📚 A-Level Physics Unit 4 Jan 2022 Formula Derivation Guide | 2022年1月单元4物理公式推导指南
This guide revisits core derivations often examined in A-Level Physics Unit 4, using the January 2022 question paper as a backdrop. Mastering these derivations deepens conceptual understanding and boosts exam performance, as many questions require you to reconstruct key results from fundamental principles. Each section below presents a critical formula, its step-by-step derivation, and the physical reasoning behind it.
本指南以 2022 年 1 月单元 4 试卷为背景，重温常考的核心公式推导。掌握这些推导能深化概念理解，提升应试表现，因为许多题目要求你从基本原理重建关键结果。以下每个小节都给出一个关键公式，并逐步推导，解释其物理依据。
1. Derivation of Centripetal Acceleration a = v²/r | 向心加速度 a = v²/r 的推导
An object moving in a circle of radius r at constant speed v experiences an acceleration towards the centre. To derive its magnitude, we analyse the velocity change over a short time Δt.
一个物体以恒定速率 v 沿半径 r 的圆周运动，必有指向圆心的加速度。为推出其大小，我们分析一小段时间 Δt 内的速度变化。
At time t, velocity v₁ is tangent to the circle. After a small angular displacement Δθ, the velocity v₂ has the same magnitude but is rotated by Δθ.
t 时刻，速度 v₁ 沿切线方向。经过一小角位移 Δθ 后，速度 v₂ 大小不变但方向旋转了 Δθ。
The change in velocity is Δv = v₂ − v₁. For small angles, the magnitude of Δv is approximately v Δθ.
速度变化量 Δv = v₂ − v₁。对于小角度，Δv 的大小近似为 v Δθ。
The arc length travelled in Δt is s = v Δt = r Δθ, so Δθ = v Δt / r.
在 Δt 内移动的弧长 s = v Δt = r Δθ，因此 Δθ = v Δt / r。
Substituting gives Δv = v × (v Δt / r) = v² Δt / r. The acceleration magnitude is a = Δv / Δt = v² / r. Using v = ωr, we also obtain a = ω²r.
代入得 Δv = v × (v Δt / r) = v² Δt / r。加速度大小 a = Δv / Δt = v² / r。利用 v = ωr，也可得 a = ω²r。
2. Derivation of the Relative Speed Formula in Elastic Collisions | 弹性碰撞中相对速度公式的推导
In a one-dimensional elastic collision, both momentum and kinetic energy are conserved. By combining these conservation laws, we can show that the relative speed of approach equals the relative speed of separation.
在一维弹性碰撞中，动量和动能均守恒。结合这两个守恒定律，我们可以证明：两物体接近的相对速度等于分离的相对速度。
For two masses m₁ and m₂ with initial velocities u₁, u₂ and final velocities v₁, v₂, momentum conservation gives m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
对质量 m₁、m₂，初速度 u₁、u₂，末速度 v₁、v₂，动量守恒写出 m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。
Kinetic energy conservation gives ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂².
动能守恒给出 ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂²。
Rearranging the momentum equation as m₁(u₁ − v₁) = m₂(v₂ − u₂) and the energy equation as m₁(u₁² − v₁²) = m₂(v₂² − u₂²), we factor the difference of squares.
将动量式改写为 m₁(u₁ − v₁) = m₂(v₂ − u₂)，动能式改写为 m₁(u₁² − v₁²) = m₂(v₂² − u₂²)，并因式分解平方差。
Dividing the energy equation by the momentum equation gives u₁ + v₁ = v₂ + u₂, which simplifies to v₂ − v₁ = −(u₂ − u₁). Hence the relative speed of separation equals the relative speed of approach.
将动能式除以动量式得 u₁ + v₁ = v₂ + u₂，化简为 v₂ − v₁ = −(u₂ − u₁)。因此分离的相对速率等于接近的相对速率。
3. Derivation of Electric Potential V = Q/(4πε₀r) | 点电荷电势 V = Q/(4πε₀r) 的推导
Electric potential V at a point is the work done per unit charge to bring a small positive test charge from infinity to that point against the electric field of a point charge Q.
电势 V 定义为将单位正试探电荷从无穷远处反抗点电荷 Q 电场力移到该点所做的功。
The force on a test charge q at a distance x from Q is F = Qq/(4πε₀x²) radially outward. The work done to move it a small displacement dx against the field is dW = −F dx.
试探电荷 q 距离 Q 为 x 时所受电场力为 F = Qq/(4πε₀x²) 沿径向向外。逆着电场力移动微小位移 dx，做功 dW = −F dx。
Integrating from x = ∞ to x = r: W = −∫∞ᵣ Qq/(4πε₀x²) dx = Qq/(4πε₀) [1/x]∞ᵣ = Qq/(4πε₀r).
