Gravitational fields are a cornerstone of the OCR A-Level Physics syllabus, linking classical mechanics with celestial motion. Understanding Newton’s law of gravitation, field strength, potential, and orbital dynamics is essential for exam success. This revision guide breaks down every key concept, equation, and application you need to master, from radial fields around point masses to the energy of satellites. Let’s explore how gravity shapes the cosmos, one clear explanation at a time.
A gravitational field is a region of space where a mass experiences a force. In A-Level Physics we model fields using field lines (or lines of force) that point in the direction a small test mass would move. For a uniform field, such as near the Earth’s surface over small distances, lines are parallel and equally spaced. For a radial field around a point mass, lines point radially inwards and spread out with distance. The concept of a field allows us to describe gravity without direct contact, unifying our understanding of terrestrial and astronomical phenomena.
Gravitational fields are vector fields; they have both magnitude and direction. The direction is always towards the centre of mass creating the field. This is why we use negative signs in radial field equations for field strength and potential, to indicate that work is done by the field when moving a mass away from the source.
Newton’s Law of Gravitation states that every particle attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. The equation is:
where F is the gravitational force, G = 6.67 × 10⁻¹¹ N m² kg⁻² (the universal gravitational constant), m₁ and m₂ are the masses, and r is the separation between their centres. The negative sign is often omitted when calculating magnitudes, but it signifies the force is attractive. For OCR exams, you must know this law by heart and be able to apply it to point masses and spherically symmetrical objects.
其中 F 是引力，G = 6.67 × 10⁻¹¹ N m² kg⁻²（万有引力常数），m₁ 和 m₂ 是质量，r 是它们中心之间的距离。计算大小时通常省略负号，但它表示力是吸引力。对于 OCR 考试，你必须熟记该定律，并能够将其应用于点质量和球对称物体。
The law is an inverse-square law, meaning if you double the distance, the force becomes one quarter. This relationship is crucial for understanding satellite orbits and the variation of gravitational field strength with altitude.
Gravitational field strength, g, is defined as the force per unit mass experienced by a small test mass placed at a point. In equation form:
引力场强度 g 定义为放置在一点的小测试质量单位质量所受的力。方程形式为：
g = F / m
Its SI unit is N kg⁻¹, which is equivalent to m s⁻², matching acceleration. Near the Earth’s surface, g ≈ 9.81 N kg⁻¹. For a radial field, the magnitude is given by:
其国际单位为 N kg⁻¹，等同于 m s⁻²，与加速度一致。接近地球表面时，g ≈ 9.81 N kg⁻¹。对于径向场，大小由下式给出：
g = G M / r²
where M is the mass of the source and r is the distance from its centre. Note this is derived from Newton’s law by putting m = test mass. The direction is towards the centre, so vector form: g = – (G M / r²) r̂, where r̂ is a unit vector pointing outward.
其中 M 是源质量，r 是到其中心的距离。需要注意的是，这由牛顿定律推导而来，令 m 为测试质量。方向指向中心，因此矢量形式：g = – (G M / r²) r̂，其中 r̂ 是指向外部的单位矢量。
Location
Formula for g
Surface of Earth (mass M, radius R)
g = G M / R²
At altitude h above surface
g = G M / (R + h)²
Inside a uniform sphere (radius R) at distance r from centre
g ∝ r (linear)
Examiners often ask you to calculate g at the height of a satellite or compare to surface value. Always convert units to metres and kilograms.
考官经常要求你计算卫星高度处的 g，或与地面值进行比较。始终将单位转换为米和千克。
4. Radial Gravitational Fields | 径向引力场
In a radial field, the field lines converge at the centre of the mass and the field strength decreases as 1/r². This is distinct from a uniform field where g is constant. For planets and stars, the field is radial outside the object. You must be comfortable sketching field lines for a point mass and indicating the direction with arrows pointing inward. The spacing between field lines indicates strength: wider spacing means weaker field.
在径向场中，场线汇聚于质量中心，场强随 1/r² 减小。这与 g 恒定的均匀场不同。对于行星和恒星，物体外部的场是径向的。你必须能够熟练地绘制点质量的场线，并用向内箭头标注方向。场线之间的间距表示强度：间距越大，场越弱。
A common exam pitfall is forgetting that gravitational field strength refers to a vector, and when adding fields from multiple masses, you must use vector addition. If two planets are aligned, the resultant g at a point is the algebraic sum considering direction.
一个常见的考试陷阱是忘记引力场强度是一个矢量，当叠加多个质量产生的场时，必须使用矢量加法。如果两颗行星对齐，某一点的合场强 g 是考虑方向后的代数和。
5. Gravitational Potential V | 引力势 V
Gravitational potential at a point is the work done per unit mass to bring a small test mass from infinity to that point. Because gravity is attractive, work is done by the field as the mass moves in, so potential is negative. The equation for radial field:
where V is in J kg⁻¹. At infinity, V = 0. As r decreases, V becomes more negative. Potential is a scalar quantity, so adding potentials from multiple masses is straightforward algebraic addition.
其中 V 的单位是 J kg⁻¹。在无穷远处，V = 0。随着 r 减小，V 变得更负。势是标量，因此多个质量的势叠加就是简单的代数相加。
You need to know how to derive field strength from potential: g = – dV/dr. For a radial field, differentiating V = -GM/r gives g = -GM/r², confirming the inverse-square relationship. The negative gradient of V gives the field strength magnitude and direction.
你需要知道如何从势导出场强：g = – dV/dr。对于径向场，对 V = -GM/r 求导得到 g = -GM/r²，证实了平方反比关系。V 的负梯度给出了场强的大小和方向。
6. Gravitational Potential Energy | 引力势能
Gravitational potential energy (U) of a system of two masses is the work done to separate them to infinity, or alternatively, the work done to bring them from infinity to their separation r. The equation is:
两个质量体系的引力势能（U）是将它们分离到无穷远所做的功，或者说是将它们从无穷远移到当前间距 r 所做的功。方程为：
U = – G M m / r
where M and m are the two masses. Like potential, it is zero at infinite separation and negative for any finite r. The relation between potential and potential energy is U = m V. This is analogous to electric potential energy.
其中 M 和 m 是两个质量。与势类似，在无穷远间距时为零，对于任何有限 r 为负值。引力势与引力势能的关系为 U = m V。这与电势能类似。
For a satellite of mass m orbiting a planet of mass M at radius r, the total mechanical energy is the sum of kinetic and potential: E = ½ m v² – G M m / r. Using orbit velocity v² = G M / r, this simplifies to E = – G M m / (2r). The negative total energy indicates a bound orbit.
对于质量为 m 的卫星在半径 r 处绕质量为 M 的行星轨道运行，总机械能为动能和势能之和：E = ½ m v² – G M m / r。利用轨道速度 v² = G M / r，这简化为 E = – G M m / (2r)。负的总能量表示束缚轨道。
7. Equipotential Surfaces | 等势面
Equipotential surfaces are surfaces of constant potential. In a radial field, they are spheres centred on the mass. No work is done moving a mass along an equipotential surface because the potential difference is zero. Field lines are always perpendicular to equipotential surfaces. The spacing of equipotentials indicates the field strength: closer surfaces mean a stronger field (greater gradient).
In a uniform field, equipotential surfaces are parallel planes perpendicular to the field lines. The potential changes linearly with distance, so ΔV = g Δh (for small height changes near Earth’s surface). OCR often asks you to sketch equipotential lines around a planet and show that they get further apart with radius.
在均匀场中，等势面是垂直于场线的平行平面。势随距离线性变化，因此 ΔV = g Δh（对于地球表面附近较小的高度变化）。OCR 经常要求你画出绕行星的等势线，并展示它们随半径增大而彼此远离。
8. Satellite Orbits & Kepler’s Laws | 卫星轨道与开普勒定律
Kepler’s three laws of planetary motion are essential for understanding orbits:
开普勒行星运动三定律对于理解轨道至关重要：
First Law (Law of Ellipses): Planets move in elliptical orbits with the Sun at one focus. For OCR, we often approximate orbits as circular.
Second Law (Equal Areas): A line joining a planet and the Sun sweeps out equal areas in equal times, implying faster motion when closer.
第二定律（面积定律）：行星与太阳的连线在相等时间内扫过相等的面积，表明距离越近运动越快。
Third Law (Harmonic Law): The square of the orbital period T is proportional to the cube of the semi-major axis r: T² ∝ r³. For circular orbits, this derives from equating gravitational force and centripetal force.
第三定律（周期定律）：轨道周期 T 的平方正比于半长轴 r 的立方：T² ∝ r³。对于圆轨道，这可由引力等于向心力推导得出。
The centripetal force for a satellite is provided by gravity: G M m / r² = m v² / r. This yields the orbital speed v = √(G M / r) and period T = 2πr / v = 2π √(r³/(G M)). This is a favourite OCR derivation.
卫星的向心力由引力提供：G M m / r² = m v² / r。由此得到轨道速度 v = √(G M / r) 和周期 T = 2πr / v = 2π √(r³/(G M))。这是 OCR 最喜欢的推导之一。
9. Geostationary Satellites | 地球同步卫星
A geostationary satellite orbits Earth directly above the equator with a period of 24 hours, matching Earth’s rotation. It remains fixed over a single point on the surface, useful for communications and weather monitoring. The required orbital radius is calculated using T = 24 h = 86400 s and M_Earth = 5.97×10²⁴ kg.
地球同步卫星在地球赤道正上方运行，周期为 24 小时，与地球自转同步。它固定在地表某一点上方，对通信和天气监测很有用。所需轨道半径通过 T = 24 小时 = 86400 秒和地球质量 M_Earth = 5.97×10²⁴ kg 计算得出。
Using Kepler’s third law: r³ = G M T² / (4π²). Substituting values gives r ≈ 4.23×10⁷ m from Earth’s centre, or an altitude of about 3.58×10⁷ m (~35,800 km). The orbit must be circular and in the equatorial plane. OCR often includes multi-step calculations involving geostationary satellites.
利用开普勒第三定律：r³ = G M T² / (4π²)。代入数值得到 r ≈ 4.23×10⁷ m（距地心），或高度约 3.58×10⁷ m（~35,800 km）。轨道必须是圆形且位于赤道平面内。OCR 经常设置涉及地球同步卫星的多步计算题。
10. Energy Considerations in Orbits | 轨道能量分析
Total energy of a satellite in a circular orbit is half its gravitational potential energy magnitude (but negative). As orbital radius increases, kinetic energy decreases (v smaller), potential energy becomes less negative (increases), and total energy increases (becomes less negative). Thus, to move a satellite to a higher orbit, work must be done, increasing its total mechanical energy. Paradoxically, the satellite slows down even though energy is added, because gravitational potential energy increases more than kinetic energy decreases.
For an elliptical orbit, total energy remains constant (if no external forces), but kinetic and potential exchange: at perihelion (closest approach), speed is highest and potential energy is most negative; at aphelion, speed is lowest and potential energy is least negative. OCR requires qualitative understanding of these energy transfers.
Escape velocity is the minimum speed an object needs to escape a planet’s gravitational field without further propulsion. It is derived by setting total energy to zero (just enough to reach infinity with zero speed): ½ m v_esc² – G M m / R = 0 →
逃逸速度是指物体无需进一步推进就能逃离行星引力场的最小速度。它通过将总能量设为零（刚好足够以零速度到达无穷远）推导得出：½ m v_esc² – G M m / R = 0 →
vesc = √(2 G M / R)
where R is the planet’s radius. Note v_esc is independent of the mass of the escaping object. For Earth, v_esc ≈ 11.2 km s⁻¹. This is √2 times the circular orbital velocity at the surface. Exam tip: don’t confuse escape velocity with orbital speed – the factor of 2 inside the root is critical.
其中 R 是行星半径。注意 v_esc 与逃逸物体的质量无关。对于地球，v_esc ≈ 11.2 km s⁻¹。这是地面圆轨道速度的 √2 倍。考试提示：不要混淆逃逸速度和轨道速度——根号内的因子 2 至关重要。
12. Summary & Key Equations | 总结与关键公式
Mastering gravitational fields requires fluency with the key equations and their relationships. Below is a quick-reference table:
掌握万有引力场需要熟练运用关键方程及其相互关系。以下是快速参考表：
Quantity
Equation
Newton’s law
F = G M m / r²
Field strength
g = F/m = G M / r²
Potential
V = – G M / r
Potential energy
U = – G M m / r
Orbital speed
v = √(G M / r)
Orbital period
T = 2π √(r³/(G M))
Total energy (circular orbit)
E = – G M m / (2r)
Escape velocity
vesc = √(2 G M / R)
g near surface
g ≈ constant, ΔV = g Δh
Always keep track of negatives for potential and energy; they indicate work done by the field. Practice multi-step problems combining these concepts, and sketch field lines and equipotentials to build intuition.
📚 Essay Writing Templates for IB and AQA Physics | IB 与 AQA 物理 Essay 写作模板
Essay-style questions in IB and AQA Physics demand structured, coherent arguments that weave together theory, calculations, and critical evaluation. Whether you are facing an IB Paper 2 long-answer problem or the AQA Paper 3 Section B essay, a clear template can transform scattered knowledge into a high-scoring response. This guide provides bilingual templates and sentence starters designed to build confidence and fluency in physics essays.
1. Overview of Essay Questions in IB Physics | IB 物理中的论述题概述
In IB Physics, extended response questions appear mainly in Paper 2 (both SL and HL) and occasionally in Paper 1. These questions require you to explain phenomena, analyse data, or evaluate a model in several connected paragraphs. The mark scheme rewards logical flow, correct use of scientific terminology, and explicit linkage to the underlying principles.
2. Overview of the AQA Physics Essay | AQA 物理 Essay 概述
The AQA A-level Physics essay (Paper 3, Section B) is a 25-mark question where you select one title from a choice of two and write an extended argument. You must draw on your knowledge from across the entire specification, link different topic areas, and demonstrate synoptic understanding. The essay is assessed for breadth of physics, relevance, and quality of written communication.
AQA A-level 物理的 essay 题（试卷三 B 部分）分值 25 分，你需从两个题目中选择一个，展开长文论述。你必须运用来自整个课纲的知识，联系不同主题领域，展现综合理解能力。评分依据物理知识的广度、相关性以及书面表达的质量。
3. The Universal Three-Part Structure | 通用的三段式结构
Both styles of essay benefit from a three-part framework: Introduction – Body – Conclusion. In the introduction, you unpack the question and outline the physics you will use. The body develops each idea in a series of paragraphs, each covering one distinct point. The conclusion synthesises the argument, states a final judgment, or links back to the real world.
Your opening paragraph should define key terms, rephrase the question, and state the physics areas you will discuss. Use phrases like: “This question concerns the concept of … which is defined as …”, “The underlying principle is …”, “I will explore this by considering … and …” Avoid simply repeating the question without adding value.
By definition, X is the rate of change of Y with respect to Z.
根据定义，X 是 Y 对 Z 的变化率。
To address this, we must consider laws A and B.
要解答此题，必须考虑定律 A 和 B。
5. PEEL Method for Body Paragraphs | 主体段落的 PEEL 法
The PEEL structure (Point, Evidence, Explanation, Link) keeps each body paragraph focused. State your point clearly, back it with physical evidence (laws, equations, data), explain how the evidence supports the point, and then link back to the question or forward to the next idea. For example: “The magnetic force does no work on a moving charge because it is always perpendicular to velocity. Evidence: F = qvB sinθ, and work done W = Fd cosφ. When φ = 90°, cos90° = 0, so no work is done. This explains why a charged particle moves in a circular path in a uniform magnetic field.”
6. Integrating Equations and Calculations | 融入方程与计算
Essays in physics are not pure text; they require equations and numerical illustrations. When introducing an equation, always state what each symbol represents and why the relationship matters. Then, if appropriate, plug in typical values to show the magnitude of an effect. Place key equations centred and bold so they stand out.
For an electron (m = 9.11×10⁻³¹ kg) travelling at 2.0×10⁷ m s⁻¹, the kinetic energy is 1.8×10⁻¹⁶ J. Such a calculation demonstrates that relativistic corrections are not yet needed at this speed.
对于电子（质量 m = 9.11×10⁻³¹ kg）以 2.0×10⁷ m s⁻¹ 运动，动能为 1.8×10⁻¹⁶ J。这一计算表明，在此速度下尚不需要相对论修正。
7. Describing Graphs and Data | 描述图表与数据
Many essay questions require you to interpret a graph or a set of data. Use systematic language: identify the variables on the axes, describe the trend (linear, exponential, inversely proportional), quote gradient or intercept values, and relate them to physical constants. For instance, “The gradient of the V–I graph gives the resistance R, which from the data is calculated as 4.7 Ω.”
Higher marks come from showing you can evaluate models, experiments, or assumptions. Use phrases such as “A limitation of this model is …”, “The assumption breaks down when …”, or “An alternative explanation could be …”. In the IB, this is often triggered by command terms like “evaluate” or “discuss”; in AQA, it lifts your essay into the top band. Always suggest improvements or state the limits of validity.
A strong conclusion summarises the key physics, explicitly answers the question, and may offer a broader implication. Avoid introducing new material. Start with “In conclusion, …” or “To summarise, …”. Then re-state how the evidence supports your main argument. For AQA essays, a final sentence linking to other topics or real-world applications is highly valued.
Here is a condensed framework you can adapt in the exam. Introduction: define key concept, state scope. Body Paragraph 1: first physics idea, equation, graphical interpretation. Body Paragraph 2: second idea, link to first, data evidence. Body Paragraph 3 (if time): evaluation or contrasting case. Conclusion: answer the question, note limitations, broader context. Practice filling this template with past paper questions.
Many students lose marks by writing everything they know without any structure, missing out on evaluation, or failing to define symbols in equations. Others write too little, leaving the essay underdeveloped. Also, avoid bullet points unless the question specifically allows them; always write in continuous prose. Finally, do not confuse precision with long-windedness—concise, physics-rich sentences earn more marks than vague generalisations.
For IB, allocate roughly 1 minute per mark for long-answer questions, and reserve 2–3 minutes for planning. For the AQA essay, you have about 35–40 minutes; spend at least 5 minutes planning and the last 5 minutes checking. During planning, list your physics ideas as a mind map, select the strongest ones, and order them intentionally. A well-planned essay flows naturally and saves time in the long run.
Welcome to this A-Level Physics revision guide on materials physics. Understanding the mechanical properties of materials, such as stress, strain, and the Young modulus, is essential for tackling exam questions on elasticity, plasticity, and material behaviour. This article will walk you through the key concepts, equations, and graphical analysis needed to excel in your exams.
Materials physics studies how solids behave when subjected to forces. It covers concepts like stiffness, strength, and ductility, which determine how a material stretches, compresses, or breaks.
Tensile stress stretches a material, while compressive stress shortens it. A-Level questions usually focus on tensile testing of wires.
拉伸应力使材料伸长，而压缩应力使其缩短。A-Level 考试通常关注金属丝的拉伸测试。
3. Young Modulus and Hooke’s Law | 杨氏模量与胡克定律
The Young modulus (E) is a measure of stiffness. It is the ratio of stress to strain in the linear elastic region: E = σ/ε. Its unit is Pa.
杨氏模量 (E) 是刚度的量度。它是线弹性区内应力与应变的比值：E = σ/ε，单位是 Pa。
Combining the definitions gives the force–extension relationship: F = (EA/L₀) ΔL. This shows that the spring constant k for a wire is k = EA/L₀.
结合定义可得到力–伸长关系：F = (EA/L₀) ΔL。这表明金属丝的弹簧常数 k 为 k = EA/L₀。
Hooke’s law states that extension is proportional to force up to the limit of proportionality. The elastic limit may be slightly beyond this point.
胡克定律指出，在比例极限内，伸长与力成正比。弹性极限可能略高于比例极限。
A material that obeys Hooke’s law is linear elastic. Beyond the elastic limit, permanent deformation occurs.
遵守胡克定律的材料是线弹性的。超过弹性极限后，会发生永久形变。
4. Stress–Strain Curves and Material Behaviour | 应力–应变曲线与材料行为
A typical stress–strain curve for a ductile metal like copper has distinct regions: a straight initial portion (Hookean), a yield point, a plastic region with strain hardening, necking, and finally fracture.
From O to P (limit of proportionality) the graph is linear. From P to E (elastic limit) the material is still elastic but may not be linear. Beyond E, plastic flow begins.
Published by TutorHao | A-Level Physics Revision Series | aleveler.com
📚 Essential Formula Derivations from Pearson’s IB Physics HL Textbook | IB 物理 HL Pearson 教材核心公式推导
Understanding the mathematical origins of key equations is fundamental to mastering IB Physics HL. The Pearson Baccalaureate textbook presents derivations that link physical concepts with rigorous algebra and calculus. This article walks through ten essential derivations, explaining each step in both English and Chinese to strengthen your conceptual and analytical skills.
Starting with the definition of constant acceleration a, we have a = (v – u)/t, where u is initial velocity, v is final velocity, and t is time. Rearranging gives the first SUVAT equation.
从恒定加速度 a 的定义出发，有 a = (v – u)/t，其中 u 为初速度，v 为末速度，t 为时间。移项得到第一个运动学方程。
v = u + at
The average velocity for uniform acceleration is v_avg = (u + v)/2. Displacement s equals average velocity multiplied by time, so s = ((u + v)/2)t. Substituting v = u + at yields s = ut + ½at².
匀加速运动的平均速度 v_avg = (u + v)/2。位移 s 等于平均速度乘以时间，因此 s = ((u + v)/2)t。代入 v = u + at 得到 s = ut + ½at²。
s = ut + ½at²
Combining the first two equations by eliminating t gives v² = u² + 2as, which is particularly useful when time is not involved.
联立前两个方程并消去 t 得到 v² = u² + 2as，这在无需时间的场景中尤其有用。
v² = u² + 2as
2. Deriving Centripetal Acceleration a = v²/r | 向心加速度公式推导
Consider an object moving at constant speed v in a circle of radius r. In a short time Δt, the object moves through a small angle Δθ. The velocity vector changes from v₁ to v₂, both of magnitude v. The change in velocity Δv points toward the centre, and its magnitude is approximately vΔθ for small angles.
考虑物体以恒定速率 v 在半径为 r 的圆周上运动。在短时间 Δt 内，物体转过小角度 Δθ。速度矢量从 v₁ 变为 v₂，两者大小均为 v。速度变化量 Δv 指向圆心，小角度下其大小约为 vΔθ。
The arc length traveled is Δs = rΔθ, and since Δs = vΔt, we have Δθ = vΔt / r. Substituting into Δv ≈ vΔθ gives Δv ≈ v²Δt / r. Acceleration a = Δv/Δt, hence a = v²/r directed radially inward.
3. Deriving Orbital Speed from Universal Gravitation | 万有引力推导轨道速度
For a satellite in a stable circular orbit, the gravitational force provides the necessary centripetal force. Newton’s law of gravitation states F = GMm/r², where M is the central mass, m is the satellite mass, and r is the orbital radius.
对于稳定圆轨道上的卫星，万有引力提供所需的向心力。牛顿万有引力定律给出 F = GMm/r²，其中 M 为中心天体质量，m 为卫星质量，r 为轨道半径。
Equating gravitational force to centripetal force mv²/r gives GMm/r² = mv²/r. Canceling m and multiplying both sides by r yields v² = GM/r, so the orbital speed is v = √(GM/r).
