AS AQA Physics Paper 1 (June 2022) Insert Analysis & Key Concepts | AS AQA 物理 2022年6月试卷1 考点精析与核心概念

📚 AS AQA Physics Paper 1 (June 2022) Insert Analysis & Key Concepts | AS AQA 物理 2022年6月试卷1 考点精析与核心概念

The AQA AS Physics Paper 1 (June 2022) insert provides essential data, equations and information that candidates must interpret correctly to succeed. This article breaks down every core topic tested in this paper, from particle physics to electricity, ensuring you understand exactly what the examiner expects.

AQA AS 物理 2022年6月试卷1的 insert 提供了考生必须正确解读的关键数据、公式和信息。本文将逐项拆解该试卷考查的每一个核心考点，从粒子物理到电学，确保你精准掌握考官的要求。

1. Particle Physics & Atomic Structure | 粒子物理与原子结构

The June 2022 paper opened with questions on fundamental particles. You must recall that protons and neutrons are composed of quarks. A proton consists of two up quarks and one down quark (uud), while a neutron consists of one up quark and two down quarks (udd). The up quark has charge +2/3 e, and the down quark has charge -1/3 e.

2022年6月试卷开篇即考查基本粒子。你必须牢记：质子和中子由夸克组成。质子由两个上夸克和一个下夸克组成（uud），中子由一个上夸克和两个下夸克组成（udd）。上夸克电荷为 +2/3 e，下夸克电荷为 -1/3 e。

For anti-particles, every particle has a corresponding anti-particle with the same mass but opposite charge. When a particle meets its anti-particle, they annihilate, producing energy in the form of photons. The minimum total energy of the photons equals the total rest energy of the particle-antiparticle pair, given by E = mc².

对于反粒子，每个粒子都有与之对应的反粒子，二者质量相同但电荷相反。当粒子与其反粒子相遇时，发生湮灭，以光子形式释放能量。光子的最小总能量等于粒子-反粒子对的静止总能量，即 E = mc²。

E = mc²

Key decay reactions to remember for AS level include beta-minus and beta-plus decay. In beta-minus decay, a neutron converts to a proton, emitting an electron and an anti-electron neutrino. In beta-plus decay, a proton converts to a neutron, emitting a positron and an electron neutrino.

AS 阶段需要牢记的关键衰变反应包括 β⁻ 衰变和 β⁺ 衰变。β⁻ 衰变中，中子转化为质子，释放电子和反电子中微子；β⁺ 衰变中，质子转化为中子，释放正电子和电子中微子。

The June 2022 paper tested conservation laws in these reactions. Always check baryon number, lepton number and charge conservation when analysing any particle interaction.

2022年6月试卷考查了这些反应中的守恒定律。分析任何粒子相互作用时，务必检查重子数、轻子数和电荷守恒。

2. Quantum Phenomena & The Photoelectric Effect | 量子现象与光电效应

The photoelectric effect is a central topic in AS AQA physics. The energy of a photon is given by E = hf, where h is Planck’s constant (6.63 × 10⁻³⁴ J s) and f is the frequency. The insert provides values for h, the electron charge and the work function of various metals.

光电效应是 AS AQA 物理的核心考点。光子的能量公式为 E = hf，其中 h 是普朗克常数（6.63 × 10⁻³⁴ J·s），f 是频率。Insert 提供了 h 值、电子电荷量以及多种金属的逸出功。

The key equation for the photoelectric effect is the Einstein photoelectric equation:

光电效应的核心方程是爱因斯坦光电方程：

hf = Φ + ½mv_max²

where f is the photon frequency, Φ is the work function of the metal, and v_max is the maximum speed of ejected electrons. The work function is the minimum energy required to liberate an electron from the metal surface.

其中 f 为光子频率，Φ 为金属的逸出功，v_max 为逸出电子的最大速度。逸出功是使电子脱离金属表面所需的最小能量。

In the June 2022 paper, candidates were asked to interpret a graph of maximum kinetic energy against frequency. The threshold frequency f₀ occurs where the graph crosses the frequency axis — at this point, the photon energy exactly equals the work function. The gradient of the straight-line graph equals Planck’s constant h.

