📚 AS Physics Unit 2: Formula Derivation from June 2022 Mark Scheme | AS物理第二单元：2022年6月评分标准公式推导
In AS Physics Unit 2 exams, derivations form a critical part of the assessment. The June 2022 mark scheme emphasized clear logical steps, starting from defining symbols, stating assumptions, and progressing from fundamental definitions to the final equation. Mastering these derivations not only secures method marks but also deepens conceptual understanding. This article revisits the key derivations commonly tested in Unit 2, aligned with the rigour expected in the 2022 marking guide.
在AS物理第二单元考试中，公式推导是评估的重要组成部分。2022年6月的评分标准强调清晰的逻辑步骤，从定义符号、阐明假设，到从基本定义推导出最终方程。掌握这些推导不仅能确保获得方法分，还能加深概念理解。本文将回顾第二单元中常考的关键推导，并按照2022年评分指南要求的严谨性进行阐述。
1. Key Requirements of the Mark Scheme | 评分标准的核心要求
The mark scheme for derivations typically awards marks for: stating the definition of the quantity (e.g. acceleration = rate of change of velocity), writing the relevant defining equation, performing algebraic manipulation, and presenting the final formula. Examiner reports from June 2022 highlighted the need to show all steps, including substitution only where necessary, and to avoid skipping algebraic details.
推导的评分标准通常对以下步骤给分：陈述物理量的定义（例如加速度 = 速度变化率），写出相关的定义方程，进行代数变换，并呈现最终公式。2022年6月的考官报告强调需要展示所有步骤，包括仅在必要时进行代入，并避免略过代数细节。
2. Derivation of v = u + at | 推导 v = u + at
Start with the definition of constant acceleration: a = (change in velocity) / time taken. If initial velocity is u and final velocity is v, then a = (v − u) / t.
从匀加速度的定义出发：a = 速度变化量 / 所用时间。设初速度为 u，末速度为 v，则 a = (v − u) / t。
Multiplying both sides by t gives at = v − u. Finally, add u to both sides to obtain the first suvat equation:
两边乘以 t 得 at = v − u。最后两边加 u，得到第一个匀加速运动方程：
v = u + at
3. Derivation of s = (u + v)t / 2 | 推导 s = (u + v)t / 2
For an object moving with uniform acceleration, the average velocity vavg = (initial velocity + final velocity) / 2. This is only valid when acceleration is constant. Displacement s = vavg × t.
对于匀加速运动的物体，平均速度 vavg = (初速度 + 末速度) / 2。这仅在加速度恒定时成立。位移 s = vavg × t。
Substituting the average velocity yields:
代入平均速度可得：
s = ½ (u + v) t
Note: explicitly state that acceleration is constant, otherwise the average velocity formula cannot be used. The June 2022 mark scheme penalised omission of this justification.
注：必须明确说明加速度恒定，否则不能使用平均速度公式。2022年6月的评分标准对缺少此说明的情况扣分。
4. Derivation of s = ut + ½ at² | 推导 s = ut + ½ at²
We now combine the two previous equations. From (1) v = u + at and (2) s = (u + v) t / 2. Substitute v from (1) into (2):
现在我们将前面两个方程组合。由 (1) v = u + at 和 (2) s = (u + v) t / 2。将 (1) 的 v 代入 (2)：
s = (u + u + at) t / 2 = (2u + at) t / 2 = ut + ½ at².
So the displacement formula without final velocity is:
因此不含末速度的位移公式为：
s = u t + ½ a t²
This derivation demands precise algebraic expansion. The mark scheme often awards a method mark for correct substitution.
这个推导要求精确的代数展开。评分标准通常对正确的代入步骤给予方法分。
5. Derivation of v² = u² + 2as | 推导 v² = u² + 2as
To eliminate time t, start from v = u + at and express t = (v − u) / a. Substitute this into s = (u + v) t / 2:
要消去时间 t，从 v = u + at 得 t = (v − u) / a。代入 s = (u + v) t / 2：
s = (u + v) / 2 × (v − u) / a = (v² − u²) / (2a).
