A-Level Physics Unit 5 Formula Derivations: Insights from the Jan21 Mark Scheme | A-Level物理第五单元公式推导：2021年1月评分方案揭秘

📚 A-Level Physics Unit 5 Formula Derivations: Insights from the Jan21 Mark Scheme | A-Level物理第五单元公式推导：2021年1月评分方案揭秘

In January 2021, A-Level Physics Unit 5 examined a range of core derivations that tested students’ ability to reason from first principles. The mark scheme reveals exactly what examiners look for: clear logical steps, correct handling of vectors, energy conservation, and the ability to link microscopic models to macroscopic observables. This article unpacks those derivations, step by step, using the same language and notation that secure marks on exam day.

2021年1月的A-Level物理第5单元，考查了一系列重要的公式推导，要求学生从基本原理出发进行逻辑推理。评分方案明确指出：清晰的推导步骤、正确的矢量与能量守恒处理，以及将微观模型与宏观观测量联系起来的能力，是得分的关键。本文将逐一拆解这些推导过程，使用与考试完全一致的语言和符号体系，帮你把评分重点变成你的得分点。

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1. Deriving the Kinetic Theory Pressure Equation | 气体动理论压强方程的推导

Begin with a single molecule of mass m moving with velocity component vₓ towards a wall of a cubic container of side length L. The molecule collides elastically, so its velocity changes from vₓ to −vₓ. The change in momentum is Δp = m(vₓ) − (−m vₓ) = 2m vₓ. The time between successive collisions with the same wall is Δt = 2L / vₓ, so the average force exerted by this molecule on the wall is F = Δp / Δt = 2m vₓ ÷ (2L / vₓ) = m vₓ² / L.

从一个质量为m的分子开始，它沿 x 方向的速度分量为 vₓ，所处容器边长为 L。分子与器壁发生弹性碰撞，速度由 vₓ 变为 −vₓ，动量变化为 Δp = 2m vₓ。两次撞击同一器壁的时间间隔为 Δt = 2L / vₓ，因此该分子对器壁的平均作用力为 F = Δp / Δt = m vₓ² / L。

Summing over all N molecules, the total force on the wall is F_total = (m / L) Σ vᵢₓ². By definition, pressure p = F_total / L² = (m / L³) Σ vᵢₓ². The mean square speed ⟨c²⟩ is defined as the average of vᵢₓ² + vᵧ² + v_z² for all molecules. Because motion is random, ⟨vₓ²⟩ = ⟨vᵧ²⟩ = ⟨v_z²⟩ = ⅓ ⟨c²⟩. Hence Σ vᵢₓ² = N ⟨vₓ²⟩ = N ⟨c²⟩ / 3. Substituting gives the fundamental equation in the mark scheme:

将所有 N 个分子的贡献求和，器壁上的总力为 F_total = (m / L) Σ vᵢₓ²。压强 p = F_total / L² = (m / L³) Σ vᵢₓ²。由于运动各向同性，⟨vₓ²⟩ = ⟨vᵧ²⟩ = ⟨v_z²⟩ = ⅓ ⟨c²⟩，故 Σ vᵢₓ² = N ⟨c²⟩ / 3。代入整理即得评分方案中要求的核心公式：

pV = ⅓ N m ⟨c²⟩

Examiners expect you to explicitly state the assumption of elastic collisions and the cubic geometry, then link mean square speeds to pressure. The mark scheme awards 4–5 marks for a complete derivation.

考官希望看到你明确写出弹性碰撞假设以及立方体几何条件，并清晰连接均方速率和压强。评分方案通常给完整推导过程 4−5 分。

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2. Connecting Kinetic Energy to Temperature | 从动能到温度的联系

The ideal gas law from experiment is pV = nRT. Comparing with pV = ⅓ N m ⟨c²⟩, we have ⅓ N m ⟨c²⟩ = nRT. Since N = n Nₐ, where Nₐ is Avogadro’s number, this becomes ⅓ N m ⟨c²⟩ = (N / Nₐ) RT. Rearranging gives the average kinetic energy per molecule: ½ m ⟨c²⟩ = (3/2) (R / Nₐ) T = (3/2) k T, where k is the Boltzmann constant.

实验得出的理想气体状态方程为 pV = nRT。将其与 pV = ⅓ N m ⟨c²⟩ 对比，可得 ⅓ N m ⟨c²⟩ = nRT。利用 N = n Nₐ（Nₐ 为阿伏伽德罗常数），代入得 ½ m ⟨c²⟩ = (3/2) (R / Nₐ) T = (3/2) k T，其中 k 为玻尔兹曼常数。

This is a favourite Jan21 exam step: the link between microscopic kinetic energy and macroscopic temperature is completed. At A-level, you need only state that k = R/Nₐ, then substitute. The mark scheme typically asks for “kinetic energy proportional to absolute temperature”.

