At the turn of the 20th century, physics stood at a crossroads. Classical mechanics, built on Newton’s laws and Maxwell’s equations, had triumphed in explaining the macroscopic world — from the orbits of planets to the propagation of light. Yet a series of puzzling experimental results defied classical explanation. The quantum revolution that followed fundamentally changed our understanding of matter and energy.
Two phenomena in particular — the photoelectric effect and wave-particle duality — shattered the classical worldview and laid the foundation for quantum mechanics. This article explores both topics in depth, following the A-Level Physics syllabus.
The Photoelectric Effect: Experimental Observations
When Heinrich Hertz first observed the photoelectric effect in 1887, he could not have anticipated the theoretical upheaval it would cause. The experiment is deceptively simple: shine light of a sufficiently high frequency onto a clean metal surface, and electrons are ejected. Yet the details of this emission defied classical wave theory.
Classical wave theory made three predictions, all of which were contradicted by experiment. First, any frequency of light should eventually eject electrons if the intensity is high enough — the wave’s energy would accumulate over time. Second, increasing the intensity of the light should increase the kinetic energy of the emitted electrons. Third, there should be a measurable time delay between illumination and electron emission, as the electron absorbs energy from the wave.
None of these predictions held. Below a certain threshold frequency f₀, no electrons were emitted regardless of intensity. Above the threshold, increasing intensity produced more electrons but did not increase their maximum kinetic energy. And emission was instantaneous, even at the lowest intensities. These results demanded a radically new explanation.
Albert Einstein’s genius lay in taking Planck’s quantization of energy — originally a mathematical trick to solve the blackbody radiation problem — and treating it as a physical reality. Einstein proposed that light consists of discrete packets of energy called photons. Each photon carries energy E = hf, where h is Planck’s constant (6.63 × 10⁻³⁴ J·s) and f is the frequency of the light.
阿尔伯特·爱因斯坦的天才之处在于，他将普朗克的能量量子化——最初只是解决黑体辐射问题的数学技巧——视为物理现实。爱因斯坦提出，光由称为光子的离散能量包组成。每个光子携带能量 E = hf，其中 h 是普朗克常数（6.63 × 10⁻³⁴ J·s），f 是光的频率。
In Einstein’s model, a single photon interacts with a single electron. The electron requires a minimum energy — called the work function φ (phi) — to escape the metal surface. Any excess photon energy becomes the electron’s kinetic energy. This yields the photoelectric equation:
This elegant equation explained all the experimental anomalies. The threshold frequency f₀ corresponds to hf₀ = φ — any photon with lower frequency simply lacks the energy to liberate an electron, regardless of how many photons arrive. Increasing intensity means more photons, hence more electrons ejected, but each photon still carries the same energy hf, so the maximum kinetic energy remains unchanged. And the one-to-one photon-electron interaction explains the instantaneous emission.
The photoelectric effect provides one of the most direct methods for measuring Planck’s constant. In the laboratory, a photoelectric cell is illuminated with monochromatic light of various known frequencies. A variable retarding potential V is applied to stop the most energetic electrons — the stopping potential V₀ at which the photocurrent drops to zero.
光电效应提供了测量普朗克常数最直接的方法之一。在实验室中，用各种已知频率的单色光照射光电管。施加可变的减速电压 V 来阻止能量最高的电子——使光电流降为零的截止电压 V₀。
The work done by the electric field in stopping an electron equals its kinetic energy: eV₀ = Ek(max). Substituting into Einstein’s equation gives eV₀ = hf − φ, which rearranges to V₀ = (h/e)f − φ/e. A graph of V₀ against f yields a straight line with gradient h/e and y-intercept −φ/e. Since the electronic charge e is known, Planck’s constant can be determined directly from the gradient.
Millikan’s famous 1916 experiment used this method and confirmed Einstein’s photoelectric equation with remarkable precision. Ironically, Millikan had set out to disprove Einstein’s photon model but ended up providing its strongest experimental support — a testament to the integrity of the scientific method.
The photoelectric effect established that light, traditionally understood as a wave, also behaves as a particle. But the symmetry of nature demanded a reciprocal question: could particles of matter, such as electrons, also exhibit wavelike behaviour?
In 1924, Louis de Broglie proposed exactly this in his PhD thesis. De Broglie suggested that any moving particle has an associated wavelength, now called the de Broglie wavelength, given by:
where p is the particle’s momentum, m is its mass, and v is its velocity. This was a breathtaking proposal — if true, it meant that electrons, protons, and even macroscopic objects had wavelengths, albeit typically far too small to detect.
其中 p 是粒子的动量，m 是其质量，v 是其速度。这是一个令人惊叹的提议——如果成立，这意味着电子、质子，甚至宏观物体都有波长，尽管通常小到无法检测。
For an electron accelerated through a potential difference of 100 V, the de Broglie wavelength is approximately 1.2 × 10⁻¹⁰ m, comparable to the spacing between atoms in a crystal. This suggested a crucial experimental test: if electrons have wavelike properties, they should produce diffraction patterns when passed through a crystal lattice, just as X-rays do.
对于一个通过 100 V 电势差加速的电子，德布罗意波长约为 1.2 × 10⁻¹⁰ 米，与晶体中原子间距相当。这提示了一个关键的实验检验：如果电子具有波的特性，当它们通过晶格时应该产生衍射图样，就像 X 射线一样。
The Davisson-Germer Experiment (1927)
The experimental confirmation of de Broglie’s hypothesis came from Davisson and Germer at Bell Labs. They were studying electron scattering from a nickel crystal when a fortunate accident occurred — their vacuum chamber broke, oxidizing the nickel sample. After annealing the nickel to remove the oxide layer, the crystal reformed into a regular lattice structure, and the scattered electrons produced a clear diffraction pattern.
The experiment showed intensity peaks at specific angles that perfectly matched the predictions of the Bragg diffraction condition (nλ = 2d sin θ), typically used for X-ray diffraction. The wavelength calculated from the diffraction pattern agreed precisely with the de Broglie wavelength for the electron’s momentum. Electrons really did behave as waves.
实验显示在特定角度出现强度峰值，与通常用于 X 射线衍射的布拉格衍射条件（nλ = 2d sin θ）完美吻合。从衍射图样计算出的波长与电子动量的德布罗意波长精确一致。电子确实表现出波的特性。
The experiment earned Davisson the Nobel Prize in Physics in 1937. Today, electron diffraction is a standard technique in materials science and structural biology, used routinely in electron microscopes and crystallography.
The double-slit experiment, first performed with light by Thomas Young in 1801, is arguably the most beautiful demonstration of wave-particle duality. When coherent light passes through two narrow slits, it produces an interference pattern of alternating bright and dark fringes on a screen — a definitive signature of wave behaviour.
In 1961, Claus Jönsson performed the experiment with electrons, and the results were stunning. When electrons were fired one at a time through the double slit, individual impacts appeared as discrete dots on the detector, consistent with particle behaviour. But over time, as thousands of electrons accumulated, the dots built up into a clear interference pattern — the hallmark of waves.
This raises a profound question: if individual electrons pass through the apparatus one at a time, what are they interfering with? The answer forces us to abandon classical intuition — each electron somehow passes through both slits simultaneously and interferes with itself. The electron is neither purely a particle nor purely a wave; it is a quantum object that exhibits properties of both, depending on how we measure it.
The orthodox interpretation of quantum mechanics, developed principally by Niels Bohr and Werner Heisenberg, is known as the Copenhagen interpretation. Central to this interpretation is Bohr’s principle of complementarity: wave and particle aspects of a quantum system are complementary — both are needed for a complete description, but they can never be observed simultaneously in the same experiment.
This is not merely a practical limitation but a fundamental feature of nature. The type of measurement we choose determines which aspect of the quantum object manifests. An apparatus designed to measure the interference pattern (e.g., a screen that records electron positions) reveals the wave nature; an apparatus designed to determine which slit each electron passes through destroys the interference pattern, revealing the particle nature.
This insight has profound implications. It means that in quantum mechanics, the observer is not a passive spectator but an active participant. The act of measurement does not simply reveal a pre-existing property — it brings that property into existence.
Electron Microscopy: Practical Applications of Wave-Particle Duality
The wave nature of electrons is not merely a philosophical curiosity — it has practical applications that have transformed science. The electron microscope exploits the short de Broglie wavelength of high-energy electrons to achieve resolving power far beyond what optical microscopes can manage.
The resolving power of a microscope is limited by diffraction, which is governed by the wavelength of the radiation used. Visible light has wavelengths around 400-700 nm, limiting optical microscopes to resolving objects no smaller than about 200 nm. In contrast, electrons accelerated through 100 kV have a de Broglie wavelength of about 0.004 nm — over 100,000 times shorter. This allows transmission electron microscopes (TEMs) to resolve individual atoms and scanning electron microscopes (SEMs) to produce detailed three-dimensional images of surfaces at the nanoscale.
A thorough understanding of the photon is essential for A-Level Physics. Despite having no rest mass, photons possess both energy and momentum. The energy of a photon is E = hf = hc/λ, where c is the speed of light. The momentum p of a photon follows from the relativistic energy-momentum relation: for a massless particle, E = pc, giving p = E/c = hf/c = h/λ.
透彻理解光子对 A-Level 物理至关重要。尽管光子没有静止质量，但它同时具有能量和动量。光子的能量为 E = hf = hc/λ，其中 c 是光速。光子的动量 p 来自相对论能量-动量关系：对于无质量粒子，E = pc，因此 p = E/c = hf/c = h/λ。
This momentum is real and measurable. When photons strike a surface, they exert radiation pressure — a phenomenon that has been proposed for solar sail propulsion in spacecraft. The Compton effect (1923), in which X-ray photons scatter from electrons with a measurable wavelength shift, provided direct confirmation of photon momentum.
这种动量是真实可测的。当光子撞击表面时会产生辐射压力——这一现象已被提议用于航天器的太阳帆推进。康普顿效应（1923 年），即 X 射线光子从电子散射时产生可测量的波长变化，直接证实了光子动量。
A common exam pitfall: the photoelectric equation Ek(max) = hf − φ uses the photon energy hf, not the photon momentum. Students sometimes confuse this with the energy of an emitted electron. Remember that the work function φ represents the minimum energy to liberate an electron from the metal surface, akin to the ionization energy of an atom but specific to the metallic bonding environment.
Wave-particle duality and the photon model provide the key to understanding atomic spectra. When an electron in an atom transitions from a higher energy level E₂ to a lower one E₁, it emits a photon whose energy equals the difference: hf = E₂ − E₁. Similarly, an atom can absorb a photon only if its energy exactly matches the gap between two energy levels.
This explains why atomic spectra consist of discrete lines rather than continuous bands — energy levels in atoms are quantized. Each element has a unique set of energy levels, giving it a characteristic emission and absorption spectrum. This is the basis of spectroscopy, one of the most powerful analytical tools in science, used in astronomy to determine the composition of stars and in forensics to identify substances.
In A-Level Physics examinations, questions on quantum phenomena typically follow certain patterns. A classic question provides a graph of stopping potential against frequency and asks you to determine Planck’s constant and the work function from the gradient and intercept. Another common style presents a table of photon wavelengths and asks whether photoemission will occur for given metals with known work functions.
Key skills tested include: converting between frequency and wavelength using c = fλ, calculating photon energy in both joules and electronvolts (1 eV = 1.60 × 10⁻¹⁹ J), applying the photoelectric equation correctly, and explaining the failure of classical wave theory to account for the experimental observations. You should also be able to calculate de Broglie wavelengths and interpret electron diffraction data.
考查的关键技能包括：使用 c = fλ 在频率和波长之间转换，以焦耳和电子伏特（1 eV = 1.60 × 10⁻¹⁹ J）两种单位计算光子能量，正确应用光电方程，以及解释经典波动理论为何无法解释实验观察结果。你还应该能够计算德布罗意波长并解释电子衍射数据。
For the highest marks, examiners look for precise language: photons interact one-to-one with electrons; the work function is the minimum energy required; kinetic energy refers specifically to the maximum kinetic energy of emitted electrons, since electrons deeper in the metal lose energy escaping. Demonstrating an understanding of these subtleties distinguishes top-grade answers.
The journey from the photoelectric effect to wave-particle duality represents one of the most significant paradigm shifts in the history of science. In the space of three decades, physicists were forced to abandon the comfortable certainty of classical determinism and embrace a reality where particles are waves, waves are particles, and measurement itself shapes what we observe.
For A-Level students, mastery of this topic requires fluency with these essential equations:
对于 A-Level 学生来说，掌握这个主题需要熟练运用以下基本方程：
E = hf = hc/λ — photon energy
Ek(max) = hf − φ — photoelectric equation
λ = h/p = h/mv — de Broglie wavelength
eV₀ = hf − φ — stopping potential relationship
p = h/λ — photon (and particle) momentum
E = hf = hc/λ — 光子能量
Ek(max) = hf − φ — 光电方程
λ = h/p = h/mv — 德布罗意波长
eV₀ = hf − φ — 截止电压关系
p = h/λ — 光子（和粒子）动量
Understanding these equations, their experimental origins, and their physical meaning provides not only exam success but a genuine appreciation of the quantum world that underpins all of modern technology — from the semiconductors in your smartphone to the lasers in fibre-optic communications.
Waves form a core part of the IB Physics syllabus, bridging fundamental concepts with more advanced applications such as interference, standing waves and the Doppler effect. Whether you are tackling Standard Level or Higher Level, mastering wave behaviour is essential for top marks on both Paper 1 and Paper 2. This guide unpacks the key ideas, essential equations and common pitfalls to help you revise efficiently.
A wave is a propagating disturbance that transfers energy without permanently displacing the medium. Key descriptors include amplitude (A), the maximum displacement from equilibrium, and wavelength (λ), the distance between two successive points in phase. The time for one complete oscillation is the period (T), and the frequency (f) counts oscillations per second.
The relationship between period and frequency is reciprocal, and wave speed is linked to wavelength and frequency. Graphs of displacement against distance (snapshot) or against time (history) for a point allow you to extract A, λ and T directly. Remember that the speed of a mechanical wave depends only on the medium, not on frequency or amplitude.
In a transverse wave, particles oscillate perpendicular to the direction of energy transfer; electromagnetic waves and waves on strings are typical examples. In a longitudinal wave, particles oscillate parallel to the propagation direction, as in sound waves. Only transverse waves can be polarised, a property exploited in IB question scenarios to distinguish wave types.
For any wave, displacement against position graphs help visualise compressions and rarefactions in longitudinal waves by mapping pressure or density variations. The wavelength remains the distance between successive compressions or crests.
The equation v = fλ is central to wave calculations. When a wave passes from one medium to another, its speed and wavelength change, but the frequency stays fixed because it is set by the source. For a string under tension, the speed depends on tension and linear density; for sound in air, on temperature.
方程 v = fλ 是波计算的核心。当波从一种介质进入另一种介质时，波速和波长发生变化，但频率保持不变，因为它由波源决定。对于张紧的弦，波速取决于张力和线密度；对于空气中的声波，取决于温度。
v = f λ and v = √(T/μ) for a stretched string
An IB problem might ask you to compute wavelength given speed and frequency, or deduce how the wavelength alters when a water wave enters shallower water. Always identify which quantity remains constant before applying the formula.
4. Wavefronts, Rays and Huygens’ Principle | 波前、射线与惠更斯原理
A wavefront joins points in phase, such as crests, and rays show the direction of energy flow, perpendicular to wavefronts. Huygens’ principle explains propagation, reflection and refraction by treating every point on a wavefront as a source of secondary spherical wavelets. The new wavefront is the envelope of these wavelets.
Understanding wavefronts is particularly useful when drawing refraction diagrams: the change in spacing of wavefronts indicates the change in speed and wavelength. These sketches often appear in IB examinations as part of qualitative questions.
5. Reflection, Refraction and Diffraction | 反射、折射与衍射
Waves obey the law of reflection: angle of incidence equals angle of reflection. Refraction occurs when waves cross a boundary into a medium of different wave speed, changing direction unless incidence is normal. Snell’s law quantifies this.
Diffraction is the spreading of a wave when it passes through a gap or around an obstacle. It is most pronounced when the aperture size is comparable to the wavelength. This fundamental behaviour underpins single-slit patterns and the resolution of optical instruments.
When two or more waves meet, the net displacement is the vector sum of individual displacements – the principle of superposition. Constructive interference occurs when waves are in phase, giving maximum amplitude; destructive interference occurs when they are out of phase by 180° (π rad), leading to cancellation.
For sustained interference patterns, sources must be coherent – that is, they maintain a constant phase difference and have the same frequency. The condition for constructive interference is a path difference Δx = nλ; for destructive interference it is Δx = (n + ½)λ, where n is an integer.
Young’s historic experiment demonstrates light interference using two narrow slits illuminated by a single coherent source. Bright fringes appear where path difference is an integer multiple of λ, dark fringes where it is a half-integer multiple. The fringe separation Δy is given by a simple geometric formula.
Here D is the distance from the slits to the screen and d is the slit separation. This equation is frequently tested, so pay attention to unit conversions and the fact that Δy is inversely proportional to d. It also allows determination of wavelength for unknown light.
其中 D 是双缝到屏幕的距离，d 是缝间距。这一方程式常被考查，注意单位转换以及 Δy 与 d 成反比。该公式还可用于测定未知光的波长。
8. Single-Slit Diffraction and Resolution | 单缝衍射与分辨率
When monochromatic light passes through a single narrow slit of width a, a central bright maximum dominates, flanked by dimmer fringes. Minima occur at angles given by a sinθ = nλ (n = 1, 2, 3…). The width of the central maximum is a measure of the spreading due to diffraction.
当单色光通过宽度为 a 的单狭缝时，形成一个占主导的中央亮纹，两侧是较暗的条纹。暗纹的位置满足 a sinθ = nλ (n = 1, 2, 3…)。中央亮纹的宽度反映了衍射引起的展宽程度。
a sinθ = nλ
Resolution of two point sources is limited by diffraction. The Rayleigh criterion states that two sources are just resolved when the first minimum of one coincides with the peak of the other. For a circular aperture of diameter b, the minimum resolvable angle is θ = 1.22λ / b; for a slit it is approximately λ / a.
Standing waves form when two identical travelling waves move in opposite directions and superpose. They exhibit nodes (zero displacement) and antinodes (maximum displacement). Common examples are waves on a plucked string or in an air column inside a pipe.
For a string fixed at both ends, the allowed wavelengths are λ = 2L / n (n = 1,2,3…), giving frequencies f = n (v / 2L). In an open pipe, identical harmonic series apply; in a closed pipe, only odd harmonics exist: f = n (v / 4L) with n = 1,3,5…
对于两端固定的弦，允许的波长为 λ = 2L / n (n = 1,2,3…)，频率为 f = n (v / 2L)。在开管中，谐波系列相同；在闭管中，仅存在奇次谐波：f = n (v / 4L)，其中 n = 1,3,5…
fₙ = n × (v / 2L) (both ends fixed or open) fₙ = n × (v / 4L) (one end closed, n odd)
10. Doppler Effect | 多普勒效应
The Doppler effect is the change in observed frequency when a source and observer move relative to each other. For sound, the observed frequency f’ increases when source and observer approach and decreases when they recede. The general formula involves the speeds of source vₛ and observer vₒ.
For electromagnetic waves, the relativistic Doppler shift is different, but IB only requires a qualitative appreciation: light from receding galaxies is redshifted (longer λ), while approaching objects produce blueshift. The Doppler effect is used in speed cameras, echocardiography and astronomy.
Polarisation is a phenomenon exclusive to transverse waves, in which oscillations are restricted to a single plane. Unpolarised light can be polarised by a filter (Polaroid). Malus’s law describes the intensity transmitted through a second polariser at angle θ to the first.
Polarisation by reflection occurs at Brewster’s angle, where reflected and refracted rays are perpendicular. The condition is tanθB = n₂ / n₁. This concept often appears in HL optics questions alongside Malus’s law.
All electromagnetic waves travel at speed c = 3.00 × 10⁸ m s⁻¹ in a vacuum and consist of oscillating electric and magnetic fields perpendicular to each other and to the direction of propagation. The spectrum, in order of decreasing wavelength, includes radio, microwave, infrared, visible, ultraviolet, X-ray and gamma radiation.
所有电磁波在真空中以速度 c = 3.00 × 10⁸ m s⁻¹ 传播，并由相互垂直且垂直于传播方向的振荡电场和磁场构成。波谱按波长递减的顺序包括无线电波、微波、红外线、可见光、紫外线、X 射线和伽马射线。
Newton’s three laws of motion form the backbone of classical mechanics and are a high-priority topic in the WJEC IGCSE Physics specification. Understanding these laws not only helps you solve numerical problems involving force, mass and acceleration but also enables you to explain a wide range of everyday phenomena, from why seatbelts are essential to how rockets launch into space. This article breaks down each law in detail, highlights key definitions, covers essential practical investigations, and points out typical exam pitfalls so you can approach any Newton’s laws question with confidence.
1. The Big Picture: Forces, Motion and Vectors | 全局视角：力、运动与矢量
Before diving into Newton’s laws, it is essential to recall that force is a vector quantity, meaning it has both magnitude and direction. In WJEC IGCSE Physics, you will often need to combine forces acting along a straight line or at right angles. The net or resultant force is the single force that has the same effect as all the original forces acting together. If the resultant force on an object is zero, the forces are balanced. If the resultant force is not zero, the forces are unbalanced and the object will accelerate in the direction of the resultant force.
Key vector concepts you will apply include drawing free-body diagrams, resolving forces into components, and understanding that acceleration is always in the direction of the resultant force. The SI unit of force is the newton (N), where 1 N is the force required to accelerate a 1 kg mass by 1 m/s².
2. Newton’s First Law: The Law of Inertia | 牛顿第一定律：惯性定律
Newton’s first law states that an object will remain at rest or continue to move at constant velocity unless acted upon by a resultant external force. This property of objects to resist changes in their state of motion is called inertia. The greater an object’s mass, the greater its inertia, and the harder it is to change its velocity.
In the WJEC exam, you might be asked to explain situations such as a passenger lurching forward when a car brakes suddenly. The person continues moving forward due to inertia while the car decelerates. Similarly, when a bus accelerates, standing passengers tend to fall backwards because their bodies resist the change in motion. A tablecloth pulled quickly from under dishes works because the dishes have high inertia and remain in place if the force from the cloth is brief enough.
A common misconception is that a continuous force is needed to keep an object moving. Newton’s first law clarifies that no force is required to maintain constant velocity; forces only cause changes in velocity (acceleration).
3. Mass, Weight and Gravitational Field Strength | 质量、重量与重力场强度
Before studying the second law, it is vital to distinguish clearly between mass and weight. Mass is a scalar quantity measuring the amount of matter in an object; it is measured in kilograms (kg) and does not change with location. Weight is a force caused by gravity acting on a mass. It is a vector, measured in newtons (N), and depends on the gravitational field strength (g). The relationship is given by the equation:
On Earth, g ≈ 9.8 m/s², but for most IGCSE calculations, you may use g = 10 m/s² if specified. On the Moon, g ≈ 1.6 m/s², so an object’s weight would be about one-sixth of its Earth weight, whereas its mass remains unchanged.
