📚 A-Level Physics: Formula Derivation Based on Unit 4 January 2021 Question Paper | A-Level物理：2021年1月U4试卷公式推导
Mastering the key derivations in the Unit 4 physics paper is essential for tackling both structured questions and synoptic challenges. This article provides step-by-step derivations for the most important formulas that appeared, or could appear, in the January 2021 Unit 4 question paper, covering further mechanics, fields, and nuclear physics.
掌握Unit 4物理试卷中的关键公式推导，对于应对结构化问题以及综合性挑战至关重要。本文针对2021年1月Unit 4试卷中出现或可能考查的最重要公式，提供逐步推导过程，内容涵盖进阶力学、场与核物理。
1. Deriving a = –ω²x for Simple Harmonic Motion | 推导简谐运动的 a = –ω²x
In simple harmonic motion (SHM), the restoring force is always directed towards the equilibrium position and its magnitude is proportional to the displacement. For a mass-spring system, this force is given by Hooke’s law: F = –kx, where k is the spring constant and x is the displacement from equilibrium.
在简谐运动中，回复力始终指向平衡位置，且其大小与位移成正比。对于弹簧-质量系统，该力由胡克定律给出：F = –kx，其中k为劲度系数，x为相对于平衡位置的位移。
Applying Newton’s second law, F = ma, we substitute the expression for the force: ma = –kx. Rearranging gives the acceleration a = –(k/m)x. Since both k and m are constants, the ratio k/m is a positive constant, which we denote as ω². Hence, we obtain the defining equation of SHM: a = –ω²x.
应用牛顿第二定律 F = ma，代入力的表达式得到 ma = –kx。整理得加速度 a = –(k/m)x。由于k和m均为常数，比值k/m是一个正常数，我们将其记作ω²。由此我们得到简谐运动的定义方程：a = –ω²x。
2. Deriving the Displacement Equation x = A cos(ωt) | 推导位移方程 x = A cos(ωt)
Starting from the acceleration equation a = d²x/dt² = –ω²x, we note that this is a second-order differential equation. A function whose second derivative is proportional to the negative of itself is the cosine function. Checking the trial solution x = A cos(ωt), where A is amplitude, we differentiate: dx/dt = –Aω sin(ωt), and d²x/dt² = –Aω² cos(ωt) = –ω²x. This satisfies the equation.
从加速度方程 a = d²x/dt² = –ω²x 出发，这实际上是一个二阶微分方程。二阶导数与自身负值成正比的函数是余弦函数。我们尝试 x = A cos(ωt) 的解，其中A为振幅。求导可得 dx/dt = –Aω sin(ωt)，再求导 d²x/dt² = –Aω² cos(ωt) = –ω²x，满足原方程。
If the oscillation starts from the equilibrium position with maximum velocity, we use the sine form x = A sin(ωt). A general solution includes a phase constant φ: x = A cos(ωt + φ). This displacement equation is fundamental for calculating velocity and energy in SHM.
若振动从平衡位置以最大速度开始，我们使用正弦形式 x = A sin(ωt)。更一般的解包含相位常数φ：x = A cos(ωt + φ)。该位移方程是计算简谐运动速度和能量的基础。
3. Deriving the Period of a Simple Pendulum | 推导单摆周期公式
For a simple pendulum of length L and bob mass m, the restoring force along the arc is F = –mg sinθ. For small angles (θ < 10°), sinθ ≈ θ in radians. The displacement along the arc is approximately x = Lθ, so θ = x/L.
对于摆长为L、摆球质量为m的单摆，沿圆弧的回复力为 F = –mg sinθ。当摆角很小（θ < 10°）时，sinθ ≈ θ（以弧度表示）。沿弧的位移可近似为 x = Lθ，因此 θ = x/L。
Substituting gives F = –mg (x/L) = –(mg/L)x. This has the same form as the mass-spring restoring force, so the effective spring constant is k = mg/L. Using the SHM period T = 2π√(m/k), we get T = 2π√(m / (mg/L)) = 2π√(L/g). The period is independent of mass and amplitude (for small angles).
代入得 F = –mg (x/L) = –(mg/L)x。这具有与弹簧回复力相同的形式，因此等效劲度系数为 k = mg/L。利用简谐运动周期公式 T = 2π√(m/k)，得到 T = 2π√(m / (mg/L)) = 2π√(L/g)。周期与质量及（小角度下的）振幅无关。
4. Deriving Centripetal Acceleration a = v²/r | 推导向心加速度 a = v²/r
Consider an object moving at constant speed v in a circular path of radius r. In a short time Δt, the object moves through a small angle Δθ. The velocity vector changes direction but not magnitude. The two velocity vectors form an isosceles triangle with the change in velocity Δv.
考虑一物体以恒定速率v在半径为r的圆形路径上运动。在很短的Δt时间内，物体转过一个小角度Δθ。速度矢量方向改变但大小不变。两个速度矢量与速度变化量Δv构成等腰三角形。
For small Δθ, the magnitude of Δv is approximately vΔθ. The radial acceleration is a = Δv/Δt = v Δθ/Δt. Since angular speed ω = Δθ/Δt and v = rω, we substitute ω = v/r to obtain a = v (v/r) = v²/r. Using ω, the same result is a = rω². This acceleration always points towards the centre.
当Δθ很小时，Δv的大小近似为 vΔθ。径向加速度为 a = Δv/Δt = v Δθ/Δt。因为角速度 ω = Δθ/Δt，且 v = rω，代入 ω = v/r 可得 a = v (v/r) = v²/r。用ω表示同样可得 a = rω²。该加速度始终指向圆心。
5. Deriving Velocities in Elastic Collisions | 推导弹性碰撞速度公式
In an elastic collision, both momentum and kinetic energy are conserved. Consider a two-body system where body 1 (mass m₁) collides head-on with stationary body 2 (mass m₂). Initial velocities: u₁ = u, u₂ = 0. Final velocities are v₁ and v₂.
在弹性碰撞中，动量和动能均守恒。考虑二体系统，物体1（质量m₁）与静止的物体2（质量m₂）发生正碰。初速度：u₁ = u，u₂ = 0。末速度分别为 v₁ 和 v₂。
Momentum conservation: m₁u = m₁v₁ + m₂v₂. Kinetic energy conservation: ½m₁u² = ½m₁v₁² + ½m₂v₂². From momentum, we write m₂v₂ = m₁(u – v₁). Substituting into the energy equation and simplifying yields v₁ = (m₁ – m₂)u / (m₁ + m₂) and then v₂ = 2m₁u / (m₁ + m₂). These derivations are vital for analysing particle collisions in Unit 4.
动量守恒：m₁u = m₁v₁ + m₂v₂。动能守恒：½m₁u² = ½m₁v₁² + ½m₂v₂²。由动量方程得 m₂v₂ = m₁(u – v₁)。代入能量方程并化简，可得 v₁ = (m₁ – m₂)u / (m₁ + m₂)，进而 v₂ = 2m₁u / (m₁ + m₂)。这些推导对于Unit 4中粒子碰撞的分析至关重要。
6. Parabolic Path of a Charged Particle in a Uniform Electric Field | 均匀电场中带电粒子的抛物线轨迹推导
A particle of charge q and mass m enters a uniform electric field E with an initial horizontal velocity v₀ perpendicular to the field. The electric force is qE vertically. There is no horizontal force, so horizontal motion is uniform: x = v₀ t.
电荷量为q、质量为m的粒子以垂直于电场的水平初速度v₀进入匀强电场E。竖向电场力为qE。水平方向无外力，因此水平运动为匀速直线运动：x = v₀ t。
Vertical acceleration is a = qE/m. Starting from zero vertical velocity, vertical displacement is y = ½ (qE/m) t². Eliminating time t using t = x/v₀, we obtain y = ½ (qE/m) (x/v₀)² = (qE/(2mv₀²)) x². This is the equation of a parabola, showing the path is parabolic, similar to projectile motion under gravity.
竖直加速度为 a = qE/m。由竖直初速为零，竖直位移为 y = ½ (qE/m) t²。利用 t = x/v₀ 消去时间，得到 y = ½ (qE/m) (x/v₀)² = (qE/(2mv₀²)) x²。这正是抛物线方程，表明运动轨迹为抛物线，类似于重力作用下的抛体运动。
7. Radius of Circular Motion in a Magnetic Field | 磁场中圆周运动半径推导
When a charged particle with charge q moves with velocity v perpendicular to a uniform magnetic field B, it experiences a magnetic force of magnitude F = qvB. This force acts as the centripetal force, causing the particle to travel in a circular arc.
当电荷量为q的粒子以速度v垂直于匀强磁场B运动时，它受到大小为 F = qvB 的磁力。该力充当向心力，使粒子做圆弧运动。
Equating magnetic force to centripetal force: qvB = mv²/r. Solving for the radius r, we derive r = mv/(qB). The period of revolution is T = 2πr/v = 2πm/(qB), which is independent of speed. This derivation is often required when analysing particle tracks in a magnetic field.
令磁力等于向心力：qvB = mv²/r。解出半径 r，得到 r = mv/(qB)。回转周期为 T = 2πr/v = 2πm/(qB)，与速度无关。在分析磁场中粒子径迹时，常需要这一推导。
8. Deriving V = V₀ e⁻ᵗ⁄ᴿᶜ for Capacitor Discharge | 推导电容放电公式 V = V₀ e⁻ᵗ⁄ᴿᶜ
During discharge through a resistor R, the capacitor’s charge Q and voltage V are related by V = Q/C. The current I = dQ/dt flows in the circuit, and Ohm’s law gives V = IR. Because the charge is decreasing, I = –dQ/dt, so Q/C = –R dQ/dt.
电容器通过电阻R放电时，电荷Q与电压V满足 V = Q/C。电路中的电流 I = dQ/dt，根据欧姆定律 V = IR。由于电荷在减少，I = –dQ/dt，因此 Q/C = –R dQ/dt。
Rearranging: dQ/Q = –dt/(RC). Integrating from initial charge Q₀ at t=0 to Q at time t gives ln(Q/Q₀) = –t/(RC). Hence Q = Q₀ exp(–t/RC). Since V ∝ Q, the voltage decays as V = V₀ exp(–t/RC). This exponential decay is a core concept in circuits and nuclear physics.
整理得 dQ/Q = –dt/(RC)。从t=0时的初始电荷Q₀积分至t时刻的Q，得到 ln(Q/Q₀) = –t/(RC)。因此 Q = Q₀ exp(–t/RC)。由于V ∝ Q，电压按 V = V₀ exp(–t/RC) 衰减。这一指数衰减是电路与核物理中的核心概念。
9. Deriving the Radioactive Decay Law N = N₀ e⁻ᴺᵗ | 推导放射性衰变定律 N = N₀ e⁻ᴺᵗ
The activity A = –dN/dt is directly proportional to the number of undecayed nuclei N: dN/dt = –λN, where λ is the decay constant. The negative sign indicates that N decreases with time.
放射性活度 A = –dN/dt 正比于未衰变核的数目N：dN/dt = –λN，其中λ为衰变常数。负号表示N随时间减少。
Separating variables: dN/N = –λ dt. Integrating both sides between limits (N₀ at t=0, N at t) yields ln(N/N₀) = –λt. Taking the exponential gives the exponential decay law N = N₀ exp(–λt). This derivation is directly analogous to capacitor discharge and is essential for understanding half-life t₁₂ = ln2/λ.
分离变量：dN/N = –λ dt。在积分限（t=0时N₀，t时刻N）之间积分得 ln(N/N₀) = –λt。取指数函数即得指数衰变定律 N = N₀ exp(–λt)。这一推导与电容放电完全类似，是理解半衰期 t₁₂ = ln2/λ 的基础。
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