📚 A2 Physics: Formula Derivations from the International Scheme of Work 9630 | A2物理：国际教学方案9630核心公式推导
In A2 Physics, understanding the mathematical origins of key relationships is essential for mastering advanced concepts. This article presents step‑by‑step derivations drawn from the International Scheme of Work 9630, covering mechanics, fields, electricity, nuclear physics and thermodynamics. Each section pairs an English explanation with its Chinese counterpart, allowing learners to build both conceptual understanding and examination confidence.
在A2物理中，理解核心关系的数学来源是掌握高级概念的关键。本文依据国际教学方案9630，精选力学、场、电学、核物理和热学中的重要公式，进行逐步推导。每一节均为英文说明与中文对照，帮助学习者建立概念理解与应试信心。
1. Centripetal Acceleration: a = v²/r = rω² | 向心加速度推导
An object moving with constant speed v in a circle of radius r undergoes continuous change in direction. In a short time Δt, the velocity vector rotates through a small angle Δθ, and the magnitude of the change in velocity is Δv ≈ vΔθ. Since Δθ = (vΔt)/r, we obtain Δv = v²Δt/r. Acceleration is defined as a = Δv/Δt, giving a = v²/r. Substituting v = rω yields the alternative form a = rω².
物体以恒定速率 v 在半径为 r 的圆周上运动时，方向不断改变。在很短的时间 Δt 内，速度矢量转过小角度 Δθ，速度变化量的大小为 Δv ≈ vΔθ。因为 Δθ = (vΔt)/r，可得 Δv = v²Δt/r。加速度定义为 a = Δv/Δt，因此 a = v²/r。代入 v = rω 即得另一形式 a = rω²。
a = v²/r = rω²
This derivation uses vector subtraction and the small‑angle approximation; it shows that centripetal acceleration always points towards the centre of the circle.
该推导运用矢量减法和小角度近似，表明向心加速度始终指向圆心。
2. Simple Harmonic Motion: a = –ω²x and Displacement Equation | 简谐运动加速度与位移关系
In SHM the restoring force is proportional to displacement: F = –kx. Applying Newton’s second law, ma = –kx, hence a = –(k/m)x. Defining the constant ω² = k/m gives the defining equation a = –ω²x. The negative sign indicates that acceleration is always directed towards the equilibrium position. Solving this second‑order differential equation yields x = A sin(ωt + φ), where A is amplitude and φ the phase constant.
简谐运动中回复力与位移成正比：F = –kx。由牛顿第二定律 ma = –kx，得 a = –(k/m)x。定义常数 ω² = k/m，即得到定义式 a = –ω²x。负号表明加速度始终指向平衡位置。求解此二阶微分方程可得 x = A sin(ωt + φ)，其中 A 为振幅，φ 为初相。
a = –ω²x x = A sin(ωt + φ)
The period T = 2π/ω follows directly from the argument of the sine function completing 2π radians in one cycle.
周期 T = 2π/ω 可直接从正弦函数的辐角在一个周期内变化 2π 得出。
3. Capacitor Discharge: Q = Q₀ e⁻ᵗ⁄ᴿᶜ | 电容放电公式推导
Consider a capacitor C discharging through a resistor R. At any instant, the p.d. across the capacitor Q/C equals the p.d. across the resistor IR. Defining current as the rate of decrease of charge, I = –dQ/dt, Kirchhoff’s loop rule gives Q/C – IR = 0? Actually around the loop: Q/C – IR = 0, so Q/C = IR. Substituting I = –dQ/dt leads to Q/C = –R dQ/dt, or dQ/dt = –Q/(RC).
考虑电容 C 通过电阻 R 放电。任一时刻，电容两端电压 Q/C 等于电阻两端电压 IR。将电流定义为电荷的减少率 I = –dQ/dt，回路电压方程为 Q/C = IR。代入 I 得到 Q/C = –R dQ/dt，即 dQ/dt = –Q/(RC)。
Separating variables and integrating from Q₀ at t = 0 to Q at time t: ∫ dQ/Q = –(1/RC) ∫ dt, giving ln(Q/Q₀) = –t/(RC). Therefore Q = Q₀ e⁻ᵗ⁄ᴿᶜ. The time constant τ = RC represents the time for the charge to fall to 1/e of its initial value.