在地球表面，g ≈ 9.8 m/s²，但在大多数IGCSE计算中，若题目说明可使用 g = 10 m/s²。月球表面的 g ≈ 1.6 m/s²，因此同一物体在月球上的重量约为地球上的六分之一，但其质量不变。
Property / 性质
Mass / 质量
Weight / 重量
Definition / 定义
Amount of matter
Gravitational force on a mass
Scalar or vector / 标量或矢量
Scalar
Vector
Unit / 单位
kilogram (kg)
newton (N)
Varies with location? / 随位置变化？
No
Yes
Measured with / 测量工具
Balance / 天平
Spring scale / 弹簧秤
In free-fall situations where only gravity acts, the weight is the resultant force, providing the acceleration due to gravity. This directly connects to Newton’s second law.
在只受重力作用的自由落体运动中，重量即为合力，产生重力加速度。这直接与牛顿第二定律相关联。
4. Newton’s Second Law: Relating Force, Mass and Acceleration | 牛顿第二定律：力、质量与加速度的关系
Newton’s second law is arguably the most important quantitative law in mechanics. It states that the acceleration of an object is directly proportional to the resultant force acting on it and inversely proportional to its mass. The direction of the acceleration is the same as the direction of the resultant force. Mathematically, this is expressed as:
Here, F is the resultant force in newtons (N), m is the mass in kilograms (kg), and a is the acceleration in metres per second squared (m/s²). Always remember that F stands for the resultant (net) force, not just any single force. If multiple forces act, you must first find the vector sum.
A typical WJEC exam question provides the thrust of an engine and the friction opposing motion. The resultant force = thrust – friction. Then, using F = m a, you can calculate acceleration. For example, a 1200 kg car experiences an engine force of 3000 N and a total resistive force of 600 N.
Resultant force = 3000 N – 600 N = 2400 N. Acceleration a = F / m = 2400 / 1200 = 2.0 m/s².
典型的WJEC试题会给出引擎推力和阻碍运动的摩擦力。合力 = 推力 – 摩擦力。然后利用 F = m a 计算加速度。例如，一辆1200 kg的汽车受到3000 N的引擎力和600 N的总阻力。
合力 = 3000 N – 600 N = 2400 N。加速度 a = F / m = 2400 / 1200 = 2.0 m/s²。
The second law also helps explain why heavy vehicles have slower acceleration for the same engine force, and why reducing mass improves a racing car’s performance. It is also used to determine the required braking force to achieve a certain deceleration.
5. Free-Body Diagrams and Resultant Force | 受力分析图与合力
Drawing clear free-body diagrams is an essential skill for applying Newton’s laws correctly. In WJEC IGCSE Physics, you are expected to represent all forces acting on an object using arrows. The length of each arrow should be proportional to the magnitude of the force, and the direction must be accurate. Common forces include weight (downwards), normal reaction (perpendicular to the surface), friction (opposing motion), tension (along a rope), and applied forces.
Draw arrows for all forces acting on that object, not forces the object exerts on other things.
If necessary, resolve forces at angles into horizontal and vertical components.
Calculate the resultant force by vector addition.
Apply F = m a along the direction of the resultant force.
分析问题时，可按以下步骤进行：
明确研究对象。
用一个点或方框代表该物体。
画出所有作用在该物体上的力（注意：不是该物体施加给其他物体的力）。
如有需要，将斜向的力分解为水平和竖直分量。
通过矢量加法求出合力。
沿合力方向应用 F = m a。
A common exam mistake is including forces that act on different objects in the same free-body diagram. Always ask: ‘Is this force acting on my chosen object?’ For a skydiver, the forces acting are weight and air resistance, not the force of the skydiver pushing on the air.
6. Newton’s Third Law: Action–Reaction Pairs | 牛顿第三定律：作用力与反作用力
Newton’s third law states that whenever two objects interact, they exert equal and opposite forces on each other. In other words, if body A exerts a force on body B, then body B simultaneously exerts a force of equal magnitude but opposite direction on body A. These two forces are called an action–reaction pair.
It is vital to note that the two forces in a third-law pair always act on different bodies. For instance, when you push against a wall, your hand exerts a force on the wall, and the wall exerts an equal and opposite force back on your hand. This is why you can feel the pressure. The forces do not cancel out because they operate on different objects. In the WJEC exam, you may be asked to identify action–reaction pairs in contexts such as a rocket launch: the rocket pushes exhaust gases downwards, and the gases push the rocket upwards.
Other classic examples include a swimmer pushing water backwards to move forwards, a bird’s wings pushing air downwards so air pushes the bird upwards, and the recoil of a gun when a bullet is fired. In each case, the pair of forces are equal in size, opposite in direction, and act on different objects.
A common pitfall is confusing third-law pairs with balanced forces. Balanced forces act on the same object and result in zero resultant force. Action–reaction forces act on two distinct objects, so they cannot cancel each other from the perspective of either object.
7. Required Practical: Investigating F = ma | 必做实验：探究 F = ma
WJEC IGCSE Physics includes an essential practical to verify Newton’s second law. The typical setup uses a dynamics trolley on a slightly sloped runway to compensate for friction, so that the trolley moves at constant velocity when given a gentle push. A known mass is hung over a pulley, providing a constant accelerating force equal to its weight (mhanging × g). As the hanging mass falls, it pulls the trolley along.
Force and acceleration (mass kept constant): Keep the total mass of the system (trolley + hanging masses) constant by transferring masses from the trolley to the hanger. Measure the acceleration for different hanging forces using a light gate and data logger or by timing how long it takes the trolley to pass between two points. Plot acceleration (y-axis) against force (x-axis); a straight line through the origin confirms a ∝ F when mass is constant.
Mass and acceleration (force kept constant): Keep the hanging mass constant (thus constant force) and change the mass on the trolley by adding slotted masses. Measure acceleration for each total mass. Plot acceleration against 1/mass; a straight line through the origin confirms a ∝ 1/m when force is constant.
该实验需探究两组关系：
力与加速度（保持质量不变）：通过将小车上的砝码转移到挂钩上，保持系统总质量不变。测量不同拉力下的加速度，可使用光电门和数据记录器，或测量小车通过两点间的时间。以加速度为纵轴，力为横轴作图；一条通过原点的直线确认质量不变时 a ∝ F。
质量与加速度（保持力不变）：保持悬挂物质量恒定（即力恒定），通过增加小车上的槽码改变小车质量。测量各总质量下的加速度。以加速度为纵轴，1/质量 为横轴作图；一条通过原点的直线确认力不变时 a ∝ 1/m。
In analysing data, calculate acceleration using the equation v = u + a t or v² = u² + 2 a s, where appropriate. Make sure to convert all masses to kg and forces to N. Typical WJEC exam questions might ask you to identify sources of error, such as friction not fully compensated, the string not being parallel to the track, or timing inaccuracies.
在数据分析中，可根据适当的公式计算加速度，如 v = u + a t 或 v² = u² + 2 a s。确保所有质量单位统一为 kg，力为 N。典型的WJEC考题可能要求你指出误差来源，例如摩擦力未完全补偿、细绳未与轨道平行，或计时不准确等。
8. Terminal Velocity and Drag Forces | 终极速度与阻力
Newton’s laws provide the perfect framework to explain terminal velocity, a concept frequently examined in WJEC IGCSE Physics. When an object falls through a fluid (a liquid or gas), it experiences a drag force that opposes its motion. For a skydiver, the main drag force is air resistance. Initially, the only force is weight, so the resultant force is downwards and the skydiver accelerates at g.
牛顿定律为解释终极速度提供了完美的框架，这也是WJEC IGCSE物理中常考的概念。当物体在流体（液体或气体）中下落时，会受到与其运动方向相反的阻力。对于跳伞者而言，主要的阻力是空气阻力。起初，只受重力作用，合力向下，跳伞者以加速度 g 加速下落。
As speed increases, air resistance increases. The resultant downward force decreases, so acceleration decreases. Eventually, air resistance equals weight. At this point, the resultant force is zero, and by Newton’s first law, the skydiver continues to fall at a constant maximum speed called terminal velocity. If the skydiver opens a parachute, the surface area increases dramatically, causing a sudden large upward drag force. This gives a resultant upward force, decelerating the skydiver until a new, much lower terminal velocity is reached.
You might also be asked to interpret velocity–time graphs for falling objects. The graph initially shows a steep curve (large acceleration), gradually levelling off to a horizontal line (terminal velocity). Key features to label include the point of parachute opening causing a rapid deceleration.
Stokes’ law is not required at IGCSE, but you should understand that drag increases with speed and depends on the shape and cross-sectional area of the object. Streamlining reduces drag, allowing higher speeds before drag balances the driving force.
9. Applying Newton’s Laws: Safety and Transport | 牛顿定律的应用：安全与交通
Newton’s laws are not just theoretical; they have life-saving applications. Seatbelts, airbags, and crumple zones in cars are all designed using the implications of the first and second laws. In a collision, the car stops abruptly, but unrestrained passengers continue moving forward due to inertia. Seatbelts provide the necessary resultant force to decelerate the passengers over a slightly longer time, reducing the force experienced because F = (change in momentum) / time.
牛顿定律并非纯理论，它在保护生命安全方面有着实际应用。汽车的安全带、安全气囊和溃缩区都是基于第一和第二定律的原理设计的。在碰撞中，汽车突然停止，而未受约束的乘客由于惯性会继续向前运动。安全带提供了所需的合力，使乘客在稍长的时间内减速，因为 F = 动量变化 / 时间，时间延长，所受力就减小。
Crumple zones at the front of a car crush on impact, increasing the time over which the car decelerates. A longer stopping time results in a smaller average force on the occupants, again by the second law via the impulse–momentum relationship. Airbags provide a soft surface that extends the deceleration time for the head and chest.
WJEC questions may ask you to explain why a heavy lorry needs a longer braking distance than a car, even at the same speed. Because kinetic energy depends on mass and the square of speed, a heavier vehicle has more energy to dissipate. With a similar maximum braking force (limited by friction), the deceleration a = F / m will be smaller for a larger mass, so the stopping distance increases. This combines Newton’s second law with work–energy principles.
WJEC考题可能要求解释为何即使是相同速度，重型卡车的刹车距离也比小汽车更长。由于动能取决于质量和速度的平方，更重的车辆具有更多需要耗散的能量。在最大制动力相似的情况下（受摩擦力限制），质量越大，减速度 a = F / m 越小，因此制动距离增加。这综合了牛顿第二定律与功能原理。
10. Common Mistakes and Exam Tips | 常见错误与应试技巧
To score top marks on Newton’s laws questions in WJEC IGCSE Physics, watch out for these common errors:
Confusing mass and weight: Always check units and contexts, especially on the Moon or other planets.
Forgetting that F is the resultant force: If a question gives thrust and friction, subtract them first before using F = m a.
Mixing up action–reaction pairs with balanced forces: Remember that third-law pairs act on different objects.
Neglecting unit conversions: Ensure mass is in kg, not grams; forces in N, not mN or kN without conversion.
Incorrectly drawing or interpreting vectors: Arrows representing forces must be correctly scaled and labelled.
Assuming acceleration is constant for terminal velocity problems: Acceleration decreases as speed increases, so motion is not uniformly accelerated.
要在WJEC IGCSE物理的牛顿定律题目中取得高分，请留意以下常见错误：
混淆质量与重量：务必检查单位和上下文，尤其是在月球或其他行星上。
忘记 F 代表合力：若题目给出推力和摩擦力，先求差再代入 F = m a。
混淆作用力–反作用力对与平衡力：牢记第三定律中的力对作用在不同物体上。
忽略单位换算：确保质量用 kg，而非 g；力用 N，在计算前完成 mN 或 kN 的换算。
矢量图绘制或解读错误：表示力的箭头必须按比例绘制并正确标注。
终极速度问题中误认为加速度恒定：随着速度增加，加速度减小，因此运动并非匀加速。
When tackling numerical questions, write down the known quantities, convert to SI, write the relevant equation, substitute carefully, and state the answer with correct units and direction where applicable. For explanation questions, use precise language: cite the specific Newton’s law, relate it to the forces involved, and describe the resulting motion.
📚 Mastering Application Questions: Oxford AQA International A Level Physics | 牛津AQA国际A Level物理应用题突破技巧
Application questions in the Oxford AQA International A Level Physics exam assess your ability to apply physics concepts to unfamiliar, real-world situations. They require more than rote recall: you must interpret scenarios, model them with appropriate principles, perform calculations, and justify your reasoning. This guide shares proven techniques to tackle these questions confidently and score top marks.
The very first step is to decode what the question is asking. Oxford AQA exam papers use specific command words such as ‘State’, ‘Describe’, ‘Explain’, ‘Calculate’, ‘Determine’, ‘Show that’, ‘Evaluate’, and ‘Suggest’. Each word indicates a distinct type of response. For example, ‘State’ requires a brief, factual answer without justification; ‘Describe’ asks for a detailed sequence of events or properties; ‘Explain’ demands a link between scientific principles and the observation.
‘Show that’ questions often involve proving a given result, typically using standard formulas. You must show every step of your working; the examiners want to see your logical flow. ‘Evaluate’ calls for a balanced discussion of advantages and disadvantages, often with a conclusion based on evidence. Highlight the command word as you read the question to stay focused on what is expected.
AQA International A Level papers embed physics in contexts like sports, medical imaging, particle accelerators, or renewable energy. To extract the physics, identify measurable quantities: speeds, forces, energies, wavelengths, etc. Then simplify the situation by deciding which assumptions to make (e.g., no air resistance, negligible friction, ideal gas behaviour). Drawing a simple diagram can help to visualise forces, energy transfers, or ray paths.
For instance, a problem about a bungee jumper involves gravitational potential energy, kinetic energy, and elastic potential energy in the rope. You need to use energy conservation with Hooke’s law. Always cross-check whether the context aligns with the principles you intend to use – for example, is the motion relativistic? (Likely not at A Level.) Practise translating descriptive text into a list of given data and symbols.
Application questions often have several parts (a), (b), (c) that build on each other. Approach them systematically: read all parts first to understand the overall story. Often part (a) asks for a basic calculation, part (b) uses that result in a new calculation, and part (c) requires interpretation or evaluation. If you get stuck on part (a), you can still attempt later parts using a sensible assumed value or by stating the formula that would be used.
When solving, label each step clearly. Use a structured approach: write the relevant formula, substitute values with units, calculate, and present the answer with the correct unit and appropriate significant figures. This not only earns method marks but also reduces careless errors. For long calculations, factor in possible intermediate rounding: keep extra digits during working and round only at the final answer.
Success in application questions hinges on your ability to select the correct principle from your toolkit. Key principles across the AQA International A Level syllabus include Newton’s laws, conservation of energy, conservation of momentum, Coulomb’s law, wave superposition, and the photoelectric effect. Practice recognising which principle is triggered by the context: a collision often implies conservation of momentum; an accelerating charge may point to electromagnetic induction.
Create a mental checklist: is there a net force? Use F = ma. Is height changing? Gravitational potential energy mgh. Is the speed of an object changing? Kinetic energy ½mv². Are there two interacting bodies? Conservation of momentum. Use the table below for quick reference:
Scenario clue
Likely Principle
Collision / explosion
Conservation of momentum
Charges stationary or moving slowly
Coulomb’s law F = kQ₁Q₂/r²
Particle in electric/magnetic field
F = qE or F = Bqv (if perpendicular)
Photoelectric emission
hf = Φ + ½mv²ₘₐₓ
制作思维清单：有无净力？用 F = ma。高度变化？重力势能 mgh。物体速率变化？动能 ½mv²。两个物体相互作用？动量守恒。可参考下表快速匹配：
Examiners value clarity. Use the GFSCAU method: Given – list the known quantities with symbols and values; Formula – state the equation you will use; Substitution – replace symbols with numbers including units; Calculation – perform the arithmetic; Answer – write the result with units; Unit/Check – verify the unit and the order of magnitude
Published by TutorHao | A-Level Physics Revision Series | aleveler.com
Kinematics is the study of how objects move, covering displacement, velocity, acceleration, and the equations that link these quantities. Mastering this topic is essential for success in Edexcel Physics, as it forms the foundation for both mechanics problems and practical analysis.
In kinematics, we distinguish between scalar quantities (magnitude only) and vector quantities (magnitude and direction). Distance and speed are scalars; displacement, velocity and acceleration are vectors.
Distance (scalar)： how much ground an object has covered; Displacement (vector)： the overall change in position with direction.
距离（标量）：物体经过路径的总长度；位移（矢量）：位置的整体变化且带方向。
Direction matters when solving problems. For example, if a car travels 100 m north then 60 m south, the distance travelled is 160 m, but the displacement is 40 m north.
2. Displacement, Velocity and Acceleration | 位移、速度与加速度
Displacement s (m) is a vector from the initial to the final position. Velocity v (m s⁻¹) is the rate of change of displacement. Acceleration a (m s⁻²) is the rate of change of velocity.
位移 s（单位米）是从初位置指向末位置的矢量。速度 v（单位 m s⁻¹）是位移的变化率。加速度 a（单位 m s⁻²）是速度的变化率。
Constant velocity means equal displacements in equal time intervals; constant acceleration means velocity changes equally in equal time intervals. Deceleration is negative acceleration relative to a chosen positive direction.
The equations linking these quantities assume constant acceleration and motion in a straight line unless otherwise stated.
联系这些量的方程假定加速度恒定且物体沿直线运动，除非另有说明。
3. SUVAT Equations Introduction | 匀加速运动方程入门
For uniformly accelerated motion, five interrelated SUVAT equations are used. The variables are s (displacement), u (initial velocity), v (final velocity), a (acceleration), and t (time).
These four are the most commonly used. Note that each includes at least one of u, v, a, s, t. Always list known and unknown variables before choosing the equation without the unknown.
To solve problems： define a positive direction; write down the five variables, marking known values and the target variable; select the SUVAT that does not include the other unknown.
解题步骤：规定正方向；列出五个变量，标出已知值和目标量；选择不含其他未知量的 SUVAT 方程。
Example: A car accelerates from 5 m s⁻¹ to 25 m s⁻¹ over a displacement of 60 m. Find acceleration. Known： u = 5, v = 25, s = 60; unknown a and t. Use v² = u² + 2as → a = (v² – u²)/(2s) = (625 – 25)/120 = 5.0 m s⁻².
示例：一辆汽车从 5 m s⁻¹ 加速到 25 m s⁻¹，位移为 60 m。求加速度。已知：u=5，v=25，s=60；未知 a 和 t。用 v² = u² + 2as → a = (25² – 5²)/(2×60) = 5.0 m s⁻²。
Watch for hidden zeroes： starting from rest (u = 0), coming to rest (v = 0), or free fall where u = 0 when dropped. Always include units.
注意隐藏的零值：从静止出发（u = 0）、停止（v = 0）、或自由落体释放瞬间 u = 0。务必携带单位。
5. Free Fall under Gravity | 重力作用下的自由落体
Near Earth’s surface, all objects experience a downward acceleration due to gravity, g = 9.81 m s⁻² (Edexcel typically uses g = 9.81 m s⁻²). In the absence of air resistance, this is constant.
在地球表面附近，所有物体都受到向下的重力加速度，g = 9.81 m s⁻²（Edexcel 通常采用 9.81 m s⁻²）。若忽略空气阻力，该加速度恒定。
For vertical motion, choose upward or downward as positive. If upward is positive, a = -g = -9.81 m s⁻². An object thrown upward will decelerate to v = 0 at maximum height before falling back.
处理竖直运动时，可规定向上或向下为正。若向上为正，则 a = -g = -9.81 m s⁻²。向上抛出的物体会减速，在最高点 v = 0，而后下落。
Example： An object is dropped from a height. u = 0, a = 9.81 m s⁻² downward. Use s = ½gt² to find the distance fallen in t seconds.
示例：物体从高处释放，u = 0，向下 a = 9.81 m s⁻²。用 s = ½gt² 求 t 秒内下落距离。
6. Motion Graphs: Displacement-Time | 位移-时间图像
A displacement-time (s-t) graph plots displacement along the chosen direction against time. The gradient gives velocity, as v = Δs/Δt.
位移-时间 (s-t) 图将选定方向上的位移对时间作图。斜率给出速度，因为 v = Δs / Δt。
The area under an s-t graph has no physical meaning. Only the gradient is relevant.
s-t 图下的面积没有物理意义，只关心斜率。
7. Motion Graphs: Velocity-Time | 速度-时间图像
A velocity-time (v-t) graph plots instantaneous velocity against time. The gradient gives acceleration; the area between the graph and time axis gives displacement.
速度-时间 (v-t) 图将瞬时速度对时间作图。斜率表示加速度；图线与时间轴围成的面积表示位移。
Constant acceleration： straight sloping line. Area = area of trapezium = ½(u+v)t, consistent with SUVAT.
匀加速：一条倾斜直线。面积 = 梯形面积 = ½(u+v)t，与 SUVAT 一致。
Non-constant acceleration： curved line, area found by counting squares or approximation.
非匀加速：曲线，面积通过数格或近似法求得。
When the line crosses the time axis, velocity changes sign, indicating a reversal of direction.
当图线穿过时间轴时，速度改变符号，表示运动方向反转。
8. Deriving and Using Graphs | 图像推导与应用
Area under a v-t graph is displacement (vector). If v is negative, the area counts as negative displacement. Total distance travelled is the sum of absolute areas.
v-t 图下的面积是位移（矢量）。若 v 为负，面积算作负位移。经过的总路程是绝对面积之和。
Acceleration could be obtained from the gradient of a v-t graph, or by drawing a tangent to a displacement-time graph. In experiments, using graphical methods often reduces random errors.
Practical skills： plot data points, draw a line of best fit, avoid forcing through origin unless justified. Use large triangles for gradient to improve precision.
实验技能：描点、画最佳拟合线，除非有理论依据，否则不强制过原点。用大三角形计算斜率以提高精确度。
9. Projectile Motion Basics | 抛体运动基础
In Edexcel Physics, projectile motion is treated as two independent perpendicular components： horizontal with constant velocity (a = 0) and vertical with uniform acceleration g downwards.
在 Edexcel 物理中，抛体运动被分解为两个相互垂直的独立分量：水平方向匀速（a = 0），竖直方向以恒定加速度 g 向下。
Horizontal： uₓ = u cos θ, sₓ = uₓ t.
水平：uₓ = u cos θ，sₓ = uₓ t。
Vertical： uᵧ = u sin θ; use SUVAT with a = -g (if upward positive). Time of flight connects both components.
竖直：uᵧ = u sin θ；使用 SUVAT，a = -g（设向上为正）。飞行时间是联系两分量的桥梁。
To find time to highest point, set vᵧ = 0. Total flight time is 2u sin θ / g for symmetric launch and landing at same level.