Multiplying both sides by 2a gives 2as = v² − u², and rearranging:
两边乘以 2a 得 2as = v² − u²，整理得：
v² = u² + 2 a s
The mark scheme rewards the correct use of algebraic identity (u+v)(v−u) = v² − u². Avoid skipping this expansion step.
评分标准鼓励正确使用代数恒等式 (u+v)(v−u) = v² − u²。避免跳过这个展开步骤。
6. Derivation of Kinetic Energy Ek = ½mv² | 推导动能 Ek = ½mv²
Consider a constant net force F acting on a mass m, causing acceleration a. Work done W = F s. By Newton’s second law, F = m a. Therefore W = m a s.
考虑作用在质量 m 上的恒定合外力 F，产生加速度 a。做功 W = F s。由牛顿第二定律 F = m a，因此 W = m a s。
Using the motion equation v² = u² + 2as with initial velocity u = 0 (starting from rest), we get a s = v² / 2. Substituting:
利用运动方程 v² = u² + 2as，并设初速度 u = 0（从静止开始），可得 a s = v² / 2。代入：
W = m × (v² / 2) = ½ m v².
This work done by the net force is equal to the kinetic energy gained. Hence:
合外力所做的功等于获得的动能。因此：
Ek = ½ m v²
If the object has an initial speed u, the work done equals the change in kinetic energy: ΔEk = ½ m v² − ½ m u². Both forms are derivable using the same logic; the June 2022 scheme accepted either, provided the link to work–energy principle was shown.
如果物体有初速度 u，则所做的功等于动能变化量：ΔEk = ½ m v² − ½ m u²。两种形式均可用相同逻辑推导；2022年6月的评分标准接受任一种，只要展示了功−能原理的联系。
7. Derivation of Gravitational Potential Energy Ep = mgh | 推导重力势能 Ep = mgh
Lifting an object of mass m through a vertical height h at constant speed requires an upward force equal to its weight mg. The work done against gravity is force × distance = mg × h.
将质量为 m 的物体匀速竖直提升高度 h，需要施加大小等于其重量 mg 的向上力。克服重力做的功为力 × 距离 = mg × h。
This work is stored as gravitational potential energy near the Earth’s surface. Taking a reference level where h = 0 gives:
在地球表面附近，该功储存为重力势能。取 h = 0 的参考面可得：
ΔEp = m g h
The derivation assumes g is constant over the height h, which is valid for small altitude changes. Always state this assumption in exams.
推导假设在高度 h 范围内 g 为常数，这对于小高度变化成立。考试时务必陈述这一假设。
8. Derivation of Elastic Potential Energy | 推导弹性势能
For a spring or wire obeying Hooke’s Law, extension ΔL is proportional to applied force F up to the limit of proportionality. The force–extension graph is a straight line through the origin.
对于遵循胡克定律的弹簧或金属丝，在比例极限内伸长量 ΔL 与施加力 F 成正比。力−伸长图是一条过原点的直线。
The work done to stretch the material equals the area under the graph, which is a triangle: Area = ½ × F × ΔL. Therefore energy stored:
拉伸材料所做的功等于图下面积，即三角形面积：½ × F × ΔL。因此储存的能量：
E = ½ F ΔL
Since F = k ΔL, where k is the spring constant, we can express the energy solely in terms of extension:
因为 F = k ΔL（k 为弹簧常数），能量可以用伸长量单独表示：
E = ½ k (ΔL)²
The June 2022 mark scheme required candidates to mention the linear relationship and the area interpretation; simply quoting the formula without the graphical justification often lost marks.
2022年6月的评分标准要求考生提及线性关系和面积解释；只机械引用公式而没有图形论证常会失分。
9. Derivation of Young Modulus Formula | 推导杨氏模量公式
Young modulus E is defined as the ratio of tensile stress to tensile strain. Stress = force per unit cross-sectional area = F / A. Strain = extension per original length = ΔL / L.
