Deriving Key AS Physics Formulas: Insights from 9630-PH01 Mark Scheme 2016 | 推导AS物理关键公式：基于9630-PH01 2016评分方案

📚 Deriving Key AS Physics Formulas: Insights from 9630-PH01 Mark Scheme 2016 | 推导AS物理关键公式：基于9630-PH01 2016评分方案

The 2016 mark scheme for 9630-PH01 International AS Physics Unit 1 places strong emphasis on deriving fundamental equations from first principles, rather than simply quoting them. Examiners consistently reward logical, step-by-step algebraic reasoning and clear substitution of definitions. This article revisits the essential formula derivations that appear across mechanics, materials, and electricity, drawing directly on the approaches validated by that mark scheme.

2016年9630-PH01国际AS物理单元1的评分方案高度重视从基本原理出发推导基础方程，而不仅仅是引用它们。考官一贯奖励逻辑清晰、逐步展开的代数推理以及明确定义的代入。本文重新梳理力学、材料学和电学中必考的关键公式推导，并直接借鉴该评分方案所认可的推导方式。

1. Deriving the SUVAT Equations from Definitions | 从定义推导匀加速运动方程

Acceleration a is defined as the rate of change of velocity: a = (v – u) / t, where u is initial velocity, v is final velocity, and t is the time interval. A simple rearrangement immediately gives the first kinematic equation: v = u + at.

加速度a定义为速度的变化率：a = (v – u) / t，其中u是初速度，v是末速度，t是时间间隔。简单变形直接得到第一个运动学方程：v = u + at。

For constant acceleration, average velocity is (u + v)/2. Displacement s is the product of average velocity and time, so s = ((u + v)/2) × t = ½(u + v)t. This relationship allows us to find displacement without knowing acceleration explicitly.

对于匀加速运动，平均速度为 (u + v)/2。位移s是平均速度与时间的乘积，因此 s = ((u + v)/2) × t = ½(u + v)t。这个关系使我们无需显式知道加速度就能求出位移。

To express s in terms of u, a, and t, substitute v = u + at into s = ½(u + v)t: s = ½(u + u + at)t = ½(2u + at)t = ut + ½at². In the 9630-PH01 mark scheme, this substitution step must be shown explicitly to earn method marks.

为了用u、a和t表示s，将v = u + at代入 s = ½(u + v)t：s = ½(u + u + at)t = ½(2u + at)t = ut + ½at²。在9630-PH01评分方案中，必须明确展示这一代入步骤才能获得方法分数。

To eliminate t from the equations, rearrange v = u + at to t = (v – u)/a and substitute into s = ½(u + v)t: s = ½(u + v) × (v – u)/a = (v² – u²)/(2a). Multiplying through by 2a yields the time-independent form v² = u² + 2as. Careful algebraic expansion is expected; errors in sign or expansion commonly lose marks.

为了从方程中消去t，将 v = u + at 变形为 t = (v – u)/a 并代入 s = ½(u + v)t：s = ½(u + v) × (v – u)/a = (v² – u²)/(2a)。两边同乘以2a得到与时间无关的形式 v² = u² + 2as。精细的代数展开是预期的；符号或展开错误通常会导致失分。

A common omission in student scripts is the failure to state that these equations apply only when acceleration is constant. The 2016 mark scheme penalises missing this condition in explanation questions.

学生答卷中常见的疏漏是没有说明这些方程仅在加速度恒定时成立。2016年评分方案在解释题中对遗漏该条件予以扣分。

2. Deriving the Impulse–Momentum Relationship from Newton’s Second Law | 从牛顿第二定律推导冲量-动量关系

Newton’s second law in its most general form states that resultant force F equals the rate of change of momentum: F = Δp/Δt, where momentum p = mv. When mass remains constant during the motion, this simplifies to F = m Δv/Δt = ma, which is the familiar form.

牛顿第二定律的最普遍形式是合力F等于动量的变化率：F = Δp/Δt，其中动量 p = mv。当运动过程中质量不变时，可简化为F = m Δv/Δt = ma，这就是大家熟悉的形式。

Rearranging Δp/Δt = F gives Δp = FΔt. The product FΔt is defined as impulse J. Hence impulse equals change in momentum: J = mv – mu. The mark scheme rewards stating that this is a vector relationship; direction must be considered when adding impulses.

重新排列 Δp/Δt = F 得到 Δp = FΔt。乘积FΔt被定义为冲量J。因此冲量等于动量的变化：J = mv – mu。评分方案奖励指出这是一个矢量关系；在合成冲量时必须考虑方向。

An alternative derivation often requested in exams starts from F = ma and a = (v – u)/t. Multiplying both sides of F = m(v – u)/t by t gives Ft = m(v – u) = mv – mu. The 2016 paper explicitly asked candidates to perform this derivation, and marks were allocated for linking the equation to Newton’s second law and clearly defining impulse.

考试中常要求的另一种

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