📚 A-Level Physics Unit 5 Question Paper Jan 2020: Formula Derivations | A-Level物理Unit 5 2020年1月试卷公式推导
In the January 2020 Unit 5 examination, students were expected to demonstrate a deep understanding of derivations ranging from radioactive decay and kinetic theory to simple harmonic motion and gravitational fields. This article reconstructs the essential derivations that underpin these topics, presenting each logical step with clarity. Mastering these derivations not only prepares you for exam-style questions but also strengthens your grasp of the underlying physics principles required at A-Level.
在2020年1月的Unit 5考试中，学生需要对放射性衰变、分子动理论、简谐运动以及引力场等专题的公式推导有深刻理解。本文重构了这些主题的关键推导过程，并清晰地展示每一步的逻辑。掌握这些推导不仅有助于应对考试类题目，还能加深你对A-Level所需物理原理的把握。
1. Radioactive Decay Law Derivation | 放射性衰变定律推导
The rate of decay of a radioactive sample is directly proportional to the number of undecayed nuclei present. We start with the observation that the activity –dN/dt ∝ N, where N is the number of nuclei. Introducing a decay constant λ gives the differential equation dN/dt = –λ N. Separating variables yields dN/N = –λ dt, and integrating both sides leads to ln N = –λ t + constant. Using the initial condition N = N₀ at t = 0, the constant becomes ln N₀. Hence, the solution is N = N₀ e⁻λᵗ, describing exponential decay.
放射性样品的衰变速率与尚未衰变的原子核数目成正比。从–dN/dt ∝ N出发，引入衰变常数λ得到微分方程dN/dt = –λ N。分离变量得dN/N = –λ dt，两边积分得到ln N = –λ t + 常数。利用初始条件t = 0时N = N₀，常数确定为ln N₀。因此解为N = N₀ e⁻λᵗ，这就是指数衰变律。
dN/dt = –λ N → N = N₀ e⁻λᵗ
2. Half-Life Equation | 半衰期方程
The half‑life T½ is the time taken for the number of nuclei to halve. Substituting N = N₀/2 into the decay law gives N₀/2 = N₀ e⁻λ T½, which simplifies to ½ = e⁻λ T½. Taking natural logarithms yields ln(½) = –λ T½, and since ln(½) = –ln 2, we obtain –ln 2 = –λ T½. Therefore, the half‑life is T½ = ln 2 / λ, independent of the initial quantity.
半衰期T½是原子核数目减半所需的时间。将N = N₀/2代入衰变律得N₀/2 = N₀ e⁻λ T½，化简为½ = e⁻λ T½。两边取自然对数得ln(½) = –λ T½，由于ln(½)= –ln 2，得到–ln 2 = –λ T½。因此半衰期为T½ = ln 2 / λ，与初始数量无关。
T½ = ln 2 / λ
3. Activity and Decay Constant | 活度与衰变常数
Activity A is defined as the number of decays per unit time, i.e. A = –dN/dt. From the decay law, –dN/dt = λ N, so A = λ N. The unit of activity is the becquerel (Bq), equivalent to one decay per second. This relation shows that activity decreases exponentially with time, following the same behaviour as N: A = λ N₀ e⁻λᵗ = A₀ e⁻λᵗ.
活度A定义为单位时间内的衰变次数，即A = –dN/dt。由衰变律得–dN/dt = λ N，因此A = λ N。活度的单位是贝克勒尔(Bq)，相当于每秒一次衰变。该关系表明活度也随时间指数衰减，遵循与N相同的规律：A = λ N₀ e⁻λᵗ = A₀ e⁻λᵗ。
A = λ N → A = A₀ e⁻λᵗ
4. Kinetic Theory: Pressure of an Ideal Gas | 分子动理论：理想气体的压强
Consider a single molecule of mass m moving with speed vₓ in a cubical container of side L. Each collision with a wall reverses the perpendicular velocity component, giving a momentum change of 2m vₓ. The time between collisions with the same wall is 2L/vₓ, so the average force on that wall is F₁ = (2m vₓ) / (2L/vₓ) = m vₓ² / L. Summing over all N molecules, the total force on one wall is F = (m/L) Σ vₓ². The pressure p is F / L², so p = (m/L³) Σ vₓ² = (m/V) Σ vₓ². Using the mean square speed ⟨vₓ²⟩ = (1/N) Σ vₓ², we get p = (N m / V) ⟨vₓ²⟩. Because motions in x, y, z directions are equally likely, ⟨v²⟩ = ⟨vₓ²⟩ + ⟨v_y²⟩ + ⟨v_z²⟩ = 3⟨vₓ²⟩, hence ⟨vₓ²⟩ = ⅓⟨c²⟩ where c = |v|. Substituting gives p = ⅓ (N m / V) ⟨c²⟩, or p = ⅓ ρ ⟨c²⟩ with density ρ = N m / V.
