📚 AS Physics Unit 1: Formula Derivation Based on Exam Report Jan 2020 | AS物理单元1：基于2020年1月考情报告的公式推导
The January 2020 AS Physics Unit 1 exam report highlighted that many candidates lost marks not because they couldn’t recall equations, but because they failed to demonstrate a clear understanding of the underlying derivations and the physical assumptions that accompany them. In this article, we will systematically derive the key formulas that appeared in the examination, paying close attention to the common pitfalls identified by examiners. Each derivation is paired with bilingual commentary to reinforce both the conceptual and linguistic precision required for top marks.
2020年1月的AS物理单元1考情报告指出，许多考生失分并非因为记不住公式，而是未能清晰展示对推导过程及伴随其物理假设的理解。本文将从零开始，系统推导本次考试涉及的核心公式，并密切关注考官指出的常见错误。每个推导均配有中英文对照的讲解，帮助你在深度理解和语言精确度上达到顶级评分标准。
1. Deriving the Equations of Uniformly Accelerated Motion | 匀加速运动方程的推导
The fundamental definition of acceleration is the rate of change of velocity: a = (v – u) / t, where u is initial velocity, v is final velocity, and t is the time taken. This can be rearranged into the familiar form v = u + at. However, the exam report noted that many candidates struggled to derive the displacement equations from first principles. Starting from the definition of average velocity for uniform acceleration, s = ((u + v) / 2) × t, we substitute v = u + at to obtain s = ut + ½at². By eliminating t, squaring the first equation, and combining it with the displacement expression, we arrive at v² = u² + 2as. The most persistent error was failing to treat velocity as a vector — candidates often lost signs when substituting values into these equations.
加速度的基本定义是速度的变化率：a = (v – u) / t，其中u为初速度，v为末速度，t为所用时间。该式可变形为我们熟悉的 v = u + at。然而，考情报告指出，许多考生在从基本原理推导位移方程时遇到困难。从匀加速度下的平均速度定义出发，s = ((u + v) / 2) × t，代入 v = u + at 即可得到 s = ut + ½at²。若消去 t，将第一式平方并与位移表达式结合，便可得到 v² = u² + 2as。最顽固的错误在于没有将速度视为矢量——考生在代入数值时常常丢掉正负号。
2. Newton’s Second Law and the Impulse-Momentum Link | 牛顿第二定律与冲量-动量关系
Newton’s second law is often stated as F = ma, but its more general form is F = Δp / Δt, where p = mv is momentum. The January 2020 report underscored that candidates who could derive impulse from this basis performed significantly better. For a constant net force, Δp = F Δt, and the impulse experienced by an object equals the change in its momentum. When mass is constant, Δp = m(v – u), leading directly to F = m(v – u)/Δt = ma. A frequent slip was forgetting that force and momentum are vectors; in collision problems, the sign of velocity must be consistently applied. Examiners advised always sketching a before-and-after diagram with labelled positive direction.
牛顿第二定律常被表述为 F = ma，但其更普适的形式为 F = Δp / Δt，其中 p = mv 为动量。2020年1月的报告强调，能够从这一基础出发推导冲量的考生表现明显更好。对于恒定的合力，Δp = F Δt，物体所受的冲量等于其动量的变化。当质量恒定不变时，Δp = m(v – u)，直接导出 F = m(v – u)/Δt = ma。一个经常出现的马虎错误是忘记力和动量都是矢量；在碰撞问题中，速度的符号必须前后统一。考官建议总是绘制标明正方向的碰撞前后示意图。
3. Work–Energy Theorem: From Forces to Kinetic Energy | 动能定理：从力到动能
Starting with the work done by a constant net force, W = F s, and substituting F = ma gives W = ma s. Using the kinematic relation v² = u² + 2as, we can express as = (v² – u²)/2. Thus, W = m × (v² – u²)/2 = ½mv² – ½mu². This shows that net work done equals the change in kinetic energy. The examiners’ report commented that candidates often confused work done by an individual force with net work. Remember: only the component of force parallel to displacement does work. Many lost marks by not resolving forces correctly before applying the theorem.
从恒定合力所做的功出发，W = F s，代入 F = ma 可得 W = ma s。利用运动学关系 v² = u² + 2as，可得 as = (v² – u²)/2。于是，W = m × (v² – u²)/2 = ½mv² – ½mu²。这表明合力做功等于动能的变化量。考官报告评论称，考生常将单个力所做的功与合力做功混淆。请记住：只有与位移平行的分力才做功。许多考生因在应用定理前未正确分解力而失分。
4. Gravitational Potential Energy from Uniform Weight | 由均匀重力推导重力势能
When an object of mass m is lifted through a vertical height h near the Earth’s surface, the lifting force must at least balance the weight mg. The minimum work done against gravity is W = F × d = mg × h = mgh. This energy is stored as gravitational potential energy, ΔEₚ = mgh. The report indicated that many candidates could state the formula but could not explain why it is only valid for uniform gravitational fields (where g is constant). A typical exam question required students to assert that the derivation assumes no significant change in g over the height h.
