📚 Essential Formula Derivations Based on OxfordAQA PH01 January 2023 Mark Scheme | 基于 OxfordAQA PH01 2023年1月评分方案的重要公式推导
In the OxfordAQA AS Physics Unit 1 (PH01) examination, a significant proportion of marks is awarded for demonstrating the ability to derive key formulae from first principles. The mark scheme often highlights the step‑by‑step logical pathway, clarity of algebraic manipulation, and correct use of definitions and fundamental concepts. This article reconstructs the essential derivations that commonly appear in such papers, based on the style of the January 2023 mark scheme. Each derivation is presented as a sequence of clear steps, with English and Chinese explanations paired to reinforce understanding.
在 OxfordAQA AS 物理单元 1 (PH01) 考试中，相当大比例的分数用于考查从基本原理推导关键公式的能力。评分方案通常会强调逐步的逻辑推导过程、清晰的代数运算以及对定义和基本概念的正确使用。本文基于 2023 年 1 月评分方案的风格，重现了试卷中常见的核心推导。每个推导都以清晰的步骤呈现，中英文说明配对，以加深理解。
1. Deriving the First Equation of Motion: v = u + at | 推导第一运动学方程：v = u + at
velocity. For uniform acceleration a, the change in velocity over time t is simply a × t.
加速度定义为速度的变化率。对于匀加速度 a，在时间 t 内速度的变化量等于 a × t。
with the definition: a = (v - u) / t, where v is final velocity and u is initial velocity.
从定义出发：a = (v - u) / t，其中 v 为末速度，u 为初速度。
Multiply both sides by t: a t = v - u.
两边同乘以 t：a t = v - u。
Add u to both sides to obtain v = u + a t. This is the first kinematic equation.
两边同时加上 u，得到 v = u + a t。这就是第一运动学方程。
2. Deriving the Displacement–Time Relation: s = ut + ½at² | 推导位移-时间关系：s = ut + ½at²
Displacement s is defined as the area under a velocity–time graph. For uniform acceleration, the graph is a straight line, and the area is the sum of a rectangle and a triangle.
位移 s 定义为速度-时间图像下的面积。对于匀加速运动，图像是一条直线，其面积由一个矩形和一个三角形组成。
locity can also be used: s = (u + v)/2 × t. Substitute v = u + at from the first equation.
也可以利用平均速度：s = (u + v)/2 × t。将第一方程 v = u + at 代入。
s = (u + u + at)/2 × t = (2u + at)/2 × t = u t + ½ a t².
s = (u + u + at)/2 × t = (2u + at)/2 × t = u t + ½ a t²。
Thus the displacement after time t under uniform acceleration is s = u t + ½ a t².
因此在匀加速下，时间 t 内的位移为 s = u t + ½ a t²。
3. Deriving v² = u² + 2as | 推导 v² = u² + 2as
This relation eliminates time from the previous equations. Start from the definitions and combine them algebraically.
这个关系式将时间从前两个方程中消去。从定义出发并进行代数组合。
We have v = u + at and s = (u + v)t / 2. Express t from the first: t = (v - u)/a.
我们有 v = u + at 和 s = (u + v)t / 2。由第一式解出 t：t = (v - u)/a。
Substitute t into the displacement equation: s = (u + v)/2 × (v - u)/a = (v² - u²) / (2a).
将 t 代入位移方程：s = (u + v)/2 × (v - u)/a = (v² - u²) / (2a)。
Multiply both sides by 2a to obtain the familiar form: v² = u² + 2 a s.
两边同乘以 2a 得到常见形式：v² = u² + 2 a s。
This equation is especially useful when time is not given or required.
当时间未给出或不需要时，这个方程特别有用。
4. Newton’s Second Law and the Impulse–Momentum Relationship | 牛顿第二定律与冲量-动量关系
F is proportional to the rate of change of momentum p. In its simplest form, F = Δp / Δt.
牛顿第二定律指出，净外力 F 与动量 p 的变化率成正比。其最简形式为 F = Δp / Δt。
nstant, Δp = m(v - u). The impulse J delivered by a constant force over time t is J = F t.
动量为 p = m v。若质量 m 恒定，则 Δp = m(v - u)。恒力在时间 t 内提供的冲量 J 为 J = F t。
Equating impulse to change in momentum: F t = m v - m u.
令冲量等于动量的变化：F t = m v - m u。
This derivation is often assessed in the mark scheme, requiring candidates to clearly state the assumption of constant mass and the vector nature of momentum.
评分方案常考查这一推导，要求考生明确说明质量恒定的假设以及动量的矢量性。
5. Deriving the Kinetic Energy Formula: Eₖ = ½mv² | 推导动能公式：Eₖ = ½mv²
g a mass m from rest (u = 0) over a displacement s. The work done by the force is W = F s.
考虑恒定的净外力 F 使质量为 m 的物体从静止 (u = 0) 开始在位移 s 上加速。力做的功为 W = F s。
Using F = m a and the kinematic relation v² = 0² + 2 a s, so a s = v² / 2.
利用 F = m a 和运动学关系 v² = 0² + 2 a s，可得 a s = v² / 2。
Therefore, W = m a × s = m × (v² / 2) = ½ m v². This work is stored as kinetic energy Eₖ.
因此，W = m a × s = m × (v² / 2) = ½ m v²。此功储存为动能 Eₖ。
xtended to an initial speed u by using v² - u² = 2as, giving W = ½ m v² - ½ m u², the work–energy connection.
