📚 A-Level Physics Insert 4 Jan22: Deriving Key Formulae | A-Level物理 Insert 4 Jan22 关键公式推导全解
The January 2022 A-Level Physics Insert (Paper 4) is the official data and formulae booklet issued with the structured questions paper. You may use it freely, but candidates who understand where each equation comes from choose the right formula faster, rearrange it with fewer errors and answer the qualitative "explain why" parts far more convincingly. This article works through the derivations of the most heavily examined formulae on that insert, step by step, with the exam technique attached to each one.
2022 年 1 月 A-Level 物理公式插页（Paper 4）是随结构化问答题一起下发的官方数据与公式表。你可以直接查用，但真正理解每个方程来源的考生，选公式更快、变形更少出错，回答 "explain why" 这类说明题时也更有说服力。本文将逐一推导该插页里最高频考点的公式，并把配套的考试技巧一并讲清。
1. What the Insert Really Is | 公式插页究竟是什么
The insert is not a textbook summary. It is a compressed reference containing roughly forty equations, a list of physical constants and a table of astronomical data. The same booklet is used in every examination series, so a derivation you learn once pays off in every paper. The examiners assume you can use these equations; what they test is whether you can select, combine and justify them.
插页并不是教材摘要，而是一份压缩的参考表：约四十个方程、一张物理常数表，以及天文数据。这本插页每考季通用，所以推导一次，终身受益。考官默认你会用这些公式，真正考查的是你能否选择、组合并论证它们。
A useful checklist before using any insert equation:
- Is the situation uniform (constant acceleration, uniform field) or varying?
情境是均匀的（匀加速、匀强场）还是变化的？ - Are the quantities scalar or vector, and have you fixed a positive direction?
量是标量还是矢量，是否已规定正方向？ - Are the units SI, and are the prefixes (k, μ, n, p) converted?
单位是否为 SI，词头（k、μ、n、p）是否换算？
that some derivations need them in combined form, such as k = 1/(4πε₀) = 8.99 × 10⁹ m F⁻¹.
插页常见常数如下；注意某些推导需要它们的组合形式，例如 k = 1/(4πε₀) = 8.99 × 10⁹ m F⁻¹。
| Constant | 常数 | Symbol | 符号 | Value | 数值 |
|---|---|---|
| Speed of light | 光速 | c | 3.00 × 10⁸ m s⁻¹ |
| Elementary charge | 元电荷 | e | 1.60 × 10⁻¹⁹ C |
| Planck constant | 普朗克常数 | h | 6.63 × 10⁻³⁴ J s |
| Gravitational constant | 引力常量 | G | 6.67 × 10⁻¹¹ N m² kg⁻² |
| Permittivity of free space | 真空介电常数 | ε₀ | 8.85 × 10⁻¹² F m⁻¹ |
| Boltzmann constant | 玻尔兹曼常数 | k | 1.38 × 10⁻²³ J K⁻¹ |
2. Kinematic Equations (SUVAT) | 运动学公式（SUVAT）
Start from the definition of acceleration: it is the rate of change of velocity. For constant acceleration a, initial velocity u, final velocity v and time t, the definition gives the first SUVAT equation directly.
从加速度的定义出发：加速度是速度的变化率。当加速度 a 恒定、初速度 u、末速度 v、历时 t 时，定义式直接给出第一个方程。
a = (v − u) / t ⟹ v = u + a t
For uniform acceleration the velocity increases linearly, so the average velocity is the midpoint value (u + v)/2. Displacement equals average velocity multiplied by time, which is the area under a v-t graph.
匀加速时速度线性增加，故平均速度为中间值 (u + v)/2。位移等于平均速度乘时间，也等于 v-t 图线下的面积。
s = ((u + v) / 2) t
ng v = u + at into this expression eliminates v and produces the second standard equation.
将 v = u + at 代入上式消去 v，得到第二个标准方程。
s = u t + ½ a t²
+ v)/2)t. The product (u + v)(v − u) = v² − u², so the a cancels the 2 in the denominator.
要求不含时间的方程，把 v = u + at 变形为 t = (v − u)/a，再代入 s = ((u + v)/2)t。因 (u + v)(v − u) = v² − u²，分母的 2 与 a 约去。
v² = u² + 2 a s
Exam note: these five equations hold only for constant acceleration in a straight line. Always state your positive direction, and remember that deceleration means a has the opposite sign to u.
考试提醒：这五个方程只适用于直线上的匀加速运动。务必声明正方向，并记住减速意味着 a 与 u 符号相反。
3. Projectile Motion and Resolving Vectors | 抛体运动与矢量分解
A projectile is simply two independent SUVAT problems sharing one time variable. Resolve the initial velocity into components: horizontal u cos θ, vertical u sin θ. Because gravity acts only vertically, the horizontal acceleration is zero and the vertical acceleration is g downwards.
抛体运动本质是两个共用同一时间的独立 SUVAT 问题。把初速度分解为水平分量 u cos θ 与竖直分量 u sin θ。因为重力只沿竖直方向作用，水平加速度为零，竖直加速度为 g 向下。
Horizontal: x = (u cos θ) t Vertical: y = (u sin θ) t − ½ g t²
θ − g t gives the time to the top, and doubling it gives the total time of flight on level ground.
