📚 AS Physics Unit 2 (Jan 2020) Formula Derivations from Mark Scheme | AS物理单元2（2020年1月）评分方案中的公式推导
In AS Physics Unit 2, understanding the derivation of key formulas is essential for both written explanations and multi-step calculations. The January 2020 mark scheme awards marks for clear logical steps, correct manipulation of symbols, and proper use of definitions. This article walks through the most important derivations that appeared or were implied in that paper, providing a reliable revision resource.
在AS物理单元2中，理解关键公式的推导对于文字解释题和多步计算题都至关重要。2020年1月的评分方案对清晰的逻辑步骤、正确的符号运算以及恰当的定义使用都有给分。本文梳理了该试卷中出现或隐含的最重要推导，为你提供一份可靠的复习资源。
1. Deriving the Wave Equation v = fλ | 推导波速公式 v = fλ
Step 1: Define frequency f as the number of complete wave cycles passing a point per second. Wavelength λ is the spatial length of one complete cycle.
步骤1：将频率 f 定义为单位时间内通过某点的完整波的个数。波长 λ 是一个完整周期的空间长度。
Step 2: The time taken for one complete cycle is the period T. By definition, T = 1/f.
步骤2：完成一个完整周期所需的时间是周期 T。根据定义，T = 1/f。
Step 3: During one period, the wave travels exactly one wavelength. Using speed = distance / time, we have v = λ / T.
步骤3：在一个周期内，波恰好传播一个波长。利用 速度 = 距离 / 时间，得到 v = λ / T。
Step 4: Substitute T = 1/f into v = λ / T to obtain v = λ ÷ (1/f) = fλ. Mark schemes often require showing this substitution step explicitly to earn the final mark.
步骤4：将 T = 1/f 代入 v = λ / T，得到 v = λ ÷ (1/f) = fλ。评分方案通常要求清晰写出代入步骤才能拿到最后的分。
2. Deriving the Equations of Motion (SUVAT) for Constant Acceleration | 匀加速直线运动方程（SUVAT）的推导
SUVAT equations are directly derived from the definitions of acceleration and average velocity. The mark scheme rewards using a velocity-time graph or algebraic substitution.
SUVAT 方程直接源于加速度和平均速度的定义。评分方案对使用速度-时间图或代数代换的方法都会给分。
Start with the definition of acceleration: a = (v – u) / t, where u is initial velocity and v is final velocity. Rearranging gives v = u + at.
从加速度的定义出发：a = (v – u) / t，其中 u 是初速度，v 是末速度。移项可得 v = u + at。
Displacement s is equal to the area under the velocity-time graph, or s = average velocity × time. For constant acceleration, average velocity = (u + v) / 2, leading to s = ½(u + v)t.
位移 s 等于速度-时间图下的面积，也即 s = 平均速度 × 时间。对于匀加速运动，平均速度 = (u + v) / 2，因此 s = ½(u + v)t。
Substituting v = u + at into s = ½(u + v)t gives s = ½(u + u + at)t = ut + ½at². This derivation yields two of the five SUVAT equations used in the January 2020 questions.
将 v = u + at 代入 s = ½(u + v)t，得到 s = ½(u + u + at)t = ut + ½at²。这个推导得到了五个SUVAT方程中的两个，也是2020年1月考题中会用到的。
3. Deriving Kinetic Energy Eₖ = ½mv² | 动能公式 Eₖ = ½mv² 的推导
Consider a constant net force F acting on an object of mass m starting from rest, causing a displacement s. The work done by the force is W = F s.
考虑一个质量为 m 的物体从静止开始受到恒定的合力 F 作用，并发生位移 s。力所做的功为 W = F s。
From Newton’s second law, F = ma. The work done becomes W = ma s. Using the motion equation v² = u² + 2as with u = 0, we have v² = 2as, so as = v²/2.
根据牛顿第二定律，F = ma。功变为 W = ma s。利用运动学方程 v² = u² + 2as，并令初速度 u = 0，可得 v² = 2as，因此 as = v²/2。
Substitute as into the work expression: W = m (v²/2) = ½mv². Since all the work is transferred to kinetic energy, Eₖ = ½mv². The mark scheme looks for the clear link between work done and gain in kinetic energy.
