📚 OCR A-Level Physics June 2023 Paper 3 Formula Derivations | OCR A-Level物理2023年6月试卷3公式推导
The OCR A-Level Physics Paper 3 (Unified Physics) consistently tests candidates’ ability to derive fundamental equations from first principles. The June 2023 paper was no exception, featuring several structured derivation questions. This article walks you through the key derivations that appeared or could have appeared, breaking down each step with clear physical reasoning. Mastering these derivations not only secures marks in Paper 3 but also deepens your understanding of the whole specification.
OCR A-Level物理试卷3（统一物理）一贯考查考生从基本原理推导重要公式的能力。2023年6月的试卷也不例外，包含了几道结构化的推导题。本文带你逐一演练那些已经出现或可能出现的关键推导，用清晰的物理逻辑拆解每一步。精通这些推导不仅能帮你拿下试卷3的分数，还能加深你对整个课程的理解。
1. Overview of Paper 3 Derivation Questions | 试卷3推导题概述
Paper 3 asks you to link different areas of the specification. A typical derivation question will give you a starting point, such as a known law or definition, and guide you through algebraic or calculus steps to reach a target formula. You must be comfortable with symbols, unit analysis, and the physical meaning of each term.
试卷3要求你联系课程的不同领域。典型的推导题会给出一个起点，例如已知的定律或定义，然后引导你通过代数或微积分步骤得到目标公式。你必须对符号、单位分析以及每一项的物理意义感到得心应手。
The June 2023 paper included derivations from mechanics, thermal physics, and fields. Success depends on clarity of layout, correct handling of vector changes, and the ability to justify approximations such as small-angle limits or steady-state assumptions.
2023年6月的试卷涵盖了力学、热物理以及场的推导。得分的关键在于清晰的书写布局、正确处理矢量变化，以及能够论证诸如小角度极限或稳态假设等近似处理。
2. Deriving Centripetal Acceleration a = v²/r | 向心加速度公式推导
Consider an object moving with constant speed v in a circle of radius r. In a short time Δt, the object moves from point A to B, subtending an angle Δθ at the centre. The velocity vector changes direction but not magnitude.
考虑一个物体以恒定速率 v 在半径为 r 的圆周上运动。在很短的时间 Δt 内，物体从 A 点运动到 B 点，在圆心处张角 Δθ。速度矢量方向改变而大小不变。
The change in velocity Δv can be drawn as the base of an isosceles triangle with sides of length v and apex angle Δθ. For small Δθ, the magnitude of Δv is approximately vΔθ.
速度变化量 Δv 可以画成一个等腰三角形的底边，两腰长为 v，顶角为 Δθ。当 Δθ 很小时，Δv 的大小近似为 vΔθ。
Δv ≈ v Δθ
The distance travelled along the arc is s = rΔθ, and since speed v = s/Δt, we have Δθ = vΔt / r.
沿弧线经历的距离为 s = rΔθ，由于速率 v = s/Δt，可得 Δθ = vΔt / r。
Δθ = (v Δt) / r
Substitute this into the expression for Δv and divide by Δt to get the acceleration a = Δv/Δt directed toward the centre.
将上式代入 Δv 的表达式，再除以 Δt 便得到方向指向圆心的加速度 a = Δv/Δt。
a = v × (v / r) = v² / r
This vector is always perpendicular to the velocity, changing only the direction, not the speed.
该矢量始终垂直于速度，仅改变运动方向而不改变速率。
3. Deriving the Kinetic Theory Equation pV = ⅓ N m ⟨c²⟩ | 气体动理论压强公式推导
Imagine a cubic box of side L containing N identical gas molecules, each of mass m, moving randomly. Focus on one molecule hitting a wall perpendicular to the x-axis.
想象一个边长为 L 的立方容器，内有 N 个相同的质量为 m 的气体分子，做无规则运动。关注一个分子撞击垂直于 x 轴的器壁。
Its x-component of velocity is cx. The change in momentum on collision with the wall is 2mcx, since the molecule rebounds elastically.
