📚 A-Level Physics: Deriving Key Formulas from the January 2022 Insert Section 5 | A-Level 物理：推导 2022年1月插入页第5节关键公式
Many A-Level Physics exams provide a data and formulae booklet, often referred to as an ‘insert’. The insert for January 2022 typically includes a section (Section 5) covering core mechanics, waves, electricity, and fields. While these formulas are given, understanding their derivations deepens conceptual grasp. This article steps through the derivations of the most essential equations likely found in that section, from SUVAT to capacitor discharge.
许多 A-Level 物理考试都会提供数据和公式手册，通常被称为“插入页”。2022年1月考试的插入页通常包含一个涵盖核心力学、波、电学和场的部分（第5节）。尽管这些公式已经给出，但理解它们的推导过程能加深概念掌握。本文逐步推导那部分最可能包含的关键方程，从 SUVAT 到电容器放电。
1. Deriving the SUVAT Equations | 推导匀加速运动方程组
Starting from the definition of constant acceleration a = (v – u) / t, we can rearrange to obtain the first equation: v = u + at. This directly links final velocity to initial velocity, acceleration, and time.
从匀加速度的定义 a = (v – u) / t 出发，我们可重新排列得到第一个方程：v = u + at。这直接将末速度与初速度、加速度和时间联系起来。
v = u + at
To find displacement, we use the fact that average velocity under constant acceleration is (u + v)/2, so s = ((u + v)/2) × t. Substituting v from the first equation gives s = ut + ½at². Squaring the first equation and substituting for t from rearranged s = ut + ½at² yields v² = u² + 2as.
为了求位移，我们利用匀加速下平均速度为 (u + v)/2，因此 s = ((u + v)/2) × t。代入第一个方程中的 v，得到 s = ut + ½at²。将第一个方程平方，并利用由 s = ut + ½at² 导出的 t 进行代换，可得 v² = u² + 2as。
s = ut + ½at²
v² = u² + 2as
2. Deriving Kinetic Energy and Work–Energy Theorem | 推导动能与功能定理
Consider a constant resultant force F acting over a displacement s. The work done is W = Fs. Using Newton’s second law F = ma and the SUVAT relation v² = u² + 2as, we solve for as = (v² – u²)/2. Then W = m × (v² – u²)/2 = ½mv² – ½mu². If the particle starts from rest, KE = ½mv².
考虑一个恒定的净力 F 作用在位移 s 上。做功为 W = Fs。利用牛顿第二定律 F = ma 和 SUVAT 关系式 v² = u² + 2as，解得 as = (v² – u²)/2。于是 W = m × (v² – u²)/2 = ½mv² – ½mu²。如果质点从静止开始，动能为 KE = ½mv²。
Eₖ = ½mv²
3. Deriving Centripetal Acceleration (a = v²/r) | 推导向心加速度 (a = v²/r)
For an object moving with constant speed v in a circle of radius r, its position vector rotates by a small angle Δθ in time Δt. The velocity vector also rotates by the same angle Δθ. The magnitude of velocity change is |Δv| ≈ v Δθ. Since Δθ = (v Δt)/r, the acceleration magnitude is a = |Δv|/Δt = v × (v/r) = v²/r. The direction is toward the centre.
对于一个以恒定速率 v 做半径为 r 的圆周运动的物体，其位置矢量在时间 Δt 内转过一个小角度 Δθ。速度矢量也转过相同的角度 Δθ。速度变化的大小为 |Δv| ≈ v Δθ。由于 Δθ = (v Δt)/r，加速度的大小为 a = |Δv|/Δt = v × (v/r) = v²/r。方向指向圆心。
a = v²/r
Using Newton’s second law, the centripetal force is F = mv²/r. Another equivalent expression using ω = v/r is F = mω²r.
应用牛顿第二定律，向心力为 F = mv²/r。利用 ω = v/r，另一个等价的表达式为 F = mω²r。
F = mv²/r = mω²r
4. Deriving Simple Harmonic Motion Acceleration | 推导简谐运动的加速度
SHM is defined by a restoring force proportional to displacement: F = -kx. From Newton’s second law, ma = -kx, giving a = -(k/m)x. Defining ω² = k/m yields the characteristic a = -ω²x. The negative sign shows the acceleration is always directed towards the equilibrium position.
