📚 A-Level Physics: Unit 4 Formula Derivations (Jan 2020 Mark Scheme) | A-Level 物理：单元4 2020年1月评分方案公式推导
Unit 4 of the A-Level Physics syllabus covers further mechanics, fields, and particles. In the January 2020 mark scheme, candidates were often required to derive key equations from first principles. This article revisits ten essential derivations that frequently appear in examinations, providing step‑by‑step reasoning in both English and Chinese. Mastering these derivations not only helps you secure high marks on structured questions but also deepens your understanding of the underlying physics.
A-Level 物理单元4涵盖进阶力学、场和粒子物理。在2020年1月的评分方案中，考生经常需要从基本原理出发推导关键公式。本文重现了考试中频繁出现的十个重要推导，并用英汉双语逐步解释。掌握这些推导不仅能帮助你在结构化题目中取得高分，还能加深你对物理本质的理解。
1. Centripetal Acceleration: a = v²/r | 向心加速度：a = v²/r
Consider an object moving with constant speed v along a circular path of radius r. Although the speed is constant, the direction of velocity changes continuously, resulting in a centripetal acceleration directed towards the centre.
考虑一个物体以恒定速率 v 沿半径为 r 的圆周运动。尽管速率不变，速度的方向时刻改变，由此产生指向圆心的向心加速度。
In a short time interval Δt, the object moves from point A to B, covering an arc length vΔt. The velocity vectors at A and B have equal magnitude v but differ in direction by angle Δθ. The change in velocity Δv is the vector difference, which points roughly toward the centre.
在很短的时间间隔 Δt 内，物体从 A 点运动到 B 点，经过弧长 vΔt。A 点和 B 点的速度矢量大小均为 v，但方向相差 Δθ 角度。速度变化量 Δv 是两个矢量的差，大致指向圆心。
For small Δθ, the magnitude of Δv is approximately v Δθ. The angle Δθ is related to the arc length by Δθ = vΔt / r. Therefore, Δv ≈ v × (vΔt / r) = v²Δt / r.
对于很小的 Δθ，Δv 的大小近似为 v Δθ。角度 Δθ 与弧长的关系为 Δθ = vΔt / r。因此 Δv ≈ v × (vΔt / r) = v²Δt / r。
The acceleration magnitude a = Δv / Δt gives a = v² / r. In terms of angular speed ω = v / r, we also have a = ω²r.
加速度大小 a = Δv / Δt，代入可得 a = v² / r。用角速度 ω = v / r 表示，也可写成 a = ω²r。
2. Coefficient of Restitution: e = (v₂ − v₁) / (u₁ − u₂) | 恢复系数：e = (v₂ − v₁) / (u₁ − u₂)
For a head‑on collision between two objects of masses m₁ and m₂, let u₁, u₂ be the initial velocities and v₁, v₂ the final velocities along a straight line. The coefficient of restitution e is defined as the ratio of relative speed after collision to relative speed before collision.
对于两个质量分别为 m₁ 和 m₂ 物体的正面碰撞，设初速度为 u₁、u₂，末速度为 v₁、v₂，沿同一直线。恢复系数 e 定义为碰撞后的相对速度与碰撞前的相对速度之比。
Newton’s experimental law states that (v₂ − v₁) = −e (u₂ − u₁). In practice, the magnitude is used: e = (v₂ − v₁) / (u₁ − u₂), assuming u₁ > u₂.
牛顿实验定律指出 (v₂ − v₁) = −e (u₂ − u₁)。实际使用时常取大小，若 u₁ > u₂，则 e = (v₂ − v₁) / (u₁ − u₂)。
Combining this with the conservation of linear momentum, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, we can derive expressions for v₁ and v₂ in terms of e. For a perfectly elastic collision, e = 1 and kinetic energy is conserved; for a perfectly inelastic collision, e = 0.
