📚 A-Level Physics Unit 4 Mark Scheme Jun22: Derivations of Key Formulas | A-Level物理第四单元2022年6月评分方案：关键公式推导
The June 2022 A‑Level Physics Unit 4 mark scheme placed strong emphasis on the ability to derive key relationships from first principles. Rather than simply quoting final equations, candidates were expected to justify each step using fundamental laws and clear logical reasoning. This article revisits the most important derivations featured in that examination series, presenting them side by side in English and Chinese to reinforce conceptual understanding.
2022 年 6 月 A‑Level 物理第四单元的评分方案突显了从基本原理推导关键关系式的能力。考生不能只是背出最终公式，而需要运用基本定律和清晰的逻辑一步步地论证。本文重访该考季中最重要的推导，并以中英对照的形式呈现，以加深对概念的理解。
1. Centripetal Acceleration Derivation | 向心加速度推导
Consider an object moving with constant speed v in a circle of radius r. In a short time Δt, the object moves through a small angle Δθ, and its velocity vector turns by the same angle while keeping magnitude unchanged.
考虑一个物体以恒定速率 v 在半径为 r 的圆周上运动。在很短的时间 Δt 内，物体转过一个小角度 Δθ，其速度矢量也转过相同角度，大小保持不变。
From the velocity triangle, the magnitude of the change in velocity is approximately Δv ≈ v Δθ. The distance travelled along the arc is Δs = r Δθ, so v = Δs/Δt = r Δθ/Δt.
由速度矢量三角形，速度变化的大小近似为 Δv ≈ v Δθ。走过的弧长 Δs = r Δθ，因此速率 v = Δs/Δt = r Δθ/Δt。
The magnitude of the centripetal acceleration is a = Δv/Δt = (v Δθ)/Δt. Using Δθ/Δt = v/r gives:
向心加速度的大小为 a = Δv/Δt = (v Δθ)/Δt。代入 Δθ/Δt = v/r 得：
a = v²/r = ω²r
The direction of acceleration is always towards the centre of the circle, as Δv points radially inward.
加速度的方向始终指向圆心，因为 Δv 沿径向向内。
2. Simple Harmonic Motion: Defining Equation and Solution | 简谐运动：定义方程与解
For a mass on a spring obeying Hooke’s law, the restoring force is F = –kx. By Newton’s second law, F = ma, hence ma = –kx, giving a = –(k/m)x.
对于遵循胡克定律的弹簧重物，回复力 F = –kx。根据牛顿第二定律 F = ma，得 ma = –kx，即 a = –(k/m)x。
Define ω² = k/m, so the defining equation of SHM becomes a = –ω²x. This is a second-order differential equation: d²x/dt² = –ω²x.
定义 ω² = k/m，则 SHM 的定义方程化为 a = –ω²x。这是一个二阶微分方程：d²x/dt² = –ω²x。
A trial solution x = A cos(ωt) satisfies the equation: dx/dt = –ωA sin(ωt), d²x/dt² = –ω²A cos(ωt) = –ω²x. Adding a phase constant φ gives the general solution:
试探解 x = A cos(ωt) 满足方程：dx/dt = –ωA sin(ωt)，d²x/dt² = –ω²A cos(ωt) = –ω²x。加入初相 φ 后得到通解：
x = A cos(ωt + φ)
Velocity and acceleration follow as v = –ωA sin(ωt + φ) and a = –ω²A cos(ωt + φ). The maximum speed is vₘₐₓ = ωA.
速度和加速度分别为 v = –ωA sin(ωt + φ)，a = –ω²A cos(ωt + φ)。最大速率 vₘₐₓ = ωA。
3. Period of a Mass‑Spring System | 弹簧振子的周期
Using the definition ω² = k/m and the relationship ω = 2π/T for periodic motion, we obtain 2π/T = √(k/m). Rearranging gives the period:
利用 ω² = k/m 以及周期运动关系 ω = 2π/T，得 2π/T = √(k/m)。整理后得到周期：
T = 2π √(m/k)
This derivation shows that the period depends only on the mass and the spring constant, not on the amplitude of oscillation.