2022年6月试卷要求考生解读最大动能对频率的图表。阈值频率 f₀ 位于图线与频率轴的交点——此时光子能量恰好等于逸出功。直线图线的斜率等于普朗克常数 h。

Common examiner comments noted that many students confused intensity with frequency. Increasing intensity increases the number of photons per second but does NOT change the maximum kinetic energy of photoelectrons — only increasing frequency (above the threshold) does that.

考官常见评语指出，许多学生混淆了强度与频率。增大强度会增加每秒光子数，但不会改变光电子的最大动能——只有增大频率（超过阈值后）才能做到。

3. Energy Levels & Line Spectra | 能级与线状光谱

Electrons in an atom occupy discrete energy levels. An electron can transition between levels by absorbing or emitting a photon whose energy exactly matches the energy difference between the two levels: ΔE = hf. The June 2022 insert contained an energy level diagram for hydrogen.

原子中的电子占据离散的能级。电子通过吸收或发射能量恰好等于两能级之差的跃迁来实现能级变化：ΔE = hf。2022年6月的 insert 中包含氢原子能级图。

When an electron falls from a higher energy level to a lower one, it emits a photon. The frequency of the emitted photon is given by:

当电子从高能级跃迁到低能级时，会发射光子。发射光子的频率由下式给出：

f = (E_higher − E_lower) / h

Ionisation occurs when an electron gains enough energy to leave the atom entirely. The ionisation energy of hydrogen from its ground state is 13.6 eV. The paper tested the ability to calculate the maximum number of spectral lines possible from a given set of energy levels. For n levels, the number of possible transitions equals n(n-1)/2.

电离发生在电子获得足够能量完全脱离原子时。氢原子从基态的电离能为 13.6 eV。试卷考查了从给定能级组计算可能光谱线最大数目的能力。对于 n 个能级，可能的跃迁数为 n(n-1)/2。

Students often forgot that the Rydberg constant or the energy values in the insert could be used to find the ionisation energy directly. Always check the insert for hydrogen energy level data before attempting spectral line calculations.

学生常常忘记可以用 insert 中的里德伯常数或能值直接求电离能。进行光谱线计算前，务必检查 insert 中是否有氢能级数据。

4. Progressive Waves & Refraction | 行波与折射

The wave section of the June 2022 paper examined travelling waves, their properties and refraction. A progressive wave transfers energy without transferring matter. Key quantities include wavelength λ, frequency f, and wave speed v, related by v = fλ.

2022年6月试卷的波部分考查了行波、波的特性及折射。行波传递能量而不传递物质。关键物理量包括波长 λ、频率 f 和波速 v，它们的关系为 v = fλ。

v = fλ

For transverse waves, the oscillations are perpendicular to the direction of energy transfer. For longitudinal waves, the oscillations are parallel to the direction of energy transfer. The insert contained data on wave speeds in different media — be prepared to calculate refractive index using:

对于横波，振动方向垂直于能量传递方向；对于纵波，振动方向平行于能量传递方向。Insert 包含不同介质中的波速数据——需准备用下式求折射率：

n = c / v

At a boundary between two media, the refractive index relates the angle of incidence to the angle of refraction through Snell’s law: n₁ sin θ₁ = n₂ sin θ₂. The insert gave refractive index values, and candidates were examined on total internal reflection at the critical angle C, where sin C = 1/n.

在两种介质的界面上，折射率通过斯涅尔定律关联入射角与折射角：n₁ sin θ₁ = n₂ sin θ₂。Insert 给出了折射率数值，考查了临界角 C 处的全内反射，其中 sin C = 1/n。

A common error in the 2022 exam was using the wrong refractive index when light travels from a rarer to a denser medium. Always identify which medium the light is entering before applying Snell’s law.

2022年考试中的常见错误是当光从疏介质进入密介质时用错折射率。应用斯涅尔定律前，务必先判断光进入的是哪种介质。

5. Superposition, Interference & Stationary Waves | 叠加、干涉与驻波

The principle of superposition states that when two waves meet, the resultant displacement at any point is the vector sum of the individual displacements. This leads to interference — constructive interference where displacements add (path difference = nλ), and destructive interference where they cancel (path difference = (n+½)λ).