简谐运动由与位移成正比的恢复力定义：F = -kx。根据牛顿第二定律，ma = -kx，得到 a = -(k/m)x。定义 ω² = k/m，就得出特征方程 a = -ω²x。负号表明加速度始终指向平衡位置。
a = -ω²x
5. Deriving Displacement Equation for SHM (x = A cos(ωt) or A sin(ωt)) | 推导简谐运动位移方程 (x = A cos(ωt) 或 A sin(ωt))
The solution to a = -ω²x is a sinusoidal function. Taking the projection of uniform circular motion, the angular displacement is θ = ωt. The x-component of the radius vector gives x = A cos(ωt + φ), where φ is the initial phase. For an SHM starting at maximum displacement, x = A cos(ωt).
方程 a = -ω²x 的解是正弦函数。通过匀速圆周运动的投影，角位移为 θ = ωt。半径矢量的 x 分量为 x = A cos(ωt + φ)，其中 φ 是初相位。对于从最大位移开始的简谐运动，x = A cos(ωt)。
x = A cos(ωt)
The corresponding velocity is v = -Aω sin(ωt), and the maximum speed is vₘₐₓ = ωA.
相应的速度为 v = -Aω sin(ωt)，最大速率为 vₘₐₓ = ωA。
v = ±ω√(A² – x²)
6. Deriving the Period of a Mass–Spring System | 推导质量-弹簧系统的周期
For a mass m on a spring of stiffness k, the restoring force is F = -kx. Using ω² = k/m from the SHM definition, and knowing that ω = 2π/T, we substitute to get 2π/T = √(k/m). Thus, the period T = 2π√(m/k).
对于劲度系数为 k 的弹簧上的质量 m，恢复力为 F = -kx。利用简谐运动定义中的 ω² = k/m，并知道 ω = 2π/T，代入可得 2π/T = √(k/m)。因此周期 T = 2π√(m/k)。
T = 2π√(m/k)
7. Deriving the Period of a Simple Pendulum | 推导单摆的周期
For a small angular displacement θ, the restoring force along the arc is -mg sinθ ≈ -mgθ. The tangential acceleration is a = -gθ. Since the displacement along the arc is x = Lθ, a = -(g/L)x. Comparing with a = -ω²x gives ω² = g/L. Therefore, T = 2π/ω = 2π√(L/g).
对于小角度位移 θ，沿弧线的恢复力为 -mg sinθ ≈ -mgθ。切向加速度为 a = -gθ。由于沿弧线的位移为 x = Lθ，有 a = -(g/L)x。与 a = -ω²x 对比，得到 ω² = g/L。因此 T = 2π/ω = 2π√(L/g)。
T = 2π√(L/g)
8. Deriving the Capacitor Discharge Equation (Q = Q₀e⁻ᵗ/ᴿᶜ) | 推导电容器放电方程 (Q = Q₀e⁻ᵗ/ᴿᶜ)
During discharge, the current I = -dQ/dt (charge leaving the plates). From Kirchhoff’s loop rule, IR = Q/C. Substituting gives -R dQ/dt = Q/C, so dQ/dt = -Q/(RC). Solving this differential equation by separation of variables: ∫ dQ/Q = -∫ dt/(RC), leading to ln Q = -t/RC + constant. Applying initial condition Q = Q₀ at t = 0 yields ln(Q/Q₀) = -t/RC, hence Q = Q₀ e⁻ᵗ/ᴿᶜ.
放电时，电流 I = -dQ/dt（电荷离开极板）。根据基尔霍夫回路定律，IR = Q/C。代入得 -R dQ/dt = Q/C，即 dQ/dt = -Q/(RC)。用分离变量法解这个微分方程：∫ dQ/Q = -∫ dt/(RC)，得到 ln Q = -t/RC + 常数。代入初始条件 t = 0 时 Q = Q₀，得出 ln(Q/Q₀) = -t/RC，因此 Q = Q₀ e⁻ᵗ/ᴿᶜ。
Q = Q₀ e⁻ᵗ/ᴿᶜ
Voltage decays as V = V₀ e⁻ᵗ/ᴿᶜ and current as I = I₀ e⁻ᵗ/ᴿᶜ, where the time constant is τ = RC.
