📚 AS Physics: Key Formula Derivations from the 9630 International Syllabus | AS 物理：9630 国际大纲关键公式推导
Understanding where formulas come from transforms physics from a memory exercise into a logical, interconnected subject. This article guides you through the essential derivations required in the Cambridge International AS Level Physics 9630 syllabus. Each step builds directly from fundamental definitions and laws, helping you to apply equations with confidence in exams.
理解公式的来源能让物理从死记硬背变成一门有逻辑、相互关联的学科。本文将带领你逐一推导剑桥国际 AS 物理 9630 大纲中的关键公式，每一步都直接源于基本定义和定律，帮助你自信地在考试中运用这些方程。
1. Uniform Acceleration Equations (SUVAT) | 匀加速运动方程
From the definition of acceleration, a = (v − u) / t, where u is initial velocity, v is final velocity, and t is time. Rearranging immediately gives the first equation: v = u + at.
由加速度的定义 a = (v − u) / t 出发，其中 u 为初速度，v 为末速度，t 为时间。简单移项就得到第一个方程：v = u + at。
Displacement s is given by average velocity multiplied by time. Because acceleration is constant, average velocity = (u + v) / 2. Substituting v from the first equation, we obtain s = (u + (u + at)) / 2 × t, which simplifies to s = ut + ½at².
位移 s 等于平均速度乘以时间。由于加速度恒定，平均速度 = (u + v) / 2。把第一个方程中的 v 代入，得到 s = (u + (u + at)) / 2 × t，化简后即为 s = ut + ½at²。
To eliminate time, square the first equation: v² = (u + at)² = u² + 2uat + a²t². Factor out 2a from the last two terms to get v² = u² + 2a(ut + ½at²). Recognising the bracket as s yields the third equation: v² = u² + 2as.
为消去时间，将第一个方程两边平方：v² = (u + at)² = u² + 2uat + a²t²。从后两项中提出 2a 得 v² = u² + 2a(ut + ½at²)。括号内正是 s，于是得到第三个方程：v² = u² + 2as。
2. Momentum and Impulse | 动量与冲量
Newton’s second law states that the net force on an object is proportional to the rate of change of its momentum: F = Δp / Δt, where momentum p = mv. For a constant mass, Δp = m(v − u), so F = m(v − u) / t = ma, which recovers the familiar F = ma.
牛顿第二定律指出，物体所受的合外力正比于其动量的变化率：F = Δp / Δt，其中动量 p = mv。对于质量不变的情况，Δp = m(v − u)，因此 F = m(v − u) / t = ma，这就回到了我们所熟悉的 F = ma。
Rearranging F = Δp / Δt gives the impulse–momentum relationship: FΔt = Δp = mv − mu. The impulse FΔt equals the change in momentum. In collisions and safety applications, this explains why extending the time of impact reduces the force.
将 F = Δp / Δt 移项可得冲量–动量关系：FΔt = Δp = mv − mu。冲量 FΔt 等于动量的变化。在碰撞和安全应用中，这就解释了为什么延长撞击时间可以减小作用力。
3. Kinetic Energy and the Work–Energy Theorem | 动能与功能原理
Work done by a constant force is W = F s. Using F = ma and the uniform acceleration equation v² = u² + 2as, we can express as = (v² − u²) / 2. Substituting gives W = m × (v² − u²) / 2 = ½mv² − ½mu². The quantity ½mv² is defined as kinetic energy E_k.
恒力做功为 W = F s。利用 F = ma 以及匀加速运动方程 v² = u² + 2as，可将 as 写成 as = (v² − u²) / 2。代入后得 W = m × (v² − u²) / 2 = ½mv² − ½mu²。其中 ½mv² 被定义为动能 E_k。
Thus, the net work done on an object equals its change in kinetic energy. This is the work–energy theorem, which is extremely useful for solving problems where acceleration is not constant.
因此，物体所受合外力做的功等于其动能的变化量。这就是功能原理，在加速度不恒定的问题中极其有用。
4. Gravitational Potential Energy near Earth’s Surface | 近地引力势能
When an object of mass m is lifted through a vertical height Δh at constant speed, the lifting force must equal the weight mg. The work done against gravity is W = F × Δh = mgΔh. This work is stored as gravitational potential energy, ΔE_p = mgΔh. The zero level can be chosen arbitrarily because only changes in potential energy are physically meaningful.