当质量为m的物体在地表附近被竖直举高h时，举力至少要与重力mg平衡。克服重力所做的最小功为 W = F × d = mg × h = mgh。这部分能量以重力势能形式储存，即 ΔEₚ = mgh。报告指出，许多考生可以写出公式，却无法解释它为何只适用于均匀重力场（g恒定）。典型的考题要求学生明确指出，该推导假设在高度h范围内g无明显变化。
5. Elastic Potential Energy: Integrating Hooke’s Law | 弹性势能：胡克定律的积分
For a spring obeying Hooke’s law, F = kx, where x is extension and k is the spring constant. The force is not constant; it grows linearly with extension. The work done stretching the spring from 0 to x is the area under the force–extension graph, a triangle of base x and height kx. Hence, elastic potential energy Eₑ = ½ × base × height = ½(kx)(x) = ½kx². Alternatively, using average force (0 + kx)/2 = ½kx, work = average force × extension = ½kx × x = ½kx². The January report flagged that many students mistakenly used F = kx as the constant force, ignoring the factor ½.
对于服从胡克定律的弹簧，F = kx，其中x为伸长量，k为劲度系数。该力并非恒定，而是随伸长量线性增长。将弹簧从0拉伸至x所做的功等于力-伸长图下的面积，即底为x、高为kx的三角形面积。因此，弹性势能 Eₑ = ½ × 底 × 高 = ½(kx)(x) = ½kx²。也可使用平均力 (0 + kx)/2 = ½kx，功 = 平均力 × 伸长量 = ½kx × x = ½kx²。一月份报告特别指出，众多学生错误地将F = kx当作恒力使用，忽略了½系数。
6. Stress, Strain and the Young Modulus Foundation | 应力、应变与杨氏模量基准
Tensile stress σ is defined as force per unit cross-sectional area: σ = F / A. Tensile strain ε is the fractional extension: ε = ΔL / L₀, where L₀ is the original length. The Young modulus E is the ratio of stress to strain within the elastic limit: E = σ / ε. Substituting the definitions yields E = (F/A) / (ΔL/L₀) = (F L₀) / (A ΔL). The examiners’ report observed that candidates frequently confused strain with extension and failed to treat stress and Young’s modulus as having the same units (Pa). They recommended always carrying out unit analysis when using these derived forms.
拉应力 σ 定义为单位横截面积上的作用力：σ = F / A。拉应变 ε 是长度的相对变化：ε = ΔL / L₀，其中L₀为原始长度。杨氏模量E是在弹性限度内应力与应变的比值：E = σ / ε。代入定义式可得 E = (F/A) / (ΔL/L₀) = (F L₀) / (A ΔL)。考官报告注意到，考生经常将应变与伸长量混淆，且未能认识到应力和杨氏模量具有相同单位（Pa）。报告建议在使用这些衍生形式时，务必进行量纲分析。
7. Fluid Pressure: Force and Column Weight | 流体压强：力与液柱重量
Consider a vertical column of liquid of height h, cross-sectional area A, and density ρ. The volume is V = A h, so its mass is m = ρ A h. The weight of this column is W = m g = ρ A h g. This weight acts on the base area A, producing a pressure p = W / A = ρ A h g / A = ρ g h. The derivation assumes the fluid is static and incompressible. Many candidates forgot that this formula gives the gauge pressure, i.e., the pressure due to the fluid alone, not the absolute pressure (which includes atmospheric pressure on the free surface).
考虑高度为h、截面积为A、密度为ρ的竖直液柱。其体积为 V = A h，因此质量 m = ρ A h。该液柱的重量为 W = m g = ρ A h g。这一重量作用在底面积A上，产生的压强为 p = W / A = ρ A h g / A = ρ g h。该推导假设流体静止且不可压缩。许多考生忘记了此公式给出的是计示压强，即仅由流体自身产生的压强，而非绝对压强（后者还包括液面的大气压强）。
8. Principle of Moments and the Equilibrium Condition | 力矩原理与平衡条件
For an object in rotational equilibrium, the sum of clockwise moments about any pivot must equal the sum of anticlockwise moments. The moment of a force is defined as the product of the force and the perpendicular distance from the pivot to the line of action: M = F d. A common derivation task in the exam was to start with the definition of moment and then apply it to a beam supported at its centre, showing that the condition for no rotation leads to F₁ d₁ = F₂ d₂. The report highlighted that candidates often measured distance to the point of force application rather than the perpendicular distance, especially when the force was applied at an angle.
