📚 Essential Formula Derivation Skills for International A-Level Physics | 国际A-Level物理核心公式推导技能
In A-Level Physics, the ability to derive formulas from fundamental principles is not just a mathematical exercise—it deepens your understanding of physical concepts and prepares you for advanced problem-solving. This article revisits key derivations from the International A-Level Science Fundamental Skills Booklet for Physics, focusing on the logical steps and algebraic manipulations that bring equations to life.
在A-Level物理中，从基本原理推导公式的能力不仅仅是数学练习，它能加深你对物理概念的理解，也是解决复杂问题的准备。本文回顾国际A-Level科学基础技能手册（物理）中的关键推导，重点关注让公式生动的逻辑步骤和代数操作。
1. Algebraic Manipulation and Equation Rearrangement | 代数操作与方程变形
Before diving into physics derivations, you must be fluent in rearranging equations. The fundamental skills booklet emphasises isolating a target variable, handling squares and roots, and substituting one expression into another. For example, Ohm’s law V = IR can be rearranged to find resistance: R = V/I, or current: I = V/R. Always perform a dimensional check to confirm the rearrangement yields consistent units.
在深入物理推导之前，你必须熟练地变形方程。基础技能手册强调分离目标变量、处理平方与根号，以及将一个表达式代入另一个。例如，欧姆定律 V = IR 可变形为求电阻 R = V/I，或求电流 I = V/R。始终进行量纲检查，以确认变形后的单位一致。
Another crucial skill is solving simultaneous equation systems. When an object undergoes constant acceleration, you may know initial velocity, time, and acceleration; rearranging v = u + at to find u, or s = ut + ½at² to extract a, involves placing one formula into the other. Mastering these manipulations means you never need to memorise every isolated form—you can always re-derive it.
另一个关键技能是求解联立方程组。当物体匀加速运动时，你可能已知初速度、时间和加速度；将 v = u + at 变形求 u，或将 s = ut + ½at² 变形提取 a，都涉及将一个公式代入另一个。掌握这些操作意味着你无需记背每一个孤立形式——你总是可以重新推导出来。
2. Deriving the Equations of Motion | 运动学公式的推导
For uniform acceleration, the definition of acceleration is a = (v – u) / t, where u is initial velocity, v is final velocity, and t is the time interval. Multiplying both sides by t and adding u gives the first suvat equation:
对于匀加速运动，加速度的定义为 a = (v – u) / t，其中 u 为初速度，v 为末速度，t 为时间间隔。两边同乘 t 并加 u 得到第一个 suvat 方程：
v = u + at
Since velocity changes linearly with time, the average velocity is the arithmetic mean of u and v: vₐᵥ = (u + v)/2. Displacement s is average velocity multiplied by time: s = vₐᵥ t = (u + v)t/2.
由于速度随时间线性变化，平均速度为 u 与 v 的算术平均值：vₐᵥ = (u + v)/2。位移 s 等于平均速度乘以时间：s = vₐᵥ t = (u + v)t/2。
Substituting v = u + at into this expression yields the second equation:
将 v = u + at 代入此式可得第二个方程：
s = ut + ½at²
To remove t, rearrange v = u + at to t = (v – u)/a and substitute into s = (u+v)t/2. After simplifying, you obtain the time-independent relation:
为消去 t，将 v = u + at 变形为 t = (v – u)/a，并代入 s = (u+v)t/2。化简后得到不含时间的方程：
v² = u² + 2as
These three equations form the backbone of kinematics. Their derivation relies only on the definition of acceleration and the concept of average velocity, making them powerful tools you can reconstruct at any time.
这三个方程构成了运动学的支柱。它们的推导仅依赖于加速度的定义和平均速度的概念，使其成为你可以随时重建的强大工具。
3. Newton’s Second Law and Momentum | 牛顿第二定律与动量
Newton’s second law is often stated as F = m a, but its more fundamental form uses momentum p = m v. The net force is the rate of change of momentum:
牛顿第二定律通常表述为 F = m a，但其更基本的形式使用动量 p = m v。合外力等于动量的变化率：
F = Δp / Δt
If the mass of the object remains constant, Δp = m Δv, so F = m Δv / Δt = m a. This derivation clarifies why the equation is valid only when mass does not change—for rockets or relativistic particles, the full momentum form must be used.
若物体质量保持不变，Δp = m Δv，则 F = m Δv / Δt = m a。此推导说明了为何该方程仅在质量不变时成立——对于火箭或相对论性粒子，必须使用完整的动量形式。
Integrating both sides over time gives the impulse–momentum theorem: F Δt = Δ(m v). Impulse equals change in momentum, which explains why a force applied for a longer duration produces a greater velocity change.
