📚 A-Level Physics Key Derivations from June 2018 Examiner’s Report | A-Level物理核心推导：2018年6月考官报告解读
The June 2018 A-Level Physics Examiner’s Report highlighted that many students lose marks not because they do not know the final formula, but because they cannot show a clear, logical derivation. Mastering derivations is essential for securing top grades in the written papers, especially in questions that explicitly ask “show that” or “derive”. This article revisits several key derivations and consolidates the examiner’s advice on how to present them effectively.
2018年6月的A-Level物理考官报告指出，许多学生丢分并非因为不知道最终公式，而是因为他们无法展示清晰、有逻辑的推导过程。掌握推导过程对于在笔试中取得高分至关重要，尤其是对于那些明确要求“证明”或“推导”的题目。本文将回顾几个重要的推导，并汇总考官对于如何有效展示推导过程的建议。
1. Importance of Derivation in A-Level Physics | A-Level物理中推导的重要性
Derivation questions test your understanding of the underlying principles, not just memory. The examiner expects you to start from fundamental definitions or laws, such as Newton’s second law or definitions of velocity and acceleration, and then mathematically manipulate them to reach the required expression.
推导题考查的是你对底层原理的理解，而不仅仅是记忆。考官期望你从基本定义或定律出发，例如牛顿第二定律或速度和加速度的定义，然后通过数学运算得出所需的表达式。
2. Examiner’s Advice from June 2018 | 2018年6月考官的忠告
The report stressed that many candidates omitted intermediate steps or used “shortcuts” that were not justified. Examiners recommended showing all algebraic steps, stating assumptions (e.g., constant acceleration, no air resistance), and clearly labelling each equation.
报告强调，许多考生省略了中间步骤，或者使用了未说明理由的“捷径”。考官建议展示所有代数步骤，陈述假设（如匀加速度、无空气阻力），并清楚地标记每个方程。
3. Deriving the SUVAT Equations | 推导匀加速运动方程
Start from the definition of acceleration: a = (v − u) / t. Rearranging gives v = u + at. This is the first equation.
从加速度的定义出发：a = (v − u) / t。整理得 v = u + at。这是第一个方程。
To derive s = ut + ½ at², use average velocity: average velocity = (u + v)/2, and displacement s = average velocity × time = ((u + v)/2) t. Substitute v = u + at: s = ((u + u + at)/2) t = (2u + at)/2 × t = ut + ½ at².
要推导 s = ut + ½ at²，使用平均速度：平均速度 = (u + v)/2，位移 s = 平均速度 × 时间 = ((u + v)/2) t。代入 v = u + at：s = ((u + u + at)/2) t = (2u + at)/2 × t = ut + ½ at²。
Then v² = u² + 2as can be derived by eliminating t: from v = u + at, t = (v − u)/a. Substitute into s = ut + ½ at²: s = u((v−u)/a) + ½ a((v−u)/a)² → multiply by 2a: 2as = 2u(v−u) + (v−u)² = 2uv − 2u² + v² − 2uv + u² = v² − u², hence v² = u² + 2as.
然后可以通过消去t推导 v² = u² + 2as：由 v = u + at 得 t = (v − u)/a。代入 s = ut + ½ at²：s = u((v−u)/a) + ½ a((v−u)/a)² → 两边乘以2a：2as = 2u(v−u) + (v−u)² = 2uv − 2u² + v² − 2uv + u² = v² − u²，因此 v² = u² + 2as。
v = u + at, s = (u+v)/2 × t, s = ut + ½ at², v² = u² + 2as
4. Deriving Kinetic Energy (½mv²) | 推导动能公式 (½mv²)
Consider a constant net force F accelerating an object of mass m from rest to speed v over displacement s. Work done = F s. Using Newton’s second law F = ma, and v² = u² + 2as with u=0 gives v² = 2as → a = v²/(2s). Then F = m × v²/(2s). Work done = (m v²/(2s)) × s = ½ m v². This work is stored as kinetic energy.
考虑恒定合外力 F 使质量为 m 的物体从静止加速到速度 v，位移为 s。做功 = F s。利用牛顿第二定律 F = ma，以及 v² = u² + 2as，u=0 得 v² = 2as → a = v²/(2s)。则 F = m × v²/(2s)。功 = (m v²/(2s)) × s = ½ m v²。这个功储存为动能。
Examiners noted that many students incorrectly assume KE = mv² without derivation. Always show the link to work–energy principle.
考官注意到，许多学生错误地假设 KE = mv² 而不加以推导。务必展示功-能原理的关联。
5. Deriving Centripetal Acceleration a = v²/r | 推导向心加速度 a = v²/r
Consider an object moving in a circle of radius r with constant speed v. In a short time Δt, the object moves from point A to B, subtending angle Δθ. The change in velocity vector Δv points toward the centre. Magnitude of Δv ≈ v Δθ (since triangle of velocities is similar to triangle of radii). Δθ = v Δt / r. Thus acceleration a = Δv/Δt = v (v Δt / r)/Δt = v²/r.
考虑一物体在半径为 r 的圆周上以恒定速率 v 运动。在短时间 Δt 内，物体从 A 点运动到 B 点，圆心角为 Δθ。速度矢量的变化 Δv 指向圆心。Δv 的大小 ≈ v Δθ（因为速度矢量三角形与半径三角形相似）。Δθ = v Δt / r。因此加速度 a = Δv/Δt = v (v Δt / r)/Δt = v²/r。
The rigorous derivation uses vector calculus, but the geometric approach is acceptable at A-Level. Always state the assumption of constant speed.
