📚 Deriving Key Formulae from the 9630 International A-Level Physics Data Booklet | 国际A-Level物理9630公式手册核心推导
The Pearson Edexcel International Advanced Level Physics Data and Formula Booklet (9630) is an essential companion for students, listing key equations without showing where they come from. Understanding the derivations deepens your grasp of physical principles and reduces reliance on rote memorisation. In this article we derive nine fundamental results that appear in that booklet, step by step, using calculus and vector reasoning appropriate for A-Level.
培生爱德思国际A-Level物理数据与公式手册（9630）是学生的重要工具，列出了关键方程却没有展示它们的来历。理解推导过程可以加深你对物理原理的掌握，减少死记硬背。在这篇文章中，我们一步步推出手册中出现的九个基本结果，采用适合A-Level的微积分与矢量推理。
1. Deriving the SUVAT Equations | 运动学方程推导
ocity u, gives ∫₀ᵗ a dt = ∫ᵤᵛ dv. Since a is constant, a t = v − u, therefore v = u + a t.
对于匀加速度 a 的运动，我们从定义 a = dv/dt 开始。分离变量并从 t=0 到 t 积分，初速度为 u，得到 ∫₀ᵗ a dt = ∫ᵤᵛ dv。由于 a 恒定，a t = v − u，因此 v = u + a t。
v = u + a t
m the average velocity form leads to v² = u² + 2 a s. These together are the SUVAT relations found in the booklet.
位移 s 是速度的积分：s = ∫₀ᵗ v dt = ∫₀ᵗ (u + a t) dt。积分得到 s = u t + ½ a t²。将第一式平方并利用平均速度替换 t，可得 v² = u² + 2 a s。这些一起就是手册中的运动学方程。
s = u t + ½ a t², v² = u² + 2 a s
2. Impulse and Momentum Conservation | 冲量与动量守恒
ing over time Δt, integrating gives Δp = F Δt, so impulse = change in momentum: F Δt = m v – m u.
牛顿第二定律的普遍形式为 F = dp/dt，其中 p = m v。对于作用时间 Δt 的恒合力，积分得 Δp = F Δt，因此冲量等于动量的变化：F Δt = m v – m u。
To arrive at conservation of momentum, consider two bodies A and B interacting with no external forces. By Newton’s third law, FAB = –FBA. During a collision, impulse on A equals –impulse on B, so ΔpA = –ΔpB, giving total momentum before = total momentum after.
要得到动量守恒，考虑两个物体 A 和 B 在无外力情况下相互作用。根据牛顿第三定律，FAB = –FBA。碰撞过程中，A 受到的冲量等于 B 受到的冲量的负值，因此 ΔpA = –ΔpB，即总初动量等于总末动量。
F = dp/dt, F Δt = Δp, Σ pbefore = Σ pafter
3. Work–Energy Theorem | 功与动能定理
and use the chain rule a = dv/dt = v dv/dx. Then W = ∫ m v dv/dx dx = ∫ᵤᵛ m v dv = ½ m v² – ½ m u². This is the work–energy theorem: net work equals change in kinetic energy.
力做的功定义为 W = ∫ F·dx。对于一维运动，代入 F = m a 并利用链式法则 a = dv/dt = v dv/dx，则 W = ∫ m v dv/dx dx = ∫ᵤᵛ m v dv = ½ m v² – ½ m u²。这就是动能定理：外力做的总功等于动能的变化。
tional potential energy GPE = m g h. These expressions are all listed in the data booklet.
当力为地表附近的重力时，做功为 m g Δh，直接导出重力势能 GPE = m g h。这些表达式都列在数据手册中。
W = F s cosθ, KE = ½ m v², GPE = m g h
4. Centripetal Acceleration a = v²/r | 向心加速度推导
object moves through angle Δθ = v Δt / r. The velocity vector changes direction by the same angle Δθ, so the magnitude of the change in velocity is |Δv| ≈ v Δθ. Hence the acceleration magnitude a = |Δv|/Δt = v (v/r) = v² / r.
一个以恒定速率 v 沿半径 r 的圆周运动的物体具有指向圆心的向心加速度。在短时间 Δt 内，物体转过角度 Δθ = v Δt / r。速度矢量的方向也改变相同的角度 Δθ，因此速度变化的大小为 |Δv| ≈ v Δθ。于是加速度大小 a = |Δv|/Δt = v (v/r) = v² / r。
a = v² / r, a = r ω²
The required centripetal force is then F = m v² / r or m r ω², as in the formula booklet.
利用 ω = Δθ/Δt = v/r 还可得到 a = r ω²。所需的向心力因此为 F = m v² / r 或 m r ω²，正如公式手册中所列。
5. Gravitational Field Strength and Potential | 引力场强度与引力势
mass: g = F/m = G M / r². This formula appears directly in the booklet for radial fields.
牛顿万有引力定律给出 F = G M m / r²。一点的引力场强 g 定义为单位质量所受的力：g = F/m = G M / r²。这个公式直接出现在手册的径向场部分。
ate the work done per unit mass in bringing a test mass from infinity to a distance r: V = – ∫∞ʳ g dr = – ∫∞ʳ (G M / r²) dr. Evaluating the integral gives V = – sign indicates that the gravitational force does work as the mass moves toward the source.
