OCR A-Level Physics June 2023 Paper 1 Formula Derivation | OCR A-Level 物理 2023年6月卷一 公式推导

📚 OCR A-Level Physics June 2023 Paper 1 Formula Derivation | OCR A-Level 物理 2023年6月卷一 公式推导

The OCR A-Level Physics Paper 1 (June 2023) included questions requiring students to derive fundamental equations from first principles. Mastering these derivations is essential for high marks. Below we present eight key derivations that mirror the exam style.

2023年6月的OCR A-Level物理试卷1包含要求从基本原理推导基本方程的题目。掌握这些推导对于获得高分至关重要。以下我们展示八个符合考试风格的关键推导。

1. Deriving s = ut + ½at² from a velocity-time graph | 由速度-时间图推导 s = ut + ½at²

Start with constant acceleration a, initial velocity u and final velocity v after time t. The average velocity under constant acceleration is v_avg = (u+v)/2.

从匀加速度 a、初速度 u 和 t 时刻的末速度 v 开始。匀加速下的平均速度为 v_avg = (u+v)/2。

From the definition of acceleration, a = (v-u)/t, so v = u + at.

由加速度定义 a = (v-u)/t，得 v = u + at。

Displacement s equals the area under a velocity-time graph: s = v_avg × t = ((u+v)/2) t.

位移 s 等于速度-时间图下的面积：s = 平均速度 × 时间 = ((u+v)/2) t。

Substitute v from above: s = ((u + u + at)/2) t = (2u + at)t / 2 = ut + ½at².

代入 v 表达式：s = ((u + u + at)/2) t = (2u + at)t / 2 = ut + ½at²。

s = ut + ½at²

This is the standard SUVAT equation for displacement with constant acceleration, which can be used directly when no final velocity is given.

这是匀加速运动的标准SUVAT位移公式，在没有给出末速度时可以直接使用。

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2. Deriving the conservation of momentum | 推导动量守恒定律

Consider two objects A (mass m₁, velocity u₁) and B (m₂, u₂) colliding and moving off with velocities v₁ and v₂. During collision, the force F_A on A by B and F_B on B by A are equal and opposite by Newton’s third law: F_A = -F_B.

考虑两个物体 A（质量 m₁，速度 u₁）和 B（m₂，u₂）发生碰撞，碰后速度分别为 v₁ 和 v₂。碰撞过程中，A 受到 B 的力 F_A 与 B 受到 A 的力 F_B 为牛顿第三定律的作用力与反作用力：F_A = -F_B。

For a force acting over a short collision time Δt, the impulse equals change in momentum: F_A Δt = m₁v₁ – m₁u₁ and F_B Δt = m₂v₂ – m₂u₂.

在短时间 Δt 内，冲量等于动量的变化：F_A Δt = m₁v₁ – m₁u₁，F_B Δt = m₂v₂ – m₂u₂。

Because F_A = -F_B, we have m₁v₁ – m₁u₁ = -(m₂v₂ – m₂u₂), which rearranges to m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.

由于 F_A = -F_B，有 m₁v₁ – m₁u₁ = -(m₂v₂ – m₂u₂)，整理得 m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。

total momentum before collision = total momentum after collision

This shows momentum is conserved in any isolated system, provided no external resultant force acts.

这表明在没有外部合外力作用的孤立系统中，动量总是守恒的。

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3. Deriving kinetic energy Eₖ = ½mv² | 推导动能公式 Eₖ = ½mv²

For constant force F accelerating a mass m from rest over displacement s, work done W = F s. Using F = ma, W = ma s.

对恒定力 F 使质量 m 从静止加速经过位移 s，做功 W = F s。用 F = ma，得 W = ma s。

From v² = u² + 2as with u=0, we have as = v²/2. Substituting gives W = m × v²/2 = ½mv².

由 v² = u² + 2as，初始静止 u=0，得 as = v²/2。代入得 W = m × v²/2 = ½mv²。

Eₖ = ½mv²

The kinetic energy stored in a moving object equals the work done to accelerate it. This derivation assumes constant resultant force, but the result is general.

