📚 Mastering Derivations from the OxfordAQA PH05 January 2022 Report | 掌握OxfordAQA PH05 2022年1月考报中的公式推导
The January 2022 examiner's report for OxfordAQA Physics 9630 Unit 5 (PH05) highlighted that many candidates lost marks not because they could not recall formulas, but because they struggled to derive them from first principles. Poor handling of integration limits, sign errors in exponential processes, and confusion between average and instantaneous quantities were common pitfalls. This article revisits the key derivations that appeared in or relate to the PH05 paper, offering step-by-step guidance, exam insights, and bilingual explanations to strengthen your understanding and exam technique.
2022年1月OxfordAQA物理9630单元5（PH05）的考官报告指出，许多考生失分并非因为记不住公式，而是因为不善于从基本原理推导公式。积分限处理不当、指数过程中的符号错误、平均值与瞬时值的混淆是常见失分点。本文重温与PH05试卷密切相关的重要公式推导，提供逐步指导、考情洞察和中英双语解释，帮助你加深理解并提升应试技巧。
1. Radioactive Decay Law | 放射性衰变定律
The decay law N = N₀ e-λt is fundamental to nuclear physics. Examiners noted that candidates often wrote down the final expression correctly but could not perform the integration from the activity equation, especially forgetting the negative sign or misplacing the constant of integration.
衰变定律 N = N₀ e-λt 是核物理的基础。考官发现考生往往能正确写出最终表达式，却无法从活度方程出发完成积分，尤其容易遗忘负号或错放积分常数。
Start with dN/dt = -λN. Separate variables: dN/N = -λ dt.
从 dN/dt = -λN 出发，分离变量：dN/N = -λ dt。
Integrate both sides: ∫N₀^N dN/N = -λ ∫_0t dt. This gives ln(N/N₀) = -λt.
两边积分：∫N₀^N dN/N = -λ ∫_0t dt，得到 ln(N/N₀) = -λt。
Exponentiate to remove the logarithm: N/N₀ = e-λt, hence N = N₀ e-λt.
取指数消去对数：N/N₀ = e-λt，因此 N = N₀ e-λt。
Common error: omitting the negative sign gives N = N₀ eλt, which erroneously predicts exponential growth.
常见错误：漏掉负号会得出 N = N₀ eλt，错误地预测指数增长。
2. Half-Life and Decay Constant | 半衰期与衰变常数
n be derived from the decay law, and the report showed that many candidates struggled to link the definition of half-life to the exponential equation.
关系式 T½ = ln2 / λ 常需从衰变定律导出，报告显示许多考生难以将半衰期定义与指数方程建立联系。
By definition, at t = T½, N = N₀ / 2. Substitute: N₀/2 = N₀ e-λ T½.
按定义，在 t = T½ 时，N = N₀ / 2。代入得：N₀/2 = N₀ e-λ T½。
Cancel N₀ and take the natural log: ln(1/2) = -λ T½, so -ln2 = -λ T½, giving T½ = ln2 / λ.
约去 N₀ 并取自然对数：ln(1/2) = -λ T½，即 -ln2 = -λ T½，得 T½ = ln2 / λ。
idates incorrectly used log base 10 or mishandled ln(1/2). Always remember ln(1/2) = -ln2.
有些考生误用常用对数，或未正确处理 ln(1/2)。务必记住 ln(1/2) = -ln2。
3. Ideal Gas Pressure Equation | 理想气体压强方程
Deriving pV = 1/3 N m
由分子运动论推导 pV = 1/3 N m
Consider a cubic box of side L. For a molecule with x-component velocity u, the momentum change per collision with a wall is 2mu.
考虑边长为 L 的立方容器。对于 x 方向速度分量为 u 的分子，每次与壁碰撞的动量变化为 2mu。
ons with the same wall is 2L/u, so the average force on that wall is 2mu / (2L/u) = mu²/L.
与同一壁碰撞的时间间隔为 2L/u，因此作用于该壁的平均力为 2mu / (2L/u) = mu²/L。
Summing over all N molecules and using mean square speed
对所有 N 个分子求和，并利用均方速率
Since pressure p = F / L² and volume V = L³, we obtain pV = 1/3 N m
因压强 p = F / L²、体积 V = L³，即得 pV = 1/3 N m
Examiners reported that some students lost marks by failing to distinguish between the speed of one molecule and the mean square speed of all molecules.
