📚 AS Physics Unit 1 Formula Derivations (Jan 2019 Mark Scheme Insights) | AS物理第一单元公式推导（2019年1月评分方案启示）
In the January 2019 AS Physics Unit 1 examination, the mark scheme placed a clear emphasis on the ability to derive fundamental equations from first principles. Rote memorisation is not enough; examiners expect you to link concepts through logical, step-by-step reasoning. This article revisits the core derivations for mechanics, materials and waves that frequently underpin exam questions. Each derivation is presented as a paired explanation to help you internalise the physics and the language needed to express it accurately.
在 2019 年 1 月的 AS 物理第一单元考试中，评分方案明确强调从基本原理推导核心方程的能力。死记硬背远远不够；考官期望你通过合乎逻辑的、循序渐进的推理将概念串联起来。本文重新梳理力学、材料和波中最常出现在考题背后的核心推导。每个推导均以中英对照的方式呈现，帮助你内化物理原理并掌握准确表达所需的语言。
1. Deriving the SUVAT Equations | 匀加速运动公式推导
The constant acceleration equations all stem from the definition of acceleration and the idea of average velocity. Starting with acceleration as the rate of change of velocity, a = (v – u) / t. Rearranging directly gives the first equation.
匀加速运动方程全都源于加速度的定义以及平均速度的概念。从加速度是速度变化率出发，a = (v – u) / t，直接整理即得第一个方程。
v = u + at
Next, we define displacement as the product of average velocity and time. When acceleration is uniform, the average velocity is ½(u + v). Substituting this into s = average velocity × t yields an expression that can be combined with the first equation.
接下来定义位移为平均速度与时间的乘积。加速度恒定时，平均速度为 ½(u + v)。将其代入 s = 平均速度 × t，得到一个可与第一式联立的表达式。
s = ½(u + v)t
Eliminating v by substituting v = u + at into the displacement equation produces the third form. Expanding the brackets gives s = ut + ½at². Similarly, eliminating t between the first two equations leads to the time-independent version.
将 v = u + at 代入位移方程，消去 v 得到第三个形式。展开括号得 s = ut + ½at²。同理，在头两个方程之间消去 t 则得到不含时间的版本。
v² = u² + 2as
2. Kinetic Energy Formula | 动能公式推导
Kinetic energy is the energy a body possesses due to its motion, and it can be derived from the work done to accelerate the body from rest. Consider a constant net force F acting on a mass m over a displacement s.
动能是物体因运动而拥有的能量，可从使物体从静止开始加速所做的功推导出来。考虑恒定的合力 F 作用于质量 m 并产生位移 s。
The work done is W = F s. Using Newton’s second law F = ma, this becomes W = ma s. For an object starting from rest (u = 0), the equation v² = u² + 2as simplifies to s = v²/(2a). Substituting into the work expression eliminates the acceleration.
做功为 W = F s。利用牛顿第二定律 F = ma，写成 W = ma s。对于从静止开始的物体（u = 0），v² = u² + 2as 简化为 s = v²/(2a)。代入功的表达式中消去了加速度。
W = ma × v²/(2a) = ½mv²
Thus the kinetic energy Eₖ gained by the body is exactly ½mv². This derivation is frequently assessed, and the mark scheme rewards clear substitution steps.
因此物体获得的动能 Eₖ 恰好就是 ½mv²。这一推导经常被考查，评分方案青睐清晰的代入步骤。
3. Gravitational Potential Energy Near the Earth’s Surface | 地表附近的重力势能推导
Lifting an object in a uniform gravitational field requires work against the weight. For a height change Δh, the minimum force needed is equal to the weight mg, applied vertically upwards.
在均匀重力场中提升物体需要克服重力做功。对于高度变化 Δh，所需的最小力等于重力 mg，方向竖直向上。
Since the force is constant and parallel to the displacement, the work done is simply force × distance: W = mgΔh. This work is stored as gravitational potential energy. In most AS contexts, Δh is written as h, giving the familiar equation.
由于力恒定且与位移平行，做功就是力 × 距离：W = mgΔh。这些功以重力势能的形式存储。在大多数 AS 语境中，Δh 写作 h，便得到熟悉的方程。
Eₚ = mgh
The reference level where h = 0 is arbitrary; only changes in potential energy are physically meaningful. Make sure you can explain why the work done against gravity is independent of the path taken.
h = 0 的参考水平面是任意选取的；只有势能的变化才具有物理意义。务必能够解释为何克服重力所做的功与路径无关。
4. Impulse–Momentum Relationship | 冲量–动量关系推导
The link between force and momentum change is one of the most powerful tools in AS mechanics. It begins with Newton’s second law expressed in terms of rate of change of momentum.
力与动量变化的联系是 AS 力学中最强大的工具之一。它始于用动量变化率表达的牛顿第二定律。
For a constant mass, the rate of change of momentum is m(v – u)/Δt, so the resultant force F = m(v – u)/Δt. Multiplying both sides by the time interval Δt gives the impulse.
对于恒定质量，动量变化率为 m(v – u)/Δt，因此合力 F = m(v – u)/Δt。两边同乘时间间隔 Δt 即得冲量。
FΔt = mv – mu
In words, impulse equals the change in momentum. This vector equation is indispensable for collision and safety applications, where forces vary but the area under a force–time graph still represents the impulse.
