributes to melting, not warming.

在加热曲线上，平坦段表示相变，此时吸收的能量用于打破键（潜热），而非提高温度。学生常忘记：冰熔化时温度保持在 0 °C，所有能量用于熔化，不升温。

Example: How much energy to melt 0.5 kg of ice at 0 °C? Use Q = mL, where L = 3.34 × 10⁵ J kg⁻¹. Q = 0.5 × 3.34 × 10⁵ = 1.67 × 10⁵ J. Do not use mcΔθ with Δθ = 0, which would erroneously give zero.

例题：熔化 0 °C 的 0.5 kg 冰需要多少能量？使用 Q = mL，L = 3.34 × 10⁵ J kg⁻¹，Q = 1.67 × 10⁵ J。不能使用 mcΔθ（Δθ = 0），否则会错误得出零。

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10. Balancing Nuclear Equations: Beta Decay Details | 平衡核方程：β 衰变的细节

In beta-minus decay, a neutron turns into a proton, emitting an electron (β⁻) and an antineutrino. Students often forget to increase the atomic number by 1 while the mass number stays the same. For beta-plus decay (positron emission), the atomic number decreases by 1. Always check conservation of charge and nucleon number.

在 β⁻ 衰变中，中子转变为质子，放出一个电子（β⁻）和一个反中微子。学生经常忘记原子序数加 1，而质量数不变。对于 β⁺ 衰变（正电子发射），原子序数减 1。务必检查电荷数和核子数的守恒。

A typical slip: ¹⁴₆C → ¹⁴₇N + ⁰_-1e is correct; mistakenly writing ¹⁴₆C → ¹⁴₇N + ⁰₊₁e or keeping the atomic number as 6 is wrong. Also, neutrinos carry energy and momentum but have negligible mass and no charge—they are often omitted in examination answers, which may lose marks.

典型错误：¹⁴₆C → ¹⁴₇N + ⁰_-1e 正确；错误写成 ¹⁴₆C → ¹⁴₇N + ⁰₊₁e 或原子序数仍为 6。此外，中微子携带能量和动量但质量可忽略，不带电——考试作答时常遗漏，可能失分。

For alpha decay: mass number drops by 4, atomic number drops by 2. The emitted alpha particle is ⁴₂He. Balancing equations is a reliable way to identify unknown daughter nuclei.

对于 α 衰变：质量数减 4，原子序数减 2。放射出的 α 粒子为 ⁴₂He。通过平衡方程可确定未知子核，是一种可靠的方法。

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11. Power, Energy and Time Units in Electrical Calculations | 电学计算中的功率、能量与时间单位

A subtle but costly error is mismatching units when calculating energy from power. The correct unit for energy is the joule when power is in watts and time in seconds. If time is given in minutes or hours, it must be converted. For kilowatt-hours: Energy (kWh) = Power (kW) × time (h).

一个细微但代价高昂的错误是在从功率计算能量时单位不匹配。当功率使用瓦特、时间使用秒时，能量的正确单位是焦耳。若时间以分钟或小时给出，必须转换。对于千瓦时：能量 (kWh) = 功率 (kW) × 时间 (h)。

Exam context: a 60 W bulb left on for 30 minutes. Energy in joules: E = 60 × (30 × 60) = 108,000 J. Often students multiply 60 × 30 directly, giving 1800 J—wrong by a factor of 60. Always check the unit required in the answer space.

考试情景：一个 60 W 的灯泡开了 30 分钟。焦耳表示的能量：E = 60 × (30 × 60) = 108,000 J。学生常直接用 60 × 30 得 1800 J——差了 60 倍。务必看清答题空要求填写的单位。

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