求达到最高点的时间，令 vᵧ = 0。在起落点高度相同的情况下，总飞行时间为 2u sin θ / g。
Remember that the horizontal component of velocity remains unchanged, so motion is parabolic.
牢记水平速度分量保持不变，因此轨迹为抛物线。
10. Common Pitfalls and Exam Tips | 常见陷阱与应试技巧
Mixing vector and scalar quantities without direction is a frequent mistake. Always define your positive direction and stick to it consistently.
混淆矢量与标量而不标明方向是常见错误。务必规定正方向并始终如一地使用。
SUVAT restrictions： only for constant acceleration in a straight line. Do not apply if acceleration changes or motion is in two dimensions unless separated into components.
SUVAT 的使用限制：仅适用于沿直线的匀加速运动。若加速度变化或运动是二维的，需分解后才可应用。
When using v² = u² + 2as, misplacing a negative sign is common. Insert values with signs (+ / -) according to chosen direction.
使用 v² = u² + 2as 时，很容易弄错负号。要根据选定的正方向带符号（+/-）代入数值。
In graphs, the phrase ‘area under graph’ means a physical quantity only for v-t (displacement) and a-t (change in velocity). Not for s-t. Read questions precisely.
Newton’s laws of motion form the foundation of classical mechanics, explaining how forces affect the motion of objects. For the GCSE OCR Physics specification, you need to understand all three laws, apply them to real‑world situations, perform calculations using F=ma, and confidently draw free‑body diagrams. This revision guide breaks down each law with clear explanations, worked examples, and exam tips to help you achieve top marks.
1. Introduction to Forces and Newton’s Laws | 力与牛顿定律导论
A force is a push or a pull that can change an object’s motion. Forces are vector quantities, meaning they have both magnitude and direction. The unit of force is the newton (N). Newton’s three laws describe the relationship between a body and the forces acting upon it, and how that body moves as a result. Understanding them is essential for explaining everything from a car accelerating to a rocket launching.
In OCR GCSE Physics, you will be expected to recall each law, apply them to unfamiliar contexts, and perform quantitative analysis. The key concepts are inertia, resultant force, mass, acceleration, and action‑reaction pairs. Let’s start with the first law.
2. Newton’s First Law (Law of Inertia) | 牛顿第一定律（惯性定律）
Newton’s first law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by a resultant force. This means if the forces on an object are balanced (resultant force = 0), its velocity will not change. A stationary object stays stationary, and a moving object continues to move at constant speed in a straight line.
This property of an object resisting a change in its motion is called inertia. Inertia is the tendency of an object to keep doing what it is already doing. The greater the mass of an object, the greater its inertia – it is harder to change its motion.
For example, when a bus suddenly stops, standing passengers lurch forward. Their bodies continue moving forward due to inertia, even though the bus has stopped. Seat belts are designed to provide a force to counteract this inertia during sudden deceleration.
Mass is a measure of inertia. In physics, mass is not the same as weight. Mass is a scalar quantity measured in kilograms (kg), and it represents how much matter an object contains. A larger mass means a larger resistance to acceleration when the same force is applied.
Think about pushing a shopping cart versus pushing a car. Both are on a flat surface, but the car has much more mass. To give it the same acceleration, you would need a much larger force. If you push with the same force, the car will accelerate much less. This directly leads us to Newton’s second law.
Newton’s second law describes what happens when a resultant force does act on an object. It states that the acceleration of an object is directly proportional to the resultant force acting on it, and inversely proportional to its mass. This is often written as the equation:
where F is the resultant force in newtons (N), m is the mass in kilograms (kg), and a is the acceleration in metres per second squared (m/s²). This equation is only valid when the mass remains constant.
其中 F 是合外力，单位为牛顿 (N)，m 是质量，单位为千克 (kg)，a 是加速度，单位为米每二次方秒 (m/s²)。该等式仅在质量恒定时才成立。
The direction of the acceleration is always the same as the direction of the resultant force. If the resultant force is zero, acceleration is zero – which is consistent with the first law. The second law allows us to calculate forces, masses, or accelerations in a huge range of problems.
Exam questions will often give you two quantities and ask you to find the third. Rearranging the formula is a key skill:
考试题目通常会给出两个量，让你求第三个量。变换公式是关键技能：
To find resultant force: F = m × a
求合外力：F = m × a
To find mass: m = F ÷ a
求质量：m = F ÷ a
To find acceleration: a = F ÷ m
求加速度：a = F ÷ m
Always ensure units are consistent: mass in kg, force in N, acceleration in m/s². If mass is given in grams or tonnes, convert first. A typical worked example:
Example: A car of mass 1200 kg accelerates at 2.5 m/s². Calculate the resultant force from the engine.
示例：一辆质量为 1200 kg 的汽车以 2.5 m/s² 的加速度行驶。计算发动机产生的合外力。
F = m × a = 1200 × 2.5 = 3000 N
The resultant force is 3000 N in the direction of acceleration. Note: in real situations, there will be resistive forces (friction, air resistance) so the driving force must exceed the total resistive forces to provide this resultant force.
A crucial application of F=ma is calculating the weight of an object. Weight is the force due to gravity acting on a mass. It is a vector quantity, directed towards the centre of the Earth (or whatever planet you are on). The weight equation is:
where W is weight in newtons (N), m is mass in kg, and g is the gravitational field strength. On Earth, g ≈ 9.8 N/kg (often rounded to 10 N/kg in some GCSE questions, but OCR typically uses 9.8 unless told otherwise).
Students often confuse mass and weight. Always remember: mass is a measure of the amount of matter and does not change with location; weight is a force and will change if g changes (e.g., on the Moon g ≈ 1.6 N/kg, so you weigh less).
学生经常混淆质量和重量。始终记住：质量是衡量物质多少的量，不随地点改变；重量是一种力，会随着 g 的变化而改变（例如，月球上 g ≈ 1.6 N/kg，所以你的重量会变小）。
7. Newton’s Third Law (Action-Reaction) | 牛顿第三定律（作用力与反作用力）
Newton’s third law states that whenever two objects interact, the forces they exert on each other are equal in magnitude and opposite in direction. This is often phrased as “for every action, there is an equal and opposite reaction.” However, “action” and “reaction” are just forces; they always come in pairs acting on two different objects.
Key point: the two forces in an action‑reaction pair act on different objects. If you push on a wall, the wall pushes back on you with an equal force. The forces are of the same type (both contact forces), equal in size, opposite in direction, but they do not cancel out because they act on different bodies.
Common example of third law: a rocket engine expels gas downwards; the gas pushes the rocket upwards. The forces are equal and opposite, but the rocket accelerates upward because the reaction force acts on the rocket.
When multiple forces act on an object, you must first find the resultant force. Draw a free‑body diagram to identify all forces. Then use F=ma in the direction of the resultant force. A common scenario is an object moving on a rough surface: driving force, friction, weight, and normal reaction are all present.
For vertical equilibrium on a horizontal surface: normal reaction = weight. For horizontal acceleration: resultant force = driving force − friction. If the driving force equals friction, resultant force is zero, so velocity is constant (first law).
Many exam questions describe a car reaching a constant speed. This means the resistive forces (drag + friction) balance the driving force, net force = 0, so acceleration = 0 and the car maintains constant velocity.
Drawing and interpreting free‑body diagrams is an essential skill. Represent the object as a dot or a box, and draw arrows representing all forces acting on it. The arrows must be labelled and their relative lengths should indicate the relative magnitudes when forces are not equal.
For a book resting on a table: downward arrow “Weight” (W = mg) and upward arrow “Normal reaction” (R) of equal length. No horizontal forces, so the book remains at rest. If you push the book horizontally, add a push force arrow and a friction arrow opposing the motion. If it moves at constant speed, push and friction are equal length.
Always remember: forces acting on the object only. Do not include forces the object exerts on other things. That’s a very common mistake.
始终记住：只画出作用在该物体上的力。不要包含该物体施加给其他物体的力。这是一个非常常见的错误。
10. Terminal Velocity and Drag | 终端速度与阻力
When an object falls through a fluid (liquid or gas), it experiences drag (air resistance or fluid resistance) opposing its motion. Drag increases with speed. At release, weight is the only force, so acceleration = g. As it speeds up, drag increases. When drag equals weight, resultant force becomes zero, and the object stops accelerating. It continues to fall at a constant speed called terminal velocity.
For a skydiver: initially she accelerates downwards. Then, as air resistance builds up, her acceleration decreases until air resistance balances weight. She reaches terminal velocity (about 200 km/h). When she opens her parachute, the surface area increases dramatically, causing a huge drag force, deceleration, and then a new, much lower terminal velocity.
This beautifully demonstrates Newton’s first and second laws: no resultant force means constant velocity, and resultant force controls acceleration.
这完美地展示了牛顿第一和第二定律：没有合力意味着恒定速度，而合力控制加速度。
11. Summary: Key Points for Exams | 概要：考试关键点
Here is a concise summary of the main ideas you must know for GCSE OCR Physics:
这是 GCSE OCR 物理你需要掌握的主要内容简要总结：
Law 定律
Statement / Formula 陈述 / 公式
Key Implication 关键含义
First Law 第一定律
Object stays at rest or moves at constant velocity if resultant force = 0
Inertia; equilibrium
Second Law 第二定律
F = m × a
Unbalanced force causes acceleration
Third Law 第三定律
Action and reaction are equal, opposite, and act on different objects
Forces always come in pairs
Also remember the weight formula W = mg, and the conditions for terminal velocity. When analysing motion, always identify all forces, calculate the resultant, and then apply the second law.
还记得重量公式 W = mg，以及达到终端速度的条件。在分析运动时，务必识别所有力，计算合力，然后应用第二定律。
12. Common Mistakes and How to Avoid Them | 常见错误及避免方法
Mistake 1: Confusing mass and weight. Remember mass is measured in kg, weight is a force measured in N. Use W = mg to convert.
错误 1：混淆质量和重量。记住质量以 kg 为单位，重量是力，以 N 为单位。用 W = mg 进行换算。
Mistake 2: Forgetting that F in F=ma is the resultant force, not just a single applied force. Always subtract opposing forces first.
错误 2：忘记 F=ma 中的 F 是合外力，而不仅仅是单个作用力。务必先减去反向的力。
Mistake 3: Thinking action‑reaction forces cancel. They act on different objects, so they cannot cancel each other out in terms of the motion of one object.
Mistake 4: Incorrect unit conversions. Always convert grams to kg (÷1000), and km/h to m/s (÷3.6) when needed for equations like F=ma or kinetic energy.
错误 4：单位换算不正确。当用到 F=ma 或动能公式时，务必把克转换为千克（÷1000），把 km/h 转换为 m/s（÷3.6）。
Mistake 5: Assuming acceleration is constant if an object moves. Constant velocity means zero acceleration, which means zero resultant force (first law).
错误 5：以为物体运动就一定有恒定的加速度。匀速意味着加速度为零，即合外力为零（第一定律）。
By carefully reading questions, drawing diagrams, and checking units, you can avoid these common pitfalls and confidently answer any Newton’s laws problem.
通过仔细阅读题目、绘制图示并检查单位，你可以避免这些常见陷阱，自信地回答任何牛顿定律的问题。
Published by TutorHao | Physics Revision Series | aleveler.com
📚 AS Physics Unit 2 Insert Jan19 Concept Breakdown | AS物理 Unit 2 插入材料概念解析
The AQA AS Physics Unit 2 (Mechanics, Materials and Waves) January 2019 insert is more than just a data sheet; it is a condensed toolkit of equations, constants, and reference diagrams. Understanding the physical principles behind each formula is essential for accurate problem-solving and conceptual analysis. This revision guide unpacks every key concept from the insert, pairing clear definitions with worked examples to build your confidence.
The insert highlights the universal wave equation v = f λ, where v is wave speed, f is frequency and λ is wavelength. This relationship holds for all wave types, including sound, light, and water waves. When a wave passes from one medium to another, its speed and wavelength change, but its frequency remains constant because it is determined by the source.
插入材料中强调普适的波动方程 v = f λ，其中 v 为波速，f 为频率，λ 为波长。该关系对所有波类型都成立，包括声波、光波和水波。当波从一种介质进入另一种介质时，波速和波长会改变，但频率保持不变，因为它由波源决定。
In calculations, rearrange the equation to find the unknown quantity. For instance, visible light has frequencies around 10¹⁴–10¹⁵ Hz. If green light of frequency 5.6 × 10¹⁴ Hz travels in vacuum (v = 3.00 × 10⁸ m s⁻¹)， its wavelength is λ =
Published by TutorHao | AS Physics Revision Series | aleveler.com
This article explores the derivation of the fundamental SUVAT equations for uniformly accelerated linear motion, a topic prominently featured in both Edexcel A Level Physics and the Pearson IB Physics HL Volume 3 course, as well as being referenced in OCR’s Table of Contents (TOC) for mechanics. By breaking down each step using definitions and graphical analysis, we demonstrate how these essential kinematic relationships emerge from first principles, equipping students with a robust conceptual toolkit for tackling problems across multiple exam boards.
本文深入探讨匀加速直线运动的基本SUVAT方程组的推导过程。这一主题在Edexcel A Level物理和Pearson IB Physics HL第三册教材中均占有核心位置，同时也出现在OCR考试局力学部分的目录(TOC)中。我们将通过定义和图像分析逐步拆解每个推导步骤，展示这些关键运动学关系如何从基本原理产生，帮助同学们建立跨考试局通用的扎实物理思维。
1. The Context of Kinematics in IB & Edexcel | IB与Edexcel运动学背景
The equations of motion for constant acceleration form the backbone of classical mechanics. In the Pearson IB Physics HL V3 textbook, the topic appears early in the ‘Mechanics’ section, closely aligned with OCR A Level specifications. Edexcel Physics also treats this as a core competency, requiring students not only to apply SUVAT but also to trace their derivation from velocity–time graphs or calculus.
匀加速运动方程是经典力学的基石。在Pearson IB物理HL第三册教材中，该主题位于“力学”部分的前端，与OCR A Level大纲高度吻合。Edexcel物理同样将其列为基本技能，不仅要求学生应用SUVAT公式，还要能够从速度–时间图或微积分出发进行推导。
2. Defining the SUVAT Variables | 定义SUVAT变量
Before any derivation, we must clearly define the five quantities involved: s is displacement (m), u is initial velocity (m s⁻¹), v is final velocity (m s⁻¹), a is constant acceleration (m s⁻²), and t is time (s). These are scalar magnitudes when motion is rectilinear; direction is embedded via sign convention, a point stressed in both IB HL and Edexcel mark schemes.
Acceleration is defined as the rate of change of velocity. For constant acceleration a, we write a = (v − u) / t. Rearranging this definition immediately yields the first SUVAT equation:
加速度定义为速度的变化率。对于恒定加速度a，可写成a = (v − u) / t。整理这一关系直接得到第一个SUVAT方程：
v = u + at
This simple algebraic step is often the starting point in Pearson’s IB V3 derivation. Note that the equation assumes a is uniform; any variation invalidates the SUVAT framework.
4. Graphical Insight from a v–t Graph | 从v–t图获得的图形洞见
Plotting velocity against time for motion with constant acceleration gives a straight line with slope a and intercept u. The area under this line between t = 0 and t = t equals the displacement s. This geometric fact provides the most intuitive path to the remaining equations.
5. Derivation of s = ut + ½ at² | 推导 s = ut + ½ at²
The total area under the v–t graph can be split into a rectangle (area u × t) and a triangle (area ½ × t × at). Summing these gives displacement: s = ut + ½ at². This derivation appears in both Edexcel and OCR mechanics materials and is explicitly illustrated in the IB HL V3 textbook.
v–t图下的总面积可分割为一个矩形(面积u × t)和一个三角形(面积½ × t × at)。两者相加得位移：s = ut + ½ at²。这一推导见于Edexcel和OCR力学资料，并在IB HL V3教材中有明确图示。
s = ut + ½at²
6. Derivation of v² = u² + 2as | 推导 v² = u² + 2as
To eliminate t, we rearrange v = u + at to obtain t = (v − u)/a and substitute into s = ut + ½ at². After simplification, s = u(v − u)/a + ½ a[(v − u)/a]² = (v² − u²)/(2a). Multiplying through by 2a gives the time‑independent equation:
为消去t，将v = u + at整理为t = (v − u)/a，代入s = ut + ½ at²。化简后得s = u(v − u)/a + ½ a[(v − u)/a]² = (v² − u²)/(2a)。两边同乘2a即得与时间无关的方程：
v² = u² + 2as
This form is particularly useful in situations where time is not given, a common trick in Edexcel exam questions.
此形式在未给出时间的问题中尤为有用，是Edexcel考题的常见考点。
7. Derivation of s = ½(u + v)t | 推导 s = ½(u + v)t
Displacement can also be expressed using the average velocity. Since acceleration is constant, average velocity = (u + v)/2. Multiplying by time yields directly s = ½(u + v)t. Alternatively, this is the area of a trapezium under the v–t graph, a perspective highlighted in the Pearson IB HL V3 OCR‑aligned section.
All four SUVAT equations are ordinarily applied as scalar equations along a chosen positive direction. Upwards, rightwards, or the direction of initial velocity can be designated positive, with vectors opposite receiving a negative sign. Consistent use of signs is essential: an upward launch with g = 9.81 m s⁻² downwards means a = −9.81 m s⁻². Both Edexcel and IB HL mark schemes penalise sign errors heavily.
所有四个SUVAT方程通常沿选定的正方向用作标量方程。向上、向右或初速度方向可设为正，反向矢量则带负号。符号必须前后一致：向上抛出时重力加速度向下，g = 9.81 m s⁻² 意味着 a = −9.81 m s⁻²。Edexcel和IB HL的评分标准都对符号错误从严扣分。
9. Worked Example: Braking Car | 应用实例：刹车问题
A car travels at 20 m s⁻¹ and brakes with a constant deceleration of 4 m s⁻². Find the stopping distance. Using v² = u² + 2as with v = 0, u = 20 m s⁻¹, a = −4 m s⁻²: 0 = 20² + 2×(−4)×s ⇒ 0 = 400 − 8s ⇒ s = 50 m. This demonstrates the efficiency of the time‑independent equation.
一辆汽车以20 m s⁻¹行驶，以4 m s⁻²恒定减速度刹车。求制动距离。使用v² = u² + 2as，v = 0, u = 20 m s⁻¹, a = −4 m s⁻²：0 = 20² + 2×(−4)×s ⇒ 0 = 400 − 8s ⇒ s = 50 m。这体现了与时间无关公式的高效性。
10. Common Mistakes and Conceptual Pitfalls | 常见错误与概念陷阱
Students often confuse displacement with distance, overlook the sign of acceleration, or apply SUVAT to non‑uniform acceleration scenarios. Another frequent error is forgetting that the equations are only valid when a is constant. In IB HL V3, a diagnostic question asks why SUVAT cannot be used for a bouncing ball during impact – the answer lies in the non‑uniform nature of the collision forces.
11. Extension: Calculus Derivation for IB HL | 拓展：IB HL的微积分推导
For IB Higher Level, calculus provides an elegant verification. Starting from a = dv/dt, integrate with constant a: ∫ dv = ∫ a dt ⇒ v = u + at. Then using v = ds/dt, integrate again: s = ∫ (u + at) dt = ut + ½ at² + constant. With s(0) = 0, we retrieve the SUVAT forms. This method reinforces the link between kinematics and calculus, a key interdisciplinary skill.
在IB高等级中，微积分提供了优美的验证。从a = dv/dt出发，对恒定a积分：∫ dv = ∫ a dt ⇒ v = u + at。再利用v = ds/dt，再次积分：s = ∫ (u + at) dt = ut + ½ at² + 常数。代入初始条件s(0) = 0，即恢复SUVAT形式。该方法强化了运动学与微积分的联系，是一项重要的跨学科技能。
12. Summary: Connecting IB, Edexcel, and OCR | 总结：连接IB、Edexcel和OCR
The four SUVAT equations – v = u + at, s = ut + ½ at², v² = u² + 2as, s = ½(u + v)t – are universal in classical motion problems. By mastering their derivations from the Pearson IB Physics HL V3 approach (which mirrors the OCR TOC progression), students simultaneously prepare for Edexcel, OCR, and IB assessments, building a versatile foundation in mechanics.
四个SUVAT方程——v = u + at, s = ut + ½ at², v² = u² + 2as, s = ½(u + v)t——在经典运动问题中普遍适用。通过掌握Pearson IB物理HL V3（其编排与OCR目录顺序一致）中的推导方法，学生可同时备战Edexcel、OCR及IB考核，在力学领域打下通用而坚实的基础。
Published by TutorHao | Physics Revision Series | aleveler.com
📚 A-Level Physics Unit 4 Jan 2022: Application Question Techniques | A-Level 物理 Unit 4 2022年1月试卷应用题技巧
Unit 4 of A-Level Physics deepens your understanding of mechanics, fields, and particle physics. The January 2022 paper tested students’ ability to apply principles to unfamiliar contexts. This guide unpacks proven techniques for tackling application questions efficiently and accurately, pairing each insight in English and Chinese to reinforce your exam strategy.
Read the stem and sub-questions carefully. Underline command words like state, explain, or calculate. A ‘state’ question requires a concise fact, while ‘explain’ demands a logical physics argument. Identify the given data, the target quantity, and any hidden assumptions.
Many Unit 4 questions combine two or more topics, such as momentum and energy. Look for the connecting physical quantity, often mass, velocity, or time. Draw a simple sketch to visualize the situation and label all relevant vectors.
Once the question is decoded, ask: which law or model governs this system? For a charged particle moving in a magnetic field, the key is F = Bqv and centripetal acceleration. For a mass on a spring, it is F = -kx and the SHM equations. Write the relevant formula(e) immediately on your answer sheet.
Application questions often mask a familiar principle with a novel scenario, such as a satellite in orbit instead of a block on a string. Practice mapping the new situation to the standard model: identify what provides the centripetal force, or what stores the energy.
Break the problem into logical stages. In a typical momentum‑energy question, first use conservation of momentum to find common velocity after collision (v = m₁u₁/(m₁+m₂) if stationary), then apply kinetic energy loss: ΔEₖ = ½m₁u₁² – ½(m₁+m₂)v². Check unit consistency at every step.
When dealing with exponential decay in capacitor discharge or radioactive decay, extract the time constant (RC or 1/λ) first. Set up the ratio carefully: V = V₀ e⁻ᵗ⁄ᴿᴼ or A = A₀ e⁻λᵗ. Use the ln form to solve for t or the half‑life.
Graph questions in Unit 4 often test area under the curve (e.g., F–x graph for work done, v–t graph for displacement) and gradient (e.g., E–t gradient for rate of change of flux). Always label axes with units and use a large triangle for gradient calculation to minimise percentage error.
For SHM energy graphs, the total energy line is horizontal. The intercepts with kinetic and potential energy curves give you the amplitude. Read off the maximum speed from the kinetic energy peak using Eₖ = ½mv²ₘₐₓ. When data is presented in a table, check for proportional relationships to decide whether to plot a straight‑line graph.