该推导表明周期仅由质量和弹簧劲度系数决定，与振幅无关。
4. Period of a Simple Pendulum | 单摆周期
For a small angular displacement θ, the restoring force along the arc is mg sin θ ≈ mg θ. The tangential acceleration is a = –g θ, and using arc length s = l θ gives a = –(g/l)s.
在小角度位移 θ 时，沿弧长的回复力为 mg sin θ ≈ mg θ。切向加速度 a = –g θ，利用弧长 s = l θ 得 a = –(g/l)s。
This has the form a = –ω²s with ω² = g/l. Thus ω = √(g/l) and the period is:
这是 a = –ω²s 的形式，其中 ω² = g/l。因此 ω = √(g/l)，周期为：
T = 2π √(l/g)
The small-angle approximation is valid for θ less than about 10°, ensuring the motion is approximately simple harmonic.
小角度近似在 θ 约小于 10° 时成立，确保运动近似为简谐运动。
5. Capacitor Discharge Equation | 电容器放电方程
During discharge, the p.d. across the capacitor equals that across the resistor: Vc = VR. Since Vc = Q/C and VR = IR, we have Q/C = IR. The current is the rate of decrease of charge, I = –dQ/dt.
放电时，电容器两端电压等于电阻两端电压：Vc = VR。由 Vc = Q/C 和 VR = IR，得 Q/C = IR。电流是电荷减少的速率，I = –dQ/dt。
Substituting gives –dQ/dt = Q/(RC), which separates to dQ/Q = –dt/(RC). Integrating from Q₀ at t=0 to Q at time t:
代入得 –dQ/dt = Q/(RC)，分离变量为 dQ/Q = –dt/(RC)。从 t=0 时 Q₀ 积分到 t 时刻的 Q：
Q = Q₀ e–t/RC
The time constant τ = RC is the time for the charge to fall to 1/e of its initial value. Voltage and current decay similarly: V = V₀ e–t/RC, I = I₀ e–t/RC.
时间常数 τ = RC，是电荷衰减到初始值 1/e 所需的时间。电压和电流也以同样方式衰减：V = V₀ e–t/RC，I = I₀ e–t/RC。
6. Capacitor Charging Equation | 电容器充电方程
When charging from a cell of e.m.f. V₀ through a resistor R, Kirchhoff’s loop rule gives V₀ = Q/C + IR. Substituting I = dQ/dt yields V₀ = Q/C + R dQ/dt.
对电动势为 V₀ 的电池通过电阻 R 给电容器充电，由基尔霍夫回路定律得 V₀ = Q/C + IR。代入 I = dQ/dt 得 V₀ = Q/C + R dQ/dt。
Rearrange to dQ/dt = (V₀C – Q)/(RC). Separating variables and integrating from Q=0 at t=0 leads to:
整理为 dQ/dt = (V₀C – Q)/(RC)。分离变量并从 t=0、Q=0 积分得：
Q = Q₀ (1 – e–t/RC)
where Q₀ = CV₀ is the maximum charge. The voltage across the capacitor rises as Vc = V₀ (1 – e–t/RC), and the charging current falls as I = (V₀/R) e–t/RC.
其中 Q₀ = CV₀ 是最大电荷量。电容器电压按 Vc = V₀ (1 – e–t/RC) 上升，充电电流按 I = (V₀/R) e–t/RC 衰减。
7. Gravitational Potential Energy | 引力势能
The gravitational force on a mass m at distance r from a mass M is F = GMm/r² (attractive). To bring m from infinity to a distance r, the work done by the external agent against the field is:
质量为 m 的物体距质量 M 为 r 时所受引力为 F = GMm/r²（吸引）。将 m 从无穷远移到距离 r 处，外力反抗引力场做的功为：
W = ∫∞r (–F) dr = ∫∞r (GMm/r²) dr = [–GMm/r]∞r = –GMm/r
This work is stored as gravitational potential energy, U = –GMm/r. The negative sign indicates that work must be done to separate the masses to infinity.
此功被储存为引力势能，U = –GMm/r。负号表示要将两物体分离至无穷远需要做功。
Near Earth’s surface, the change ΔU = mgΔh is obtained by expanding U around r = Rₑ and keeping the linear term.