严格推导使用矢量微积分，但在A-Level中几何方法是可以接受的。务必说明速率恒定的假设。
6. Deriving Projectile Range on Horizontal Ground | 推导水平地面上抛体射程
For a projectile launched at speed u at angle θ to horizontal, resolve: ux = u cos θ, uy = u sin θ. Time of flight t = 2 uy / g = 2u sin θ / g. Horizontal range R = ux × t = (u cos θ) × (2u sin θ / g) = (u² × 2 sin θ cos θ)/g = (u² sin 2θ)/g. Maximum range occurs at θ = 45°.
对于以速度 u、与水平成 θ 角发射的抛体，分解：ux = u cos θ, uy = u sin θ。飞行时间 t = 2 uy / g = 2u sin θ / g。水平射程 R = ux × t = (u cos θ) × (2u sin θ / g) = (u² × 2 sin θ cos θ)/g = (u² sin 2θ)/g。最大射程出现在 θ = 45°。
The June 2018 report highlighted that many students forgot to double the time to peak, using only uy/g. Always derive time of flight from vertical motion.
2018年6月的报告指出，许多学生忘记将到达顶点的时间翻倍，只用了 uy/g。务必从竖直运动推导飞行时间。
7. Deriving Capacitor Discharge Equation Q = Q₀ e−t/(RC) | 推导电容放电方程 Q = Q₀ e−t/(RC)
During discharge, current I = −dQ/dt, and I = V/R, also V = Q/C. So −dQ/dt = Q/(RC). Separating variables: dQ/Q = −1/(RC) dt. Integrate both sides: ln Q = −t/(RC) + constant. At t=0, Q=Q₀, so constant = ln Q₀. Thus ln(Q/Q₀) = −t/(RC), or Q = Q₀ e−t/(RC).
放电过程中，电流 I = −dQ/dt，同时 I = V/R，且 V = Q/C。因此 −dQ/dt = Q/(RC)。分离变量：dQ/Q = −1/(RC) dt。两边积分：ln Q = −t/(RC) + 常数。t=0 时 Q=Q₀，故常数 = ln Q₀。因此 ln(Q/Q₀) = −t/(RC)，即 Q = Q₀ e−t/(RC)。
Examiners expected clear separation of variables and explicit limits of integration. Missing the negative sign is a common mistake.
考官期望清晰的变量分离和明确的积分限。遗漏负号是一个常见错误。
8. Deriving F = ma from Momentum | 由动量推导 F = ma
Newton’s second law: Force = rate of change of momentum. Momentum p = mv. So F = dp/dt = d(mv)/dt. If mass is constant, F = m dv/dt = m a. This derivation is often required when linking laws.
牛顿第二定律：力 = 动量变化率。动量 p = mv。因此 F = dp/dt = d(mv)/dt。如果质量恒定，F = m dv/dt = m a。这个推导在联系各定律时经常需要。
The report emphasized that candidates should state the assumption of constant mass; otherwise, the product rule must be used.
报告强调，考生应说明质量恒定的假设；否则必须使用乘积法则。
9. Deriving the Period of a Simple Pendulum (T = 2π√(L/g)) | 推导单摆周期 T = 2π√(L/g)
For small angles, restoring force F = −mg sin θ ≈ −mg θ. Using arc displacement x = Lθ, acceleration a = − (g/L) x. This is SHM with ω² = g/L. Period T = 2π/ω = 2π √(L/g).
对于小角度，回复力 F = −mg sin θ ≈ −mg θ。利用弧位移 x = Lθ，加速度 a = − (g/L) x。这是简谐运动，ω² = g/L。周期 T = 2π/ω = 2π √(L/g)。
Examiners noted that credit is lost if students omit the small-angle approximation. Show that sin θ ≈ θ is valid when θ is in radians and small.
考官指出，如果学生遗漏了小角度近似，就会失分。应展示当 θ 以弧度为单位且很小时，sin θ ≈ θ 成立。
10. Deriving the Ideal Gas Equation from Kinetic Theory | 从分子动理论推导理想气体方程
Consider a single molecule in a cubic box of side L. Change in momentum per collision on one wall = 2mvx. Time between collisions = 2L/vx. Force on wall = (2mvx) / (2L/vx) = mvx²/L. For N molecules, total force = (m/L) Σ vx². Pressure p = Force/Area = (m/L³) Σ vx² = (m/V) Σ vx². With average velocity squared and using Σ v² = 3 Σ vx², pV = (1/3) N m ⟨v²⟩. Then with mean kinetic energy = ½ m ⟨v²⟩ = (3/2) kT, we get pV = NkT or pV = nRT.
考虑边长为 L 的立方体中的一个分子。每次与壁碰撞的动量变化 = 2mvx。碰撞间隔时间 = 2L/vx。对壁的力 = (2mvx) / (2L/vx) = mvx²/L。对 N 个分子，总力 = (m/L) Σ vx²。压强 p = 力/面积 = (m/L³) Σ vx² = (m/V) Σ vx²。利用速度平方平均值，并由 Σ v² = 3 Σ vx²，得 pV = (1/3) N m ⟨v²⟩。再由平均动能 = ½ m ⟨v²⟩ = (3/2) kT，可得 pV = NkT 或 pV = nRT。
Although this derivation is not always required in full, the examiner’s report indicated that understanding the steps helps in answering conceptual questions.
尽管这个推导并不总是需要完整给出，但考官报告指出，理解
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