📚 AS Physics Paper 2 Markscheme January 2018 Formula Derivations | AS物理 Paper 2 2018年1月评分方案 公式推导
This article examines the key formula derivations that featured in the AS Physics Paper 2 markscheme from January 2018. Understanding how each equation is built from fundamental principles is essential for mastering the subject and performing well in examination questions that probe deeper than simple recall.
本文探讨2018年1月AS物理Paper 2评分方案中出现的关键公式推导。理解每一个方程如何从基本原理建立起来，对于掌握这门学科、在考查深层理解而非简单记忆的考试中取得好成绩至关重要。
1. Derivation of Linear Motion Equations | 直线运动方程推导
The equation v = u + at comes directly from the definition of acceleration as the rate of change of velocity. Rearranging a = (v – u)/t gives the first SUVAT equation.
方程 v = u + at 直接来源于加速度的定义——速度的变化率。将 a = (v – u)/t 重新整理就得到第一个SUVAT方程。
v = u + at
To find displacement, we use average velocity. For uniform acceleration, average velocity = (u + v)/2, so displacement s = average velocity × time, leading to s = (u + v)t / 2.
求位移时我们使用平均速度。在匀加速运动下，平均速度 = (u + v)/2，因此位移 s = 平均速度 × 时间，得到 s = (u + v)t / 2。
s = (u + v)t / 2
Substituting v = u + at into this displacement equation eliminates v and yields s = ut + ½at². Markschemes reward clear algebraic steps and identification of the substitution.
将 v = u + at 代入位移方程就可消去 v，得出 s = ut + ½at²。评分方案中对清晰的代数步骤和替换过程予以认可。
s = ut + ½at²
Combining the equations to eliminate t produces v² = u² + 2as. This derivation is a favourite in Paper 2 as it connects the fundamental definitions with algebraic manipulation.
将方程联立并消去 t 就得到 v² = u² + 2as。这个推导将基本定义与代数处理结合起来，是Paper 2中常见的考点。
v² = u² + 2as
2. Derivation of Kinetic Energy Formula | 动能公式推导
Kinetic energy is derived from the work done by a resultant force. Starting from Work = Force × displacement and using Newton’s second law F = ma, the work done on an object accelerating from rest is W = ma × s.
动能是由合力所做的功推导出来的。从 功 = 力 × 位移 出发，并利用牛顿第二定律 F = ma，对一个从静止加速的物体所做的功为 W = ma × s。
Using v² = u² + 2as with u = 0 gives as = v²/2. Substituting into W = m × as yields W = ½mv². The markscheme expects students to state that this work is stored as kinetic energy.
利用 v² = u² + 2as 并令 u = 0，得到 as = v²/2。代入 W = m × as 即得 W = ½mv²。评分方案期望学生明确此功以动能形式储存。
Ek = ½mv²
This simple derivation was required in the January 2018 paper where candidates had to justify the kinetic energy expression rather than simply quote it.
这个简明的推导曾出现在2018年1月的试卷中，考生需要论证动能表达式而不仅仅是直接引用它。
3. Derivation of Gravitational Potential Energy | 重力势能推导
The change in gravitational potential energy near the Earth’s surface is derived from the work done against gravity. Lifting an object of mass m through a vertical height h requires a force equal to its weight mg.
地表附近重力势能的变化来源于克服重力所做的功。将质量为 m 的物体垂直提升高度 h 需要的力等于它的重量 mg。
Work done = force × distance moved in the direction of the force, so W = mg × h. Since this work is stored as gravitational potential energy, we write ΔEp = mgh.
功 = 力 × 沿力方向移动的距离，因此 W = mg × h。由于此功以重力势能的形式储存，我们写成 ΔEp = mgh。
ΔEp = mgΔh
Markschemes often award marks for recognising that this holds only for uniform gravitational fields where g is constant.
评分方案中经常会因考生认识到此式仅适用于均匀重力场（g 为常数）而给分。
4. Derivation of Power as Force × Velocity | 功率为力乘速度的推导
Power is defined as the rate of doing work. For a constant force F moving an object at constant velocity v, the distance covered in time t is s = vt.
功率定义为做功的速率。对一个使物体以恒定速度 v 运动的恒力 F 而言，在时间 t 内经过的距离为 s = vt。
Work done by the force is W = F × s = F × vt. Therefore, power P = W/t = (Fvt)/t = Fv, giving the useful expression P = Fv.
力所做的功为 W = F × s = F × vt。因此，功率 P = W/t = (Fvt)/t = Fv，得到实用的 P = Fv 表达式。
P = Fv
This relationship frequently appears in questions about vehicles moving at top speed, and the derivation from first principles was expected in the January 2018 markscheme.
这一关系经常出现在关于车辆以最高速度运动的问题中，2018年1月的评分方案期望从基本原理出发进行推导。
5. Derivation of Centripetal Acceleration | 向心加速度推导
For an object moving in a circle of radius r at constant speed v, we consider the change in velocity vector over a short time Δt. The magnitude of the velocity remains v, but the direction changes.
对于以恒定速率 v 在半径为 r 的圆周上运动的物体，我们考虑很短时间 Δt 内速度矢量的变化。速度的大小保持为 v，但方向在改变。
By vector subtraction, the change in velocity points towards the centre, and for small Δθ the magnitude of the change is Δv = vΔθ. Angular displacement Δθ = (vΔt)/r, so Δv = v²Δt/r.
