📚 International AS-Level Physics Example Responses PH01 Unit 1: Formula Derivations | PH01 单元 1 示例作答：公式推导
Mastering formula derivations is essential for success in the International AS Physics Unit 1 (PH01) exam. Understanding where key equations come from not only helps you remember them but also enables you to apply them correctly in unfamiliar contexts. This article walks you through the step‑by‑step derivation of the most important mechanics and materials formulas required for the Edexcel IAL Physics specification.
掌握公式推导是在国际 AS 物理第一单元 (PH01) 考试中取得好成绩的关键。理解关键方程的来源不仅能帮助你记忆，还能使你在不熟悉的场景中正确应用它们。本文将带你逐步推导爱德思 IAL 物理大纲中力学和材料部分最重要的公式。
1. Introduction to Derivation Skills | 推导技能入门
Derivations in physics rely on clear definitions, algebraic manipulation, and sometimes graphical interpretation. Always start by writing down the fundamental definitions or laws you are allowed to use. Then work logically towards the target formula, showing every step. Examiners award marks for correct reasoning, not just the final answer.
物理中的推导依赖于清晰的定义、代数运算，有时还需要图像解释。始终从写下你可以使用的基本定义或定律开始，然后逻辑严谨地朝着目标公式推进，展示每一步。考官对正确的推理给分，而不仅仅是看最终答案。
2. Deriving the First Equation of Motion (v = u + at) | 推导第一个运动方程 (v = u + at)
Start with the definition of uniform acceleration: acceleration is the rate of change of velocity. If an object starts with initial velocity u and accelerates uniformly at a for a time t, the change in velocity is a × t. Therefore the final velocity v is u plus the change. So v = u + at.
从匀加速度的定义出发：加速度是速度的变化率。如果一个物体以初速度 u 开始，以加速度 a 均匀加速时间 t，速度的变化量为 a × t。因此末速度 v 等于 u 加上这个变化量，即 v = u + at。
a = (v − u) / t → v = u + at
从加速度定义式重新整理：a = (v − u) / t，两边乘以 t 再移项即得到 v = u + at。
3. Deriving the Second Equation of Motion (s = ut + ½at²) | 推导第二个运动方程 (s = ut + ½at²)
Displacement s is given by average velocity multiplied by time. For uniform acceleration, average velocity = (initial velocity + final velocity)/2. Substituting v = u + at gives average velocity = (u + u + at)/2 = u + ½at. Multiplying by t yields s = ut + ½at².
位移 s 由平均速度乘以时间给出。对于匀加速度，平均速度 = (初速度 + 末速度)/2。代入 v = u + at 得到平均速度 = (u + u + at)/2 = u + ½at。再乘以时间 t 便得到 s = ut + ½at²。
s = ( (u+v)/2 ) × t = (u + (u+at))/2 × t = ut + ½at²
另一种方法：用速度‑时间图下的面积。梯形面积 = (u+v)t/2，同样结果。
4. Deriving the Third Equation of Motion (v² = u² + 2as) | 推导第三个运动方程 (v² = u² + 2as)
Eliminate t from the first two equations. From v = u + at, we have t = (v − u)/a. Substitute into s = ut + ½at². After algebraic simplification, you obtain v² = u² + 2as. This equation is useful when time is not known.
从前两个方程中消去 t。由 v = u + at 得 t = (v − u)/a。代入 s = ut + ½at²，经代数化简后得到 v² = u² + 2as。这个方程在不知道时间时非常有用。
s = u(t) + ½a(t)² = u((v−u)/a) + ½a((v−u)/a)² → v² = u² + 2as
5. Kinetic Energy Formula (Eₖ = ½mv²) | 动能公式 (Eₖ = ½mv²)
Consider a constant net force F accelerating a mass m from rest to speed v over a distance s. Work done = F × s. Using F = ma and from v² = u² + 2as with u=0 we get a = v²/(2s). Then work done = m × (v²/(2s)) × s = ½mv². This work is stored as kinetic energy.
考虑一个恒定的合力 F 将质量为 m 的物体从静止加速到速度 v，位移为 s。做的功 = F × s。应用 F = ma，并由 v² = u² + 2as 令 u=0 得 a = v²/(2s)。因此功 = m × (v²/(2s)) × s = ½mv²。这个功就储存为动能。
W = Fs = ma × s = m × (v²/(2s)) × s = ½mv² → Kinetic energy = ½mv²
6. Change in Gravitational Potential Energy (ΔEₚ = mgΔh) | 重力势能的变化 (ΔEₚ = mgΔh)
To lift an object of mass m through a vertical height Δh at constant speed, the lifting force must equal the weight mg. Work done = force × distance = mg × Δh. This work increases the gravitational potential energy. Therefore ΔEₚ = mgΔh.