质量为 m 的物体被匀速竖直提升 Δh 时，提升力必须等于重力 mg。克服重力做的功为 W = F × Δh = mgΔh。这部分功储存为引力势能，ΔE_p = mgΔh。零势能面可以任意选取，因为只有势能的变化才有物理意义。
The derivation assumes g is constant, which is valid only close to the Earth’s surface. In the 9630 syllabus, this formula is used for all terrestrial mechanics problems.
该推导假设 g 恒定，这仅在地球表面附近成立。在 9630 大纲中，所有地面力学问题都使用此公式。
5. Electrical Power: P = IV, P = I²R, P = V²/R | 电功率公式
Power is the rate of energy transfer. When a charge Q moves through a potential difference V, the energy transferred is W = QV. Current I is defined as Q/t, so power P = W/t = (QV)/t = IV. This gives the fundamental electrical power equation.
功率是能量转移的速率。当电荷 Q 通过电势差 V 时，传递的能量为 W = QV。电流 I 定义为 Q/t，因此功率 P = W/t = (QV)/t = IV，这就是电功率的基本公式。
For a resistor, Ohm’s law states V = IR. Substituting V into P = IV gives P = I(IR) = I²R. Alternatively, substituting I = V/R gives P = (V/R)V = V²/R. All three forms are equivalent for ohmic conductors.
对于电阻器，欧姆定律给出 V = IR。将 V 代入 P = IV 得 P = I(IR) = I²R。或者，将 I = V/R 代入得 P = (V/R)V = V²/R。对于欧姆导体，这三种形式是等价的。
6. Resistors in Series and Parallel | 串并联电阻
In a series circuit, the same current I flows through each resistor. The total potential difference V across the combination is the sum of individual p.d.s: V = V₁ + V₂ + V₃ + … Using Ohm’s law, V = IR_total and each V_n = I R_n. Substitution gives IR_total = I(R₁ + R₂ + R₃ + …). Cancelling I yields R_total = R₁ + R₂ + R₃ + …
在串联电路中，每个电阻流过相同的电流 I。组合两端的总电势差 V 等于各电阻电势差之和：V = V₁ + V₂ + V₃ + …。利用欧姆定律，V = IR_total，且每个 V_n = I R_n。代入得 IR_total = I(R₁ + R₂ + R₃ + …)。消去 I 即得 R_total = R₁ + R₂ + R₃ + …
In a parallel circuit, each resistor experiences the same potential difference V. The total current I is the sum of branch currents: I = I₁ + I₂ + I₃ + … Using I = V / R_total and each I_n = V / R_n, we have V / R_total = V / R₁ + V / R₂ + V / R₃ + … Dividing by V gives the reciprocal formula: 1 / R_total = 1 / R₁ + 1 / R₂ + 1 / R₃ + …
在并联电路中，各电阻两端电势差相同，均为 V。总电流 I 等于各支路电流之和：I = I₁ + I₂ + I₃ + …。利用 I = V / R_total 及各支路 I_n = V / R_n，可得 V / R_total = V / R₁ + V / R₂ + V / R₃ + …。除以 V 即得倒数和公式：1 / R_total = 1 / R₁ + 1 / R₂ + 1 / R₃ + …
7. Wave Equation: v = fλ | 波动方程推导
A wave travels one wavelength λ in one period T. Speed is distance over time, so v = λ / T. Since frequency f = 1/T, substituting gives v = fλ. This relationship is fundamental to all wave phenomena, from sound to electromagnetic waves.
一个周期 T 内，波传播一个波长 λ。速度等于距离除以时间，因此 v = λ / T。又因频率 f = 1/T，代入后得 v = fλ。这个关系对于所有波动现象——从声波到电磁波——都是基本的。
In the AS syllabus this equation is used for waves on strings, water waves and light. It is particularly important when discussing refraction, where frequency remains constant and wavelength changes cause a change in speed.
