A-Level Physics PH05 Report on Exams Jun22: Mastering Formula Derivations | A-Level 物理 PH05 2022年6月考试报告：掌握公式推导

📚 A-Level Physics PH05 Report on Exams Jun22: Mastering Formula Derivations | A-Level 物理 PH05 2022年6月考试报告：掌握公式推导

The June 2022 examiners’ report for PH05 (Fields and Further Mechanics) highlighted that many students lost marks not because they couldn’t recall final formulas, but because they failed to present clear, logical derivations from first principles. Understanding how key results are obtained is essential for tackling multi-step questions and for avoiding sign errors. This article revisits ten fundamental derivations that frequently appear in the PH05 paper, breaking each one into manageable steps.

2022年6月的PH05（场与进阶力学）考官报告指出，许多学生失分并非记不住最终公式，而是无法从基本原理出发给出清晰、逻辑严密的推导。理解关键结果的来源对于应对多步计算题和避免正负号错误至关重要。本文重温十个在PH05试卷中反复出现的基本推导，将每个推导拆解为易于掌握的步骤。

1. Circular Motion: Deriving Centripetal Acceleration | 圆周运动：向心加速度推导

The examiners noted that candidates often state a = v²/r without showing how it arises from the geometry of velocity change. The derivation starts by considering an object moving with constant speed v in a circle of radius r. After a very short time interval Δt, the velocity vector has turned through a small angle Δθ, and its change Δv points towards the centre of the circle.

考官指出，考生常常写下 a = v²/r 却不展示它如何从速度变化的几何关系得出。推导从一物体以恒定速率 v 在半径为 r 的圆周上运动开始。经过一段极短的时间 Δt 后，速度矢量转过一个小角度 Δθ，其变化量 Δv 指向圆心。

Using vector subtraction, the magnitude of Δv is approximately equal to the arc length of the velocity circle: Δv ≈ v Δθ. Since the object’s position changes by Δθ = (v Δt)/r, we substitute to obtain Δv ≈ v · (v Δt / r) = v² Δt / r. Acceleration is the rate of change of velocity: a = Δv / Δt = v² / r.

利用矢量减法，Δv 的大小近似等于速度圆的弧长：Δv ≈ v Δθ。由于物体的位置角度变化为 Δθ = (v Δt)/r，代入可得 Δv ≈ v · (v Δt / r) = v² Δt / r。加速度是速度的变化率：a = Δv / Δt = v² / r。

Taking the limit Δt → 0 makes the earlier approximation exact. The direction of acceleration is always radially inward, and it can also be expressed in terms of angular speed ω: a = r ω².

取极限 Δt → 0 后上述近似变为精确。加速度的方向始终沿径向指向圆心，它也可用角速度 ω 表示为 a = r ω²。

a = v² / r = r ω²

2. Simple Harmonic Motion: Deriving Displacement Equation | 简谐运动：位移方程推导

In SHM questions, candidates were sometimes unable to connect the defining condition a ∝ −x with the sinusoidal solution. Starting from a = −ω² x, where ω is the angular frequency, one can use the projection of uniform circular motion to generate the displacement–time graph. Imagine a point moving round a reference circle of radius A with angular speed ω.

在简谐运动题目中，考生有时无法将定义条件 a ∝ −x 与正弦形式的解联系起来。从 a = −ω² x 出发（ω 为角频率），可以利用匀速圆周运动的投影产生位移–时间图像。设想一个点在半径为 A 的参考圆上以角速度 ω 运动。

If the point starts from the positive x-axis, its x-coordinate is x = A cos θ, and θ = ω t. Therefore, the displacement is x = A cos(ω t). Differentiating twice with respect to time gives velocity v = −A ω sin(ω t) and acceleration a = −A ω² cos(ω t), which is exactly −ω² x.

若该点从正 x 轴出发，它的 x 坐标为 x = A cos θ，且 θ = ω t。因此位移为 x = A cos(ω t)。对时间求导两次分别得到速度 v = −A ω sin(ω t) 和加速度 a = −A ω² cos(ω t)，这正好就是 −ω² x。

The phase constant φ may be added to match initial conditions, so the general solution is x = A cos(ω t + φ). The maximum speed is A ω and maximum acceleration is A ω².

可添加相位常数 φ 来匹配初始条件，因此通解为 x = A cos(ω t + φ)。最大速率为 A ω，最大加速度为 A ω²。

x = A cos(ω t + φ)

3. Gravitational Potential: Integrating Newton’s Law | 引力势：对牛顿定律积分

Many candidates lost marks by forgetting the negative sign in V = −GM / r. Gravitational potential is defined as the work done per unit mass by an external agent in bringing a small mass from infinity to a point, or equivalently the negative of the work done by the field. The gravitational force is F = −G M m / r², where the minus sign indicates attraction towards the central mass M.

许多考生因为忘记 V = −GM / r 中的负号而失分。引力势的定义是将一个小质量从无穷远处移动到某点过程中外力对单位质量所做的功，亦即引力场做功的负值。万有引力为 F = −G M m / r²，其中负号表示指向中心质量 M 的吸引力。

If mass m moves from infinity to a distance r, the work done by gravity (taking the radial inward direction carefully) is W = ∫_∞^r F dr = ∫_∞^r (−G M m / r²) dr. Evaluating the integral gives W = G M m / r. The change in potential energy is ΔU = −W, so with U = 0 at infinity we obtain U = −G M m / r. Potential V = U / m = −G M / r.

如果质量 m 从无穷远移动到距离 r 处，引力所做的功（小心处理径向向内方向）为 W = ∫_∞^r F dr = ∫_∞^r (−G M m / r²) dr。计算积分得到 W = G M m / r。势能变化为 ΔU = −W，因此取无穷远处 U = 0 即得 U = −G M m / r。势 V = U / m = −G M / r。

V = −G M / r

Examiners frequently observed confusion between potential and potential energy – remember to divide by the test mass.

考官经常发现学生混淆势和势能——记得要除以检验质量。

4. Electric Field and Potential Gradient | 电场与电势梯度

The relation E = −dV / dr is a common multi-step derivation. Consider a small positive charge q moved a distance Δx against a uniform electric field E. The work done by the external force is q E Δx, and this work equals the increase in electrical potential energy, q ΔV. However, moving opposite to the field increases potential, so ΔV = +E Δx, or more generally, ΔV = −E Δx for movement in the direction of the field.

E = −dV / dr 这一关系是常见的多步推导。考虑一个小正电荷 q 逆着均匀电场 E 移动一段距离 Δx。外力做功为 q E Δx，此功等于电势能的增加量 q ΔV。但逆着电场移动时电势升高，因此 ΔV = +E Δx，或者更一般地，顺着电场方向移动时 ΔV = −E Δx。

Thus, E = −ΔV / Δx. For a non-uniform field, taking the limit gives the instantaneous gradient: E = −dV / dr. The negative sign shows that the field points in the direction of decreasing potential.

于是 E = −ΔV / Δx。对于非均匀电场，取极限得到瞬时梯度：E = −dV / dr。负号表明电场指向电势降低的方向。

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