令万有引力等于向心力 mv²/r，得到 GMm/r² = mv²/r。消去 m 并两边乘以 r 得 v² = GM/r，因此轨道速度 v = √(GM/r)。
v = √(GM / r)
4. Deriving the Ideal Gas Pressure Equation | 理想气体压强公式推导
Imagine a cubical container of side L containing N gas molecules, each of mass m, moving with root-mean-square speed cᵣₘₛ. Consider one molecule striking a wall perpendicular to the x-axis. Its velocity component changes from +uₓ to -uₓ, so the momentum change per collision is 2muₓ.
想象一个边长为 L 的立方体容器，内装 N 个质量均为 m 的气体分子，均方根速率为 cᵣₘₛ。考虑一个分子撞击垂直于 x 轴的器壁：其速度分量从 +uₓ 变为 -uₓ，每次碰撞动量改变为 2muₓ。
The time between collisions with the same wall is 2L/uₓ. Average force exerted by this molecule is F = Δp/Δt = (2muₓ)/(2L/uₓ) = muₓ²/L. Summing over all N molecules and noting the mean square speeds are equal in all three directions (uₓ² = uᵧ² = u_z² = c²/3), total force on the wall is F_total = Nm(c²/3)/L.
与同一器壁的两次碰撞间隔时间为 2L/uₓ。该分子施加的平均力为 F = Δp/Δt = (2muₓ)/(2L/uₓ) = muₓ²/L。对所有 N 个分子求和，并注意三个方向均方速度相等（uₓ² = uᵧ² = u_z² = c²/3），器壁总受力 F_total = Nm(c²/3)/L。
Pressure p is force per unit area A = L², so p = (Nm(c²/3)/L)/L² = Nmc²/(3L³) = (1/3)ρc², where ρ = Nm/V is density.
压强 p 是单位面积受力 A = L²，故 p = (Nm(c²/3)/L)/L² = Nmc²/(3L³) = (1/3)ρc²，其中 ρ = Nm/V 为密度。
p = ⅓ ρ ⟨c²⟩
5. Deriving the Charging Equation for an RC Circuit | RC 电路充电方程推导
In a series RC circuit connected to a battery of emf ε, Kirchhoff’s loop rule gives ε – IR – Q/C = 0, where Q is the charge on the capacitor and I = dQ/dt is the current. Substituting I yields ε – R dQ/dt – Q/C = 0.
在串联 RC 电路连接至电动势 ε 的电池时，基尔霍夫回路定律给出 ε – IR – Q/C = 0，其中 Q 为电容器电荷，I = dQ/dt 为电流。代入 I 得 ε – R dQ/dt – Q/C = 0。
Rearrange to separate variables: dQ/dt = (εC – Q)/(RC). Integration from Q = 0 at t = 0 gives ∫₀ᵠ dQ/(εC – Q) = ∫₀ᵗ dt/(RC), resulting in -ln(εC – Q) + ln(εC) = t/(RC). Simplify to Q = εC(1 – e^{-t/RC}).
The product RC is the time constant τ, representing the time taken for charge to reach about 63% of its final value Q₀ = εC.
乘积 RC 为时间常数 τ，表示电荷达到最终值 Q₀ = εC 的约 63% 所需的时间。
6. Deriving the SHM Differential Equation and Solution | 简谐运动微分方程及解推导
Simple harmonic motion occurs when the restoring force is proportional to displacement and opposite in direction: F = -kx. By Newton’s second law, ma = -kx. Acceleration a is the second derivative of displacement x, so m d²x/dt² = -kx.
当回复力与位移成正比且方向相反时，物体做简谐运动：F = -kx。由牛顿第二定律，ma = -kx。加速度 a 是位移 x 的二阶导数，因此 m d²x/dt² = -kx。
Defining angular frequency ω = √(k/m), the differential equation becomes d²x/dt² = -ω²x. The general solution is x = A cos(ωt) + B sin(ωt) or equivalently x = x₀ sin(ωt + φ).
定义角频率 ω = √(k/m)，微分方程化为 d²x/dt² = -ω²x。其通解为 x = A cos(ωt) + B sin(ωt)，或等效写为 x = x₀ sin(ωt + φ)。
x = A sin(ωt + φ)
Velocity and acceleration are then v = dx/dt = Aω cos(ωt + φ) and a = dv/dt = -Aω² sin(ωt + φ) = -ω²x, confirming the defining relationship.
速度和加速度则为 v = dx/dt = Aω cos(ωt + φ) 以及 a = dv/dt = -Aω² sin(ωt + φ) = -ω²x，验证了定义关系。
7. Deriving the Doppler Effect for Sound | 多普勒效应公式（声波）推导
When a source emitting sound of frequency f moves towards a stationary observer at speed v_s, the wavelength in front of the source is compressed. In source’s frame, the wave travels at speed c, but in one period T = 1/f, the source moves v_s T. The observed wavelength λ’ = cT – v_s T = (c – v_s)/f.
当声源以速度 v_s 向静止观察者移动并发出频率为 f 的声波时，声源前方的波长被压缩。在声源参考系中，波速为 c，但在一个周期 T = 1/f 内声源移动 v_s T。观察到的波长 λ’ = cT – v_s T = (c – v_s)/f。
Using the wave equation f’ = c/λ’, the observed frequency is f’ = c f/(c – v_s). This implies a higher pitch as the source approaches. For a stationary source and moving observer, the formula becomes f’ = f (c + v_o)/c for an observer moving toward the source.
应用波速公式 f’ = c/λ’，观察频率为 f’ = c f/(c – v_s)。声源靠近时音调变高。对于静止声源和运动观察者，观察者朝向声源运动时公式为 f’ = f (c + v_o)/c。
f’ = f (c ± v_o)/(c ∓ v_s)
8. Deriving the Radius of a Charged Particle in a Magnetic Field | 磁场中带电粒子半径推导
A charged particle q moving with speed v perpendicular to a uniform magnetic field B experiences a magnetic force F = qvB. This force acts as the centripetal force, F = mv²/r. Equating gives qvB = mv²/r, hence r = mv/(qB).
带电粒子 q 以速度 v 垂直于匀强磁场 B 运动时，受到洛伦兹力 F = qvB。该力充当向心力 F = mv²/r。令两式相等得 qvB = mv²/r，因此 r = mv/(qB)。
r = mv / (qB)
The radius is proportional to momentum and inversely proportional to both charge and magnetic field strength. The period T of the circular motion is T = 2πr/v = 2πm/(qB), which is independent of speed.
半径与动量成正比，与电荷和磁感应强度成反比。圆周运动的周期 T = 2πr/v = 2πm/(qB)，与速度无关。
9. Deriving Electric Potential Energy and Potential | 电势能和电势的推导
For two point charges Q and q separated by distance r, the Coulomb force is F = kQq/r². The work done to bring q from infinity to a distance r against this force is W = ∫∞ʳ F dr = ∫∞ʳ kQq/r² dr = -[kQq/r]∞ʳ = kQq/r. This work is stored as electric potential energy U.
两点电荷 Q 与 q 相距 r 时，库仑力为 F = kQq/r²。将 q 从无穷远处移到距 Q 为 r 的位置时克服库仑力做功 W = ∫∞ʳ F dr = ∫∞ʳ kQq/r² dr = -[kQq/r]∞ʳ = kQq/r。该功储存为电势能 U。
U = k Qq / r
The electric potential V at a point is defined as the potential energy per unit charge: V = U/q = kQ/r. For a uniform electric field E between parallel plates separated by distance d, potential difference ΔV = Ed arises from integrating E along the displacement.
电场中某点的电势 V 定义为单位电荷的电势能：V = U/q = kQ/r。对于间距为 d 的平行板间的匀强电场 E，电势差 ΔV = Ed 来自沿位移积分 E 的结果。
10. Deriving the Single-Slit Diffraction Minimum Condition | 单缝衍射极小条件推导
When light of wavelength λ passes through a single slit of width a, each point across the slit acts as a secondary wavelet source. Consider two rays from the top and the centre of the slit that meet at angle θ on a distant screen. The path difference between these two rays is (a/2)sinθ.
波长为 λ 的光通过宽度为 a 的单缝时，缝上每一点都可看作次级子波源。考虑来自缝顶端和缝中心的两条光线，它们以角度 θ 在远处屏幕相遇。这两条光线的光程差为 (a/2)sinθ。
If this path difference equals λ/2, the two waves arrive out of phase and cancel. Similarly, the entire slit can be paired up to produce complete destructive interference. This leads to the condition for the first minimum: a sinθ = λ. In general, minima occur at a sinθ = nλ, where n = 1, 2, 3,…
Welcome to this focused revision guide on capacitance, tailored to the Edexcel A-Level Physics specification. We will break down the key concepts, equations, graphs, and pitfalls you need to master for your exam. Whether you are tackling multiple-choice questions or the extended written paper, a solid grasp of capacitor behaviour is essential.
A capacitor is a device designed to store electric charge and hence energy. In its simplest form, it consists of two parallel conducting plates separated by an insulator or dielectric. When connected to a source of potential difference, charge builds up on the plates — one plate gains electrons (negative), and the other loses electrons (positive).
The capacitance C of a capacitor is defined as the charge stored per unit potential difference across the plates. The defining equation is C = Q / V, where Q is the magnitude of the charge on one plate and V is the p.d. between the plates. The SI unit is the farad (F), which is equivalent to C V⁻¹. A 1 F capacitor stores 1 C of charge when the p.d. across it is 1 V.
电容 C 的定义是单位电压下储存的电荷量。定义式为 C = Q / V，其中 Q 是单块极板的电荷量，V 是极板间的电势差。国际单位是法拉 (F)，相当于 C V⁻¹。一个 1 F 的电容在 1 V 电压下可储存 1 C 的电荷。
2. Factors Affecting Capacitance | 影响电容的因素
For a parallel-plate capacitor, the capacitance is determined by three factors: the area of overlap of the plates A, the separation between them d, and the permittivity of the dielectric material ε between the plates. The relationship is C = εA / d. This shows that capacitance increases with larger plate area and decreases with greater plate separation.
对于平行板电容器，电容由三个因素决定：极板的正对面积 A、极板间距 d，以及极板间电介质的介电常数 ε。关系式为 C = εA / d。这表明电容随极板面积增大而增大，随间距增大而减小。
The permittivity ε = ε₀εᵣ, where ε₀ is the permittivity of free space (8.85 × 10⁻¹² F m⁻¹) and εᵣ is the relative permittivity (dielectric constant) of the material. A dielectric with a high εᵣ increases the capacitance significantly compared to a vacuum. This is because the dielectric becomes polarised, reducing the effective electric field between the plates and allowing more charge to be stored for the same p.d.
3. Charging a Capacitor in a DC Circuit | 直流电路中的电容器充电过程
When an uncharged capacitor is connected in series with a resistor and a battery, charge does not instantaneously reach its maximum. The p.d. across the capacitor increases gradually according to the equation V = V₀(1 − e^{-t/RC}), where V₀ is the supply voltage and RC is the time constant. The current starts at a maximum I₀ = V₀/R and decays exponentially to zero.
During charging, electrons flow onto one plate and off the other, creating an increasing p.d. that opposes the driving voltage. The rate of charging is highest at the start and slows down as the capacitor approaches full charge. After a time equal to RC, the capacitor voltage reaches roughly 63% of V₀; after 5RC, it is considered fully charged (over 99%).
When a charged capacitor is disconnected from the battery and connected across a resistor, it discharges. The voltage, charge, and current all decay exponentially according to the formula X = X₀ e^{-t/RC}, where X can be V, Q, or I. The initial discharge current is I₀ = V₀/R, where V₀ is the initial capacitor voltage.
当充满电的电容器脱离电池并并联一个电阻时，它会放电。电压、电荷和电流都按 X = X₀ e^{-t/RC} 的指数规律衰减，其中 X 可代表 V、Q 或 I。初始放电电流为 I₀ = V₀/R，其中 V₀ 是电容器初始电压。
The product RC, known as the time constant τ, has units of seconds. In one time constant, the voltage falls to about 37% of its initial value. The decay is exponential because the rate of discharge is proportional to the remaining charge: dQ/dt = -Q/RC. The negative sign indicates a decrease.
5. Time Constant τ and Its Significance | 时间常数 τ 及其意义
The time constant τ = RC is a crucial parameter in both charging and discharging curves. It tells you how quickly a capacitor charges or discharges. A larger RC means a slower response. This is particularly important in timing circuits, filters, and smoothing applications.
Graphically, τ can be found from a voltage-time graph. For discharging, it is the time taken for V to fall to V₀/e, or equivalently to about 37% of V₀. One can also draw a tangent at t=0; the intercept on the time axis equals τ. In charging, τ is the time to reach 63% of the final value. Many exam questions require you to determine τ from a graph or to sketch the curve.
Work is done to charge a capacitor, and this work is stored as electric potential energy in the electric field between the plates. The energy can be expressed in three equivalent forms: E = ½ QV = ½ CV² = ½ Q²/C. The factor ½ arises because the average p.d. during charging is half the final p.d.
给电容器充电需要做功，这些功以电势能的形式储存在极板间的电场中。能量有三种等价表达式：E = ½ QV = ½ CV² = ½ Q²/C。之所以有系数 ½，是因为充电过程中的平均电压是最终电压的一半。
This stored energy can be released rapidly, as in a camera flash or a defibrillator. In a camera flash, a capacitor is slowly charged and then quickly discharged through the flash lamp, producing a bright, short pulse of light. The power during discharge can be very high even though the total energy is modest.
The rules for combining capacitors are the reverse of those for resistors. For capacitors in parallel, the total capacitance is simply the sum of individual capacitances: C_total = C₁ + C₂ + … . This is because the p.d. across each is the same, and charges add up. The parallel combination stores more charge.
For capacitors in series, the reciprocal of the total capacitance equals the sum of reciprocals: 1/C_total = 1/C₁ + 1/C₂ + … . Consequently, the total capacitance is always smaller than the smallest individual capacitance. In series, the charge on each capacitor is identical, but the p.d.s add up.
Inserting a dielectric between the plates does more than just increase capacitance. The dielectric material contains polar molecules that align, to some extent, with the external electric field. This creates an internal electric field opposing the applied field, reducing the net field. For a given charge, the potential difference decreases, which by C = Q/V means capacitance increases.
在极板间加入电介质不仅仅是增加电容。电介质材料含有极性分子，它们会在一定程度上顺着外电场排列。这会产生一个与外电场方向相反的内电场，从而削弱净电场。对于给定电荷量，电势差降低，由 C = Q/V 可知电容增大。
Dielectrics also prevent direct electrical contact between plates and can withstand high voltages without breakdown. The maximum electric field a dielectric can tolerate is its dielectric strength. Exceeding this causes a spark or permanent damage. Mica, ceramics, and plastics are common dielectric materials.
9. Graphical Analysis of Charging and Discharging | 充放电的图像分析
The voltage-time, charge-time, and current-time graphs for both charging and discharging are standard exam content. You must be able to sketch these graphs accurately, label axes, and mark the time constant, initial and final values. For charging V and Q, the graph starts at zero and rises asymptotically towards V₀ or Q₀. For current, it falls from I₀ to zero.
充电和放电的电压-时间、电荷-时间以及电流-时间图像是标准的考试内容。你必须能够准确地绘制这些图像，标注坐标轴，并标明时间常数、初始值和终值。对于充电的 V 和 Q，图像从零开始并渐近上升至 V₀ 或 Q₀。电流则从 I₀ 衰减到零。
Discharging graphs for V, Q, and I all have identical exponential decay shapes, starting at a maximum and decaying towards zero. Tangents at t=0 can be used to find τ graphically. In a log-linear graph of ln(V) vs t, the gradient equals -1/RC, which is a common practical method to determine capacitance.
放电的 V、Q 和 I 图像具有完全相同的指数衰减形状，从最大值开始衰减到零。在 t=0 处的切线可用于从图像求 τ。在 ln(V) 对 t 的半对数图中，斜率等于 -1/RC，这是一种常用的测量电容的实验方法。
10. Exponential Equations and Natural Logarithms | 指数方程与自然对数
The discharge equation V = V₀ e^{-t/RC} can be linearised by taking natural logarithms: ln(V) = ln(V₀) – t/RC. This has the form y = mx + c with gradient m = -1/RC and intercept ln(V₀). This relationship allows experimental determination of RC, and hence C, from a set of voltage-time data.
放电方程 V = V₀ e^{-t/RC} 可以通过取自然对数线性化：ln(V) = ln(V₀) – t/RC。这具有 y = mx + c 的形式，斜率 m = -1/RC，截距 ln(V₀)。利用这一关系，可从一组电压-时间数据中实验测定 RC，进而得出 C。
Similar manipulations apply for charge and current. A graph of ln(Q) against time also yields a straight line. Remember that ln(e) = 1, and simplifying requires familiarity with log rules. In exams, you might be given a table of data and asked to plot a graph to find time constant or an unknown resistance.
11. Capacitors in DC and AC Circuits | 直流与交流电路中的电容
In a DC circuit, after the transient charging/discharging period, a fully charged capacitor blocks direct current — it acts as an open circuit. However, in an AC circuit, the capacitor continuously charges and discharges as the polarity alternates, allowing an alternating current to flow. The opposition to AC is called capacitive reactance Xc = 1/(2πfC).
Capacitive reactance decreases as frequency or capacitance increases. This property makes capacitors useful in filter circuits: low-pass, high-pass, and band-pass filters. In A-Level, detailed AC theory may not be required, but understanding the distinction between DC and AC behaviour is important.
12. Common Exam Pitfalls and Practical Techniques | 常见考点与实验技巧
One frequent misconception is confusing the charging and discharging equations. Remember: charging uses (1 – e^{-t/RC}), while discharging uses e^{-t/RC}. Also, do not forget that the time constant is in seconds, so resistance in Ω and capacitance in F. Be careful with unit prefixes (μF, nF, kΩ).
In practical work, a data logger or oscilloscope is often used. When determining τ from a graph, students sometimes use the wrong percentage (63% for charging, 37% for discharging). A better method is to measure the time for V to halve, then use t₁/₂ = RC ln 2 ≈ 0.693RC. This avoids reading errors.
Finally, always check whether the question expects you to derive a formula (starting from Q=CV and Kirchhoff’s laws) or simply apply it. Practice using the linearised equation ln(V) = ln(V₀) – t/RC for data analysis, as this is a common assessment task in Edexcel practical papers.
Even students who grasp the main ideas often lose marks in GCSE OCR Physics by falling into predictable traps. This article highlights ten of the most widespread mistakes and misconceptions, explains why they are wrong, and shows you how to avoid them. Read carefully – each point pairs an English explanation with its Chinese equivalent so that you can reinforce correct understanding in both languages.
Many students incorrectly use ‘mass’ and ‘weight’ as if they were the same thing. A common mistake is to state that a 5 kg mass ‘weighs 5 kg’ or that weight is measured in kilograms.
Mass is the amount of matter in an object, measured in kilograms (kg) and is a scalar quantity. It does not change no matter where you are in the universe. Weight is a force caused by gravity, measured in newtons (N) and is a vector quantity. Weight depends on the gravitational field strength g (around 9.8 N/kg on Earth). The correct relationship is W = m × g.
质量是物体所含物质的多少，以千克（kg）为单位，是标量。无论你在宇宙中的哪个地方，它都不会改变。重量是由重力引起的力，以牛顿（N）为单位，是矢量。重量取决于重力场强度 g（地球表面约为 9.8 N/kg）。正确的公式是 W = m × g。
For example, your mass is 50 kg on both Earth and the Moon, but your weight is about 490 N on Earth and only about 80 N on the Moon. In your exam, always check the units and write weight in newtons.
A frequent misconception is that speed and velocity are interchangeable. Students often give a velocity answer but omit the direction, thereby making it speed, or they calculate speed but claim it is velocity.
Speed is a scalar: it tells you how fast something is moving (e.g. 30 m/s). Velocity is a vector: it tells you both the speed and the direction (e.g. 30 m/s north). If an object changes direction but maintains the same speed, its velocity changes. This distinction is essential for questions on motion graphs and Newton’s laws.
速率是标量：它告诉你物体运动的快慢（例如 30 m/s）。速度是矢量：它告诉你运动的快慢和方向（例如 30 m/s 向北）。如果物体改变方向但保持相同的速率，其速度是变化的。这一区别对于运动图像和牛顿定律的问题至关重要。
Typical exam trap: describing velocity using only a magnitude with no direction will lose you a mark. Always state a direction when the question asks for velocity.
典型考试陷阱：只用一个大小而没有方向来描述速度会让你丢分。当题目问速度时，一定要说明方向。
3. Misunderstanding Resultant Forces and Newton’s First Law | 误解合力和牛顿第一定律
Many pupils think that a constant forward force is needed to keep an object moving at a steady speed. This directly contradicts Newton’s First Law.
许多学生认为物体要保持匀速运动就需要一个恒定的向前的力。这直接违背了牛顿第一定律。
Newton’s First Law states that an object will remain at rest or move with constant velocity unless a resultant force acts on it. If an object is moving at constant velocity, all forces are balanced and the resultant force is zero. A driving force may still be present, but it is exactly cancelled by resistive forces such as friction or air resistance.
For example, a car cruising at a steady 70 mph on a motorway experiences a forward engine force exactly balanced by total resistive forces. Resultant force is zero, so acceleration is zero.
In exam responses, always link constant velocity to balanced forces, not to a ‘driving force being bigger’.
在考试回答中，始终将匀速运动与平衡的力联系起来，而不是与“驱动力更大”联系起来。
4. Newton’s Third Law Pairs | 牛顿第三定律中的力对
Students can usually recite ‘to every action there is an equal and opposite reaction’, but they frequently misidentify the force pair. A classic error is to say that a book on a table has a Third Law pair made up of its weight and the normal reaction.
Weight and normal reaction are not a Newton’s Third Law pair because they act on the same object. A Third Law pair must act on two different bodies and be of the same type. The correct pair for weight is: the Earth pulls the book down (weight), and the book pulls the Earth up with an equal gravitational force. The pair for the normal force is: the table pushes the book up, and the book pushes the table down.
Always check: are the forces acting on different bodies? Are they the same type (both gravitational, both electrostatic, etc.)? Then move on to magnitude and direction.
5. Electrical Circuits: Series and Parallel Misconceptions | 电路：串联和并联的误解
A common belief is that ‘current gets used up’ as it passes through a circuit. In a series circuit, students often think that the current is larger before a lamp than after it.
Current is the rate of flow of charge. In a series circuit, the current is the same at every point. Charge carriers are not consumed; they simply transfer energy to the components. What changes around a circuit is potential difference (voltage), which is shared across components in series, not current.
In parallel circuits, potential difference across each loop is the same, but current splits at junctions. The total current entering a junction equals the total current leaving it.
在并联电路中，每个支路两端的电势差相同，但电流在节点处分流。流入节点的总电流等于流出节点的总电流。
For describing observations: never write ‘current is used up’. Instead, say ‘energy is transferred from the charge carriers to the component’.
描述观察现象时：永远不要写“电流被用掉了”。应该说“能量从电荷载流子传递给了元件”。
6. Energy Transfers and Sankey Diagrams | 能量转移和桑基图
Pupils often draw Sankey diagrams with the useful output arrow wider than the input arrow, or they show the input splitting into useful and wasted without understanding the width proportionality.
