AS Physics Unit 1 – Formula Derivations from Jan 2020 Mark Scheme | AS物理单元1

📚 AS Physics Unit 1 – Formula Derivations from Jan 2020 Mark Scheme | AS物理单元1 – 2020年1月评分方案公式推导

Mastering formula derivations is essential for top marks in AS Physics Unit 1. The January 2020 mark scheme rewards candidates who clearly state definitions, show algebraic substitutions, and link concepts to physical principles. This article revisits the most frequently tested derivations, breaking down each step as an examiner would expect, so you can turn qualitative understanding into precise, credit-worthy solutions.

掌握公式推导是AS物理单元1取得高分的关键。2020年1月的评分方案奖励那些清晰写出定义、展示代数替换步骤并将概念与物理原理联系起来的考生。本文回顾了最常考的推导，按照阅卷官期望的方式分解每一步，帮助你把定性理解转化为精准、能得分的解答。

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

All four SUVAT equations follow from the definition of constant acceleration a = (v − u) ∕ t. Rearranging instantly gives v = u + at. For displacement, you must recognise that average velocity is (u + v) ∕ 2 when acceleration is uniform. Multiply by time to obtain s = (u + v)t ∕ 2. Substituting v from the first equation yields s = ut + ½at². Finally, eliminating t between v = u + at and s = (u + v)t ∕ 2 produces v² = u² + 2as. The mark scheme expects you to write the defining equation for acceleration and to show the algebraic elimination clearly.

所有四个匀加速方程都源自恒定加速度的定义 a = (v − u) ∕ t 。整理后立即得到 v = u + at 。对于位移，你需要认识到当加速度均匀时平均速度为 (u + v) ∕ 2 。乘以时间得到 s = (u + v)t ∕ 2 。将第一个方程中的 v 代入即可得到 s = ut + ½at² 。最后，从 v = u + at 和 s = (u + v)t ∕ 2 中消去 t ，可推出 v² = u² + 2as 。评分方案要求你写出加速度的定义方程，并清晰地展示代数消元过程。

v = u + at → s = (u + v)t ∕ 2 → s = ut + ½at² → v² = u² + 2as

2. Resolving Weight on an Inclined Plane | 斜面上重力的分解

When an object rests on a slope at angle θ to the horizontal, its weight mg acts vertically downwards. Construct a right‑angled triangle with mg as the hypotenuse. The angle between the weight vector and the normal to the plane equals θ. Therefore the component perpendicular to the plane is mg cos θ and the component parallel to the slope is mg sin θ. Examiners look for a correctly labelled force diagram and a statement that the two components replace the single weight vector.

当物体静止在与水平面成 θ 角的斜面上时，其重力 mg 竖直向下。以 mg 为斜边构建直角三角形。重力矢量与斜面法线之间的夹角等于 θ 。因此垂直于斜面的分量为 mg cos θ ，平行于斜面的分量为 mg sin θ 。阅卷官期望看到正确标注的受力图，以及说明这两个分量替代了单一的重力矢量。

3. Derivation of Young Modulus from Stress and Strain | 从应力与应变推导杨氏模量

Young modulus E is defined as the ratio of tensile stress to tensile strain. Stress = F ∕ A and strain = ΔL ∕ L, where L is the original length. Substitution gives E = (F ∕ A) ÷ (ΔL ∕ L) = FL ∕ AΔL. In the linear region, Hooke’s law F = kΔL applies, but the mark scheme often expects you to start from the definition rather than simply quoting the final result. An alternative form using spring constant is E = kL ∕ A, obtained by substituting F = kΔL into the modulus equation.

