📚 Formula Derivation Techniques from OxfordAQA PH05 Mark Scheme | 基于牛津AQA PH05评分方案的公式推导要点
In OxfordAQA A-Level Physics Unit 5 (PH05), candidates are frequently asked to derive key equations from first principles. The January 2023 mark scheme reveals the precise logical steps, assumptions, and algebraic manipulations that earn full marks. Understanding these derivation patterns not only secures marks in structured questions but also deepens conceptual mastery of fields, oscillations, and circuits. This article unpacks the essential derivations highlighted in the PH05 Final MS Jan23, pairing each method with clear commentary so you can reproduce them confidently under exam conditions.
在牛津AQA A-Level物理单元5 (PH05) 中，考生经常被要求从基本原理出发推导关键方程。2023年1月的评分方案揭示了获得满分的精确逻辑步骤、假设条件和代数操作。理解这些推导模式不仅有助于在结构化问题中得分，还能加深对场、振荡和电路等概念的理解。本文拆解了PH05 Final MS Jan23中强调的基本推导，并为每种方法配上清晰的解说，使你能够在考试中自信地重现它们。
1. Centripetal Acceleration Derivation | 向心加速度推导
Consider an object moving with constant speed v in a circular path of radius r. In a short time interval Δt, the position vector sweeps out a small angle Δθ, and the velocity vector changes direction by the same angle. The change in velocity Δv points towards the centre, and its magnitude is approximately vΔθ.
考虑一个物体以恒定速率 v 在半径为 r 的圆形路径上运动。在很短的时间间隔 Δt 内，位置向量扫过一个小角度 Δθ，速度向量的方向变化了相同的角度。速度的变化量 Δv 指向圆心，其大小近似为 vΔθ。
Using the relationship between arc length and angle, the distance moved along the circle is vΔt = rΔθ. Thus Δθ = vΔt / r. Substituting this into Δv gives Δv = v × (vΔt / r) = v²Δt / r.
使用弧长与角度的关系，沿圆周运动的距离为 vΔt = rΔθ。因此 Δθ = vΔt / r。将其代入 Δv 得到 Δv = v × (vΔt / r) = v²Δt / r。
Acceleration is the rate of change of velocity, so a = Δv / Δt = v² / r. The direction is radially inward, so we can write the vector form as a = − v²/r r̂.
加速度是速度的变化率，所以 a = Δv / Δt = v² / r。方向径向向内，因此可以写成矢量形式 a = − v²/r r̂。
a = v² / r
The mark scheme explicitly rewards stating the small-angle approximation and showing the vector triangle for velocity change. Always mention that the speed remains constant, so only direction changes.
评分方案明确奖励指出小角度近似并画出速度变化矢量三角形。务必说明速率保持恒定，因此只有方向变化。
2. Simple Harmonic Motion Differential Equation | 简谐运动微分方程
For a body of mass m attached to a spring with spring constant k, Hooke’s law gives the restoring force as F = −kx. Applying Newton’s second law yields ma = −kx, so aceleration is proportional to displacement and oppositely directed.
对于附着在劲度系数为 k 的弹簧上的质量为 m 的物体，胡克定律给出回复力 F = −kx。应用牛顿第二定律得到 ma = −kx，因此加速度与位移成正比且方向相反。
By definition, acceleration is the second derivative of displacement: a = d²x/dt². Substituting gives the defining SHM equation.
根据定义，加速度是位移对时间的二阶导数：a = d²x/dt²。代入后得到简谐运动的定义方程。
d²x / dt² = − (k / m) x
Introducing ω² = k/m produces the standard form d²x/dt² = −ω²x. The general solution is x = A cos(ωt + φ), where ω = 2πf. The PH05 mark scheme often expects you to verify that this solution satisfies the differential equation by differentiation.
引入 ω² = k/m 得到标准形式 d²x/dt² = −ω²x。通解为 x = A cos(ωt + φ)，其中 ω = 2πf。PH05评分方案通常期望你通过微分验证该解满足微分方程。
Velocity is dx/dt = −ωA sin(ωt + φ) and acceleration is d²x/dt² = −ω²A cos(ωt + φ) = −ω²x, confirming consistency. The maximum speed is ωA at the equilibrium position.
速度为 dx/dt = −ωA sin(ωt + φ)，加速度为 d²x/dt² = −ω²A cos(ωt + φ) = −ω²x，验证了一致性。最大速率出现在平衡位置，为 ωA。
3. Gravitational Potential in a Radial Field | 径向引力场的引力势
Gravitational potential V at a point is defined as the work done per unit mass to bring a small test mass from infinity to that point. The gravitational force on a mass m due to a mass M is F = GMm/r², directed towards M.