该推导可推广到有初速度 u 的情况，利用 v² - u² = 2as，得到 W = ½ m v² - ½ m u²，即功与能的联系。
6. Elastic Potential Energy for a Spring Obeying Hooke’s Law: Eₚ = ½FΔx | 服从胡克定律的弹簧的弹性势能：Eₚ = ½FΔx
it of proportionality, the force F is directly proportional to the extension Δx: F = k Δx.
对于弹簧或任何处于比例极限内的弹性材料，力 F 与伸长量 Δx 成正比：F = k Δx。
The work done to stretch the spring is the area under the force–on graph, which is a triangle. Hence the energy stored is Eₚ = ½ × base × height = ½ F Δx.
拉伸弹簧所做的功等于力-伸长图下方的面积，该形状为三角形。因此储存的能量为 Eₚ = ½ × 底 × 高 = ½ F Δx。
½ k (Δx)². The mark scheme often rewards stating the condition that Hooke’s law is obeyed.
代入 F = k Δx 得到另一种形式 Eₚ = ½ k (Δx)²。评分方案通常奖励说明满足胡克定律的条件。
7. Young Modulus: Deriving the Formula E = σ/ε | 杨氏模量：推导公式 E = σ/ε
The Young modulus E quantifies the stiffness of a material. It is defined as tensile stress σ divided by tensile strain ε, provided the material is under elastic deformation.
杨氏模量 E 量化材料的刚度。其定义为拉伸应力 σ 与拉伸应变 ε 之比，前提是材料处于弹性形变。
Stress σ = F / A, where F is the applied force and A is the original cross‑sectional area. Strain ε = ΔL / L₀, the extension per unit original length.
应力 σ = F / A，其中 F 为施加的力，A 为原始横截面积。应变 ε = ΔL / L₀，即单位原始长度的伸长量。
/ A) / (ΔL / L₀) = (F L₀) / (A ΔL). The unit of E is the pascal (Pa), equivalent to N m⁻².
因此 E = (F / A) / (ΔL / L₀) = (F L₀) / (A ΔL)。E 的单位是帕斯卡 (Pa)，等同于 N m⁻²。
n also be rearranged to ΔL = (F L₀) / (A E), which is often used in experimental analysis.
该公式也可改写为 ΔL = (F L₀) / (A E)，常用于实验分析。
8. Work Done by a Constant Force and the Work–Energy Theorem | 恒力做功与功能定理
When a constant force F acts on an object moving through a displacement s, the work done W is the component of force along the displacement times the displacement.
当一个恒力 F 作用在物体上并使其发生位移 s，所做的功 W 等于力在位移方向的分量与位移的乘积。
the angle between F and s is θ, then W = F s cosθ. This is the general definition of work.
若 F 与 s 之间的夹角为 θ，则 W = F s cosθ。这是功的普遍定义。
The work–energy theorem states that the net work done on an object equals its change in kinetic energy: Wnet = ΔEₖ = ½ m v² - ½ m u².
功能定理指出，作用在物体上的净功等于其动能的变化：Wnet = ΔEₖ = ½ m v² - ½ m u²。
wton’s second law and the kinematic equation v² = u² + 2 a s, assuming constant net force.
该定理可通过结合牛顿第二定律和运动学方程 v² = u² + 2 a s 推导得出，假设净力恒定。
9. Wave Speed, Frequency and Wavelength: v = fλ | 波速、频率与波长：v = fλ
The wave speed v is the distance travelled by a wavefront per unit time. One full wavelength λ corresponds to one complete cycle.
波速 v 是波前在单位时间内传播的距离。一个完整的波长 λ 对应一个完整的周期。
f is the number of cycles per second, so the time for one cycle (the period T) is T = 1/f.
频率 f 是每秒的周期数，因此一个周期的时间（周期 T）为 T = 1/f。
wave advances by one wavelength. Hence speed = distance / time = λ / T = λ × (1/T) = f λ.
在一个周期内，波前进一个波长。因此速度 = 距离 / 时间 = λ / T = λ × (1/T) = f λ。
This simple but fundamental relationship is expected to be derived or applied confidently in the AS examination.
这个简单但基础的关系式在 AS 考试中要求能够自信地进行推导或应用。
10. Deriving Snell’s Law from the Wave Speed Ratio | 从波速比推导斯涅尔定律
Snell’s law describes the refraction of a wave at a boundary between two media. It can be derived by considering wavefronts and wave speeds v₁ and v₂.
斯涅尔定律描述了波在两种介质界面处的折射。可以通过考虑波前和波速 v₁、v₂ 来推导。
When a plane wavefront strikes a boundary at an angle, the side that enters the new medium first changes speed, causing the wavefront to bend.
当平面波前以一定角度入射到界面时，先进入新介质的部分速度发生变化，导致波前弯曲。
₂ / v₂, where θ₁ and θ₂ are the angles of incidence and refraction measured to the normal.
根据波前的几何关系，有 sinθ₁ / v₁ = sinθ₂ / v₂，其中 θ₁ 和 θ₂ 分别为入射角和折射角，均相对于法线测量。
/ v, where c is the speed of light in a vacuum. This is the standard form of Snell’s law.
整理后得到 n₁ sinθ₁ = n₂ sinθ₂，其中折射率 n = c / v，c 为真空中的光速。这就是斯涅尔定律的标准形式。
The mark scheme frequently tests the ability to link wave speed, refractive index and the bending of rays.
评分方案经常考查联系波速、折射率与光线弯曲的能力。
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