In a correct Sankey diagram, the width of the input arrow represents the total energy supplied. The width of the useful output arrow(s) and the wasted energy arrow(s) must add up to the width of the input arrow. No energy is lost; it is only transferred to less useful stores, often thermal energy of the surroundings.
Efficiency = useful output energy transfer ÷ total input energy transfer. This ratio can also be worked out directly from the widths. Never claim efficiency > 1, and remember that wasted energy reduces efficiency but is always present.
Many mistakes arise from confusing the mass being heated with the mass of the heater, or from using the temperature change in degrees Celsius without converting correctly. Students also often forget to convert grams to kilograms when using the standard unit J/(kg·°C).
The specific heat capacity equation is ΔE = m c Δθ. m is the mass of the substance being heated (in kg), c is the specific heat capacity (J/(kg·°C)), and Δθ (often written ΔT) is the temperature change in °C or K. Always write the temperature change, not the absolute temperature.
比热容方程为 ΔE = m c Δθ。m 是被加热物质的质量（以 kg 为单位），c 是比热容（J/(kg·°C)），Δθ（常写作 ΔT）是温度变化，单位为 °C 或 K。始终要写温度变化，而不是绝对温度。
Double-check your substitution: if a question gives mass in grams, convert to kilograms (÷1000). If it provides the power of a heater and the time, energy supplied = power × time.
A frequent slip is to mix up frequency and period when using the wave equation v = f λ. Some students use T (time period) in place of f, or they use the wavelength incorrectly, perhaps measuring the distance from a crest to a trough.
一个常见错误是在使用波速公式 v = f λ 时混淆频率和周期。有些学生用 T（周期）代替 f，或者他们错误地使用了波长，例如可能测量了从波峰到波谷的距离。
The correct wave equation is wave speed = frequency × wavelength. Frequency f is in hertz (Hz), wavelength λ is in metres (m). Remember that frequency is the reciprocal of the period: f = 1/T. Wavelength is the distance between two consecutive identical points, e.g. from crest to crest, not crest to trough.
When measuring from oscilloscope traces, read the time base correctly to find period, then calculate frequency. Always state the unit of speed as m/s.
从示波器波形上测量时，要正确读取时基来找出周期，然后计算频率。始终以 m/s 为单位给出波速。
v = f λ and f = 1 / T
9. Radioactive Decay and Half-life Graphs | 放射性衰变和半衰期图像
Misreading the graph axes is the most common error. Students may read the half-life as the time for the count rate to drop to half the displayed maximum, even if that maximum is not the initial count. Others assume that after two half-lives the sample becomes non-radioactive.
Half-life is the time taken for the number of unstable nuclei (or the count rate) to halve from its original value. Always start from the initial activity on the graph. After one half-life, activity halves; after two, it quarters; after three, it becomes one-eighth. Radioactivity never truly reaches zero, but background levels are eventually reached.
When interpreting decay graphs, draw horizontal lines from half the current value to the curve, then read the time interval. Repeat for different points to confirm consistency.
10. Momentum and Conservation of Momentum | 动量与动量守恒
Students frequently treat momentum as a force or forget its vector nature when calculating totals. In collisions, they may add momenta as scalars without considering direction, especially in recoil scenarios.
Momentum p = m × v, a vector quantity. The unit is kg m/s. The conservation law states that the total momentum before an event equals the total momentum after, provided no external resultant force acts. Direction must be taken into account: assign a positive direction and treat movements in opposite directions as negative momentum values.
动量 p = m × v，是矢量。单位为 kg m/s。动量守恒定律指出，在没有外部合力作用的情况下，事件前的总动量等于事件后的总动量。必须考虑方向：规定正方向，并把相反方向的运动处理为负的动量值。
For explosion or recoil problems: total momentum initially is zero, so the momentum of one part after the explosion is equal in magnitude and opposite in direction to the other part. Always set up a positive direction and write momentum equations accordingly.
Quantum physics introduces the idea that light and matter can behave as both waves and particles. In GCSE AQA Physics, you will explore the photon model, the photoelectric effect, energy levels in atoms, and how these ideas explain absorption and emission spectra.
Electromagnetic radiation, including visible light, can be described as a stream of wave packets or ‘quanta’ called photons. Each photon carries a discrete amount of energy that depends only on the frequency of the radiation.
A photon is a massless bundle of electromagnetic energy. The energy of a photon is directly proportional to its frequency, with the constant of proportionality being Planck’s constant h.
光子是无质量的电磁能量束。光子的能量正比于其频率，比例常数是普朗克常数 h。
The photon model successfully explains phenomena that the classical wave theory cannot, such as the photoelectric effect. It reveals the particle-like behaviour of light.
光子模型成功解释了经典波动理论无法解释的现象，例如光电效应。它揭示了光的粒子性行为。
E = hf
where h = 6.63 × 10⁻³⁴ J·s (Planck’s constant). Since c = fλ for all electromagnetic waves, the photon energy can also be expressed in terms of wavelength:
其中 h = 6.63 × 10⁻³⁴ J·s（普朗克常数）。由于所有电磁波都满足 c = fλ，光子能量也可以用波长表示为：
E = hc / λ
2. Photon Energy Equation | 光子能量方程
Using E = hf or E = hc/λ, you can calculate the energy of any photon if you know its frequency or wavelength. Remember that frequency is measured in hertz (Hz) and wavelength in metres (m).
利用 E = hf 或 E = hc/λ，只要知道频率或波长，就可以计算任何光子的能量。记住，频率的单位是赫兹 (Hz)，波长的单位是米 (m)。
For example, a photon of red light with a wavelength of 700 nm has energy:
These energy values are extremely small, so physicists often use a more convenient unit, the electronvolt (eV), which will be introduced next.
这些能量值非常小，因此物理学家经常使用更方便的单位——电子伏特 (eV)，将在下节引入。
3. The Electronvolt (eV) | 电子伏特 (eV)
One electronvolt is defined as the energy transferred when an electron moves through a potential difference of one volt. It is a very small unit of energy suitable for atomic and quantum scales.
To convert from joules to electronvolts, divide the energy in joules by 1.60 × 10⁻¹⁹. Conversely, multiply the value in eV by 1.60 × 10⁻¹⁹ to get joules.
要将焦耳转换为电子伏特，只需将焦耳数除以 1.60 × 10⁻¹⁹。反过来，将 eV 值乘以 1.60 × 10⁻¹⁹ 即可得到焦耳。
Using the previous red-light photon example: 2.84 × 10⁻¹⁹ J ÷ 1.60 × 10⁻¹⁹ = 1.78 eV. Photon energies in the visible range are typically a few electronvolts.
The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency shines on it. This effect provides strong evidence for the photon model.
Key observations from experiments: electrons are emitted only if the incident frequency exceeds a certain threshold frequency, no matter how intense the light is. Increasing intensity simply increases the number of emitted electrons, not their kinetic energy.
This cannot be explained by classical wave theory, which predicts that energy from waves would accumulate over time and eventually eject electrons at any frequency. The photon model explains it instantly.
5. Work Function and Threshold Frequency | 功函数与阈值频率
Each metal has a characteristic work function (Φ), which is the minimum energy required to remove an electron from the surface of the metal. The work function is usually given in electronvolts.
The threshold frequency f₀ is the minimum frequency of radiation needed to cause photoemission. It is related to the work function by:
阈值频率 f₀ 是引发光电发射所需的最小辐射频率。它与功函数的关系是：
Φ = h f₀
If the photon energy is less than the work function (hf < Φ), no electrons will be emitted, regardless of the light intensity.
如果光子能量小于功函数 (hf < Φ)，那么无论光强多强，都不会有电子发射。
For example, sodium has a work function of about 2.3 eV. Therefore the threshold frequency is f₀ = Φ / h ≈ 5.5 × 10¹⁴ Hz, which lies in the visible spectrum.
When a photon with energy hf greater than the work function strikes a metal, the excess energy appears as kinetic energy of the emitted electron. The maximum kinetic energy Eₖₘₐₓ is given by the Einstein photoelectric equation:
Here, Eₖₘₐₓ is the maximum kinetic energy because some energy may be lost as the electron moves through the metal. The equation shows that Eₖₘₐₓ depends only on frequency, not on intensity.
If we plot maximum kinetic energy against frequency, the gradient of the line equals Planck’s constant h, and the intercept on the frequency axis gives the threshold frequency f₀.
7. Atomic Models: From Plum Pudding to Rutherford | 原子模型：从葡萄干布丁到卢瑟福
Early models of the atom were simple. J.J. Thomson’s ‘plum pudding’ model (1897) suggested that the atom consisted of a sphere of positive charge with negative electrons embedded inside it, like plums in a pudding.
Rutherford’s gold foil experiment (1911) changed this view. A beam of alpha particles was directed at a thin gold foil. Most passed straight through, but a small number were deflected through large angles.
Rutherford concluded that the atom had a tiny, dense, positively charged nucleus where most of the mass was concentrated, with electrons orbiting at relatively large distances. The nuclear model replaced the plum pudding model.
Rutherford’s nuclear model could not explain why electrons did not spiral into the nucleus. Niels Bohr adapted the model by proposing that electrons orbit only in certain allowed ‘energy levels’ or shells.
Each energy level corresponds to a fixed amount of energy. The lowest energy level is called the ground state (n = 1). Higher levels are excited states. Electrons can move between levels by absorbing or emitting photons.
The energy of the emitted or absorbed photon equals the difference between two energy levels:
发射或吸收的光子能量等于两个能级之间的能量差：
ΔE = E₂ – E₁ = hf
This explains why atoms produce line spectra rather than continuous spectra.
这解释了为什么原子产生线状光谱而非连续光谱。
9. Absorption and Emission Spectra | 吸收与发射光谱
When electrons in an atom absorb energy, they jump to higher energy levels. This produces an absorption spectrum: a continuous spectrum with dark lines corresponding to absorbed wavelengths.
When excited electrons fall back to lower levels, they emit photons of specific energies, producing an emission spectrum: a series of bright coloured lines on a dark background.
Each element has a unique set of energy levels, so its line spectrum acts as a ‘fingerprint’. This is used in astronomy to identify the chemical composition of stars and in laboratory analysis.
When an electron gains enough energy to completely escape from the atom, the atom becomes a positively charged ion. The minimum energy required to remove an electron from the ground state is the ionisation energy.
Ionisation can occur by absorbing a photon with energy equal to or greater than the ionisation energy, or through collisions with fast-moving particles like other electrons.
电离可以通过吸收能量等于或大于电离能的光子发生，也可以通过与其他快速运动的粒子（如电子）碰撞发生。
Excitation, on the other hand, occurs when an electron moves to a higher energy level but does not leave the atom. The atom is then in an excited state, which is unstable and often decays back to a lower level within about 10⁻⁸ s.
Fluorescence is a process in which a substance absorbs high-energy (often ultraviolet) photons and re-emits lower-energy visible photons. The atom’s electrons are excited to higher levels, then cascade down in steps.
Because the emitted photons have less energy than the absorbed ones, the wavelength increases. This is used in fluorescent lights, security markers, and TV screens.
由于发射光子的能量低于吸收光子，波长变长。这被应用于日光灯、防伪标记和电视屏幕中。
In a fluorescent tube, mercury vapour emits ultraviolet radiation when excited by an electric current. This UV light is absorbed by a phosphor coating inside the tube, which then fluoresces visible light.
在日光灯管中，汞蒸气被电流激发后发出紫外辐射。这些紫外光被管内荧光粉涂层吸收，然后荧光发出可见光。
12. Summary of Key Equations | 关键方程总结
Make sure you are confident using the core equations of quantum physics. They all link photon energy, work function, kinetic energy, and energy-level differences.
请确保你熟练运用量子物理的核心方程。它们都将光子能量、功函数、动能和能级差联系在一起。
Use the table below for a quick review. Remember to convert units correctly, especially between joules and electronvolts.
使用下表快速回顾。切记正确转换单位，尤其是焦耳与电子伏特之间的转换。
Equation / 方程
Meaning / 含义
E = hf
Photon energy / 光子能量
E = hc/λ
Photon energy from wavelength / 由波长求光子能量
Φ = h f₀
Work function and threshold frequency / 功函数与阈值频率
Eₖₘₐₓ = hf – Φ
Maximum kinetic energy of photoelectrons / 光电子最大动能
ΔE = hf = E₂ – E₁
Energy difference between levels / 能级间能量差
Published by TutorHao | Physics Revision Series | aleveler.com
Diffraction is a fundamental wave phenomenon that occurs when light encounters an obstacle or a slit. Understanding diffraction is essential for IB and OCR physics students, as it not only explains the behavior of light beyond geometrical optics but also underpins many modern technologies such as spectroscopy and high-resolution imaging.
Diffraction refers to the spreading of waves as they pass through an aperture or around an obstacle. The amount of bending depends on the ratio of the wavelength to the size of the opening. For light, significant diffraction occurs when the slit width is comparable to the wavelength, typically a few micrometres. If the aperture is much larger than the wavelength, the wave travels in straight lines and geometrical optics applies.
Huygens’ principle states that every point on a wavefront acts as a source of secondary spherical wavelets. The new wavefront is the envelope of these wavelets. When the wavefront encounters an aperture, the wavelets along the edges propagate into the geometrical shadow, explaining diffraction. The superposition of these secondary wavelets gives rise to the interference pattern observed on a screen.
When monochromatic light passes through a narrow single slit of width a, a diffraction pattern appears on a distant screen. It consists of a broad central bright fringe (the central maximum) flanked by alternating dark and bright fringes of decreasing intensity. The intensity I as a function of angle θ can be described by the expression I = I₀ (sinβ / β)², where β = (π a sinθ) / λ and I₀ is the intensity at the centre. Most exam questions focus on the positions of dark fringes and the angular width of the central maximum.
当单色光通过宽度为 a 的窄单缝时，远处的屏幕上会出现衍射图样。它包括一个宽阔的中央亮纹（中央极大），两侧是强度逐渐减弱的交替明暗条纹。强度 I 随角度 θ 的变化可以表示为 I = I₀ (sinβ / β)²，其中 β = (π a sinθ) / λ，I₀ 是中心处的强度。大多数考题关注暗纹的位置和中央明纹的角宽度。
The positions of dark fringes (minima) are given by the single-slit equation:
a sinθ = mλ , m = ±1, ±2, ±3, …
Note that m = 0 corresponds to the central maximum, not a dark fringe. For small angles, the angular width of the central maximum is approximately Δθ = 2λ / a, which shows that a narrower slit produces a wider central maximum.
4. Factors Affecting the Diffraction Pattern | 影响衍射图样的因素
The single-slit diffraction pattern is highly dependent on the slit width a and the wavelength λ. When a decreases, the angle θ for the first minimum (m=1) increases, so the pattern spreads out. Similarly, a larger wavelength λ broadens the pattern. This is why red light (longer λ) diffracts more than blue light (shorter λ) through the same slit. In exam scenarios, you may be asked to sketch how the pattern changes when the slit width is halved or when white light is used.
单缝衍射图样很大程度上取决于缝宽 a 和波长 λ。当 a 减小时，第一极小（m=1）对应的角度 θ 增大，因此图样发散。同样，较大的波长 λ 也会使图样变宽。这就是为什么红光（λ 较长）通过相同狭缝时比蓝光（λ 较短）衍射更明显。在考试中，你可能会被要求画出当缝宽减半或使用白光时图样的变化。
5. Diffraction Grating and the Grating Equation | 衍射光栅与光栅方程
A diffraction grating consists of a large number of equally spaced parallel slits (or grooves). The distance between adjacent slits is d, called the grating spacing. When monochromatic light is incident on the grating, constructive interference produces very sharp and bright peaks called principal maxima at angles given by the grating equation:
d sinθ = nλ , n = 0, ±1, ±2, …
Here n is the order of diffraction. The zeroth order (n=0) is the straight-through beam, and higher orders appear symmetrically on either side. The grating produces much sharper lines compared to a double slit because multiple slits reinforce the interfering beams.
6. Angular Dispersion and Resolving Power of a Grating | 光栅的角色散与分辨本领
The angular dispersion measures how much the angular position of a maximum changes with wavelength. For a grating, it is derived from differentiating the grating equation: dθ/dλ = n / (d cosθ). This means better dispersion is achieved with a higher order n and a smaller grating spacing d. The resolving power R is the ability to separate two closely spaced wavelengths, λ and λ + Δλ, defined as R = λ / Δλ. For a grating, R = nN, where N is the total number of illuminated lines (or slits). A higher number of lines and a higher diffraction order increase the resolving power.
角色散衡量最大值的角位置随波长的变化率。对光栅而言，由光栅方程微分得到 dθ/dλ = n / (d cosθ)。这意味着更高的级次 n 和更小的光栅常数 d 能获得更好的色散。分辨本领 R 是指区分两个非常接近的波长 λ 和 λ + Δλ 的能力，定义为 R = λ / Δλ。对于光栅，R = nN，其中 N 是被照射的总线条数（或狭缝数）。线条数越多，衍射级次越高，分辨本领越强。
7. Double-Slit Interference with Diffraction Effects | 双缝干涉与衍射效应
In an ideal Young’s double-slit experiment, the slits are treated as infinitely narrow, producing equally bright interference fringes. In reality, each slit has a finite width a, and its single-slit diffraction pattern modulates the interference fringes. The resulting intensity pattern is the product of the double-slit interference term and the single-slit diffraction envelope. The interference maxima still occur at d sinθ = mλ (m = 0, ±1, ±2, …) but their brightness is scaled by the single-slit intensity function. When a maximum of the interference coincides with a minimum of the diffraction envelope, that fringe vanishes — this is known as a missing order.
A missing order occurs when the path difference condition for an interference maximum, d sinθ = mλ, coincides with the condition for a diffraction minimum, a sinθ = nλ (n = ±1, ±2, …). Dividing the two equations gives the condition for a missing order: d / a = m / n, where m and n are integers. For example, if the slit separation d is twice the slit width a (d = 2a), then interference orders m = 2, 4, 6, … correspond to diffraction minima n = 1, 2, 3, …, so every even-order bright fringe disappears.
当干涉极大的路径差条件 d sinθ = mλ 与衍射极小的条件 a sinθ = nλ (n = ±1, ±2, …) 重合时，就出现缺级。将两式相除，可得缺级条件：d / a = m / n，其中 m 和 n 都是整数。例如，如果缝间距 d 是缝宽 a 的两倍（d = 2a），那么干涉级次 m = 2, 4, 6, … 对应衍射极小 n = 1, 2, 3, …，因此每个偶数级亮纹都会消失。
9. Diffraction by a Circular Aperture: the Airy Disk | 圆孔衍射与艾里斑
When light passes through a circular aperture of diameter D, the diffraction pattern is characterized by a central bright region called the Airy disk, surrounded by concentric dark and bright rings. The angular position of the first dark ring is given by sinθ ≈ 1.22λ / D. For small angles, θ ≈ 1.22λ / D. This result is extremely important in optical instruments because the finite size of the Airy disk limits the resolution of the system.
当光通过直径为 D 的圆孔时，衍射图样由一个称为艾里斑的中央亮区以及围绕它的同心明暗环组成。第一暗环的角位置由 sinθ ≈ 1.22λ / D 给出。对于小角度，θ ≈ 1.22λ / D。这一结果在光学仪器中极其重要，因为艾里斑的有限大小限制了系统的分辨率。
10. Rayleigh Criterion and Optical Resolution | 瑞利判据与光学分辨率
The Rayleigh criterion provides a standard for when two adjacent point sources can be resolved. For a circular aperture, two sources are said to be just resolvable when the central maximum of one diffraction pattern falls exactly on the first minimum of the other. The minimum resolvable angular separation is thus θR = 1.22λ / D. This principle applies to telescopes, microscopes, and even the human eye. To improve resolution, you can increase the aperture diameter D or use a shorter wavelength λ (e.g., electron microscopes).
瑞利判据提供了一个判断两个相邻点光源能否被分辨的标准。对于圆孔，当一个衍射图样的中央极大刚好落在另一个衍射图样的第一极小时，就说这两个点光源刚好能被分辨。因此最小可分辨角为 θR = 1.22λ / D。这一原理适用于望远镜、显微镜，甚至人眼。要提高分辨率，可以增大孔径直径 D 或使用更短的波长 λ（例如电子显微镜）。
11. Common Exam Pitfalls and Tips | 常见考试陷阱与技巧
Students often confuse the single-slit equation (a sinθ = mλ) with the grating equation (d sinθ = nλ). Remember that in the single-slit pattern, the central maximum is at m = 0 and is the brightest, whereas the grating zeroth order is simply the undiffracted beam. When sketching double-slit patterns with finite slit widths, always draw the diffraction envelope limiting the fringe heights. Also, check whether the question asks for angular separation or linear separation on a screen, and be ready to use the small-angle approximation sinθ ≈ tanθ ≈ θ (in radians) if the distance to the screen is large.
学生经常混淆单缝方程 (a sinθ = mλ) 与光栅方程 (d sinθ = nλ)。请记住，在单缝图样中，中央极大位于 m = 0 且最亮，而光栅零级只是未被衍射的光束。在绘制有限缝宽的双缝图样时，一定要画出限制条纹高度的衍射包络。此外，要检查题目所问的是角间距还是屏幕上的线性间距，如果距离屏幕很远，准备好使用小角度近似 sinθ ≈ tanθ ≈ θ（弧度制）。
12. Summary and Key Formulae | 总结与关键公式
Concept
Formula
Notes
Single-slit minima
a sinθ = mλ
m = ±1, ±2, …; central max at m=0
Grating maxima
d sinθ = nλ
n = 0, ±1, ±2, …
Double-slit (ideal)
d sinθ = mλ
m = 0, ±1, ±2, …
Circular aperture (1st min)
sinθ ≈ 1.22λ / D
D = diameter
Rayleigh criterion
θR = 1.22λ / D
Just
Published by TutorHao | IB Physics Revision Series | aleveler.com
📚 A-Level Physics Unit 3 Jan 2019 Past Paper: Concept Analysis | A-Level 物理 Unit 3 2019年1月真题概念解析
This article unpacks the core practical skills and concepts tested in the Edexcel A-Level Physics Unit 3 January 2019 past paper. Unit 3, ‘Practical Skills in Physics I’, assesses your ability to plan, analyse, and evaluate experiments. It does not require you to recall a specific required practical from memory, but rather to apply your understanding of measurement, uncertainty, graphing, and critical evaluation to unfamiliar scenarios. The Jan 2019 paper focused heavily on topics such as determining the resistivity of a wire, investigating the behaviour of a thermistor, and analysing motion using light gates. By dissecting the key ideas behind typical questions, you will strengthen your ability to tackle any Unit 3 paper with confidence.