考官报告指出，部分学生因未能区分单个分子的速率与所有分子的均方速率而失分。
4. Gravitational Potential Energy | 引力势能
against the gravitational field is a classic step that the report showed candidates often mishandled by ignoring the sign or choosing the wrong integration limits.
从克服引力场做功出发推导 U = -GMm/r 是经典步骤，报告显示考生常因忽略符号或选错积分限而出错。
l agent must apply an opposite force, so the work done by that agent is ∫_∞^r (GMm/r²) dr.
将质量 m 从无穷远处移至点 r 所做功为 W = ∫_∞^r F dr，其中引力 F = GMm/r² 方向指向中心。外部施力者需施加反向力，因此其做功为 ∫_∞^r (GMm/r²) dr。
Evaluate: W = GMm [ -1/r ]_∞^r = -GMm/r - ( -GMm/∞ ) = -GMm/r.
计算：W = GMm [ -1/r ]_∞^r = -GMm/r - ( -GMm/∞ ) = -GMm/r。
This work is stored as potential energy, so U = -GMm/r, with zero defined at infinity.
此项功储存为势能，故 U = -GMm/r，并定义无穷远处为零势能。
ert U = GMm/r, forgetting the negative sign that indicates the bound nature of the system.
常见错误是写成 U = GMm/r，忘记了体现系统束缚性的负号。
5. Simple Harmonic Motion Displacement Equation | 简谐运动的位移方程
defining equation a = -ω²x, nor could they clearly link circular motion projection to SHM.
考官评论说，学生虽能熟练使用 x = A cos(ωt)，但许多人无法从定义式 a = -ω²x 出发推导，也不能清晰地将圆周运动的投影与简谐运动联系起来。
sing the right boundary condition (e.g., maximum displacement at t=0) gives x = A cos(ωt).
由 a = d²x/dt² = -ω²x，我们识别出一个二阶微分方程，其通解为 x = A sin(ωt) + B cos(ωt)。选择适当的初始条件（如 t=0 时位移最大），即得 x = A cos(ωt)。
Alternatively, imagine a particle moving in uniform circular motion with radius A and angular speed ω. Its projection onto the x-axis moves as A cos(ωt).
另一种思路：设想一个质点以半径 A 和角速度 ω 做匀速圆周运动，该质点在 x 轴上的投影运动即为 A cos(ωt)。
ferentiating this gives velocity v = -Aω sin(ωt) and acceleration a = -Aω² cos(ωt) = -ω²x.
对此求导得速度 v = -Aω sin(ωt)，加速度 a = -Aω² cos(ωt) = -ω²x。
The report noted that some candidates incorrectly used sine instead of cosine without stating the phase constant, leading to inconsistent initial conditions.
报告指出，一些考生在不声明相位常数的情况下误用正弦代替余弦，导致初始条件前后矛盾。
6. Capacitor Discharge | 电容器放电
The derivation of Q = Q₀ e-t/RC e sign of the rate of charge flow. The report stressed the importance of using I = -dQ/dt.
放电电容器的公式 Q = Q₀ e-t/RC 的推导是常见的错误来源，特别是关于电荷流量速率的符号。报告强调了使用 I = -dQ/dt 的重要性。
ging through a resistor, the stored charge Q decreases, so dQ/dt is negative. By Kirchhoff' voltage law, the p.d. across the capacitor equals the p.d. across the resistor: Q/C = IR.
对于通过电阻放电的电容器，所储电荷 Q 减少，因此 dQ/dt 为负。由基尔霍夫电压定律，电容两端电压等于电阻两端电压：Q/C = IR。
e leaves the capacitor, so I = -dQ/dt. Substitute: Q/C = -R dQ/dt, giving dQ/dt = -Q/(RC).
但电路中的电流 I 正是电荷离开电容器的速率，故 I = -dQ/dt。代入得：Q/C = -R dQ/dt，即 dQ/dt = -Q/(RC)。
Separate variables: dQ/Q = -dt/(RC). Integrate: ∫Q₀^Q dQ/Q = -1/(RC) ∫_0t dt, giving ln(Q/Q₀) = -t/(RC), so Q = Q₀ e-t/RC.
分离变量：dQ/Q = -dt/(RC)。积分：∫Q₀^Q dQ/Q = -1/(RC) ∫
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