结合动量守恒 m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂，可以用 e 导出 v₁ 和 v₂ 的表达式。完全弹性碰撞时 e = 1，动能守恒；完全非弹性碰撞时 e = 0。
3. Capacitor Discharge: Q = Q₀ e⁻t/RC | 电容器放电：Q = Q₀ e⁻t/RC
When a charged capacitor of capacitance C discharges through a resistor R, the rate of charge loss is proportional to the charge remaining on the plates. By Kirchhoff’s voltage law, the p.d. across the capacitor equals the p.d. across the resistor: Q/C = IR.
当一个电容 C 已充电的电容器通过电阻 R 放电时，电荷减少的速率与极板上剩余的电荷成正比。根据基尔霍夫电压定律，电容器两端的电势差等于电阻两端的电势差：Q/C = IR。
Since current I is the rate of decrease of charge, I = −dQ/dt. Substituting gives −dQ/dt × R = Q/C, so dQ/dt = −Q/(RC).
因为电流 I 是电荷减少的速率，I = −dQ/dt。代入得 −dQ/dt × R = Q/C，因此 dQ/dt = −Q/(RC)。
This is a first‑order differential equation whose solution is Q(t) = Q₀ e⁻t/RC, where Q₀ is the initial charge at t = 0. The time constant τ = RC is the time taken for the charge to fall to 1/e of its initial value.
这是一个一阶微分方程，其解为 Q(t) = Q₀ e⁻t/RC，其中 Q₀ 是 t = 0 时的初始电荷。时间常数 τ = RC 是电荷衰减到初始值的 1/e 所需的时间。
The same exponential decay describes the current I = I₀ e⁻t/RC and the p.d. V = V₀ e⁻t/RC during discharge.
同样的指数衰减规律也适用于放电过程中的电流 I = I₀ e⁻t/RC 和电压 V = V₀ e⁻t/RC。
4. Uniform Electric Field: E = V/d | 匀强电场：E = V/d
Consider two parallel conducting plates separated by a distance d, with a potential difference V maintained between them. The work done by the electric field in moving a charge q from one plate to the other is W = qV.
考虑两块相距为 d 的平行导体板，板间维持电势差 V。电场将电荷 q 从一块板移动到另一块板所做的功为 W = qV。
The force on the charge due to a uniform electric field of strength E is F = qE. If the charge moves along the field lines from the positive to the negative plate, the work done is also F × d = qE d.
电荷在强度为 E 的匀强电场中所受的力为 F = qE。如果电荷沿电场线从正极板移动到负极板，做功也可以表示为 F × d = qE d。
Equating the two expressions for work, qV = qE d, gives the relation E = V/d. This formula is valid only for a uniform field, such as that between parallel plates with small separation compared to their area.
令两种做功表达式相等，qV = qE d，即可得到 E = V/d。该公式只适用于匀强电场，例如板间距远小于板面积的平行板之间的电场。
5. Hall Voltage: VH = B I / (n q t) | 霍尔电压：VH = B I / (n q t)
In a thin conducting slab of thickness t, carrying a current I and placed in a perpendicular magnetic field B, charge carriers experience a magnetic force FB = q v B, where v is the drift velocity. This force deflects carriers to one side, creating a transverse electric field.
在一厚度为 t 的薄导电片中，通以电流 I，并置于垂直磁场 B 中，电荷载流子受到磁力 FB = q v B 的作用，其中 v 是漂移速度。这个力将载流子偏转到一侧，形成一个横向电场。
The resulting electric field exerts an opposing force FE = q EH = q (VH / w), where w is the slab width. At equilibrium, the magnetic force equals the electric force: q v B = q VH / w, so VH = v B w.
产生的电场施加反向力 FE = q EH = q (VH / w)，w 为片的宽度。平衡时，磁力与电场力相等：q v B = q VH / w，因此 VH = v B w。
Current I is given by I = n A v q, where n is the number density of charge carriers and A = w t is the cross‑sectional area. Hence v = I / (n w t q). Substituting this into the expression for VH yields VH = (B w) × [I / (n w t q)] = B I / (n q t).