分离变量并从 t=0 时 Q₀ 积分至 t 时刻 Q：∫ dQ/Q = –(1/RC) ∫ dt，得 ln(Q/Q₀) = –t/(RC)，因此 Q = Q₀ e⁻ᵗ⁄ᴿᶜ。时间常数 τ = RC 表示电荷衰减至初始值 1/e 所需的时间。
Q = Q₀ e⁻ᵗ⁄ᴿᶜ
4. Radioactive Decay Law: N = N₀ e⁻λᵗ | 放射性衰变定律
The activity A of a radioactive sample is proportional to the number of undecayed nuclei N: A = λN. Since activity is the rate of decay, A = –dN/dt (decay reduces N). Equating gives dN/dt = –λN. This is a first‑order differential equation identical in form to the capacitor discharge equation.
放射性样品的活度 A 与未衰变原子核数目 N 成正比：A = λN。因为活度即衰变率，A = –dN/dt（衰变使 N 减少）。联立得 dN/dt = –λN。这是一阶微分方程，形式与电容放电方程相同。
Integration yields ln(N/N₀) = –λt, hence N = N₀ e⁻λᵗ. The decay constant λ is related to half‑life T₁/₂ by λ = ln2 / T₁/₂.
积分得 ln(N/N₀) = –λt，故 N = N₀ e⁻λᵗ。衰变常量 λ 与半衰期 T₁/₂ 的关系为 λ = ln2 / T₁/₂。
N = N₀ e⁻λᵗ λ = ln2 / T₁/₂
5. Faraday’s Law and Induced EMF: ε = –dΦ/dt | 法拉第电磁感应定律
Faraday’s law states that the magnitude of the induced emf in a circuit is equal to the rate of change of magnetic flux linkage. For a single loop, ε = –dΦ/dt. The minus sign indicates Lenz’s law: the induced current opposes the change in flux. If a coil has N turns, the flux linkage is NΦ and ε = –N dΦ/dt.
法拉第定律指出，回路中感应电动势的大小等于磁通匝链数的变化率。对单匝线圈，ε = –dΦ/dt。负号表示楞次定律：感应电流阻碍磁通量的变化。若线圈有 N 匝，磁通匝链数为 NΦ，则 ε = –N dΦ/dt。
A special case is motional emf: a conductor of length l moving perpendicularly across a uniform magnetic field B with speed v induces emf ε = Blv. This can be derived from ε = dΦ/dt where Φ = BA = Blx, giving ε = Bl dx/dt = Blv.
特例是动生电动势：长度为 l 的导体以速度 v 垂直切割匀强磁场 B 时，感应电动势 ε = Blv。可由 ε = dΦ/dt 导出：Φ = BA = Blx，故 ε = Bl dx/dt = Blv。
ε = –N dΦ/dt (ε = Blv)
6. Ideal Gas Pressure: p = ⅓ ρ ⟨c²⟩ | 理想气体压强推导
Consider a single molecule of mass m moving with velocity component vₓ perpendicular to a wall of area A in a cubic container of side L. It collides elastically, changing momentum by Δp = 2mvₓ, and the time between collisions with the same wall is 2L/vₓ. The average force on the wall due to this molecule is F₁ = Δp/Δt = mvₓ²/L.
考虑单个质量为 m 的气体分子，以垂直于面积为 A 的器壁的速度分量 vₓ 在边长为 L 的立方容器中运动。弹性碰撞使动量改变 Δp = 2mvₓ，两次碰撞同一壁的时间间隔为 2L/vₓ。该分子对器壁的平均作用力为 F₁ = Δp/Δt = mvₓ²/L。
For N molecules, the total force is F = (m/L) ∑ vₓᵢ². Defining the mean square speed in the x‑direction ⟨vₓ²⟩ = (1/N) ∑ vₓᵢ², and using the fact that ⟨c²⟩ = ⟨vₓ²⟩ + ⟨v_y²⟩ + ⟨v_z²⟩ = 3⟨vₓ²⟩ for random motion, we obtain F = (m/L) N ⟨c²⟩/3. Pressure p = F/A = F/L² = (1/3)(Nm/V)⟨c²⟩. Since Nm/V is density ρ, p = ⅓ ρ ⟨c²⟩.
