📚 OxfordAQA PH01 Formula Derivations (Jun 2023 Mark Scheme Insights) | OxfordAQA PH01 公式推导（2023年6月评分方案精要）
Mastering derivations of key formulas is essential for high marks in OxfordAQA AS Physics Unit 1 (PH01). The June 2023 mark scheme reveals that examiners look for clear logical steps, correct use of definitions, and proper algebraic manipulation. This article covers the most important derivations likely to appear in PH01, helping you understand and reproduce them accurately under exam conditions.
掌握关键公式的推导对于在 OxfordAQA 物理 AS 单元1（PH01）中取得高分至关重要。2023年6月的评分方案显示，考官看重清晰的逻辑步骤、正确使用定义以及规范的代数运算。本文涵盖了PH01中最可能出现的核心推导，帮助你深入理解并在考试中准确再现。
1. Deriving the SUVAT Equations | 推导匀加速运动方程
The SUVAT equations describe uniformly accelerated motion in a straight line. They are derived from the definitions of acceleration and average velocity, assuming acceleration is constant.
匀加速运动方程描述了直线上的匀加速运动。它们由加速度和平均速度的定义推导得出，前提是加速度恒定。
Start with the definition of acceleration: a = (v − u) / t. Rearranging gives the first equation.
从加速度的定义出发：a = (v − u) / t。移项即得第一方程。
v = u + at
Since velocity changes linearly with time, the average velocity is (u + v) / 2. Displacement s equals average velocity multiplied by time.
由于速度随时间线性变化，平均速度为 (u + v) / 2。位移 s 等于平均速度乘以时间。
s = ½(u + v)t
Substitute v = u + at into the displacement equation to obtain an expression without the final velocity.
将 v = u + at 代入位移方程，可得到不含末速度的表达式。
s = ut + ½at²
To eliminate t, rearrange v = u + at to t = (v − u) / a, and substitute into s = ½(u + v)t. Simplification yields the fourth equation.
为消去 t，将 v = u + at 改写为 t = (v − u) / a，并代入 s = ½(u + v)t，化简可得第四方程。
v² = u² + 2as
These four equations are the foundation for solving any constant acceleration problem in PH01.
这四个方程是解决 PH01 中任何匀加速问题的基础。
2. Kinetic Energy Formula from Work Done | 由做功推导动能公式
The formula for kinetic energy (KE) is derived by calculating the work done to accelerate an object from rest to speed v.
动能公式的推导通过计算将物体从静止加速到速度 v 所需做的功来完成。
Work done W = F × s. For constant net force F = ma. If initial speed u = 0, then from v² = u² + 2as we get s = v² / (2a).
做功 W = F × s。恒定的净力 F = ma。若初速度 u = 0，由 v² = u² + 2as 可得 s = v² / (2a)。
W = (ma) × (v² / (2a)) = ½mv²
This work is fully converted into kinetic energy, so KE = ½mv². The derivation holds for any constant net force and is central to energy conservation problems.
这部分功全部转化为动能，因此 KE = ½mv²。该推导适用于任何恒净力，是能量守恒问题的核心。
3. Conservation of Momentum from Newton’s Third Law | 由牛顿第三定律推导动量守恒
Momentum is conserved in all isolated systems as a direct consequence of Newton’s laws of motion.
作为牛顿运动定律的直接结果，动量在所有孤立系统中守恒。
During a collision, object A exerts force F on object B, while B exerts force –F on A (Newton’s third law). The collision time Δt is identical for both forces.
碰撞过程中，物体 A 对 B 施加力 F，同时 B 对 A 施加力 –F（牛顿第三定律）。两力作用时间 Δt 相同。
Impulse on A = –FΔt, Impulse on B = FΔt
Impulse equals change in momentum Δp. Therefore, Δpₐ = –FΔt and Δpᵦ = FΔt, so Δpₐ + Δpᵦ = 0. Total momentum before and after the collision is unchanged.
冲量等于动量的变化 Δp。因此 Δpₐ = –FΔt 且 Δpᵦ = FΔt，故 Δpₐ + Δpᵦ = 0。碰撞前后总动量不变。
4. Elastic Potential Energy Stored in a Spring | 弹簧弹性势能推导
When a material obeys Hooke’s law, the force is directly proportional to extension. The energy stored is found from the area under the force–extension graph.