本文深入解析了爱德思 A-Level 物理 Unit 3 2019年1月真题所考查的核心实验技能与概念。Unit 3 ‘物理实验技能 I’ 旨在评估你计划、分析和评估实验的能力。它不要求你死记硬背某个特定的规定实验，而是要求你将测量、不确定度、图表和批判性评价的理解应用到陌生的情境中。2019年1月的试卷重点考查了诸如测定导线电阻率、研究热敏电阻特性以及使用光闸分析运动等主题。通过剖析这些典型问题背后的关键思想，你将更有信心应对任何 Unit 3 试卷。
1. Understanding Absolute and Percentage Uncertainty | 理解绝对不确定度与百分不确定度
Every measurement in physics has an associated uncertainty. The absolute uncertainty for a single reading using a digital instrument is usually taken as ± the resolution. For an analogue scale, it is ± half the smallest scale division. If you take multiple repeat readings, the absolute uncertainty can be estimated as ± half the range of the repeats. For example, if you measure a length with a metre ruler (resolution 1 mm), the absolute uncertainty in a single reading is ±1 mm. If you measure the length of a wire as 0.845 m, you might record it as (0.845 ± 0.001) m. Percentage uncertainty is calculated as (absolute uncertainty / measured value) × 100%. This concept was crucial in the Jan 2019 paper where you needed to combine uncertainties from length, diameter, resistance, and voltage to find the uncertainty in resistivity.
2. Reading Vernier Calipers and Micrometer Screw Gauges | 游标卡尺与螺旋测微器的读数
The Jan 2019 paper included questions requiring you to read a micrometer screw gauge and a vernier caliper accurately. For a micrometer, the main scale typically reads to 0.5 mm, and the rotating thimble gives an additional 0.01 mm precision. You must add the main scale reading to the thimble reading, being careful not to double-count the 0.5 mm divisions. A common error is misreading the half-millimetre mark; if the thimble edge has passed the half-mark but the mark is still partially visible, check if the thimble reading is close to 0 or 50. For vernier calipers, the main scale gives millimetres, and the vernier scale aligns at the tenths of a millimetre. A typical reading might be 3.46 cm, where the vernier mark ‘6’ coincides perfectly with a main scale mark. Always record the reading with the correct precision and zero error correction.
3. Distinguishing Precision, Accuracy, and Sensitivity | 区分精密度、准确度和灵敏度
Many candidates confuse these terms, and the Jan 2019 paper tested this distinction in the evaluation sections. Precision relates to the spread of repeated measurements; a set of readings with very little variation is precise, even if they are not close to the true value. Accuracy describes how close a measurement is to the accepted or true value. A precise but inaccurate result suggests a systematic error. Sensitivity is the smallest change in the quantity being measured that an instrument can detect. For example, a thermistor circuit can be made more sensitive by increasing the voltage of the power supply or using a more sensitive ammeter. In the thermistor investigation, you may have been asked how to increase the sensitivity of the temperature measurement.
4. Plotting Graphs and Drawing Error Bars | 绘制图表与画出误差棒
In the resistivity question, you were required to plot a graph of resistance R against length L, and then use the gradient to find resistivity. The Jan 2019 paper expected you to choose sensible scales that use more than half the graph paper, label axes with quantities and units, and plot points accurately. Error bars represent the absolute uncertainty in each measurement. On the resistance vs length graph, horizontal error bars represent the uncertainty in length (±1 mm or half the range), while vertical error bars represent the uncertainty in resistance (± the resolution of the ohmmeter or half the range). The line of best fit should pass through as many error bars as possible. You also need to draw the worst acceptable line (steepest or shallowest) to determine the uncertainty in the gradient.
在电阻率题目中，你需要绘制电阻 R 与长度 L 的关系图，然后用斜率求电阻率。2019年1月的试卷希望你选择能利用一半以上坐标纸的合理刻度，用物理量和单位标注坐标轴，并准确描点。误差棒代表每个测量值的绝对不确定度。在电阻与长度的关系图中，水平误差棒表示长度不确定度（±1 mm 或半极差），垂直误差棒表示电阻不确定度（± 欧姆表分辨率或半极差）。最佳拟合线应尽可能多地穿过误差棒。你还需要画出最差可接受直线（最陡或最平缓）以确定斜率的不确定度。
5. Determining Gradient and Its Uncertainty | 确定斜率及其不确定度
The gradient is calculated from the line of best fit using a large triangle whose vertices are on the line. Do not use plotted data points. The formula is gradient = (y₂ − y₁) / (x₂ − x₁). In the Jan 2019 resistivity experiment, the gradient of the R vs L graph gives R/L, and resistivity ρ = gradient × cross-sectional area A. The uncertainty in the gradient is found by drawing the steepest and shallowest lines that still pass through the error bars. Then, Δgradient = (gradient_steepest − gradient_shallowest) / 2. The percentage uncertainty in the gradient is then (Δgradient / gradient) × 100%. This uncertainty is combined with the uncertainty in the wire’s cross-sectional area to find the overall uncertainty in resistivity.
6. Calculating Cross-Sectional Area and Propagating Uncertainties | 计算横截面积与不确定度传递
The wire’s diameter d was measured using a micrometer, yielding a value such as (0.274 ± 0.001) mm. The cross-sectional area A = πd²/4. Calculation of A must account for the square of the diameter. The percentage uncertainty in d is (%Ud). Since A depends on d², the percentage uncertainty in A is 2 × %Ud. For instance, if d = 0.274 mm, %Ud = (0.001/0.274) × 100% ≈ 0.36%, so %UA = 0.73%. This propagation rule appears regularly. Then, the percentage uncertainty in resistivity %Uρ = %U_gradient + %UA. The absolute uncertainty in ρ is then (%Uρ / 100) × ρ. In the Jan 2019 paper, you had to compare your experimental resistivity value with the accepted value for the wire material (e.g., constantan) using percentage difference.
导线的直径 d 使用螺旋测微器测量，得到如 (0.274 ± 0.001) mm 的值。横截面积 A = πd²/4。计算 A 时必须考虑直径的平方。d 的百分不确定度为 (%Ud)。由于 A 取决于 d²，A 的百分不确定度为 2 × %Ud。例如，若 d = 0.274 mm，%Ud = (0.001/0.274) × 100% ≈ 0.36%，因此 %UA = 0.73%。这一传递规则经常出现。然后，电阻率的百分不确定度 %Uρ = %U_斜率 + %UA。ρ 的绝对不确定度为 (%Uρ / 100) × ρ。在2019年1月的试卷中，你需要利用百分差将实验电阻率值与导线材料（如康铜）的公认值进行比较。
7. Calculating Percentage Difference and Assessing Accuracy | 计算百分差与评估准确度
Percentage difference is used to judge how close your experimental result is to the known value. It is defined as |(experimental value − accepted value)| / accepted value × 100%. If this percentage difference is less than your calculated experimental percentage uncertainty, the result is considered accurate within the limits of the experiment, because the accepted value lies within the error bars. If the percentage difference is much larger, systematic errors or mistakes are likely present. In the Jan 2019 resistivity practical, a percentage difference of, say, 5% compared with a total uncertainty of 8% would indicate good agreement.
8. Evaluating Experimental Procedures and Identifying Limitations | 评估实验步骤与识别局限性
The Jan 2019 paper asked you to evaluate the method used to determine resistivity. Common limitations include: the wire may not have a uniform cross-section along its length; kinks in the wire can affect the measured length; the resistance of connecting leads and contact resistance at crocodile clips introduce a systematic error; temperature rise due to current heating may change resistance. These factors cause the experimental resistivity to be unreliable. When evaluating, you must link the limitation to the specific measurement it affects. For instance, if the wire is not straight when measuring its length, the recorded length is greater than the true length, leading to a systematic overestimation of resistivity.
9. Suggesting Improvements and Controlling Variables | 提出改进措施与控制变量
For each limitation identified, you must propose a realistic and practical improvement. To ensure uniform cross-section, you could measure the diameter at several orientations along the wire and calculate an average diameter. To eliminate zero error on the micrometer, check the reading when fully closed and subtract this value. To minimise heating, use a low current or switch the circuit on only when taking readings. Contact resistance can be reduced by using soldered connections or cleaning the wire ends with emery paper. You should also mention how to better control other variables: keep room temperature constant, support the wire so it does not sag, and use large distances between measuring points to reduce length uncertainty. These suggestions were core to the evaluation questions in the Jan 2019 paper.
10. Using Light Gates and Timing to Determine Velocity | 使用光闸与计时来确定速度
Another context in the Jan 2019 paper involved a dynamics experiment where a glider passed through light gates on an air track. A light gate measures the time for an interrupting card of known length to break the beam. The instantaneous velocity is calculated using v = d / t, where d is the length of the card and t is the interruption time. To reduce uncertainty, you should use a longer card to increase the interruption time, but not so long that the velocity changes significantly during the interruption. The most accurate velocity is obtained when the light gate is positioned at the intended measurement point. Multiple readings help identify random errors.
2019年1月试卷的另一个情境涉及一个动力学实验，其中滑块在气垫导轨上通过光闸。光闸测量一个已知长度的遮光片阻断光束的时间。瞬时速度使用 v = d / t 计算，其中 d 为遮光片长度，t 为阻断时间。为了减小不确定度，应使用较长的遮光片以延长阻断时间，但不应过长以至于在阻断期间速度发生显著变化。将光闸放置在预定的测量点上可获得最准确的速度。多次读数有助于识别随机误差。
11. Investigating Thermistor Characteristics and Logarithmic Analysis | 研究热敏电阻特性与对数分析
In the thermistor question, you likely measured resistance R at different temperatures T and investigated the relationship R = R₀ e^(B/T). Taking natural logarithms gives ln R = ln R₀ + B(1/T). Thus, a graph of ln R against 1/T produces a straight line whose gradient equals B and intercept equals ln R₀. You must convert temperature to Kelvin (K = °C + 273.15) and calculate 1/T. The Jan 2019 paper tested your ability to plot such a graph with correct labelling and to extract the constant B. The uncertainty in ln R is computed as Δ(ln R) = ΔR / R, which follows from calculus. This allows you to add error bars to the log plot.
在热敏电阻题目中，你很可能在不同温度 T 下测量了电阻 R，并研究了关系式 R = R₀ e^(B/T)。取自然对数得到 ln R = ln R₀ + B(1/T)。因此，绘制 ln R 与 1/T 的关系图会得到一条直线，其斜率等于 B，截距等于 ln R₀。你必须将温度转换为开尔文 (K = °C + 273.15) 并计算 1/T。2019年1月的试卷考查了你绘制这种具有正确定标的图表并提取常数 B 的能力。ln R 的不确定度可根据 Δ(ln R) = ΔR / R 计算得出，这源自微积分。这使你能够在双对数图上添加误差棒。
12. Drawing Valid Conclusions and Comparing Results | 得出有效结论与比较结果
When writing a conclusion, state your experimental result and its uncertainty, then compare with the accepted value using percentage difference. If the two values overlap within the combined uncertainties, you can claim that the experiment supports the theoretical relationship. If not, identify the dominant factor causing the discrepancy, such as a systematic error that was not eliminated. In the Jan 2019 evaluation, you might conclude that the resistivity of the wire is similar to constantan within experimental errors, or that the value of B for the thermistor matches the manufacturer’s data. Never forget to cite evidence from your graph or calculations to back up your statement.
撰写结论时，陈述你的实验值及其不确定度，然后使用百分差与公认值进行比较。如果两个值在组合不确定度范围内重叠，你可以宣称实验支持理论关系。如果不重叠，则找出导致差异的主要因素，例如一个未能消除的系统误差。在2019年1月的评估中，你可能会得出导线电阻率在实验误差范围内与康铜相似，或者热敏电阻的 B 值与制造商数据匹配。永远不要忘记引用图表或计算中的证据来支持你的陈述。
Published by TutorHao | Physics Revision Series | aleveler.com
📚 IGCSE AQA Physics: Work and Energy | IGCSE AQA 物理：功与能量考点精讲
In IGCSE AQA Physics, the concepts of work and energy are fundamental to understanding how forces cause changes in motion and how energy is transferred and conserved. This revision guide covers the key definitions, formulas, units, and common exam applications, from work done by a force to kinetic and gravitational potential energy, power, and efficiency.
Work is done whenever a force causes an object to move through a distance. For work to be done, the force must have a component in the direction of the displacement. If the force is perpendicular to the movement, no work is done – for example, when holding a heavy bag still, you may feel tired but no mechanical work is done on the bag because there is no movement in the direction of the force.
The amount of work done is calculated using the equation: work done = force × distance moved in the direction of the force. This can be written as W = F d, where W is work done in joules (J), F is the force in newtons (N), and d is the distance in metres (m).
做功的大小用以下公式计算：功 = 力 × 沿力方向移动的距离。写作 W = F d，其中 W 是功，单位为焦耳 (J)；F 是力，单位为牛顿 (N)；d 是距离，单位为米 (m)。
W = F d
One joule is defined as the work done when a force of one newton moves an object one metre in the direction of the force.
一焦耳定义为：一牛顿的力使物体沿力的方向移动一米所做的功。
3. Units of Work and Energy | 功和能量的单位
Work and all forms of energy are measured in joules (J). This is the SI unit for energy. Sometimes kilojoules (kJ) are used, where 1 kJ = 1000 J. The joule is named after James Prescott Joule, who demonstrated the equivalence of mechanical work and heat energy.
Gravitational potential energy (GPE) is the energy an object possesses due to its height above the ground. The change in GPE when an object is lifted is equal to the work done against gravity. The formula is ΔEₚ = m g h, where m is mass (kg), g is gravitational field strength (9.8 N/kg on Earth, often rounded to 10 N/kg in IGCSE problems), and h is the change in height (m).
重力势能 (GPE) 是物体因离地高度而具有的能量。物体被抬高时重力势能的变化等于克服重力所做的功。公式为 ΔEₚ = m g h，其中 m 为质量 (kg)，g 为重力场强度（地球表面为 9.8 N/kg，IGCSE 题目中通常取 10 N/kg），h 为高度变化 (m)。
ΔEₚ = m g h
When an object falls, it loses GPE and gains kinetic energy, assuming no air resistance. The equation also assumes the gravitational field is uniform.
物体下落时，若无空气阻力，重力势能减少，动能增加。此公式也假设重力场是均匀的。
5. Kinetic Energy | 动能
Kinetic energy is the energy of a moving object. Any object with mass m and speed v has kinetic energy given by the equation Eₖ = ½ m v². The unit is joules (J). Doubling the mass doubles the kinetic energy; doubling the speed quadruples the kinetic energy because it depends on v².
动能是运动物体的能量。任何质量为 m、速度为 v 的物体都具有动能，公式为 Eₖ = ½ m v²，单位为焦耳 (J)。质量加倍，动能加倍；速度加倍，动能变为原来的四倍，因为它取决于 v²。
Eₖ = ½ m v²
This relationship is frequently tested: if the speed of a car increases from 10 m/s to 20 m/s, its kinetic energy increases by a factor of 4, provided mass stays the same.
这个关系经常被考查：如果汽车速度从 10 m/s 增加到 20 m/s，在质量不变的情况下，其动能增加为原来的 4 倍。
6. Conservation of Energy | 能量守恒
The principle of conservation of energy states that energy cannot be created or destroyed; it can only be transferred from one form to another or moved from one place to another. In a closed system, the total energy is constant. For a falling object (ignoring air resistance), the decrease in GPE equals the increase in kinetic energy.
In real situations, some energy is always transferred as heat due to friction or air resistance, but the total energy is still conserved.
在实际情况下，由于摩擦或空气阻力，总有一部分能量转化为热能，但总能量仍然守恒。
7. Work-Energy Principle | 功能原理
The work-energy principle states that the net work done on an object is equal to the change in its kinetic energy. If a resultant force does work on an object, it speeds up; if the object does work against a force (e.g., braking), it slows down. This links the concepts of force, distance, and energy change directly.
For example, to find the braking force needed to stop a car, you can calculate the car’s initial kinetic energy and set work done = force × braking distance.
Power is the rate at which work is done or energy is transferred. The formula is P = W / t or P = ΔE / t, where P is power in watts (W), W is work done or energy transferred in joules (J), and t is time in seconds (s). One watt is one joule per second.
功率是做功或能量转化的速率。公式为 P = W / t 或 P = ΔE / t，其中 P 是功率，单位为瓦特 (W)；W 是功或能量转化量，单位为焦耳 (J)；t 是时间，单位为秒 (s)。一瓦特等于每秒一焦耳。
P = W / t
Another useful form combines force and velocity: when a constant force moves an object at a steady speed, power can be calculated as P = F v, where v is the velocity in m/s. This is derived from P = (F d) / t = F v.
另一个实用形式结合了力和速度：当恒定力使物体匀速运动时，功率可以用 P = F v 计算，其中 v 是速度 (m/s)。这是由 P = (F d) / t = F v 推导而来的。
9. Efficiency | 效率
Efficiency measures how much of the energy input is converted into useful output energy. It is given by the ratio: efficiency = (useful energy output ÷ total energy input) × 100%. Efficiency can also be expressed in terms of power: efficiency = (useful power output ÷ total power input) × 100%.
No real machine is 100% efficient; some energy is always wasted, usually as heat due to friction. IGCSE problems may ask you to identify wasted energy and calculate efficiency from given data.
When an object slides across a surface, work is done against friction. This work converts kinetic energy (or some input energy) into thermal energy, raising the temperature of the surfaces. The work done against friction is equal to the frictional force multiplied by the distance moved: W_friction = F_friction × d.
Understanding this is important for braking distance and energy dissipation questions. The kinetic energy lost is transferred to heat in the brakes and tyres.
理解这一点对于刹车距离和能量耗散问题非常重要。损失的动能转化为刹车和轮胎中的热量。
11. Summary and Common Mistakes | 总结与常见错误
Key points to remember: work requires both force and movement in the same direction; energy is always conserved; GPE = mgh, KE = ½mv²; power is rate of energy transfer. Common mistakes include forgetting to square the velocity in the KE equation, confusing mass and weight, using cm or km instead of metres, and forgetting to convert to joules. In efficiency calculations, always check whether you need a decimal or a percentage, and ensure the useful output never exceeds the input.
📚 Electric Fields for GCSE CCEA Physics | GCSE CCEA 物理：电场 考点精讲
Electric fields are a core topic in the GCSE CCEA Physics specification. Understanding how charges interact without direct contact, how field lines represent forces, and how potential difference links to energy is essential for success in the exam. This guide breaks down every key concept, from atomic charge to practical applications, with clear explanations and paired English–Chinese summaries to support revision.
All matter is made of atoms. An atom contains a positively charged nucleus, consisting of protons and neutrons, surrounded by negatively charged electrons. Protons carry a charge of +1.6 × 10⁻¹⁹ C, electrons carry −1.6 × 10⁻¹⁹ C, and neutrons have zero charge. An object becomes charged when it gains or loses electrons.
所有物质都由原子组成。原子包含带正电的原子核（由质子和中子组成），核外有带负电的电子。质子带 +1.6 × 10⁻¹⁹ C 的电荷，电子带 −1.6 × 10⁻¹⁹ C，中子不带电。物体获得或失去电子后就会带电。
The unit of electric charge is the coulomb (C). In GCSE problems, you will often use the fact that like charges repel and unlike charges attract. This fundamental rule explains the behaviour of charged objects in electric fields.
An electric field is a region around a charged object where a force is exerted on another charged object. The field extends in all directions and gets weaker with distance. Even though the charges are not touching, the field transmits the force. This is an example of a non-contact force.
In CCEA exams, you must be able to state that an electric field is created by any charged object. The direction of the field is defined as the direction a positive test charge would move if placed in the field.
Field lines represent the strength and direction of an electric field. The lines always start on positive charges and end on negative charges. The closer the lines are together, the stronger the field.
电场线表示电场的强度和方向。电场线总是从正电荷出发，终止于负电荷。线越密，电场越强。
For an isolated positive point charge, field lines radiate outward in straight lines. For an isolated negative point charge, field lines point inward. Between two oppositely charged parallel plates, the field lines are parallel and evenly spaced, showing a uniform field.
Key exam tip: arrows on field lines show the direction of force on a positive charge. Never draw field lines crossing each other.
考试重要提示：电场线上的箭头表示正电荷受力的方向。绝对不要画出交叉的电场线。
4. Uniform and Non-Uniform Fields | 匀强电场与非匀强电场
A uniform electric field has the same strength and direction at every point. This occurs between two parallel plates with opposite charges. A non-uniform field, such as that around a point charge, varies in strength and direction with position.
In a uniform field, the force on a charge is constant. In a non-uniform field, the force changes as the charge moves. CCEA often asks students to sketch and compare these two types of fields.
Electric field strength (E) is defined as the force (F) per unit charge (Q). The formula is:
E = F / Q
电场强度 (E) 定义为单位电荷所受的力 (F)。公式为：
E = F / Q
E is measured in newtons per coulomb (N/C). If a charge of 3 C experiences a force of 6 N, the field strength is 2 N/C. This formula is especially useful for calculations involving uniform fields between parallel plates.
E 的单位是牛每库仑 (N/C)。如果一个 3 C 的电荷受到 6 N 的力，则电场强度为 2 N/C。这个公式在涉及平行板间匀强电场的计算中特别有用。
Note: In a uniform field, E can also be expressed as E = V / d, where V is the potential difference between the plates and d is the separation. You will be expected to use both forms.
注意：在匀强电场中，E 也可表示为 E = V / d，其中 V 是两板间的电势差，d 是板间距离。需要掌握这两种表达形式。
6. Electric Potential Difference | 电势差
Electric potential difference (V) is the work done per unit charge in moving a charge between two points in an electric field. The relation is:
V = W / Q
电势差 (V) 是将单位电荷从电场中一点移动到另一点所做的功。关系式为：
V = W / Q
Here W is the work done in joules (J), Q is the charge in coulombs (C), and V is in volts (V). 1 volt means 1 joule of work is done per coulomb of charge moved.
When a charge moves in the direction of the electric field, work is done by the field and the charge loses potential energy. Against the field, work must be done on the charge, increasing its potential energy.
7. Static Electricity and Charging by Friction | 静电与摩擦起电
Static electricity is the build-up of electric charge on the surface of an insulator. When two insulating materials are rubbed together, electrons are transferred from one material to the other. The material that gains electrons becomes negatively charged; the one that loses electrons becomes positively charged.
Examples include rubbing a polythene rod with a cloth (rod becomes negative, cloth positive) and rubbing a perspex rod with a cloth (rod becomes positive, cloth negative). CCEA exam questions often describe these practical demonstrations.
Charged objects can attract small neutral objects, like pieces of paper, due to induced charges on the neutral object’s surface.
带电物体能吸引轻小不带电物体（如纸屑），这是因为中性物体表面产生感应电荷。
8. Dangers and Uses of Static Electricity | 静电的危害与应用
Static electricity can be hazardous. A spark from a charged object can ignite flammable gases or vapours, causing explosions. This is why tankers carrying fuel must be earthed during transfer, and why hospitals use anti-static flooring.
On the other hand, static electricity has many useful applications. Electrostatic precipitators in chimneys remove smoke particles; photocopiers and laser printers use static charges to transfer toner onto paper; spray painting uses charged paint droplets for even coverage.
You should be able to describe at least one use and one danger in detail for the exam.
在考试中，应能详细描述至少一种应用和一种危害。
9. Electric Fields and Conductors | 电场与导体
In a conductor, electric charges (electrons) are free to move. When a charged object is brought near a conductor, the free electrons move within the conductor, causing one end to become negatively induced and the other positively induced. This is electrostatic induction.
If the conductor is then earthed, charges can flow to or from the earth, leaving the conductor with a net charge. This process is used to charge objects by induction without direct contact.
For a conductor placed inside an external electric field, the field inside the conductor becomes zero once electrostatic equilibrium is reached. This is why metal cages (Faraday cages) can shield against electric fields.