对于 N 个分子，总力 F = (m/L) ∑ vₓᵢ²。定义 x 方向方均速率 ⟨vₓ²⟩ = (1/N) ∑ vₓᵢ²，并利用随机运动中 ⟨c²⟩ = ⟨vₓ²⟩ + ⟨v_y²⟩ + ⟨v_z²⟩ = 3⟨vₓ²⟩，得 F = (m/L) N ⟨c²⟩/3。压强 p = F/A = F/L² = (1/3)(Nm/V)⟨c²⟩。因 Nm/V 即密度 ρ，故 p = ⅓ ρ ⟨c²⟩。
p = ⅓ ρ ⟨c²⟩ = ⅓ (Nm/V) ⟨c²⟩
7. Gravitational Potential: V = –GM/r | 引力势推导
Gravitational potential 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 is F = GMm/r², directed towards the central mass M. Work done by an external agent against gravity when moving from infinity to a distance r is W = –∫∞ʳ (GMm/r²) dr = –[–GMm/r]∞ʳ = –GMm/r. Dividing by m gives potential V = –GM/r.
引力势定义为将单位质量从无穷远处移至该点外力所做的功。万有引力 F = GMm/r²，指向中心质量 M。从无穷远移到距离 r 处，外力克服引力做功 W = –∫∞ʳ (GMm/r²) dr = –[–GMm/r]∞ʳ = –GMm/r。除以 m 得势 V = –GM/r。
The negative sign indicates that work is done by the gravitational field as the mass approaches. The potential is zero at infinity.
负号表明当质量靠近时引力场做正功。无穷远处势能为零。
Vgrav = –GM/r
8. Electric Potential due to a Point Charge: V = kQ/r | 点电荷电势
By analogy with gravity, the electric potential at a distance r from a point charge Q is the work done per unit positive charge brought from infinity. The electrostatic force is F = kQq/r², where k = 1/(4πε₀). The work done is W = –∫∞ʳ (kQq/r²) dr = kQq/r. Dividing by q yields V = kQ/r. The potential is positive if Q is positive, and the zero is at infinity.
与引力类比，距离点电荷 Q 为 r 处的电势等于将单位正电荷从无穷远移至该点外力所做的功。静电力 F = kQq/r²，其中 k = 1/(4πε₀)。做功 W = –∫∞ʳ (kQq/r²) dr = kQq/r。除以 q 得 V = kQ/r。若 Q 为正，电势为正；零势点在无穷远。
Velec = kQ/r = Q/(4πε₀r)
9. RMS Values in AC: Irms = I₀/√2 | 交流电有效值推导
Alternating current varies sinusoidally: I = I₀ sin ωt. The power dissipated in a resistor R is P = I²R = I₀²R sin² ωt. Over a full cycle, the average value of sin² ωt is ½. Therefore the average power ⟨P⟩ = ½ I₀²R. The rms (root‑mean‑square) current is defined as the equivalent direct current that delivers the same average power: Irms²R = ⟨P⟩. Hence Irms²R = ½ I₀²R, giving Irms = I₀/√2. The same relation holds for voltage: Vrms = V₀/√2.
交流电流按正弦变化：I = I₀ sin ωt。电阻 R 上的功率为 P = I²R = I₀²R sin² ωt。在一个完整周期内，sin² ωt 的平均值为 ½。故平均功率 ⟨P⟩ = ½ I₀²R。有效值（方均根值）定义为产生相同平均功率的等效直流电流：Irms²R = ⟨P⟩。因此 Irms²R = ½ I₀²R，得 Irms = I₀/√2。电压同样有 Vrms = V₀/√2。
Irms = I₀/√2 Vrms = V₀/√2
10. Energy Stored in a Capacitor: E = ½ QV = ½ CV² | 电容器储能
During charging, the work done to move a small additional charge dq onto the capacitor when the potential difference is V = q/C is dW = V dq = (q/C) dq. Integrating from q = 0 to Q gives the total stored energy: E = ∫₀ᑫ (q/C) dq = [q²/(2C)]₀ᑫ = Q²/(2C). Using Q = CV gives the equivalent forms E = ½ QV = ½ CV².
充电过程中，当电容器两端电压为 V = q/C 时，迁移微小电荷 dq 所做的功为 dW = V dq = (q/C) dq。从 q = 0 积分至 Q，得总储能：E = ∫₀ᑫ (q/C) dq = [q²/(2C)]₀ᑫ = Q²/(2C)。利用 Q = CV 可得等价形式 E = ½ QV = ½ CV²。
E = ½ QV = ½ CV² = Q²/(2C)
Published by TutorHao | Physics Revision Series | aleveler.com
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