该分子的速度 x 分量为 cx。由于发生弹性碰撞，与器壁碰撞时的动量变化为 2mcx。
The time between successive collisions with the same wall is 2L / cx. Hence the average force exerted by this one molecule on that wall is F = (change in momentum) / time = 2mcx ÷ (2L / cx) = m cx² / L.
与同一器壁连续碰撞的时间间隔为 2L / cx。因此，这一个分子对该器壁施加的平均力为 F = (动量变化) / 时间 = 2mcx ÷ (2L / cx) = m cx² / L。
Summing over all N molecules, the total force on the wall is F_total = (m/L) Σ cx². Since all directions are equivalent, we use the mean square speed ⟨c²⟩ and the fact that ⟨c²⟩ = ⟨cx²⟩ + ⟨cy²⟩ + ⟨cz²⟩ = 3⟨cx²⟩.
对所有 N 个分子求和，器壁上的总力为 F_total = (m/L) Σ cx²。由于各个方向等价，我们使用均方速率 ⟨c²⟩，并有 ⟨c²⟩ = ⟨cx²⟩ + ⟨cy²⟩ + ⟨cz²⟩ = 3⟨cx²⟩。
Thus Σ cx² = N⟨cx²⟩ = (N/3)⟨c²⟩. Pressure p = force per unit area = F_total / L².
因此 Σ cx² = N⟨cx²⟩ = (N/3)⟨c²⟩。压强 p = 作用在单位面积上的力 = F_total / L²。
p = (m/L) × (N/3)⟨c²⟩ / L² = ⅓ (N m ⟨c²⟩) / L³
Since volume V = L³, we arrive at the celebrated result:
由于体积 V = L³，我们得到著名结论：
pV = ⅓ N m ⟨c²⟩
4. Deriving the Capacitor Discharge Equation Q = Q₀ e–t/RC | 电容器放电方程推导
Consider a capacitor of capacitance C discharging through a resistor R. At any instant, the charge on the capacitor is Q, the p.d. across it is V = Q/C, and the current in the circuit is I = –dQ/dt (negative because charge decreases).
考虑一个电容 C 通过电阻 R 放电。在任意时刻，电容器上的电荷为 Q，其两端的电压为 V = Q/C，电路中的电流为 I = –dQ/dt（负号是因为电荷在减少）。
From Ohm’s law for the resistor, V = IR. Substituting gives Q/C = –R dQ/dt.
由电阻的欧姆定律 V = IR，代入得 Q/C = –R dQ/dt。
dQ/dt = –Q / (RC)
This is a first-order differential equation. Separate variables and integrate:
这是一阶微分方程。分离变量并积分：
∫ dQ / Q = – ∫ dt / (RC)
Carrying out the integration yields ln Q = –t/(RC) + constant. Applying the initial condition that at t=0, Q=Q₀ gives ln Q₀ = constant.
积分得 ln Q = –t/(RC) + 常数。利用初始条件 t=0 时 Q=Q₀，可得 ln Q₀ = 常数。
ln (Q / Q₀) = –t / (RC)
Exponentiating both sides produces the exponential decay law:
两边取指数得到指数衰减规律：
Q = Q₀ e–t/(RC)
The time constant RC is the time for the charge to fall to 1/e of its initial value.
时间常数 RC 是电荷降至初始值 1/e 所需的时间。
5. Deriving the Gravitational Potential V = –GM/r | 引力势公式推导
Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point. The force per unit mass (field strength) is g = GM/r² directed towards the centre of the mass M.