For uniform fields, V = Ed is valid only if d is measured parallel to the field lines. A charged particle moving parallel to the field undergoes constant acceleration; use suvat equations. Remember electron volt conversions: 1 eV = 1.6×10⁻¹⁹ J.
在匀强电场中，V = Ed 仅在沿电场线方向测量 d 时成立。沿电场方向运动的带电粒子做匀加速运动，可使用运动学方程。记住电子伏特换算：1 eV = 1.6×10⁻¹⁹ J。
In radial field problems, combine Coulomb’s law F = kQq/r² with electric potential V = kQ/r. The potential is a scalar, so superposition is algebraic. When an electron moves from a negative to a positive plate, its change in electric potential energy ΔU = qΔV is negative, converting to kinetic energy.
在辐射状电场问题中，将库仑定律 F = kQq/r² 与电势 V = kQ/r 结合使用。电势是标量，叠加时直接代数求和。当电子从负极板移向正极板时，其电势能变化 ΔU = qΔV 为负，并转化为动能。
6. Circular Motion and Centripetal Force | 圆周运动与向心力
Always draw a free‑body diagram showing real forces (tension, weight, normal reaction). The resultant of these forces towards the centre is the centripetal force m×a = mv²/r or mrω². In vertical circles, speed is not constant; combine energy conservation mv²ₜₒₚ/2 + mg(2r) = mv²₋ₒₜₜₒₘ/2 to find the minimum speed at the top: vₜₒₚ = √(gr).
When a satellite changes orbit, gravitational force provides centripetal force: GMm/r² = mv²/r, giving v = √(GM/r). Notice that orbital speed decreases with radius. For geostationary satellites, set period T = 24 hours and use ω = 2π/T to find r.
当卫星变轨时，万有引力提供向心力：GMm/r² = mv²/r，可导出 v = √(GM/r)。注意轨道速度随半径增大而减小。对于地球同步卫星，设周期 T = 24小时，用 ω = 2π/T 求出轨道半径。
7. Simple Harmonic Motion (SHM) | 简谐运动问题
Define the equilibrium position clearly. Measure displacement x from there. The acceleration is always a = -ω²x. In a mass‑spring system, ω = √(k/m), and for a pendulum, ω = √(g/L). Use the reference circle to visualise displacement x = A cos(ωt) or x = A sin(ωt).
明确定义平衡位置，并从那里开始测量位移 x。加速度始终满足 a = -ω²x。在弹簧振子中，ω = √(k/m)；对于单摆，ω = √(g/L)。用参考圆来可视化位移 x = A cos(ωt) 或 x = A sin(ωt)。
Questions often ask for the time to travel between two positions. Use ωt = cos⁻¹(x/A) or sin⁻¹(x/A). Pay attention to the mode of clock start; if timing begins at maximum displacement, use cos. If from equilibrium moving positively, use sin.
题目常要求计算在两位置之间运动所需的时间。利用 ωt = cos⁻¹(x/A) 或 sin⁻¹(x/A)。注意时间的起点：若从最大位移处开始计时，用 cos 形式；若从平衡位置向正方向开始，用 sin 形式。
8. Electromagnetic Induction and Lenz’s Law | 电磁感应与楞次定律
Identify what is changing: B, A, or θ. The induced emf magnitude is ε = N ΔΦ/Δt, where Φ = BA cos θ. For a conductor moving across field lines, ε = Blv. Use Lenz’s law to predict direction: the induced current creates a flux that opposes the change in flux.
先判断什么在变：B、A 还是 θ。感应电动势的大小为 ε = N ΔΦ/Δt，其中 Φ = BA cos θ。对于切割磁感线的导体，ε = Blv。用楞次定律判断方向：感应电流产生的磁通量会阻碍磁通量的变化。
Plotting flux‑time graph to extract the emf‑time graph requires gradient analysis. The induced emf is the negative gradient of the Φ–t graph. For a coil rotating in a uniform field, the emf is sinusoidal: ε = NBAω sin(ωt). The peak emf occurs when the plane of the coil is parallel to the field.
9. Particle Tracks and Magnetic Fields | 粒子径迹与磁场分析
A charged particle moving in a magnetic field provides a visible track in a detector. The radius of curvature r is given by r = p/(Bq), where p = mv. A high‑momentum particle produces a larger radius and less curvature. If the track spirals, the particle is losing energy, usually by ionisation.
带电粒子在磁场中运动时会在探测器中留下可观测的径迹。曲率半径 r 由 r = p/(Bq) 给出，其中 p = mv。高动量粒子的径迹半径更大，弯曲程度更小。若径迹呈螺旋状，说明粒子正在损失能量，通常是通过电离方式。
Use Fleming’s left‑hand rule to determine the sign of charge from the direction of curvature in a known magnetic field. The direction of the force is perpendicular to both v and B. Two particles of equal charge and mass will have identical curvatures for the same momentum, allowing identification.
利用弗莱明左手定则，根据已知磁场方向下的弯曲方向来判断电荷正负。力的方向垂直于 v 和 B 两者。两个带等量同种电荷、质量相等的粒子，若动量相同，径迹曲率将完全相同，这可用于粒子鉴别。
10. Unit and Significant Figure Discipline | 单位与有效数字规范
Convert all quantities to SI units before substituting: lengths in m, mass in kg, time in s. For charge, 1 μC = 10⁻⁶ C. Distance in cm must become 10⁻² m. Write the unit of the final answer explicitly, and use prefixes like MHz or kN if appropriate.
Match the number of significant figures to the least precise data given. If the data include 2.0 s and 0.500 m, your answer should have 2 or 3 s.f., not 5. Carry full precision in intermediate steps, then round at the end. Show a clear substitution line to gain method marks even if arithmetic slips.
Unit 4 papers are mark‑heavy with tight time constraints. Allocate roughly 1 minute per mark. Start with the question you find easiest to secure quick marks. For a difficult calculation, bullet‑point the relevant equations and values to partially answer even if you cannot finish.
If you are stuck on a part for more than 2 minutes, mark it and move on. Later parts often give hints, or the data you need is in the stem. Use the front formula sheet smartly: check which forms of standard equations are provided and avoid re‑deriving them.
Mixing up left‑hand and right‑hand rules is a classic error: left‑hand for motor effect (force on current), right‑hand for dynamo effect (induced emf). Also, don’t forget to square the velocity in centripetal force or kinetic energy; missing the square is a frequent slip.
In circular motion, the force equation must refer to the radius of the circle, not necessarily the length of a string if it’s an inclined circle. For capacitors in series and parallel, the rules are opposite to resistors: charge is same in series, voltage is same in parallel. Always verify the sign of ΔV in energy calculations.
This set of last-minute revision notes distils the essential concepts, key formulas and common exam pitfalls from the Edexcel A-Level Physics specification. Each section pairs concise English explanations with Chinese translations to reinforce understanding across mechanics, waves, electricity, fields, quantum physics, thermodynamics and oscillations.
The four SUVAT equations (v = u + at, s = ut + ½at², s = ½(u+v)t, v² = u² + 2as) are only valid when acceleration is constant. Always define the positive direction before substituting values.
四个 SUVAT 方程 (v = u + at, s = ut + ½at², s = ½(u+v)t, v² = u² + 2as) 仅在加速度恒定时有效。代入数值前必须先规定正方向。
For projectiles, split the initial velocity into horizontal (u cosθ) and vertical (u sinθ) components. The horizontal motion has zero acceleration, while the vertical acceleration is g = 9.81 m s⁻² downward.
对于抛体运动，将初速度分解为水平分量 (u cosθ) 和竖直分量 (u sinθ)。水平方向加速度为零，竖直方向加速度为向下的 g = 9.81 m s⁻²。
Newton’s second law in vector form ΣF = ma must be applied by resolving all forces along chosen axes. Action–reaction pairs act on different bodies and are equal in magnitude but opposite in direction.
牛顿第二定律的矢量形式 ΣF = ma 需要沿选定轴分解所有力。作用力与反作用力作用在不同物体上，大小相等、方向相反。
v² = u² + 2as
2. Momentum, Energy and Power | 动量、能量与功率
Linear momentum p = mv is a vector quantity. The impulse FΔt equals the change in momentum Δp. In collisions, total momentum is conserved provided no external resultant force acts.
线动量 p = mv 是矢量。冲量 FΔt 等于动量的变化 Δp。只要无外合力作用，碰撞中总动量守恒。
Kinetic energy Eₖ = ½mv², gravitational potential energy ΔEₚ = mgΔh. The work done by a force is W = Fd cosθ, where θ is the angle between force and displacement.
Power is the rate of energy transfer: P = W/t. For a constant force moving at constant velocity, P = Fv. Efficiency = (useful energy output)/(total energy input).
Elastic collisions conserve kinetic energy; inelastic collisions do not, although momentum is still conserved. Always check if a collision is perfectly inelastic (objects stick together).
3. Materials: Stress, Strain and Young Modulus | 材料：应力、应变与杨氏模量
Hooke’s law states that extension ΔL is proportional to the applied force F, up to the limit of proportionality: F = kΔL. The spring constant k depends on the material and dimensions.
胡克定律指出，在比例极限内，伸长量 ΔL 与作用力 F 成正比：F = kΔL。弹簧常数 k 取决于材料和尺寸。
Tensile stress σ = F/A, tensile strain ε = ΔL/L. The Young modulus E = σ/ε, with units N m⁻² or Pa. It measures a material’s stiffness and is independent of sample dimensions.
拉应力 σ = F/A，拉应变 ε = ΔL/L。杨氏模量 E = σ/ε，单位为 N m⁻² 或 Pa。它衡量材料的刚度，与样品尺寸无关。
The force–extension graph shows an initial linear region, then an elastic limit beyond which plastic deformation occurs. Area under the graph gives work done (elastic strain energy = ½FΔx).
4. Waves: Wave Equation, Polarisation and Refraction | 波动：波方程、偏振与折射
The wave equation v = fλ links speed, frequency and wavelength. Transverse waves have oscillations perpendicular to energy transfer; longitudinal waves have oscillations parallel.
波方程 v = fλ 将波速、频率和波长联系起来。横波的振动方向与能量传递方向垂直；纵波的振动方向平行于传播方向。
Polarisation can only occur for transverse waves. A polarising filter transmits only the component of the wave parallel to its transmission axis, reducing intensity according to Malus’s law: I = I₀ cos²θ.
偏振只有横波才能发生。偏振片只透过与透振轴平行的分量，光强按马吕斯定律 I = I₀ cos²θ 减小。
Refraction is described by Snell’s law: n₁ sinθ₁ = n₂ sinθ₂. The refractive index n = c/v, where c is the speed of light in vacuum. Total internal reflection occurs when the angle of incidence exceeds the critical angle.
折射遵循斯涅尔定律：n₁ sinθ₁ = n₂ sinθ₂。折射率 n = c/v，c 为真空光速。当入射角大于临界角时发生全内反射。
v = fλ
5. Superposition, Interference and Stationary Waves | 叠加、干涉与驻波
When two coherent waves meet, superposition leads to constructive interference (path difference nλ) or destructive interference (path difference (n+½)λ). Phase difference is key.
In Young’s double-slit experiment, fringe spacing Δx = λD/a, where D is slit-to-screen distance and a is slit separation. For a diffraction grating, d sinθ = nλ gives maxima angles.
Stationary waves form when two identical waves travel in opposite directions. Nodes have zero displacement, antinodes have maximum amplitude. For a string fixed at both ends, λₙ = 2L/n.
驻波由两列相同的波相向传播形成。波节位移为零，波腹振幅最大。对于两端固定的弦，λₙ = 2L/n。
6. Electricity: Current, Resistance and Circuits | 电学：电流、电阻与电路
Current I = ΔQ/Δt. Ohm’s law V = IR holds for ohmic conductors at constant temperature. Resistivity ρ = RA/L, linking resistance to material and geometry.
电流 I = ΔQ/Δt。欧姆定律 V = IR 在恒温下适用于欧姆导体。电阻率 ρ = RA/L，将电阻与材料和几何尺寸联系起来。
Kirchhoff’s first law: ΣI into a junction = ΣI out (charge conservation). Second law: ΣEMF = Σpd around any closed loop (energy conservation). Use these to solve multi-loop circuits.
A potential divider gives V_out = V_in (R₂/(R₁+R₂)). EMF ε = I(R+r), terminal pd V = ε – Ir, where r is internal resistance. Plotting V against I yields gradient –r and intercept ε.
分压器输出电压 V_out = V_in (R₂/(R₁+R₂))。电动势 ε = I(R+r)，端电压 V = ε – Ir，其中 r 为内阻。作 V-I 图，斜率为 –r，截距为 ε。
P = I²R = V²/R
7. Quantum Physics: Photoelectric Effect, Energy Levels & de Broglie | 量子物理：光电效应、能级与德布罗意波
The photoelectric effect cannot be explained by wave theory. Photons carry energy E = hf = hc/λ. Electrons are emitted only if hf > Φ (work function), with maximum kinetic energy K_max = hf – Φ.
Stopping potential V_s relates to K_max by eV_s = K_max. Threshold frequency f₀ = Φ/h. The graph of K_max vs frequency gives slope h and x-intercept f₀.
Electrons in atoms exist in discrete energy levels. Emission or absorption of a photon occurs when an electron transitions, with hf = |E₂ – E₁|. Ionisation energy is the energy to remove an electron from ground state.
de Broglie wavelength λ = h/p shows wave–particle duality. Electrons can be diffracted, e.g. by graphite, providing evidence for matter waves.
德布罗意波长 λ = h/p 体现了波粒二象性。电子可被石墨等晶体衍射，为物质波提供证据。
8. Circular Motion and Gravitational Fields | 圆周运动与引力场
For an object moving in a circle at constant speed, centripetal acceleration a = v²/r = ω²r, and centripetal force F = mv²/r = mω²r. The force is always directed towards the centre.
物体匀速圆周运动时，向心加速度 a = v²/r = ω²r，向心力 F = mv²/r = mω²r。该力始终指向圆心。
Newton’s law of gravitation: F = GMm/r². Gravitational field strength g = F/m, and for a point mass g = GM/r². Gravitational potential V = –GM/r, and g = –dV/dr.
万有引力定律：F = GMm/r²。引力场强度 g = F/m，对于质点 g = GM/r²。引力势 V = –GM/r，且 g = –dV/dr。
Kepler’s third law T² ∝ r³ for planets moving around the Sun can be derived from equating gravitational and centripetal forces. Satellites in geostationary orbit have T = 24 h and orbit above the equator.
Coulomb’s law: F = kQq/r², where k = 1/(4πε₀). Electric field strength E = F/q; for a point charge E = Q/(4πε₀r²). In a uniform field between parallel plates, E = V/d.
库仑定律：F = kQq/r²，其中 k = 1/(4πε₀)。电场强度 E = F/q；点电荷电场 E = Q/(4πε₀r²)。平行板间的匀强电场 E = V/d。
Electric potential V = Q/(4πε₀r) for a point charge. Work done in moving a charge q through ΔV is W = qΔV. Equipotential surfaces are perpendicular to field lines.
点电荷的电势 V = Q/(4πε₀r)。移动电荷 q 经过电势差 ΔV 做的功为 W = qΔV。等势面与电场线垂直。
A capacitor stores charge Q = CV. For a parallel-plate capacitor, C = ε₀A/d. Energy stored = ½QV = ½CV² = ½Q²/C. Time constant τ = RC governs exponential charging and discharging: Q = Q₀ e⁻ᵗ/ʳᶜ.
10. Magnetic Fields and Electromagnetic Induction | 磁场与电磁感应
A current-carrying wire in a magnetic field experiences a force F = BIL sinθ (Fleming’s left-hand rule). A moving charge experiences F = Bqv sinθ; circular motion results if v ⟂ B.
载流导线在磁场中受力 F = BIL sinθ（弗莱明左手定则）。运动电荷受力 F = Bqv sinθ；若 v ⟂ B，电荷做圆周运动。
Magnetic flux Φ = BA cosθ, flux linkage = NΦ. Faraday’s law: induced EMF ε = –d(NΦ)/dt. Lenz’s law states the induced current opposes the change that produced it.
磁通量 Φ = BA cosθ，磁通匝链数 = NΦ。法拉第定律：感应电动势 ε = –d(NΦ)/dt。楞次定律指出，感应电流的方向总是阻碍引起感应的变化。
A transformer changes voltage according to Vₛ/Vₚ = Nₛ/Nₚ. For an ideal transformer, power input equals power output: IₚVₚ = IₛVₛ. Efficiency is reduced by eddy currents and flux leakage.
📚 A-Level Physics: Jun 18 Examiner Report 5 – Formula Derivation Insights | A-Level物理：2018年6月考官报告5 – 公式推导启示
The June 2018 examiner report for Paper 5 (Practical Skills) highlights a recurring weakness: many candidates struggle to manipulate basic equations into the linear forms required for graphical analysis. This article unpacks the derivation errors flagged by examiners, using the classic capacitor discharge experiment as a core example, and provides a step‑by‑step guide to mastering formula derivation under exam conditions.
Paper 5 (Practical Skills) of the A‑Level Physics examination assesses the ability to design, analyse, and evaluate experiments. The June 2018 examiner report noted that a significant number of candidates lost marks not because they misunderstood the physics, but because they could not reliably derive the linear relationship needed to plot a straight‑line graph from a raw exponential or power‑law equation.
Examiners specifically mentioned that when an equation like V = V0e–t/(RC) appeared, many responses showed incorrect algebraic steps, misplacement of the natural logarithm, or confusion between the dependent and independent variables for a straight‑line plot.
考官特别提到，当出现 V = V0e–t/(RC) 这样的方程时，许多答卷显示出错误的代数步骤、自然对数位置不当，或者混淆了直线图中的因变量与自变量。
2. Common Mistake: Misinterpreting the Exponential Decay | 常见错误：误解指数衰减
The discharge of a capacitor through a fixed resistor follows the equation V = V0e–t/(RC), where V is the potential difference at time t, V0 is the initial p.d., R is resistance, and C is capacitance. A typical mistake is to attempt to plot V against t directly and force a straight line, which obviously fails because the relationship is exponential.
电容器通过固定电阻放电遵循方程 V = V0e–t/(RC)，其中 V 是 t 时刻的电压，V0 是初始电压，R 是电阻，C 是电容。一个典型错误是试图直接绘制 V‑t 图并强行拟合直线，这显然失败，因为关系是指数型的。
The examiner report identified that students often wrote ln(V) = ln(V0) – t/(RC) but then misidentified the term ln(V0) as the gradient or placed t/(RC) incorrectly on the y‑axis. Understanding the structure of y = mx + c is essential before taking any logarithms.
考官报告发现，学生通常能写出 ln(V) = ln(V0) – t/(RC)，但随后将 ln(V0) 误认为斜率，或将 t/(RC) 错误地放在 y 轴上。在进行任何对数运算之前，必须理解 y = mx + c 的结构。
3. Step‑by‑Step Derivation of the Linear Form | 线性形式的逐步推导
Start with the exponential decay law: V = V0e–t/(RC). Take the natural logarithm of both sides: ln(V) = ln(V0 · e–t/(RC)). Apply the logarithm product rule: ln(V) = ln(V0) + ln(e–t/(RC)). Since ln(ex) = x, this simplifies to ln(V) = ln(V0) – t/(RC).
Rearrange to match y = mx + c: here y = ln(V), x = t, gradient m = –1/(RC), and y‑intercept c = ln(V0). This rearrangement is exactly what the examiner expected to see clearly stated.
整理使之匹配 y = mx + c：这里 y = ln(V)，x = t，斜率 m = –1/(RC)，纵截距 c = ln(V0)。这样的整理正是考官期望明确写出的步骤。
4. Identifying Variables for a Straight‑Line Graph | 识别直线图变量
A common failure was to plot ln(V) against t but label axes incorrectly or misinterpret the physical meaning of the gradient. The independent variable is time t (horizontal axis), and the dependent variable is ln(V) (vertical axis). The examiner emphasised that candidates should always state these clearly in their plan.
If a student plots V against t on log‑linear paper, they must still identify that the gradient of the straight line equals –1/(RC). The report advised that when using log‑linear graph paper, the derivation must be adapted to common logarithms: log10(V) = log10(V0) – t/(RC·ln(10)). Many lost marks by skipping this conversion.
5. Calculating the Time Constant from Gradient | 从斜率计算时间常数
The time constant τ = RC can be determined directly from the gradient m of the ln(V) versus t graph. Since m = –1/(RC), it follows that RC = –1/m. The examiner commented that too many candidates left the answer as a negative value or forgot to take the reciprocal.
时间常数 τ = RC 可直接从 ln(V)‑t 图的斜率 m 求得。因为 m = –1/(RC)，所以 RC = –1/m。考官评论称，太多考生将答案保留为负值，或忘记取倒数。
For a graph plotted with log10(V) against t, the relationship becomes gradient = –1/(RC · ln(10)). Then RC = –1/(gradient × ln(10)). Examiners recommended that candidates always verify that the derived RC has dimensions of time (seconds) as a quick check.
6. Handling Units and Significant Figures in Derivation | 推导中的单位与有效数字处理
In the derivation process, students often neglected to carry units through the algebra. For instance, writing “RC = 5.0” without seconds caused ambiguity. The examiner report noted that clear unit propagation is part of a rigorous derivation and is rewarded in the mark scheme.
When computing ln(V), note that V has units of volts. The logarithm of a quantity with units is mathematically tricky; strictly, one should divide V by a unit reference, e.g. ln(V / V). In A‑Level physics, it is acceptable to take ln of a numerical value in volts as long as the constant ln(V0) compensates. Exam reports remind students to state that V and V0 are in the same units.
计算 ln(V) 时，注意 V 的单位为伏特。对有单位的量取对数在数学上需要谨慎：严格来说，应将 V 除以一个单位参考，例如 ln(V / V)。在 A‑Level 物理中，只要常数 ln(V0) 补偿，取电压数值的自然对数是可以接受的。考官报告提醒学生应声明 V 和 V0 使用相同单位。
7. Examiner’s Comments on Algebraic Manipulation | 考官对代数运算的评语
The June 2018 report highlighted that examiners are looking for a logical flow of algebraic steps, not just the final expression. Writing “ln V = ln V0 – t/RC” without showing the application of logarithm rules was sometimes penalised when the subsequent gradient interpretation was wrong.
2018年6月的报告强调，考官看重的是逻辑流畅的代数步骤，而不仅仅是最终表达式。在后续斜率解释错误时，如果只是写出“ln V = ln V0 – t/RC”而未展示对数法则的应用，有时会被扣分。
Examiners also cautioned about sign errors. If a candidate mistakenly derives ln(V) = ln(V0) + t/(RC), then the gradient becomes positive, contradicting the physics of exponential decay. Checking the physical reasonableness of the sign is a valuable habit.