在最高点竖直速度瞬时为零，由 0 = u sin θ − g t 得到上升时间，倍之即为水平地面上的总飞行时间。
t(top) = u sin θ / g H = u² sin² θ / (2 g) R = u² sin 2θ / g
tity 2 sin θ cos θ = sin 2θ. It shows the maximum range on level ground occurs at θ = 45°.
射程公式由总飞行时间代入水平方程并使用 2 sin θ cos θ = sin 2θ 得出，可见水平地面上最大射程出现在 θ = 45°。
These projectile results are not always printed in the insert, so being able to re-derive them in ten seconds is a genuine advantage in Paper 4.
这些抛体结论在插页中未必给出，能在十秒内重新推导出来，是 Paper 4 的实打实优势。
4. Newton's Second Law, Momentum and Impulse | 牛顿第二定律、动量与冲量
Newton's second law in its most general form states that the resultant force equals the rate of change of momentum. When the mass is constant, Δp/Δt becomes m Δv/Δt, which is ma.
牛顿第二定律最普遍的形式是：合外力等于动量的变化率。当质量恒定时，Δp/Δt 化为 m Δv/Δt，即 ma。
F = Δp / Δt = Δ(m v) / Δt and for constant m: F = m a
Rearranging gives the impulse-momentum equation, the reason airbags and crumple zones work: extending the collision time reduces the average force for the same change in momentum.
变形即得冲量—动量方程，这正是安全气囊与溃缩区有效的原理：在动量变化相同的情况下，延长碰撞时间可减小平均作用力。
F Δt = Δp = m v − m u
Conservation of momentum can be derived from Newton's third law. During a collision two bodies exert equal and opposite forces on each other for the same time, so the impulses are equal and opposite. One body's momentum gain equals the other's loss, so total momentum is unchanged.
动量守恒可由牛顿第三定律推出。碰撞中两物体相互施加等大反向的力，且作用时间相同，故冲量等大反向：一方动量的增加等于另一方动量的减少，总动量不变。
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
In an elastic collision kinetic energy is also conserved; in an inelastic collision it is not, and the "missing" energy becomes thermal energy, sound or deformation.
弹性碰撞中动能同样守恒；非弹性碰撞中动能不能守恒，损失的动能转化为热能、声能或形变能。
5. Circular Motion and Centripetal Acceleration | 圆周运动与向心加速度
Angular speed is defined as the angle swept per unit time, and since one full revolution is 2π radians in period T, the link between linear and angular speed follows directly from the definition of the radian.
角速度定义为单位时间扫过的角度。一整圈为 2π 弧度、周期为 T，由弧度的定义即可得到线速度与角速度的关系。
ω = 2π / T and v = r ω
To derive the acceleration, consider two positions separated by a small angle δθ. The velocity vectors have the same length v but different directions; drawing them tail to tail shows the change in velocity Δv is directed towards the centre and has magnitude v δθ.
推导加速度时，取相隔小角度 δθ 的两个位置。速度矢量大小同为 v 但方向不同；把它们首尾相接可见速度变化量 Δv 指向圆心，大小约为 v δθ。
t = δθ/ω gives the magnitude of the centripetal acceleration, and combining it with Newton's second law gives the centripetal force.
除以时间间隔 δt = δθ/ω 即得向心加速度大小；再结合牛顿第二定律得到向心力。
a = v ω = v² / r = ω² r F = m v² / r = m ω² r
Key exam point: the centripetal force is not a new force. It is the resultant of real forces such as tension, friction, gravity or the normal contact force, and it always points to the centre of the circle.
关键考点：向心力不是一种新的力，而是张力、摩擦力、重力或法向接触力等真实力的合力，方向始终指向圆心。
6. Simple Harmonic Motion | 简谐运动
Simple harmonic motion is defined by acceleration proportional to displacement and directed towards the equilibrium position. The constant of proportionality is ω², where ω is the angular frequency.
简谐运动的定义是：加速度与位移成正比，方向指向平衡位置。比例常数为 ω²，其中 ω 是角频率。
a = − ω² x
its projection has displacement x = A cos ωt, and differentiating twice gives a = −A ω² cos ωt = −ω²x.
该关系可由匀速圆周运动在直径上的投影推出。若质点以角速度 ω 在半径 A 的圆上运动，其投影位移为 x = A cos ωt，两次求导得 a = −A ω² cos ωt = −ω²x。
locity-displacement relation, and energy conservation gives the same result independently.
位移求一次导得速度；用 sin² ωt + cos² ωt = 1 消去时间得速度—位移关系，用能量守恒也可独立得到同一结果。
v = ± ω √A² − x² v(max) = A ω T = 2π / ω
For a mass on a spring, the restoring force is −kx, so ma = −kx gives ω = √k/m. For a small-angle pendulum, the restoring component mg sin θ ≈ mg θ = mg x/L gives ω = √g/L.
弹簧振子的回复力为 −kx，由 ma = −kx 得 ω = √k/m。小角度单摆的
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