杨氏模量 E 定义为拉伸应力与拉伸应变之比。应力 = 单位横截面积的力 = F / A。应变 = 伸长量 / 原长 = ΔL / L。
Therefore:
因此：
E = stress / strain = (F / A) / (ΔL / L) = (F L) / (A ΔL)
This formula is used to determine Young modulus experimentally by measuring F, L, A and ΔL. In derivation questions, always start with the definition of stress and strain. Some mark schemes also require stating the SI unit of E as pascal (Pa) or N m⁻².
该公式用于通过测量 F、L、A 和 ΔL 实验测定杨氏模量。在推导题中，务必从应力和应变的定义入手。有些评分标准还要求写出 E 的国际单位帕斯卡 (Pa) 或 N m⁻²。
10. Derivation of Conservation of Momentum | 推导动量守恒
For two objects A and B colliding, Newton’s third law states they exert equal and opposite forces on each other: FA = −FB. The contact time Δt is identical for both.
对于碰撞的两个物体 A 和 B，牛顿第三定律指出它们施加给对方大小相等、方向相反的力：FA = −FB。接触时间 Δt 相同。
Impulse = force × time = change in momentum: F Δt = Δp. Hence for A and B:
冲量 = 力 × 时间 = 动量变化量：F Δt = Δp。因此对于 A 和 B：
ΔpA = FA Δt, ΔpB = FB Δt = −FA Δt = −ΔpA.
Thus the total change in momentum ΔpA + ΔpB = 0. Momentum is conserved:
因此总动量变化量 ΔpA + ΔpB = 0。动量守恒：
mA uA + mB uB = mA vA + mB vB
This derivation assumes no external resultant force acts on the system during the collision. The June 2022 examiners expected candidates to state this condition explicitly.
该推导假设碰撞过程中系统不受合外力作用。2022年6月的考官期望考生明确陈述这一条件。
11. Common Mistakes and Mark Scheme Insights | 常见错误与评分要点
The June 2022 mark scheme revealed recurrent errors: using average velocity for non-uniform acceleration without justification, confusing displacement with distance in suvat derivations, and omitting the factor of ½ in energy derivations. Always derive from basic definitions rather than memorised sequences. Labels like “by definition”, “assuming constant acceleration”, and “using Newton’s second law” are essential for securing method marks. For algebraic derivations, show expansion steps—do not jump from (u+v)(v-u) to v² − u² without the intermediate line.
2022年6月的评分标准揭示了反复出现的错误：未经证明就为非匀加速度使用平均速度，在匀加速运动方程推导中混淆位移和路程，以及在能量推导中遗漏因子 ½。务必从基本定义出发而非死记顺序。“根据定义”“假设加速度恒定”“运用牛顿第二定律”等标签对于获得方法分至关重要。代数推导时，要展示展开步骤——勿直接从 (u+v)(v-u) 跳至 v² − u² 而缺少中间行。
Another critical point: place the derivation in a clear logical structure—define symbols, state assumptions, write the foundational equation, perform manipulation, and present the final formula with correct units where applicable. This mirrors the mark allocation in the actual scheme.
另一个关键点：将推导置于清晰的逻辑结构中——定义符号，陈述假设，写出基本方程，进行代数操作，并给出最终公式（必要时附上正确单位）。这与实际评分方案中的分值分配相符。
12. Summary | 总结
Deriving formulas in AS Physics Unit 2 is a skill that rewards precision and systematic thinking. By internalising the patterns shown above and aligning your working with the June 2022 mark scheme expectations, you can convert these questions into reliable sources of marks. Practise writing out each derivation fully, from definition to final expression, and self-check against the standard steps.
在AS物理第二单元中，推导公式是一项奖励精确性和系统思维的技能。通过内化上述模式，并使你的解题过程符合2022年6月评分标准的期望，你可以将这些题目转化为可靠的得分点。练习完整写出每一个推导，从定义到最终表达式，并对照标准步骤自我检查。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply