📚 Deriving Key Formulas from the A-Level Physics Data and Formula Booklet (Jan 2018) | A-Level 物理数据与公式手册（2018年1月）核心公式推导
The A-Level Physics Data and Formula Booklet provides a concise collection of essential equations. However, memorisation alone is not sufficient; understanding their derivations deepens conceptual insight and problem-solving skills. This article derives several key formulas from the booklet, covering mechanics, fields, electricity, and waves.
A-Level 物理数据与公式手册提供了考试必需的核心方程，但单纯记忆并不足够；理解它们的推导过程有助于深化概念和提升解题能力。本文从该手册中选取若干关键公式进行推导，涵盖力学、场、电学和波。
1. Equations of Uniformly Accelerated Motion | 匀加速运动方程
We start from the definition of acceleration: a = (v − u) / t, where u is initial velocity, v is final velocity, and t is time. Rearranging gives the first equation: v = u + a t.
从加速度的定义 a = (v − u) / t 出发，其中 u 为初速度，v 为末速度，t 为时间。重排得到第一个方程：v = u + a t。
Displacement s equals the area under a velocity–time graph. For uniform acceleration the graph is a straight line, so the area is a trapezium: s = (u + v) t / 2.
位移 s 等于速度–时间图下的面积。对于匀加速运动，该图为一条直线，其面积为梯形面积：s = (u + v) t / 2。
Substituting v = u + a t gives s = (u + u + a t) t / 2, which simplifies to the second equation:
将 v = u + a t 代入得 s = (u + u + a t) t / 2，化简得到第二个方程：
s = u t + ½ a t²
To eliminate t, square the first equation: v² = (u + a t)² = u² + 2 a u t + a² t². But from the second equation, a t² = 2(s − u t). Substituting and simplifying yields the third equation:
为消去 t，将第一个方程平方：v² = (u + a t)² = u² + 2 a u t + a² t²。由第二个方程知 a t² = 2(s − u t)。代入并化简即得第三个方程：
v² = u² + 2 a s
2. Newton’s Second Law and Impulse | 牛顿第二定律与冲量
Newton’s second law states that the resultant force equals the rate of change of momentum: F = dp / dt. For constant mass, p = m v, so F = d(m v)/dt = m (dv/dt) = m a.
牛顿第二定律指出合外力等于动量的变化率：F = dp / dt。当质量恒定时，p = m v，故 F = d(m v)/dt = m (dv/dt) = m a。
Impulse J is the integral of force over time: J = ∫ F dt. Since F dt = dp, the impulse equals the change in momentum:
冲量 J 是力对时间的积分：J = ∫ F dt。由于 F dt = dp，冲量等于动量的变化：
J = Δp = m(v − u)
This relationship is especially useful in collisions, where the force may vary but the total impulse can be found from the area under a force–time graph.
这一关系在碰撞问题中尤为有用——尽管力可能变化，但总冲量可由力–时间图下的面积求得。
3. Work–Energy Theorem | 功能定理
Work done by a constant force is W = F s cosθ. For motion along the line of the force, cosθ = 1. Using F = m a and the equation v² = u² + 2 a s, we rearrange a s = (v² − u²) / 2.
恒力做功定义为 W = F s cosθ。当位移与力同向时 cosθ = 1。利用 F = m a 及 v² = u² + 2 a s，重排得 a s = (v² − u²) / 2。
Substituting gives W = m × (v² − u²) / 2 = ½ m v² − ½ m u². This is the kinetic energy change of the particle. Hence the net work done equals the change in kinetic energy.
代入得 W = m × (v² − u²) / 2 = ½ m v² − ½ m u²，这正是该质点的动能变化。因此合外力做功等于动能的变化。
4. Projectile Motion Path Equation | 抛体运动轨迹方程
Consider a projectile launched with speed u at angle θ to the horizontal. Resolving gives horizontal component ux = u cosθ and vertical component uy = u sinθ.
考虑一物体以速率 u、与水平成 θ 角抛出。分解初速度得水平分量 ux = u cosθ，竖直分量 uy = u sinθ。
Horizontal motion is uniform: x = u cosθ × t. Vertical motion has constant acceleration −g: y = u sinθ × t − ½ g t².