这也是 Jan21 试卷中常见的考察点：完成微观动能与宏观温度之间的关联。在A-Level中，只需说明 k = R/Nₐ 并代入即可。评分标准往往看是否判断出“平均动能与绝对温度成正比”。

½ m ⟨c²⟩ = ³⁄₂ k T

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3. Deriving the Ideal Gas Law from Kinetic Theory | 从动理论推导理想气体定律

Starting from pV = ⅓ N m ⟨c²⟩ and the average kinetic energy expression ½ m ⟨c²⟩ = ³⁄₂ k T, eliminate ½ m ⟨c²⟩. Multiply the kinetic energy by 2: m ⟨c²⟩ = 3 k T. Substituting into the pressure equation yields pV = ⅓ N × 3 k T = N k T. Using N = n Nₐ, we obtain pV = n Nₐ k T = n R T, which is exactly the ideal gas law. This shows that the kinetic theory model is consistent with empirical gas behaviour.

从 pV = ⅓ N m ⟨c²⟩ 和 ½ m ⟨c²⟩ = ³⁄₂ k T 出发，消去 m ⟨c²⟩。将动能表达式乘以2得 m ⟨c²⟩ = 3 k T，代入压强方程得 pV = ⅓ N × 3 k T = N k T。再利用 N = n Nₐ 和 R = Nₐ k，便可得到 pV = n R T。这一步完整证明了动理论模型与实验结果的一致性。

The mark scheme explicitly rewards the logical flow that connects pV = ⅓ N m ⟨c²⟩ to the kinetic energy result, and then to pV = nRT. Candidates often lose marks by missing the multiplication step or forgetting to define Nₐ.

评分方案会奖励从 pV = ⅓ N m ⟨c²⟩ 到动能关系、再到 pV = nRT 的逻辑链条。考生常因跳过乘以2的步骤或未说明 Nₐ 而丢分。

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4. Radioactive Decay Law and Half-Life Derivation | 放射性衰变定律与半衰期推导

Radioactive decay is a random process where the number of nuclei decaying per unit time is proportional to the number present: dN/dt = −λ N, where λ is the decay constant. Separating variables gives dN/N = −λ dt. Integrating both sides: ∫(1/N) dN = −λ ∫ dt ⇒ ln N = −λ t + C. At t = 0, N = N₀, so C = ln N₀. Therefore ln(N / N₀) = −λ t, which exponentiates to the familiar decay law:

放射性衰变是一个随机过程，单位时间内衰变的原子核数正比于现存核数目：dN/dt = −λ N（λ 为衰变常量）。分离变量得 dN/N = −λ dt，积分得 ln N = −λ t + C。代入初始条件 t = 0 时 N = N₀，得 C = ln N₀，于是 ln(N / N₀) = −λ t，指数化后得到熟知的形式：

N = N₀ e^{−λ t}

To find half-life T₁/₂, set N = N₀ / 2 at t = T₁/₂: N₀/2 = N₀ e^{−λ T₁/₂} ⇒ e^{−λ T₁/₂} = 1/2 ⇒ −λ T₁/₂ = ln(1/2) = −ln 2. Hence T₁/₂ = ln 2 / λ. The Jan21 mark scheme insists on showing the natural logarithm manipulation explicitly.

求半衰期 T₁/₂ 时，令 t = T₁/₂ 时 N = N₀/2，代入得 e^{−λ T₁/₂} = 1/2，取对数得 λ T₁/₂ = ln 2，所以 T₁/₂ = ln 2 / λ。Jan21 评分方案强调必须完整展示自然对数运算过程。

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5. Deriving Activity from Decay Constant | 从衰变常数推导活度

Activity A is defined as the number of decays per unit time, which is exactly the magnitude of the decay rate: A = |dN/dt|. Using the decay equation dN/dt = −λ N, we immediately have A = λ N. Substituting N = N₀ e^{−λ t} gives the exponential decay of activity: A = λ N₀ e^{−λ t} = A₀ e^{−λ t}, where A₀ = λ N₀ is the initial activity. This derivations is often worth 2 marks in the Jan21 mark scheme — one for stating A = λ N and one for linking to the exponential form.