电压按 V = V₀ e⁻ᵗ/ᴿᶜ 衰减，电流按 I = I₀ e⁻ᵗ/ᴿᶜ 衰减，其中时间常数 τ = RC。
9. Deriving Gravitational Potential (V = -GM/r) | 推导引力势 (V = -GM/r)
Gravitational potential is defined as work done per unit mass to bring a test mass from infinity to a point. The gravitational force is F = GMm/r². Work done against gravity for a small displacement dr is dW = (GMm/r²) dr. Integrating from infinity to r: W = ∫∞ʳ (GMm/r²) dr = GMm [-1/r]∞ʳ = -GMm/r. Dividing by m gives potential V = -GM/r.
引力势定义为单位质量的试验质量从无穷远处移动到某一点所做的功。引力为 F = GMm/r²。对抗引力进行一个微小位移 dr 所做的功为 dW = (GMm/r²) dr。从无穷远积分到 r：W = ∫∞ʳ (GMm/r²) dr = GMm [-1/r]∞ʳ = -GMm/r。除以 m 得到势 V = -GM/r。
V = -GM/r
10. Deriving the Force on a Current-Carrying Wire in a Magnetic Field | 推导载流导线在磁场中所受的力
A current I is a flow of charge. Each charge carrier experiences Lorentz force F = qvB sinθ. For a wire of length L with n charge carriers per unit volume, total charge moving is nAL q. Total force F = (nAL q)vB sinθ. Since current I = nAqv, the force becomes F = BIL sinθ. When the wire is perpendicular to the field (θ = 90°), F = BIL.
电流 I 是电荷的流动。每个载流子受到洛伦兹力 F = qvB sinθ。对于长为 L、单位体积内有 n 个载流子的导线，移动的总电荷为 nAL q。总力 F = (nAL q)vB sinθ。因为电流 I = nAqv，力变为 F = BIL sinθ。当导线与磁场垂直时 (θ = 90°)，F = BIL。
F = BIL sinθ
This derivation unifies microscopic and macroscopic views, and the formula itself appears prominently in Section 5 of the standard A-level physics insert.
这一推导联系了微观和宏观视角，而该公式本身在标准 A-Level 物理插入页第5节中占有突出地位。
11. Deriving the Transformer Equation | 推导变压器方程
An ideal transformer assumes no flux leakage. The primary coil creates a changing magnetic flux Φ, which links the secondary coil. Faraday’s law gives induced emfs: Vₚ = Nₚ dΦ/dt and Vₛ = Nₛ dΦ/dt. Dividing the two equations yields Vₛ/Vₚ = Nₛ/Nₚ. For 100% efficiency, input power equals output power, so Vₚ Iₚ = Vₛ Iₛ, leading to Iₛ/Iₚ = Nₚ/Nₛ.
理想变压器假设没有磁通量泄漏。初级线圈产生变化的磁通量 Φ，该磁通量穿过次级线圈。法拉第定律给出感应电动势：Vₚ = Nₚ dΦ/dt 和 Vₛ = Nₛ dΦ/dt。两式相除得到 Vₛ/Vₚ = Nₛ/Nₚ。对于100%效率，输入功率等于输出功率，即 Vₚ Iₚ = Vₛ Iₛ，得出 Iₛ/Iₚ = Nₚ/Nₛ。
Vₛ / Vₚ = Nₛ / Nₚ
12. Deriving the Equation for Uniform Electric Field (E = V/d) | 推导匀强电场方程 (E = V/d)
For two parallel plates separated by distance d with potential difference V, the work done to move a charge q across the gap is W = qV. By definition of electric field, W = Fd = qEd. Equating the two expressions: qV = qEd, thus E = V/d. This shows the link between field strength and potential gradient and is a foundational formula in Section 5 of many data inserts.
对于相距 d、电势差为 V 的两块平行板，移动电荷 q 穿越间隙所做的功为 W = qV。根据电场定义，W = Fd = qEd。令两式相等：qV = qEd，因此 E = V/d。这显示了场强与电势梯度之间的关系，是许多数据插入页第5节中的一个基础公式。
E = V/d
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