为了推导引力势 V，我们计算将检验质量从无穷远处移动到距离 r 处每单位质量所做的功：V = – ∫∞ʳ g dr = – ∫∞ʳ (G M / r²) dr。积分得到 V = – G M / r。负号表示检验质量向场源移动时引力做正功。
g = G M / r², V = – G M / r, ΔU = m ΔV
6. Simple Harmonic Motion (SHM) Equations | 简谐运动方程推导
SHM is defined by a restoring force proportional to displacement: F = – k x, and so acceleration a = – (k/m) x. Setting ω² = k/m yields the defining differential equation a = – ω² x, or d²x/dt² = – ω² x. The solution is x = A cos(ω t + φ), where A is amplitude and φ the phase constant.
简谐运动定义为回复力与位移成正比：F = – k x，因此加速度 a = – (k/m) x。令 ω² = k/m 便得到定义性的微分方程 a = – ω² x，即 d²x/dt² = – ω² x。其解为 x = A cos(ω t + φ)，其中 A 为振幅，φ 为初相。
Differentiating once gives velocity v = dx/dt = – A ω sin(ω t + φ), and twice gives a = – A ω² cos(ω t + φ) = – ω² x. From these we obtain the energy expressions: kinetic energy Ek = ½ m v² = ½ m A² ω² sin²(ω t + φ), elastic potential energy Ep = ½ k x² = ½ m ω² A² cos²(ω t + φ), so total energy E = ½ m ω² A², constant.
求导一次得到速度 v = dx/dt = – A ω sin(ω t + φ)，再求导得 a = – A ω² cos(ω t + φ) = – ω² x。由此可得能量表达式：动能 Ek = ½ m v² = ½ m A² ω² sin²(ω t + φ)，弹性势能 Ep = ½ k x² = ½ m ω² A² cos²(ω t + φ)，因此总能量 E = ½ m ω² A²，保持不变。
x = A cos(ω t), v = – A ω sin(ω t), a = – ω² x, ω = 2π f
7. Energy Stored in a Capacitor | 电容器储存的能量
dq is dW = V dq = (q / C) dq. Integrating from q = 0 to the final charge Q yields W = ½ Q² / C. Using Q = C V, this can be expressed as E = ½ C V² = ½ Q V.
给电容器充电需要对不断升高的电势差搬运电荷做功。任一时刻 V = q / C，增加微小电荷 dq 所做的功为 dW = V dq = (q / C) dq。从 q = 0 积分到最终电荷 Q 得 W = ½ Q² / C。利用 Q = C V 可改写成 E = ½ C V² = ½ Q V。
E = ½ C V² = ½ Q V = ½ Q² / C
ts the time constant τ = R C for discharge, derived from the differential equation dQ/dt = – Q/(R C).
手册还列出了放电的时间常数 τ = R C，它源自微分方程 dQ/dt = – Q/(R C)。
8. Faraday’s Law and Motional EMF | 法拉第定律与动生电动势
the induced emf in a coil equals the negative rate of change of magnetic flux linkage: ε = – cos θ. The minus sign shows that the induced emf opposes the change in flux (Lenz’s law).
法拉第电磁感应定律指出，线圈中感生电动势等于磁通匝链数的负变化率：ε = – N dΦ/dt，其中磁通量 Φ = B A cos θ。负号表示感生电动势总是反抗磁通量的变化（楞次定律）。
r sweeps out area ΔA = l v Δt, so the change in flux is ΔΦ = B l v Δt. The magnitude of the induced emf across the conductor is therefore ε = B l v. If motion is at angle θ to the field, the component v sin θ appears, giving ε = B l v sin θ.
一个简单的特例是长为 l 的直导体在垂直于匀强磁场 B 的方向上以速度 v 运动。在时间 Δt 内，导体扫过的面积 ΔA = l v Δt，因此磁通量的变化为 ΔΦ = B l v Δt。导体两端感生电动势的大小就是 ε = B l v。若运动方向与磁场夹角为 θ，则有效分量为 v sin θ，得 ε = B l v sin θ。
ε = – N dΦ/dt, ε = B l v sin θ
9. Kinetic Theory Derivation of Pressure | 气体动理论压强推导
Consider an ideal gas inside a cube of side L. A single molecule of mass m moving with velocity component vx collides elastically with a wall; the change in momentum is Δp = 2 m vx. The time between successive collisions with the same wall is 2 L / vx, so the average force on that wall is F = Δp / Δt = m vx g over N molecules and using the definition of pressure P = total force / L² gives P = N m
考虑一个边长为 L 的立方体容器内的理想气体。一个质量为 m 的分子以速度分量 vx 与器壁发生弹性碰撞，动量变化为 Δp = 2 m vx。与同一器壁连续两次碰撞的时间间隔为 2 L / vx，因此对该壁的平均作用力为 F = Δp / Δt = m vx² / L。对所有 N 个分子求和并利用压强定义 P = 总力 / L² 得到 P = N m
In random motion,
在无规则运动中，
P = ⅓ ρ
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