8. Extensions: Deriving Half‑Life from the Exponential | 扩展：从指数关系推导半衰期
While the report focused on the linearisation, a related derivation that often appears is the formula for half‑life T½. Set V = V0/2, then V0/2 = V0e–T½/(RC). Cancel V0: 1/2 = e–T½/(RC). Take ln: ln(1/2) = –T½/(RC). Since ln(1/2) = –ln(2), we get T½ = RC ln(2).
Examiners noted that mixing up half‑life with the time constant τ = RC was a common mistake. The half‑life is about 0.693 × RC. Deriving it from first principles demonstrates deeper understanding.
9. Common Pitfalls with Logarithmic Axes | 对数坐标轴常见陷阱
When instructed to plot ln(V) against t, some candidates still attempted to plot raw V on a logarithmic scale without taking logs. The examiner report warned that this leads to a non‑linear curve on standard graph paper or requires using log‑linear paper, which must be explicitly justified.
当要求绘制 ln(V)‑t 图时，一些考生仍试图在对数刻度上绘制原始 V 而不取对数。考官报告警告说，这会导致在标准坐标纸上画出非直线，或需要使用半对数纸，而此举必须明确说明其理由。
If using log‑linear paper, the gradient formula changes subtly. A clear derivation for the gradient in terms of RC must be shown. Many lost marks by simply stating “gradient = –1/RC” without acknowledging the base‑10 logarithm conversion.
10. Practice Problems from Past Papers | 来自历年试卷的练习题
To solidify these skills, attempt the following typical tasks: (i) Given a set of (t, V) data, derive the equation for a straight‑line graph and state the quantities to be plotted. (ii) From a graph of ln(V) vs. t, the gradient is –0.25 s–1; calculate RC. (iii) Show that the time for the voltage to fall to 1/e of its initial value is exactly τ.
Solutions: (ii) RC = –1/(–0.25) = 4.0 s. (iii) Set V = V0/e, then 1/e = e–t/(RC), taking ln gives –1 = –t/(RC), so t = RC. These short derivations are exactly the kind of algebraic fluency the examiner expects.
解答：(ii) RC = –1/(–0.25) = 4.0 s。(iii) 令 V = V0/e，则 1/e = e–t/(RC)，取对数得 –1 = –t/(RC)，所以 t = RC。这些简短推导正是考官期望的代数流利度。
11. Tips for Mastering Formula Derivation | 掌握公式推导的技巧
Always write the raw physical law first. Identify the target linear form y = mx + c before touching logarithms. Take logarithms step‑by‑step, stating each rule used (e.g. product rule, power rule).
始终先写下原始物理定律。在动用对数之前，先明确目标线性形式 y = mx + c。逐步取对数，并说明每一步用到的法则（如乘积法则、幂法则）。
After deriving the linear equation, explicitly map each term: “y = ln(V), x = t, m = –1/(RC), c = ln(V0)”. This explicit mapping is what examiners look for in high‑scoring scripts. Practice with different relationships, such as power laws of the form y = kxn, where logs give ln(y) = ln(k) + n ln(x).
推导出线性方程后，明确对应各项：“y = ln(V)，x = t，m = –1/(RC)，c = ln(V0)”。这种明确的对应是考官在满分答案中寻找的。练习不同关系，如 y = kxn 形式的幂律，取对数得 ln(y) = ln(k) + n ln(x)。
12. Conclusion: Lessons from the Examiner Report | 结论：考官报告带来的启示
The June 2018 examiner report for Paper 5 underscores that formula derivation is a skill that bridges theoretical understanding and practical analysis. By learning to linearise exponential and power‑law relationships systematically, students not only secure marks in the practical paper but also strengthen their grasp of fundamental physics.
2018年6月 Paper 5 的考官报告强调，公式推导是连接理论理解与实验分析的桥梁技能。通过学会系统地将指数关系和幂律关系线性化，学生不仅能确保在实验卷中得分，还能加深对基础物理的掌握。
Consistent practice with clear algebraic steps, unit handling, and sign checks transforms this examiner‑identified weakness into a reliable strength.
通过清晰的代数步骤、单位处理和符号检查的持续练习，可以将考官指出的薄弱环节转变为可靠的优势。
Published by TutorHao | Physics Revision Series | aleveler.com
📚 A-Level Physics: Formula Derivations from Unit 5 Jan 22 Paper | A-Level 物理：2022年1月单元5试卷公式推导
In the January 2022 A-Level Physics Unit 5 examination, candidates were challenged to derive and apply several fundamental equations from thermodynamics, nuclear physics and astrophysics. This article steps through those derivations with clarity, supporting students who wish to master the reasoning behind each relationship.
1. Kinetic Theory Derivation of Ideal Gas Equation | 理想气体方程的分子动理论推导
Consider a cube of side L containing N identical particles, each of mass m. A single particle moves with velocity components (vₓ, v_y, v_z). Its x-component of momentum change when striking a wall is 2 m vₓ.
考虑一个边长为 L 的立方体，内有 N 个相同的粒子，每个质量为 m。一个粒子以速度分量 (vₓ, v_y, v_z) 运动。它撞击器壁时，动量的 x 分量变化为 2 m vₓ。
The time between successive collisions with the same wall is 2 L / vₓ, so the average force exerted by this particle on the wall is F = (2 m vₓ) / (2 L / vₓ) = m vₓ² / L.
同一壁面两次碰撞之间的时间为 2 L / vₓ，因此该粒子对器壁的平均作用力为 F = (2 m vₓ) / (2 L / vₓ) = m vₓ² / L。
Summing over all N particles and dividing by the wall area A = L² gives the pressure: P = (1 / L²) (m / L) Σ vₓ² = (m / V) Σ vₓ², where V = L³. Using the mean square speed ⟨v²⟩ = (vₓ² + v_y² + v_z²) and isotropy ⟨vₓ²⟩ = ⟨v_y²⟩ = ⟨v_z²⟩ = ⅓ ⟨v²⟩, we obtain P = (1/3) (N m / V) ⟨v²⟩.
对所有 N 个粒子求和，并除以壁面积 A = L² 得到压强：P = (1 / L²) (m / L) Σ vₓ² = (m / V) Σ vₓ²，其中 V = L³。利用均方速率 ⟨v²⟩ = (vₓ² + v_y² + v_z²) 以及各向同性 ⟨vₓ²⟩ = ⟨v_y²⟩ = ⟨v_z²⟩ = ⅓ ⟨v²⟩，可得 P = (1/3) (N m / V) ⟨v²⟩。
Introduce the average translational kinetic energy: ⟨Eₖ⟩ = ½ m ⟨v²⟩. The kinetic theory links this to absolute temperature via ⟨Eₖ⟩ = (3/2) k T, where k is the Boltzmann constant. Substituting gives P V = N k T, and with the mole concept N = n N_A, k N_A = R, we arrive at the ideal gas equation P V = n R T.
引入平均平动动能：⟨Eₖ⟩ = ½ m ⟨v²⟩。分子动理论将它与热力学温度联系起来，即 ⟨Eₖ⟩ = (3/2) k T，其中 k 为玻尔兹曼常数。代入得 P V = N k T，再利用摩尔概念 N = n N_A，k N_A = R，最终导出理想气体方程 P V = n R T。
2. Radioactive Decay Law and Half-Life Derivation | 放射性衰变定律与半衰期推导
The activity A of a sample is the number of decays per unit time, which is proportional to the number of undecayed nuclei N: A = -dN/dt = λ N, where λ is the decay constant.
样品的活度 A 是单位时间的衰变次数，与未衰变核的数目 N 成正比：A = -dN/dt = λ N，其中 λ 为衰变常量。
Separating variables and integrating from N₀ (at t = 0) to N gives ∫_{N₀}^{N} dN / N = -λ ∫_{0}^{t} dt, leading to ln (N / N₀) = -λ t. Rearranging yields the exponential decay law N = N₀ e^(-λ t).
Half-life T₁/₂ is the time when N = N₀ / 2. Substituting gives ½ = e^(-λ T₁/₂); taking natural logarithms produces T₁/₂ = ln 2 / λ. This shows the inverse relation between half-life and decay constant.
半衰期 T₁/₂ 是 N = N₀ / 2 的时刻。代入得 ½ = e^(-λ T₁/₂)，取自然对数后得到 T₁/₂ = ln 2 / λ。这表明半衰期与衰变常量成反比。
3. Gravitational Potential Energy and Escape Velocity | 引力势能与逃逸速度
The gravitational force between two point masses M and m separated by distance r is F = G M m / r². To bring m from infinity to a distance r against this force, work must be done.
两个质点 M 和 m 相距 r 时的引力为 F = G M m / r²。要克服此力将 m 从无穷远移动到距离 r 处，需要做功。
The work done by an external agent is W = ∫_{∞}^{r} (G M m / x²) dx = [- G M m / x]_{∞}^{r} = – G M m / r. This work is stored as gravitational potential energy U, so U = – G M m / r. The negative sign indicates a bound system.
外力做功为 W = ∫_{∞}^{r} (G M m / x²) dx = [- G M m / x]_{∞}^{r} = – G M m / r。此功储存为引力势能 U，故 U = – G M m / r。负号表示系统处于束缚态。
For an object to escape a planet’s gravity from its surface (radius R), its kinetic energy must equal the magnitude of the potential energy: ½ m vₑₛ꜀² = G M m / R. Cancelling m and solving gives escape velocity vₑₛ꜀ = √(2 G M / R).
物体要从行星表面（半径 R）脱离引力束缚，其动能必须等于势能的绝对值：½ m vₑₛ꜀² = G M m / R。消去 m 并求解得逃逸速度 vₑₛ꜀ = √(2 G M / R)。
Simple harmonic motion (SHM) arises when the restoring force is proportional to the displacement from equilibrium and directed opposite to it: F = – k x. Newton’s second law gives m a = – k x, so a = – (k/m) x.
当回复力与离开平衡位置的位移成正比且方向相反时，便产生简谐运动（SHM）：F = – k x。根据牛顿第二定律，m a = – k x，因此 a = – (k/m) x。
Defining the angular frequency ω = √(k/m), we have a = – ω² x. This is the defining equation of SHM. Its general solution can be written as x = A cos(ω t + φ), where A is the amplitude and φ the phase constant.
定义角频率 ω = √(k/m)，则有 a = – ω² x。这正是简谐运动的定义方程。其通解可写作 x = A cos(ω t + φ)，其中 A 为振幅，φ 为初相位。
Differentiating twice confirms the acceleration: v = dx/dt = – ω A sin(ω t + φ) and a = d²x/dt² = – ω² A cos(ω t + φ) = – ω² x. Thus the motion satisfies the SHM condition.
两次求导可验证加速度：v = dx/dt = – ω A sin(ω t + φ)，a = d²x/dt² = – ω² A cos(ω t + φ) = – ω² x。因此运动满足简谐运动条件。
5. Energy Transformations in Simple Harmonic Motion | 简谐运动中的能量转化
The kinetic energy of a mass–spring system in SHM is Eₖ = ½ m v² = ½ m ω² A² sin²(ω t + φ). The potential energy stored in the spring is Eₚ = ½ k x² = ½ m ω² A² cos²(ω t + φ), using k = m ω².
弹簧振子在做简谐运动时，动能为 Eₖ = ½ m v² = ½ m ω² A² sin²(ω t + φ)。弹簧的弹性势能为 Eₚ = ½ k x² = ½ m ω² A² cos²(ω t + φ)，其中利用了 k = m ω²。
The total mechanical energy is E_total = Eₖ + Eₚ = ½ m ω² A² [sin²(ω t + φ) + cos²(ω t + φ)] = ½ m ω² A². This shows that the total energy is constant and proportional to the square of the amplitude.
系统总机械能为 E_total = Eₖ + Eₚ = ½ m ω² A² [sin²(ω t + φ) + cos²(ω t + φ)] = ½ m ω² A²。可见总能量守恒，且与振幅的平方成正比。
At maximum displacement, energy is entirely potential; at equilibrium, energy is entirely kinetic. The continuous interchange illustrates energy conservation in an isolated oscillator.
6. Thermal Energy Transfer and Specific Latent Heat | 热能传递与比潜热
When a substance changes temperature without changing phase, the energy transferred ΔQ is related to the temperature change Δθ by ΔQ = m c Δθ, where c is the specific heat capacity. This is a direct consequence of the definition of c.
当物质温度变化而物态不变时，传递的能量 ΔQ 与温度变化 Δθ 的关系为 ΔQ = m c Δθ，其中 c 为比热容。这是比热容定义的直接结果。
During a phase change at constant temperature, the energy supplied goes into breaking intermolecular bonds rather than raising kinetic energy. The energy needed per unit mass is the specific latent heat L, so ΔQ = m L.
在恒定温度下发生相变时，所提供能量用于打破分子间键，而非增加动能。单位质量所需的能量称为比潜热 L，因此 ΔQ = m L。
These relations can be combined with power P = ΔQ / Δt or electrical methods (P = V I) to determine c or L experimentally. In a typical exam problem, students might derive L from a cooling curve or from the gradient of a temperature–time graph.
这些关系式可与功率 P = ΔQ / Δt 或电学方法 (P = V I) 结合，通过实验测定 c 或 L。典型考题中，学生可能需要从冷却曲线或温度–时间图的斜率推导 L。
7. Stellar Luminosity and the Stefan–Boltzmann Law | 恒星光度与斯特藩–玻尔兹曼定律
A star radiates energy from its photosphere, which can be treated as a black body. The Stefan–Boltzmann law states that the power emitted per unit area is J = σ T⁴, where σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ and T is the surface temperature.
If the star has radius R, its surface area is 4πR², so its luminosity L (total power output) is L = 4πR² σ T⁴. This equation links measurable quantities and allows astronomers to estimate stellar radii if luminosity and temperature are known.
若恒星半径为 R，表面积为 4πR²，则其光度 L（总输出功率）为 L = 4πR² σ T⁴。该方程将可测量量联系起来，若已知光度和温度，天文学家便可估算恒星半径。
A derivation of peak wavelength from Wien’s displacement law λ_max T = constant is often paired with the Stefan–Boltzmann law in astrophysics questions. No explicit derivation of the T⁴ dependence is required at A-Level, but students should be able to use L ∝ R² T⁴ in proportional reasoning.
维恩位移定律 λ_max T = 常数 常与斯特藩–玻尔兹曼定律一同出现在天体物理题中。A-Level不要求推导 T⁴ 依赖关系，但学生应能运用 L ∝ R² T⁴ 进行比例推理。
8. Nuclear Binding Energy and Mass Defect | 核结合能与质量亏损
Experimental measurements show that the mass of an atomic nucleus is less than the sum of the masses of its separate protons and neutrons. This difference is the mass defect Δm.
实验测量显示，原子核的质量小于其独立质子和中子质量之和。这一差值即为质量亏损 Δm。
According to Einstein’s mass–energy equivalence, the binding energy that holds the nucleus together is E_binding = Δm c², where c is the speed of light in vacuum. This energy represents the work required to separate the nucleus into its individual nucleons.
根据爱因斯坦质能等价关系，束缚原子核的结合能为 E_binding = Δm c²，其中 c 为真空光速。该能量代表将原子核拆分成单个核子所需的功。
To find the binding energy per nucleon, divide E_binding by the mass number A. The curve of binding energy per nucleon against A peaks around iron-56, explaining the stability of nuclei and the release of energy in fusion and fission.
要计算每个核子的结合能，可将 E_binding 除以核子数 A。每个核子的结合能随 A 变化的曲线在铁-56附近达到峰值，这解释了原子核的稳定性以及聚变和裂变中能量的释放。
When a nucleus undergoes decay or reaction, the difference in total binding energy between products and reactants is released as kinetic energy of the products or as photons. This underpins the energy calculations in nuclear physics exam questions.
Interference of light is a foundational topic in wave optics, revealing the wave nature of light through the superposition of coherent waves. A thorough understanding of interference patterns, conditions, and quantitative analysis is essential for success in IB Physics and Edexcel A Level Physics examinations. This guide covers all key concepts, common pitfalls, and formula derivations you need to master.
光的干涉是波动光学的基础课题，通过相干波的叠加揭示了光的波动性。透彻理解干涉图样、条件及定量分析对于在 IB 物理和 Edexcel A Level 物理考试中取得成功至关重要。本篇指南涵盖了你需要掌握的所有核心概念、常见错误以及公式推导。
1. What is Interference of Light? | 什么是光的干涉？
Interference occurs when two or more coherent light waves superpose in space, resulting in a new intensity distribution. According to the principle of superposition, the resultant displacement at any point is the vector sum of the individual displacements. If the waves arrive in phase, they interfere constructively, producing a bright fringe; if they arrive half a wavelength out of phase, destructive interference yields a dark fringe.
The intensity distribution is not simply a sum of individual intensities but depends on the phase relationship. For two identical sources, the intensity varies from zero (dark) to four times the intensity of a single source (bright) where constructive interference occurs. This energy redistribution is a hallmark of interference.
Stable and observable interference patterns require coherent sources. Coherence implies that the light waves maintain a constant phase relationship over time. The three main conditions are: (i) the sources must have the same frequency (monochromaticity); (ii) the phase difference must remain constant (temporal coherence); and (iii) the waves should have parallel or nearly parallel polarisation for maximum contrast.
In practice, achieving coherence often involves dividing a single wavefront or amplitude, as exemplified by Young’s double-slit or Michelson interferometer. Lasers are highly coherent sources because their stimulated emission produces waves with identical frequency and locked phases. Ordinary thermal sources require spatial filtering, such as a narrow single slit, to improve coherence.
Young’s double-slit experiment is the classic demonstration of light interference. Monochromatic light first passes through a narrow single slit to create an approximate point source of coherent wavefronts. This wave then falls on two parallel slits S₁ and S₂ separated by a distance d, acting as secondary coherent sources. Beyond the double slit, a screen is placed at a large distance L to observe the interference pattern.
杨氏双缝实验是光干涉的经典演示。单色光首先通过一条窄单缝，形成一个近似的点相干波前源。该波前随后照射到相距为 d 的两条平行狭缝 S₁ 和 S₂ 上，作为次级相干光源。在双缝后较远距离 L 处放置一块屏，用以观察干涉图样。
The resulting pattern consists of a series of bright and dark fringes parallel to the slits. The central maximum (zero-order fringe) is located where the path difference from the two slits is zero. Moving away from the centre, alternating maxima and minima appear, labelled n = ±1, ±2, … The intensity of the bright fringes gradually decreases due to the single-slit diffraction envelope.
形成的图样由一系列平行于狭缝的明暗条纹组成。中央明纹（零级）位于两缝光程差为零处。从中心向外，交替出现明暗条纹，标记为 n = ±1, ±2, …。由于单缝衍射包络的影响，亮条纹的强度逐渐减弱。
4. Derivation of Fringe Spacing Formula | 条纹间距公式的推导
The fringe separation Δx is a critical measurement. For a point P at a distance x from the central axis, the path difference between waves from S₁ and S₂ is approximately d sinθ. Using the small-angle approximation sinθ ≈ tanθ = x/L, the path difference becomes d × (x/L). Constructive interference (bright fringe) occurs when
d × (x/L) = nλ
leading to the position of the n-th bright fringe: xₙ = nλL/d. Hence the fringe separation (distance between adjacent bright or dark fringes) is
Δx = λL / d
条纹间距 Δx 是一个关键测量量。对于偏离中央轴线距离 x 的点 P，S₁ 和 S₂ 的光程差近似为 d sinθ。利用小角度近似 sinθ ≈ tanθ = x/L，光程差可写为 d × (x/L)。相长干涉（亮条纹）的条件为
d × (x/L) = nλ
由此得到第 n 级亮纹位置：xₙ = nλL/d。因此条纹间距（相邻明纹或暗纹间的距离）为
Δx = λL / d
This derivation assumes L ≫ d and that the maxima are viewed at small angles. The formula reveals that Δx increases with wavelength and screen distance, and decreases with slit separation. It enables experimental determination of light wavelength and is frequently examined in IB Edexcel practical-based questions.
该推导假设 L ≫ d 且在小角度下观察极大值。该公式表明 Δx 随波长和屏距增大而增大，随缝距增大而减小。它可用于实验测定光波波长，并在 IB Edexcel 基于实验的考题中频繁出现。
5. Path Difference and Interference Orders | 光程差与干涉级次
Constructive interference arises when the path difference δ between the two waves is an integer multiple of the wavelength: δ = nλ, where n = 0, 1, 2, … (order number). Destructive interference corresponds to δ = (n + ½)λ. The phase difference Δφ is directly proportional to path difference:
Δφ = (2π/λ) × δ
Thus a path difference of λ corresponds to a phase shift of 2π radians.
It is essential to distinguish between geometrical path length and optical path length when a medium of refractive index n is present: optical path = n × geometrical path. This concept is particularly important in thin film interference, where a phase change of π (equivalent to λ/2) may occur upon reflection at an interface from lower to higher refractive index.
当存在折射率为 n 的介质时，必须区分几何路径与光程：光程 = n × 几何路径。这一概念在薄膜干涉中尤为重要，因为光在从低折射率到高折射率界面反射时，可能会发生 π 的相位突变（相当于 λ/2 光程差）。
6. Thin Film Interference: Principles | 薄膜干涉原理
Thin film interference results from partial reflections at the upper and lower boundaries of a thin film, such as a soap bubble or an oil layer on water. The two reflected waves travel different optical path lengths before recombining. For a film of thickness t and refractive index n, the optical path difference for near-normal incidence is approximately 2nt, but phase changes on reflection must be accounted for.
薄膜干涉源于薄膜（如肥皂泡或水面油膜）上下边界部分反射光的叠加。两束反射光在重新汇合前经历了不同的光程。对于厚度为 t、折射率为 n 的薄膜，在近垂直入射下，光程差约为 2nt，但还必须考虑反射时的相位变化。
Reflection at an interface from a medium of lower refractive index to one of higher refractive index introduces a phase reversal of π (an effective λ/2 shift). If the film is surrounded by air (n_air = 1), the light reflecting from the top surface undergoes a phase reversal, whereas the bottom reflection may or may not, depending on the substrate. This determines whether constructive or destructive interference occurs for specific wavelengths.
For a film in air with one phase reversal, the condition for constructive interference in reflected light is 2nt = (m + ½)λ, and for destructive interference 2nt = mλ (m = 0,1,2…). This explains why soap bubbles appear coloured: varying thickness t gives rise to interference maxima for different wavelengths across the visible spectrum.
7. Anti-reflection Coatings and Applications | 增透膜及其应用
Anti-reflection coatings utilise destructive interference to minimise reflected light from glass surfaces. A thin layer of material with refractive index n_coating less than that of glass (n_coating < n_glass) is deposited. Both reflections (air–coating and coating–glass) undergo phase reversals because each reflection is from lower to higher index. Thus the net phase difference from reflections is zero, and the condition for destructive interference in reflected light becomes 2n_coating t = (m + ½)λ.
