📚 The Derivation of Key Formulas in PH01 Insert | PH01插入页关键公式推导
This article walks you through the logical derivation of every major formula supplied in the PH01 International AS Physics Insert. Understanding where these equations come from will deepen your grasp of mechanics, energy, momentum and materials, and prepare you to use them confidently in unfamiliar contexts.
本文将带你逐一推导国际 AS 物理 PH01 试卷插入页中提供的每一个重要公式。理解这些方程的来源将加深你对力学、能量、动量和材料的掌握，并帮助你在陌生情境中自信地使用它们。
1. Acceleration and the First Kinematic Equation | 加速度与第一个运动学方程
Acceleration is defined as the rate of change of velocity. If an object speeds up from an initial velocity u to a final velocity v in a time interval t, the constant acceleration a is given by a = (v − u) / t. Multiplying both sides by t and adding u yields the first equation of motion for uniform acceleration.
加速度定义为速度的变化率。若物体在时间间隔 t 内从初速度 u 加速至末速度 v，则恒定加速度 a 为 a = (v − u) / t。两边同乘 t 并加上 u 即得匀加速运动的第一个运动方程。
v = u + a t
2. Displacement from Average Velocity | 从平均速度求位移
For constant acceleration, velocity changes linearly with time. The average velocity is therefore (u + v)/2. Since displacement equals average velocity multiplied by time, we obtain s = ½(u + v) t. Substituting v = u + a t into this expression gives the more familiar form s = u t + ½ a t².
在恒定加速度下，速度随时间线性变化，因此平均速度为 (u + v)/2。位移等于平均速度乘以时间，故得 s = ½(u + v) t。将 v = u + a t 代入该式，就得到了更常见的形式 s = u t + ½ a t²。
s = ½ (u + v) t
s = u t + ½ a t²
3. Eliminating Time to Relate v, u, a and s | 消去时间联系 v、u、a 和 s
Sometimes time t is not known, so we eliminate it by rearranging v = u + a t to give t = (v − u)/a and substituting into s = ½(u + v) t. This yields v² = u² + 2 a s. This relation is particularly useful when time is the unknown quantity.
有时时间 t 未知，我们可以从 v = u + a t 解出 t = (v − u)/a，代入 s = ½(u + v) t 即可消去时间，得到 v² = u² + 2 a s。当时间为未知量时，该关系式格外有用。
v² = u² + 2 a s
4. Newton’s Second Law in Terms of Momentum | 用动量表述的牛顿第二定律
Newton’s second law is often written F = m a. If we write acceleration as a = Δv / Δt, then F = m Δv / Δt. For a constant mass m, the change in momentum Δp is m Δv, so F = Δp / Δt. This form shows that the resultant force equals the rate of change of momentum.
牛顿第二定律常写作 F = m a。将加速度写为 a = Δv / Δt，则 F = m Δv / Δt。对于恒定质量 m，动量的变化 Δp 为 m Δv，于是 F = Δp / Δt。此形式表明合外力等于动量的变化率。
F = Δp / Δt
Momentum p itself is defined as the product of mass and velocity.
动量 p 本身定义为质量与速度的乘积。
p = m v
5. Impulse Equals Change in Momentum | 冲量等于动量的变化
The impulse of a force is the force multiplied by the time for which it acts. From F = Δp / Δt, multiplying both sides by Δt gives impulse = F Δt = Δp. This is the impulse–momentum theorem and explains why a brief, large force can cause the same change in motion as a smaller force acting over a longer time.
力的冲量是力与其作用时间的乘积。由 F = Δp / Δt，两边同乘 Δt 即得冲量 = F Δt = Δp。这就是冲量－动量定理，它解释了一个短暂的大力为何能产生与长期小力相同的运动变化。
F Δt = Δp
6. Work and the Work–Kinetic Energy Theorem | 功与功－动能定理
When a constant force F moves an object a distance s in the direction of the force, the work done is W = F s. Using F = m a and the kinematic relation 2 a s = v² − u², we can express work as W = m a s = ½ m (v² − u²) = ½ m v² − ½ m u². This shows that the work done on an object equals its change in kinetic energy, leading to the definition Eₖ = ½ m v².
