📚 Deriving the A-Level Physics Insert (Unit 1, Jan 2021) Formulas | A-Level物理插入页（单元1，2021年1月）公式推导
The A-Level Physics data and formulae insert for Unit 1 (January 2021) is a vital companion for students on exam day. It lists essential constants, wave equations, particle relations, mechanics and materials formulas. This article walks through the derivations of many key equations found in that insert, showing how each arises from fundamental definitions and principles. Gaining this deeper understanding will help you recall and apply them with confidence.
A-Level物理单元1（2021年1月）的数据与公式插入页是考试日学生不可缺少的帮手。它列出了基本常数、波动方程、粒子关系式、力学和材料公式。本文逐一推导了该插入页中的许多关键公式，展示它们如何从基本定义和原理中产生。掌握这些推导将有助于你自信地回忆和应用它们。
1. Photon Energy and the Wave Equation | 光子能量与波动方程
The insert gives the fundamental relation between photon energy E, frequency f and Planck’s constant h: E = hf. Because light travels at speed c, its frequency and wavelength are connected by c = fλ. Substituting f = c/λ directly yields the alternative form E = hc/λ. This shows how a photon’s energy is inversely proportional to its wavelength, explaining why ultraviolet photons are more energetic than infrared ones.
插入页给出了光子能量 E、频率 f 与普朗克常数 h 的基本关系：E = hf。由于光速为 c，光的频率与波长满足 c = fλ。代入 f = c/λ 即得另一种形式 E = hc/λ。这表明光子能量与波长成反比，解释了为何紫外光子比红外光子能量更高。
c = fλ → E = hf = hc/λ
2. de Broglie Wavelength | 德布罗意波长
The wave–particle duality concept is captured in the de Broglie equation on the insert: λ = h/p. Here p = mv is the linear momentum. The derivation originally came from combining Einstein’s E = hf with the relativistic energy–momentum relation for a photon; for a particle of momentum p, the associated wavelength is simply λ = h/p. For a non‑relativistic electron accelerated through a potential difference, the wavelength becomes λ = h/√(2meV).
波粒二象性体现在插入页的德布罗意公式中：λ = h/p。其中 p = mv 是线动量。这一推导最初来自把爱因斯坦的 E = hf 与光子的相对论能量–动量关系结合；对于动量为 p 的粒子，其伴随波长就是 λ = h/p。对于经电势差加速的非相对论电子，波长可表为 λ = h/√(2meV)。
λ = h/p = h/(mv)
3. Einstein’s Photoelectric Equation | 爱因斯坦光电效应方程
In the photoelectric effect, a photon gives all its energy hf to a single electron. Some energy φ (the work function) is needed to escape the metal; any surplus appears as the electron’s maximum kinetic energy Eₖ(max). Hence hf = φ + Eₖ(max). The stopping potential Vₛ relates to this kinetic energy by Eₖ(max) = eVₛ, giving hf = φ + eVₛ. These two forms, both present on the insert, allow calculation of Planck’s constant and work function from shift data.
在光电效应中，一个光子把它所有的能量 hf 给予一个电子。一部分能量 φ（逸出功）用于从金属中逸出；剩余部分表现为电子的最大动能 Eₖ(max)。因此 hf = φ + Eₖ(max)。截止电压 Vₛ 与该动能的关系是 Eₖ(max) = eVₛ，于是 hf = φ + eVₛ。插入页上的这两种形式，可通过截止电压数据求得普朗克常数和逸出功。
hf = φ + ½ m v²_max = φ + eVₛ
4. The Electronvolt and Energy Conversion | 电子伏特与能量转换
The insert lists the conversion factor 1 eV = 1.60 × 10⁻¹⁹ J. This follows directly from the definition of the electronvolt: the kinetic energy gained by an electron when it is accelerated through a potential difference of 1 volt. Since the work done on a charge q by a voltage V is W = qV, substituting q = e and V = 1 V gives 1 eV = e × 1 V = 1.60 × 10⁻¹⁹ C × 1 V = 1.60 × 10⁻¹⁹ J.
插入页给出了换算因子 1 eV = 1.60 × 10⁻¹⁹ J。这直接来自电子伏特的定义：一个电子经1伏电势差加速后获得的动能。由于电压 V 对电荷 q 做功为 W = qV，代入 q = e 和 V = 1 V，即得 1 eV = e × 1 V = 1.60 × 10⁻¹⁹ C × 1 V = 1.60 × 10⁻¹⁹ J。
Eₖ = eV → 1 eV = 1.60 × 10⁻¹⁹ J
5. Momentum and Impulse | 动量与冲量
Momentum is defined as p = m v. The impulse F Δt delivered by a resultant force changes the momentum: F Δt = Δp. This relationship is Newton’s second law in its most general form F = Δp/Δt. When mass is constant, Δp = m Δv, leading to the familiar F = m a. The insert collects these as p = mv and F = Δ(mv)/Δt.
动量定义为 p = m v。合力施加的冲量 F Δt 会改变动量：F Δt = Δp。这就是牛顿第二定律的最普遍形式 F = Δp/Δt。当质量不变时，Δp = m Δv，便得到熟悉的 F = m a。插入页将这些总结为 p = mv 和 F = Δ(mv)/Δt。
p = mv ; F = Δp/Δt = ma
6. Newton’s Second Law Derivation | 牛顿第二定律的推导
Starting from F = Δp/Δt and considering a constant mass m, we have Δp = m(v – u). Therefore F = m(v – u)/Δt = m a because acceleration a = (v – u)/Δt. This explains the insert’s listing of F = ma as a special case. Additionally, for a falling object, the weight is simply W = mg, obtained by setting a = g.
