📚 AS Physics Unit 2 (Jan 19) Mark Scheme: Formula Derivation Breakdown | AS 物理 单元2 2019年1月评分方案 公式推导精析
In the January 2019 AS Physics Unit 2 examination, many of the high-mark questions required candidates to derive key formulas from first principles. The mark scheme reveals a clear emphasis on logical steps, correct use of definitions, and precise handling of algebraic manipulations. This article unpacks the most important derivations featured in that paper, explaining each step in a way that builds deep conceptual understanding for revision.
在2019年1月AS物理单元2考试中，许多高分题都要求考生从基本原理推导关键公式。评分方案清晰地展示了对逻辑步骤、正确定义以及精确代数处理的重视。本文拆解了该试卷中最重要的几个推导，以帮助考生在复习中建立深刻的概念理解。
1. Impulse from Newton’s Second Law | 从牛顿第二定律推导冲量
Newton’s second law in its most useful form for mechanics problems links resultant force to the rate of change of momentum. Starting with F = ma and the definition of acceleration, we can quickly arrive at the impulse–momentum relationship that is frequently tested.
在力学问题中，牛顿第二定律最实用的形式将合力与动量变化率联系起来。从F = ma和加速度的定义出发，我们可以迅速得到经常考查的冲量–动量关系。
Write force as F = m × (v − u)/t, where u and v are initial and final velocities. Multiply both sides by time t to obtain Ft = mv − mu. The left side is impulse, the right side is change in momentum. The exam expects candidates to state that this derivation assumes constant mass and a constant resultant force over the interval.
将力写作F = m × (v − u)/t，其中u和v是初速度和末速度。两边同乘以时间t得到Ft = mv − mu。左边是冲量，右边是动量的变化量。考试期望考生说明这一推导假设质量恒定且合力在时间间隔内保持不变。
FΔt = Δp = mv − mu
2. Conservation of Momentum Derivation | 动量守恒推导
The principle of conservation of momentum comes directly from Newton’s third law and the definition of impulse. In a system of two colliding bodies A and B, the force that A exerts on B is equal in magnitude and opposite in direction to the force B exerts on A.
动量守恒原理直接来源于牛顿第三定律和冲量的定义。在一个由两个碰撞物体A和B组成的系统中，A对B施加的力与B对A施加的力大小相等、方向相反。
During the collision time Δt, impulse on A is FBAΔt and impulse on B is FABΔt. Since FBA = −FAB, the impulses are equal and opposite. Therefore the change in momentum of A is equal and opposite to the change in momentum of B, so total momentum remains constant.
在碰撞时间Δt内，A所受冲量为FBAΔt，B所受冲量为FABΔt。因为FBA = −FAB，冲量大小相等、方向相反。因此A的动量变化量与B的动量变化量等大反向，总动量保持恒定。
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
3. Kinetic Energy Formula | 动能公式推导
Kinetic energy is the energy a body possesses due to its motion. The familiar expression Eₖ = ½mv² can be derived by considering the work done by a constant resultant force to accelerate a mass from rest to velocity v.
动能是物体由于运动而具有的能量。常见的表达式Eₖ = ½mv²可以通过考虑一个恒定合力将物体从静止加速到速度v所做的功来推导。
Work done W = F × s. Using F = ma and the kinematic relation v² = u² + 2as, and setting u = 0, we have s = v²/(2a). Substituting gives W = ma × (v²/(2a)) = ½mv². This work becomes the kinetic energy. The same derivation works for any change of velocity: W = Fs = ½m(v² − u²).
做功W = F × s。利用F = ma和运动学关系v² = u² + 2as，并令u = 0，得到s = v²/(2a)。代入得W = ma × (v²/(2a)) = ½mv²。这个功转化为动能。同样的推导适用于任意速度变化：W = Fs = ½m(v² − u²)。
Eₖ = ½mv²
4. Elastic Potential Energy of a Spring | 弹簧的弹性势能
A material obeying Hooke’s law stores energy when stretched or compressed. The elastic potential energy Eₑ can be derived by averaging the force required to extend the spring, or by using the area under the force–extension graph.
遵守胡克定律的材料在拉伸或压缩时储存能量。弹性势能Eₑ可以通过平均拉伸弹簧的力，或利用力–伸长图下的面积来推导。
Hooke’s law states F = kx. The work done to stretch the spring from 0 to extension x is the area of a triangle under the F–x graph: W = ½ × base × height = ½ × x × kx = ½kx². This energy is stored as Eₑ. The mark scheme often requires candidates to explain why the average force ½F is used.
