A-Level Physics Unit 4 Jan22 Mark Scheme: Key Formula Derivations | A-Level物理第四单元Jan22评分方案：关键公式推导

📚 A-Level Physics Unit 4 Jan22 Mark Scheme: Key Formula Derivations | A-Level物理第四单元Jan22评分方案：关键公式推导

The January 2022 Unit 4 mark scheme rewards students who can derive fundamental formulas from first principles. Whether you are tackling circular motion, simple harmonic motion, gravitational fields, electric fields, capacitors, magnetic fields, or nuclear physics, the ability to show where an equation comes from is essential for top grades. This article walks through the key derivations that appeared in the Jan22 assessment, explaining each step clearly so you can reproduce them with confidence.

2022年1月第四单元的评分方案对能够从基本原理推导出重要公式的学生给予奖励。无论你面对的是圆周运动、简谐运动、引力场、电场、电容器、磁场还是核物理，“展示推导过程”的能力对于取得高分至关重要。本文逐一讲解Jan22试卷中出现的核心推导，清晰解释每一步，让你能够自信地重现这些推导。

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1. Deriving Centripetal Acceleration | 推导向心加速度

Consider an object moving at constant speed v in a circle of radius r. In a short time Δt the object moves from point A to point B through a small angle Δθ. The velocity vector rotates by the same angle Δθ, so the change in velocity Δv is directed towards the centre of the circle.

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，转过小角度 Δθ。速度矢量也转过相同的角度 Δθ，因此速度的变化量 Δv 指向圆心。

From the vector triangle, the magnitude of Δv is approximately vΔθ. The acceleration a is the rate of change of velocity: a = Δv / Δt = vΔθ / Δt = vω, where ω = Δθ / Δt is the angular velocity.

由矢量三角形可知，Δv 的大小近似为 vΔθ。加速度 a 是速度的变化率：a = Δv / Δt = vΔθ / Δt = vω，其中 ω = Δθ / Δt 是角速度。

Using the relationship v = ωr from the definition of radian measure, we substitute ω = v / r to obtain the familiar expressions for centripetal acceleration:

利用弧度制定义得出的关系式 v = ωr，代入 ω = v / r，我们得到向心加速度的常见表达式：

a = v² / r = ω² r

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2. Relating Angular Velocity and Linear Speed | 角速度与线速度的关系

In circular motion, the angle θ in radians is defined as the arc length s divided by the radius r: θ = s / r. Differentiating with respect to time t gives dθ / dt = (1/r) ds / dt.

在圆周运动中，弧度制角度 θ 定义为弧长 s 除以半径 r：θ = s / r。对时间 t 求导，得到 dθ / dt = (1/r) ds / dt。

The rate of change of angle is angular velocity ω, and the rate of change of arc length is linear speed v. Therefore:

角度的变化率是角速度 ω，弧长的变化率是线速度 v。因此：

v = ω r

This simple equation is fundamental for linking rotational and translational quantities, and it appears repeatedly in Jan22 mark scheme answers involving gears, pulleys, and orbits.

这个简单的方程是联系转动量与平动量的基础，在Jan22评分方案的齿轮、滑轮及轨道相关问题中反复出现。

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3. Simple Harmonic Motion: Displacement–Time Equation | 简谐运动：位移–时间方程

Simple harmonic motion (SHM) can be modelled as the projection of uniform circular motion onto a diameter. If a particle moves around a reference circle of radius A with constant angular velocity ω, its displacement x from the equilibrium position is the horizontal component of the radius vector.

简谐运动可模拟为匀速圆周运动在直径上的投影。若一质点以恒定角速度 ω 在半径为 A 的参考圆上运动，则它相对于平衡位置的位移 x 就是半径矢量的水平分量。

At time t = 0, let the radius make an angle φ with the horizontal axis. Then x = A cos(ωt + φ). If timing starts when the particle is at maximum positive displacement, φ = 0 and we have:

在 t = 0 时，设半径与水平轴的夹角为 φ。则 x = A cos(ωt + φ)。若从质点位于最大正位移时开始计时，则 φ = 0，于是：

x = A cos(ωt)

From this equation, velocity and acceleration can be obtained by successive differentiation: v = –Aω sin(ωt) and a = –Aω² cos(ωt) = –ω² x.