在 AS 大纲中，该方程用于弦上的波、水波和光波。在讨论折射时它尤其重要，因为折射过程中频率保持不变，波长的变化导致波速的改变。
8. Young’s Double-Slit Fringe Spacing: Δx = λD / d | 杨氏双缝条纹间距
Consider monochromatic light of wavelength λ passing through two slits separated by a distance d, producing an interference pattern on a screen at distance D (with D ≫ d). For a bright fringe at position x from the centre, the path difference to the two slits is d sin θ, where θ ≈ x/D for small angles.
考虑波长为 λ 的单色光通过相距为 d 的两条狭缝，在距离为 D 的屏幕上形成干涉图样（且 D ≫ d）。对于距中心 x 处的亮条纹，到达两缝的光程差为 d sin θ，在小角度下 θ ≈ x/D。
Constructive interference occurs when path difference = nλ. For the first order (n=1), d sin θ = λ. Using the small-angle approximation sin θ ≈ tan θ = x/D, we obtain d (x/D) = λ, so x = λD/d. The distance between adjacent bright fringes, Δx, is thus λD/d.
相干加强发生在光程差等于 nλ 时。对于一级亮纹（n=1），有 d sin θ = λ。利用小角度近似 sin θ ≈ tan θ = x/D，得到 d (x/D) = λ，即 x = λD/d。因此相邻亮纹的间距 Δx = λD/d。
This derivation is a core part of the waves section. It allows the wavelength of light to be determined from measurable distances.
该推导是波动章节的核心内容，通过测量可测距离就可以求出光波波长。
9. Centripetal Acceleration: a = v² / r and a = rω² | 向心加速度
An object moving at constant speed v in a circle of radius r travels from point A to point B in a short time Δt. The change in velocity Δv points towards the centre. The two velocity vectors and the displacement chord AB form two similar isosceles triangles.
物体以恒定速率 v 沿半径为 r 的圆周运动，在很短的时间 Δt 内从 A 点运动到 B 点。速度的变化量 Δv 指向圆心。两个速度矢量与弦 AB 构成两个相似的等腰三角形。
From the velocity triangle, Δv / v = chord AB / r. For small Δt, chord AB ≈ arc length vΔt. Therefore Δv / v = vΔt / r, giving Δv / Δt = v² / r. As Δt → 0, Δv/Δt is the instantaneous acceleration a, so a = v² / r.
从速度三角形可得 Δv / v = 弦 AB / r。当 Δt 很小时，弦长 AB ≈ 弧长 vΔt。于是 Δv / v = vΔt / r，移项得 Δv / Δt = v² / r。当 Δt → 0 时，Δv/Δt 就是瞬时加速度 a，因此 a = v² / r。
Using the relationship v = rω, where ω is angular velocity, we obtain the alternative form a = (rω)² / r = rω².
利用 v = rω（ω 为角速度），可得到向心加速度的另一种形式：a = (rω)² / r = rω²。
10. Simple Harmonic Motion: Displacement Equation | 简谐运动位移方程
For a particle moving with uniform circular motion, its projection onto one diameter performs simple harmonic motion (SHM). If the circular motion has radius A and angular speed ω, the horizontal displacement is x = A cos(ωt) (or x = A sin(ωt), depending on the starting point).
做匀速圆周运动的质点在直径上的投影做简谐运动（SHM）。若圆周运动的半径为 A、角速度为 ω，则水平位移为 x = A cos(ωt)（或 x = A sin(ωt)，与起始点有关）。
The defining equation of SHM states that acceleration a is proportional to displacement and directed towards the equilibrium position: a = −ω²x. Differentiating x = A cos(ωt) twice confirms this: v = −Aω sin(ωt), and a = −Aω² cos(ωt) = −ω²x.
简谐运动的定义方程指出加速度 a 与位移成正比且指向平衡位置：a = −ω²x。将 x = A cos(ωt) 对时间求导两次即可验证：v = −Aω sin(ωt)，a = −Aω² cos(ωt) = −ω²x。
From this, maximum speed v_max = ωA and maximum acceleration a_max = ω²A. These derivations build a bridge between circular motion and SHM, which is essential for analysing pendulums and mass–spring systems.
由此可得最大速率 v_max = ωA 及最大加速度 a_max = ω²A。这些推导在圆周运动与简谐运动之间架起了桥梁，是分析单摆和弹簧振子系统的关键。
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