杨氏模量 E 定义为拉伸应力与拉伸应变的比值。应力 = F ∕ A ，应变 = ΔL ∕ L ，其中 L 为原长。代入可得 E = (F ∕ A) ÷ (ΔL ∕ A) = FL ∕ AΔL 。在线性区域，胡克定律 F = kΔL 成立，但评分方案往往期望你从定义出发，而不是只写出最终结果。通过将 F = kΔL 代入模量方程，可得到用弹簧常数表示的形式 E = kL ∕ A 。

4. Deriving Wave Speed v = f λ from First Principles | 从基本原理推导波速 v = fλ

Consider a wave of period T and wavelength λ. In one complete period the wave travels a distance of one wavelength. Hence the speed is v = distance ∕ time = λ ∕ T. Since frequency f = 1 ∕ T, substitution gives v = f λ. This derivation is short but requires explicit mention of the definition of period and the relationship between frequency and period. The mark scheme penalises candidates who just write the final equation without linking steps.

考虑周期为 T 、波长为 λ 的波。在一个完整周期内，波传播的距离是一个波长。因此波速为 v = 距离 ∕ 时间 = λ ∕ T 。由于频率 f = 1 ∕ T ，代入可得 v = f λ 。该推导简短，但需要明确提及周期的定义以及频率与周期的关系。评分方案会扣减那些只写出最终方程而未展示推导步骤的考生的分数。

5. Impulse and the Change in Momentum | 冲量与动量的变化

Newton’s second law in its original form states that net force equals the rate of change of momentum: F = Δp ∕ Δt. For a constant mass, Δp = m(v − u). Thus F = m(v − u) ∕ Δt. Multiplying both sides by Δt gives F Δt = mv − mu. The left side is defined as impulse. The mark scheme often accepts either starting point—F = ma with a = (v − u) ∕ t or the momentum form—provided the reasoning is consistent. Always define impulse explicitly if the question uses the term.

牛顿第二定律的原始形式指出，合力等于动量的变化率：F = Δp ∕ Δt 。对于质量不变的情况，Δp = m(v − u) 。因此 F = m(v − u) ∕ Δt 。两边同乘 Δt 得到 F Δt = mv − mu 。等式左边定义为冲量。评分方案通常接受任意一种出发点——用 F = ma 且 a = (v − u) ∕ t 或动量形式——只要推理连贯。如果题目使用了冲量一词，务必明确定义。

6. Kinetic Energy Derived from Work Done | 从做功推导动能

For a constant net force F acting over a displacement s and doing work W = Fs. Using Newton’s second law F = ma and the SUVAT relation v² = u² + 2as, work becomes W = ma × s. When the object starts from rest, u = 0, so s = v² ∕ (2a). Substituting gives W = m a × v² ∕ (2a) = ½mv². This work is stored as kinetic energy. The crucial mark‑scheme point is correctly linking work, acceleration, and the kinematic equation without omitting steps.

对于恒定的合力 F 作用一段位移 s ，做功 W = Fs 。利用牛顿第二定律 F = ma 和匀加速关系式 v² = u² + 2as ，功可写为 W = ma × s 。当物体从静止开始时，u = 0 ，因此 s = v² ∕ (2a) 。代入得 W = m a × v² ∕ (2a) = ½mv² 。这些功以动能形式储存。评分方案的关键在于正确地将功、加速度和运动学方程联系起来，不遗漏步骤。

7. Pressure in a Fluid Column: p = ρgh | 液柱压强公式 p = ρgh

Consider a vertical column of liquid of height h, cross‑sectional area A, and density ρ. The mass of the liquid is m = ρV = ρAh. Its weight is mg = ρAhg. Pressure at the base due to the liquid alone is force per unit area: p = F ∕ A = (ρAhg) ∕ A = ρgh. The derivation assumes the liquid is static and incompressible, and that atmospheric pressure adds to the total pressure. A common exam tip is to draw the column and label the forces clearly before writing the algebraic steps.