叠加原理指出：当两列波相遇时，任意一点处的合位移等于各波位移的矢量和。这导致干涉——相长干涉处位移相加（光程差 = nλ），相消干涉处相互抵消（光程差 = (n+½)λ）。

The June 2022 paper included questions on Young’s double-slit experiment. When coherent light passes through two narrow slits, an interference pattern of bright and dark fringes forms on a screen. The fringe spacing w is given by:

2022年6月试卷包含杨氏双缝实验的问题。当相干光通过两条窄缝时，屏幕上形成明暗相间的干涉条纹。条纹间距 w 由下式给出：

w = λD / s

where λ is the wavelength of light, D is the distance from the slits to the screen, and s is the slit separation. The insert provided slit separation values, and candidates had to calculate wavelength from measured fringe spacing.

其中 λ 为光波长，D 为双缝到屏幕的距离，s 为双缝间距。Insert 提供了双缝间距的数值，考生需要用测得的条纹间距来求波长。

Stationary (standing) waves form when two waves of equal frequency and amplitude travel in opposite directions. Nodes are points of zero displacement, and antinodes are points of maximum displacement. For a string fixed at both ends, the fundamental frequency has wavelength λ = 2L, where L is the string length.

驻波由两列等频、等幅、反向传播的波叠加形成。波节是位移为零的点，波腹是位移最大的点。对于两端固定的弦，基频的波长为 λ = 2L，其中 L 为弦长。

6. Kinematics & Motion in a Straight Line | 运动学与直线运动

The mechanics section of the paper tested the SUVAT equations of motion for objects moving with constant acceleration. The insert provided the standard equations, but you must know when to apply each one. The five equations are:

试卷的力学部分考查了匀加速运动中物体的 SUVAT 运动学方程。Insert 提供了标准方程，但你必须知道何时使用哪个方程。五个方程为：

v = u + at
s = ut + ½at²
s = ½(u + v)t
v² = u² + 2as
s = vt − ½at²

where u is initial velocity, v is final velocity, a is acceleration, t is time, and s is displacement. For projectile motion, treat horizontal and vertical components independently. The horizontal component of velocity remains constant (ignoring air resistance), while the vertical component accelerates at g = 9.81 m s⁻².

其中 u 为初速度，v 为末速度，a 为加速度，t 为时间，s 为位移。对于抛体运动，将水平与竖直分量独立处理。水平速度分量保持不变（忽略空气阻力），竖直分量以 g = 9.81 m s⁻² 加速。

In the 2022 paper, a projectile question required students to use the insert’s value of g and calculate the time of flight from the vertical motion first, then use this time to find the horizontal range. A frequent mistake was to use the resultant velocity instead of components — always resolve velocity into perpendicular components.

2022年试卷中的抛体问题要求考生使用 insert 中的 g 值，先从竖直运动求出飞行时间，再用该时间求水平射程。常见错误是使用了合速度而非分量——务必把速度分解为垂直分量。

Velocity-time graphs are also examined: the gradient gives acceleration, the area under the graph gives displacement, and a horizontal line indicates constant velocity. Displacement-time graphs: the gradient gives velocity, and a curved line indicates acceleration.

速度-时间图像也会考查：斜率给出加速度，图像下面积为位移，水平线表示匀速。位移-时间图像：斜率给出速度，曲线表示加速运动。

7. Newton’s Laws, Momentum & Impulse | 牛顿定律、动量与冲量

Newton’s three laws were examined in the context of simple dynamics problems. Newton’s second law has two forms: F = ma for constant mass, and F = Δ(mv)/Δt for the general case where mass may change. The insert provided the relationship between force and momentum change.

试卷在简单动力学问题中考查了牛顿三定律。牛顿第二定律有两种形式：质量恒定时的 F = ma，以及质量可能变化时的普遍形式 F = Δ(mv)/Δt。Insert 提供了力与动量变化之间的关系。

F = ma and F = Δp/Δt

Momentum p is defined as p = mv. The principle of conservation of momentum states that in an isolated system, the total momentum before a collision equals the total momentum after the collision. For elastic collisions, kinetic energy is conserved; for inelastic collisions, it is not.