考虑一个质量为m的分子以速度分量vₓ在边长为L的立方容器中运动。每次与器壁碰撞，垂直分量反向，动量改变为2m vₓ。与同一壁的两次碰撞间的时间为2L/vₓ，因此作用于该壁的平均力为F₁ = (2m vₓ) / (2L/vₓ) = m vₓ² / L。对全部N个分子求和，施加于一面壁的总力为F = (m/L) Σ vₓ²。压强p = F / L²，故p = (m/L³) Σ vₓ² = (m/V) Σ vₓ²。引入均方速率⟨vₓ²⟩ = (1/N) Σ vₓ²，得p = (N m / V) ⟨vₓ²⟩。由于x、y、z方向运动等概率，⟨v²⟩ = ⟨vₓ²⟩ + ⟨v_y²⟩ + ⟨v_z²⟩ = 3⟨vₓ²⟩，因此⟨vₓ²⟩ = ⅓⟨c²⟩，其中c为速度大小。代入后得到p = ⅓ (N m / V) ⟨c²⟩，或者用密度ρ = N m / V表示为p = ⅓ ρ ⟨c²⟩。
p = ⅓ (N m / V) ⟨c²⟩ = ⅓ ρ ⟨c²⟩
5. Mean Kinetic Energy and Absolute Temperature | 平均动能与绝对温度
The ideal gas equation in terms of number of molecules is pV = N k T, where k is the Boltzmann constant. Equating this with the pressure derived from kinetic theory, N k T = ⅓ N m ⟨c²⟩. Cancelling N and rearranging gives ½ m ⟨c²⟩ = (3/2) k T. The left‑hand side is precisely the mean translational kinetic energy of a molecule. Thus, the absolute temperature of an ideal gas is a direct measure of the average random kinetic energy of its particles.
用分子数表示的理想气体方程为pV = N k T，其中k为玻尔兹曼常数。令此式与动理论推导的压强相等，得N k T = ⅓ N m ⟨c²⟩。消去N并整理，即得½ m ⟨c²⟩ = (3/2) k T。等式左边正是分子的平均平动动能。因此，理想气体的绝对温度是其粒子平均无规则动能的直接量度。
½ m ⟨c²⟩ = (3/2) k T
6. Ideal Gas Equation pV = NkT | 理想气体状态方程 pV = NkT
Combining the kinetic pressure expression p = ⅓ (N/V) m ⟨c²⟩ with the kinetic‑energy–temperature relation ½ m ⟨c²⟩ = (3/2) k T yields p = ⅓ (N/V) · 2·(½ m ⟨c²⟩) = ⅓ (N/V) · 2·(3/2) k T = (N/V) k T. Multiplying both sides by V recovers the familiar form pV = N k T. This derivation bridges microscopic mechanics and macroscopic thermodynamics, showing that macroscopic state variables emerge from molecular motion.
将动理论压强表达式p = ⅓ (N/V) m ⟨c²⟩与动能-温度关系½ m ⟨c²⟩ = (3/2) k T结合，得到p = ⅓ (N/V) · 2·(½ m ⟨c²⟩) = ⅓ (N/V) · 2·(3/2) k T = (N/V) k T。两边同乘V即得常见的pV = N k T。该推导在微观力学与宏观热力学之间架起了桥梁，表明宏观状态量源于分子运动。
pV = N k T
7. Simple Harmonic Motion: Defining Equation | 简谐运动：定义方程
Simple harmonic motion (SHM) occurs when the restoring force on an object is directly proportional to its displacement from equilibrium and acts in the opposite direction. For a mass–spring system, Hooke’s law gives F = –k x. Using Newton’s second law, F = m a, we obtain m a = –k x, or a = –(k/m) x. Defining the constant ω² = k/m, the acceleration becomes a = –ω² x. This differential equation defines SHM and can be written as d²x/dt² = –ω² x.