10. Lightning and Atmospheric Electricity | 闪电与大气静电
Lightning is a dramatic natural example of electric fields. In thunderclouds, ice crystal collisions separate charges: positive charges accumulate at the top of the cloud, negatives at the base. The intense electric field between the cloud and the ground or between cloud regions leads to a rapid discharge – a lightning strike.
A lightning conductor provides a safe path for the charge to earth, protecting tall buildings. This is an application of the attraction of charges and the movement of charge through a conductor.
避雷针为电荷流入大地提供安全通道，保护高大建筑物。这是电荷吸引和电荷通过导体移动原理的应用。
Exam questions may link static electricity to everyday phenomena, so be prepared to explain lightning using charge separation and electric field action.
考试题可能将静电与日常现象联系起来，要做好用电荷分离和电场作用解释闪电的准备。
11. Summary of Key Equations | 关键公式总结
Memorise these equations for your CCEA exam:
Electric field strength: E = F / Q (N/C)
电场强度：E = F / Q (N/C)
Potential difference (energy): V = W / Q (V or J/C)
电势差（能量）：V = W / Q (V 或 J/C)
Uniform field between plates: E = V / d (N/C or V/m)
平行板间匀强电场：E = V / d (N/C 或 V/m)
Be careful with units: force in newtons, charge in coulombs, energy in joules, distance in metres, potential difference in volts. Show all steps in calculations to gain full marks.
When drawing field lines, always start from positive and end on negative with arrows. Do not forget to label arrows. In uniform fields, ensure lines are straight, parallel, and equally spaced.
Do not confuse electric field strength E with energy E. Use context and unit checks. Also, remember that a charge moving in the direction of the field loses potential energy – in exam, link to V = W / Q correctly.
不要混淆电场强度 E 与能量 E。根据上下文和单位加以区分。还要记住，电荷沿电场方向移动时电势能减少——考试时要正确关联 V = W / Q。
Finally, practice calculations that combine E = F / Q and E = V / d, as CCEA often asks students to derive one from the other for a given uniform field scenario.
最后，多练习结合 E = F / Q 和 E = V / d 的计算题，CCEA 常要求考生在给定的匀强电场情境下，由一个公式推导出另一个。
Published by TutorHao | GCSE CCEA Physics Revision Series | aleveler.com
📚 Essential Physics for Cambridge IGCSE: Key Formula Derivations | 剑桥IGCSE物理核心公式推导
Mastering IGCSE Physics is not just about memorising equations — it’s about understanding where they come from. In this article, we systematically derive the most essential formulas you will encounter in the Cambridge IGCSE syllabus, using clear logical steps and fundamental principles. By seeing how each relationship is built, you’ll gain deeper insight and the confidence to tackle both calculation and explanation questions in your exams.
1. Deriving the SUVAT Equations from a Velocity–Time Graph | 从速度-时间图推导运动学方程
Consider an object moving with uniform acceleration. Its velocity–time graph is a straight line with initial velocity u, final velocity v, and time taken t. The gradient of the line is acceleration a, so we can immediately write a = (v – u) / t, which rearranges to:
考虑一个匀加速运动的物体。其速度-时间图像是一条直线，初速度为u，末速度为v，用时为t。该直线的斜率即为加速度a，因此可直接写出a = (v – u) / t，整理得：
v = u + at
The total displacement s is the area under the graph. This area can be split into a rectangle of area ut and a triangle of area ½ (v – u)t. Using v – u = at, the total area becomes s = ut + ½ at².
总位移s等于图像下的面积。该面积可分为一个面积为ut的矩形和一个面积为½ (v – u)t的三角形。将v – u = at代入，总面积变为s = ut + ½ at²。
s = ut + ½ at²
We can also express displacement using the average velocity. For uniform acceleration, average velocity = (u+v)/2, so s = (u+v)t/2. Combining this with v = u + at to eliminate t yields the third equation:
我们还可以用平均速度表示位移。对于匀加速运动，平均速度 = (u+v)/2，因此s = (u+v)t/2。将此式与v = u + at联立消去t，即得到第三个方程：
v² = u² + 2as
2. Newton’s Second Law as F = ma | 牛顿第二定律 F = ma 的推导
Newton’s second law states that the net force on an object is proportional to the rate of change of its momentum. Momentum p = mv, so for constant mass, force F = (mv – mu) / t = m(v – u) / t. Since (v – u)/t = a, we obtain the familiar form:
This derivation emphasises that force is directly linked to acceleration, not just motion, and it is the foundation for almost all mechanics calculations in the IGCSE syllabus.
这一推导强调了力与加速度直接相关，而不仅仅是运动，它是IGCSE课程中几乎所有力学计算的基础。
3. Work Done and the Kinetic Energy Formula | 功与动能公式的推导
Work done by a constant force over a displacement s is W = F s. If this force accelerates a mass from rest, we can substitute F = ma and v² = 2as (since u = 0), giving a = v²/(2s). Thus W = m × v²/(2s) × s = ½ m v². This energy transferred is stored as kinetic energy:
恒力作用一段位移s所做的功为W = F s。如果这个力使一个物体从静止开始加速，代入F = ma以及v² = 2as（因为u=0），可得a = v²/(2s)。因此W = m × v²/(2s) × s = ½ m v²。这部分转移的能量即储存为动能：
KE = ½ m v²
If the object already had initial speed u, the same method using v² = u² + 2as gives the net work done as the change in kinetic energy: W = ½ m (v² – u²).
若物体已有初速度u，使用v² = u² + 2as同样可推导出净功等于动能的变化量：W = ½ m (v² – u²)。
4. Gravitational Potential Energy near Earth’s Surface | 地球表面附近的重力势能
When an object of weight mg is lifted through a vertical height Δh, the work done against gravity is W = force × distance = mg Δh. This work is stored as gravitational potential energy. Assuming the field is uniform, we define:
IGCSE questions often ask you to equate GPE gained to kinetic energy lost in a falling object, leading to mgh = ½ m v², which allows calculation of impact speed.
IGCSE题目经常要求你将物体下落时获得的重力势能与失去的动能等同起来，即mgh = ½ m v²，从而计算撞击速度。
5. Density and Pressure in Fluids | 密度与液体压强
Density ρ is defined as mass per unit volume: ρ = m / V. To derive fluid pressure at depth h, consider a column of liquid of cross-sectional area A. The weight of the column is mg = (ρ × A × h) g. This weight acts on area A, producing pressure:
密度ρ定义为单位体积的质量：ρ = m / V。为推导深度h处的液体压强，考虑一截面积为A的液柱。液柱的重量为mg = (ρ × A × h) g。该重量作用在面积A上，产生的压强为：
P = F / A = ρgh
The total pressure at a point also includes atmospheric pressure on the surface, but the formula above gives the pressure due to the liquid alone.
某点的总压强还需加上液面的大气压，但上述公式给出了仅由液体产生的压强。
6. Ohm’s Law and Defining Resistance | 欧姆定律与电阻定义
For many conductors at constant temperature, the current I is directly proportional to the potential difference V across it. The constant of proportionality is the resistance R, defined as R = V / I. While Ohm’s law is an experimental observation, rearranging gives the universally applicable definition:
对于许多恒温下的导体，电流I与其两端的电势差V成正比。比例常数即为电阻R，定义为R = V / I。尽管欧姆定律是实验观察结果，但整理后可得到普遍适用的定义：
V = I R
Resistance can also be expressed in terms of physical properties: R = ρ L / A, where ρ is resistivity, L length, and A cross-sectional area. This formula is derived from the idea that resistance increases with length and decreases with area.
电阻还可用物理特性表示为：R = ρ L / A，其中ρ为电阻率，L为长度，A为横截面积。该公式的思路是电阻随长度增加而增大，随截面积增大而减小。
7. Resistors in Series and Parallel | 串联和并联电阻的公式推导
For resistors in series, the same current flows through each. The total p.d. is the sum of individual p.d.s: V_total = V₁ + V₂ + V₃. Applying V = IR to each gives I R_total = I R₁ + I R₂ + I R₃, so:
串联电阻中流过相同的电流。总电压等于各电阻电压之和：V_total = V₁ + V₂ + V₃。对每个电阻应用V = IR得I R_total = I R₁ + I R₂ + I R₃，因此：
R_total = R₁ + R₂ + R₃
For parallel resistors, the p.d. across each branch is the same. The total current is the sum of branch currents: I_total = I₁ + I₂ + I₃. Using I = V/R, we have V / R_total = V/R₁ + V/R₂ + V/R₃, which simplifies to:
These derivations are simple but powerful in circuit analysis.
这些推导虽简单，但在电路分析中十分有力。
8. Electrical Power and Energy Dissipated | 电功率与消耗的电能
Power is the rate of doing work: P = W / t. In an electrical component, the work done to move charge Q through a p.d. V is W = V Q. Current I is charge per unit time: I = Q / t, so Q = I t. Substituting gives:
功率是做功的速率：P = W / t。在电器元件中，移送电荷Q通过电势差V所做的功为W = V Q。电流I是单位时间流过的电荷：I = Q / t，故Q = I t。代入得：
P = V I
Using Ohm’s law, we can replace V or I to obtain two more useful forms:
利用欧姆定律，可替换V或I得到另两种有用形式：
P = I² R and P = V² / R
These are essential when analysing heating effects in resistors.
这些公式在分析电阻的热效应时至关重要。
9. The Transformer Equation | 变压器公式
An ideal transformer operates on the principle of electromagnetic induction. The alternating flux links both primary and secondary coils, inducing e.m.f.s proportional to their respective turns. For 100% efficiency, input power equals output power: V_P × I_P = V_S × I_S. Moreover, the voltage ratio equals the turns ratio:
This is derived from Faraday’s law: the induced e.m.f. in a coil is proportional to the rate of change of flux linkage, and with the same flux linking both coils, the ratio holds.
A wave’s speed, frequency and wavelength are connected by a simple relationship. Frequency f is the number of complete waves passing a point per second; wavelength λ is the distance travelled by one complete wave. In one second, f waves pass a fixed point, so the distance covered by the wave front is f × λ. Therefore, the speed is:
This equation applies to all waves — sound, water, light — and is fundamental in understanding phenomena like refraction.
该方程适用于所有波——声波、水波、光波——是理解折射等现象的基础。
11. Efficiency of Energy Transfers | 能量传递的效率
No device is perfect; some energy is always wasted as heat or sound. Efficiency measures how much of the input energy is converted to useful output. Expressed either in terms of energy or power:
Efficiency = (Useful energy output / Total energy input) × 100%
Efficiency = (Useful power output / Total power input) × 100%
This simple ratio is derived directly from the principle of conservation of energy and is vital for evaluating real-world systems like motors and lamps.
这个简单的比值直接来自能量守恒原理，对于评估电动机和灯泡等实际系统至关重要。
12. Hooke’s Law and the Spring Constant | 胡克定律与弹簧常数
For many elastic materials, the extension x is directly proportional to the applied force F, provided the elastic limit is not exceeded. This proportionality is Hooke’s law, written as:
对于许多弹性材料，只要未超过弹性极限，伸长量x与施加的力F成正比。这一比例关系即为胡克定律，写作：
F = k x
The constant k is the spring constant (stiffness). While this is an experimental law, the energy stored in a stretched spring can be derived from the area under the force–extension graph (a triangle), giving Elastic potential energy = ½ F x = ½ k x². This derivation mirrors the kinetic energy approach.
常数k为弹簧常数（劲度系数）。虽然这是一条实验定律，但通过力-伸长图下方三角形的面积可推导出弹簧储存的能量：弹性势能 = ½ F x = ½ k x²。这一推导与动能公式的方法类似。
Published by TutorHao | Physics Revision Series | aleveler.com
Particle physics in the IGCSE WJEC Physics syllabus focuses on the fundamental constituents of atoms, the nature of radioactivity, and how nuclear processes can be described, measured and applied. Mastering these ideas is essential not only for the exam but also for understanding the physical world at its smallest scale. This article walks you through every key point, from atomic structure and isotopes to fission, fusion and radiation safety.
An atom consists of a small, dense nucleus surrounded by electrons in shells. The nucleus contains protons and neutrons (collectively called nucleons). Protons carry a positive charge, neutrons are neutral, and electrons carry a negative charge. Almost all the mass of the atom is concentrated in the nucleus, while the electrons occupy most of the volume.
The relative masses and charges of these subatomic particles are fundamental data you must memorise. The proton and neutron each have a relative mass of 1, while the electron has a relative mass of approximately 1/1836 (often taken as negligible). In terms of charge, the proton has a relative charge of +1, the electron –1, and the neutron 0.
The atomic number (Z) is the number of protons in the nucleus. It defines the element. The mass number (A) is the total number of protons and neutrons (nucleons) in the nucleus. An atom is represented as ᴬZX, where X is the chemical symbol.
原子序数（Z）是原子核中的质子数，它决定了元素种类。质量数（A）是原子核中质子和中子（核子）的总数。一个原子通常表示为 ᴬZX，其中 X 是元素符号。
For a neutral atom, the number of electrons equals the number of protons. The number of neutrons is given by A – Z. In WJEC IGCSE, you will frequently need to calculate neutron numbers from given data or identify an isotope.
对于中性原子，电子数等于质子数。中子数的计算为 A – Z。在 WJEC IGCSE 考试中，你经常需要根据给定的数据计算中子数，或者识别同位素。
3. Isotopes | 同位素
Isotopes are atoms of the same element (same atomic number Z) but with different mass numbers (A), meaning they have different numbers of neutrons. For example, carbon-12 (¹²₆C) and carbon-14 (¹⁴₆C) are isotopes. They share identical chemical properties because their electron arrangements are the same, but their physical properties such as mass and nuclear stability may differ.
同位素是同一元素（原子序数 Z 相同）但质量数（A）不同的原子，即它们的中子数不同。例如，碳-12（¹²₆C）和碳-14（¹⁴₆C）是同位素。由于它们的电子排布相同，因此具有完全相同的化学性质，但质量、核稳定性等物理性质可能不同。
Some isotopes are stable, while others are radioactive (radioisotopes). Radioisotopes undergo decay to become more stable, emitting radiation in the process. Understanding isotopes is essential for interpreting nuclear equations and applications like carbon dating.
Radioactive decay is the spontaneous, random emission of radiation from an unstable nucleus. The nucleus may emit an alpha particle, a beta particle, or gamma radiation. Decay is not affected by external conditions such as temperature or pressure. The process changes the nucleus into a different nuclide, often of a different element.
WJEC requires you to describe the nature, penetration and ionising ability of the three types of radiation. Alpha radiation consists of helium nuclei (₂⁴He), beta radiation is high-speed electrons (₋₁⁰e), and gamma radiation is an electromagnetic wave of very short wavelength. All three originate from the nucleus.
5. Properties of Alpha, Beta and Gamma | α、β 和 γ 辐射的性质
Alpha particles are highly ionising but have low penetrating power; they can be stopped by a sheet of paper or a few centimetres of air. Beta particles are moderately ionising and can travel through air for about one metre; they are stopped by a few millimetres of aluminium. Gamma rays are weakly ionising but extremely penetrating; thick lead or several metres of concrete are needed to reduce them significantly.
In electric and magnetic fields, alpha (positive) and beta (negative) particles are deflected in opposite directions; gamma rays are not deflected. Alpha particles show the least deflection because of their larger mass. These deflection experiments provide evidence for the charges and masses of the emissions.
Common detectors include the Geiger–Müller (GM) tube, photographic film, and cloud chambers. A GM tube connected to a counter gives a count rate, usually in counts per second. When investigating absorption or half-life, the background count must be subtracted to obtain the corrected count rate.
For thickness monitoring in industry, a radiation source and detector are placed on opposite sides of a material. The amount of radiation passing through indicates the thickness. Beta emitters are often used for thin metal foils, while gamma emitters are used for thicker materials.
Background radiation is the low-level radiation that is always present in the environment. It comes from natural sources such as cosmic rays, radon gas from rocks, and radioactive isotopes in the ground and food, as well as from artificial sources like medical X‑rays and fallout from nuclear weapons testing.
背景辐射是环境中始终存在的低强度辐射。它来自宇宙射线、岩石释放的氡气、土壤和食物中的放射性同位素等天然源，也来自医疗 X 光、核武器试验沉降物等人工源。
When measuring the activity of a radioactive sample, background radiation must be accounted for. The corrected count rate = total count rate − background count rate. The unit of radioactive activity is the becquerel (Bq), where 1 Bq = 1 decay per second.
Half-life is the time taken for half the radioactive nuclei in a sample to decay, or equivalently, the time taken for the activity (or count rate) to fall to half its initial value. It is a constant for a given isotope and cannot be changed by physical or chemical means.
You may be asked to calculate half-life from a decay graph, or predict the activity remaining after several half-lives. For example, after n half-lives, the fraction remaining is (½)ⁿ. This is used in carbon‑14 dating to estimate the age of archaeological samples.
考试可能会要求你根据衰变曲线计算半衰期，或预测若干半衰期后的剩余活度。例如，经过 n 个半衰期后，剩余的比例为 (½)ⁿ。碳‑14 定年法正是利用这一点来估计考古样品的年龄。
Activity after n half-lives = initial activity × (½)ⁿ
9. Nuclear Equations | 核方程
Nuclear equations show the changes in atomic and mass numbers during decay. In alpha decay, the mass number decreases by 4 and the atomic number decreases by 2. For example, uranium-238 undergoes alpha decay to thorium-234: ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He.
In beta-minus decay, a neutron turns into a proton and an electron (the beta particle) is emitted. The mass number stays the same while the atomic number increases by 1. An antineutrino is also emitted but is not required in IGCSE equations. Example: ¹⁴₆C → ¹⁴₇N + ⁰₋₁e.
Gamma emission usually accompanies alpha or beta decay, representing the loss of excess energy. It does not change the atomic or mass number. Nuclear equations must balance in terms of both total mass number and total atomic number on each side.
Nuclear fission is the splitting of a large, unstable nucleus (such as uranium-235 or plutonium-239) into two smaller, more stable daughter nuclei, accompanied by the release of two or three neutrons and a huge amount of energy. It can occur spontaneously or be induced by bombarding the nucleus with a neutron.
The neutrons released can go on to cause further fissions, leading to a chain reaction. In a nuclear reactor, control rods (e.g., boron) absorb excess neutrons to keep the reaction controlled. The energy released is used to heat water into steam, which drives turbines to generate electricity.
Nuclear fusion is the joining of two light nuclei to form a heavier nucleus, releasing even more energy per unit mass than fission. Fusion powers the Sun and other stars, where hydrogen nuclei fuse to form helium under extremely high temperature and pressure.
On Earth, achieving controlled fusion is extremely difficult because positively charged nuclei repel each other. Scientists are developing fusion reactors that use magnetic confinement or laser-driven implosion to achieve the necessary conditions. Fusion produces far less long-lived radioactive waste than fission.
Radiation has many beneficial uses: medical imaging (gamma cameras), cancer treatment (radiotherapy), sterilising medical equipment (gamma rays), and tracers in industry and medicine (beta/gamma emitters). The choice of isotope depends on half-life and penetrating ability needed for the job.
However, ionising radiation can damage living cells, causing mutations or killing them. High doses can lead to radiation sickness and cancer. Safety precautions include using shielding, minimising exposure time, keeping distance from sources, and wearing protective clothing. Radioactive waste must be stored safely for thousands of years.
📚 AS Physics Paper 4 Report on Exams: Concept Analysis | AS 物理 Paper 4 考试报告概念解析
Cambridge AS Physics Paper 4 (Advanced Practical Skills) is a demanding practical examination that assesses your ability to handle apparatus, record and present data, analyse results, and evaluate experimental procedures. Examiner reports reveal consistent areas where candidates lose marks – from basic table design to uncertainty propagation. This article unpacks the key concepts you must master, using insights from recent exam reports, to help you approach Paper 4 with confidence.
剑桥 AS 物理 Paper 4（高级实验技能）是一场要求很高的实验考试，重点考察操作仪器、记录呈现数据、分析结果以及评价实验步骤的能力。考官报告揭示了考生反复丢分的常见领域——从基础的表格设计到不确定度的传递。本文结合近期考试报告中的见解，解析你必须掌握的核心概念，帮助你自信应对 Paper 4。
1. Understanding Paper 4: Advanced Practical Skills | 理解 Paper 4：高级实验技能
Paper 4 lasts 2 hours and consists of two compulsory practical questions, each worth 20 marks. The exam tests three skill areas: Manipulation, Measurement and Observation (MMO); Presentation of Data and Observations (PDO); and Analysis, Conclusions and Evaluation (ACE). Examiner reports stress that you must demonstrate competence in all three, not just one or two. Time management is crucial – spend roughly one hour per question, leaving a few minutes to check for omitted units or mislabelled axes.
Paper 4 考试时间为 2 小时，包含两道必做实验题，每题 20 分。考试评估三项技能：操作、测量与观察 (MMO)；数据与观测结果的呈现 (PDO)；以及分析、结论与评估 (ACE)。考官报告强调，你必须在三个方面都展现能力，而不是只擅长一两个。时间管理至关重要——每题约分配一小时，留下几分钟检查遗漏的单位或轴标签。
Examiners often note that candidates rush the final evaluation section, losing easy ACE marks. Plan to reach the evaluation with at least 15 minutes to spare. Familiarise yourself with the mark scheme: typically, MMO includes careful measurement and sensible repeat readings; PDO covers table layout, significant figures, and graph plotting; ACE requires calculating gradients, intercepts, uncertainties, and suggesting improvements.
2. Handling Practical Equipment and Measurements | 实验设备操作与测量
Confident use of instruments like vernier calipers, micrometer screw gauges, stopwatches, voltmeters, ammeters, and metre rules is assumed. For each measurement, you must record the raw value to the instrument’s precision and state the absolute uncertainty. A common rule is ± half the smallest scale division for analogue instruments and ± the smallest digit for digital ones, though exam reports accept ± the precision if justified. Repeat readings reduce random error, and the absolute uncertainty from repeats is typically half the range.
Parallax error must be avoided by positioning your eye perpendicular to the scale, and you should mention this in your procedure if relevant. When using a stopwatch, human reaction time introduces an uncertainty of about 0.2 s to 0.3 s – do not claim a precision of 0.01 s just because the display shows it. Examiner reports frequently flag candidates who record stopwatch readings to 0.01 s; they should be given to 0.1 s at best.
必须通过将眼睛垂直于刻度来避免视差，并在实验步骤中提及这一点（如果相关）。使用秒表时，人反应时间会引入约 0.2 s 至 0.3 s 的不确定度——不要因为显示屏显示 0.01 s 就声称精度是 0.01 s。考官报告经常指出，考生将秒表读数记到 0.01 s；最好只记录到 0.1 s。
For electrical circuits, always check the zero error of meters and mention that you tapped the meter to reduce sticking. When measuring length, use fiducial markers and repeat at different orientations to average out irregularities. All these good practices count towards MMO marks and make your data more reliable.
3. Designing Experiments and Controlling Variables | 设计实验与控制变量
Paper 4 often includes a question where you must describe how to investigate a relationship. Your answer should state the independent variable, dependent variable, and all variables to be kept constant, with practical methods of control. For example, when investigating how the period of a pendulum depends on length, you would say ‘vary the length l by adjusting the thread, measure l with a metre rule, and keep the mass of the bob and amplitude constant.’
