📚 Formula Derivations from OxfordAQA 9630 PH03 January 2022 Report | 牛津AQA 9630 PH03 2022年1月报告中的公式推导
The January 2022 OxfordAQA Physics Unit 3 (PH03) exam report highlighted that many candidates lost marks not because they failed to recall final equations, but because they struggled with step-by-step derivations, vector manipulation, and applying fundamental definitions. This article revisits key formula derivations from the specification, focusing on areas flagged in the exam report, providing clear reasoning and bilingual explanations to deepen understanding and avoid common pitfalls.
2022年1月牛津AQA物理第三单元(PH03)考试报告指出，许多考生失分并非因为记不住最终公式，而是在分步推导、矢量操作和应用基本定义方面遇到困难。本文重新梳理考纲中的关键公式推导，重点针对考试报告中指出的薄弱环节，提供清晰的推理和双语解释，以加深理解并避免常见错误。
1. Derivation of Centripetal Acceleration (a = v²/r) | 向心加速度公式推导
locity vectors at P and Q have magnitude v and are tangents. The change in velocity Δv = vQ − vP is obtained by vector subtraction. Since the speed is constant, the triangle formed by vP, vQ and Δv is isosceles.
考虑一个物体以恒定速率 v 沿半径为 r 的圆周运动。在很短时间内 Δt 内，物体从点 P 运动到点 Q，在圆心处转过角度 Δθ。P 点和 Q 点的速度矢量大小均为 v，方向分别沿切线。速度变化量 Δv = vQ − vP 通过矢量减法得到。由于速率不变，由 vP、vQ 和 Δv 构成的三角形是等腰三角形。
/ v) = v²/r. Vectorially, Δv points toward the centre, so the acceleration is centripetal.
当 Δθ 很小时，Δv 的大小近似为 vΔθ（弧度制）。通过的弧长为 rΔθ，所用时间 Δt = rΔθ / v。加速度定义为 a = Δv / Δt。代入可得 a = (vΔθ) / (rΔθ / v) = v²/r。由于 Δv 指向圆心，加速度是向心的。
a = v² / r or a = ω² r (since v = ωr)
a = v² / r 或 a = ω² r （因为 v = ωr）
treatment, or mistakenly using the distance travelled as the change in velocity. Some students also confused centripetal acceleration with tangential acceleration.
报告中指出的常见错误是未使用矢量处理而直接假设 Δv = v − v = 0，或错误地将路程当作速度变化量。部分学生还混淆了向心加速度与切向加速度。
2. Derivation of Gravitational Potential (V = –GM/r) | 引力势推导
s from r to r + dr (increasing r), an external force must do work dW = F dr = (GMm/r²) dr.
引力势 V 定义为将单位质量从无穷远移至该点过程中引力场所做的功。对于点质量 M，作用于检验质量 m 的引力为 F = GMm/r²，方向指向 M。要使检验质量从 r 移到 r + dr（增加 r），外力需做功 dW = F dr = (GMm/r²) dr。
Integrating from infinity (where V = 0) to a radial distance r: V = – ∫∞^r (GM/r²) dr = – [–GM/r]∞^r = – ign indicates that work is done by the field as the mass moves closer, lowering potential.
从无穷远处（V = 0）积分到距离 r 处：V = – ∫∞^r (GM/r²) dr = – [–GM/r]∞^r = –GM/r。负号表示当检验质量靠近时引力场做正功，势能降低。
V = –GM / r
The January examiners reported that many students omitted the negative sign or attempted to derive V without referring to infinity as the reference point. Remember that potential is always negative in a gravitational field, reaching zero only at infinite separation.
考官报告指出，许多学生遗漏负号，或者未以无穷远为零参考点去推导。务必记住，在引力场中势总是负值，只有在无限远处才为零。
3. Energy Stored in a Capacitor (E = ½CV²) | 电容器储存的能量推导
dq when the pd across the plates is v is dW = v dq. Since v = q/C, we have dW = (q/C) dq.
当电容为 C 的电容器充电至电势差 V 时，极板上的电荷 q 从 0 增加到 Q = CV。当极板间电势差为 v 时，增加微量电荷 dq 所需做的功为 dW = v dq。由于 v = q/C，可得 dW = (q/C) dq。
Total energy stored is the integral of dW from q = 0 to q = Q: E = ∫₀Q (q/C) dq = (1/C)[½q²]₀Q = ½ Q²/C. Substituting Q = CV gives E = ½ CV².
