AS Physics Insert 1 Jan22: Formula Derivations | AS物理公式册1（2022年1月）公式推导

📚 AS Physics Insert 1 Jan22: Formula Derivations | AS物理公式册1（2022年1月）公式推导

The AS Physics insert (Data and Formulae sheet) for the January 2022 exam series lists key equations that students are expected to use. However, rote memorisation is far less effective than understanding how these formulas arise. In this article, we derive many of the core relationships from first principles, covering mechanics, materials, electricity and quantum phenomena. By tracing the logical steps behind each equation, you will build a deeper mastery of the subject and be better prepared to tackle unfamiliar problems.

AS物理考试册（2022年1月）列出了学生需要使用的关键公式。然而，死记硬背远不如理解这些公式的来龙去脉有效。本文从基本原理出发，推导力学、材料、电学和量子现象中的许多核心关系式。通过追溯每个方程背后的逻辑步骤，你将建立起对学科更深入的掌握，并能更好地应对陌生问题。

1. Deriving v = u + at from Acceleration | 由加速度定义推导 v = u + at

Acceleration a is defined as the rate of change of velocity. For uniform acceleration, a = (v – u) / t. Rearranging gives v = u + at.

加速度 a 定义为速度的变化率。对于匀加速，a = (v – u) / t。移项得 v = u + at。

v = u + at

2. Deriving s = ut + ½at² from a v–t Graph | 由 v–t 图推导 s = ut + ½at²

For uniform acceleration, the velocity–time graph is a straight line. The displacement s is the area under the graph. This area is a trapezium with parallel sides u and v at t=0 and t=t: area = ½ (u + v) t. Substituting v = u + at yields s = ½ (u + u + at) t = ut + ½at².

对于匀加速运动，速度–时间图是一条直线。位移 s 是图线下的面积。该面积是一个梯形，平行边分别为 t=0 时的 u 和 t 时的 v：面积 = ½ (u + v) t。将 v = u + at 代入，得到 s = ½ (u + u + at) t = ut + ½at²。

s = ut + ½at²

3. Deriving v² = u² + 2as by Eliminating t | 消去时间 t 推导 v² = u² + 2as

From v = u + at we get t = (v – u)/a. Insert this into s = ut + ½at². Then s = u((v-u)/a) + ½a((v-u)/a)². Simplify: s = (uv – u²)/a + ½a(v² – 2uv + u²)/a² = (uv – u²)/a + (v² – 2uv + u²)/(2a). Combine over denominator 2a: 2uv – 2u² + v² – 2uv + u² = v² – u², so s = (v² – u²)/(2a), hence v² = u² + 2as.

由 v = u + at 得 t = (v – u)/a。将其代入 s = ut + ½at²。则 s = u((v-u)/a) + ½a((v-u)/a)²。化简：s = (uv – u²)/a + ½a(v² – 2uv + u²)/a² = (uv – u²)/a + (v² – 2uv + u²)/(2a)。通分合并分子：2uv – 2u² + v² – 2uv + u² = v² – u²，因此 s = (v² – u²)/(2a)，移项得 v² = u² + 2as。

v² = u² + 2as

4. Newton’s Second Law: F = ma from Momentum | 由动量变化推导 F = ma

The resultant force on an object equals the rate of change of its momentum: F = Δp/Δt. If mass m is constant, Δp = m(v – u), so F = m(v – u)/Δt = ma because a = (v – u)/Δt. Thus F = ma.

物体所受的合外力等于其动量的变化率：F = Δp/Δt。若质量 m 恒定，则 Δp = m(v – u)，故 F = m(v – u)/Δt = ma，因为 a = (v – u)/Δt。因此 F = ma。

F = ma

5. Work Done and Kinetic Energy: Eₖ = ½mv² | 功与动能：Eₖ = ½mv²

Consider a constant resultant force F acting on a mass m initially at rest over displacement s. Using v² = u² + 2as with u=0 gives v² = 2as, so a = v²/(2s). Then F = ma = m v²/(2s). The work done W = Fs = (m v²/(2s)) × s = ½mv². This is the kinetic energy gained, so Eₖ = ½mv².

考虑恒定的合外力 F 作用在初速为零的质量 m 上，位移为 s。由 v² = u² + 2as，u=0 得 v² = 2as，故 a = v²/(2s)。于是 F = ma = m v²/(2s)。所做的功 W = Fs = (m v²/(2s)) × s = ½mv²。这就是获得的动能，因此 Eₖ = ½mv²。

Eₖ = ½mv²

6. Gravitational Potential Energy: ΔEₚ = mgΔh | 重力势能变化：ΔEₚ = mgΔh

When lifting an object of mass m at constant speed through a vertical height Δh, the force needed equals the weight mg. Work done W = force × distance = mg × Δh. This work is stored as gravitational potential energy, so ΔEₚ = mgΔh.

当以恒定速度将质量为 m 的物体竖直提升 Δh 高度时，所需的力等于重力 mg。所做的功 W = 力 × 距离 = mg × Δh。该功储存为重力势能，因此 ΔEₚ = mgΔh。

ΔEₚ = mgΔh

7. Hooke’s Law and Elastic Potential Energy | 胡克定律与弹性势能

For a spring obeying Hooke’s law, F = kΔL, where k is the spring constant. The force varies linearly with extension, so the average force during stretching is ½F. Work done = average force × extension = ½FΔL. Substituting F = kΔL gives E = ½k(ΔL)². This is the elastic potential energy stored.

对于满足胡克定律的弹簧，F = kΔL，k 为劲度系数。力与伸长量成线性关系，拉伸过程中的平均力为 ½F。所做的功 = 平均力 × 伸长量 = ½FΔL。代入 F = kΔL 得 E = ½k(ΔL)²。这就是储存的弹性势能。

F = kΔL, E = ½FΔL = ½k(ΔL)²

8. Resistivity Formula R = ρL/A | 电阻率公式 R = ρL/A

Experiment shows that for a given material, resistance R is proportional to length L and inversely proportional to cross‑sectional area A. Introducing the constant of proportionality ρ (resistivity) gives R = ρL/A. This definition is the basis for comparing different conductors.

实验表明，对于给定的材料，电阻 R 与长度 L 成正比，与横截面积 A 成反比。引入比例常数 ρ（电阻率），即得 R = ρL/A。这一定义是比较不同导体的基础。

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AS Physics Insert 1 Jan22: Formula Derivations | AS物理公式册1（2022年1月）公式推导