📚 A-Level Physics Unit 4 Mark Scheme Jan21 Formula Derivation | A-Level物理单元4 2021年1月评分方案公式推导
The January 2021 Edexcel A-Level Physics Unit 4 paper tested a wide range of derivations essential for a deep understanding of mechanics, fields, and particles. Mark schemes reward clear logical steps, correct use of physical laws, and precise algebra. This article revisits key derivations often required in such examinations, with full bilingual explanations to help you master the methods expected by examiners.
2021年1月爱德思A-Level物理第四单元考试涵盖了力学、场和粒子物理中必须掌握的推导过程。评分方案奖励清晰的逻辑步骤、正确运用物理定律以及精准的代数运算。本文将重新梳理此类考试中常考的关键公式推导，配合完整的中英双语讲解，帮助同学们掌握考官所期待的推导方法。
1. Centripetal Acceleration: a = v²/r | 向心加速度推导
Consider an object moving at constant speed v in a circle of radius r. In a short time Δt, the object moves from point P to Q through an angle Δθ. The velocity vectors at P and Q have the same magnitude v but different directions. The change in velocity Δv points approximately towards the centre. From the vector triangle, Δv = vΔθ (for small Δθ).
考虑一个物体以恒定速率v在半径为r的圆上运动。在短时间Δt内，物体从P点运动到Q点，转过角度Δθ。P点和Q点的速度大小相等，但方向不同，速度变化量Δv的方向近似指向圆心。由速度矢量三角形可得，对于小角度Δθ，Δv = vΔθ。
Since Δθ = vΔt/r, we have Δv = v(vΔt/r) = v²Δt/r. Acceleration a = Δv/Δt = v²/r. The direction of this acceleration is towards the centre of the circle, justifying the expression for centripetal acceleration.
因为Δθ = vΔt/r，所以Δv = v(vΔt/r) = v²Δt/r。加速度a = Δv/Δt = v²/r。该加速度方向指向圆心，这就是向心加速度公式的推导。
2. Deriving the Period of Circular Motion: T = 2πr/v | 圆周运动周期推导
From the definition of speed, distance travelled in one revolution is the circumference 2πr. Time taken is the period T. Hence v = 2πr/T, which rearranges to T = 2πr/v.
根据速度的定义，物体运动一周的路程为周长2πr，所需时间为周期T。因此v = 2πr/T，整理可得T = 2πr/v。
Combining with a = v²/r, one can also express the period in terms of centripetal acceleration: T = 2π√(r/a) or for gravitational orbits T² ∝ r³ (Kepler’s third law).
结合向心加速度公式a = v²/r，还可以用加速度表示周期：T = 2π√(r/a)。对于引力轨道，这直接导致T² ∝ r³（开普勒第三定律）。
3. Simple Harmonic Motion: a = -ω²x Displacement Derivation | 简谐运动位移公式推导
Simple harmonic motion (SHM) is defined by the condition that acceleration is directly proportional to displacement from equilibrium and always directed opposite to displacement: a = -kx. For SHM, the solution to the differential equation d²x/dt² = -ω²x is x = A cos(ωt) or x = A sin(ωt), where ω is the angular frequency. Differentiating twice confirms this: v = dx/dt = -Aω sin(ωt) and a = d²x/dt² = -Aω² cos(ωt) = -ω²x.
简谐运动定义为加速度与离开平衡位置的位移成正比且方向相反：a = -kx。对于SHM，微分方程d²x/dt² = -ω²x的解为x = A cos(ωt)或x = A sin(ωt)，其中ω为角频率。对位移公式求导两次即可验证：v = dx/dt = -Aω sin(ωt)，a = d²x/dt² = -Aω² cos(ωt) = -ω²x。
The mark scheme often expects the linking of maximum acceleration amax = ω²A and the relationship between ω, period T, and frequency f: ω = 2π/T = 2πf.
评分方案通常要求关联最大加速度amax = ω²A，以及ω与周期T、频率f的关系：ω = 2π/T = 2πf。
4. Gravitational Field Strength: g = GM/r² from Newton’s Law | 引力场强度公式推导
Newton’s law of gravitation states that the force between two point masses M and m separated by distance r is F = GMm/r². The gravitational field strength g at a point is defined as the force per unit mass acting on a small test mass placed at that point: g = F/m.
牛顿万有引力定律指出，相距r的两个质点M和m之间的引力为F = GMm/r²。引力场强度g定义为在该点放置一个小检验质量时单位质量所受的力：g = F/m。
Substituting for F gives g = (GMm/r²)/m = GM/r². This expression is valid for the field outside a spherical mass, treating the mass as if concentrated at its centre. In a radial field, g follows an inverse square law.
代入F的表达式得g = (GMm/r²)/m = GM/r²。该公式适用于球形质量外部的引力场，将质量视为集中于球心。在径向场中，g遵循平方反比律。
5. Electric Field Strength: E = kQ/r² and E = F/q | 电场强度公式推导
For a point charge Q, Coulomb’s law gives the force on a test charge q as F = kQq/r², where k = 1/(4πε₀). Electric field strength E is defined as the force per unit positive charge: E = F/q. Hence E = kQ/r². This shows that the electric field due to a point charge also obeys the inverse square law.
对于点电荷Q，库仑定律给出检验电荷q所受的力为F = kQq/r²，其中k = 1/(4πε₀)。电场强度E定义为单位正电荷所受的力：E = F/q。因此E = kQ/r²，表明点电荷产生的电场同样遵循平方反比律。
In a uniform electric field between parallel plates, E = V/d is derived from the work done moving a charge between plates: W = Fd = qEd, and also W = qV, equating gives E = V/d.