电流 I 由 I = n A v q 给出，n 为载流子数密度，A = w t 为横截面积。因此 v = I / (n w t q)。将此式代入 VH 表达式中，得到 VH = (B w) × [I / (n w t q)] = B I / (n q t)。
The Hall voltage is thus proportional to the product of magnetic field and current, and inversely proportional to the charge carrier density and slab thickness. Its sign indicates the type of charge carriers.
因此，霍尔电压与磁感应强度和电流的乘积成正比，与载流子密度和板厚度成反比。其正负符号可指示载流子类型。
6. Radius of Curvature in a Magnetic Field: r = m v / (B q) | 磁场中的曲率半径：r = m v / (B q)
When a charged particle of mass m, charge q, and speed v enters a uniform magnetic field of flux density B perpendicularly, it experiences a magnetic force F = B q v. This force is always perpendicular to the velocity, so it acts as a centripetal force.
当一个质量为 m、电荷为 q、速率为 v 的带电粒子垂直进入磁感应强度为 B 的匀强磁场时，它会受到磁力 F = B q v 的作用。这个力始终垂直于速度方向，因此充当向心力。
The centripetal force required for circular motion is Fc = m v² / r, where r is the radius of the circular path. Equating the two forces: B q v = m v² / r.
圆周运动所需的向心力为 Fc = m v² / r，r 为圆轨道半径。令两力相等：B q v = m v² / r。
Rearranging gives r = m v / (B q). The radius is directly proportional to the particle’s momentum m v and inversely proportional to both the magnetic field strength and the charge. The period T of the circular motion is independent of velocity: T = 2π m / (B q).
整理可得 r = m v / (B q)。半径与粒子的动量 m v 成正比，与磁场强度和电荷成反比。圆周运动的周期 T 与速度无关：T = 2π m / (B q)。
7. Faraday’s Law and AC Generator EMF: ε = −N dΦ/dt | 法拉第定律与交流发电机电动势：ε = −N dΦ/dt
Faraday’s law of electromagnetic induction states that the induced emf in a coil of N turns is equal to the negative rate of change of magnetic flux linkage: ε = −N dΦ/dt. The negative sign reflects Lenz’s law.
法拉第电磁感应定律指出，N 匝线圈中的感应电动势等于磁链变化率的负值：ε = −N dΦ/dt。负号反映了楞次定律。
In a simple AC generator, a coil rotates with constant angular speed ω in a uniform magnetic field B. If the coil area is A and at t = 0 its plane is perpendicular to the field, the flux through one turn is Φ = B A cos(ωt).
在简单交流发电机中，线圈在匀强磁场 B 中以恒定角速度 ω 旋转。若线圈面积为 A，且 t = 0 时线圈平面与磁场垂直，则单匝的磁通量为 Φ = B A cos(ωt)。
The induced emf for N turns is ε = −N d/dt [B A cos(ωt)] = N B A ω sin(ωt). The peak emf ε₀ occurs when sin(ωt) = 1, giving ε₀ = N B A ω. This sinusoidal variation is the basis of AC electricity generation.
N 匝线圈的感应电动势为 ε = −N d/dt [B A cos(ωt)] = N B A ω sin(ωt)。峰值电动势 ε₀ 出现在 sin(ωt) = 1 时，即 ε₀ = N B A ω。这种正弦变化是交流电发电的基础。
To increase the induced emf, one can increase the number of turns, the magnetic field strength, the coil area, or the angular frequency of rotation.
要增大感应电动势，可以增加线圈匝数、磁场强度、线圈面积或旋转角频率。
8. Energy Stored in a Capacitor: E = ½ C V² | 电容器储存的能量：E = ½ C V²
To charge a capacitor, work must be done to separate charges on the plates against the growing potential difference. Suppose a small amount of charge dq is moved through a potential difference V. The incremental work done is dW = V dq.
给电容器充电时，需要克服不断增大的电势差将电荷分离到极板上。设移动微小电荷量 dq 通过电势差 V，所做的元功为 dW = V dq。
Since V = q / C at any instant, we have dW = (q / C) dq. Integrating from zero initial charge to the final charge Q gives the total energy stored: E = ∫₀Q (q / C) dq = (1/2) Q² / C.