由该方程，通过连续求导可得到速度和加速度：v = –Aω sin(ωt)，a = –Aω² cos(ωt) = –ω² x。

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4. Acceleration in SHM and the Defining Equation | SHM中的加速度与定义方程

A hallmark of simple harmonic motion is that the restoring force, and therefore the acceleration, is directly proportional to the displacement from equilibrium and always directed towards it. This condition is written as:

简谐运动的标志是回复力（因而加速度）与偏离平衡位置的位移成正比，并始终指向平衡位置。该条件写为：

a = – ω² x

The negative sign indicates that acceleration and displacement are in opposite directions. The Jan22 mark scheme frequently tests whether you can recognise this relationship and use it to prove that a system is executing SHM.

负号表示加速度与位移方向相反。Jan22评分方案经常考查你是否能识别这一关系，并用它来证明某个系统在做简谐运动。

For a mass–spring system, Hooke’s law gives F = –k x. Since F = m a, we have a = –(k/m) x. Comparing with a = –ω² x yields ω² = k/m, so the period is T = 2π √(m/k). Similarly, for a simple pendulum of length L, the restoring force leads to ω² = g/L.

对于质量-弹簧系统，胡克定律给出 F = –k x。由 F = m a 得 a = –(k/m) x。与 a = –ω² x 对比可得 ω² = k/m，因此周期 T = 2π √(m/k)。类似地，对长度为 L 的单摆，回复力推导出 ω² = g/L。

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5. Newton’s Law of Gravitation and Surface Gravity | 万有引力定律与表面重力

Newton’s law of gravitation states that any two point masses m₁ and m₂ separated by a distance r attract each other with a force:

牛顿万有引力定律指出，任意两个相距 r 的质点 m₁ 和 m₂ 彼此吸引的力为：

F = G m₁ m₂ / r²

When one of the masses is a planet of mass M and radius R, and the second mass is a small object of mass m on its surface, the distance r is effectively R. The gravitational force on the object is its weight m g. Equating the two expressions:

当其中一个质量是质量为 M、半径为 R 的行星，另一个质量是行星表面质量为 m 的小物体时，距离 r 就是 R。物体所受引力即为其重力 m g。令两式相等：

G M m / R² = m g ⇒ g = G M / R²

This derivation, frequently required in Jan22 papers, shows how surface gravity depends on the mass and radius of the planet, not on the small object’s mass.

这一推导在Jan22试卷中经常要求写出来，它表明表面重力加速度取决于行星的质量和半径，而与物体的质量无关。

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6. Gravitational Potential Energy Derivation | 引力势能推导

Gravitational potential energy U is the work done by an external agent to bring a mass m from infinity to a point a distance r from the centre of a mass M, without acceleration. The force overcome is the gravitational attraction F = G M m / r².

引力势能 U 是将质量 m 从无穷远处匀速移至距质量 M 中心距离为 r 处时外力所做的功。克服的力为引力 F = G M m / r²。

Since the force varies with distance, we integrate: U = ∫ from ∞ to r of F dr′. Taking the outward direction as positive, the displacement dr′ is in the negative direction, so the work done by the external force is ∫ from ∞ to r (–G M m / r′²) dr′ = [G M m / r′] from ∞ to r = G M m / r – 0. Hence the potential energy is:

由于力随距离变化，需要进行积分：U = ∫_{∞}^{r} F dr′。规定向外为正方向，位移 dr′ 为负方向，外力做功为 ∫_{∞}^{r} (–G M m / r′²) dr′ = [G M m / r′]_{∞}^{r} = G M m / r – 0。因此势能为：

U = – G M m / r

The negative sign appears because we define the potential energy to be zero at infinity; as the mass moves closer, the field does positive work and the potential energy becomes negative. The Jan22 mark scheme often asks for the full integration steps.

负号的出现是因为我们规定无穷远处势能为零；随着质量靠近，引力场做正功，势能变为负值。Jan22评分方案时常要求写出完整的积分步骤。

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7. Electric Field Strength from Potential Gradient | 电势梯度求电场强度

In a uniform electric field, the work done by the field in moving a positive charge q from one plate to another separated by distance d is W = q E d. This work is also equal to the loss in electric potential energy q ΔV, where ΔV is the potential difference between the plates.

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