将 as 代入功的表达式：W = m (v²/2) = ½mv²。因为所有功都转化为动能，所以 Eₖ = ½mv²。评分方案会关注做功与动能增加之间的清晰关联。
4. Deriving Power as P = Fv | 功率公式 P = Fv 的推导
Power is defined as the rate of doing work: P = W / t. For a constant force F moving an object through a small displacement Δs in time Δt, the work done is W = F Δs.
功率被定义为做功的快慢：P = W / t。对于一个恒定的力 F，在时间 Δt 内使物体发生微小位移 Δs，所做的功为 W = F Δs。
Therefore, P = (F Δs) / Δt = F (Δs / Δt). Since Δs / Δt is the instantaneous velocity v, we get P = Fv. This derivation appears in questions requiring calculation of force from power and speed.
因此，P = (F Δs) / Δt = F (Δs / Δt)。由于 Δs / Δt 是瞬时速度 v，故得到 P = Fv。这一推导常见于需要根据功率和速度求力的问题中。
The mark scheme often accepts this derivation, but also expects students to note that the force and velocity must be in the same direction for this scalar product form to hold.
评分方案通常接受这个推导，但也期望学生指出，力和速度必须在同一直线上，才能使用这个标量乘积形式。
5. Deriving Young’s Modulus E = (F/A) / (ΔL/L₀) | 杨氏模量 E = (F/A) / (ΔL/L₀) 的推导
Young’s modulus is defined as the ratio of tensile stress to tensile strain. Stress σ is the force applied per unit cross-sectional area: σ = F / A.
杨氏模量定义为拉伸应力与拉伸应变的比值。应力 σ 是单位横截面积上施加的力：σ = F / A。
Strain ε is the fractional extension, i.e. the extension ΔL divided by the original length L₀: ε = ΔL / L₀.
应变 ε 是长度的相对变化，即伸长量 ΔL 除以原长 L₀：ε = ΔL / L₀。
Thus, Young’s modulus E = σ / ε = (F / A) / (ΔL / L₀). Rearranging, the force-extension relationship becomes F = (E A / L₀) ΔL, which is the spring constant form often used in experiments.
因此，杨氏模量 E = σ / ε = (F / A) / (ΔL / L₀)。整理后，力-伸长关系变为 F = (E A / L₀) ΔL，这正是在实验中经常用到的弹簧常数形式。
In the January 2020 paper, one mark was awarded for stating the definitions of stress and strain, and another for substituting them into the ratio correctly.
在2020年1月的试卷中，写出应力和应变的定义可得到一分，而正确将它们代入比值又可得一分。
6. Deriving the Refractive Index n = c / v | 折射率 n = c / v 的推导
Refractive index n of a medium is defined as the ratio of the speed of light in a vacuum c to the speed of light in that medium v. This definition is the starting point.
介质的折射率 n 定义为光在真空中的速度 c 与光在该介质中的速度 v 之比。这个定义是推导的起点。
From Snell’s law, n₁ sin θ₁ = n₂ sin θ₂. Taking n₁ = 1 for vacuum, and if the angle of incidence and refraction are known, n can be found. However, the derivation of n = c/v comes directly from wave theory.
根据斯涅尔定律，n₁ sin θ₁ = n₂ sin θ₂。若真空的 n₁ = 1，且已知入射角和折射角，则可求出 n。但 n = c/v 的推导直接来自波动理论。
When a wave enters a new medium, its frequency f remains constant, but its speed and wavelength change. Since v = fλ, v ∝ λ. The refractive index is also equal to λ₀ / λ, where λ₀ is the wavelength in vacuum. Combining gives n = c/v.
当波进入新介质时，频率 f 保持不变，但速度和波长改变。因为 v = fλ，所以 v ∝ λ。折射率也等于真空波长 λ₀ 与介质波长 λ 之比。结合即得 n = c/v。
The mark scheme often requires students to explain why frequency is constant across boundaries to justify the relationship.
评分方案常要求学生解释为什么频率在边界上保持不变，以便证明这个关系。
7. Deriving Stress σ = F/A and Strain ε = ΔL/L₀ from First Principles | 由基本原理推导应力 σ = F/A 和应变 ε = ΔL/L₀
These definitions are foundational in materials physics. Stress is a measure of internal forces acting within a deformable body per unit area, defined so that larger forces or smaller areas produce larger stress.