因为在任意时刻 V = q / C，所以有 dW = (q / C) dq。从初始零电荷积分到最终电荷量 Q，得到储存的总能量：E = ∫₀Q (q / C) dq = (1/2) Q² / C。
Using the relationship Q = C V, the energy can be expressed in other familiar forms: E = ½ C V² = ½ Q V. This energy is stored in the electric field between the plates.
利用关系式 Q = C V，能量可以表示为其他熟知的形式：E = ½ C V² = ½ Q V。该能量储存在两极板之间的电场中。
9. Simple Harmonic Motion: a = − ω² x | 简谐运动：a = − ω² x
An object executes simple harmonic motion (SHM) if its acceleration a is directly proportional to its displacement x from the equilibrium position and always directed towards it. The defining equation is a = − ω² x, where ω is the angular frequency.
如果一个物体的加速度 a 与其离开平衡位置的位移 x 成正比且方向总是指向平衡位置，则该物体做简谐运动（SHM）。其定义方程为 a = − ω² x，其中 ω 是角频率。
For a spring‑mass system, the restoring force is F = −k x, and by Newton’s second law m a = −k x, giving a = −(k/m) x. Hence ω² = k/m and the period T = 2π√(m/k).
对于弹簧‑振子系统，回复力为 F = −k x，根据牛顿第二定律 m a = −k x，得 a = −(k/m) x。因此 ω² = k/m，周期 T = 2π√(m/k)。
A standard solution to the differential equation d²x/dt² = − ω² x is x = A cos(ωt + φ), where A is the amplitude and φ is the phase constant. The velocity is v = dx/dt = −A ω sin(ωt + φ).
微分方程 d²x/dt² = − ω² x 的一个标准解为 x = A cos(ωt + φ)，其中 A 为振幅，φ 为初相。速度为 v = dx/dt = −A ω sin(ωt + φ)。
It follows that the maximum speed vmax = A ω, occurring at the equilibrium position, and the maximum acceleration amax = A ω², occurring at the extreme points. The total energy in SHM is the sum of kinetic and potential energies, which remains constant.
由此可得最大速度 vmax = A ω，出现在平衡位置；最大加速度 amax = A ω²，出现在极端点。简谐运动中的总能量为动能与势能之和，且保持恒定。
10. Mass‑Energy Equivalence in Particle Physics: ΔE = Δm c² | 粒子物理中的质能等价：ΔE = Δm c²
In nuclear and particle reactions, the total relativistic energy of a system is conserved, but mass may be converted into energy and vice‑versa. Einstein’s famous equation E = m c² relates rest mass m to rest energy.
在核反应和粒子反应中，系统的总相对论能量守恒，但质量可以转化为能量，反之亦然。爱因斯坦著名的方程 E = m c² 将静止质量 m 与静止能量联系起来。
If the total mass of the products is less than the total mass of the reactants, the mass defect Δm is released as kinetic energy of the products or as radiation: ΔE = Δm c². This principle explains the energy released in fission and fusion.
如果产物的总质量小于反应物的总质量，质量亏损 Δm 会以产物动能或辐射的形式释放出来：ΔE = Δm c²。这一原理解释了裂变和聚变中释放的能量。
For instance, in electron‑positron annihilation, the rest mass of the electron and positron (each 9.11×10⁻³¹ kg) is entirely converted into two photons, each carrying energy E = me c² ≈ 0.511 MeV.
例如，在电子‑正电子湮灭中，电子和正电子的静止质量（各为 9.11×10⁻³¹ kg）完全转化为两个光子，每个光子的能量为 E = me c² ≈ 0.511 MeV。
In A‑level calculations, masses are often expressed in atomic mass units u, where 1 u = 1.66×10⁻²⁷ kg, and the energy equivalent of 1 u is 931.5 MeV. Understanding this derivation allows candidates to compute energy changes from mass differences confidently.
在 A‑Level 计算中，质量常用原子质量单位 u 表示，1 u = 1.66×10⁻²⁷ kg，而 1 u 的能量当量为 931.5 MeV。理解这一推导后，考生便能自信地根据质量差计算能量变化。
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