动量 p 定义为 p = mv。动量守恒原理指出：在孤立系统中，碰撞前的总动量等于碰撞后的总动量。弹性碰撞中动能守恒；非弹性碰撞中动能不守恒。

Impulse is defined as the product of the average force and the time for which it acts: impulse = FΔt. This equals the change in momentum. On a force-time graph, the area under the graph equals the impulse. The June 2022 paper asked candidates to calculate the impulse from a force-time graph and use it to determine the final velocity of an object.

冲量定义为平均力与其作用时间的乘积：冲量 = FΔt，等于动量的变化。在力-时间图像中，图下面积等于冲量。2022年6月试卷要求考生从力-时间图像求冲量，并用它确定物体的末速度。

Examiner reports highlighted that students often forgot that impulse is a vector quantity — direction matters. Also, in collision problems, always define a positive direction and assign signs consistently throughout the calculation.

考官报告指出，学生经常忘记冲量是矢量——方向很重要。此外，在碰撞问题中，务必规定正方向并在整个计算中一致地赋予符号。

8. Work, Energy & Power | 功、能与功率

Work done is defined as the product of the force and the distance moved in the direction of the force: W = Fs cos θ. When the force acts in the direction of motion, θ = 0° and W = Fs. The unit of work is the joule (J). The insert in the June 2022 paper gave mechanical work and energy equations explicitly.

功定义为力与力方向上的位移之积：W = Fs cos θ。当力沿运动方向作用时，θ = 0°，W = Fs。功的单位是焦耳（J）。2022年6月试卷的 insert 明确给出了力学功和能量方程。

Kinetic energy and gravitational potential energy are given by:

动能和重力势能由下式给出：

Ek = ½mv²
Ep = mgh

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In a closed system, the total energy remains constant. For an object falling freely, gravitational potential energy converts to kinetic energy, and assuming no air resistance: mgh = ½mv².

能量守恒原理指出：能量不能凭空创造或消灭，只能从一种形式转化为另一种形式。在封闭系统中，总能量保持不变。对自由落体，重力势能转化为动能，假设无空气阻力：mgh = ½mv²。

Power is the rate of doing work or transferring energy: P = W/t = Fv. The paper included a question on the power output of a motor lifting a load at constant speed. Remember that the useful power output equals the rate of increase of gravitational potential energy, and efficiency equals (useful output power ÷ input power) × 100%.

功率是做功或转移能量的速率：P = W/t = Fv。试卷包含电机匀速提升负载的功率输出问题。记住：有用输出功率等于重力势能增加率，效率等于（有用输出功率 ÷ 输入功率）× 100%。

9. Materials & Elasticity | 材料与弹性

The materials section examined Hooke’s law, elastic strain energy, and the elastic properties of solids. Hooke’s law states that extension is directly proportional to applied force, provided the limit of proportionality is not exceeded: F = kx, where k is the spring constant.

材料部分考查了胡克定律、弹性应变能以及固体的弹性性质。胡克定律指出：在比例极限内，伸长量与所受外力成正比：F = kx，其中 k 为劲度系数。

F = kx

Elastic strain energy is the energy stored in a deformed elastic material. For a spring obeying Hooke’s law, the elastic potential energy is the area under the force-extension graph:

弹性应变能是变形弹性材料中储存的能量。对于遵守胡克定律的弹簧，弹性势能等于力-伸长量图下的面积：

E = ½kx²

Tensile stress and strain are defined as stress = F/A and strain = x/L. The Young modulus is the ratio of tensile stress to tensile strain: E = (F/A) ÷ (x/L) = FL/Ax. The June 2022 insert included a stress-strain graph for a metal wire, requiring candidates to identify the elastic limit, yield point and ultimate tensile strength.