储存的总能量即 dW 从 q = 0 到 Q 的积分：E = ∫₀Q (q/C) dq = (1/C)[½q²]₀Q = ½ Q²/C。代入 Q = CV 得到 E = ½ CV²。
E = ½ CV² = ½ QV = ½ Q² / C
CV², missing the factor of ½. This often arose from incorrectly assuming a constant pd during the whole charging process rather than using integration or average voltage.
考试反馈显示，相当多的考生写成 E = CV²，丢了因子 ½。这通常是因为他们错误地假设整个充电过程中电势差恒定，而没有使用积分或平均电压的思路。
4. Radius of Curvature in a Magnetic Field (r = mv/Bq) | 磁场中带电粒子的偏转半径推导
nces a magnetic force F = Bqv (where Bqv is the magnitude, with direction given by Fleming' lways acts perpendicular to velocity, providing the centripetal force for circular motion.
电荷量为 q 的带电粒子以速度 v 垂直于磁通密度 B 的匀强磁场运动，受到磁力作用 F = Bqv（大小，方向由弗莱明左手定则确定）。该力始终垂直于速度，充当圆周运动的向心力。
agnetic force to centripetal force: Bqv = mv²/r. Solving for radius r gives r = mv / (Bq).
对于圆周运动：向心力 = mv²/r。令磁力等于向心力：Bqv = mv²/r。求解半径 r 得 r = mv / (Bq)。
r = mv / Bq
v/r, forgetting the square on v. The report also warned against sign errors from not handling vector directions properly, though magnitude derivation usually suffices for numeric answers.
许多学生错误地使用了 Bqv = mv/r，忘记了 v 的平方。报告还提醒注意由于矢量方向处理不当带来的符号错误，不过数值计算时通常只需要大小推导。
5. Capacitor Discharge Equation (V = V₀ e⁻ᵗ⁄ᴿᶜ) | 电容器放电方程推导
For a capacitor discharging through a resistor R, the current I = – tor is V = Q/C, and the same pd appears across the resistor: V = IR. Combining gives Q/C = – R (dQ/dt). Rearranging: dQ/dt = – Q/(RC).
电容器通过电阻 R 放电时，电流 I = – dQ/dt（负号表示电荷减少）。电容器两端电势差为 V = Q/C，该电势差同时施加在电阻上：V = IR。联立得 Q/C = – R (dQ/dt)。整理为 dQ/dt = – Q/(RC)。
This is a first-order differential equation. Separating variables: dQ/Q = – dt/(RC). Integrating both sides: ln Q = – t/(RC) + constant. At t = 0, Q = Q₀ = CV₀, so constant = ln Q₀. Hence ln(Q/Q₀) = – t/(RC). Converting to exponential form: Q = Q₀ e⁻ᵗ⁄ᴿᶜ.
这是一阶微分方程。分离变量：dQ/Q = – dt/(RC)。两边积分：ln Q = – t/(RC) + 常数。当 t = 0 时，Q = Q₀ = CV₀，得常数 = ln Q₀。因此 ln(Q/Q₀) = – t/(RC)。转化为指数形式：Q = Q₀ e⁻ᵗ⁄ᴿᶜ。
Since V is proportional to Q (V = Q/C), the pd also decays exponentially: V = V₀ e⁻ᵗ⁄ᴿᶜ.
由于 V 与 Q 成正比（V = Q/C），电势差同样按指数衰减：V = V₀ e⁻ᵗ⁄ᴿᶜ。
V = V₀ e⁻ᵗ⁄ᴿᶜ and I = I₀ e⁻ᵗ⁄ᴿᶜ
The exam report noted that candidates often failed to show the integration step or omitted the constant of integration, losing method marks. Some also misapplied the initial conditions.
考试报告指出，考生经常未展示积分步骤或遗漏积分常数，从而丢失方法分。也有人错误地使用初始条件。
6. Acceleration in Simple Harmonic Motion (a = – ω²x) | 简谐运动的加速度推导
adius A with constant angular speed ω. The projection onto a diameter gives displacement x = A cos(ωt) or x = A sin(ωt).
简谐运动定义为回复力（及加速度）与离开平衡位置的位移成正比且指向平衡位置的运动。从圆周运动定义出发：考虑一个粒子以恒定角速度 ω 在半径为 A 的圆上运动，其在直径上的投影给出位移 x = A cos(ωt) 或 x = A sin(ωt)。
Velocity is v = dx/dt = – Aω sin(ωt). Acceleration is a = dv/dt = – Aω² cos(ωt) = – ω² x. This matches the SHM condition a ∝ – x, with the constant of proportionality being ω².