通过矢量减法，速度的变化指向圆心，且当 Δθ 很小时其大小为 Δv = vΔθ。角位移 Δθ = (vΔt)/r，因此 Δv = v²Δt/r。
Acceleration is Δv/Δt, giving a = v²/r. The markscheme awards credit for clear vector diagrams and the use of small-angle approximation where Δθ is small.
加速度为 Δv/Δt，得到 a = v²/r。评分方案对清晰的矢量图以及在 Δθ 很小时使用小角近似会给予分数。
a = v²/r
6. Derivation of Resistivity Equation | 电阻率方程推导
Resistance R of a wire is found to be directly proportional to its length L and inversely proportional to its cross-sectional area A. The constant of proportionality is the resistivity ρ.
实验发现导线的电阻 R 与长度 L 成正比，与横截面积 A 成反比。比例常数就是电阻率 ρ。
Thus, R ∝ L/A, and introducing resistivity gives R = ρL/A. Deriving this formula from microscopic principles is not required at AS level, but candidates must be able to rearrange and use it.
因此 R ∝ L/A，引入电阻率即得 R = ρL/A。AS阶段不要求从微观原理推导该公式，但考生必须能够变换和使用它。
R = ρL/A
The January 2018 markscheme accepted explanations based on the idea that longer conductors provide more collisions for charge carriers, and wider conductors allow easier flow.
2018年1月的评分方案接受基于以下思想的解释：更长的导体为电荷载流子提供更多碰撞机会，而更宽的导体则使流动更容易。
7. Derivation of EMF and Internal Resistance | 电动势和内阻推导
A source of electromotive force (emf) ε does work on charges. When current I flows, some energy is dissipated inside the source due to its internal resistance r. The terminal potential difference V is less than ε.
电动势源 ε 对电荷做功。当电流 I 流过时，源内部由于内阻 r 会消耗部分能量。端电压 V 小于 ε。
Energy conservation gives: energy per unit charge produced by source = energy per unit charge used in external resistance + internal resistance. Thus ε = V + Ir, where V = IR for the external resistor.
能量守恒给出：源提供的每单位电荷能量 = 外电阻消耗的每单位电荷能量 + 内阻消耗的。因此 ε = V + Ir，其中外电阻满足 V = IR。
ε = I(R + r)
Markschemes look for the idea of ‘lost volts’ and the fact that Ir represents the internal energy dissipation. A clear circuit diagram labelling ε, r, and R is essential.
评分方案关注“损耗电压”的概念，以及 Ir 代表内部能量消耗这一事实。清晰的电路图并标出 ε、r 和 R 至关重要。
8. Derivation of Wave Speed Equation | 波速方程推导
The fundamental wave equation links wave speed v, frequency f, and wavelength λ. One can derive it from the definitions: frequency is the number of cycles per second, and wavelength is the distance per cycle.
基本波动方程将波速 v、频率 f 和波长 λ 联系起来。可以从定义导出：频率是每秒的周期数，波长是每个周期的距离。
Distance travelled in one second = number of cycles per second × distance per cycle, so v = f × λ.
一秒钟内传播的距离 = 每秒周期数 × 每个周期的距离，因此 v = f × λ。
v = fλ
This simple logic was required in the 2018 paper when explaining the relationship between the quantities without simply stating the formula.
2018年的试卷要求用这种简单逻辑解释各物理量之间的关系，而不仅仅是写出公式。
9. Derivation of Young’s Modulus from Hooke’s Law | 从胡克定律推导杨氏模量
Hooke’s law for a wire states that tension F is proportional to extension ΔL. To make this a material property, stress F/A and strain ΔL/L₀ are used, where L₀ is the original length.
金属丝的胡克定律指出拉力 F 与伸长量 ΔL 成正比。为了使之成为材料属性，引入应力 F/A 和应变 ΔL/L₀，其中 L₀ 为原长。
Young’s modulus is defined as the ratio of tensile stress to tensile strain within the proportionality limit, so E = (F/A) ÷ (ΔL/L₀) = FL₀ / AΔL.
杨氏模量定义为在比例极限内拉伸应力与拉伸应变之比，因此 E = (F/A) ÷ (ΔL/L₀) = FL₀ / AΔL。
E = FL₀ / AΔL
The markscheme rewards candidates who explain that this is independent of the dimensions of the wire and characterises the material.
评分方案鼓励考生解释该量独立于细丝尺寸，是材料的特征。
10. Derivation of Range of a Projectile | 抛体射程推导
For a projectile launched from ground level at speed u and angle θ to the horizontal, the horizontal and vertical components are ux = u cosθ and uy = u sinθ.
对于从地面以速率 u、与水平成 θ 角发射的抛体，水平和竖直分量分别为 ux = u cosθ 和 uy = u sinθ。
Time of flight is found from vertical motion: when the projectile returns to the ground, vertical displacement = 0. Using s = uyt + ½(-g)t², we get t = 2u sinθ / g.
飞行时间由竖直运动求得：当抛体回到地面时，竖直位移为 0。利用 s = uyt + ½(-g)t²，得到 t = 2u sinθ / g。
Horizontal range R = ux × t = u cosθ × (2u sinθ / g) = u² sin2θ / g, using the identity 2 sinθ cosθ = sin2θ.
水平射程 R = ux × t = u cosθ × (2u sinθ / g) = u² sin2θ / g，这里使用了恒等式 2 sinθ cosθ = sin2θ。
R = u² sin2θ / g
The markscheme for January 2018 accepted this derivation and often required candidates to comment on the symmetry of trajectory and the condition for maximum range.
2018年1月的评分方案认可这一推导，并经常要求考生就轨迹的对称性以及最大射程的条件进行论述。
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