要以恒定速度将质量为 m 的物体竖直提升 Δh，提升力必须等于重力 mg。做的功 = 力 × 距离 = mg × Δh。这个功增加了重力势能，所以 ΔEₚ = mgΔh。
7. Impulse and Change in Momentum (FΔt = Δp) | 冲量与动量变化 (FΔt = Δp)
Newton’s second law can be written as F = Δp/Δt, where p = mv is momentum. Rearranging gives FΔt = Δp, which is the impulse–momentum theorem. For a constant force, impulse equals the change in momentum of an object.
牛顿第二定律可以写作 F = Δp/Δt，其中 p = mv 是动量。移项得 FΔt = Δp，这就是冲量‑动量定理。对于恒定力，冲量等于物体动量的变化量。
F = Δp / Δt = (mv − mu) / Δt → FΔt = mv − mu
8. Conservation of Linear Momentum | 线性动量守恒
When two objects interact, Newton’s third law states that the forces they exert on each other are equal and opposite. If the net external force on a system is zero, the total change in momentum of the system is zero. Hence total momentum before collision equals total momentum after collision.
当两个物体相互作用时，牛顿第三定律指出它们彼此施加的力大小相等、方向相反。如果系统所受合外力为零，系统总动量的变化就为零，因此碰撞前总动量等于碰撞后总动量。
For a two‑body collision: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. This can be derived by noting that F₁₂ = –F₂₁, so m₁a₁ = –m₂a₂, which leads to m₁(v₁−u₁) + m₂(v₂−u₂) = 0.
对于两体碰撞：m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。这可以通过 F₁₂ = –F₂₁ 推出，即 m₁a₁ = –m₂a₂，进而得到 m₁(v₁−u₁) + m₂(v₂−u₂) = 0。
9. Hooke’s Law and Elastic Potential Energy | 胡克定律与弹性势能
For a spring obeying Hooke’s law, the force F = kx, where x is the extension and k is the spring constant. The work done in stretching the spring from 0 to x is the area under the force‑extension graph, which is a triangle. Work = ½ × force × extension = ½ × kx × x = ½kx².
对于遵守胡克定律的弹簧，力 F = kx，其中 x 是伸长量，k 是劲度系数。将弹簧从 0 拉伸到 x 所做的功等于力‑伸长图下的面积，即一个三角形。功 = ½ × 力 × 伸长 = ½ × kx × x = ½kx²。
Elastic potential energy stored = ½kx²
This energy is recoverable when the spring returns to its original length, assuming the elastic limit is not exceeded.
只要不超过弹性极限，这些能量在弹簧恢复原长时可以重新释放。
10. Young Modulus and Stress‑Strain Relationship | 杨氏模量与应力‑应变关系
Young modulus E is defined as stress/strain for a material under elastic deformation. Stress σ = F/A, where F is the applied force and A is the cross‑sectional area. Strain ε = ΔL/L, where ΔL is the extension and L is the original length. Therefore E = (F/A) / (ΔL/L) = FL / (AΔL).
杨氏模量 E 定义为材料在弹性形变时的应力与应变之比。应力 σ = F/A，F 是施加的力，A 是横截面积。应变 ε = ΔL/L，ΔL 是伸长量，L 是原长。因此 E = (F/A) / (ΔL/L) = FL / (AΔL)。
This relationship can be rearranged to find the extension of a wire: ΔL = FL / (AE). The equation is often tested in the PH01 exam, especially when combined with Hooke’s law and the spring constant k = EA/L.
这个关系可以重新整理以求出金属丝的伸长：ΔL = FL / (AE)。该方程在 PH01 考试中经常出现，特别是与胡克定律以及 k = EA/L 结合考查时。
11. Exam Tips for Showing Derivations | 展示推导的考试技巧
In the exam, always state the assumption(s) you are making, e.g. ‘constant acceleration’, ‘no air resistance’, or ‘elastic limit not exceeded’. Write the starting formula clearly, then carry out each algebraic step. If you get stuck, check the units – correct derivations must be dimensionally consistent.
在考试中，始终说明你所做的假设，例如“匀加速度”、“无空气阻力”或“不超过弹性极限”。清晰地写出起始公式，然后逐步进行代数运算。如果遇到困难，检查单位——正确的推导必须在量纲上一致。
Practise writing derivations without looking at notes; this builds deep understanding and helps you answer ‘explain’ or ‘show that’ questions with confidence.
练习不参考笔记写推导，这能建立深刻的理解，并帮助你自信地回答“解释”或“证明”类题目。
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