在地表附近，通过对 U 在 r = Rₑ 处展开并保留线性项，得到变化量 ΔU = mgΔh。
8. Electric Potential of a Point Charge | 点电荷的电势
The electric field due to a point charge Q is E = kQ/r² radially outward. The potential difference V between a point at distance r and infinity is the work done per unit test charge moving against the field:
点电荷 Q 的电场为 E = kQ/r² 沿径向向外。距离 r 处与无穷远之间的电势差 V，是将单位检验电荷反抗电场移动所做的功：
V = – ∫∞r E dr = – ∫∞r (kQ/r²) dr = [kQ/r]∞r = kQ/r
For a positive Q, the potential is positive and decreases as 1/r. The absolute potential at infinity is taken as zero.
对于正的 Q，电势为正，并按 1/r 递减。定义无穷远处的电势为零。
9. Adiabatic Process for an Ideal Gas | 理想气体的绝热过程
In an adiabatic process, no heat enters or leaves the system (dQ = 0). The first law of thermodynamics gives dU = –dW. For an ideal gas, dU = nCᵥ dT and dW = p dV.
绝热过程中系统与外界无热交换（dQ = 0）。热力学第一定律给出 dU = –dW。对理想气体，dU = nCᵥ dT，dW = p dV。
Thus nCᵥ dT = –p dV. Using the ideal gas equation pV = nRT, we can replace dT by differentiating: p dV + V dp = nR dT. After eliminating dT and using R = Cp – Cᵥ and γ = Cp/Cᵥ, we obtain:
因此 nCᵥ dT = –p dV。利用理想气体方程 pV = nRT，通过微分代换 dT：p dV + V dp = nR dT。消去 dT 并利用 R = Cp – Cᵥ 及 γ = Cp/Cᵥ，可得：
pVγ = constant
Equivalently, TVγ–1 = constant and Tγ p1–γ = constant. This derivation is a standard requirement in the Unit 4 mark scheme when analysing Carnot cycles or expansions.
等价的表达式有 TVγ–1 = 常数 以及 Tγ p1–γ = 常数。这一推导在 Unit 4 评分方案分析卡诺循环或膨胀过程中是常见要求。
10. Total Energy in Simple Harmonic Motion | 简谐运动的总能量
The kinetic energy of the oscillating mass is K = ½ m v². Using v = –ωA sin(ωt + φ) gives K = ½ m ω² A² sin²(ωt + φ).
振子的动能为 K = ½ m v²。代入 v = –ωA sin(ωt + φ) 得 K = ½ m ω² A² sin²(ωt + φ)。
The potential energy stored in the spring is U = ½ k x² = ½ m ω² A² cos²(ωt + φ), since k = m ω².
弹簧储存的势能为 U = ½ k x² = ½ m ω² A² cos²(ωt + φ)，因为 k = m ω²。
Adding the two, sin² + cos² = 1, so the total mechanical energy is constant:
两者相加，sin² + cos² = 1，因此总机械能守恒：
E = K + U = ½ m ω² A²
This result is independent of time and shows that the energy is proportional to the square of the amplitude.
该结果与时间无关，表明能量与振幅的平方成正比。
11. Motional E.M.F. Derivation (Faraday’s Law) | 动生电动势推导（法拉第定律）
A conductor of length l moving with velocity v perpendicular to a uniform magnetic field B cuts magnetic flux. In time Δt, the area swept is l v Δt, so the change in flux is ΔΦ = B l v Δt.
长度为 l 的导体以速度 v 垂直穿过匀强磁场 B 时切割磁感线。在时间 Δt 内，扫过的面积为 l v Δt，因此磁通量变化为 ΔΦ = B l v Δt。
According to Faraday’s law, the induced e.m.f. is the rate of change of flux linkage: ε = – dΦ/dt. For a single conductor, this simplifies to:
根据法拉第定律，感应电动势等于磁链变化率的负值：ε = – dΦ/dt。对于单根导体，简化为：
ε = B l v
The direction of the induced e.m.f. is given by Fleming’s right-hand rule or Lenz’s law, which accounts for the negative sign in the formal law.
感应电动势的方向由弗莱明右手定则或楞次定律判定，对应公式中的负号。
Published by TutorHao | Physics Revision Series | aleveler.com
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