增透膜利用相消干涉来减少玻璃表面的反射光。在玻璃上沉积一层折射率小于玻璃的薄层材料（n_涂层 < n_玻璃）。由于两次反射（空气–涂层和涂层–玻璃）都是从光疏到光密介质，均发生相位突变，因此反射引起的净相位差为零，反射光相消干涉的条件变为 2n_涂层 t = (m + ½)λ。
For a single-layer coating at normal incidence and minimum thickness (m=0), the optical thickness must be λ/4: n_coating t = λ/4. Such coatings are widely used in camera lenses, spectacles, and solar cells to enhance transmission. Conversely, high-reflection coatings can be designed using constructive interference of reflected waves by stacking layers with alternating refractive indices.
对于单层增透膜在正入射且最小厚度（m=0）时，光学厚度需为 λ/4：n_涂层 t = λ/4。这类镀膜广泛应用于相机镜头、眼镜镜片和太阳能电池中以增强透光率。反之，通过交替折射率的叠层设计，可利用反射光的相长干涉制成高反射膜。
8. Interference with White Light and Colours | 白光干涉与色彩
When white light (a continuous spectrum) is used in a double-slit or thin film experiment, each wavelength produces its own interference pattern. At the central maximum, all wavelengths undergo constructive interference path difference zero, resulting in a white central fringe. Away from the centre, the fringe pattern is a rainbow-like spectrum because red light (longer λ) produces wider fringe spacing than blue light (shorter λ).
In thin films, white light interference creates vivid colour patterns visible in soap bubbles and oil slicks. The observed colour at a given point corresponds to the wavelengths that interfere constructively for that local film thickness. Because thickness variations are gradual, bands of colour appear. Higher-order fringes may overlap, causing colours to wash out.
White light interference is also used in practical tests of optical flatness: a thin air wedge between a flat glass and a test surface produces straight, parallel fringes; any irregularities indicate surface deviations of the order of fractions of a wavelength
Published by TutorHao | IB Physics Revision Series | aleveler.com
Nuclear physics is a cornerstone of the IB and CCEA A‑Level Physics specifications, exploring the structure of the atomic nucleus, the forces that hold it together, and the energy released in nuclear transformations. This article distills the essential concepts—from the strong nuclear force and binding energy to radioactive decay, fission, and fusion—into a clear, bilingual revision guide. Each section pairs English explanations with precise Chinese translations, equipping students with the clarity and confidence needed for exam success.
The atom consists of a tiny, dense nucleus containing protons and neutrons (nucleons), surrounded by electrons in discrete energy levels. Rutherford’s alpha‑particle scattering experiment revealed that most of the atom’s mass and all its positive charge reside in a nucleus roughly 10⁻¹⁵ m across, while the atom itself is about 10⁻¹⁰ m in size. This model replaced the earlier ‘plum pudding’ picture and forms the basis for understanding nuclear stability.
原子由一个微小、致密的原子核和核外分层排布的电子构成，原子核内含质子和中子（统称核子）。卢瑟福的 α 粒子散射实验表明，原子的绝大部分质量与全部正电荷集中在直径约 10⁻¹⁵ m 的原子核中，而整个原子的尺度约为 10⁻¹⁰ m。这一模型取代了早期的“葡萄干布丁”图像，为理解核稳定性奠定了基础。
2. Nucleon Number, Proton Number and Isotopes | 核子数、质子数与同位素
The proton number Z defines the element, while the nucleon number A is the total number of protons and neutrons. Isotopes are atoms of the same element (same Z) with different numbers of neutrons, hence different A. Chemical properties are virtually identical, but nuclear stability can vary dramatically. A nuclide is represented as AZX, for example 146C.
质子数 Z 决定元素种类，而核子数 A 是质子与中子总数。同位素是质子数相同但中子数不同（因而 A 不同）的原子。它们的化学性质几乎完全相同，但核稳定性可能差异巨大。一种核素记为 AZX，例如 146C。
3. The Strong Nuclear Force | 强核力
The strong nuclear force binds nucleons together, overcoming the electrostatic repulsion between protons. It is an extremely short‑range attractive force (effective up to about 3–4 fm) that acts equally between proton–proton, neutron–neutron, and proton–neutron pairs. At very small separations (below ~0.5 fm), the force becomes repulsive, preventing nucleons from collapsing into one another. The balance between the strong force and Coulomb repulsion determines nuclear stability.
强核力将核子束缚在一起，克服质子间的静电排斥。它是一种极短程吸引力（有效范围约 3–4 fm），作用于质子–质子、中子–中子、质子–中子对时强度相等。在极小的间距下（约 0.5 fm 以下），力变为排斥，阻止核子坍缩。强核力与库仑斥力的平衡决定了原子核的稳定性。
4. Mass Defect and Binding Energy | 质量亏损与结合能
The mass of a nucleus is always less than the sum of the masses of its individual nucleons. This mass defect Δm is converted into binding energy Eb upon formation of the nucleus, according to Einstein’s equation Eb = Δmc². Binding energy represents the work required to separate a nucleus into its constituent nucleons. A larger binding energy per nucleon indicates a more stable nucleus; iron‑56 (⁵⁶Fe) has the highest binding energy per nucleon, about 8.8 MeV.
5. Radioactive Decay and the Decay Constant | 放射性衰变与衰变常量
Unstable nuclei emit radiation to become more stable. The three main types are alpha (α) decay (emission of a helium nucleus, 42He), beta (β⁻) decay (a neutron converts to a proton, emitting an electron and an antineutrino), and gamma (γ) emission (release of high‑energy photons). The decay constant λ (unit s⁻¹) is the probability that a given nucleus decays per unit time. The activity A of a sample is A = λN, where N is the number of undecayed nuclei.
不稳定的原子核通过辐射来趋向稳定。三种主要类型是：α 衰变（释放氦核 42He）、β⁻ 衰变（中子转变为质子，释放电子与反中微子）和 γ 辐射（释放高能光子）。衰变常量 λ（单位 s⁻¹）是单个核在单位时间内发生衰变的概率。样品的活度 A = λN，其中 N 为未衰变核的数目。
6. Exponential Decay Law and Half‑Life | 指数衰变律与半衰期
Radioactive decay follows an exponential law: N = N₀e–λt, where N₀ is the initial number of nuclei. The half‑life T½ is the time for half the nuclei to decay, related to λ by T½ = ln2 / λ. Activity A also decreases exponentially: A = A₀e–λt. The decay curve is characterised by a constant half‑life, independent of the initial quantity. This property is used in radiometric dating.
7. Nuclear Reactions and Conservation Laws | 核反应与守恒定律
In any nuclear reaction, the total nucleon number and total charge (proton number) are conserved. Energy, momentum, and lepton number (where applicable) are also conserved. A typical nuclear reaction is written as a + X → Y + b + Q, where Q is the energy released (Q‑value). Q can be calculated from the mass difference before and after the reaction: Q = (Σmreactants – Σmproducts)c². Exothermic reactions have Q > 0.
在任何核反应中，总核子数与总电荷（质子数）均守恒。能量、动量以及轻子数（若适用）也守恒。典型的核反应可写为 a + X → Y + b + Q，其中 Q 为释放的能量（Q 值）。Q 可由反应前后的质量差计算：Q = (Σm反应物 – Σm产物)c²。放热反应中 Q > 0。
8. Nuclear Fission | 核裂变
Fission occurs when a heavy nucleus (e.g., uranium‑235) captures a slow neutron and splits into two lighter daughter nuclei, releasing two or three further neutrons and a large amount of energy (≈200 MeV per fission). The energy comes from the difference in binding energy per nucleon between the parent and the fragments. A chain reaction is sustained if at least one neutron from each fission induces another fission; this principle underlies nuclear reactors and atomic bombs. Control rods and moderators manage the neutron population in a reactor.
Fusion is the combining of light nuclei (e.g., deuterium and tritium) to form a heavier nucleus, accompanied by a large energy release. The energy output per unit mass can exceed that of fission. Fusion requires extremely high temperatures (≈10⁸ K) to overcome the Coulomb barrier between the positively charged nuclei. In stars, fusion powers the luminosity through reactions like the proton‑proton chain. On Earth, magnetic confinement (tokamak) and inertial confinement are being pursued for controlled fusion power.
10. Mass‑Energy Equivalence in Nuclear Processes | 核过程中的质能等价
The equivalence E = mc² is not only used to calculate binding energy but also to account for the energy released or absorbed in any nuclear transformation. The change in mass Δm directly corresponds to the energy change: 1 u (unified atomic mass unit) of mass is equivalent to 931.5 MeV of energy. Students must be able to convert between atomic mass units and MeV/c² and to compute Q‑values from given atomic masses, taking care to include electron masses if using nuclear rather than atomic masses.
11. The Standard Model and Fundamental Particles | 标准模型与基本粒子
The IB and CCEA syllabi touch on the quark model of hadrons. Protons (uud) and neutrons (udd) consist of up and down quarks. The strong force between nucleons is a residual effect of the colour force between quarks, mediated by gluons. Beta decay is explained at the quark level: a down quark changes into an up quark, emitting a W⁻ boson that subsequently decays into an electron and an antineutrino. This deeper picture connects nuclear physics to particle physics.
Always distinguish between atomic mass and nuclear mass when calculating mass defect. Use consistent units: convert all masses to u or kg, and energies to J or eV as appropriate. Remember that activity is proportional to the number of undecayed nuclei, and the half‑life is a statistical property; never say that exactly half the nuclei decay in one half‑life for a small sample. Practice sketching binding energy per nucleon curves and marking the peaks. In fusion and fission arguments, focus on the change in binding energy per nucleon rather than the absolute energy of the nuclei.
计算质量亏损时，务必区分原子质量与核质量。使用一致的单位：将所有质量转换为 u 或 kg，能量转换为 J 或 eV。记住活度与未衰变核数目成正比，半衰期是一种统计性质；对于小样本，切勿说恰好一半的核在一个半衰期内衰变。练习绘制比结合能曲线并标出峰值。在论证裂变与聚变时，重点关注比结合能的变化，而非原子核的绝对能量。
Published by TutorHao | Physics Revision Series | aleveler.com
Electric fields lie at the heart of many A-Level Physics topics, from atomic structure to circuits. In this OCR revision guide, we break down every essential concept, equation and experiment you need to master for the exam. Whether it’s Coulomb’s law, uniform fields between parallel plates or Millikan’s oil drop, you’ll find clear explanations and real exam focus.
An electric field is a region of space where a stationary charged particle experiences an electric force. The field is produced by source charges and is a vector quantity, having both magnitude and direction at every point.
We define the direction of an electric field as the direction of the force that a small positive test charge (+q) would experience if placed at that point. This convention makes field lines point away from positive charges and toward negative charges.
Coulomb’s law gives the magnitude of the force between two point charges Q₁ and Q₂ separated by distance r in a vacuum: directly proportional to the product of the charges and inversely proportional to the square of the distance.
The constant ε₀ is the permittivity of free space, ε₀ ≈ 8.85 × 10⁻¹² F m⁻¹. The forces are attractive if the charges have opposite signs and repulsive if they have the same sign.
常数ε₀是自由空间介电常数，ε₀ ≈ 8.85 × 10⁻¹² F m⁻¹。电荷异号时力为吸引，同号时为排斥。
3. Electric Field Strength | 电场强度
Electric field strength (E) at a point is the force per unit positive charge experienced by a small test charge placed at that point:
电场强度（E）定义为置于该点的小检验电荷每单位正电荷所受的力：
E = F / q
Its SI unit is N C⁻¹ (equivalent to V m⁻¹). E is a vector; its direction is the same as the force on a positive test charge.
其国际单位是 N C⁻¹（相当于 V m⁻¹）。E是矢量，方向与正检验电荷受力方向相同。
For a point charge Q, the field strength at a distance r is:
对于点电荷Q，距离r处的电场强度为：
E = (1 / (4π ε₀)) × (Q / r²)
In a uniform electric field between two parallel plates, the field strength is constant and given by the potential difference V and plate separation d:
在两平行板之间的匀强电场中，场强恒定，与电势差V和板间距d的关系为：
E = V / d
4. Electric Field Lines | 电场线
Field lines are a visual tool to represent electric fields. They begin on positive charges and end on negative charges, never forming closed loops. The tangent to a line at any point shows the direction of the field, and the density of lines indicates the field strength.
Key rules: field lines do not cross; they are perpendicular to the surface of a conductor at equilibrium; and in a uniform field they appear as equally spaced parallel lines.
5. Uniform Electric Fields and Parallel Plates | 匀强电场与平行板
A uniform electric field has the same magnitude and direction everywhere in a region. This is closely approximated by two parallel conducting plates with a constant potential difference V across them, separated by distance d.
The field strength is E = V/d, and the field direction is from the positive plate (higher potential) to the negative plate (lower potential). Charged particles in such a field experience a constant electric force F = qE.
场强大小为 E = V/d，方向从正极板（高电势）指向负极板（低电势）。处于该场中的带电粒子受到恒定的电场力 F = qE。
This arrangement is used in particle accelerators, ink-jet printers and cathode-ray tubes to deflect charged beams.
这种装置用于粒子加速器、喷墨打印机和阴极射线管，以偏转带电束流。
6. Electric Potential and Potential Energy | 电势与电势能
Electric potential (V) at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point, without any change in kinetic energy. It is a scalar quantity measured in volts (V), where 1 V = 1 J C⁻¹.
电势（V）是指将小正检验电荷从无穷远移至该点所做的功（每单位正电荷），且过程中动能不变。它是标量，单位为伏特（V），1 V = 1 J C⁻¹。
For a point charge Q, the potential at distance r is:
对于点电荷Q，距离r处的电势为：
V = (1 / (4π ε₀)) × (Q / r)
The electric potential energy (U) of a charge q at a point where the potential is V is U = qV. When charges move through a potential difference ΔV, the change in potential energy is ΔU = q ΔV.
电荷q在电势为V处的电势能为 U = qV。当电荷通过电势差ΔV时，电势能的变化为 ΔU = q ΔV。
7. Relationship Between Field and Potential | 场与电势的关系
In a non-uniform field, the electric field strength is the negative of the potential gradient:
在非匀强电场中，电场强度是电势梯度的负值：
E = − dV / dr
This means the field points in the direction of steepest decrease in potential. For a uniform field, this reduces to E = −ΔV/Δd, so the magnitude is simply ΔV/d with direction from higher to lower potential.
这意味着场的方向指向电势下降最快的方向。对于匀强电场，这简化为 E = −ΔV/Δd，因此大小就是 ΔV/d，方向由高电势指向低电势。
Understanding this link helps you move between E–r and V–r graphs for point charges and uniform fields.
理解这一联系有助于你在点电荷和匀强电场的 E–r 图和 V–r 图之间转换。
8. Equipotential Surfaces | 等势面
An equipotential surface is a surface on which the electric potential is the same everywhere. No work is required to move a charge along an equipotential surface because the potential difference is zero.
等势面是上面所有点电势都相同的面。由于电势差为零，电荷沿等势面移动时不需要做功。
Equipotential surfaces are always perpendicular to electric field lines. Around an isolated point charge they are concentric spheres; between uniform parallel plates they are planes parallel to the plates.
9. Motion of Charged Particles in Electric Fields | 带电粒子在电场中的运动
When a charged particle enters a uniform electric field at right angles, its motion mimics that of a projectile in a gravitational field. The constant electric force produces a constant acceleration in the field direction, while velocity parallel to the plates remains unchanged.
For an electron (charge −e) injected with speed v₀ into a field of strength E over a horizontal length L, the vertical deflection y is:
对于以速度v₀射入电场E中的电子（电荷−e），水平长度为L时，垂直偏转量y为：
y = ½ (eE / m) (L / v₀)²
The angular deflection θ satisfies tanθ = (eEL) / (m v₀²). These relationships are vital for understanding devices like oscilloscopes.
偏转角θ满足 tanθ = (eEL) / (m v₀²)。这些关系对于理解示波器等设备至关重要。
10. Millikan’s Oil Drop Experiment | 密立根油滴实验
Millikan’s experiment determined the fundamental unit of charge, e, by balancing tiny charged oil drops between parallel plates. When the drop is stationary, the electric force qE equals the weight mg minus the upthrust (often negligible).
Millikan found that all charges were integer multiples of e ≈ 1.60 × 10⁻¹⁹ C, proving charge quantisation. Students must be able to explain the experimental procedure and calculate q and e from given data.
密立根发现所有电荷都是 e ≈ 1.60 × 10⁻¹⁹ C 的整数倍，证明了电荷的量子化。考生必须能够解释实验步骤并根据数据计算q和e。
11. Comparison of Electric and Gravitational Fields | 电场与引力场的类比
Electric and gravitational fields share many similarities, but also have crucial differences. The table below highlights the key comparisons relevant to OCR exams.
电场与引力场有许多相似之处，但也有关键区别。下表突出了与OCR考试相关的重要对比。
Property
Gravitational Field
Electric Field
Source
Mass
Charge
Force law
F = G m₁m₂ / r² (always attractive)
F = (1/4π ε₀) Q₁Q₂ / r² (attractive or repulsive)
Field strength
g = F/m (N kg⁻¹)
E = F/q (N C⁻¹)
Potential
V_g = −G M / r (scalar)
V = (1/4π ε₀) Q / r (scalar, can be + or −)
Uniform field
g ≈ constant near Earth’s surface
E = V/d between parallel plates
Both obey inverse-square laws and have equipotential surfaces perpendicular to field lines. However, only electric fields can be shielded and can exert forces on stationary particles without requiring a mass.
Mastering electric fields requires fluency with the following equations. Practice applying them to different scenarios, such as oil drop problems, deflection tubes and radial field graphs.
Deflection (electron): y = ½ (eE/m) (L/v₀)² | 偏转（电子）：y = ½ (eE/m) (L/v₀)²
Millikan’s condition: q = mgd / V | 密立根条件：q = mgd / V
Always remember to include directions for vectors, use SI units and treat potential as a scalar superposition. With this structured revision, you are now equipped to tackle any OCR electric fields question with confidence.
Dynamics is the study of forces and motion. In the CIE IGCSE Physics syllabus, this topic bridges key concepts from scalars and vectors to Newton’s laws, momentum, and terminal velocity. Understanding dynamics is essential for solving real-world problems and for success in the Paper 2 and Paper 4 examinations. This article will guide you through the core points, common graphs, and key equations you must master.
A scalar quantity has magnitude only, such as distance, speed, mass, energy, and time. A vector quantity has both magnitude and direction, such as displacement, velocity, acceleration, force, and momentum. When adding vectors in the same direction, simply sum their magnitudes; for opposite directions, subtract them. For forces at right angles, use Pythagoras’ theorem or scale diagrams to find the resultant vector.
Always include the direction when stating a vector answer. For example, a resultant force might be ‘5 N at 37° to the horizontal’ or ’10 m/s due east’. In exam questions, forgetting the direction loses marks on vector calculations.
表述矢量答案时一定要包含方向。例如，合力可能是“5 N，与水平方向夹角 37°”或“10 m/s 正东”。考试中如果忘记注明方向，矢量计算题会被扣分。
2. Speed, Velocity, and Acceleration | 速度、速率与加速度
Speed is the distance travelled per unit time; it is a scalar. Average speed = total distance ÷ total time. Velocity is speed in a given direction; it is a vector. Acceleration is the rate of change of velocity: a = (v − u) ÷ t, measured in m/s². Deceleration is negative acceleration. Constant velocity means both constant speed and constant direction.
In many CIE questions, you are given initial and final velocities along with a time interval. Remember that acceleration is a vector: if an object slows down while moving forward, its acceleration is in the opposite direction to its motion. When calculating average speed for a journey with different segments, use total distance over total time, not the arithmetic mean of the speeds.
3. Distance-Time and Speed-Time Graphs | 距离-时间图与速率-时间图
A distance-time graph shows how an object’s distance changes over time. The gradient of a distance-time graph gives the speed. A horizontal line means the object is stationary. A straight, sloping line indicates constant speed; a curve means the speed is changing (accelerating or decelerating). A speed-time graph plots speed against time. The gradient gives the acceleration, and the area under the graph gives the distance travelled.
When tackling graph interpretation questions, always check the axes labels first. Many students confuse distance-time and speed-time graphs. Practice calculating gradients and areas, and remember that for a speed-time graph, the total distance may be found by counting squares under a curve if the graph is non-linear.
Newton’s First Law states that an object remains at rest or moves with constant velocity unless acted upon by a resultant force. This property is called inertia: the tendency of an object to resist changes in its velocity. The greater an object’s mass, the greater its inertia and the harder it is to accelerate or decelerate. A passenger in a car lurching forward when the car brakes is demonstrating inertia.
In exam contexts, ‘resultant force’ or ‘net force’ is the vector sum of all forces acting. If the resultant force is zero, the object is at equilibrium: it may be stationary or moving with constant velocity. This is important when analysing terminal velocity or objects on a slope with balanced forces.
Newton’s Second Law relates resultant force, mass, and acceleration: F = m × a. Force is measured in newtons (N), mass in kilograms (kg), and acceleration in m/s². One newton is the force required to accelerate 1 kg by 1 m/s². The acceleration is directly proportional to the resultant force and inversely proportional to the mass. This equation is fundamental to dynamics and must be used with consistent SI units.
牛顿第二定律联系了合力、质量和加速度：F = m × a。力的单位是牛顿（N），质量的单位是千克（kg），加速度的单位是 m/s²。一牛顿相当于使 1 kg 物体产生 1 m/s² 加速度所需的力。加速度与合外力成正比，与质量成反比。该方程是动力学的基础，必须使用统一的 SI 单位进行计算。
When several forces act on an object, first find the resultant force by vector addition. Then apply F=ma. In CIE problems, you may need to resolve forces along a slope or calculate the tension in a string connecting two masses. Always identify the direction of positive motion when setting up equations.
Mass is a scalar quantity measuring the amount of matter in an object; it is constant everywhere and measured in kilograms. Weight is the gravitational force acting on a mass, a vector directed towards the centre of the planet. Weight W = m × g, where g is the gravitational field strength (on Earth, g ≈ 9.8 N/kg, often rounded to 10 N/kg in CIE calculations). Weight changes with location, for example on the Moon where g is smaller.
质量是标量，衡量物体所含物质的多少；它在任何地方都保持不变，单位为千克。重量是作用于质量上的引力，是矢量，方向指向地心。重量 W = m × g，其中 g 为引力场强度（地球上 g ≈ 9.8 N/kg，CIE 计算中常取 10 N/kg）。重量随位置变化，例如在月球上 g 较小，重量也较轻。
A common misconception is confusing mass and weight. Remember: if you take a 1 kg object to the Moon, its mass remains 1 kg, but its weight becomes about 1.6 N. In multiple-choice questions, expect to distinguish between mass and weight using definitions or units.
常见的误解是混淆质量与重量。记住：你把 1 kg 的物体带上月球，它的质量仍是 1 kg，但重量变为约 1.6 N。选择题中，常需要根据定义或单位区分质量与重量。
7. Free Fall and Terminal Velocity | 自由落体与终端速度
When an object falls freely under gravity with no air resistance, it accelerates at g (constant acceleration). In the presence of air resistance, the resultant force decreases as speed increases because air resistance rises with speed. Eventually, air resistance equals the weight, the resultant force becomes zero, and the object falls at a constant terminal velocity. A skydiver experiences this: acceleration from jump until air drag balances weight.