胡克定律指出F = kx。将弹簧从0拉伸到伸长量x所做的功是F–x图下三角形的面积：W = ½ × 底 × 高 = ½ × x × kx = ½kx²。这个能量被储存为Eₑ。评分方案常常要求考生解释为什么使用平均力½F。
Eₑ = ½kx²
5. Work Done and Energy Transfer in Mechanics | 力学中的做功与能量转移
Work done is defined as the product of force and displacement in the direction of the force, W = Fs cosθ. When the force is parallel to displacement, W = Fs. This definition forms the bridge between force–distance calculations and energy changes.
功被定义为力与力方向上位移的乘积，W = Fs cosθ。当力与位移平行时，W = Fs。这一定义构成了力–距离计算与能量变化之间的桥梁。
In a situation where a lifting force counteracts weight, the work done against gravity becomes gravitational potential energy. For a mass m raised through height h, W = mgh, so ΔEp = mgh. The Jan 19 Unit 2 paper often embeds this in multi-step problems.
在升力抵消重力的情况下，克服重力所做的功转化为重力势能。对于一个被提升高度h的质量m，W = mgh，所以ΔEp = mgh。2019年1月的单元2试卷常将这一点融入多步骤问题中。
W = F × s and ΔEp = mgh
6. The Young Modulus Equation | 杨氏模量方程
The Young modulus E is a measure of a material’s stiffness, defined as the ratio of tensile stress to tensile strain within the linear elastic region. The formula that appears in the mark scheme is E = (F/A) / (ΔL/L), which rearranges to a very useful laboratory form.
杨氏模量E是衡量材料刚度的量，定义为线弹性区域内拉伸应力与拉伸应变的比值。评分方案中出现的公式是E = (F/A) / (ΔL/L)，它可以重排为一种非常有用的实验形式。
Stress = F/A, strain = ΔL/L. Therefore E = F L / (A ΔL). Candidates are expected to know that ΔL is the extension, L is the original length, A is the cross-sectional area. A typical derivation question might ask you to show that the spring constant k of a wire is given by k = EA/L.
应力 = F/A，应变 = ΔL/L。因此E = F L / (A ΔL)。考生应知道ΔL是伸长量，L是原长，A是横截面积。典型的推导题可能会要求你证明一根金属丝的弹簧常数k由k = EA/L给出。
E = stress/strain = (F/A) / (ΔL/L) → F = (EA/L) ΔL
7. Refractive Index and Critical Angle | 折射率与临界角
Snell’s law governs the refraction of light at a boundary between two media. When light passes from a medium of refractive index n₁ into a medium of lower index n₂, the outgoing ray bends away from the normal. At a certain angle of incidence, the angle of refraction becomes 90°.
斯涅尔定律描述了光在两种介质界面处的折射。当光从折射率为n₁的介质进入折射率较低的n₂介质时，出射光线会偏离法线。在某个入射角下，折射角变为90°。
Setting θ₂ = 90° in n₁ sin θ₁ = n₂ sin θ₂ gives n₁ sin θc = n₂ × 1, so sin θc = n₂/n₁. If the second medium is air (n₂ = 1), this simplifies to sin θc = 1/n. A mark scheme point often highlights the need to mention that total internal reflection occurs only when n₁ > n₂ and the angle of incidence exceeds the critical angle.
在n₁ sin θ₁ = n₂ sin θ₂中设θ₂ = 90°，得到n₁ sin θc = n₂ × 1，因此sin θc = n₂/n₁。若第二介质是空气（n₂ = 1），则简化为sin θc = 1/n。评分方案常强调需指出全内反射仅发生在n₁ > n₂且入射角大于临界角时。
sin θc = n₂ / n₁ (for n₁ > n₂)
8. The Diffraction Grating Formula | 衍射光栅公式
A diffraction grating produces a pattern of bright maxima when light passes through many equally spaced slits. The condition for constructive interference is that the path difference between adjacent slits equals an integer multiple of the wavelength.
当光通过许多等间距的狭缝时，衍射光栅会产生明亮的极大图案。相长干涉的条件是相邻狭缝之间的光程差等于波长的整数倍。
If the slit separation is d, the angle θ to the nth order maximum satisfies d sin θ = nλ. To derive this, consider a right-angled triangle with hypotenuse d and opposite side equal to the path difference. The mark scheme rewards linking the geometry clearly to the wavelength. The number of slits per metre N relates to d by d = 1/N.