运动物体储存的动能等于加速它所做的功。推导假设合外力恒定，但结果具有普遍性。

If the mass had initial speed u, the work done is change in KE: W = ½mv² – ½mu².

若物体有初速度 u，则功等于动能变化量：W = ½mv² – ½mu²。

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4. Deriving elastic potential energy E = ½kx² | 推导弹性势能公式 E = ½kx²

Hooke’s law states that the force F needed to extend or compress a spring by an amount x is F = kx, where k is the spring constant.

胡克定律指出，拉伸或压缩弹簧 x 所需的力为 F = kx，k 是劲度系数。

As force varies linearly from 0 to kx, the average force during extension is F_avg = ½kx. Work done = average force × extension = (½kx) × x = ½kx².

由于力从 0 线性增加到 kx，伸长过程中的平均力为 F_avg = ½kx。功 = 平均力 × 伸长量 = (½kx) × x = ½kx²。

E_elastic = ½kx²

This work is stored as elastic potential energy. Graphically, it represents the area under the force-extension graph.

这个功就以弹性势能形式储存。在图上，它表示力-伸长量图下的面积。

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5. Deriving centripetal acceleration a = v²/r | 推导向心加速度 a = v²/r

An object moving at constant speed v in a circle of radius r travels from point A to B in a short time Δt. The change in velocity Δv points towards the centre. The arc AB has length vΔt, and the angle Δθ equals arc length/radius = vΔt/r.

物体以恒定速率 v 在半径为 r 的圆上运动，短时间 Δt 内从点 A 到点 B。速度变化 Δv 指向圆心。弧长 AB 为 vΔt，角度 Δθ = 弧长/半径 = vΔt/r。

Magnitude of velocity change Δv = v Δθ (since the velocity vectors form an isosceles triangle with small angle). Thus Δv = v × (vΔt/r) = v²Δt/r.

速度变化大小 Δv = v Δθ（因为速度矢量构成小角度的等腰三角形）。所以 Δv = v × (vΔt/r) = v²Δt/r。

Acceleration a = Δv/Δt = (v²Δt/r) / Δt = v²/r directed towards the centre.

加速度 a = Δv/Δt = (v²Δt/r) / Δt = v²/r，方向指向圆心。

a = v²/r

This centripetal acceleration is always perpendicular to velocity and is required for any circular motion.

该向心加速度始终垂直于速度，是所有圆周运动所必需的。

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6. Deriving energy stored in a capacitor E = ½CV² | 推导电容器储存能量 E = ½CV²

When charging a capacitor of capacitance C, the p.d. V builds up from 0 to the supply voltage. The charge stored Q = CV.

对电容为 C 的电容器充电时，电势差 V 从 0 上升到电源电压。储存的电荷量 Q = CV。

The work done to move a small charge dq at potential v is v dq. Since v = q/C, total work stored as electrical energy is ∫ from 0 to Q of (q/C) dq = [q²/(2C)]₀ᵆ = Q²/(2C).

将微量电荷 dq 在电势 v 下移动所做的功为 v dq。因 v = q/C，储存的总电能 = ∫₀ᵆ (q/C) dq = [q²/(2C)]₀ᵆ = Q²/(2C)。

Substituting Q = CV gives E = ½CV². Equivalent forms are ½QV and Q²/(2C).

代入 Q = CV 得 E = ½CV²。等价形式有 ½QV 和 Q²/(2C)。

E = ½CV²

The factor ½ arises because the average p.d. during charging is V/2.

因子 ½ 的出现是由于充电过程中的平均电势差为 V/2。

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7. Deriving resistivity R = ρL/A | 推导电阻率公式 R = ρL/A

Resistance R of a uniform conductor is proportional to its length L and inversely proportional to its cross-sectional area A. Introducing the constant of proportionality ρ (resistivity) gives R = ρL/A.