这些定义是材料物理的基础。应力是衡量可变形体内部单位面积上作用内力的量，其定义使得越大的力或越小的面积产生越大的应力。
Therefore, tensile stress σ = F / A, where F is the applied force normal to the cross-section, and A is the original cross-sectional area. Strain is a dimensionless measure of deformation, given by the ratio of extension to original length: ε = ΔL / L₀.
因此，拉伸应力 σ = F / A，其中 F 是垂直于横截面的施加力，A 是原横截面积。应变是一个无量纲的形变量度，由伸长量与原长之比给出：ε = ΔL / L₀。
In the January 2020 mark scheme, simply writing these formulas without explanation only receives partial credit; the physical meaning must be stated.
在2020年1月的评分方案中，仅写出这些公式而不加解释只能得到部分分数；必须说明其物理意义。
8. Deriving the Period of a Simple Pendulum (Small Angle Approximation) | 单摆周期公式的推导（小角度近似）
For a simple pendulum of length L, the restoring force on the bob is F = -mg sin θ. For small angles (θ ≈ 0), sin θ ≈ θ. The angular displacement θ is related to the arc length x by x = Lθ.
对于摆长为 L 的单摆，摆锤受到的回复力为 F = -mg sin θ。对小角度（θ ≈ 0），sin θ ≈ θ。角位移 θ 与弧长 x 的关系为 x = Lθ。
The equation of motion becomes ma = -mg (x/L), so a = -(g/L)x. This is a simple harmonic motion (SHM) equation with ω² = g/L, where ω = 2π/T.
运动方程变为 ma = -mg (x/L)，即 a = -(g/L)x。这是简谐运动（SHM）方程，其中角频率 ω² = g/L，而 ω = 2π/T。
Therefore, (2π/T)² = g/L ⇒ T = 2π √(L/g). The mark scheme rewards identifying the SHM condition and correctly substituting the angular frequency.
因此，(2π/T)² = g/L ⇒ T = 2π √(L/g)。评分方案对识别出SHM条件并正确代入角频率的步骤给分。
9. Deriving the Acceleration from a Velocity-Time Graph | 从速度-时间图推导加速度
The average acceleration a between two points on a v-t graph is defined as the gradient of the straight line connecting those points: a = Δv / Δt.
速度-时间图上两点之间的平均加速度 a 定义为连接这两点的直线斜率：a = Δv / Δt。
If the graph is a curve, the instantaneous acceleration is the gradient of the tangent at a specific time. In the 2020 paper, students had to draw a tangent and compute its gradient to find the acceleration at t = 0.5 s.
如果图线是曲线，则瞬时加速度是特定时刻处切线的斜率。在2020年的试卷中，学生需要画出切线并计算其斜率，以求出 t = 0.5 s 时的加速度。
The mark scheme required a large triangle (covering at least half the graph) and correct reading of Δv and Δt values to minimise uncertainty.
评分方案要求绘制一个大的三角形（至少覆盖图的一半）并正确读取 Δv 和 Δt 的值，以减小误差。
10. Deriving the Relationship Between Critical Angle and Refractive Index | 临界角与折射率关系的推导
When light travels from a denser medium (refractive index n) into a rarer medium (air, n=1), at the critical angle θ𝒸, the angle of refraction is 90°.
当光从光密介质（折射率 n）射向光疏介质（空气，n=1）时，在临界角 θ𝒸 处，折射角为 90°。
Applying Snell’s law: n sin θ𝒸 = 1 × sin 90°. Since sin 90° = 1, we obtain sin θ𝒸 = 1/n. This derivation is a one-step process but frequently appears in mark schemes.
应用斯涅尔定律：n sin θ𝒸 = 1 × sin 90°。由于 sin 90° = 1，得到 sin θ𝒸 = 1/n。这个推导只有一步，但在评分方案中经常出现。
Examiners also expect recall of the fact that total internal reflection occurs only when the incident angle is greater than the critical angle and the light is moving from denser to rarer medium.
考官还期望能回忆起：只有当入射角大于临界角且光从光密介质射向光疏介质时，才会发生全内反射。
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