活度 A 定义为单位时间内的衰变次数，即衰变率的绝对值：A = |dN/dt|。由衰变方程 dN/dt = −λ N 直接可得 A = λ N。将 N = N₀ e^{−λ t} 代入，得到活度随时间的指数衰减规律：A = λ N₀ e^{−λ t} = A₀ e^{−λ t}（其中 A₀ = λ N₀ 为初始活度）。在 Jan21 评分方案中，这个推导通常占 2 分——1分写出 A = λ N，1分完成指数形式的转换。

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6. Escape Velocity Derivation from Energy Conservation | 通过能量守恒推导逃逸速度

To escape a planet’s gravitational field, a projectile must have enough initial kinetic energy to do work against gravity to reach infinity. The gravitational potential energy at radial distance r from the planet’s centre is U = −G M m / r. At infinity, both kinetic and potential energy become zero. Conservation of total mechanical energy gives: ½ m v² − G M m / r = 0. Solving for v yields the escape speed:

要逃离行星的引力场，抛射体需拥有足够的初始动能以克服引力做功到达无穷远。距离行星中心 r 处的引力势能为 U = −G M m / r，无穷远处动能和势能均为零。根据机械能守恒：½ m v² − G M m / r = 0，解得逃逸速度：

v_esc = √(2 G M / r)

At the planet’s surface, r = R, and g = G M / R², so an alternative form is v_esc = √(2 g R). The Jan21 mark scheme often asks candidates to derive both forms and to state the energy conservation principle explicitly.

在行星表面 r = R，且 g = G M / R²，因此逃逸速度也可写成 v_esc = √(2 g R)。Jan21 评分要求考生能够推导这两种形式，并明确写出能量守恒原理。

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7. Deriving Kepler’s Third Law for Circular Orbits | 圆形轨道开普勒第三定律推导

For a satellite in a circular orbit of radius r, the gravitational force provides the centripetal force: G M m / r² = m v² / r, so v² = G M / r. The orbital period is T = 2π r / v, hence v = 2π r / T. Substituting v² = 4π² r² / T² into the force equation gives G M / r = 4π² r² / T², which rearranges to:

对于半径为 r 的圆形轨道，万有引力提供向心力：G M m / r² = m v² / r，故 v² = G M / r。轨道周期 T = 2π r / v，因此 v = 2π r / T，代入得 G M / r = 4π² r² / T²，整理可得：

T² = (4π² / G M) r³

This is Kepler’s third law for circular motion: T² ∝ r³. The proportionality constant depends only on the central mass. In the Jan21 astrophysics option, this derivation carries 3 marks and must include the substitution step for v.

这就是圆形轨道下的开普勒第三定律：T² ∝ r³，比例常数仅与中心天体质量有关。在 Jan21 天体物理选考中，这一推导占 3 分，必须体现 v 的替换步骤。

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8. Derivation of the Gravitational Potential Formula | 引力势公式的推导

Gravitational potential V at a point is defined as the work done per unit mass in bringing a small test mass from infinity to that point. The force on a unit mass at distance r from mass M is G M / r² directed towards M. The work done by an external agent moving from r’ = ∞ to r is W = ∫_{∞}^{r} (−G M / r’²) dr’ (the negative sign ensures the force is overcome). Evaluating the integral: W = [G M / r’]_{∞}^{r} = G M / r − 0. Since potential is negative of work done by the gravitational field, we get V = −G M / r.

引力势 V 定义为将单位质量从无穷远处移动到该点外力克服引力所做的功。在距离 M 为 r 处，单位质量所受引力为 G M / r² 指向 M。外力做功为 W = ∫_{∞}^{r} (G M / r’²) dr’（或直接积分引力做功为负值）。计算积分得 W = [−G M / r’]_{∞}^{r} = −G M / r − 0，因此引力势为 V = −G M / r。

The Jan21 mark scheme specifically looks for the correct limits and the negative sign. Candidates must also explain that the potential is negative because work is done by the field as an object approaches.

Jan21 评分方案非常看重积分上下限的设置以及负号的出现，考生还需说明势能为负是因为当物体靠近时引力场做正功。

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9. Deriving Binding Energy per Nucleon from Mass Defect | 由质量亏损推导比结合能

The mass of a nucleus is always less than the sum of the masses of its individual protons and neutrons. This mass defect Δm is converted into binding energy via E = Δm c². In nuclear physics, masses are given in unified atomic mass units u, where 1 u = 1.661 × 10⁻²⁷ kg and c² expressed in MeV becomes 931.5 MeV / u. To find the binding energy per nucleon, compute E_binding = Δm × 931.5 MeV then divide by the nucleon number A.

原子核的质量总是小于其内部各核子质量之和，这个质量差 Δm 称为质量亏损，对应的结合能为 E = Δm c²。核物理中质量常以原子质量单位 u 给出，1 u = 1.661 × 10⁻²⁷ kg，c² 等价于 931.5 MeV/u。比结合能的计算为：E_binding = Δm × 931.5 MeV / A。

In full derivation form, the mark scheme expects you to write Δm = Z m_p + (A − Z) m_n − m_nucleus, then multiply by c². When using data tables, always convert mass defect to kg if using E = Δm c² in joules, or use the MeV/u shortcut.

在完整推导中，评分方案期待写出 Δm = Z m_p + (A − Z) m_n − m_nucleus，再乘以 c²。使用数据表时，可根据需要将质量亏损转化为 kg 后用焦耳表达，或直接使用 MeV/u 转换因子。

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