均匀导体的电阻 R 与其长度 L 成正比，与横截面积 A 成反比。引入比例常数 ρ（电阻率），得 R = ρL/A。

Formally, using R = V/I and the microscopic relation E = V/L, J = I/A, and E = ρ J (where ρ is resistivity), we obtain V/L = ρ (I/A) → V/I = ρL/A → R = ρL/A.

形式上，用 R = V/I 及微观关系 E = V/L、J = I/A 以及 E = ρ J（ρ 为电阻率），可得 V/L = ρ (I/A) → V/I = ρL/A → R = ρL/A。

R = ρL/A

This equation is vital in understanding how material and geometry affect electric resistance.

该公式对理解材料和几何形状如何影响电阻至关重要。

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8. Deriving the ideal gas pressure equation p = ⅓ρ⟨c²⟩ | 推导理想气体压强公式 p = ⅓ρ⟨c²⟩

Consider a cubic box of side L containing N molecules of mass m. A molecule with x-component velocity u_x collides elastically with a wall, changing momentum by 2mu_x. The time between collisions with the same wall is 2L/u_x, so force = rate of change of momentum = mu_x²/L.

考虑边长为 L 的立方盒中有 N 个质量为 m 的分子。一个 x 方向分速度为 u_x 的分子与器壁弹性碰撞，动量变化量为 2mu_x。与同一器壁碰撞的时间间隔为 2L/u_x，故力 = 动量变化率 = mu_x²/L。

Summing over all N molecules, total force on wall = (m/L) Σu_x². Since by isotropy Σu_x² = Σu_y² = Σu_z² = ⅓Σc², where c is the molecular speed. Pressure p = Force/Area = (m/L × N⟨c²⟩/3) / L² = ⅓ (Nm/V) ⟨c²⟩.

对 N 个分子求和，器壁总受力 = (m/L) Σu_x²。由各向同性，Σu_x² = Σu_y² = Σu_z² = ⅓Σc²，c 为分子速率。压强 p = 力/面积 = (m/L × N⟨c²⟩/3) / L² = ⅓ (Nm/V) ⟨c²⟩。

Since density ρ = Nm/V, we obtain p = ⅓ρ⟨c²⟩. Combining with pV = NkT gives the link ⟨½mc²⟩ = (3/2)kT.

由密度 ρ = Nm/V，得 p = ⅓ρ⟨c²⟩。结合 pV = NkT 可得平均动能的联系 ⟨½mc²⟩ = (3/2)kT。

p = ⅓ρ⟨c²⟩

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9. Deriving wave speed v = fλ | 推导波速公式 v = fλ

A wave of frequency f completes one full oscillation in period T = 1/f. In that time, the wave travels exactly one wavelength λ.

频率为 f 的波在周期 T = 1/f 内完成一次全振动。这段时间内波正好传播了一个波长 λ。

Speed v is distance over time, so v = λ / T = λ × (1/T) = fλ.

速度 v 等于距离除以时间，因此 v = λ / T = λ × (1/T) = fλ。

v = fλ

This simple yet powerful equation connects the speed, frequency and wavelength of any progressive wave.

这个简单却强大的公式将行波的波速、频率和波长联系在一起。

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10. Deriving the diffraction grating equation d sinθ = nλ | 推导衍射光栅方程 d sinθ = nλ

For a diffraction grating with slit spacing d, light of wavelength λ incident normally produces maxima when path difference between adjacent slits is an integer multiple of λ.

对于缝间距为 d 的衍射光栅，波长为 λ 的光垂直入射时，相邻狭缝光程差为波长 λ 的整数倍时产生极大。

The extra distance travelled by a ray from an adjacent slit at angle θ to the normal is d sinθ. Constructive interference occurs when d sinθ = nλ, where n is the order number.

相对于法线成 θ 角的相邻狭缝光线额外走过的距离为 d sinθ。当 d sinθ = nλ 时发生相长干涉，n 为级数。

d sinθ = nλ

This derivation shows how the geometry of the grating directly relates the angle of a bright fringe to the wavelength.

该推导展示了光栅几何如何直接将亮纹角度与波长联系起来。

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