当物体所受的回复力与其离开平衡位置的位移成正比且方向相反时，物体便做简谐运动（SHM）。对于弹簧振子，胡克定律给出F = –k x。应用牛顿第二定律F = m a，得m a = –k x，即a = –(k/m) x。定义常数ω² = k/m，加速度即写为a = –ω² x。该微分方程定义了简谐运动，亦可表示为d²x/dt² = –ω² x。
a = –ω² x or d²x/dt² = –ω² x
8. Displacement–Time Solution for SHM | 简谐运动的位移–时间解
The equation d²x/dt² = –ω² x is a second‑order linear differential equation whose general solution is x = A cos(ω t + φ) or equivalently x = A sin(ω t + φ). The constants A (amplitude) and φ (phase constant) are determined by initial conditions. Starting from x = A cos(ω t), velocity is v = dx/dt = –A ω sin(ω t), and acceleration is a = dv/dt = –A ω² cos(ω t) = –ω² x, confirming it satisfies the SHM equation. The system undergoes sinusoidal oscillations with angular frequency ω and period T = 2π/ω.
方程d²x/dt² = –ω² x是一个二阶线性微分方程，其通解为x = A cos(ω t + φ)或等价形式x = A sin(ω t + φ)。常数A（振幅）和φ（初相）由初始条件决定。以x = A cos(ω t)为例，速度为v = dx/dt = –A ω sin(ω t)，加速度为a = dv/dt = –A ω² cos(ω t) = –ω² x，验证其满足简谐运动方程。系统以角频率ω作正弦振荡，周期T = 2π/ω。
x = A cos(ω t) → v = –A ω sin(ω t) → a = –A ω² cos(ω t)
9. Gravitational Potential Energy Derivation | 引力势能推导
The gravitational potential energy U at a distance r from a mass M is obtained by considering the work done against gravity to bring a test mass m from infinity to that point. The gravitational force is F = –(G M m / r²) r̂, where the negative sign indicates attraction. Choosing the potential energy at infinity to be zero, the change in potential energy is ΔU = –∫ F·dr. Taking the radial outward path, the work done by the gravitational force is ∫_{∞}^{r} –(G M m / r²) dr = [G M m / r]_{∞}^{r} = G M m / r. Since ΔU = U(r) – U(∞) = –(work done by field), we obtain U(r) = –G M m / r. This negative value reflects the bound nature of the system.
距离质量M为r处的引力势能U，通过将检验质量m从无穷远移至该点时克服引力做的功来求得。引力为F = –(G M m / r²) r̂，负号表示吸引。选无穷远处势能为零，势能的变化量为ΔU = –∫ F·dr。沿径向向外路径，引力所做的功为∫_{∞}^{r} –(G M m / r²) dr = [G M m / r]_{∞}^{r} = G M m / r。由于ΔU = U(r) – U(∞) = –(场力做功)，故得U(r) = –G M m / r。负值反映出系统的束缚特性。
U = –G M m / r
10. Escape Velocity Derivation | 逃逸速度推导
Escape velocity is the minimum speed needed for an object to leave a planet’s gravitational field without further propulsion, i.e. to reach infinite distance with zero final kinetic energy. Using energy conservation: total energy at the surface (kinetic + potential) must equal total energy at infinity (zero). Thus, ½ m v² + (–G M m / R) = 0. Solving for v gives v² = 2 G M / R, so the escape velocity is v_esc = √(2 G M / R). This expression is independent of the object’s mass and depends only on the planet’s mass M and radius R.
逃逸速度是物体脱离行星引力场无需继续推进所需的最小速度，即到达无穷远处时动能恰好为零。利用能量守恒：表面处的总能量（动能加势能）必须等于无穷远处的总能量（设为零）。因此，½ m v² + (–G M m / R) = 0。解出v得v² = 2 G M / R，于是逃逸速度为v_esc = √(2 G M / R)。该表达式与物体质量无关，仅取决于行星的质量M与半径R。
v_esc = √(2 G M / R)
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