某点的引力势 V 定义为将单位质量的小测试质量从无穷远处移至该点所做的功。质量 M 对质量 m 的引力为 F = GMm/r²，方向指向 M。
Because force varies with distance, work done over a small displacement dr away from M is dW = −F dr = − (GMm/r²) dr. The negative sign appears because the external agent works against the attractive force when moving away.
由于力随距离变化，在远离 M 方向上的微小位移 dr 所做的功为 dW = −F dr = − (GMm/r²) dr。负号是因为向外移动时外力要克服引力做功。
Integrating from r = ∞ to r = R gives the total work per unit mass:
从 r = ∞ 到 r = R 积分得到单位质量的总功：
V = − GM / R
The mark scheme values the clear statement of the integration limits and handling of the sign. Gravitational potential energy of a mass m at that point is U = mV = − GMm/R.
评分方案看重清晰地陈述积分限和处理符号。该点处质量 m 的引力势能为 U = mV = − GMm/R。
4. Energy Stored in a Capacitor | 电容器储存的能量
Charging a capacitor from 0 to a final charge Q requires work because overcoming the electrostatic repulsion. At any instant when the p.d. is v, moving a small additional charge dq requires work dW = v dq.
将电容器从0充电到最终电荷 Q 需要做功，因为必须克服静电排斥。在任何时刻当电势差为 v 时，移动少量额外电荷 dq 需要做功 dW = v dq。
Using the capacitor equation q = Cv, we write v = q/C. Substituting gives dW = (q/C) dq. Integrating from 0 to Q yields:
利用电容器方程 q = Cv，可写 v = q/C。代入得 dW = (q/C) dq。从0到 Q 积分得到：
W = ½ Q² / C = ½ C V² = ½ Q V
The PH05 January 2023 mark scheme explicitly requires showing the substitution and integration steps. Also explain that this energy is stored in the electric field between plates.
PH05 2023年1月的评分方案明确要求展示代入和积分步骤。同时解释该能量储存在两极板间的电场中。
5. Radioactive Decay Law Derivation | 放射性衰变定律推导
The activity A of a radioactive sample is proportional to the number of undecayed nuclei N. Hence the decay rate is A = −dN/dt = λN, where λ is the decay constant. The negative sign reflects a decrease in N.
放射性样品的活度 A 与未衰变核的数量 N 成正比。因此衰变率为 A = −dN/dt = λN，其中 λ 是衰变常量。负号表示 N 在减少。
Separating variables gives dN/N = −λ dt. Integrating both sides between N₀ at t = 0 and N at time t:
分离变量得到 dN/N = −λ dt。在 t = 0 时 N = N₀ 和 t 时刻 N 之间积分：
∫N₀N dN/N = −λ ∫0t dt → ln(N/N₀) = −λ t
Exponentiating gives the exponential decay law N = N₀ e−λt. The mark scheme emphasises correctly identifying limits and the natural logarithm manipulation.
取指数得到指数衰变定律 N = N₀ e−λt。评分方案强调要正确识别积分限并进行自然对数运算。
Half-life T½ is derived by setting N = N₀/2: ln(½) = −λ T½ → T½ = ln 2 / λ. Activity follows the same law: A = A₀ e−λt.
令 N = N₀/2 可推导半衰期 T½：ln(½) = −λ T½ → T½ = ln 2 / λ。活度遵循相同规律：A = A₀ e−λt。
6. Induced EMF from Faraday’s Law | 法拉第定律与感生电动势
Faraday’s law states that the magnitude of induced emf in a coil is directly proportional to the rate of change of magnetic flux linkage. For a coil of N turns, flux linkage is NΦ, and the emf is ε = − d(NΦ)/dt.
法拉第定律指出，线圈中感生电动势的大小与磁通链的变化率成正比。对于 N 匝线圈，磁通链为 NΦ，电动势为 ε = − d(NΦ)/dt。
The negative sign encapsulates Lenz’s law: the induced current flows in a direction to oppose the change producing it. If flux changes uniformly, ε = −N ΔΦ/Δt.
负号体现了楞次定律：感生电流的方向总是阻碍引起它的变化。如果磁通量均匀变化，则 ε = −N ΔΦ/Δt。
In the PH05 assessment, you may need to derive the emf for a rod moving in a magnetic field. For a conductor of length L moving with velocity v perpendicular to a field B, the induced emf across ends is ε = BLv. Derivation: in time Δt, area swept is LvΔt, flux cut = B × (LvΔt), so emf = ΔΦ/Δt = BLv. Always state that the motion is perpendicular to both field and length.