考虑一个高为 h 、截面积为 A 、密度为 ρ 的竖直液柱。液体的质量为 m = ρV = ρAh 。其重量为 mg = ρAhg 。仅由液体引起的底部压强为力除以面积：p = F ∕ A = (ρAhg) ∕ A = ρgh 。推导假设液体是静止且不可压缩的，并且大气压会叠加在总压强上。常见的考试技巧是先画出液柱并清晰地标注各力，再进行代数推导。

8. Deriving Total Resistance for Series and Parallel Circuits | 串联与并联总电阻的推导

For resistors in series, the same current I flows through each. The total p.d. V = V₁ + V₂ + … = IR₁ + IR₂ + … = I(R₁ + R₂ + …). Hence R_total = V ∕ I = R₁ + R₂ + …. For resistors in parallel, the p.d. V is the same across each branch. The total current I = I₁ + I₂ + … = V ∕ R₁ + V ∕ R₂ + … . Therefore 1 ∕ R_total = I ∕ V = 1 ∕ R₁ + 1 ∕ R₂ + …. The mark scheme expects explicit use of Kirchhoff’s current law for the parallel case and conservation of energy for series.

对于串联电阻，相同的电流 I 流过每个电阻。总电压 V = V₁ + V₂ + … = IR₁ + IR₂ + … = I(R₁ + R₂ + …) 。因此 R_total = V ∕ I = R₁ + R₂ + … 。对于并联电阻，每个支路两端的电压 V 相同。总电流 I = I₁ + I₂ + … = V ∕ R₁ + V ∕ R₂ + … 。所以 1 ∕ R_total = I ∕ V = 1 ∕ R₁ + 1 ∕ R₂ + … 。评分方案期望在并联情况下明确使用基尔霍夫电流定律，串联时使用能量守恒。

9. Propagation of Uncertainties for Addition and Multiplication | 加减与乘除运算中不确定度的传递

When adding or subtracting quantities, absolute uncertainties add. If Q = A + B or A − B, then ΔQ = ΔA + ΔB. When multiplying or dividing, percentage (or fractional) uncertainties add. For Q = AB or A ∕ B, %ΔQ = %ΔA + %ΔB. These rules are derived from worst‑case combinations. A simplified proof for multiplication: the maximum value of AB is (A + ΔA)(B + ΔB) ≈ AB + AΔB + BΔA, so absolute uncertainty ≈ BΔA + AΔB. Dividing by AB gives the fractional sum. The mark scheme rewards candidates who show the steps or at least state the correct combination rule and apply it.

当进行加减运算时，绝对不确定度相加。若 Q = A + B 或 A − B ，则 ΔQ = ΔA + ΔB 。当进行乘除运算时，百分比（或相对）不确定度相加。对于 Q = AB 或 A ∕ B ，%ΔQ = %ΔA + %ΔB 。这些规则源于最不利组合。乘法的简单证明：AB 的最大值为 (A + ΔA)(B + ΔB) ≈ AB + AΔB + BΔA ，因此绝对不确定度约为 BΔA + AΔB 。除以 AB 即得相对值之和。评分方案会给展示推导步骤或至少正确陈述组合规则并加以应用的考生加分。

10. Conditions for Equilibrium and the Principle of Moments | 平衡条件与力矩原理

For a body to be in static equilibrium, the resultant force must be zero in all directions and the resultant moment about any point must be zero. The moment of a force is defined as force × perpendicular distance from the pivot. To derive the second condition, consider a uniform beam supported at its centre. If a weight W is placed at a distance d from the pivot, the clockwise moment is Wd. For balance, an equal anticlockwise moment must be provided. Algebraically, Σ clockwise moments = Σ anticlockwise moments. The mark scheme expects candidates to state that the sum of the moments is zero, not just that “clockwise = anticlockwise” without context.

要使物体处于静态平衡，所有方向上的合力必须为零，且绕任意点的合力矩必须为零。力矩定义为力 × 到支点的垂直距离。为了推导第二个条件，考虑一根支于中心的均匀横梁。如果在距支点 d 处放置一个重物 W ，则顺时针力矩为 Wd 。为达到平衡，必须提供一个等大的逆时针力矩。代数上表示为 Σ 顺时针力矩 = Σ 逆时针力矩。评分方案期望考生说明力矩之和为零，而不是仅写出“顺时针 = 逆时针”而不结合情境。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

AS Physics Unit 1 – Formula Derivations from Jan 2020 Mark Scheme | AS物理单元1 – 2020年1月评分方案公式推导