在平行板间的匀强电场中，E = V/d的推导基于移动电荷做功：W = Fd = qEd，同时W = qV，联立即得E = V/d。
6. Capacitor Discharge: Exponential Decay Derivation | 电容器放电指数衰减推导
When a capacitor of capacitance C discharges through a resistor R, the current I at any instant is I = -dQ/dt. By definition, V = Q/C and also V = IR, so Q/C = -R dQ/dt. This first-order differential equation dQ/dt = -Q/(RC) integrates to Q = Q₀ e^(-t/RC).
当电容C通过电阻R放电时，任意时刻的电流为I = -dQ/dt。根据定义，V = Q/C且V = IR，所以Q/C = -R dQ/dt。这个一阶微分方程dQ/dt = -Q/(RC)积分后得到Q = Q₀ e^(-t/RC)。
From Q = CV, the voltage decay is V = V₀ e^(-t/RC) and current decay I = I₀ e^(-t/RC). The time constant τ = RC appears naturally. The mark scheme often asks for the derivation of half-life t₁/₂ = RC ln 2.
由Q = CV可得电压衰减规律V = V₀ e^(-t/RC)，电流衰减I = I₀ e^(-t/RC)。时间常数τ = RC自然出现。评分方案常要求推导半衰期t₁/₂ = RC ln 2。
7. Magnetic Force on a Moving Charge: F = Bqv and Circular Path Radius | 运动电荷在磁场中的受力与半径推导
A charge q moving with velocity v perpendicular to a uniform magnetic field B experiences a force given by Fleming’s left-hand rule: F = Bqv sinθ. When θ = 90°, F = Bqv. This force acts as a centripetal force, causing the charge to follow a circular path: Bqv = mv²/r.
电荷q以速度v垂直于匀强磁场B运动时，受到洛伦兹力F = Bqv sinθ。当θ = 90°时，F = Bqv。这个力充当向心力，使电荷做圆周运动：Bqv = mv²/r。
Rearranging gives the radius of the path:
r = mv/(Bq)
. The period of revolution is T = 2πr/v = 2πm/(Bq), independent of speed – a key result used in cyclotron design.
整理可得轨道半径公式：
r = mv/(Bq)
。回转周期T = 2πr/v = 2πm/(Bq)，与速度无关，这是回旋加速器设计的关键。
8. Mass–Energy Equivalence: E = mc² in Nuclear Reactions | 质能方程与核反应推导
Einstein’s mass–energy relation E = mc² states that mass can be converted into energy and vice versa. In nuclear decays, the mass defect Δm – the difference between the mass of a nucleus and the sum of the masses of its separate nucleons – is equivalent to the binding energy. For a reaction, the energy released Q = Δm c².
爱因斯坦质能方程E = mc²表明质量与能量可以相互转化。在核衰变中，质量亏损Δm——原子核质量与组成它的核子质量之和的差值——等价于结合能。对于核反应，释放的能量Q = Δm c²。
The mark scheme expects unit conversions: 1 u (atomic mass unit) = 931.5 MeV/c². Derivation: 1 u = 1.6605×10⁻²⁷ kg, so E = (1.6605×10⁻²⁷)(2.9979×10⁸)² J ≈ 1.492×10⁻¹⁰ J. Converting to eV gives 931.5 MeV.
评分方案期望能够进行单位换算：1 u（原子质量单位）= 931.5 MeV/c²。推导：1 u = 1.6605×10⁻²⁷ kg，故E = (1.6605×10⁻²⁷)(2.9979×10⁸)² J ≈ 1.492×10⁻¹⁰ J。转换为电子伏即得931.5 MeV。
9. Conservation Laws in Particle Interactions: Momentum and Charge | 粒子相互作用中的守恒律推导
In particle physics, all interactions must obey conservation of charge, baryon number, lepton number, and energy–momentum. For example, in β⁻ decay: n → p + e⁻ + ν̄ₑ, charge is conserved: 0 = +1 -1 +0. Baryon number: 1 = 1 + 0 + 0. Lepton number: 0 = 0 + 1 -1. The Q-value is derived from mass differences.
在粒子物理中，所有相互作用都必须满足电荷数、重子数、轻子数和能量–动量守恒。例如β⁻衰变：n → p + e⁻ + ν̄ₑ，电荷守恒：0 = +1 -1 +0。重子数：1 = 1 + 0 + 0。轻子数：0 = 0 + 1 -1。反应能Q值由质量差推导。
The mark scheme in Unit 4 often includes questions requiring students to apply these conservation rules to unfamiliar particle decays, deduce unknown particles, and calculate kinetic energy sharing between products using conservation of momentum.
第四单元评分方案经常包含要求学生运用这些守恒规则推断未知粒子，并利用动量守恒计算生成物之间动能分配的题目。
10. Stress, Strain and the Young Modulus: E = σ/ε | 应力应变与杨氏模量推导
Stress σ is defined as force per unit cross-sectional area: σ = F/A. Strain ε is the extension per unit original length: ε = ΔL/L₀. For a material obeying Hooke’s law within the limit of proportionality, stress is proportional to strain: σ = E ε, where E is the Young modulus.
应力σ定义为单位横截面积上的力：σ = F/A。应变ε定义为单位原长的伸长量：ε = ΔL/L₀。对于在比例极限内服从胡克定律的材料，应力与应变成正比：σ = E ε，其中E为杨氏模量。
Combining with the spring constant k: F = kΔL, and using the definitions, one can derive E = (kL₀)/A. This derivation links macroscopic stiffness to microscopic material property.
结合弹簧常数k：F = kΔL，并运用上述定义，可以推导出E = (kL₀)/A。这一推导将宏观刚度与微观材料属性联系起来。
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