在PH05的考核中，你可能需要推导导体棒在磁场中运动产生的电动势。对于长度为 L、以速度 v 垂直于磁场 B 运动的导体，两端产生的感生电动势为 ε = BLv。推导：在 Δt 时间内，扫过的面积为 LvΔt，切割的磁通量为 B × (LvΔt)，因此 ε = ΔΦ/Δt = BLv。务必说明运动方向同时垂直于磁场和长度。
7. Capacitor Discharge Equation | 电容器放电方程
For a capacitor discharging through a resistor, the current I is the rate of decrease of charge: I = −dq/dt. Ohm’s law gives V = IR, and the capacitor equation is q = CV. Combining these yields V = q/C = −R dq/dt.
对于通过电阻放电的电容器，电流 I 是电荷减少的速率：I = −dq/dt。欧姆定律给出 V = IR，电容器方程为 q = CV。联立得到 V = q/C = −R dq/dt。
Rearranging: dq/dt = − q/(RC). This is a first-order differential equation. Separating variables and integrating from Q₀ at t=0 to q at time t:
整理得：dq/dt = − q/(RC)。这是一阶微分方程。分离变量并从 t=0 时的 Q₀ 积分到时间 t 的 q：
∫Q₀q dq/q = − (1/RC) ∫0t dt → ln(q/Q₀) = − t/RC
Exponential form: q = Q₀ e−t/RC. The product RC is the time constant τ. Similarly, voltage decays as V = V₀ e−t/RC. The half-life for discharge is t½ = RC ln 2.
指数形式：q = Q₀ e−t/RC。乘积 RC 是时间常数 τ。类似地，电压衰减为 V = V₀ e−t/RC。放电的半衰期为 t½ = RC ln 2。
The mark scheme rewards clear identification of differential relationship, separation of variables, and substitution of limits. Remember that charging follows a similar logic but with a different differential equation.
评分方案奖励明确识别微分关系、分离变量和代入积分限。记住充电过程遵循类似逻辑，但使用不同的微分方程。
8. Kinetic Energy and Momentum Relationship | 动能与动量的关系
While seemingly simple, the derivation of Ek = p²/(2m) is frequently required in nuclear and particle physics contexts. Start with momentum p = mv and kinetic energy Ek = ½ mv².
这个看似简单，但在核物理和粒子物理中经常需要推导 Ek = p²/(2m)。从动量 p = mv 和动能 Ek = ½ mv² 出发。
Solve for v from momentum: v = p/m. Substitute into kinetic energy:
由动量解出 v：v = p/m。代入动能：
Ek = ½ m (p/m)² = p² / (2m)
This form is used when dealing with non-relativistic collisions and particle scattering. The PH05 mark scheme sometimes uses it to connect stopping potential to momentum in photoelectric emission or to discuss ion energies in a mass spectrometer.
该形式用于处理非相对论碰撞和粒子散射。PH05评分方案有时会利用它把停止电势与光电发射中的动量联系起来，或在质谱仪中讨论离子能量。
9. Ideal Gas Law from Kinetic Theory | 从分子动理论推导理想气体定律
The pressure exerted by an ideal gas on a container wall can be derived by considering the change in momentum of molecules colliding elastically. Assume N molecules of mass m in a cubic box of side L, moving randomly with average speed c.
理想气体对容器壁施加的压强可以通过考虑分子弹性碰撞的动量变化来推导。假设在边长为 L 的立方盒中有 N 个质量为 m 的分子，以平均速率 c 做无规则运动。
For one molecule striking a wall perpendicularly, momentum change is 2mcx (taking the x-component). The time between collisions on that wall is 2L/cx. Force on wall from that molecule is momentum change per unit time = mcx²/L.
对一个分子垂直撞击一个壁，动量变化为 2mcx（取 x 分量）。与该壁碰撞的时间间隔为 2L/cx。该分子对壁的作用力为单位时间动量变化 = mcx²/L。
Summing over all N molecules and using the average of squared velocity components,
对所有 N 个分子求和，并使用速度分量的平方平均值
pV = ⅓ N m
Comparing with the experimental gas law pV = nRT and using total mass M = Nm, we deduce that average translational kinetic energy is proportional to absolute temperature: ½ m
与实验气体定律 pV = nRT 比较，并利用总质量 M = Nm，可推导出平均平动动能与绝对温度成正比：½ mPublished by TutorHao | Physics Revision Series | aleveler.com
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