引力势是指将单位质量的小检验物体从无穷远处移至该点所做的功。单位质量的力（场强）为 g = GM/r²，方向指向质量 M 的中心。
Work done against the gravitational field when moving a distance dr away from M is dW = –g dr = – (GM/r²) dr (the negative sign indicates work is done against the field when moving outward). To bring mass from infinity to a distance r, we integrate:
远离 M 移动 dr 时，克服引力场做的功为 dW = –g dr = – (GM/r²) dr（负号表示向外移动时克服场做功）。要将质量从无穷远带到距离 r 处，需积分：
V = ∫∞r – (GM / r²) dr = GM [1/r]∞r
Evaluating the brackets gives V = GM (1/r – 1/∞) = GM/r. However, by convention, potential at infinity is zero, and the field is attractive, so the potential is increasingly negative as we approach M. Thus the correct expression with sign convention is:
计算括号内的值，得 V = GM (1/r – 1/∞) = GM/r。但按照惯例，无穷远处的势为零，且引力场是吸引的，所以越靠近 M，势就越负。因此，加上符号约定后正确的表达式为：
V = – GM / r
This shows that work must be done to remove a mass from the gravitational influence of M.
这表明要将一个质量从 M 的引力影响下移开，外界必须做功。
6. Deriving Escape Velocity v_esc = √(2GM/r) | 逃逸速度推导
An object can escape a planet’s gravitational field if its kinetic energy at the surface equals or exceeds the magnitude of the gravitational potential energy (taking zero at infinity).
如果物体在行星表面的动能等于或大于该处引力势能的绝对值（取无穷远处为零），它就能脱离行星的引力场。
Set E_k + E_p ≥ 0, where E_p = –GMm/r. At the threshold, total mechanical energy is zero:
令 E_k + E_p ≥ 0，其中 E_p = –GMm/r。在临界状态下，总机械能为零：
½ m v_esc² – GMm / r = 0
The mass m cancels, and solving for v_esc:
质量 m 可以消去，解出 v_esc：
½ v_esc² = GM / r → v_esc = √(2GM / r)
This derivation assumes no atmospheric drag and that the planet is the only source of gravity. It links the concepts of field and potential directly to a measurable speed.
该推导假设没有大气阻力且行星是唯一的引力源。它将场与势的概念直接与可测量的速度联系起来。
7. Deriving the Lens Formula 1/f = 1/u + 1/v | 透镜公式推导
Using similar triangles formed by a thin converging lens, we can relate object distance u, image distance v, and focal length f. Consider the two principal rays: one through the optical centre, undeviated, and one parallel to the axis that passes through the focus.
利用薄凸透镜形成的相似三角形，我们可以将物距 u、像距 v 和焦距 f 联系起来。考虑两条主光线：一条通过光心不偏折，另一条平行于主轴并穿过焦点。
For a real object and real image, the triangle involving the object height h and its image height h’ gives magnification m = h’/h = v/u. Another pair of similar triangles involving the focal point gives m = (v – f)/f.
当实物成实像时，包含物高 h 和像高 h’ 的三角形给出放大率 m = h’/h = v/u。围绕着焦点的另一组相似三角形则给出 m = (v – f)/f。
Equating the two expressions for m:
令两个 m 的表达式相等：
v / u = (v – f) / f
Cross-multiply: v f = u v – u f. Rearrange to obtain u v = f v + f u. Divide through by u v f to isolate the reciprocals:
交叉相乘：v f = u v – u f。移项得 u v = f v + f u。两边同时除以 u v f 以得到倒数形式：
1/f = 1/u + 1/v
This sign convention is for the real-is-positive convention used in many textbooks. The derivation shows the power of geometry in wave optics.
这个符号约定采用的是许多教科书中的“实正虚负”规定。该推导展示了几何在波动光学中的威力。
8. Deriving the Electrical Power P = I²R | 电功率推导
When a charge Q moves through a potential difference V, the work done on it is W = Q V. In a resistor, this energy is dissipated as heat.
当电荷 Q 通过电势差 V 时，对它做的功为 W = Q V。在电阻中，这部分能量以热的形式耗散。
Power is the rate of doing work: P = dW/dt. Since V is constant for a steady circuit, P = d(QV)/dt = V dQ/dt = V I, because current I = dQ/dt.