拉应力与应变定义为：应力 = F/A，应变 = x/L。杨氏模量是拉应力与拉应变之比：E = (F/A) ÷ (x/L) = FL/Ax。2022年6月 insert 包含金属丝的应力-应变图，要求考生识别弹性极限、屈服点和抗拉强度极限。

A common source of confusion was distinguishing between elastic limit and limit of proportionality. The limit of proportionality is where the force-extension graph stops being linear; the elastic limit is where permanent deformation begins. For a metal obeying Hooke’s law up to the limit of proportionality, these coincide, but for rubber they do not.

一个常见的混淆点是区分弹性极限与比例极限。比例极限是力-伸长量图不再线性的位置；弹性极限是开始发生永久形变的位置。对于遵守胡克定律至比例极限的金属，二者重合；但对于橡胶则不然。

10. Electric Current & Resistance | 电流与电阻

The electricity section of the paper covered current, potential difference, resistance and the factors affecting resistance. Electric current is the rate of flow of charge: I = ΔQ/Δt, measured in amperes. The insert may provide the elementary charge e = 1.60 × 10⁻¹⁹ C.

试卷的电学部分涵盖电流、电势差、电阻以及影响电阻的因素。电流是电荷流动的速率：I = ΔQ/Δt，单位为安培。Insert 可能提供元电荷 e = 1.60 × 10⁻¹⁹ C。

I = ΔQ/Δt and V = IR

For metallic conductors at constant temperature, current is directly proportional to potential difference — this is Ohm’s law. The resistance of a conductor depends on its length, cross-sectional area and resistivity: R = ρL/A. The insert gave values of resistivity, and candidates calculated resistance for different wire dimensions.

对恒温下的金属导体，电流与电势差成正比——这就是欧姆定律。导体的电阻取决于其长度、横截面积和电阻率：R = ρL/A。Insert 给出了电阻率值，考生需对不同尺寸的导线计算电阻。

R = ρL / A

The I-V characteristic graphs were also examined. A resistor obeying Ohm’s law gives a straight line through the origin. A filament lamp gives a curved characteristic because the resistance increases with temperature. A diode conducts in one direction only, with very high resistance in reverse bias. The paper tested reading values from these graphs to calculate resistance at specific points.

I-V 特性曲线图也被考查。遵守欧姆定律的电阻给出过原点的直线。白炽灯的特性曲线为曲线，因为电阻随温度升高而增大。二极管仅单向导通，反向偏置时电阻极高。试卷考查了从这些图读取数值以计算特定点的电阻。

Examiner feedback from the June 2022 session indicated that students frequently forgot to convert units before substituting into R = ρL/A. Always convert length to metres and cross-sectional area to square metres.

2022年6月考季的考官反馈指出，学生在代入 R = ρL/A 前经常忘记进行单位换算。务必把长度换算为米，横截面积换算为平方米。

11. Circuits, EMF & Internal Resistance | 电路、电动势与内阻

For series circuits, the total resistance is the sum of individual resistances: R_total = R₁ + R₂ + R₃. For parallel circuits, the reciprocal of total resistance equals the sum of reciprocals: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃. The current is the same through components in series; the potential difference is divided. In parallel, the potential difference is the same across each branch; the current is divided.

串联电路中，总电阻等于各电阻之和：R_total = R₁ + R₂ + R₃。并联电路中，总电阻的倒数等于各电阻倒数之和：1/R_total = 1/R₁ + 1/R₂ + 1/R₃。串联元件中电流处处相同，电势差按电阻分配；并联支路两端电势差相同，电流按支路分配。

The electromotive force (EMF) of a source is the energy transferred per unit charge in driving charge around a complete circuit. The terminal potential difference of a cell equals the EMF minus the lost volts due to internal resistance:

电源的电动势（EMF）是驱动电荷沿完整电路运行一周时单位电荷所获得的能量。电池的端电压等于电动势减去内阻导致的lost volts（电压损耗）：

E = I(R + r) or V = E − Ir

where E is the emf, I is the current, R is the external resistance, r is the internal resistance, and V is the terminal potential difference. The June 2022 paper required candidates to use a graph of terminal voltage against current to determine both the emf (y-intercept) and internal resistance (negative of the gradient).