Paper 4 常包含一道要求描述如何研究某个关系的题目。你的答案应说明自变量、因变量以及所有需要保持不变的变量，并给出具体控制方法。例如，研究单摆周期与摆长的关系时，应写“通过调节线长改变长度 l，用来尺测量 l，并保持摆锤质量和振幅不变”。
Describe steps in a logical order, using clear instructions such as ‘set up the circuit as shown in the diagram’ or ‘release the ball from the same height using a ruler as a guide.’ Examiners value precautions like ‘ensure the ruler is vertical using a plumb line’ or ‘use a set square to align the meter rule perpendicular to the bench.’ State the range of values you will use and suggest at least six different readings spread evenly across the range.
Avoid vague language like ‘do the experiment carefully’ – instead, specify what careful means in context. The ACE section for such a question may ask you to identify a safety precaution or an additional improvement, so build these into your plan.
4. Recording Data in Tables: Best Practices | 表格记录数据的最佳实践
Examiner reports consistently highlight that many candidates lose PDO marks because of poorly constructed tables. Every column must have a clear heading consisting of the quantity and its unit separated by a slash, e.g., d/cm, T/s, V/V. Do not put units in the body of the table, only in the heading. The independent variable should appear in the leftmost column, with dependent variables and processed columns to the right.
Raw data must be recorded with a consistent number of decimal places appropriate to the measuring instrument. If repeated readings are taken, show all individual readings and then include a column for the mean. Never process data before recording raw values – a column like ‘R = V/I’ is calculated and should appear only after the raw V and I columns.
A model table might look like: h/cm | t₁/s | t₂/s | t_mean/s. If you suspect a reading is anomalous, still include it in the table but circle it and add a comment; do not delete it. Examiners also check that all raw data in the table match the values you later plot on the graph.
5. Calculating Uncertainties: Absolute and Percentage | 不确定度计算：绝对与百分比
You must be able to determine the absolute uncertainty Δx for any measurement. For a single reading, use the instrument precision. For repeated readings, Δx = (max – min)/2, unless half the range is smaller than the instrument precision, in which case use the precision. The percentage uncertainty is (Δx / mean) × 100%. When a quantity is measured directly once, estimate the uncertainty sensibly – for a metre rule, it might be ±2 mm if alignment is tricky.
A common mistake is to confuse absolute and percentage uncertainties when comparing the significance of errors. Percentage uncertainties let you compare the relative sizes of uncertainties from different sources. For instance, a ±0.1 s uncertainty in a time of 20 s is 0.5%, while a ±0.1 cm uncertainty in a length of 10 cm is 1%. The length measurement contributes more to the overall uncertainty.
常见错误是在比较误差的大小时混淆绝对不确定度和百分比不确定度。百分比不确定度让你比较不同来源不确定度的相对大小。例如，20 s 中 ±0.1 s 的百分比不确定度为 0.5%，而 10 cm 中 ±0.1 cm 则是 1%。长度的测量对总不确定度的贡献更大。
When data are processed, you must know the rules for combining uncertainties. For addition or subtraction, absolute uncertainties add: if Z = X – Y, then ΔZ = ΔX + ΔY. For multiplication or division, percentage uncertainties add: if Z = X × Y or Z = X / Y, then %Z = %X + %Y. For a power, multiply the percentage uncertainty by the power: if Z = X², %Z = 2 × %X.
当数据处理时，你必须知道结合不确定度的规则。对于加减运算，绝对不确定度相加：若 Z = X – Y，则 ΔZ = ΔX + ΔY。对于乘除运算，百分比不确定度相加：若 Z = X × Y 或 Z = X / Y，则 %Z = %X + %Y。对于幂次，将百分比不确定度乘以指数：若 Z = X²，%Z = 2 × %X。
6. Plotting Graphs and Drawing Lines of Best Fit | 绘制图表与最佳拟合线
Most Paper 4 questions require a graph. Use the full grid provided; do not shrink your graph into a corner. Label each axis with the quantity and unit, e.g., ‘l / cm’ or ‘T² / s²’. Choose scales that are simple multiples (1, 2, 5, 10, etc.) to make plotting and reading easy. Plot data points with small crosses (×) or dots with circles, and, if required, add error bars in one direction. The exam report often warns against using tiny dots that become invisible or thick marks that are imprecise.
After plotting the points, draw a single straight line of best fit (or a smooth curve if a linear relationship is not expected). The line should have a balance of points on either side, and you must ignore obvious anomalous points. Do not force the line through the origin unless the theory demands it or the question instructs you to. The examiner’s report notes that a common error is to join the first and last points; a best-fit line is determined visually, not by two points.
If an intercept is expected, extend your line to meet the axis and read the value carefully. Show clearly on the graph how you obtained any values used for gradient calculation by drawing a large triangle with coordinates marked.
7. Determining Gradient and Intercept with Uncertainties | 计算斜率与截距及其不确定度
To find the gradient, choose two points on your line of best fit that are at least half the graph’s length apart, and read their coordinates. Use the formula:
gradient = (y₂ – y₁) / (x₂ – x₁)
要计算斜率，在最佳拟合线上选择相距至少半个图长的两点，读取它们的坐标。使用公式：
斜率 = (y₂ – y₁) / (x₂ – x₁)
Never use data points from the table unless they lie exactly on the line. To determine the uncertainty in the gradient, draw the steepest and shallowest acceptable lines (worst-fit lines) that still pass through most of the error bars or data points. Calculate the maximum and minimum gradients from these lines. The uncertainty Δm is:
Similarly, the intercept c and its uncertainty can be found from the line extremes. Many candidates forget to give absolute or percentage uncertainty for the gradient, even when the question specifically asks for it. Always state your final gradient to the same number of significant figures as the data allows, typically 2 or 3 s.f.
类似地，截距 c 及其不确定度可以从极端线求得。许多考生在问题明确要求时，也忘了给出斜率的绝对不确定度或百分比不确定度。始终以数据允许的相同有效数字位数给出最终斜率值，通常为 2 或 3 位有效数字。
8. Propagation of Errors in Calculations | 计算中的误差传递
When you use measured quantities in a formula, the final result carries a combined uncertainty. For example, the resistance R is calculated from R = V / I. If voltmeter reading V = 2.50 V ± 0.01 V and ammeter reading I = 0.40 A ± 0.01 A, then %R = %V + %I = (0.01/2.50)×100% + (0.01/0.40)×100% = 0.4% + 2.5% = 2.9%. Therefore, R = 6.25 Ω ± 2.9%, so absolute uncertainty ΔR = 6.25 × 0.029 = 0.18 Ω. Report as R = 6.3 Ω ± 0.2 Ω.
当你将测量量代入公式时，最终结果会带有合成不确定度。例如，电阻 R 通过 R = V / I 计算。若电压表读数 V = 2.50 V ± 0.01 V，电流表读数 I = 0.40 A ± 0.01 A，则 %R = %V + %I = (0.01/2.50)×100% + (0.01/0.40)×100% = 0.4% + 2.5% = 2.9%。因此 R = 6.25 Ω ± 2.9%，绝对不确定度 ΔR = 6.25 × 0.029 ≈ 0.18 Ω。报告为 R = 6.3 Ω ± 0.2 Ω。
In cases where a quantity is subtracted, absolute uncertainties add. If the extension e = l – l₀, with l = 24.5 cm ± 0.1 cm and l₀ = 12.0 cm ± 0.1 cm, then Δe = 0.2 cm. The percentage uncertainty in e becomes large if the difference is small, which is why examiners encourage you to measure the difference directly if possible.
在相减的情况下，绝对不确定度相加。若伸长量 e = l – l₀，其中 l = 24.5 cm ± 0.1 cm，l₀ = 12.0 cm ± 0.1 cm，则 Δe = 0.2 cm。如果差值很小，e 的百分比不确定度会很大，因此考官鼓励尽可能直接测量差值。
Examiners’ reports show that students often forget to propagate uncertainties or do so incorrectly by adding absolute uncertainties where percentages are required, or vice versa. Always identify the form of the equation before combining uncertainties.
9. Evaluating Experiments and Identifying Limitations | 实验评估与局限性识别
The evaluation section (ACE) asks you to assess the reliability of your results and suggest improvements. Start by commenting on the scatter of points about the line of best fit. If points lie close to the line with small scatter, the results are precise and random errors are small. Identify the largest source of uncertainty, usually by comparing percentage uncertainties of measured quantities. For instance, ‘The percentage uncertainty in time (4%) is larger than that in length (1%), so the time measurement contributes most to the overall error.’
Discuss systematic errors that may have affected the experiment, such as zero error on a meter, friction in a pulley, or neglecting air resistance. Propose realistic improvements: ‘Use a light gate and data logger to reduce reaction time error’ or ‘Take more readings and increase the range of the independent variable to improve gradient reliability.’
Do not suggest vague improvements like ‘do the experiment more carefully’. Specific, practical suggestions earn marks. Also, comment on whether the graph supports the theoretical relationship. If plotting a derived quantity like T² against l yields a straight line through the origin, the relationship T² ∝ l is verified.
10. Common Pitfalls in Paper 4 and How to Avoid Them | Paper 4 常见陷阱与避免方法
Based on examiner reports, here are the most frequent errors and how to steer clear of them. First, misaligning table headings – always use the format quantity/unit. Second, recording processed data in a raw data column; keep raw and calculated data separate. Third, choosing awkward scales that waste graph paper or make plotting difficult; stick to multiples of 2, 5, or 10. Fourth, confusing absolute and percentage uncertainty when discussing dominant error sources.
Fifth, omitting units from the graph axes or using an incorrect unit conversion. Sixth, drawing a line that doesn’t have equal numbers of points on each side. Seventh, failing to indicate how the gradient was obtained – always draw a large triangle and show the coordinates used. Eighth, quoting a gradient to too many significant figures or giving an uncertainty that doesn’t match the precision.
Ninth, forgetting to circle anomalous points and neglecting to comment on them in the evaluation. Tenth, giving an improvement that is irrelevant or unrealistic – for example, ‘use a laser’ when a simple optical pin would suffice. Always root your suggestions in the context of the experiment you have just performed.
Reviewing these pitfalls before the exam will make you alert to the precise standards examiners expect. With targeted practice, you can significantly improve your Paper 4 performance.
考前回顾这些陷阱将使你对考官的明确标准保持警醒。通过有针对性的练习，你可以显著提高 Paper 4 的成绩。
Published by TutorHao | Physics Revision Series | aleveler.com
📚 A-Level Physics Unit 5 Formula Derivations: Insights from the Jan21 Mark Scheme | A-Level物理第五单元公式推导：2021年1月评分方案揭秘
In January 2021, A-Level Physics Unit 5 examined a range of core derivations that tested students’ ability to reason from first principles. The mark scheme reveals exactly what examiners look for: clear logical steps, correct handling of vectors, energy conservation, and the ability to link microscopic models to macroscopic observables. This article unpacks those derivations, step by step, using the same language and notation that secure marks on exam day.
1. Deriving the Kinetic Theory Pressure Equation | 气体动理论压强方程的推导
Begin with a single molecule of mass m moving with velocity component vₓ towards a wall of a cubic container of side length L. The molecule collides elastically, so its velocity changes from vₓ to −vₓ. The change in momentum is Δp = m(vₓ) − (−m vₓ) = 2m vₓ. The time between successive collisions with the same wall is Δt = 2L / vₓ, so the average force exerted by this molecule on the wall is F = Δp / Δt = 2m vₓ ÷ (2L / vₓ) = m vₓ² / L.
从一个质量为m的分子开始，它沿 x 方向的速度分量为 vₓ，所处容器边长为 L。分子与器壁发生弹性碰撞，速度由 vₓ 变为 −vₓ，动量变化为 Δp = 2m vₓ。两次撞击同一器壁的时间间隔为 Δt = 2L / vₓ，因此该分子对器壁的平均作用力为 F = Δp / Δt = m vₓ² / L。
Summing over all N molecules, the total force on the wall is F_total = (m / L) Σ vᵢₓ². By definition, pressure p = F_total / L² = (m / L³) Σ vᵢₓ². The mean square speed ⟨c²⟩ is defined as the average of vᵢₓ² + vᵧ² + v_z² for all molecules. Because motion is random, ⟨vₓ²⟩ = ⟨vᵧ²⟩ = ⟨v_z²⟩ = ⅓ ⟨c²⟩. Hence Σ vᵢₓ² = N ⟨vₓ²⟩ = N ⟨c²⟩ / 3. Substituting gives the fundamental equation in the mark scheme:
Examiners expect you to explicitly state the assumption of elastic collisions and the cubic geometry, then link mean square speeds to pressure. The mark scheme awards 4–5 marks for a complete derivation.
2. Connecting Kinetic Energy to Temperature | 从动能到温度的联系
The ideal gas law from experiment is pV = nRT. Comparing with pV = ⅓ N m ⟨c²⟩, we have ⅓ N m ⟨c²⟩ = nRT. Since N = n Nₐ, where Nₐ is Avogadro’s number, this becomes ⅓ N m ⟨c²⟩ = (N / Nₐ) RT. Rearranging gives the average kinetic energy per molecule: ½ m ⟨c²⟩ = (3/2) (R / Nₐ) T = (3/2) k T, where k is the Boltzmann constant.
实验得出的理想气体状态方程为 pV = nRT。将其与 pV = ⅓ N m ⟨c²⟩ 对比，可得 ⅓ N m ⟨c²⟩ = nRT。利用 N = n Nₐ（Nₐ 为阿伏伽德罗常数），代入得 ½ m ⟨c²⟩ = (3/2) (R / Nₐ) T = (3/2) k T，其中 k 为玻尔兹曼常数。
This is a favourite Jan21 exam step: the link between microscopic kinetic energy and macroscopic temperature is completed. At A-level, you need only state that k = R/Nₐ, then substitute. The mark scheme typically asks for “kinetic energy proportional to absolute temperature”.
这也是 Jan21 试卷中常见的考察点：完成微观动能与宏观温度之间的关联。在A-Level中，只需说明 k = R/Nₐ 并代入即可。评分标准往往看是否判断出“平均动能与绝对温度成正比”。
½ m ⟨c²⟩ = ³⁄₂ k T
3. Deriving the Ideal Gas Law from Kinetic Theory | 从动理论推导理想气体定律
Starting from pV = ⅓ N m ⟨c²⟩ and the average kinetic energy expression ½ m ⟨c²⟩ = ³⁄₂ k T, eliminate ½ m ⟨c²⟩. Multiply the kinetic energy by 2: m ⟨c²⟩ = 3 k T. Substituting into the pressure equation yields pV = ⅓ N × 3 k T = N k T. Using N = n Nₐ, we obtain pV = n Nₐ k T = n R T, which is exactly the ideal gas law. This shows that the kinetic theory model is consistent with empirical gas behaviour.
从 pV = ⅓ N m ⟨c²⟩ 和 ½ m ⟨c²⟩ = ³⁄₂ k T 出发，消去 m ⟨c²⟩。将动能表达式乘以2得 m ⟨c²⟩ = 3 k T，代入压强方程得 pV = ⅓ N × 3 k T = N k T。再利用 N = n Nₐ 和 R = Nₐ k，便可得到 pV = n R T。这一步完整证明了动理论模型与实验结果的一致性。
The mark scheme explicitly rewards the logical flow that connects pV = ⅓ N m ⟨c²⟩ to the kinetic energy result, and then to pV = nRT. Candidates often lose marks by missing the multiplication step or forgetting to define Nₐ.
评分方案会奖励从 pV = ⅓ N m ⟨c²⟩ 到动能关系、再到 pV = nRT 的逻辑链条。考生常因跳过乘以2的步骤或未说明 Nₐ 而丢分。
4. Radioactive Decay Law and Half-Life Derivation | 放射性衰变定律与半衰期推导
Radioactive decay is a random process where the number of nuclei decaying per unit time is proportional to the number present: dN/dt = −λ N, where λ is the decay constant. Separating variables gives dN/N = −λ dt. Integrating both sides: ∫(1/N) dN = −λ ∫ dt ⇒ ln N = −λ t + C. At t = 0, N = N₀, so C = ln N₀. Therefore ln(N / N₀) = −λ t, which exponentiates to the familiar decay law:
放射性衰变是一个随机过程，单位时间内衰变的原子核数正比于现存核数目：dN/dt = −λ N（λ 为衰变常量）。分离变量得 dN/N = −λ dt，积分得 ln N = −λ t + C。代入初始条件 t = 0 时 N = N₀，得 C = ln N₀，于是 ln(N / N₀) = −λ t，指数化后得到熟知的形式：
N = N₀ e−λ t
To find half-life T₁/₂, set N = N₀ / 2 at t = T₁/₂: N₀/2 = N₀ e−λ T₁/₂ ⇒ e−λ T₁/₂ = 1/2 ⇒ −λ T₁/₂ = ln(1/2) = −ln 2. Hence T₁/₂ = ln 2 / λ. The Jan21 mark scheme insists on showing the natural logarithm manipulation explicitly.
5. Deriving Activity from Decay Constant | 从衰变常数推导活度
Activity A is defined as the number of decays per unit time, which is exactly the magnitude of the decay rate: A = |dN/dt|. Using the decay equation dN/dt = −λ N, we immediately have A = λ N. Substituting N = N₀ e−λ t gives the exponential decay of activity: A = λ N₀ e−λ t = A₀ e−λ t, where A₀ = λ N₀ is the initial activity. This derivations is often worth 2 marks in the Jan21 mark scheme — one for stating A = λ N and one for linking to the exponential form.
活度 A 定义为单位时间内的衰变次数，即衰变率的绝对值：A = |dN/dt|。由衰变方程 dN/dt = −λ N 直接可得 A = λ N。将 N = N₀ e−λ t 代入，得到活度随时间的指数衰减规律：A = λ N₀ e−λ t = A₀ e−λ t（其中 A₀ = λ N₀ 为初始活度）。在 Jan21 评分方案中，这个推导通常占 2 分——1分写出 A = λ N，1分完成指数形式的转换。
6. Escape Velocity Derivation from Energy Conservation | 通过能量守恒推导逃逸速度
To escape a planet’s gravitational field, a projectile must have enough initial kinetic energy to do work against gravity to reach infinity. The gravitational potential energy at radial distance r from the planet’s centre is U = −G M m / r. At infinity, both kinetic and potential energy become zero. Conservation of total mechanical energy gives: ½ m v² − G M m / r = 0. Solving for v yields the escape speed:
要逃离行星的引力场，抛射体需拥有足够的初始动能以克服引力做功到达无穷远。距离行星中心 r 处的引力势能为 U = −G M m / r，无穷远处动能和势能均为零。根据机械能守恒：½ m v² − G M m / r = 0，解得逃逸速度：
v_esc = √(2 G M / r)
At the planet’s surface, r = R, and g = G M / R², so an alternative form is v_esc = √(2 g R). The Jan21 mark scheme often asks candidates to derive both forms and to state the energy conservation principle explicitly.
在行星表面 r = R，且 g = G M / R²，因此逃逸速度也可写成 v_esc = √(2 g R)。Jan21 评分要求考生能够推导这两种形式，并明确写出能量守恒原理。
7. Deriving Kepler’s Third Law for Circular Orbits | 圆形轨道开普勒第三定律推导
For a satellite in a circular orbit of radius r, the gravitational force provides the centripetal force: G M m / r² = m v² / r, so v² = G M / r. The orbital period is T = 2π r / v, hence v = 2π r / T. Substituting v² = 4π² r² / T² into the force equation gives G M / r = 4π² r² / T², which rearranges to:
对于半径为 r 的圆形轨道，万有引力提供向心力：G M m / r² = m v² / r，故 v² = G M / r。轨道周期 T = 2π r / v，因此 v = 2π r / T，代入得 G M / r = 4π² r² / T²，整理可得：
T² = (4π² / G M) r³
This is Kepler’s third law for circular motion: T² ∝ r³. The proportionality constant depends only on the central mass. In the Jan21 astrophysics option, this derivation carries 3 marks and must include the substitution step for v.
这就是圆形轨道下的开普勒第三定律：T² ∝ r³，比例常数仅与中心天体质量有关。在 Jan21 天体物理选考中，这一推导占 3 分，必须体现 v 的替换步骤。
8. Derivation of the Gravitational Potential Formula | 引力势公式的推导
Gravitational potential V at a point is defined as the work done per unit mass in bringing a small test mass from infinity to that point. The force on a unit mass at distance r from mass M is G M / r² directed towards M. The work done by an external agent moving from r’ = ∞ to r is W = ∫_{∞}^{r} (−G M / r’²) dr’ (the negative sign ensures the force is overcome). Evaluating the integral: W = [G M / r’]_{∞}^{r} = G M / r − 0. Since potential is negative of work done by the gravitational field, we get V = −G M / r.
引力势 V 定义为将单位质量从无穷远处移动到该点外力克服引力所做的功。在距离 M 为 r 处，单位质量所受引力为 G M / r² 指向 M。外力做功为 W = ∫_{∞}^{r} (G M / r’²) dr’（或直接积分引力做功为负值）。计算积分得 W = [−G M / r’]_{∞}^{r} = −G M / r − 0，因此引力势为 V = −G M / r。
The Jan21 mark scheme specifically looks for the correct limits and the negative sign. Candidates must also explain that the potential is negative because work is done by the field as an object approaches.
9. Deriving Binding Energy per Nucleon from Mass Defect | 由质量亏损推导比结合能
The mass of a nucleus is always less than the sum of the masses of its individual protons and neutrons. This mass defect Δm is converted into binding energy via E = Δm c². In nuclear physics, masses are given in unified atomic mass units u, where 1 u = 1.661 × 10⁻²⁷ kg and c² expressed in MeV becomes 931.5 MeV / u. To find the binding energy per nucleon, compute E_binding = Δm × 931.5 MeV then divide by the nucleon number A.
原子核的质量总是小于其内部各核子质量之和，这个质量差 Δm 称为质量亏损，对应的结合能为 E = Δm c²。核物理中质量常以原子质量单位 u 给出，1 u = 1.661 × 10⁻²⁷ kg，c² 等价于 931.5 MeV/u。比结合能的计算为：E_binding = Δm × 931.5 MeV / A。
In full derivation form, the mark scheme expects you to write Δm = Z m_p + (A − Z) m_n − m_nucleus, then multiply by c². When using data tables, always convert mass defect to kg if using E = Δm c² in joules, or use the MeV/u shortcut.
在完整推导中，评分方案期待写出 Δm = Z m_p + (A − Z) m_n − m_nucleus，再乘以 c²。使用数据表时，可根据需要将质量亏损转化为 kg 后用焦耳表达，或直接使用 MeV/u 转换因子。
Published by TutorHao | Physics Revision Series | aleveler.com
📚 Tackling PH05 Applied Questions: Key Strategies for International A-Level Physics | 攻克PH05应用题：国际A-Level物理关键策略
Applied problems in PH05 International A-Level Physics require not only a solid grasp of theoretical concepts but also the ability to translate real-world situations into mathematical models, manipulate equations confidently, and interpret data from graphs and tables. This article presents a systematic approach to mastering the problem-solving skills essential for success, covering everything from initial reading strategies to final answer checks.
Before writing down any equations, read the question thoroughly at least twice. Identify exactly which physical quantities are given, which are being asked for, and any implicit assumptions such as “negligible air resistance” or “uniform field”. Underline key numerical values and take note of their units.