若狭缝间距为d，第n级极大的角度θ满足d sin θ = nλ。为推导这一点，考虑一个以d为斜边、对边等于光程差的直角三角形。评分方案奖励将几何关系与波长明确联系起来。每米狭缝数N与d的关系为d = 1/N。
d sin θ = nλ (where d = 1/N)
9. Einstein’s Photoelectric Equation | 爱因斯坦光电方程
The photoelectric effect provides evidence for the particle nature of light. Einstein’s explanation gives a maximum kinetic energy for emitted electrons that depends on photon energy and the work function φ of the metal.
光电效应为光的粒子性提供了证据。爱因斯坦的解释给出了发射电子的最大动能，该动能取决于光子能量和金属的功函数φ。
Energy conservation requires hf = φ + Ek max. The work function φ is the minimum energy needed to liberate an electron; any leftover photon energy becomes kinetic energy. The mark scheme often asks candidates to derive the threshold wavelength or to explain why intensity affects current but not maximum kinetic energy.
能量守恒要求hf = φ + Ek max。功函数φ是释放电子所需的最小能量；剩余的光子能量转化为动能。评分方案常要求考生推导阈值波长，或解释为什么光强影响电流而不影响最大动能。
hf = φ + ½ m vmax²
10. Combining Resistors and Electrical Power | 电阻组合与电功率
While not exclusively a derivation, the home tuition context frequently tests the ability to derive effective resistance for series and parallel combinations, as well as the heating effect formulas. The Jan 19 mark scheme included similar algebraic treatments.
虽然不是纯推导题，但辅导中常考查推导串联和并联组合有效电阻的能力，以及热效应公式。2019年1月的评分方案包含了类似的代数处理。
For series: V = V₁ + V₂, and current I is common. Using V = IR gives IRT = IR₁ + IR₂, so RT = R₁ + R₂. For parallel: I = I₁ + I₂, with common p.d. V. Using I = V/R gives V/RT = V/R₁ + V/R₂, hence 1/RT = 1/R₁ + 1/R₂. Power dissipation P = IV = I²R = V²/R can be linked back to energy transferred per unit time.
串联：V = V₁ + V₂，电流I相同。代入V = IR得到IRT = IR₁ + IR₂，因此RT = R₁ + R₂。并联：I = I₁ + I₂，电压V相同。代入I = V/R得到V/RT = V/R₁ + V/R₂，因此1/RT = 1/R₁ + 1/R₂。功率耗散P = IV = I²R = V²/R可追溯到单位时间转移的能量。
Series: RT = R₁ + R₂ Parallel: 1/RT = 1/R₁ + 1/R₂
11. The Lens Formula and Magnification (if applicable) | 透镜公式与放大率（如适用）
Depending on the specification, Unit 2 may include basic optics. The thin lens formula 1/f = 1/u + 1/v and magnification m = v/u can be derived from similar triangles in a ray diagram. The mark scheme rewards accurate sign conventions and the correct placement of real and virtual images.
根据考纲的不同，单元2可能包含基础光学。薄透镜公式1/f = 1/u + 1/v和放大率m = v/u可由光线图中的相似三角形导出。评分方案奖励准确的符号规则以及实像和虚像的正确定位。
By considering a ray through the centre of the lens and a ray parallel to the axis, the two pairs of similar triangles yield hi/ho = v/u and hi/ho = (v−f)/f. Equating gives 1/f = 1/u + 1/v. Real‑is‑positive convention must be stated clearly.
通过考虑一条过透镜中心的光线和一条平行于主轴的光线，两组相似三角形给出hi/ho = v/u和hi/ho = (v−f)/f。等式联立得到1/f = 1/u + 1/v。必须清晰说明实正符号规则。
1/f = 1/u + 1/v and m = v/u
12. Practical Application: From Mark Scheme to Exam Success | 实际应用：从评分方案到考试成功
Reviewing the January 2019 mark scheme highlights that full marks in derivation questions are rarely awarded for just quoting the final formula. Examiners look for a clear statement of the physical law used, substitution of defining equations, algebraic simplification, and a concluding statement of the result with its limitations.
回顾2019年1月的评分方案可以看出，推导题很少仅凭引用最终公式就给出满分。考官看重是否清晰地陈述所用的物理定律、代入定义方程、代数简化，以及总结结果并说明其局限性。
When practising, always write down the fundamental principle (e.g. Newton II, energy conservation, wave interference condition) as your starting point. Then show how you manipulate the equations step by step. Finally, present the derived formula in a box or on a separate line, just as we have done above, to ensure the examiner sees the required outcome.
在练习时，始终写下基本原理（例如牛顿第二定律、能量守恒、波的干涉条件）作为起点。然后展示如何一步步处理方程。最后，将推导出的公式写在框内或单独一行，正如上文所示，以确保考官看到所需的结果。
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