其中 E 为电动势，I 为电流，R 为外电阻，r 为内阻，V 为端电压。2022年6月试卷要求考生利用端电压对电流的图像来确定电动势（y 截距）和内阻（斜率的负值）。

Students often struggled with circuit problems where resistors were arranged in a combination of series and parallel. A reliable strategy is: simplify the circuit step by step, replace each parallel combination with a single equivalent resistance, then redraw the simplified circuit before calculating currents and potential differences.

学生在处理串并联组合电路问题时经常遇到困难。可靠策略是：逐步化简电路，把每个并联组合替换为等效电阻，重画简化电路，再计算电流和电势差。

For the 2022 exam specifically, the insert contained a circuit with a voltmeter of finite resistance. When a voltmeter is connected in parallel with a resistor, it draws current and changes the circuit behaviour. Calculate the combined resistance of the voltmeter and resistor in parallel before finding the total circuit resistance.

针对2022年考试，insert 中包含一个带有有限内阻电压表的电路。当电压表与电阻并联时，它本身会分流并改变电路行为。计算总电路电阻前，须先求电压表与电阻并联后的等效电阻。

12. Exam Strategy & Common Pitfalls | 应试策略与常见失误

Based on examiner reports from the June 2022 AQA AS Physics Paper 1, several recurring errors cost candidates valuable marks. First, many students failed to quote numerical answers to the correct number of significant figures — AQA expects answers consistent with the data provided, typically 2 or 3 significant figures. Second, when using the insert’s data, always check the units stated: the insert may give energy in eV rather than joules, and you must convert appropriately (1 eV = 1.60 × 10⁻¹⁹ J).

根据 2022年6月 AQA AS 物理试卷1的考官报告，以下几个重复出现的错误让考生丢了不少分数。第一，许多学生没有按正确的有效数字位数给出数值答案——AQA 期望答案与所给数据一致，通常为2或3位有效数字。第二，使用 insert 中的数据时，务必检查其单位：insert 可能以 eV 而非焦耳给出能量，你必须进行适当换算（1 eV = 1.60 × 10⁻¹⁹ J）。

For multiple-choice questions in Section A, AQA uses a negative-marking formula in some versions of the paper — check whether incorrect answers lose marks. In June 2022, the multiple-choice section awarded 1 mark for each correct answer with no negative marking for wrong answers, but you should always verify this on your specific paper.

对于 A 部分的选择题，某些版本的试卷可能采用倒扣分公式——检查错误答案是否扣分。2022年6月的试卷选择题部分每题1分，答错不扣分，但你应在自己手中的具体试卷上核实这一点。

In written answers, always define the physical quantities you introduce, state units with every numerical value, and show your full working for calculation questions. Marks are available for method even when the final answer is incorrect. When interpreting the insert’s data, avoid common errors such as confusing wavelength with frequency readings, or reading the scale of the graph incorrectly by not using the smallest division.

在文字题回答中，始终定义你引入的物理量，每个数值都要注明单位，并在计算题中展示完整过程。即使最终答案错误，方法正确也能得分。解读 insert 数据时，避免常见错误，如混淆波长与频率读数，或未使用最小刻度读取图表的量程。

Finally, practise converting between prefixes — the insert frequently uses k, M, m, μ, n and p (10³, 10⁶, 10⁻³, 10⁻⁶, 10⁻⁹, 10⁻¹²). Write out the full equation, substitute with units, and cancel units to check dimensional consistency. If the units do not simplify to the expected SI unit, you have made an error somewhere.

最后，多加练习前缀之间的换算——insert 中经常使用 k、M、m、μ、n 和 p（10³、10⁶、10⁻³、10⁻⁶、10⁻⁹、10⁻¹²）。写出完整方程，代入带单位的数值，并消去单位检验量纲一致性。如果单位不能化简为预期的 SI 单位，说明某处出了差错。

Published by TutorHao | Physics Revision Series | aleveler.com

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AS AQA Physics Paper 1 (June 2022) Insert Analysis & Key Concepts | AS AQA 物理 2022年6月试卷1 考点精析与核心概念