当恒力 F 使物体沿力的方向移动距离 s 时，所做的功为 W = F s。利用 F = m a 及运动学关系 2 a s = v² − u²，可将功表示为 W = m a s = ½ m (v² − u²) = ½ m v² − ½ m u²。这表明对物体做的功等于其动能的变化，由此定义动能 Eₖ = ½ m v²。
W = F s
Eₖ = ½ m v²
7. Gravitational Potential Energy Gained | 增加的重力势能
Lifting an object of mass m through a vertical height Δh requires work against gravity. The force needed is equal to the weight m g, and the distance moved is Δh. Hence the work done is W = m g Δh. This energy is stored as gravitational potential energy, so ΔEₚ = m g Δh.
将质量为 m 的物体举升竖直高度 Δh 需要克服重力做功。所需力等于其重量 m g，移动距离为 Δh，故做功 W = m g Δh。此能量以重力势能形式储存，故 ΔEₚ = m g Δh。
ΔEₚ = m g Δh
8. Power: The Rate of Doing Work | 功率：做功的速率
Power is the rate at which work is done or energy is transferred: P = W / t. If a constant force F moves its point of application at constant velocity v in the direction of the force, then W = F s and s = v t, giving P = F v. This derivation also works for instantaneous power when velocity is not constant.
功率是做功或能量转移的速率：P = W / t。若恒力 F 的作用点以恒定速度 v 沿力的方向移动，则由 W = F s 与 s = v t 可得 P = F v。此推导同样适用于速度非恒定时的瞬时功率。
P = W / t
P = F v
9. Efficiency of Energy Transfer | 能量传递的效率
Efficiency measures how well a system converts input energy into useful output energy. It is defined as the ratio of useful output power (or energy) to total input power (or energy), often expressed as a percentage.
效率衡量系统将输入能量转化为有用输出能量的能力。它定义为有用输出功率（或能量）与总输入功率（或能量）之比，常以百分比表示。
efficiency = (useful output / total input) × 100%
10. Stress and Strain Defined | 应力与应变的定义
When a material is subjected to a force, stress describes the internal resistance per unit cross-sectional area. It is defined simply as σ = F / A. Strain measures the fractional extension produced and has no units: ε = ΔL / L, where L is the original length.
当材料受力时，应力描述单位横截面积上的内部阻力，简单定义为 σ = F / A。应变则衡量产生的相对伸长量，无单位：ε = ΔL / L，其中 L 为原长。
σ = F / A
ε = ΔL / L
11. Young Modulus from Stress and Strain | 由应力应变求杨氏模量
For many materials, stress is proportional to strain up to the limit of proportionality. The ratio of stress to strain in this linear region is a constant called the Young modulus E. Substituting the expressions for stress and strain, we obtain E = (F / A) / (ΔL / L) = F L / A ΔL. This formula is the one most often used in calculations involving wires and rods.
对许多材料而言，在比例极限以内应力与应变成正比。在此线性区域内，应力与应变之比为一常数，称为杨氏模量 E。代入应力和应变的表达式可得 E = (F / A) / (ΔL / L) = F L / A ΔL。该公式最常用于涉及金属丝与杆的计算。
E = σ / ε
E = F L / A ΔL
12. Elastic Strain Energy for a Stretched Wire | 拉伸线材的弹性应变能
If a wire obeys Hooke’s law, the force required to stretch it increases linearly from zero to F. The work done by the average force (F/2) over an extension ΔL is therefore W = ½ F ΔL. This work is stored as elastic strain energy inside the material, so ΔE = ½ F ΔL. By substituting F = k ΔL for a spring, the familiar ½ k x² is obtained.
若一条金属丝服从胡克定律，拉伸它所需的力从零线性增大至 F。平均力 (F/2) 在伸长 ΔL 下所做的功因此为 W = ½ F ΔL。此功以弹性应变能的形式储存在材料内部，故 ΔE = ½ F ΔL。若对弹簧代入 F = k ΔL，即得常见的 ½ k x²。
ΔE = ½ F ΔL
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