也就是说，冲量等于动量的变化。这个矢量方程在碰撞和安全应用中不可或缺，即使力是变化的，力–时间图下的面积依然代表冲量。
5. Elastic Potential Energy Stored in a Spring | 弹簧中储存的弹性势能推导
An ideal spring obeys Hooke’s law: the extension x is proportional to the applied force, so F = kx, where k is the spring constant. The work done in stretching the spring is not a simple Fx because the force increases from zero.
理想弹簧遵循胡克定律：伸长量 x 与施加的力成正比，即 F = kx，其中 k 是弹簧常数。拉伸弹簧所做的功并非简单的 Fx，因为力是从零开始逐渐增大的。
The work done equals the area under the force–extension graph, which is a triangle of base x and height F = kx. The area is therefore ½ × base × height = ½ × x × kx.
所做的功等于力–伸长图下方的面积，这是一个底为 x、高为 F = kx 的三角形。因此面积为 ½ × 底 × 高 = ½ × x × kx。
Eₑₗ = ½kx²
This elastic potential energy is stored in the spring and is recoverable. The derivation assumes that the elastic limit is not exceeded, so Hooke’s law remains valid throughout the extension.
这一弹性势能储存在弹簧中且可恢复。推导假设未超过弹性限度，因此胡克定律在伸长全程均有效。
6. Power as the Product of Force and Velocity | 功率作为力与速度的乘积推导
Power is defined as the rate of doing work. When a constant force F moves an object through a small displacement Δs in a time Δt, the work done is FΔs, provided the force is parallel to the displacement.
功率定义为做功的快慢。当恒力 F 使物体在时间 Δt 内产生微小位移 Δs，且力平行于位移，则所做的功为 FΔs。
Dividing the work by the time interval gives the average power: P = FΔs / Δt = F v, where v is the constant speed (or instantaneous speed for a very short interval). This relationship explains why a car climbing a hill at constant power must reduce its speed.
将功除以时间间隔得到平均功率：P = FΔs / Δt = F v，其中 v 为恒定速度（或很短时间内的瞬时速度）。这一关系解释了为何汽车以恒定功率爬坡时必须减速——需要更大的力来克服重力分量。
P = Fv
The mark scheme often expects you to state the condition: the force must be in the direction of motion. For non-parallel forces, use the component of force along the displacement.
评分方案通常要求你说明条件：力必须沿运动方向。若力不平行，则应使用沿位移方向的分力。
7. The Wave Equation v = fλ | 波速公式 v = fλ 推导
Waves transfer energy without transferring matter. The speed of a wave is the distance travelled by a point of constant phase, such as a crest, per unit time. During one period T, the wave advances by exactly one wavelength λ.
波传递能量而不传递物质。波速是恒定相位的点（例如波峰）在单位时间内传播的距离。在一个周期 T 内，波恰好前进一个波长 λ。
Thus the wave speed v = distance / time = λ / T. Frequency f is the reciprocal of the period, f = 1/T. Substituting gives the universal wave equation.
因此波速 v = 距离 / 时间 = λ / T。频率 f 是周期的倒数，f = 1/T。代入后得到普适的波速公式。
v = fλ
This equation holds for all types of progressive waves – transverse and longitudinal – provided the medium is uniform. In the exam, you must be able to derive it quickly from the definitions of period and wavelength.
该方程对所有类型的行波——横波和纵波——均成立，前提是介质均匀。考试中你必须能够从周期和波长的定义快速完成推导。
8. Young Modulus Formula Derivation | 杨氏模量公式推导
The Young modulus E quantifies the stiffness of a material and is defined as the ratio of tensile stress to tensile strain, at least within the limit of proportionality. This definition leads to a practical formula that links measurable quantities.
杨氏模量 E 衡量材料的刚度，定义为拉伸应力与拉伸应变之比，至少在比例极限内如此。这个定义导出了一个联系可测量量的实用公式。
Stress σ is force per unit cross-sectional area, σ = F/A. Strain ε is fractional extension, ε = ΔL/L, where L is the original length. Therefore the Young modulus is E = σ / ε = (F/A) / (ΔL/L).
应力 σ 是单位横截面积所受的力，σ = F/A。应变 ε 是伸长量与原长之比，ε = ΔL/L，其中 L 为原长。因此杨氏模量为 E = σ / ε = (F/A) / (ΔL/L)。
Re-arranging the fraction gives a form that is particularly useful when using force–extension graphs: the gradient is k, and substituting k = EA/L links the microscopic stiffness to macroscopic behaviour.
将分数整理后得到一种在使用力–伸长图时特别有用的形式：斜率即 k，而 k = EA/L 将微观刚度与宏观行为建立了联系。
E = FL / (A ΔL)
The stress–strain curve obtained in a tensile test allows the determination of E from the initial linear gradient. Understanding this derivation helps you interpret why a longer wire extends more for the same force, as predicted by ΔL = FL/(AE).
通过拉伸试验获得的应力–应变曲线，根据初始线性段的斜率可以确定 E。理解这一推导有助于你解释为何在相同力作用下较长的金属丝伸长更多，正如 ΔL = FL/(AE) 所预测的那样。
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