对于处于转动平衡的物体，关于任意支点，顺时针力矩之和必须等于逆时针力矩之和。力矩被定义为力与从支点到力作用线的垂直距离的乘积：M = F d。考试中常见的推导题是从力矩定义出发，将其应用于一中点支撑的横梁，证明无转动的条件为 F₁ d₁ = F₂ d₂。报告强调，考生经常测量到力作用点的距离，而非垂直距离，尤其是当力以某一角度施加时。
9. Projectile Motion: Deriving Range on a Horizontal Plane | 抛体运动：水平面上射程的推导
For a projectile launched from ground level with speed u at an angle θ to the horizontal, the initial vertical component is u sinθ. Taking upward as positive and using s = ut + ½at² vertically, the time of flight T is found by setting vertical displacement to zero: 0 = (u sinθ) T – ½g T², giving T = (2 u sinθ) / g (ignoring the t=0 solution). The horizontal range R = horizontal velocity × time = (u cosθ) × T = (u cosθ) × (2 u sinθ / g) = (u² × 2 sinθ cosθ) / g = (u² sin2θ) / g. The examiners pointed out that a frequent mistake was to assume the time of flight is the time to maximum height, inadvertently halving the range. Another error was not checking that the landing point was at the same vertical level.
对于从地面以速度u、与水平方向成θ角发射的抛体，初始竖直分量为 u sinθ。取向上为正，使用竖直方向上的s = ut + ½at²，令竖直位移为零求得飞行时间T：0 = (u sinθ) T – ½g T²，得到 T = (2 u sinθ) / g（舍去t=0的解）。水平射程 R = 水平速度 × 时间 = (u cosθ) × T = (u cosθ) × (2 u sinθ / g) = (u² × 2 sinθ cosθ) / g = (u² sin2θ) / g。考官指出，一个常见错误是将飞行时间误认为是到达最高点的时间，从而不经意间将射程减半。另一个错误是没有检验落地点是否与出发点位于同一竖直高度。
10. Centre of Mass and Stability: Deriving the Toppling Condition | 质心与稳定性：倾倒条件的推导
The centre of mass of a uniform regular solid is at its geometric centre. For an object to be stable, the line of action of its weight must fall inside its base area. The January 2020 report included a derivation task where candidates had to show, using moments, that a block of width w and height h tilted at an angle θ will topple when tan θ = w / h. By drawing the weight vector through the centre of mass and requiring it to pass through the edge of the base, the critical angle is determined by simple trigonometry: tan θ = (w/2) / (h/2) = w/h. The most common error was misplacing the pivot point or failing to use the correct dimensions.
均匀规则固体的质心位于其几何中心。物体要保持稳定，其重力的作用线必须落在底面积之内。2020年1月报告中有一道推导题，要求考生利用力矩证明，宽为w、高为h的物块倾斜角度θ时，当 tan θ = w / h 即会倾倒。通过画出经过质心的重力矢量，并要求它通过底座边缘，临界角可由简单三角函数求得：tan θ = (w/2) / (h/2) = w/h。最常见的错误在于支点位置判断有误，或使用了错误的边长尺寸。
11. Retaining Precision: A Common Theme in the Report | 保持精度：报告中的共性主题
Throughout the derivation questions in January 2020, examiners repeatedly emphasised the importance of algebraic consistency. Candidates often lost marks by failing to show intermediate steps, mishandling minus signs, or substituting numerical values too early. The report recommended that students practise derivations by starting with definitions in words, translating them into symbols, and only inserting numbers at the very end. This method minimises rounding errors and demonstrates a higher level of physical insight.
在2020年1月所有的推导题中，考官反复强调了代数一致性的重要性。考生常常因未展示中间步骤、符号处理不当或过早代入数值而丢分。报告建议学生从文字定义开始推导，将其转换为符号，只到最后才代入数字。这种方法能最大限度地减少舍入误差，并展示更高层次的物理洞见。
12. Sharpening Exam Technique with the Mark Scheme Logic | 结合评分逻辑提升应试技巧
The exam report also highlighted that marks are explicitly allocated for stating the relevant physical law (e.g., “by Newton’s second law”), naming the assumption (e.g., “air resistance is negligible”), and carrying out a clear algebraic manipulation. Simply jumping to the final formula rarely earned full credit. By embedding your derivation in a logical structure — define terms, write the law, apply to the situation, manipulate algebraically, state the result — you mirror the mark scheme and safeguard against losing easy marks.
考情报告还指出，评分方案会明确为相关物理定律的陈述（例如“由牛顿第二定律”）、假设的说明（例如“忽略空气阻力”）以及清晰的代数推导步骤分配分数。仅仅跳到最终公式几乎不可能拿到满分。将推导嵌入一种逻辑结构——定义变量、写出定律、应用到情境中、进行代数处理、陈述结果——你就会贴合评分方案，避免丢失本可得手的基础分。
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