Look for instruction words like “calculate”, “show that”, “estimate”, “compare”, or “explain”. A “show that” question often provides the final result, so your working must be clear enough to convince the examiner that you derived it step by step.
Draw a simple labelled diagram if the situation involves geometry or motion. This visual representation can reveal relationships between lengths, angles, velocities, or forces that are not immediately obvious from the text alone.
2. Unit Conversion and Dimensional Analysis | 单位换算与量纲分析
Always convert all data into standard SI units before plugging numbers into formulas: metres, kilograms, seconds, joules, pascals, etc. For example, convert 25 cm to 0.25 m, and 1.2 g cm⁻³ to 1200 kg m⁻³. Failing to do so is one of the most frequent causes of lost marks.
在代入公式之前，务必将所有数据转换为标准国际单位：米、千克、秒、焦耳、帕斯卡等。例如，将 25 cm 转换为 0.25 m，将 1.2 g cm⁻³ 转换为 1200 kg m⁻³。忽略单位换算是导致失分的最常见原因之一。
In particle and nuclear physics, the electronvolt (eV) is widely used. Remember: 1 eV = 1.60 × 10⁻¹⁹ J. When dealing with mass–energy equivalence, you may need to convert atomic mass units: 1 u = 931.5 MeV = 1.66 × 10⁻²⁷ kg. Practise these conversions until they become automatic.
在粒子与核物理中，电子伏特 (eV) 使用广泛。请牢记：1 eV = 1.60 × 10⁻¹⁹ J。处理质能等价关系时，可能需要换算原子质量单位：1 u = 931.5 MeV = 1.66 × 10⁻²⁷ kg。反复练习这些换算，直至熟练自如。
Perform a quick dimensional check by substituting base units into your final expression. For instance, if an equation is supposed to give an energy, the right-hand side must reduce to kg m² s⁻². If it gives kg m s⁻¹ instead, you have made a mistake.
将基本单位代入最终表达式，进行快速量纲检查。例如，若某方程应得出能量，等号右侧的量纲必须化简为 kg m² s⁻²。如果结果是 kg m s⁻¹，就说明出错了。
3. Algebraic Manipulation and Formula Rearrangement | 代数运算与公式变形
Be completely comfortable rearranging equations before substituting numbers. Many applied problems become much simpler if you first isolate the unknown symbol. For example, from Eₖ = ½mv², rearranging to v = √(2Eₖ/m) avoids messy arithmetic with numbers too early.
务必能从容地在代入数值前变换公式。先将未知量符号分离出来，许多应用题就会简单很多。例如，从 Eₖ = ½mv² 变形为 v = √(2Eₖ/m)，可避免过早陷入繁琐的数值运算。
When the same variable appears in multiple terms, factor it out. For instance, if you have F – mg = ma, solve for a by writing a = (F/m) – g. This approach reduces the chance of sign errors and clarifies the physical meaning of each term.
当同一变量出现在多个项中时，请将其因式分解出来。例如，对于 F – mg = ma，求解 a 时可写成 a = (F/m) – g。这种方法能降低符号出错的可能性，并使每项的物理意义更加清晰。
Use substitution for simultaneous equations. In mechanics, you often need to combine kinematic equations and Newton’s second law. Write down all relevant equations first, then systematically eliminate unwanted variables algebraically.
Applied questions frequently ask you to extract information from a graph. Identify the type of graph: linear, exponential, inverse, or logarithmic. Check the axes labels carefully – a plot of v² against s, for example, has a gradient equal to 2a.
应用题经常要求从图像中提取信息。首先要判断图像类型：线性、指数、反比或对数。仔细检查坐标轴标签——例如，v² 对 s 的图像，其斜率等于 2a。
When determining a gradient, use a large triangle whose vertices are far apart on the drawn line, never use plotted data points directly. Show your working by writing Δy/Δx = (y₂ − y₁)/(x₂ − x₁) and quoting the unit of the gradient.
For area under a curve, remember the physical significance: the area of a force–time graph gives impulse, a velocity–time graph gives displacement, and a pressure–volume graph gives work done. If you are asked to estimate the area, divide the shape into countable squares or simple geometric strips.
Many PH05 questions test your ability to make reasonable estimates. Rather than searching for an exact number, think in terms of powers of ten. For instance, the radius of an atomic nucleus is roughly 10⁻¹⁵ m, the mass of a proton is about 10⁻²⁷ kg, and the typical binding energy per nucleon is 8 MeV.
When estimating, break a complex quantity into smaller, manageable pieces. To estimate the kinetic energy of an ejected particle, start with a known rest mass and an approximate velocity expressed as a fraction of the speed of light, then apply the relativistic energy formula if required.
Always state your assumptions clearly and evaluate whether your final figure makes sense physically. A common exam instruction is “Comment on the reasonableness of your answer” – you must be prepared to compare it with a typical scale.
Complex applied problems can be turned into a series of simpler sub-problems. Read the question and decide where the physical situation can be split – often before and after a collision, or between two different stages of motion.
Set out your calculation in clear logical blocks with headings such as “Stage 1: Acceleration”, “Stage 2: Projectile motion”. This not only helps you avoid confusion but also enables the examiner to award method marks even if a numerical slip occurs later.
Use a “know–find–plan” strategy: list what you Know, what you need to Find, and then outline your Plan. This simple structure prevents you from jumping straight into an equation that contains two unknowns.
Conservation of momentum, (mass–)energy, and charge are among the most powerful tools in physics. In nuclear reactions and particle decays, always check that the total charge and baryon number are the same on both sides of the equation before calculating energy releases.
For an inelastic collision where only momentum is conserved, write: m₁u₁ + m₂u₂ = (m₁+m₂)v. Avoid the common mistake of also assuming kinetic energy is conserved – in an inelastic collision kinetic energy is not conserved; some is transformed into internal energy, sound, or heat.
In thermal physics, the First Law of Thermodynamics (Q = ΔU + W) is your guide to energy balance. Define the system clearly, adopt a consistent sign convention (e.g., work done by the gas is positive), and use it to track energy transfers in a cycle.
8. Tackling Nuclear and Astrophysics Questions (PH05 Specific) | 攻克核物理与天体物理问题（PH05特辑）
In nuclear physics, you are frequently asked to calculate the energy released in a reaction using ΔE = Δmc². Always find the total mass of the reactants and the total mass of the products separately, then subtract to get Δm. Convert Δm from atomic mass units to kg or directly to MeV using 1 u = 931.5 MeV.
Binding energy per nucleon graphs are a typical source of exam questions. Remember that the peak lies around iron-56. Fusing nuclei lighter than iron releases energy, while fission of nuclei heavier than iron releases energy. Be prepared to read values from the graph and compute total binding energies.
For astrophysics, the Hertzsprung–Russell diagram is central. You should be confident reading luminosity, temperature, and spectral class axes, and linking a star’s position to its life cycle stage. When given absolute magnitude and apparent magnitude, use m − M = 5 log(d/10) to find distance d in parsecs.
在天体物理中，赫罗图至关重要。你应当能熟练读取光度、温度和光谱型坐标轴，并将恒星的位置与其生命周期阶段联系起来。当给出绝对星等和视星等时，使用 m − M = 5 log(d/10) 求得以秒差距为单位的距离 d。
Questions about the Hubble law require v = H₀ d. Ensure you convert distance into Mpc if H₀ is given in km s⁻¹ Mpc⁻¹, and pay attention to the unit of velocity. Sometimes you need to estimate the age of the Universe as 1/H₀, but remember to convert units to seconds first.
关于哈勃定律的问题需要 v = H₀ d。若 H₀ 的单位为 km s⁻¹ Mpc⁻¹，请确保将距离转换为 Mpc，并注意速度的单位。有时你需要将宇宙年龄估计为 1/H₀，但切记先将单位转换为秒。
9. Common Pitfalls and How to Avoid Them | 常见陷阱及避免方法
One of the biggest traps is using the wrong sign for gravitational potential energy or potential. In radial fields, V = −GM/r, and the potential is negative. When calculating escape velocity, set ½mv² + (−GMm/r) = 0, not ½mv² = GMm/r.
Misreading prefixes such as nano (n = 10⁻⁹), micro (μ = 10⁻⁶), and pico (p = 10⁻¹²) often leads to a factor-of-1000 error. Write the conversion explicitly in your working rather than relying on mental arithmetic.
Students frequently confuse the definitions of half-life and decay constant. The relationship is λ = ln 2 / T½. When a problem asks for the activity after a certain time, use A = A₀ e⁻λ t, and be careful to express t in the same time unit as λ.
学生经常混淆半衰期与衰变常量的定义。它们的关系为 λ = ln 2 / T½。当问题要求计算特定时间后的活度时，应使用 A = A₀ e⁻λ t，并注意 t 的时间单位须与 λ 一致。
10. Answer Checking Strategies | 答案检查策略
Once you have a numerical answer, ask yourself: is this value physically plausible? For example, a car’s acceleration of 150 m s⁻² is clearly unrealistic, whereas 3.5 m s⁻² makes sense. If the answer is implausible, scan your working for arithmetic slips.
得到数值答案后，问自己：这个值在物理上合理吗？例如，一辆汽车的加速度为 150 m s⁻² 显然不切实际，而 3.5 m s⁻² 则合理。如果答案不合理，快速扫描计算过程，寻找算术错误。
Substitute your answer back into the original equation(s) to verify that both sides match. In multi-step problems, check the intermediate values against the physical scenario – e.g., intermediate velocities should not exceed the speed of light.
If time permits, re-read the question to ensure you have answered exactly what was asked. Pay special attention to the required format – “give your answer in MeV”, “state the direction”, or “express to two significant figures”. Missing these details can lose easy marks.
📚 GCSE AQA Physics: Past Paper Analysis | GCSE AQA 物理：历年真题解析
Mastering GCSE AQA Physics requires thorough preparation, and one of the most effective tools is past paper practice. By analysing previous exams, students can identify recurring question patterns, understand mark schemes, and develop time management skills. This guide provides an in-depth analysis of how to extract maximum value from past papers, covering key topics, common mistakes, and strategies for success.
Past papers are not just revision tools; they are a window into the examiner’s expectations. AQA exam papers follow a consistent structure and style, with certain topics appearing year after year. Regularly working through past papers helps you become familiar with the command words used, such as ‘describe’, ‘explain’, and ‘evaluate’.
Furthermore, past papers enable you to practise applying knowledge to unfamiliar contexts, which is a key skill tested at GCSE level. Questions often combine multiple topics, so using past papers improves your ability to link concepts together and build coherent answers.
Mark schemes reveal precisely what examiners award marks for. After completing a paper, always review the mark scheme to understand the specific points needed. This reduces the risk of losing marks due to incomplete or vague answers, which is a common problem for many students.
2. Effective Strategies for Using Past Papers | 高效利用真题的策略
To get the most out of past papers, start by attempting a full paper under timed conditions without any notes. This simulates the real exam environment and highlights your weaker areas. After marking your answers, spend time analysing every mistake instead of simply noting the score.
Create a topic-by-topic error log. For any question you got wrong or partially right, write down the topic, the mistake, and the correct concept. Revisit these errors regularly to ensure understanding is durable. Using examiners’ reports alongside past papers provides further insight into typical errors made by candidates.
Rotate through different years of papers, and reserve the most recent papers for final mock exams. This approach allows you to track progress and ensures you are not memorising answers but truly understanding the underlying physics principles.
3. Common Pitfalls and How to Avoid Them | 常见错误与规避方法
One of the most frequent mistakes is misreading the command word. For instance, a question asking you to ‘explain’ requires a logical cause-and-effect chain, whereas ‘describe’ only needs a statement of what happens. Mark schemes consistently penalise answers that do not match the command word.
Another common issue is neglecting units or significant figures in calculations. AQA physics mark schemes explicitly award marks for correct units and appropriate significant figures. Always check that your final answer includes the proper unit, such as m/s or N, and is given to the same significant figures as the data provided.
另一个常见问题是忽略计算中的单位或有效数字。AQA 物理评分方案明确为正确的单位和适当的有效数字给分。务必检查你的最终答案是否包含了正确的单位，如 m/s 或 N，并与所给数据的有效数字位数一致。
Many students lose easy marks by drawing incomplete or poorly labelled diagrams. When a question asks for a circuit diagram or a force arrow, use a ruler, label components clearly, and show direction. A well-presented diagram can often gain full marks even if the written explanation is not perfect.
Mechanics questions frequently appear in both paper 1 and paper 2. You can expect to calculate speed, acceleration, resultant force, and momentum. A typical question might show a velocity-time graph and ask for the distance travelled; remember that the area under the graph represents displacement.
Newton’s second law is fundamental, and the equation
F = m a
must be applied correctly. Pay attention to whether the forces are balanced or unbalanced. In free-body diagrams, identify all forces, including weight, normal contact force, friction, and tension, and ensure arrows are proportional.
Momentum conservation is often tested with collision or explosion scenarios. Use the equation
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
and always specify a positive direction. Past papers show that students often lose marks by forgetting that momentum is a vector quantity.
动量守恒经常在碰撞或爆炸场景中考查。使用公式
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
并始终规定一个正方向。历年真题表明，学生们常因忘记动量是矢量而失分。
5. Electricity Questions Unpacked | 电学题目剖析
Circuit analysis is a staple of AQA physics exams. You will be required to calculate resistance using Ohm’s law, and combine resistors in series or parallel. A common past paper question gives a circuit diagram with one unknown resistor and asks you to determine its value from voltage and current readings.
must be used carefully. In series circuits, current is the same everywhere; in parallel, voltage is the same across branches. Many mark schemes specifically award marks for stating these rules before applying the equation.
Power and energy calculations often combine multiple equations such as
P = I V
and
E = P t
. When a question asks about energy transferred in a component, remember to convert time to seconds and check if the component is in series or parallel, as this affects current and voltage values.
Questions on waves typically require you to recall the wave equation
v = f λ
and apply it to both transverse and longitudinal waves. Past papers frequently ask for the calculation of wavelength or frequency from a given diagram showing a wave on a grid.
关于波的题目通常要求你回忆波动方程
v = f λ
并将其应用于横波和纵波。真题经常要求根据一幅展示在网格上的波形图，计算波长或频率。
When interpreting oscilloscope traces, measure the time period carefully and know that frequency is the reciprocal of the period. AQA mark schemes reward clear working: showing the conversion from timebase and volts/div settings earns method marks.
Refraction and reflection diagrams must be drawn accurately with a ruler, and all angles clearly labelled. Remember that the normal is a dashed line perpendicular to the surface. In past papers, students often confuse the angle of incidence and the angle of refraction, so label each angle to avoid ambiguity.
This topic appears across both papers and often involves comparing renewable and non-renewable energy resources. Exam questions may ask you to evaluate the environmental and economic impacts of different energy types, such as fossil fuels, nuclear, wind, and solar power.
Energy transfer calculations using gravitational potential energy and kinetic energy are common. You must be fluent with
Eₖ = ½ m v²
and
ΔEₚ = m g Δh
. In past papers, diagrams of falling objects or roller coasters often serve as contexts for these equations.
利用重力势能和动能的能量转移计算很常见。你必须熟练掌握
Eₖ = ½ m v²
和
ΔEₚ = m g Δh
。在真题中，落体或过山车的简图常常作为这些公式的情境。
Efficiency is another key concept, calculated as useful output over total input. AQA often frames these questions with Sankey diagrams or numerical examples. Remember that efficiency can be expressed as a percentage and that no real device is 100% efficient due to energy dissipation.
Nuclear physics questions require knowledge of the atom’s structure and the properties of alpha, beta, and gamma radiation. AQA mark schemes expect you to describe the ionising power and penetrating ability of each radiation type, as well as their applications in medicine and industry.
Radioactive decay equations frequently appear. You may need to complete nuclear equations such as
²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He
ensuring that both mass numbers and atomic numbers balance. Use the periodic table to identify unknown elements from their atomic number.
放射性衰变方程经常出现。你可能需要完成核方程，例如
²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He
并确保质量数和原子序数均守恒。利用元素周期表，根据原子序数识别未知元素。
Half-life problems can be graphical or numerical. When using a decay curve, draw lines on the graph to show how you find the half-life. For numerical questions, use the systematic halving method, and write down each step to reduce careless errors, as examiners’ reports mention this as a frequent weakness.
Beyond basic mechanics, AQA tests understanding of stopping distance, which consists of thinking distance and braking distance. Past paper questions often provide a graph or table of data and ask you to explain how speed, road conditions, or alcohol affect these distances.
Hooke’s law is a required practical, and exam questions will link directly to this experiment. You need to know that
F = k x
and be able to interpret force-extension graphs. Past papers show that many students fail to convert extension from centimetres to metres, which leads to incorrect spring constant values.
Moments and levers are also common. Use the principle of moments: sum of clockwise moments equals sum of anticlockwise moments. When drawing force arrows on a lever diagram, be precise about the pivot position and the perpendicular distance. Units of moments are newton-metres (Nm).
10. Required Practicals in Exam Questions | 必做实验在考题中的体现
AQA physics exams heavily assess the required practicals, not just recalling procedures but also analysing results and identifying sources of error. For example, the resistance of a wire practical often appears with a question about why the current must be kept constant to avoid heating effects.
When describing a method, use precise sequential steps and mention all the equipment. The mark scheme may award marks for saying ‘record the ammeter reading’ rather than just ‘measure current’. Always refer to the specific instruments used in the practical.
Graph skills from practicals are frequently tested: drawing a line of best fit, calculating gradient, and intercepts. The gradient of a distance–time graph gives speed, while the gradient of a velocity–time graph gives acceleration. Show your working for gradient calculation clearly to earn full method marks.
The 6-mark questions at the end of many past papers require a structured answer. AQA mark schemes use a levels-based approach: Level 1 (1–2 marks) for basic statements, Level 2 (3–4 marks) for cohesive but incomplete logic, and Level 3 (5–6 marks) for a fully detailed, coherent argument that uses correct scientific terminology.
Plan your answer before writing. Use a framework such as PEEL (Point, Evidence, Explanation, Link) to ensure every part of your answer addresses the question. For a ‘compare’ question, clearly state similarities and differences with supporting data. For an ‘explain’ question, build a causal chain step by step.
Incorporate relevant equations and make qualitative predictions based on them. For instance, if explaining why a filament lamp’s resistance increases, refer to the ions vibrating more as temperature rises, causing more electron collisions. Use key terms like ‘resistance’, ‘temperature’, ‘ions’ and ‘electrons’ for full credit.
Many marks depend on accurate graphing and interpretation. When plotting data, use a sharp pencil, label axes with quantities and units, and choose scales that use at least half the graph paper. In past papers, students frequently lose marks by using non-linear scales or omitting the origin unless specified.
Describing trends requires precision. Instead of saying ‘as speed increases, thinking distance goes up’, write ‘thinking distance is directly proportional to speed because the line passes through the origin and is straight’. Back up every trend description with evidence from the graph.
Calculating gradients and intercepts often uses a large triangle on the graph. Remember to read the coordinates of two points on the line of best fit, not from the data table. Write the gradient as
gradient = (y₂ – y₁) / (x₂ – x₁)
and include the unit, such as m/s for a distance–time graph.
Mastering GCSE OCR Physics requires a solid grasp of key principles and the ability to apply them confidently to a variety of problems. This article presents detailed, step-by-step worked examples drawn from core topics such as mechanics, electricity, energy, waves and nuclear physics. Each example is paired with clear explanations in both English and Chinese to support bilingual revision and deepen your problem-solving skills.
Newton’s second law states that the acceleration of an object is proportional to the net force acting on it and inversely proportional to its mass. The relationship is written as:
F = m a
牛顿第二定律指出，物体的加速度与作用在其上的净力成正比，与质量成反比。该关系写作：
F = m a
where F is the net force in newtons (N), m is mass in kilograms (kg) and a is acceleration in metres per second squared (m/s²).
其中 F 是净力（牛顿，N），m 是质量（千克，kg），a 是加速度（米每平方秒，m/s²）。
Example: A car of mass 1200 kg experiences a constant net forward force of 3000 N. Calculate its acceleration and determine how long it takes to reach 15 m/s from rest.
例题：一辆质量为 1200 kg 的汽车受到 3000 N 的恒定净力。计算其加速度，并求从静止加速到 15 m/s 所需的时间。
Solution: Rearranging F = m a gives a = F / m = 3000 N / 1200 kg = 2.5 m/s².
解答：由 F = m a 得 a = F / m = 3000 N / 1200 kg = 2.5 m/s²。
Using the kinematic equation v = u + a t, with initial velocity u = 0, final velocity v = 15 m/s and a = 2.5 m/s²:
t = v / a = (15 m/s) / (2.5 m/s²) = 6 s
使用运动学方程 v = u + a t，初速度 u = 0，末速度 v = 15 m/s，a = 2.5 m/s²：
t = v / a = (15 m/s) / (2.5 m/s²) = 6 s
The car accelerates at 2.5 m/s² and reaches the target speed in 6 seconds.
汽车以 2.5 m/s² 加速，用 6 秒达到目标速度。
2. Work, Energy and Efficiency | 功、能量与效率
When a force moves an object through a distance, work is done. For lifting an object, the work done against gravity is equal to the gain in gravitational potential energy:
W = m g h
当力使物体移动一段距离时，就做了功。对于提升物体，克服重力做的功等于增加的重力势能：
W = m g h
where m is mass, g is gravitational field strength (10 N/kg on Earth) and h is the height raised.
其中 m 为质量，g 为引力场强（地球取 10 N/kg），h 为提升高度。
Example: An electric motor lifts a 50 kg load vertically through 12 m. The motor receives 8000 J of electrical energy. Calculate the useful work done on the load and the efficiency of the motor.
例题：一台电动机将 50 kg 的重物竖直提升 12 m。电动机接收 8000 J 的电能。计算对重物做的有用功和电动机的效率。
Solution: Useful work done W = m g h = 50 kg × 10 N/kg × 12 m = 6000 J.
解答：有用功 W = m g h = 50 kg × 10 N/kg × 12 m = 6000 J。
The motor is 75% efficient, meaning 25% of the input energy is dissipated as heat or sound.
电动机效率为 75%，意味着 25% 的输入能量以热或声音的形式散失。
3. Circuit Rules and Ohm’s Law | 电路规律与欧姆定律
In a series circuit, current is the same everywhere, and the total potential difference (voltage) from the battery is shared across components. Ohm’s law links voltage, current and resistance:
V = I R
在串联电路中，各处电流相同，电池提供的总电压被各元件分担。欧姆定律将电压、电流与电阻联系起来：
V = I R
Example: A 12 V battery is connected in series with a 4 Ω resistor and a 6 Ω resistor. Determine the total current and the voltage across each resistor.
例题：一个 12 V 电池与一个 4 Ω 和一个 6 Ω 的电阻器串联。求总电流及各电阻两端的电压。
Example: How much energy is required to heat 2.0 kg of water from 20 °C to 80 °C? The specific heat capacity of water is 4200 J/(kg °C).
例题：将 2.0 kg 水从 20 °C 加热到 80 °C 需要多少能量？水的比热容为 4200 J/(kg °C)。
Solution: Δθ = 80 °C − 20 °C = 60 °C. Q = m c Δθ = 2.0 kg × 4200 J/(kg °C) × 60 °C = 504,000 J (or 504 kJ).