对时间积分两边得到冲量–动量定理：F Δt = Δ(m v)。冲量等于动量的变化，这解释了为何力作用的时间越长，产生的速度变化越大。
In collision problems, applying conservation of momentum often requires deriving expressions from F = Δp/Δt and Newton’s third law. For two bodies, F₁₂ = –F₂₁ implies Δp₁/Δt = –Δp₂/Δt, hence Δp₁ + Δp₂ = 0—total momentum is conserved.
在碰撞问题中，应用动量守恒通常需要从 F = Δp/Δt 和牛顿第三定律进行推导。对于两个物体，F₁₂ = –F₂₁ 意味着 Δp₁/Δt = –Δp₂/Δt，因此 Δp₁ + Δp₂ = 0——总动量守恒。
4. Work, Energy and the Work-Energy Theorem | 功、能与动能定理
Work done by a constant force is defined as W = F s cosθ, where s is the displacement and θ is the angle between force and displacement. To link work to kinetic energy, consider a net force F acting along the direction of motion. Using F = m a and the kinematic formula v² = u² + 2as, rearrange to a s = (v² – u²)/2.
恒力做功定义为 W = F s cosθ，其中 s 为位移，θ 为力与位移的夹角。为将功与动能联系起来，考虑沿运动方向的合外力 F。利用 F = m a 和运动学公式 v² = u² + 2as，变形得 a s = (v² – u²)/2。
Substituting into W = F s = m a s gives:
代入 W = F s = m a s 得：
W = m (v² – u²) / 2 = ½m v² – ½m u²
This shows that the net work done on an object equals its change in kinetic energy (ΔK). The expression ½m v² is therefore defined as kinetic energy. This derivation makes it clear that kinetic energy is not an arbitrary concept but a direct consequence of Newton’s laws and kinematics.
这表明对物体所做的净功等于其动能的变化（ΔK）。因此 ½m v² 被定义为动能。此推导清楚地表明，动能并非一个任意的概念，而是牛顿定律和运动学的直接结果。
For gravitational potential energy near Earth’s surface, lifting an object of mass m by height h against gravity requires work W = m g h. This work is stored as potential energy ΔU = m g h, assuming no kinetic change. The conservation of mechanical energy follows when only conservative forces do work.
对于地球表面附近的重力势能，将质量为 m 的物体举高 h 对抗重力需要做功 W = m g h。假设动能不变，此功储存为势能 ΔU = m g h。当只有保守力做功时，机械能守恒便随之成立。
5. Centripetal Acceleration for Circular Motion | 圆周运动向心加速度推导
An object moving at constant speed v in a circle of radius r continually changes direction. In a short time Δt, it sweeps an angle Δθ = (v Δt) / r. The velocity vector rotates by the same angle Δθ. The change in velocity Δv points toward the centre, and its magnitude is approximately:
物体以恒定速率 v 在半径为 r 的圆周上运动，方向不断改变。在短时间 Δt 内，它扫过的角度为 Δθ = (v Δt) / r。速度矢量转过相同的角度 Δθ。速度变化量 Δv 指向圆心，其大小近似为：
|Δv| ≈ v Δθ = v (v Δt / r) = v² Δt / r
Dividing by Δt gives the magnitude of the instantaneous acceleration:
除以 Δt 得到瞬时加速度的大小：
a = v² / r
The direction is radially inward—hence ‘centripetal’. Using angular velocity ω = v/r, this can also be written as a = ω² r. This geometric derivation is preferred in the skills booklet because it avoids calculus while reinforcing vector reasoning.
方向沿半径向内——因此称为“向心”。利用角速度 ω = v/r，也可写成 a = ω² r。基础技能手册中更推荐这种几何推导，因为它避免了微积分，同时强化了矢量推理。
The centripetal force is then given by F = m a = m v² / r. This force is not a new type of force but the net force required to maintain circular motion; it could be tension, gravity, or friction.
向心力则由 F = m a = m v² / r 给出。这个力不是一种新的力，而是维持圆周运动所需的合外力；它可以是张力、重力或摩擦力。
6. Gravitational Field Strength from Newton’s Law | 从万有引力推导重力场强
Newton’s law of universal gravitation states that two point masses attract each other with a force:
牛顿万有引力定律指出，两个质点以如下力相互吸引：
F = G M m / r²
where G is the gravitational constant, M and m are the masses, and r is their separation. The gravitational field strength g at a point is defined as the force per unit mass experienced by a small test mass placed there: g = F/m.