功率是做功的速率：P = dW/dt。对于稳态电路，V 恒定，所以 P = d(QV)/dt = V dQ/dt = V I，因为电流 I = dQ/dt。
Using Ohm’s law V = I R for a purely resistive component, we substitute to get two alternative forms:
对纯电阻元件应用欧姆定律 V = I R，代入可得到另外两种形式：
P = V I = I² R = V² / R
The form P = I²R is particularly useful for calculating thermal losses in transmission lines, as it shows the importance of reducing current to improve efficiency.
形式 P = I²R 在计算输电线热损耗时特别有用，因为它表明降低电流对提升效率的重要性。
9. Deriving the Impulse-Momentum Relationship | 冲量动量关系推导
Newton’s second law can be expressed in terms of momentum: Force is equal to the rate of change of momentum, F = dp/dt. When a constant resultant force acts for a time Δt, the impulse J = F Δt.
牛顿第二定律可以用动量表述：力等于动量的变化率，F = dp/dt。当一个恒定的合力作用了 Δt 时间，冲量 J = F Δt。
Integrate F dt over the time interval:
对时间间隔积分 F dt：
J = ∫ F dt = ∫ dp = Δp = p_final – p_initial
This shows that impulse equals the change in momentum, an extremely useful principle in collision and safety applications where forces vary rapidly.
这表明冲量等于动量的变化，这一定理在碰撞和安全应用中极为有用，因为那些情形中力变化很快。
For a constant mass m, we recover the familiar F Δt = m(v – u), with u initial speed and v final speed.
对于恒定质量 m，我们便可恢复熟悉的形式 F Δt = m(v – u)，其中 u 为初速度，v 为末速度。
10. Common Pitfalls and Tips for Derivation Questions | 推导题常见陷阱与技巧
One common mistake is losing track of vector directions. Always draw a diagram and label the positive direction before starting the algebra. Another is forgetting to justify the small-angle approximation when using sinθ ≈ θ or Δθ being small.
常见错误之一是矢量方向混乱。务必先画示意图并标明正方向，再开始代数推导。另一个是忘记在使用 sinθ ≈ θ 或 Δθ 很小时论证小角度近似的合理性。
In thermal derivations, distinguish clearly between capital N (number of molecules) and small n (number of moles). Also, many students confuse ⟨c²⟩ with (⟨c⟩)² – the mean square speed is not the square of the mean speed.
在热学推导中，要清楚区分大写 N（分子数）和小写 n（摩尔数）。此外，许多学生将均方速率 ⟨c²⟩ 与平均速率的平方 (⟨c⟩)² 混淆——均方速率并非平均速率的平方。
For calculus derivations, show the separation of variables and limits explicitly. Never jump from a differential equation to the final solution without showing the integration step, even if the result is given in the formula booklet.
对于微积分推导，要清晰地展示变量分离和积分限。不要从微分方程直接跳到最终解而省略积分步骤，即便公式表里给出了结果。
11. Practice Derivation from June 2023 Context | 2023年6月真题推导练习
In the June 2023 Paper 3, one structured question asked candidates to derive the period of a simple pendulum, T = 2π √(L/g). Starting from the restoring force for small amplitudes, the component of weight along the arc is mg sinθ ≈ mgθ. The displacement along the arc is x = Lθ, so the restoring force is –(mg/L)x.
在2023年6月试卷3中，一道结构化题目要求考生推导单摆的周期 T = 2π √(L/g)。从小振幅的回复力出发，重力沿弧线的分量为 mg sinθ ≈ mgθ。沿弧线的位移为 x = Lθ，所以回复力为 –(mg/L)x。
Comparing with simple harmonic motion F = –kx, we identify the effective spring constant k = mg/L. The angular frequency ω = √(k/m) = √(g/L), and since T = 2π/ω, we obtain T = 2π √(L/g).
与简谐运动 F = –kx 对比，可识别出等效劲度系数 k = mg
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