当材料遵循胡克定律时，力与伸长量成正比。储存的能量可由力–伸长量图下的面积求得。
Hooke’s law: F = kx, where k is the spring constant and x is the extension. The graph of F against x is a straight line through the origin.
胡克定律：F = kx，其中 k 为劲度系数，x 为伸长量。F 对 x 的图像为一条过原点的直线。
Work done = average force × extension = ½F x
Substituting F = kx gives Eₑₗ = ½kx². Expressed in load and extension, E = ½FΔL.
代入 F = kx 得 Eₑₗ = ½kx²。用载荷与伸长量表示则为 E = ½FΔL。
5. Gravitational Potential Energy Change (ΔEₚ = mgΔh) | 重力势能变化公式推导
In a uniform gravitational field near Earth’s surface, the change in gravitational potential energy is simply the work done against gravity.
在地表附近的均匀引力场中，重力势能的变化即为克服重力所做的功。
Weight = mg. Lifting an object vertically through height Δh requires force mg over distance Δh.
重力 = mg。将物体垂直提升高度 Δh 需要力 mg 作用距离 Δh。
ΔEₚ = force × vertical distance = mgΔh
This derivation assumes g is constant and the object moves without acceleration; it is the foundation of all energy interchange with height.
该推导假设 g 为常量且物体无加速运动；它是所有与高度有关的能量转换的基础。
6. Deriving the Range of a Projectile | 推导抛体射程公式
Projectile motion is analysed by separating horizontal and vertical components. The range equation follows from the independence of these directions.
抛体运动通过分解水平与竖直分量来分析。射程公式正是基于这两个方向的独立性得出的。
For a projectile launched at speed u at an angle θ to the horizontal: initial horizontal velocity = u cosθ, initial vertical velocity = u sinθ.
对于以速度 u、仰角 θ 发射的抛体：初始水平速度为 u cosθ，初始竖直速度为 u sinθ。
Time of flight T is found from vertical displacement = 0: using s = ut + ½at² with a = –g gives 0 = (u sinθ)T − ½gT², hence T = 2u sinθ / g.
飞行时间 T 由竖直位移为 0 求得：利用 s = ut + ½at²，a = –g，得 0 = (u sinθ)T − ½gT²，故 T = 2u sinθ / g。
Horizontal range R = horizontal velocity × time = (u cosθ) × (2u sinθ / g) = (u² × 2 sinθ cosθ) / g.
水平射程 R = 水平速度 × 时间 = (u cosθ) × (2u sinθ / g) = (u² × 2 sinθ cosθ) / g。
R = u² sin2θ / g
Maximum range occurs at θ = 45°, and the formula assumes no air resistance.
最大射程出现在 θ = 45°，且公式假设无空气阻力。
7. Young Modulus and Strain Energy Density | 杨氏模量与应变能密度
The Young modulus describes the stiffness of a material, linking stress and strain in the elastic region.
杨氏模量描述材料的刚度，关联弹性区域内的应力与应变。
Definition: E = stress / strain = (F/A) / (ΔL/L). Rearranging gives F = (EA/L)ΔL, which is the force–extension relationship for a wire.
定义：E = 应力 / 应变 = (F/A) / (ΔL/L)。移项得 F = (EA/L)ΔL，即金属丝的力–伸长关系式。
The energy stored per unit volume (strain energy density) under elastic deformation is the area under the stress–strain graph.
弹性形变下单位体积储存的能量（应变能密度）是应力–应变图下的面积。
Strain energy density = ½ × stress × strain = ½E ε²
This derivation connects the macroscopic work done to the microscopic elastic energy of bonds, and it is directly tested in PH01 material questions.
该推导将宏观做功与微观键弹性能联系起来，在 PH01 材料题中直接考查。
8. Deriving Centripetal Acceleration a = v²/r | 推导向心加速度公式
An object moving at constant speed in a circle experiences an acceleration towards the centre. The magnitude can be derived using vector geometry.
