OCR A-Level Physics June 2023 Mark Scheme 1: Formula Derivations | OCR A-Level物理2023年6月评分标准1：公式推导

📚 OCR A-Level Physics June 2023 Mark Scheme 1: Formula Derivations | OCR A-Level物理2023年6月评分标准1：公式推导

In the OCR A-Level Physics examination, Paper 1 often tests the ability to derive fundamental equations from first principles. This article revisits key derivations that appeared in the June 2023 mark scheme, using clear step-by-step reasoning. Mastering these derivations strengthens conceptual understanding and prepares students for high-mark questions.

在OCR A-Level物理考试中，Paper 1经常考查从基本原理推导基本方程的能力。本文回顾了2023年6月评分标准中出现的关键推导，采用清晰的逐步推理。掌握这些推导可以加深概念理解，并为学生应对高分题目做好准备。

1. Deriving the First Equation of Motion | 推导第一运动方程

Acceleration is defined as the rate of change of velocity. If an object has initial velocity u, constant acceleration a, and final velocity v after time t, then by definition:

加速度定义为速度的变化率。如果一个物体具有初速度u、恒定加速度a，在时间t后末速度为v，那么根据定义：

a = (v − u) / t

Rearranging this gives the well-known form used in the OCR mark scheme:

将其重排即得OCR评分标准中使用的熟知形式：

v = u + a t

This equation assumes no displacement information is required and is valid only for uniform acceleration in a straight line.

该方程假设不需要位移信息，且仅适用于匀加速直线运动。

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2. Deriving the Displacement Equation | 推导位移方程

Displacement is the area under a velocity–time graph. For uniform acceleration, the graph is a straight line from u to v. The average velocity is (u + v)/2, so displacement s can be expressed as:

位移是速度-时间图线下的面积。对于匀加速运动，图线是从u到v的直线。平均速度为(u + v)/2，因此位移s可表示为：

s = (u + v)/2 × t

Substituting v = u + a t from the first equation yields:

代入第一方程 v = u + a t 得到：

s = (u + u + a t)/2 × t = (2u + a t)t / 2 = u t + ½ a t²

This derivation was required in a 4-mark question on the June 2023 Paper 1, where candidates had to show the intermediate steps.

在2023年6月Paper 1的一道4分题中，要求考生展示该推导的中间步骤。

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3. Deriving the Velocity-Displacement Relation | 推导速度-位移关系式

substitute into the displacement equation. Starting with v = u + a t and s = u t + ½ a t²:

为从前两个方程中消去时间t，我们将第一方程重排为t = (v − u)/a，并代入位移方程。从v = u + a t和s = u t + ½ a t²出发：

s = u[(v − u)/a] + ½ a [(v − u)/a]²

Simplifying gives:

简化得到：

2as = 2u(v − u) + (v − u)² = 2uv − 2u² + v² − 2uv + u² = v² − u²

Thus, the classic kinematic relation is derived:

于是得到经典的动力学关系式：

v² = u² + 2 a s

The mark scheme awards marks for correct algebraic manipulation and explicitly stating that acceleration is constant.

评分标准对正确的代数运算和明确说明加速度恒定给予分数。

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4. Deriving the Work–Kinetic Energy Theorem | 推导动能定理

net force F acting on a mass m over displacement s. The work done is W = F s. Using Newton' F = m a and the kinematic equation v² = u² + 2 a s, we can connect work to kinetic energy:

考虑一个恒定合力F作用在质量m上，位移为s。做功为W = F s。利用牛顿第二定律F = m a和运动学方程 v² = u² + 2 a s，可以将功与动能联系起来：

W = m a × s and a s = (v² − u²)/2

Therefore:

因此：

W = m (v² − u²)/2 = ½ m v² − ½ m u²

This shows that the net work done equals the change in kinetic energy. In the OCR mark scheme, explicit mention of the constant force assumption is required for full marks.

这表明合力做的功等于动能的变化量。在OCR评分标准中，获得满分需要明确提及恒定力的假设。

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5. Deriving Time of Flight for a Projectile | 推导抛体飞行时间

lic path, the time to rise to maximum height equals the time to fall. Using v = u + a t vertically, with final vertical velocity 0 at the peak and acceleration −g:

对于以初速度u、与水平方向成θ角发射的抛体，速度的竖直分量为 u sinθ。根据抛物线路径的对称性，上升到最大高度的时间等于下落时间。在竖直方向使用 v = u + a t，峰顶竖直末速度为0，加速度为−g：

0 = u sinθ − g tₘₐₓ → tₘₐₓ = u sinθ / g

The total time of flight T is twice this value:

总飞行时间T为此值的两倍：

T = 2 u sinθ / g

The June 2023 paper rewarded this derivation when candidates identified that vertical motion is independent of horizontal motion and applied the correct sign convention for g.

2023年6月的试卷奖励了这一推导，考生需指明竖直运动与水平运动独立，并对g使用正确的符号约定。

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6. Deriving Centripetal Acceleration a = v²/r | 推导向心加速度a = v²/r

tion continuously. Consider a small time Δt and small angle Δθ. The change in velocity Δv = v Δθ (from vector triangle). Thus acceleration magnitude a = v Δθ / Δt. Angular speed ω = Δθ / Δt = v / r, so:

一个物体在半径为r的圆形路径上以恒定速率v运动，周期T内行进距离2πr，因此 v = 2πr / T。速度矢量方向不断改变。考虑微小时间Δt和微小角度Δθ。速度变化量Δv = v Δθ（来自矢量三角形）。因此加速度大小 a = v Δθ / Δt。角速度 ω = Δθ / Δt = v / r，因而：

a = v × (v / r) = v² / r

²r) / T². The OCR mark scheme accepts both geometric and algebraic derivations, as long as the vector nature of velocity change is acknowledged.

或者使用周期：a = (4π²r) / T²。OCR评分标准接受几何和代数推导，只要承认速度变化的矢量性质。

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7. Deriving the Formula for Resistors in Parallel | 推导并联电阻公式

R₂. Total current I = I₁ + I₂. The equivalent resistance Rₚ satisfies V = I Rₚ. Therefore:

对于两个并联电阻R₁和R₂，其两端电压V相同。由欧姆定律，电流为 I₁ = V / R₁，I₂ = V / R₂。总电流 I = I₁ + I₂。等效电阻Rₚ满足 V = I Rₚ。因此：

I = V / R₁ + V / R₂ = V (1/R₁ + 1/R₂)

Substituting I = V / Rₚ gives:

代入 I = V / Rₚ 得到：

1 / Rₚ = 1 / R₁ + 1 / R₂

This derivation was part of a practical circuit question; marks were given for correctly linking current conservation and potential difference.

此推导是实践电路题的一部分，正确联系电流守恒与电势差可获得分数。

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8. Deriving Energy Stored in a Capacitor | 推导电容器储存的能量

to the other when the p.d. is v is v Δq. Since v = q / C, total work done is the area under the q–v graph, which is a triangle of height V and base Q:

考虑一个电容为C的电容器充电至电压V。储存的电荷为 Q = C V。当极板间电压为v时，将微小增量电荷Δq从一个极板移动到另一个极板所做的功为 v Δq。因为 v = q / C，总功为q–v图线下的面积，是一个高为V、底为Q的三角形面积：

W = ½ Q V

Using Q = C V, this becomes:

利用 Q = C V，可得：

W = ½ C V²

This energy is stored in the electric field. The OCR examiner expected candidates to explain why the factor ½ appears (because the average p.d. during charging is V/2).

该能量储存在电场中。OCR考官期望考生解释为何出现½因子（因为充电过程中的平均电压为V/2）。

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