积分从 x = ∞ 到 r：W = −∫∞ᵣ Qq/(4πε₀x²) dx = Qq/(4πε₀) [1/x]∞ᵣ = Qq/(4πε₀r)。
Since potential V = W/q, we obtain V = Q/(4πε₀r).
因电势 V = W/q，故得 V = Q/(4πε₀r)。
4. Derivation of Energy Stored in a Capacitor E = ½CV² | 电容器储存能量 E = ½CV² 的推导
Charging a capacitor requires work to move charge against the growing potential difference across the plates. The total energy stored can be found by integrating the work done in small increments.
给电容器充电需要克服极板间逐渐升高的电势差移送电荷。通过对微小增量做功进行积分，可求得总储存能量。
When the capacitor holds charge q, the potential difference is V = q/C. Adding a further small charge dq requires work dW = V dq = (q/C) dq.
当电容器已充有电荷 q 时，其电压为 V = q/C。再充电荷量 dq 需做功 dW = V dq = (q/C) dq。
Integrating from q = 0 to Q: W = ∫₀ᵠ (q/C) dq = ½ Q²/C.
积分 q 从 0 到 Q：W = ∫₀ᵠ (q/C) dq = ½ Q²/C。
Using Q = CV, this becomes E = ½ CV², which equals the area under the voltage–charge graph.
代入 Q = CV，得 E = ½ CV²，该值等于电压–电荷图像下的面积。
5. Derivation of Magnetic Force on a Moving Charge F = Bqv | 运动电荷受磁场力 F = Bqv 的推导
The force on a current-carrying conductor in a magnetic field can be related to the motion of individual charge carriers, leading to the expression for the force on a single charge.
磁场对载流导线的作用力可归结于其中单个载流子的运动，从而推出单个电荷所受磁力的表达式。
Consider a straight wire of length L carrying current I, placed perpendicular to a uniform magnetic field B. The force on the wire is F = BIL.
考虑一段长为 L 的直导线，通有电流 I，垂直置于匀强磁场 B 中。导线受力 F = BIL。
Current is I = nAqv, where n is the number of charge carriers per unit volume, A is cross-sectional area, q is the charge on a carrier, and v is the drift speed.
电流 I = nAqv，式中 n 为单位体积载流子数，A 为截面积，q 为每个载流子电荷量，v 为漂移速率。
Substituting: F = B (nAqv) L. The total number of carriers in the wire is N = nAL, so the force per carrier is f = F/N = Bqv.
代入得 F = B (nAqv) L。导线中载流子总数 N = nAL，因此单个载流子受力 f = F/N = Bqv。
If the charge moves at an angle θ to the field, the general form is F = Bqv sinθ.
若电荷运动方向与磁场成 θ 角，则一般式为 F = Bqv sinθ。
6. Derivation of Induced EMF and Faraday’s Law ε = −N ΔΦ/Δt | 感生电动势与法拉第定律 ε = −N ΔΦ/Δt 的推导
Faraday’s law links the induced EMF in a coil to the rate of change of magnetic flux. A simple case of a moving conductor in a uniform field provides clear physical insight.
法拉第定律将线圈中的感生电动势与磁通量变化率联系起来。一根在匀强磁场中运动的导体提供了清晰的物理图像。
Consider a conducting rod of length l sliding at speed v along two parallel rails, completing a circuit in a magnetic field B perpendicular to the plane. The area swept out per unit time is l v.
设想一根长度为 l 的导体棒，以速度 v 沿两条平行导轨滑动，形成回路，磁场 B 垂直于导轨平面。单位时间扫过的面积为 l v。
The magnetic flux Φ = BA, so the change in flux in time Δt is ΔΦ = B × (l v Δt).
磁通量 Φ = BA，因此 Δt 时间内的通量变化为 ΔΦ = B × (l v Δt)。
By the principle of conservation of energy, the induced EMF equals the work done per unit charge, giving ε = Blv. Since ΔΦ = Blv Δt, we obtain ε = ΔΦ/Δt.
由能量守恒，感生电动势等于单位电荷所做的功，即有 ε = Blv。因 ΔΦ = Blv Δt，故 επ = ΔΦ/Δt。
For a coil of N turns, the flux linkage is NΦ, so ε = −N ΔΦ/Δt. The negative sign (Lenz’s law) indicates the EMF opposes the change in flux.
对于 N 匝线圈，磁链为 NΦ，因此 ε = −N ΔΦ/Δt。负号（楞次定律）表示电动势反抗磁通量的变化。
7. Derivation of Escape Velocity v = √(2GM/R) | 逃逸速度 v = √(2GM/R) 的推导
Escape velocity is the minimum speed needed for an object to completely break free from a planet’s gravitational field without further propulsion.
逃逸速度是物体无需额外推进就能完全挣脱行星引力场的最小初速度。
The work done against gravity to move a mass m from the planet’s surface (radius R) to infinity is W = ∫∞ᵣ (GMm/x²) dx = GMm/R.
将质量 m 从行星表面（半径 R）移至无穷远反抗引力做功 W = ∫∞ᵣ (GMm/x²) dx = GMm/R。
For the object to just escape, its initial kinetic energy must equal this work: ½mv² = GMm/R.
要使物体刚好逃脱，其初始动能必须等于该功：½mv² = GMm/R。
Solving gives v = √(2GM/R). The escape velocity is independent of the escaping mass.
解得 v = √(2GM/R)。逃逸速度与逃逸物体的质量无关。
8. Derivation of Ideal Gas Pressure P = ⅓ρ⟨c²⟩ | 理想气体压强 P = ⅓ρ⟨c²⟩ 的推导
The kinetic theory model connects the microscopic motion of gas molecules to the macroscopic pressure. Assuming perfectly elastic collisions with container walls, we can derive the pressure formula.
气体动理论模型将气体分子的微观运动与宏观压强联系起来。假设分子与器壁发生完全弹性碰撞，可推导压强公式。
Consider a single molecule of mass m moving with x-component velocity u towards a wall of area A. Collision reverses momentum, giving a change 2mu.
考虑一个质量为 m 的分子，以 x 方向速度分量 u 撞向面积为 A 的器壁。碰撞使动量反向，改变量为 2mu。
The time between successive collisions with that wall is Δt = 2L/u, where L is the container length along x. Average force on the wall from this molecule: f = 2mu / (2L/u) = mu²/L.
该分子连续两次撞击同一器壁的时间间隔为 Δt = 2L/u，L 是容器沿 x 方向的长度。它对壁的平均作用力 f = 2mu / (2L/u) = mu²/L。
For N molecules, the total force is F = (N/L) m⟨u²⟩, where ⟨u²⟩ is the mean square of x-components. Pressure P = F/A = (N/V) m⟨u²⟩.
对于 N 个分子，总力 F = (N/L) m⟨u²⟩，⟨u²⟩ 是 x 分量平方的均值。压强 P = F/A = (N/V) m⟨u²⟩。
Because ⟨c²⟩ = ⟨u²⟩ + ⟨v²⟩ + ⟨w²⟩ = 3⟨u²⟩, we have P = ⅓ (N/V) m⟨c²⟩ = ⅓ρ⟨c²⟩, where ρ = Nm/V is the density.
因为 ⟨c²⟩ = ⟨u²⟩ + ⟨v²⟩ + ⟨w²⟩ = 3⟨u²⟩，所以 P = ⅓ (N/V) m⟨c²⟩ = ⅓ρ⟨c²⟩，其中 ρ = Nm/V 为密度。
9. Derivation of Radius of Curvature in Magnetic Field r = mv/(Bq) | 磁场中偏转半径 r = mv/(Bq) 的推导
A charged particle moving perpendicularly to a uniform magnetic field experiences a centripetal force, causing circular motion. Equating magnetic force to the required centripetal force yields the orbit radius.
带电粒子垂直射入匀强磁场时，磁场力提供向心力使其做圆周运动。令磁力等于所需向心力，即可得出轨道半径。
Magnetic force magnitude Fₘ = Bqv. For circular motion at speed v and radius r, centripetal force F_c = mv²/r.
磁力大小 Fₘ = Bqv。对于速度为 v、半径为 r 的圆周运动，向心力 F_c = mv²/r。
Setting the magnetic force as the centripetal force: Bqv = mv²/r.
由磁场力提供向心力：Bqv = mv²/r。
Cancelling v and rearranging gives r = mv/(Bq). This radius is often called the gyroradius.
约去 v 并整理得 r = mv/(Bq)。该半径常称为回旋半径。
If the momentum p = mv is known, r = p/(Bq), which is useful in particle identification in detectors.
若已知动量 p = mv，则有 r = p/(Bq)，这在探测器粒子鉴别中非常有用。
10. Derivation of de Broglie Wavelength λ = h/p | 德布罗意波长 λ = h/p 的推导
De Broglie proposed that all matter has a wave-like nature, with wavelength inversely proportional to momentum. This can be motivated by analogy with photons and verified by electron diffraction.
德布罗意提出所有物质具有波动性，波长与动量成反比。这可通过类比光子得到启示，并由电子衍射实验证实。
For a photon, energy E = hf = hc/λ and momentum p = E/c = h/λ, so λ = h/p.
对光子，能量 E = hf = hc/λ，动量 p = E/c = h/λ，因而 λ = h/p。
Extending this relation to particles, the de Broglie wavelength is λ = h/p = h/(mv).
将此关系推广到实物粒子，得德布罗意波长 λ = h/p =
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