以恒定速率做圆周运动的物体具有指向圆心的加速度。其大小可通过矢量几何推导。
In a short time Δt, the object moves through angle Δθ = vΔt / r. The change in velocity vector Δv points towards the centre.
在短暂时间 Δt 内，物体转过角度 Δθ = vΔt / r。速度矢量的变化量 Δv 指向圆心。
For small Δθ, |Δv| ≈ v Δθ. Then acceleration a = |Δv| / Δt ≈ v (vΔt / r) / Δt = v²/r.
对于小角度 Δθ，|Δv| ≈ v Δθ。因此加速度 a = |Δv| / Δt ≈ v (vΔt / r) / Δt = v²/r。
a = v² / r
This derivation uses the small-angle approximation and is essential for explaining circular motion in PH01 mechanics.
该推导使用了小角近似，是解释 PH01 力学中圆周运动的关键。
9. Einstein’s Photoelectric Equation | 爱因斯坦光电方程推导
The photoelectric effect provides evidence for the particle nature of light. The equation follows from conservation of energy.
光电效应为光的粒子性提供了证据。该方程由能量守恒推导而来。
A photon of frequency f carries energy E = hf (h is Planck’s constant). When it strikes a metal surface, this energy is transferred to a single electron.
频率为 f 的光子携带能量 E = hf（h 为普朗克常数）。当它撞击金属表面时，该能量转移给单个电子。
The minimum energy needed to liberate an electron is the work function φ. Any surplus energy becomes the electron’s maximum kinetic energy.
释放一个电子所需的最小能量是功函数 φ。多余的能量转化为电子的最大动能。
hf = φ + ½mv²ₘₐₓ
This derivation is fundamental to quantum physics and frequently examined in PH01.
该推导是量子物理的基础，在 PH01 中经常考查。
10. Relating Kinetic Energy and Momentum | 动能与动量的关系推导
Kinetic energy and momentum are closely linked; deriving their relationship is straightforward but immensely useful in problem-solving.
动能与动量紧密相连；推导它们的关系简单直接，但在解题时极其实用。
Momentum p = mv. Kinetic energy KE = ½mv². Express v in terms of p: v = p/m.
动量 p = mv。动能 KE = ½mv²。用 p 表示 v：v = p/m。
KE = ½ m (p/m)² = p² / (2m)
Conversely, p = √(2mKE). This relation avoids velocity altogether and is often used in collision or particle physics problems.
反之，p = √(2mKE)。该关系完全避开了速度，常用于碰撞或粒子物理问题中。
11. Terminal Velocity from Balanced Forces | 由平衡力求终端速度
An object falling through a fluid reaches terminal velocity when the resultant force is zero. This simple derivation is often required in PH01.
物体在流体中下落，当合力为零时达到终端速度。这一简单推导在 PH01 中常常需要。
At terminal speed, weight mg equals the total upward force. If buoyancy is negligible, net upward force is due to drag.
在终端速度下，重力 mg 等于向上的总力。若浮力可忽略，净向上力来自阻力。
For a resistive force proportional to v² (e.g., D = ½CρAv²), setting mg = ½CρAv² gives terminal velocity vₜ = √(2mg / (CρA)).
对于与 v² 成正比的阻力（如 D = ½CρAv²），令 mg = ½CρAv² 可得终端速度 vₜ = √(2mg / (CρA))。
The exam may ask for the derivation using a given drag law; the principle remains: net force zero ⇒ a = 0 ⇒ constant velocity.
考试可能要求利用给定的阻力定律进行推导；原理不变：净合力为零 ⇒ a = 0 ⇒ 恒定速度。
12. Efficiency as a Derived Quantity | 效率的推导量
Efficiency is not a derivation from first principles but a ratio that requires clear understanding of input and output energies or powers. PH01 expects you to derive efficiency expressions from system data.
效率并非从基本原理推导，而是一个需要清晰理解输入与输出能量或功率的比率。PH01 期望你能由系统数据推演出效率表达式。
Efficiency = (useful output) / (total input). In mechanical systems, this may be expressed as output power / input power, or as the ratio of gravitational potential gained to work done.
效率 = （有用输出）/（总输入）。在机械系统中，可表示为输出功率 / 输入功率，或增加的重力势能与所做功的比值。
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