解答：Δθ = 80 °C − 20 °C = 60 °C。 Q = m c Δθ = 2.0 kg × 4200 J/(kg °C) × 60 °C = 504,000 J（即 504 kJ）。
Thus, over half a million joules of energy are needed to achieve this temperature rise.
因此，需要超过 50 万焦耳的能量来实现这个温升。
5. Half-Life Calculations | 半衰期计算
The half-life of a radioactive isotope is the time taken for half of its unstable nuclei to decay, or for the count rate to fall by half. After n half-lives, the remaining quantity is (1/2)n of the original.
放射性同位素的半衰期是指一半的不稳定原子核发生衰变，或者计数率减半所需的时间。经过 n 个半衰期后，剩余量是原来的 (1/2)n。
Example: A radioactive sample has an initial count rate of 800 counts per minute and a half-life of 2 days. Predict the count rate after 6 days.
例题：某放射性样品的初始计数率为 800 次/分钟，半衰期为 2 天。预测 6 天后的计数率。
Solution: Number of half-lives elapsed = total time / half-life = 6 days / 2 days = 3.
解答：经过的半衰期次数 = 总时间 / 半衰期 = 6 days / 2 days = 3。
After 1 half-life: count rate = 800 / 2 = 400 counts/min. After 2 half-lives: 400 / 2 = 200 counts/min. After 3 half-lives: 200 / 2 = 100 counts/min.
Therefore, after 6 days the count rate will be 100 counts per minute.
因此，6 天后计数率将为 100 次/分钟。
6. Wave Equation Applications | 波动方程应用
For all waves, the speed v is related to frequency f and wavelength λ by the wave equation:
v = f λ
对所有波，波速 v 与频率 f 和波长 λ 的关系由波动方程给出：
v = f λ
Example: A water wave has a frequency of 5 Hz and a wavelength of 0.4 m. Calculate the wave speed. If the wave speed remains the same but the frequency is increased to 10 Hz, what is the new wavelength?
📚 Common Misconceptions in GCSE Edexcel Physics | GCSE Edexcel 物理常见误区
Many students find GCSE Physics challenging not because the ideas are too difficult, but because their everyday intuition clashes with scientific principles. Unlearning these deeply held misconceptions is often the biggest hurdle on the road to top grades. This article unpacks the most common errors that Edexcel examiners see year after year, clarifying the correct physics in both English and Chinese so that you can avoid the same traps.
A classic misconception: mass and weight are interchangeable. In everyday language we say ‘I weigh 50 kilograms’, but in physics that statement is technically wrong. Mass is a measure of the amount of matter in an object, measured in kilograms (kg). It does not change wherever you go in the universe. Weight is the force due to gravity acting on that mass, measured in newtons (N). On Earth, weight = mass × gravitational field strength (W = m g), where g ≈ 10 N/kg (or 9.8 N/kg for Edexcel).
经典误区：把质量和重量混为一谈。日常中我们会说“我称起来 50 公斤”，但物理上这个说法严格来说是不对的。质量衡量的是物体所含物质的多少，单位是千克 (kg)，无论你身处宇宙何处都不会改变。重量则是作用在该质量上的重力，单位是牛顿 (N)。在地球上，重量 = 质量 × 重力场强度 (W = m g)，其中 g ≈ 10 N/kg（Edexcel 也使用 9.8 N/kg）。
If you take a 1 kg mass to the Moon, its mass remains 1 kg, but its weight drops to about 1.6 N because the Moon’s g is only about 1.6 N/kg. Edexcel questions often ask for the weight of an object given its mass; students mistakenly write ’50 kg’ instead of ‘500 N’. Remember: scales measure weight by measuring the force on a spring, even if the display shows ‘kg’.
如果你把 1 kg 的物体带到月球上，它的质量仍然是 1 kg，但重量会降到大约 1.6 N，因为月球的 g 只有约 1.6 N/kg。Edexcel 考题经常给出质量让你求重量；学生经常错误地写上“50 kg”而不是“500 N”。请记住：弹簧秤是通过测量弹簧受力来测重量的，即使显示屏显示的是“kg”。
2. A Force Is Needed to Keep Things Moving | 物体运动需要力来维持
It seems obvious: if you stop pushing a trolley, it stops. This observation, dating back to Aristotle, leads many students to believe that a constant force is required for constant velocity. Newton’s first law says otherwise: an object will remain at rest or move with a constant velocity unless acted upon by a resultant external force. The trolley stops because friction (an opposing force) is acting.
In space, where there is virtually no friction, a spacecraft with engines off continues moving at constant velocity. Edexcel exam questions frequently test this by showing a graph of velocity: a resultant force causes acceleration (change in velocity), not constant speed. If the forward driving force equals the backward resistive force, the resultant force is zero and the object moves at constant speed.
3. Current Gets ‘Used Up’ in a Circuit | 电流在电路中被“消耗掉”
A very persistent error: thinking that current is somehow used up as it passes through components. In a series circuit, the current is the same at all points. The number of charges flowing per second remains constant. What does get ‘used up’ or transferred is energy, not current.
In the water analogy, current is like the flow rate of water in a closed pipe loop – you cannot lose flow rate in one section unless there is a leak. Energy is like the pressure that drops across a water wheel. The examiner may ask you to state the ammeter reading at different points in a series circuit: if you say it is smaller after passing through a lamp, you lose the mark.
4. Voltage Is the Same as Energy, or Flows Around | 电压等同于能量，或者“流动”
Students often use the terms voltage and current interchangeably, or say ‘voltage flows through a wire’. Voltage (potential difference) does not flow; it is a measure of the energy transferred per unit charge between two points. The unit is the volt (V), where 1 V = 1 J/C.
学生经常把电压和电流两个术语混用，或者说“电压流过导线”。电压（电势差）并不流动；它衡量的是两点之间每单位电荷所转移的能量。单位是伏特 (V)，1 V = 1 J/C。
Think of voltage as the ‘push’ or ‘electrical pressure difference’ from a battery or power supply. In parallel circuits, the voltage across each branch is the same as the supply voltage, something many students find surprising. They expect voltage to split like current does in series, but voltage splits only across components in series.
Drop a hammer and a feather in air, and the hammer hits first. But this is due to air resistance, not gravity. Galileo’s insight was that in the absence of air resistance, all objects fall with the same acceleration due to gravity, g. The Edexcel specification expects you to know that g is constant near the Earth’s surface, about 10 m/s², independent of mass.
在空气中同时扔下锤子和羽毛，锤子先落地。但这是空气阻力的缘故，而非重力。伽利略的洞见是：如果没有空气阻力，所有物体都以相同的重力加速度 g 下落。Edexcel 大纲要求你知道，在地球表面附近 g 是一个常数，约 10 m/s²，与物体的质量无关。
For a falling object with no air resistance, the equation v² − u² = 2 a s applies, with a = g. All objects, regardless of mass, follow the same kinematics. Even in photographs or videos showing the Apollo 15 astronaut dropping a hammer and feather on the Moon, both hit the ground together. In exam questions, you must ignore air resistance unless explicitly told otherwise.
在没有空气阻力的情况下，下落物体的运动满足公式 v² − u² = 2 a s，其中 a = g。不论质量大小，所有物体都遵循相同的运动学规律。即便在阿波罗 15 号宇航员在月球上同时丢下锤子和羽毛的视频中，两者也是同时落地的。考试时，除非题目明确说明要考虑空气阻力，否则你应当忽略它。
6. Objects That Are Moving Must Have a Resultant Force Acting on Them | 运动的物体一定受到合外力的作用
Closely related to misconception #2, this emerges when students link force directly to motion rather than to acceleration. A car cruising at a steady 70 mph on a motorway has zero resultant force because the driving force precisely balances friction and air resistance. If there were a resultant force, the car would accelerate (speed up or slow down), not maintain constant speed.
When you see ‘terminal velocity’ questions with falling objects, the object initially accelerates due to weight being greater than air resistance. As air resistance builds up, it eventually equals weight; resultant force becomes zero, and the object falls at constant terminal velocity. The presence of motion does not imply the presence of a resultant force.
Conservation of energy is a fundamental principle: energy can be transferred usefully, stored, or dissipated, but it cannot be created or destroyed. A common mistake is to say ‘energy is used up’ in a circuit or in a mechanical process, as if energy vanishes. What really happens is that energy is transferred to less useful stores, usually thermal energy of the surroundings.
In a light bulb, electrical energy transfers to light and thermal energy. Even the light eventually heats up the walls of the room, increasing internal energy of the surroundings. Edexcel expects you to describe energy transfers in terms of stores: kinetic, gravitational potential, elastic potential, thermal, chemical, nuclear, electrostatic, magnetic. Avoid saying ‘heat energy’, use ‘thermal energy’ or ‘internal energy’.
8. Mistaking Reflection for Refraction and Vice Versa | 混淆反射和折射
Students can label the wrong angle on diagrams or think that refraction is just ‘light bouncing’. Reflection occurs when a wave strikes a surface and bounces back, obeying the law of reflection: angle of incidence = angle of reflection, measured from the normal. Refraction is the bending of light as it passes from one medium to another of different optical density, and it occurs because the speed of light changes.
For a ray entering glass from air, it bends towards the normal because light slows down. When it leaves the glass, it speeds up and bends away from the normal. If the ray strikes along the normal, no bending occurs but the speed still changes. Edexcel also requires you to know that wavelength changes during refraction, but frequency stays the same.
9. The Total Resistance in a Parallel Circuit Is Higher | 并联电路的总电阻更大
Adding a resistor in parallel provides an additional path for current, so the total resistance of the circuit decreases. Many students incorrectly add resistances for parallel branches in the same way as for a series circuit. For two resistors R₁ and R₂ in parallel, 1/R_total = 1/R₁ + 1/R₂. This always yields a total less than even the smallest individual resistor.
Think of a supermarket: adding an extra checkout till reduces the overall ‘resistance’ to the flow of shoppers. In Edexcel required practicals, you might measure current and voltage to calculate resistance. If you connect an extra bulb in parallel, the total circuit current increases because the total resistance decreases, even though the supply voltage stays the same.
10. Nuclear Fission and Fusion Are the Same | 核裂变和核聚变是一回事
Despite both releasing energy from nuclear changes, fission and fusion are opposite processes. Fission is the splitting of a large, unstable nucleus (e.g. uranium-235 or plutonium-239) into two smaller nuclei, along with two or three neutrons and a large amount of energy. Fusion is the joining of two light nuclei (e.g. isotopes of hydrogen) to form a heavier nucleus, releasing even more energy. Both involve mass being converted to energy according to E = m c².
尽管两者都通过核变化释放能量，但裂变和聚变是相反的过程。裂变是一个大质量的不稳定核（如铀-235 或钚-239）分裂成两个较小的核，同时释放出两到三个中子以及巨大的能量。聚变是两个轻核（如氢的同位素）结合形成一个较重的核，释放出更大的能量。两者都涉及根据 E = m c² 质量转化为能量。
Edexcel students should know that fusion is the process powering the Sun and other stars, requiring very high temperatures and pressures to overcome electrostatic repulsion between positively charged nuclei. Fission occurs in nuclear power stations, where control rods absorb excess neutrons and moderators slow them down to sustain a chain reaction. A common exam error is stating that fusion happens in reactors today.
11. All Power Equations Can Be Used Interchangeably | 所有功率公式都可以随意混用
While energy transferred = power × time (E = P t) is always valid for any energy transfer, electrical power equations must be used carefully. P = I V is universally true for any component. P = I² R and P = V² / R are derived from it using V = I R, but this means they are only valid if the component obeys Ohm’s law (constant resistance). For a filament bulb or diode where resistance changes, using P = I² R with a single resistance value is incorrect unless the instantaneous values are used.
虽然能量转移公式 E = P t 对任何能量传递过程都成立，但电功率公式的使用必须谨慎。P = I V 对任何元件都普遍适用。P = I² R 和 P = V² / R 是从 P = I V 结合 V = I R 推导出来的，这意味着它们仅当元件遵循欧姆定律（电阻恒定）时才成立。对于灯丝灯泡或二极管等电阻会变化的情况，除非使用瞬时值，否则用单个电阻值代入 P = I² R 是错误的。
An Edexcel exam question may give you a current – voltage graph for a non-ohmic conductor and ask for power at a particular point. You must read I and V from the graph and use P = I V, not try to calculate R and then use P = I² R (unless you first calculate the correct R at that specific point). Sticking to P = I V keeps you safe.
Edexcel 的考题可能会给出非欧姆导体的电流 — 电压图像，然后让你求出某一点的功率。你必须从图像上读取 I 和 V，并使用 P = I V，而不是先算出 R 然后代入 P = I² R（除非你已经算出了该特定点的正确电阻值）。坚持使用 P = I V 最为稳妥。
12. The Direction of Current Is the Same as Electron Flow | 电流方向与电子流动方向相同
Conventional current is defined as the direction in which positive charges would move, from the positive terminal to the negative terminal of a battery. In reality, in metallic wires, only negatively charged electrons are free to move, and they drift from negative to positive. Students often swap these when labelling arrows on circuit diagrams or describing flow. Edexcel expects you to know this and to draw arrows for conventional current.
In electrolysis, both positive and negative ions carry charge, and current is still defined as the net movement of positive charge. When answering questions about magnetic fields around a wire, always use the right-hand grip rule with conventional current direction, not electron flow, to determine the direction of the magnetic field lines.
Circular motion is a fundamental topic in both IB and WJEC physics syllabi, describing the motion of an object along a circular path. Mastering angular quantities, centripetal acceleration, and the forces that cause such motion is vital for solving problems involving loops, banked curves, satellites, and conical pendulums. This guide breaks down every key concept with clear explanations, derived equations using only Unicode symbols, and exam-focused tips to help you avoid common pitfalls.
An object is in circular motion if its trajectory forms a circle. In uniform circular motion, the object’s speed (magnitude of velocity) remains constant, yet the velocity vector changes continuously because its direction is always tangent to the circle. This continuous change in velocity implies the presence of an acceleration directed towards the centre, known as centripetal acceleration.
The motion can be described using linear quantities (displacement s, velocity v) along the circumference, or angular quantities (angular displacement Δθ, angular velocity ω) measured in radians. The two sets are linked by the radius r of the circle.
描述这种运动时，既可以用沿圆周的线量（弧长 s、线速度 v），也可以用使用弧度制表示的角量（角位移 Δθ、角速度 ω）。这两套物理量通过圆半径 r 相互联系。
2. Angular Displacement, Velocity and Speed | 角位移、角速度与线速度
Angular displacement Δθ is the angle subtended at the centre when an object moves along an arc. It must be measured in radians for the fundamental relationships to be valid. One radian is the angle for which arc length equals the radius.
The instantaneous angular velocity ω is the rate of change of angular displacement: ω = Δθ/Δt. The linear speed v of the object is related to ω by the equation:
Since v is the magnitude of the tangential velocity, its direction is always perpendicular to the radius. The arc length s travelled in time t is s = rθ (with θ in radians), and differentiating this with respect to time gives v = rω. This connection is essential for converting between linear and angular descriptions in all circular motion problems.
因为 v 是切线方向速度的大小，它的方向始终与半径垂直。在弧度制下，弧长 s = rθ，两边对时间求导即得 v = rω。这一联系是所有圆周运动问题中在角度描述与线量描述之间转换的核心。
3. Period and Frequency | 周期与频率
The period T is the time taken for one complete revolution. Frequency f is the number of revolutions per second, so f = 1/T. Angular velocity can be expressed in terms of T and f:
ω = 2π/T = 2πf
周期 T 是物体完成一整圈运动所需的时间。频率 f 是每秒转动的圈数，满足 f = 1/T。角速度可以用 T 和 f 表示为：ω = 2π/T = 2πf。
Substituting these into v = rω gives the useful relationships:
v = 2πr/T = 2πrf
将其代入 v = rω，得到常用关系式：v = 2πr/T = 2πrf。
When solving problems, always check whether the system rotates with a constant period (e.g. a satellite) or whether the period varies. Many exam questions ask for the velocity or acceleration at a specific point in the motion, requiring you to link f or T to the linear quantities.
解题时要分清系统是否以恒定周期旋转（例如卫星），还是周期在变化。许多考题会要求你计算某一点的线速度或加速度，这时就要将 f 或 T 与线量关联起来。
4. Centripetal Acceleration – Magnitude and Direction | 向心加速度的大小和方向
For an object moving with constant speed v in a circle of radius r, the magnitude of the centripetal acceleration aₙ is:
aₙ = v²/r = ω²r
物体以恒定速率 v 在半径为 r 的圆周上运动时，向心加速度 aₙ 的大小为：aₙ = v²/r = ω²r。
The direction of aₙ is always towards the centre of the circle, perpendicular to the instantaneous velocity. Since the acceleration is perpendicular to velocity, it changes the direction of the velocity without changing its magnitude. This is why uniform circular motion still requires a resultant centripetal force.
In terms of period T, the magnitude can also be written as aₙ = 4π²r/T². This form is particularly helpful when analysing planetary motion or objects with a known rotation period.
利用周期 T 可将向心加速度写为 aₙ = 4π²r/T²。在分析行星运动或已知旋转周期的问题时，这种形式特别有用。
5. Derivation of Centripetal Acceleration | 向心加速度的推导
Consider an object moving at constant speed v in a circle. Over a small time interval Δt, the object moves from point A to B, sweeping out an angle Δθ. The change in velocity Δv = v•B – v•A is directed towards the centre. By geometry, the triangle formed by velocities and the triangle formed by the radii are similar, giving |Δv|/v = Δs/r, where Δs is the chord length. For small Δt, chord length ≈ arc length vΔt.
设想物体以恒定速率 v 做圆周运动。在很短的时间 Δt 内，物体从 A 点运动到 B 点，转过的角度为 Δθ。速度的变化量 Δv = v_B – v_A 指向圆心。由几何关系，速度矢量三角形与半径三角形相似，因此有 |Δv|/v = Δs/r，其中 Δs 是弦长。Δt 很小时，弦长可近似为弧长 vΔt。
Using the approximate relation Δs ≈ vΔt, we obtain |Δv| ≈ (v × vΔt)/r. Dividing by Δt gives the magnitude of acceleration:
This derivation shows why the acceleration points radially inward and why its magnitude is v²/r. The same reasoning, expressed in terms of ω, yields aₙ = ω²r. Both forms appear regularly in IB and WJEC exam questions, so understanding the underlying geometry helps you remember which formula to use.
6. Centripetal Force: It’s Not a New Force | 向心力：并非新的力
Newton’s second law applied to the centripetal acceleration gives the centripetal force Fₙ required to keep an object moving in a circle:
Fₙ = m aₙ = mv²/r = mω²r
将牛顿第二定律用于向心加速度，即得到维持圆周运动所需的向心力 Fₙ：Fₙ = m v²/r = m ω²r。
Many students mistakenly treat centripetal force as a distinct “new” force, like gravity or tension. In reality, centripetal force is simply the name given to the resultant force directed towards the centre of the circle. It is provided by one or more real forces – tension in a string, friction on a flat track, the normal component on a banked curve, or gravitational attraction for a satellite.
When drawing free-body diagrams, never include a separate arrow labelled “centripetal force”. Instead, identify the physical forces present and then equate their radial component to mv²/r or mω²r. This approach will keep you safe from the common “centrifugal force” distraction in exams.
When an object rotates in a horizontal circle, the net radial inward force provides the centripetal force. A classic example is a ball on a string swung in a horizontal circle with negligible air resistance; the tension T acts radially inward, so T = mv²/r.
物体在水平面内做圆周运动时，指向圆心的径向合力提供向心力。一个典型例子是忽略空气阻力时绳子拴着的小球在水平面旋转：绳子拉力 T 指向圆心，因此 T = mv²/r。
Another common case is a mass resting on a rough turntable. Static friction f supplies the centripetal force up to its maximum value fₘₐₓ = μₛ m g. The condition for the mass not to slip is:
mv²/r ≤ μₛ m g ⇒ v ≤ √(μₛ g r)
另一种常见情况是粗糙转盘上的物体。静摩擦力 f 提供向心力，其最大值为 f_max = μₛ m g。物体不滑动的条件为：mv²/r ≤ μₛ m g，即 v ≤ √(μₛ g r)。
In such problems, always identify the force that acts towards the centre. If friction acts perpendicular to the instantaneous velocity, it does no work, so speed can stay constant even without a motor. Exam questions often ask for the minimum coefficient of friction or the maximum angular velocity before slipping occurs.
For an object attached to a string and whirled in a vertical circle, tension varies with position. At the top, both tension and weight point downwards, contributing to the centripetal force:
At the bottom, tension acts upward while weight acts downward, so:
T_bottom – mg = mv²_bottom / r
在最低点，拉力向上而重力向下，有：T_bottom – mg = mv²_bottom / r。
For a string to remain taut at the top, the speed must be large enough that the required centripetal force is at least mg. The critical minimum speed at the top is v_min = √(gr). If v_top < √(gr), the string goes slack and the object leaves the circular path. For a rigid rod, however, the rod can exert a compressive force, so the speed at the top can be as low as zero.
The same principles apply to a car moving over a convex hill. At the top of the hill, the normal reaction N is less than the weight:
mg – N = mv²/r
汽车过拱桥顶部也是类似情形。在拱顶，支持力 N 小于重力：mg – N = mv²/r。速度越大，N 越小，当 v = √(gr) 时 N = 0，车辆将失去路面支撑。
9. Banked Tracks and Conical Pendulum | 倾斜轨道与圆锥摆
A banked curve allows a vehicle to negotiate a corner at a certain speed without relying on friction. The horizontal component of the normal reaction provides the centripetal force. Resolving forces for a bank angle θ and neglecting friction gives:
N sinθ = mv²/r
N cosθ = mg
Dividing the two equations yields the ideal banking angle condition:
A conical pendulum consists of a mass attached to a string tracing out a horizontal circle at constant angular velocity. The string sweeps out a cone. The same force resolution applies, with the radius r = L sinθ, where L is the string length. The period of rotation is independent of the mass and is given by T = 2π√(L cosθ/g), a classic result often examined.
圆锥摆是通过绳子将小球系住，小球在水平面内做匀速圆周运动，悬线扫出圆锥面的形状。受力分解与倾斜轨道相同，半径满足 r = L sinθ（L 为绳长）。旋转周期与摆球质量无关，T = 2π√(L cosθ/g)，这是一个经典考点。
10. Gravitational Force as Centripetal – Orbits and Satellites | 万有引力提供向心力 – 轨道与卫星
For a satellite of mass m in a circular orbit about a planet of mass M, the gravitational attraction supplies the centripetal force:
G M m / r² = mv²/r
质量为 m 的卫星在质量 M 的行星周边做圆周轨道运动时，万有引力提供向心力：G M m/r² = mv²/r。
Cancelling m gives the orbital speed:
v = √(GM/r)
这表明轨道速度与卫星质量无关：v = √(GM/r)。
Using v = 2πr/T, the orbital period T satisfies Kepler’s third law:
T² = (4π²/GM) r³
The period increases with orbital radius. Geostationary satellites have a period of 24 hours and an orbital radius of about 42,200 km from the Earth’s centre. Questions often require you to derive the orbital radius from T or vice versa, and to compare the speeds and periods of satellites at different altitudes.
利用 v
Published by TutorHao | IB Physics Revision Series | aleveler.com