从 F = Δp/Δt 出发并考虑质量 m 恒定，有 Δp = m(v – u)。因此 F = m(v – u)/Δt = m a，因为加速度 a = (v – u)/Δt。这解释了插入页将 F = ma 列为一个特例。此外，对下落物体，令 a = g 即得重量 W = mg。
F = Δp/Δt = mΔv/Δt = ma
7. Kinetic Energy Formula | 动能公式
Kinetic energy can be derived from the work done by a constant net force. Work W = F s. Using F = ma and the kinematic relation v² = u² + 2as (which rearranges to s = (v² – u²)/(2a)), we substitute: W = ma × (v² – u²)/(2a) = ½ m v² – ½ m u². Hence the change in kinetic energy is ΔEₖ = ½ m (v² – u²), and for a body starting from rest, Eₖ = ½ m v².
动能可以从恒定合力所做的功推导出来。功 W = F s。利用 F = ma 和运动学关系式 v² = u² + 2as（变形为 s = (v² – u²)/(2a)），代入得 W = ma × (v² – u²)/(2a) = ½ m v² – ½ m u²。因此动能的改变量是 ΔEₖ = ½ m (v² – u²)，对从静止开始的物体，Eₖ = ½ m v²。
Eₖ = ½ m v²
8. Work Done by a Force at an Angle | 力在夹角下所做的功
When a constant force F acts at an angle θ to the displacement s, only the component F cosθ along the displacement does work. Thus W = (F cosθ) × s, which is written compactly as W = Fs cosθ. If the force is perpendicular to the motion (θ = 90°), no work is done. This general expression, on the insert, reduces to W = Fs when the force and displacement are parallel.
当恒力 F 与位移 s 的夹角为 θ 时，只有沿位移方向的分量 F cosθ 做功。因此 W = (F cosθ) × s，简写为 W = Fs cosθ。如果力垂直于运动方向（θ = 90°），则不做功。这个通用表达式（在插入页上）在力与位移平行时退化为 W = Fs。
W = F s cosθ
9. Power as Force × Velocity | 功率等于力乘以速度
Power is the rate of doing work: P = ΔW/Δt. If a constant force F acts along the direction of motion, the small work done in time Δt is ΔW = F Δs. Therefore P = F Δs/Δt = F v, where v is the instantaneous velocity. This relation appears on the insert as P = Fv and is extremely useful for vehicle and conveyor belt problems.
功率是做功的速率：P = ΔW/Δt。若恒力 F 沿着运动方向作用，在 Δt 时间内做的微功为 ΔW = F Δs。因此 P = F Δs/Δt = F v，其中 v 是瞬时速度。这一关系（即插入页上的 P = Fv）在车辆和传送带问题中极为有用。
P = ΔW/Δt = F Δs/Δt = Fv
10. Gravitational Potential Energy Change | 重力势能变化
Lifting an object of weight mg through a vertical height Δh requires work against gravity. The minimum lifting force equals mg, so the work done is W = mg × Δh. This work is stored as gravitational potential energy, giving ΔEₚ = mgΔh. This expression, often written on the insert, assumes g is constant near the Earth’s surface.
将重为 mg 的物体竖直举高 Δh，需要克服重力做功。最小举力等于 mg，因此做功 W = mg × Δh。这些功储存为重力势能，即 ΔEₚ = mgΔh。这一表达式（常见于插入页）假定地表附近 g 为常量。
ΔEₚ = mgΔh
11. Hooke’s Law and Elastic Strain Energy | 胡克定律与弹性应变能
Hooke’s law states that the extension ΔL of a spring is proportional to the applied force, within the elastic limit: F = k ΔL, where k is the spring constant. The work done in stretching the spring is the area under the force–extension graph, a triangle, so E = ½ F ΔL. Substituting F = k ΔL gives the strain energy forms E = ½ k (ΔL)². These appear on the insert as E = ½ F ΔL and the related Young modulus energy.
胡克定律指出，在弹性限度内弹簧的伸长量 ΔL 与所加力成正比：F = k ΔL，其中 k 是劲度系数。拉伸弹簧所做的功等于力–伸长图下的面积（三角形），因此 E = ½ F ΔL。代入 F = k ΔL 即得应变能形式 E = ½ k (ΔL)²。插入页上包括 E = ½ F ΔL 和相关杨氏模量能量形式。
F = kΔL ; E = ½ FΔL = ½ k(ΔL)²
12. Young Modulus and Stress–Strain Relationship | 杨氏模量及应力–应变关系
The insert provides the definitions of stress and strain: stress = F/A, strain = ΔL/L. The Young modulus E is the ratio of tensile stress to tensile strain within the elastic region: E = stress / strain = (F/A) / (ΔL/L), which simplifies to E = FL / AΔL. This is not a derived law but a definition; however, combining it with the elastic strain energy gives the energy stored per unit volume as ½ × stress × strain.
插入页给出应力和应变的定义：应力 = F/A，应变 = ΔL/L。杨氏模量 E 是弹性区域内拉伸应力与拉伸应变之比：E = 应力 / 应变 = (F/A) / (ΔL/L)，简化得 E = FL / AΔL。这不是推导出的定律，而是一个定义；不过将它和弹性应变能结合可得单位体积储存的能量为 ½ × 应力 × 应变。
E = (F/A) / (ΔL/L) = FL / AΔL
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