其中 G 为引力常量，M 和 m 为质量，r 为它们之间的距离。引力场强 g 定义为置于该点的小测试质量所受的力与其质量之比：g = F/m。
Substituting the gravitational force expression gives:
代入引力表达式得：
g = G M / r²
This formula shows that the field strength depends only on the source mass M and the distance r. Near Earth’s surface, r ≈ R_E (Earth’s radius), so g ≈ G M_E / R_E² ≈ 9.81 N kg⁻¹. This derivation links the abstract gravitational constant to the familiar acceleration of free fall.
该公式表明场强仅取决于源质量 M 和距离 r。在地球表面附近，r ≈ R_E（地球半径），因此 g ≈ G M_E / R_E² ≈ 9.81 N kg⁻¹。此推导将抽象的引力常量与熟悉的自由落体加速度联系了起来。
In orbit problems, setting centripetal force equal to gravitational force, m v² / r = G M m / r², allows you to derive orbital speed v = √(G M / r). Such derivations are typical of the skills booklet, merging two fundamental principles.
在轨道问题中，令向心力等于引力，即 m v² / r = G M m / r²，可推导出轨道速率 v = √(G M / r)。此类推导是基础技能手册的典型内容，融合了两个基本原理。
7. Electric Field Strength and Potential Gradient | 电场强度与电势梯度
For a uniform electric field between two parallel plates, the field strength E is defined as the force per unit charge: E = F/q. When a charge q moves from one plate to the other, the work done by the field is W = F d = q E d, where d is the plate separation. This work also equals the loss in electrical potential energy, which is q V, with V being the potential difference between the plates.
对于两平行板间的匀强电场，电场强度 E 定义为单位电荷所受的力：E = F/q。当电荷 q 从一板移动到另一板，电场所做的功为 W = F d = q E d，其中 d 为板间距。此功也等于电势能的减少量，即 q V，V 为两板间的电势差。
Equating the two expressions for work:
令两个功的表达式相等：
q E d = q V ⇒ E = V / d
This simple derivation is often tested. It also introduces the concept of potential gradient: in a uniform field, E is the negative of the spatial rate of change of potential. For non-uniform fields, the relation generalises to E = –dV/dr.
这个简单的推导经常被考查。它还引入了电势梯度的概念：在匀强电场中，E 等于电势随空间变化率的负值。对于非匀强电场，此关系推广为 E = –dV/dr。
Understanding this derivation helps explain why the unit of electric field can be V m⁻¹ as well as N C⁻¹. It also underpins the energy method for solving particle motion in electric fields.
理解此推导有助于解释为何电场强度单位既可以是 V m⁻¹，也可以是 N C⁻¹。它也为用能量方法求解带电粒子在电场中的运动奠定了基础。
8. Resistors in Series and Parallel | 电阻的串联与并联公式推导
Resistors in series share the same current. By Ohm’s law, the voltage across each resistor is V₁ = I R₁, V₂ = I R₂, and so on. The total voltage supplied is the sum of individual voltages: V_total = I R₁ + I R₂ + … = I (R₁ + R₂ + …). Hence the equivalent resistance is:
串联的电阻器流过相同的电流。根据欧姆定律，每个电阻两端的电压为 V₁ = I R₁、V₂ = I R₂，以此类推。总电压等于各电压之和：V_total = I R₁ + I R₂ + … = I (R₁ + R₂ + …)。因此等效电阻为：
R_total = R₁ + R₂ + R₃ + …
For resistors in parallel, the voltage across each branch is the same V. The current through each resistor is I₁ = V / R₁, I₂ = V / R₂, etc. The total current supplied is the sum: I_total = V/R₁ + V/R₂ + … = V (1/R₁ + 1/R₂ + …). Since I_total = V / R_total, it follows that:
对于并联电阻器，各支路两端电压相同为 V。通过每个电阻的电流为 I₁ = V / R₁、I₂ = V / R₂ 等。总电流为各支路电流之和：I_total = V/R₁ + V/R₂ + … = V (1/R₁ + 1/R₂ + …)。又因为 I_total = V / R_total，得到：
1 / R_total = 1 / R₁ + 1 / R₂ + 1 / R₃ + …
For two resistors in parallel, this simplifies to R_total = (R₁ R₂) / (R₁ + R₂). These derivations, rooted in conservation of charge (current) and energy (voltage), are fundamental in circuit analysis.
对于两个并联电阻，可简化为 R_total = (R₁ R₂) / (R₁ + R₂)。这些推导植根于电荷守恒（电流）和能量守恒（电压），是电路分析的基础。
Being able to re-derive these formulas ensures you can handle more complex networks, such as series-parallel combinations, without relying solely on memorised shortcuts.
能够重新推导这些公式，可以确保你在处理串并联组合等更复杂的网络时，不完全依赖记忆的捷径。
Published by TutorHao | Physics Revision Series | aleveler.com
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