物体在没有空气阻力的情况下自由下落时，会以 g 做匀加速运动。在有空气阻力时，随着速度增大，空气阻力增加，合力减小。最终，空气阻力等于重量，合力为零，物体便以恒定的终端速度下落。跳伞运动员就会经历这一过程：从跳下开始加速，直到空气阻力与重力平衡。
On a speed-time graph for a skydiver, the curve starts with a steep slope equal to g, then the slope decreases as air resistance grows, finally flattening to a horizontal line at terminal velocity. When the parachute opens, the sudden increase in area greatly increases air resistance, causing rapid deceleration until a new, lower terminal velocity is reached.
跳伞者的速率-时间图曲线起初有接近于 g 的陡峭斜率，然后随着空气阻力增加而斜率渐缓，最终在终端速度处变为水平。降落伞打开时，面积急剧增大，空气阻力骤增，导致快速减速，直至达到一个新的、较低的终端速度。
8. Forces and Elasticity: Hooke’s Law | 力与弹性：胡克定律
When a spring is stretched, the extension is directly proportional to the applied force, provided the elastic limit is not exceeded. This is Hooke’s Law: F = k × x, where k is the spring constant (N/m). The spring constant measures stiffness: a steep force-extension graph indicates a stiff spring. Beyond the elastic limit, the spring deforms plastically and does not return to its original length when the force is removed.
拉伸弹簧时，只要未超过弹性限度，伸长量与施加的力成正比。这就是胡克定律：F = k × x，其中 k 是弹簧常数（N/m）。弹簧常数衡量劲度：力-伸长图中斜率越陡，弹簧越“硬”。超过弹性限度后，弹簧发生塑性变形，撤去外力后无法恢复原长。
In the CIE practical component, you may investigate Hooke’s Law by adding masses to a spring and measuring extension. Plot force against extension to obtain a straight line through the origin. The gradient of this line is the spring constant k. Remember to record extension = stretched length − original length.
Momentum p is the product of mass and velocity: p = m × v. It is a vector with units kg m/s. The conservation of momentum is a key principle, but first, understand impulse. Impulse is the change in momentum: F × t = Δp = mv − mu. This explains why crumple zones and airbags reduce injury: increasing the time of impact reduces the average force for a given momentum change.
动量 p 是质量与速度的乘积：p = m × v。它是矢量，单位是 kg m/s。动量守恒是一个关键原理，但首先要理解冲量。冲量是动量的变化量：F × t = Δp = mv − mu。这解释了为何溃缩区和安全气囊能降低伤害：对于给定的动量变化，增加碰撞时间可减小平均受力。
In CIE examination questions, you may need to calculate the force using impulse data or explain safety features in terms of impulse. A common application: a cricket fielder catching a ball moves hands backward to increase contact time, reducing the force exerted on the hands. Always use the same direction convention when dealing with momentum changes.
10. Conservation of Momentum and Collisions | 动量守恒与碰撞
The total momentum of a closed system remains constant before and after a collision or explosion, provided no external resultant force acts. This is the principle of conservation of momentum. For two colliding objects, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, where u and v are initial and final velocities. In an explosion, the two parts move apart such that their total momentum remains zero.
在没有合外力作用的封闭系统中，碰撞或爆炸前后的总动量保持不变。这就是动量守恒原理。对于两个碰撞物体，m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂，其中 u 和 v 分别为初、末速度。爆炸时，裂开的两部分向相反方向运动，总动量保持为零。
CIE questions often involve recoil velocities: a stationary cannon fires a cannonball, and the cannon recoils. Because initial momentum is zero, the final momenta of cannonball and cannon must be equal in magnitude but opposite in direction. Always assign one direction as positive. Also note that momentum is a vector, so take direction carefully in two-dimensional collisions (though IGCSE typically focuses on one-dimensional cases).
Understanding these dynamics essentials will give you confidence in tackling numerical problems and graph-based questions. Practice converting units, drawing free-body diagrams, and applying F=ma and momentum principles. Consistent practice with past papers is the best way to master dynamics for the CIE IGCSE Physics exam.
📚 A-Level Physics: Key Points on Alternating Current | A-Level 物理：交流电 考点精讲
Alternating current (AC) is a fundamental concept in A-Level Physics, appearing in topics ranging from circuit analysis to electromagnetic induction. Unlike direct current (DC), AC reverses direction periodically, giving rise to unique quantities like peak, root-mean-square (rms) values, and phase relationships. Mastering these ideas is essential for tackling questions on transformers, rectification, and power dissipation. This article reviews the key learning points for AC, structured to align with typical A-Level specifications, and provides clear explanations to build confidence.
Alternating current is a flow of electric charge that periodically reverses direction. Mathematically, it is often represented as a sinusoidal function of time: i(t) = I₀ sin(ωt) or v(t) = V₀ sin(ωt), where I₀ and V₀ are the peak current and peak voltage, ω is the angular frequency (ω = 2πf), and f is the frequency in hertz. The instantaneous value changes continuously from positive to negative, completing one full cycle in a period T = 1/f.
2. Peak, Peak-to-Peak and Instantaneous Values | 峰值、峰峰值与瞬时值
The peak value (V₀ or I₀) is the maximum magnitude of the alternating quantity. The peak-to-peak value is the total swing from positive peak to negative peak, i.e., 2V₀. The instantaneous value is the value at any specific instant of time, given by the sinusoidal equation. For a mains supply of 230 V rms in the UK, the peak voltage is approximately 325 V (since V₀ = √2 × Vrms).
The rms value of an AC is the equivalent DC value that would produce the same heating effect in a resistor. For a sinusoidal waveform, Vrms = V₀ / √2 and Irms = I₀ / √2. These relationships are derived by averaging the square of the instantaneous values over a full cycle and then taking the square root. Rms quantities are used in power calculations: P = IrmsVrms for a purely resistive load.
When AC flows through components like capacitors or inductors, the voltage and current may not peak at the same time; there is a phase difference. The phase angle φ describes this shift: in a purely capacitive circuit, current leads voltage by 90° (π/2 rad); in a purely inductive circuit, current lags voltage by 90°. In a resistor, they are in phase (φ = 0). Understanding phase is crucial for analysing LCR circuits and power factor.
In a purely resistive circuit, the instantaneous voltage and current are directly proportional according to Ohm’s law: v(t) = i(t) R. The waveforms are in phase, and the power dissipated is P = Irms² R = Vrms² / R. The average power over a full cycle is constant, unlike reactive components where energy is stored and returned.
在纯电阻电路中，瞬时电压和电流根据欧姆定律成正比：v(t) = i(t) R。波形同相，耗散功率为 P = Irms² R = Vrms² / R。整个周期内的平均功率是恒定的，这与电抗性元件存储和返回能量不同。
6. AC in a Pure Capacitor | 纯电容中的交流电
For a capacitor, the current is proportional to the rate of change of voltage: i(t) = C dv/dt. With v = V₀ sin(ωt), differentiation gives i = ωCV₀ cos(ωt), showing that current leads voltage by 90°. The capacitive reactance is XC = 1 / (ωC) = 1 / (2πfC). Reactance decreases with increasing frequency, so capacitors conduct AC more easily at high frequencies. No net power is dissipated in an ideal capacitor.
对于电容器，电流与电压的变化率成正比：i(t) = C dv/dt。对于 v = V₀ sin(ωt)，微分可得 i = ωCV₀ cos(ωt)，表明电流超前电压 90°。容抗为 XC = 1 / (ωC) = 1 / (2πfC)。容抗随频率增高而减小，因此电容器在高频下更容易导通交流电。理想电容器不消耗净功率。
7. AC in a Pure Inductor | 纯电感中的交流电
In an inductor, the back emf opposes changes in current, leading to a voltage that is proportional to the rate of change of current: v = L di/dt. For i = I₀ sin(ωt), v = ωLI₀ cos(ωt); voltage leads current by 90°. Inductive reactance is XL = ωL = 2πfL. Reactance increases with frequency, so inductors block high-frequency AC while allowing DC to pass. Ideal inductors also dissipate zero average power.
在电感器中，反电动势阻碍电流的变化，因此电压与电流的变化率成正比：v = L di/dt。对于 i = I₀ sin(ωt)，v = ωLI₀ cos(ωt)；电压超前电流 90°。感抗为 XL = ωL = 2πfL。感抗随频率增加而增大，因此电感器阻碍高频交流电而允许直流通过。理想电感器也消耗零平均功率。
8. Impedance and Phasor Diagrams | 阻抗与相量图
Impedance Z combines resistance and reactance in an AC circuit and is defined as Z = Vrms / Irms. For a series LCR circuit, Z = √(R² + (XL − XC)²). The phase angle φ is given by tan φ = (XL − XC) / R. Phasor diagrams represent these quantities as rotating vectors, with the angle between voltage and current phasors equal to φ. They are powerful tools for solving AC circuit problems without differentiation.
阻抗 Z 综合了交流电路中的电阻和电抗，定义为 Z = Vrms / Irms。对于串联 LCR 电路，Z = √(R² + (XL − XC)²)。相位角 φ 由 tan φ = (XL − XC) / R 给出。相量图将这些量表示为旋转矢量，电压与电流相量之间的夹角等于 φ。它们是解决交流电路问题而不需微积分的有效工具。
9. Transformers – Principle and Equations | 变压器 – 原理与方程
A transformer uses electromagnetic induction to change the magnitude of an alternating voltage. It consists of two coils wound on a common iron core. For an ideal transformer with no energy losses, Vs / Vp = Ns / Np = Ip / Is. The turns ratio determines whether the output is stepped up or stepped down. Real transformers experience losses due to resistance heating, eddy currents, and hysteresis, which can be calculated from efficiency = (output power / input power) × 100%.
10. Rectification – Half-Wave and Full-Wave | 整流 – 半波与全波
Rectification converts AC into DC. In half-wave rectification, a single diode allows current to pass only during one half of the cycle, producing a pulsating DC with a large ripple. In full-wave rectification, a bridge rectifier (four diodes) inverts the negative half-cycles, giving an output that uses both halves. The average DC output voltage for a full-wave rectifier is Vdc = (2/π) V₀, which is higher than the half-wave value (1/π) V₀.
The output of a rectifier still fluctuates. A smoothing capacitor connected in parallel charges during voltage peaks and discharges through the load when the voltage drops, reducing ripple. The time constant τ = RLC determines the discharge rate; a larger capacitance or load resistance gives smoother output. The ripple voltage depends on the load current and capacitance: ΔV ≈ Iload / (2fC) for full-wave.
12. Power in AC Circuits and Power Factor | 交流电路功率与功率因数
The true power (average power) in an AC circuit is P = VrmsIrms cos φ, where cos φ is the power factor. It accounts for the phase difference: only the in-phase component of current does useful work. For pure resistors, cos φ = 1; for pure inductors or capacitors, cos φ = 0 and no net power is transferred. Apparent power is VrmsIrms, and reactive power is associated with energy storage. Improving the power factor is important in mains electricity distribution.
交流电路中的有功功率（平均功率）为 P = VrmsIrms cos φ，其中 cos φ 是功率因数。它考虑了相位差：只有电流的同相分量做有用功。对于纯电阻，cos φ = 1；对于纯电感或纯电容，cos φ = 0，没有净功率传输。视在功率为 VrmsIrms，无功功率与能量存储相关。提高功率因数在电力输送中很重要。
Published by TutorHao | Physics Revision Series | aleveler.com
In A2 Physics, mastering the most commonly examined topics is essential for achieving top grades. This article distills the key concepts, formulas, and typical pitfalls across core areas such as circular motion, gravitational fields, oscillations, thermal physics, electromagnetism, quantum phenomena, and nuclear/particle physics. Each section pairs crisp explanations in English and Chinese, ensuring bilingual learners can reinforce understanding while tackling exam-style reasoning.
Circular motion involves an object moving along a circular path at constant angular speed or with changing speed. The key descriptors are angular displacement θ (in radians), angular velocity ω = Δθ/Δt, period T, and frequency f. Centripetal acceleration a = v²/r = ω²r is always directed towards the centre, and the centripetal force F = mω²r = mv²/r is the resultant force responsible for this acceleration—not an extra ‘force’.
v = ωr — linear speed equals angular speed times radius. / 线速率等于角速率乘以半径。
a = v²/r = ω²r — centripetal acceleration. / 向心加速度。
F = mv²/r = mω²r — centripetal force. / 向心力。
In vertical circles, minimum speed at the top requires mg = mv²/r (when tension is zero). For a mass on a string, v_min = √(gr) at the top. / 竖直圆周运动中，顶部最小速率需满足mg = mv²/r（拉力为零）。对于绳系小球，顶部v_min = √(gr)。
Common pitfall: confusing centripetal with centrifugal; only centripetal force is real in an inertial frame. / 常见误区：向心力与离心力混淆；在惯性系中只有向心力是真实的。
F_c = mω²r = mV²/r
2. Gravitational Fields | 引力场
Newton’s law of gravitation states that the force between two point masses is F = Gm₁m₂/r². The gravitational field strength g at a point is the force per unit mass, g = F/m = GM/r² for a point mass or outside a spherical mass. Gravitational potential V = −GM/r is always negative, representing work done per unit mass to bring a mass from infinity.
g = GM/r² — field strength at distance r from centre. / 距离中心r处的场强。
V = −GM/r — gravitational potential. / 引力势。
Escape velocity: v_esc = √(2GM/r) or v_esc = √(2gr) at surface. / 逃逸速度：v_esc = √(2GM/r) 或在地表 v_esc = √(2gr)。
Kepler’s third law: T² ∝ r³ for circular orbits, derived by equating gravitational force to centripetal force: GMm/r² = mω²r ⇒ T² = (4π²/GM)r³. / 开普勒第三定律：对于圆轨道，T² ∝ r³，由引力提供向心力导出：GMm/r² = mω²r ⇒ T² = (4π²/GM)r³。
Geostationary orbit: period 24 hours, orbits above equator at radius ~42 300 km. / 地球同步轨道：周期24小时，位于赤道上空约42 300 km。
T² = (4π²/GM) r³
3. Simple Harmonic Motion | 简谐运动
Simple harmonic motion (SHM) is oscillatory motion where acceleration is directly proportional to displacement from equilibrium and always directed towards it: a = −ω²x. This leads to sinusoidal variations: x = x₀ sin ωt or x = x₀ cos ωt, v = ωx₀ cos ωt = ±ω√(x₀² − x²), and a = −ω²x₀ sin ωt = −ω²x. Energy continuously interchanges between kinetic and potential, total energy E = ½mω²x₀².
简谐运动是一种加速度与平衡位置位移成正比且始终指向平衡位置的振动：a = −ω²x。其位移随时间正弦变化：x = x₀ sin ωt 或 x = x₀ cos ωt，速度v = ωx₀ cos ωt = ±ω√(x₀² − x²)，加速度a = −ω²x₀ sin ωt = −ω²x。能量在动能与势能间持续转化，总能量E = ½mω²x₀²。
Examples: mass-spring system T = 2π√(m/k), simple pendulum T = 2π√(l/g) for small angles. / 实例：弹簧振子T = 2π√(m/k)，单摆小角度T = 2π√(l/g)。
Velocity: v = ± ω √(x₀² − x²), maximum at equilibrium (x=0). / 速度：v = ± ω √(x₀² − x²)，平衡位置（x=0）最大。
Damping reduces amplitude over time; critical damping returns to equilibrium in the shortest time without oscillation. / 阻尼使振幅逐渐减小；临界阻尼在无振荡的情况下最短时间回到平衡。
Resonance occurs when driving frequency ≈ natural frequency, leading to maximum amplitude. / 受迫振动频率接近固有频率时发生共振，振幅达到最大。
The ideal gas equation is pV = nRT = NkT, linking pressure p, volume V, and thermodynamic temperature T. The kinetic theory models gas pressure as arising from molecular collisions: pV = ⅓ N m , where is the mean square speed. Combining with pV = nRT gives the translational kinetic energy per particle: <½ m c²> = (3/2)kT. The internal energy of an ideal gas depends solely on temperature.
理想气体状态方程为pV = nRT = NkT，将压强p、体积V和热力学温度T联系起来。分子动理论认为气体压强源自分子碰撞：pV = ⅓ N m ，其中为均方速率。结合pV = nRT可得每个分子的平均平移动能：<½ m c²> = (3/2)kT。理想气体的内能仅取决于温度。
pV = nRT — n is number of moles. / n为摩尔数。
pV = NkT — N is number of molecules, k = Boltzmann constant. / N为分子数，k为玻尔兹曼常数。
Mean kinetic energy per particle: = ½ m = (3/2)kT. / 每分子平均动能： = ½ m = (3/2)kT。
Avogadro’s law: equal volumes of gases at same T and p contain equal numbers of molecules. / 阿伏伽德罗定律：同温同压下，相同体积的气体含有相同数目的分子。
When applying, temperature must be in kelvin. / 使用时温度必须用开尔文。
pV = ⅓ N m
5. Thermodynamics: First Law and Processes | 热力学：第一定律与过程
The first law of thermodynamics is expressed as ΔU = Q − W, where ΔU is the increase in internal energy, Q is the heat supplied to the system, and W is the work done by the system. It is crucial to track signs: work done ON the system is −W. For ideal gases, ΔU depends only on temperature change: ΔU = n C_V ΔT.
热力学第一定律表示为ΔU = Q − W，其中ΔU是内能增量，Q是系统吸收的热量，W是系统对外做的功。符号需特别注意：外界对系统做功为 −W。对于理想气体，ΔU仅取决于温度变化：ΔU = n C_V ΔT。
Isobaric (constant p): W = pΔV, Q = ΔU + pΔV = n C_P ΔT. / 等压过程：W = pΔV，Q = ΔU + pΔV = n C_P ΔT。
Isochoric (constant V): W = 0, Q = ΔU = n C_V ΔT. / 等容过程：W = 0，Q = ΔU = n C_V ΔT。
Cyclic processes: net ΔU = 0, net work done equals net heat supplied. / 循环过程：净ΔU=0，净功等于净热量。
p-V diagrams: area under curve gives work done by gas; clockwise cycles are heat engines. / p-V 图：曲线下面积表示气体做的功；顺时针循环为热机。
ΔU = Q − W
6. Capacitors | 电容器
A capacitor stores charge and energy in an electric field. Capacitance C = Q/V is measured in farads. For a parallel-plate capacitor, C = ε₀εᵣ A/d. Energy stored U = ½QV = ½CV² = ½ Q²/C. In DC circuits, charging and discharging follow exponential curves: Q = Q₀(1 − e^(−t/RC)) for charging, Q = Q₀ e^(−t/RC) for discharging, with time constant τ = RC.
C = ε₀εᵣ A/d — increasing plate area or reducing separation raises capacitance. / 增大板面积或减小板间距可提高电容。
Time constant τ = RC: after one time constant, charge falls to 37% of initial during discharge, or rises to 63% during charging. / 时间常数τ=RC：放电时经过一个τ，电荷降为原来的37%，充电时升至63%。
Current and voltage also decay/rise exponentially. For discharge: I = I₀ e^(−t/RC), V = V₀ e^(−t/RC). / 电流与电压同样指数变化。放电：I = I₀ e^(−t/RC)，V = V₀ e^(−t/RC)。
Dielectric effect: insertion of dielectric (εᵣ > 1) increases capacitance and energy stored for a given voltage. / 介质效应：插入介电体（εᵣ > 1）提高电容和给定电压下的储能。
Common pitfall: confusing series (1/C_eq = 1/C₁ + 1/C₂) and parallel (C_eq = C₁ + C₂) combinations. / 常见误区：串联(1/C_eq = 1/C₁ + 1/C₂)与并联(C_eq = C₁ + C₂)混淆。
U = ½ CV²
7. Magnetic Fields: Forces and Hall Effect | 磁场：力与霍尔效应
Magnetic fields exert forces on moving charges and current-carrying conductors. Force on a conductor: F = BIL sinθ, where θ is angle between B and current. Force on a single charge: F = BQv sinθ. The direction is given by Fleming’s left-hand rule. The Hall effect arises when a current-carrying slab in a transverse magnetic field develops a Hall voltage V_H = B I / (n q t), where n is charge carrier density and t is thickness.
磁场对运动电荷和载流导体施加力。导线受力：F = BIL sinθ，θ是B与电流方向的夹角。单电荷受力：F = BQv sinθ。方向由弗莱明左手定则判断。霍尔效应中，载流薄片在横向磁场中产生霍尔电压V_H = B I / (n q t)，其中n为载流子密度，t为薄片厚度。
F = BIL sinθ — applies when the field is uniform. / 适用于均匀磁场。
Circular motion of charged particle in uniform B: magnetic force provides centripetal force ⇒ BQv = mv²/r, so r = mv/(BQ). / 带电粒子在匀强磁场中的圆周运动：磁力提供向心力⇒BQv = mv²/r，r = mv/(BQ)。
Velocity selector: crossed E and B fields allow particles with speed v = E/B to pass undeflected. / 速度选择器：正交的电场与磁场使速度v = E/B的粒子无偏转通过。
Hall probe measures magnetic flux density; V_H ∝ B. / 霍尔探头测量磁感应强度；V_H ∝ B。
For a current loop, torque τ = B I A N sinθ. / 载流线圈力矩τ = B I A N sinθ。
F = BQv sinθ
8. Electromagnetic Induction | 电磁感应
Faraday’s law states that the induced e.m.f. in a circuit equals the rate of change of magnetic flux linkage: ε = −N (dΦ/dt). Lenz’s law gives the minus sign: induced current flows to oppose the change in flux. Applications include generators, transformers, and induction braking. Transformers follow V_s/V_p = N_s/N_p and, for an ideal transformer, I_p V_p = I_s V_s.
Magnetic flux Φ = B A cosθ; flux linkage = NΦ. / 磁通量Φ = B A cosθ；磁通量链 = NΦ。
Ways to induce e.m.f.: move magnet relative to coil, change area, rotate coil, or change B. / 产生感应电动势的方法：磁铁与线圈相对运动、改变面积、旋转线圈或改变B。
AC generator: rotating coil gives ε = B A N ω sin ωt. / 交流发电机：旋转线圈产生ε = B A N ω sin ωt。
Eddy currents: circulating currents in bulk conductors causing heating and braking; reduced by laminations. / 涡流：块状导体中的环流，导致发热和制动；通过叠片减少涡流。
Self-inductance: ε = −L (dI/dt), energy stored = ½ L I². / 自感：ε = −L (dI/dt)，储存能量 = ½ L I²。
ε = −N dΦ/dt
9. Alternating Currents | 交流电
Alternating current (AC) varies sinusoidally: I = I₀ sin ωt, V = V₀ sin ωt. The root-mean-square (r.m.s.) value is the effective DC equivalent: I_rms = I₀/√2, V_rms = V₀/√2. Power in resistive circuits is P = I_rms V_rms = I_rms² R. Rectification using diodes converts AC to pulsating DC; smoothing with capacitors reduces ripple.