水平方向为匀速直线运动：x = u cosθ × t。竖直方向有恒定加速度 −g：y = u sinθ × t − ½ g t²。
Eliminate t by writing t = x / (u cosθ). Substituting into the y-equation yields the trajectory:
消去时间 t，由 t = x / (u cosθ) 代入 y 的方程得到轨迹方程：
y = x tanθ − g x² / (2 u² cos²θ)
This is a parabola, confirming the projectile’s path is parabolic in the absence of air resistance.
这是一个抛物线方程，证实了忽略空气阻力时抛体的路径是抛物线。
5. Centripetal Acceleration | 向心加速度
For an object moving at constant speed v in a circle of radius r, consider two velocity vectors separated by a small time interval Δt. The object turns through an angle Δθ. The magnitude of the velocity change is Δv ≈ v Δθ (for small angles).
对于以恒定速率 v 在半径为 r 的圆周上运动的物体，取间隔很小时段 Δt 的两个速度矢量。物体转过的角度为 Δθ。当角度很小时，速度变化的大小 Δv ≈ v Δθ。
The distance travelled along the arc is Δs = r Δθ, so Δθ = Δs / r. Substituting gives Δv ≈ v (Δs / r).
沿弧走过的距离为 Δs = r Δθ，故 Δθ = Δs / r。代入得 Δv ≈ v (Δs / r)。
Acceleration a = Δv / Δt ≈ (v Δs) / (r Δt) = v² / r. In terms of angular speed ω = v / r, a = ω² r.
加速度 a = Δv / Δt ≈ (v Δs) / (r Δt) = v² / r。用角速率 ω = v / r 表示，a = ω² r。
a = v² / r = ω² r
6. Gravitational Field Strength | 引力场强度
Newton’s law of gravitation states that the force between two point masses M and m separated by distance r is F = G M m / r².
牛顿万有引力定律给出两质点 M 和 m 间距 r 时的引力为 F = G M m / r²。
Gravitational field strength g is defined as the force per unit mass on a small test mass: g = F / m. Hence near a spherical mass M,
引力场强度 g 定义为作用于小检验质量的单位质量所受引力：g = F / m。因此在一球形质量 M 附近，
g = G M / r²
The gravitational potential V at a point is the work done per unit mass to bring a test mass from infinity to that point. Integrating −g dr gives V = −G M / r. This is why gravitational potential is negative.
某点的引力势 V 是将单位检验质量从无穷远移至该点过程中引力做的功。对 −g dr 积分即得 V = −G M / r，因此引力势为负值。
7. Coulomb’s Law and Uniform Electric Field | 库仑定律与匀强电场
Coulomb’s law for the electrostatic force between two point charges Q and q is F = k Q q / r², where k = 1/(4πε0).
两点电荷 Q 与 q 间的静电力遵循库仑定律 F = k Q q / r²，其中 k = 1/(4πε0)。
Electric field strength E is defined as F / q for a small positive test charge, giving the radial field E = k Q / r².
电场强度 E 定义为对小的正检验电荷 F / q，由此得点电荷的径向电场 E = k Q / r²。
For a uniform electric field between two parallel plates, the field is constant. Work done moving a charge q across a distance d is W = q E d. Since potential difference V = W / q, we obtain V = E d, or E = V / d.
在平行板间的匀强电场中，场强处处相等。将电荷 q 移动 d 距离所做的功为 W = q E d。因电势差 V = W / q，得 V = E d，即 E = V / d。
E = V / d (uniform field)
8. Energy Stored by a Capacitor | 电容器储存的能量
When a capacitor of capacitance C is charged, the charge Q and the potential difference V across it are related by Q = C V. The work done to add a small additional charge dq when the potential is V is dW = V dq = (q / C) dq.
电容为 C 的电容器充电时，电荷量 Q 与两极间电势差 V 满足 Q = C V。当电势为 V 时增加微量电荷 dq 因克服电场力所做的功为 dW = V dq = (q / C) dq。
Integrating from 0 to Q gives the total work stored as electric potential energy:
从 0 积分到 Q 得到储存的总电能为：
W = ∫0Q (q / C) dq = ½ Q² / C = ½ Q V =
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