速度为 v = dx/dt = – Aω sin(ωt)。加速度为 a = dv/dt = – Aω² cos(ωt) = – ω² x。这符合简谐运动的条件 a ∝ – x，比例常数为 ω²。
a = – ω² x
Alternatively, using Newton's second law for a mass-spring system: F = – kx = ma → a = – (k/m) x, giving ω² = k/m, so T = 2π/ω = 2π√m/k. This derivation also appeared in the Jan 2022 paper, and examiners noted that students often failed to link ω to the physical properties k and m.
或者利用弹簧振子的牛顿第二定律：F = – kx = ma → a = – (k/m) x，得 ω² = k/m，从而 T = 2π/ω = 2π√m/k。该推导也在2022年1月试卷中出现，考官指出学生经常未能将 ω 与物理量 k 和 m 联系起来。
7. Derivation of Escape Velocity (v = √2GM/R) | 逃逸速度推导
Escape velocity is the minimum speed needed for an object to escape a planet' conservation of energy: total energy at surface = total energy at infinity. At the planet's surface (radius R, mass M), kinetic energy = ½mv², gravitational potential energy = – GMm/R. At infinity, kinetic energy = 0, potential energy = 0.
逃逸速度是物体从行星表面逃离引力场、最终到达无穷远处且动能为零所需的最小速度。根据能量守恒：表面总能量 = 无穷远处总能量。在行星表面（半径 R，质量 M），动能 = ½mv²，引力势能 = – GMm/R。无穷远处动能 = 0，势能 = 0。
Therefore, ½mv² – GMm/R = 0 + 0. Cancelling m: ½v² = GM/R → v = √2GM/R.
因此，½mv² – GMm/R = 0 + 0。消去 m：½v² = GM/R → v = √2GM/R。
vesc = √2GM / R
ing the factor of 2) or forgot that gravitational potential is negative, which led to sign errors in the energy balance.
报告显示，一些学生错误地设为 ½mv² = GMm/R（遗漏因子 2），或者忘记了引力势为负值，导致能量平衡中出现符号错误。
8. Electric Field Strength from Potential Gradient (E = – dV/dr) | 由电势梯度求电场强度
πε₀r). The electric field strength E is defined as the negative gradient of potential: E = – dV/dr. Taking the derivative: dV/dr = – Q/(4πε₀r²). Hence E = Q/(4πε₀r²), which is the familiar Coulomb's law expression.
在点电荷 Q 的径向电场中，距离 r 处的电势为 V = Q/(4πε₀r)。电场强度 E 定义为电势的负梯度：E = – dV/dr。求导得：dV/dr = – Q/(4πε₀r²)，因此 E = Q/(4πε₀r²)，即熟知的库仑定律表达式。
istance d with pd V, the potential changes linearly: E = V/d. This is a special case where |dV/dr| = V/d and direction is from higher to lower potential.
对于匀强电场，比如相距为 d、电势差为 V 的平行板之间的场，电势线性变化：E = V/d。这是 |dV/dr| = V/d 且方向由高电势指向低电势的特殊情况。
E = – dV / dr ; for uniform field E = V / d
or failed to include the negative sign when discussing direction, particularly in non-uniform fields.
2022年1月报告提到，考生经常错误地写成 E = V × d，或在讨论方向时遗漏负号，尤其是在非匀强电场中。
9. Period of a Mass-Spring System (T = 2π√m/k) | 弹簧振子周期推导
g of stiffness constant k, the restoring force when displaced by x from equilibrium is F = – kx. By Newton's second law: – kx = m a. Since a = dv/dt = d²x/dt², this gives m (d²x/dt²) = – kx, or d²x/dt² = – (k/m) x.
对于连接在刚度为 k 的弹簧上的质量 m，当偏离平衡位置 x 时，回复力为 F = – kx。根据牛顿第二定律：– kx = m a。由于 a = dv/dt = d²x/dt²，得 m (d²x/dt²) = – kx 或 d²x/dt² = – (k/m) x。
This has the same form as a = – ntify ω² = k/m. The period T is related to angular frequency by T = 2π/ω. Substituting ω = √k/m yields T = 2π√m/k.
这与 a = – ω²x 的形式相同，因此可认定 ω² = k/m。周期 T 与角频率的关系为 T = 2π/ω。代入 ω = √k/m 得到 T = 2π√m/k。
T = 2π √m / k
he expression for ω in terms of k and m, leaving the derivation incomplete. Links to the defining equation of SHM are essential.
考官强调，许多学生一开始就写下简谐运动解 x = A cos(ωt)，但无法推导出用 k 和 m 表示的 ω 表达式，导致推导不完整。将推导与简谐运动定义方程联系至关重要。
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