交流电按正弦变化：I = I₀ sin ωt，V = V₀ sin ωt。均方根值(r.m.s.)是等效直流值：I_rms = I₀/√2，V_rms = V₀/√2。纯电阻电路功率P = I_rms V_rms = I_rms² R。二极管整流将交流变为脉动直流，电容滤波减小纹波。
Peak, peak-to-peak, and r.m.s. values: r.m.s. is most relevant for power calculations. / 峰值、峰峰值和均方根值：功率计算常用均方根值。
Half-wave rectification: one diode, output only positive halves. / 半波整流：一个二极管，仅输出正半周。
Full-wave rectification: diode bridge, both halves become positive. / 全波整流：二极管桥，正负半周均变为正向。
Smoothing capacitor: larger C gives smaller ripple; time constant RC >> T. / 滤波电容：C越大纹波越小；时间常数RC远大于周期T。
Reactance: inductive X_L = 2πfL, capacitive X_C = 1/(2πfC); phase differences in L and C circuits. / 电抗：感抗X_L = 2πfL，容抗X_C = 1/(2πfC)；存在相位差。
V_rms = V₀/√2
10. Quantum Physics | 量子物理
Photon model: light consists of photons with energy E = h f = h c/λ. The photoelectric effect demonstrates that electrons are emitted only if photon energy exceeds the work function φ. Einstein’s equation: h f = φ + ½ m v²_max. Stopping potential V_s gives ½ m v²_max = e V_s. Threshold frequency f₀ = φ/h. This evidence supports the particle nature of light.
光子模型：光由光子组成，能量E = h f = h c/λ。光电效应表明，只有光子能量大于逸出功φ时才能打出电子。爱因斯坦方程：h f = φ + ½ m v²_max。遏止电势V_s满足½ m v²_max = e V_s。极限频率f₀ = φ/h。这为光的粒子性提供了证据。
E = h f — Planck’s constant h = 6.63 × 10⁻³⁴ J s. / 普朗克常数。
Photon momentum: p = h/λ. / 光子动量：p = h/λ。
Energy levels in atoms: discrete energies; electrons jump by absorbing/emitting photons ΔE = h f = E₂ − E₁. / 原子能级分立；电子通过吸收或辐射光子跃迁，ΔE = h f = E₂ − E₁。
De Broglie wavelength: λ = h/p = h/(mv) — every moving particle has a wave nature. / 德布罗意波长：λ = h/p = h/(mv)，所有运动粒子具有波动性。
Spectra: emission line spectra correspond to transitions between energy levels; absorption spectra show dark lines. / 光谱：发射线谱对应能级跃迁；吸收光谱显示暗线。
Nuclear structure: nucleus contains protons and neutrons; mass number A, atomic number Z. The strong nuclear force binds nucleons. Mass defect and binding energy: E = Δm c²; binding energy per nucleon indicates stability, peaking around iron-56. Radioactive decay follows N = N₀ e^(−λt), activity A = λN; half-life t₁/₂ = ln 2/λ. Fission of heavy nuclei and fusion of light nuclei release energy because they move the products toward higher binding energy per nucleon.
Alpha decay: nucleus emits ⁴₂He; beta-minus: n → p + e⁻ + ν̄ₑ; beta-plus: p → n + e⁺ + νₑ. / α衰变放出⁴₂He；β⁻衰变：n → p + e⁻ + 反电子中微子；β⁺衰变：p → n + e⁺ + 中微子。
Exponential decay law: N = N₀ e^(−λt). / 指数衰变律。
Half-life: time for half the nuclei to decay; useful for dating. / 半衰期：一半原子核衰变所需时间；用于年代测定。
Activity A = λN, units becquerel (Bq). / 活度A = λN，单位贝克勒尔(Bq)。
Fission: chain reaction controlled by neutrons; fusion requires high temperature and pressure. / 裂变：链式反应由中子控制；聚变需要高温高压。
Mass–energy equivalence: 1 u = 931.5 MeV. / 质能等价：1 u = 931.5 MeV。
ΔE = Δm c²
12. Particle Physics and Optional Highlights | 粒子物理与选修聚焦
The Standard Model classifies fundamental particles into quarks (up, down, charm, strange, top, bottom) and leptons (electron, muon, tau, and their neutrinos). Hadrons are composite: baryons (3 quarks, e.g. proton uud) and mesons (quark–antiquark). Conservation laws (charge, baryon number, lepton number, strangeness) govern interactions. The four fundamental forces are mediated by gauge bosons: photon (electromagnetic), W⁺/W⁻/Z⁰ (weak), gluons (strong), and graviton (gravity – not in Standard Model).
Optional topics such as astrophysics and medical physics appear frequently. In astrophysics, Hubble’s law v = H₀ d, stellar luminosity, and the Hertzsprung–Russell diagram are key. Distance measurements: parallax p (arcsec) → d (pc) = 1/p. In medical physics, X-ray attenuation I = I₀ e^(−μx), ultrasound imaging using acoustic impedance
Published by TutorHao | Physics Revision Series | aleveler.com
In IGCSE Edexcel Physics, experimental investigations are at the heart of understanding how physical principles are discovered and verified. Student Book 2 provides a structured approach to designing, carrying out, and analysing experiments, ensuring students develop the practical skills essential for both the written examination and further scientific study. This article explores the key investigations, methods, and analytical techniques covered in the book, equipping you with a robust framework for experimental physics.
Every investigation begins with a thorough risk assessment. Student Book 2 emphasises identifying hazards such as hot surfaces, electrical shocks, or heavy falling masses, and implementing control measures like using heat-proof mats, low-voltage power supplies, and safety screens. A well-designed experiment includes clear independent, dependent, and control variables, ensuring that the results are both valid and reproducible.
Accurate measurements are fundamental in physics. The book teaches how to use instruments such as rulers, vernier calipers, micrometers, stopwatches, ammeters, and voltmeters, paying close attention to parallax error and zero error. Every measurement carries an uncertainty, typically taken as ± half the smallest scale division for digital instruments or ± the reading error for analogue ones. Students learn to record values as (reading ± uncertainty) and propagate uncertainties through calculations.
3. Investigating Density of Regular and Irregular Solids | 探究规则与不规则固体的密度
To find the density (ρ) of a regular solid, measure its mass using a digital balance and calculate its volume from geometric dimensions (e.g., for a cube V = L × W × H). For an irregular solid, the displacement method is used: lower the object into a measuring cylinder partially filled with water and record the rise in water level. Density is then calculated using ρ = m / V. Sources of error include trapped air bubbles and water splashes.
要找出规则固体的密度 (ρ)，先使用电子天平测量其质量，通过几何尺寸计算体积（例如长方体 V = 长 × 宽 × 高）。对于不规则固体，采用排水法：将物体放入部分装有水的量筒中，记录水面上升的高度。然后使用公式 ρ = m / V 计算密度。误差来源包括附着的气泡和水花飞溅。
4. Newton’s Second Law: Force and Acceleration | 牛顿第二定律：力与加速度
A classic investigation uses a trolley on a friction-compensated ramp, pulled by a falling mass over a pulley. By keeping the total mass of the system constant and varying the accelerating force (by transferring slotted masses from the trolley to the hanger), students plot acceleration (a) against force (F). The graph should be a straight line through the origin, verifying F = ma. The gradient equals 1/(total mass). A second experiment keeps the force constant and varies the mass, showing a hyperbolic relationship (a ∝ 1/m).
一个经典探究实验使用放置在补偿摩擦的斜面上的小车，通过滑轮被下落的砝码拉动。保持系统的总质量不变，改变加速力（将槽码从小车转移到挂盘上），学生绘制加速度 (a) 与力 (F) 的关系图。图像应是一条通过原点的直线，验证 F = ma。斜率等于 1/(总质量)。第二个实验保持力不变而改变质量，显示双曲线关系 (a ∝ 1/m)。
5. Ohm’s Law and Resistance of a Wire | 欧姆定律与导线电阻
The relationship between voltage (V) and current (I) for a metallic conductor at constant temperature is explored by varying a power supply and recording corresponding ammeter and voltmeter readings. A graph of V against I produces a straight line, demonstrating Ohm’s Law (V = IR). To investigate how resistance depends on length, use a long resistance wire and measure voltage drop across different lengths while keeping current constant. The resistance R = V/I is proportional to length L.
恒温下金属导体两端电压 (V) 与电流 (I) 的关系，通过改变电源电压并记录相应的安培表和伏特表读数来进行探究。V-I 图是一条直线，证明欧姆定律 (V = IR)。要探究电阻如何随长度变化，使用一段长的电阻丝，在保持电流恒定的情况下测量不同长度上的电压降。电阻 R = V/I 与长度 L 成正比。
6. Determining the Acceleration of Free Fall (g) | 测定自由落体加速度 (g)
A common method involves dropping a steel ball-bearing from a known height and measuring the time of fall using a trapdoor and electronic timer. The equation h = ½ g t² can be used; plotting h against t² yields a straight line with gradient ½ g. Another approach uses a pendulum: measure the period T for different lengths L, then use T² = (4π²/g) L, plotting T² against L to find g from the gradient. Both experiments require careful timing and minimising air resistance.
常见方法是让一个钢球从已知高度落下，利用活板门和电子计时器测量下落时间。使用方程 h = ½ g t²；绘制 h 与 t² 的图，得到一条直线，斜率为 ½ g。另一种方法使用单摆：测量不同摆长 L 下的周期 T，利用 T² = (4π²/g) L，绘制 T² 与 L 的关系图，由斜率求出 g。这两个实验都要求精确计时并尽量减小空气阻力。
7. Refraction and Snell’s Law | 折射与斯涅尔定律
Using a ray box, a glass block, and a protractor, students measure angles of incidence (i) and refraction (r) for light passing from air into glass. By plotting sin i against sin r, a straight line through the origin confirms Snell’s Law: n₁ sin i = n₂ sin r, where the gradient gives the refractive index of glass. Multiple readings help reduce random error. The critical angle is also investigated by reversing the ray to travel from glass to air.
使用光线盒、玻璃砖和量角器，学生测量光从空气进入玻璃时的入射角 (i) 和折射角 (r)。绘制 sin i 与 sin r 的关系图，若得到通过原点的直线，则证实斯涅尔定律：n₁ sin i = n₂ sin r，斜率即为玻璃的折射率。多次读数有助于减少随机误差。还可通过倒置光路，让光线从玻璃射向空气，研究临界角。
8. Hooke’s Law for a Spring | 弹簧的胡克定律
A helical spring is suspended with a pointer and a metre rule. Known masses are added and the extension (e) is measured from the original length. Plotting force (F = mg) against extension yields a straight line up to the limit of proportionality, verifying F = k e, where k is the spring constant. Beyond the elastic limit, the spring deforms plastically. Students should avoid exceeding the elastic limit to ensure repeatable results.
将一个螺旋弹簧悬挂起来，并配以指针和米尺。添加已知质量，测量相对于原长的伸长量 (e)。绘制力 (F = mg) 与伸长量的图，在比例极限内得到一条直线，验证 F = k e，其中 k 是弹性系数。超过弹性极限后，弹簧发生塑性形变。学生应注意不要超过弹性极限以保持结果可重复。
9. Specific Heat Capacity of a Solid | 固体的比热容
A metal block (usually aluminium) is electrically heated with a known power for a measured time. The energy supplied E = P × t = I V t. The temperature rise Δθ is recorded, and the specific heat capacity c is calculated from E = m c Δθ. To improve accuracy, insulation is used to reduce heat loss to the surroundings, and the block is stirred to ensure uniform temperature. The final value is compared with the accepted value (e.g., for aluminium, ~900 J/(kg °C)).
使用已知功率的电加热器对一个金属块（通常为铝）加热已知时间。供给的能量 E = P × t = I V t。记录温度的升高 Δθ，然后通过 E = m c Δθ 计算比热容 c。为提高准确性，使用隔热材料以减少热量散失，并搅拌金属块以确保温度均匀。最终值与公认值（如铝约 900 J/(kg °C)）进行比较。
10. Data Analysis and Graphical Skills | 数据分析与图示技能
All experiments require systematic data recording in tables with appropriate units and headings. Graphs are plotted with labelled axes, sensible scales, and best-fit lines. Student Book 2 teaches how to interpret the gradient and y-intercept to extract physical constants. For non-linear relationships, students may linearise the data, for example by squaring or taking reciprocals, to test proportionalities. Calculating percentage difference between experimental and accepted values helps evaluate the experiment’s accuracy.
所有实验都需要在表格中系统记录数据，表格应有合适的单位和标题。作图时要标注坐标轴、选择合理的刻度并绘制最佳拟合线。学生用书 2 教授如何通过斜率和 y 截距提取物理常数。对于非线性关系，学生可以将数据线性化，例如通过平方或取倒数，以检验正比关系。计算实验值与公认值之间的百分差有助于评估实验的准确性。
11. Writing a Laboratory Report | 撰写实验报告
A complete report includes an aim, hypothesis, equipment list, method, results, analysis, conclusion, and evaluation. The evaluation critically reflects on sources of error, suggests improvements, and discusses whether the results support the hypothesis. Student Book 2 stresses using correct scientific terminology and presenting calculations clearly, including the propagation of uncertainties where relevant.
12. Common Pitfalls and Tips for Success | 常见陷阱与成功技巧
Many students lose marks by not stating control variables, failing to record readings to an appropriate number of significant figures, or misinterpreting graphs. Always repeat measurements and calculate averages to minimise random error; check for zero errors before use. In investigations involving heat, minimise heat loss and insulate apparatus. For electrical circuits, avoid loose connections and use components within their ratings. Finally, when evaluating results, quantify the reliability using range bars or percentage differences.
📚 Deriving Kinetic Energy Equation from AS Physics Unit 1 Mark Scheme Jun22 | AS物理单元1 分值方案公式推导：动能方程
In the June 2022 AS Physics Unit 1 examination, one of the high-mark questions required candidates to derive the kinetic energy formula from fundamental principles. This question tested the ability to connect Newton’s second law, the concept of work done by a resultant force, and the equations of uniform acceleration. Understanding how these building blocks fit together is essential for mastering mechanics at this level.
1. The Examination Context and Mark Allocation | 考试背景与分值分配
The question appeared in Section B of the paper and carried 6 marks, with the mark scheme explicitly rewarding clear logical steps. Candidates were expected to start by stating the definition of work done, then progress through substitution and algebraic manipulation. The final expression for kinetic energy had to be presented in the standard form Eₖ = ½mv².
Work done = force × distance moved in direction of force
功 = 力 × 沿力方向移动的距离
2
F = ma from Newton’s second law
根据牛顿第二定律 F = ma
3
Use of v² = u² + 2as to eliminate acceleration
利用 v² = u² + 2as 消去加速度
4
Correct substitution to obtain W = ½mv² – ½mu²
正确代换得出 W = ½mv² – ½mu²
5
Identification of kinetic energy gain
识别动能增量
6
Final neat statement Eₖ = ½mv²
最终简洁的表达式 Eₖ = ½mv²
2. Foundation: Work Done by a Constant Force | 基础：恒力做的功
In mechanics, work is done when a force moves an object through a displacement. The definition is strictly scalar: work done W = F s cosθ. For the derivation, we consider the simplest case where the force is applied in the direction of motion, so θ = 0° and cos 0° = 1, giving W = F s.
在力学中，当一个力使物体发生位移时就做了功。该定义为标量：功 W = F s cosθ。为了推导，我们考虑最简单的情形——力沿着运动方向施加，所以 θ = 0°，cos 0° = 1，得到 W = F s。
The resultant force is the net forward push that causes acceleration. When friction or other resistive forces are absent, the resultant force equals the applied force. The mark scheme rewarded stating this assumption explicitly, as it shows awareness that kinetic energy change equals net work.
3. Newton’s Second Law in Symbolic Form | 符号形式的牛顿第二定律
Newton’s second law states that the acceleration a of an object is directly proportional to the resultant force F and inversely proportional to its mass m: F = ma. This vector equation can be applied in one dimension for linear motion.
牛顿第二定律指出，物体的加速度 a 与合力 F 成正比，与其质量 m 成反比：F = ma。这一矢量方程可用于一维直线运动。
The mass m is assumed constant, as required in classical mechanics. By substituting F = ma into the work equation, we obtain W = (ma) × s. This step merges dynamics with energy concepts, a pivotal moment that the mark scheme highlighted as ‘algebraic substitution mark’.
质量 m 假设为常数，符合经典力学要求。将 F = ma 代入功的方程，得到 W = (ma) × s。这一步将动力学与能量概念融合，是评分方案中强调的“代数代换分”。
4. Recalling the Uniform Acceleration Equations | 回顾匀加速运动方程
The derivation hinges on the selection of the correct kinematic equation. For a body starting with initial velocity u and accelerating uniformly to final velocity v over displacement s, the appropriate equation is:
推导的关键在于选择正确的运动学方程。对于一个以初速度 u 开始、匀加速到末速度 v、位移为 s 的物体，合适的方程为：
v² = u² + 2as
This equation links velocities, acceleration, and displacement directly, making it ideal for eliminating a. Candidates often confuse this with s = ut + ½at², which would not directly give the desired energy expression without further steps.
该方程直接联系了速度、加速度和位移，非常适合消去 a。考生常将其与 s = ut + ½at² 混淆，后者在没有额外步骤的情况下无法直接给出所需的能量表达式。
5. Rearranging to Express ‘as’ | 变形以表达 as
We need the product a × s to appear in our work equation. Rearranging v² = u² + 2as gives:
我们需要乘积 a × s 出现在功的方程中。变形 v² = u² + 2as 得到：
2as = v² – u²
Therefore, as = (v² – u²)/2. This rearrangement is a simple algebraic manipulation, but the mark scheme required it to be shown clearly. An alternative route is to solve for a and multiply by s, but the direct rearrangement saves a line of working.
因此，as = (v² – u²)/2。这一变形是简单的代数处理，但评分方案要求清晰展示。另一种方法是先求 a 再乘以 s，但直接变形可节省一步书写。
6. Substituting into the Work Expression | 代入功的表达式中
We now replace the product a s in W = m × a × s with the expression derived from kinematics:
现在我们将运动学导出的表达式代入 W = m × a × s 中的乘积 a s：
W = m × ( (v² – u²)/2 )
Multiplying through by m yields W = m(v² – u²)/2. This can be separated into two terms: W = ½mv² – ½mu². The mark scheme insisted on showing this factorization explicitly to earn the manipulation mark.
乘以 m 后得到 W = m(v² – u²)/2。可以拆分为两项：W = ½mv² – ½mu²。评分方案坚持要求明确展示这一因式分解才能获得处理分。
7. Physical Interpretation: Kinetic Energy Change | 物理解释：动能的变化
The expression ½mv² represents a form of energy dependent solely on mass and instantaneous speed. The term ½mu² is the corresponding energy at the initial state. Hence, the net work done by the resultant force equals the change in kinetic energy, ΔEₖ.
If the initial kinetic energy is taken as zero (object at rest, u = 0), the work done to accelerate the object to speed v is exactly ½mv². This defines the kinetic energy stored in a moving body. The mark scheme accepted statements like ‘W = ΔEₖ = final Eₖ – initial Eₖ’.
In SI units, mass m is in kilograms (kg) and speed v in metres per second (m s⁻¹). Consequently, kinetic energy has units of kg m² s⁻², which is equivalent to the joule (J). The mark scheme often required candidates to check dimensional consistency for the final mark if the question asked for units.
在国际单位制中，质量 m 以千克 (kg) 为单位，速度 v 以米每秒 (m s⁻¹) 为单位。因此，动能的单位是 kg m² s⁻²，等同于焦耳 (J)。如果题目要求写出单位，评分方案通常要求考生检查量纲一致性以获得最后分数。
9. Common Pitfalls Highlighted by the Mark Scheme | 评分方案强调的常见错误
Many scripts lost marks by omitting the crucial step of stating F = ma, jumping directly from work to kinematics without linking through resultant force. Others incorrectly used v² = u² + 2as but then rearranged to a = (v – u)/t, losing the connection to displacement. The mark scheme penalised missing steps that broke the logical chain.
许多答卷因遗漏陈述 F = ma 的关键步骤而失分，跳过了合力环节直接从功到运动学。另一些考生虽然正确使用了 v² = u² + 2as，但却变形为 a = (v – u)/t，失去了与位移的联系。评分方案对破坏逻辑链的跳步予以扣分。
Forgetting to define work done as force × displacement.
忘记将功定义为力乘以位移。
Using the wrong kinematic equation (e.g., s = ut + ½at²) and getting stuck.
使用错误的运动学方程（如 s = ut + ½at²）而陷入困境。
Not separating ½mv² and ½mu² when required by the question context.
未按题目要求分开写出 ½mv² 和 ½mu²。
10. Practice Scenario: Applying the Derivation in a Problem | 练习情境：在问题中应用该推导
A typical follow-up question asks: ‘A car of mass 1200 kg accelerates from rest to 15 m s⁻¹. Using the derived expression, calculate its kinetic energy and state the work done by the engine assuming no friction.’ The solution requires Eₖ = ½ × 1200 × (15)² = 135 000 J. The work done is 135 kJ, reinforcing the result.
一道典型的后续问题是：“一辆质量 1200 kg 的汽车从静止加速到 15 m s⁻¹。利用推导出的表达式，计算汽车的动能，并假设无摩擦时指出发动机做的功。”解：Eₖ = ½ × 1200 × (15)² = 135 000 J。所做的功为 135 kJ，强化了这一结果。
Such questions test not only recall of the formula but a deep understanding that kinetic energy equals the work done to achieve that speed. The mark scheme for the Jun22 paper awarded marks for correct substitution and unit conversion, mirroring the derivation’s logic.
11. Extending to Variable Forces and Graph Analysis | 拓展至变力与图像分析
Although the derivation assumes a constant resultant force, the concept of kinetic energy holds for variable forces as well. In later topics, the area under a force–displacement graph represents work done, and the same energy relationship emerges through integration. The AS mark scheme occasionally includes a graph evaluation where students must recognise that the work done equals the change in kinetic energy regardless of force constancy.
For variable forces, one cannot use v² = u² + 2as directly because acceleration is not uniform; however, the principle that net work transfers energy remains unchanged. This deeper idea is a bridge to A2 studies.
The kinetic energy derivation is a classic example of synoptic thinking in physics, combining definitions, laws, and algebra. Mastering it equips students to tackle similar derivations, such as gravitational potential energy or elastic potential energy. The mark scheme rewards a step-by-step logical flow, so practising writing out the derivation with annotated steps is an effective revision technique.
Revisiting past paper mark schemes reveals that examiners look for precise language, clear algebraic justification, and correct unit handling. Students should aim to reproduce the derivation from memory, then check against the official scheme to identify any gaps.