📚 AQA A-level Physics June 2018 Paper 5 Walkthrough | AQA A-level 物理 2018年6月 试卷5 解析
The June 2018 AQA A-level Physics examination series is a valuable resource for students seeking to understand the style, depth, and mark-scheme expectations of the current specification (7408). Although the official AQA exam series consists of Paper 1, Paper 2 and Paper 3, many revision programmes and tutoring packs label a supplementary practice paper as “Paper 5”. This walkthrough is based on that supplementary practice material and covers the core topics, worked solutions, and common misconceptions that appear in the June 2018 papers.
2018年6月AQA A-level物理考试系列是学生理解现行考纲（7408）的题型、深度和评分标准的宝贵资源。虽然官方AQA考试系列包含试卷1、试卷2和试卷3，但许多复习课程和辅导资料将补充练习卷命名为“试卷5”。本解析基于该补充练习材料，涵盖2018年6月试卷中出现的核心知识点、解答步骤和常见误解。
1. Exam Structure and Mark Scheme | 试卷结构与评分标准
The AQA A-level Physics specification is assessed through three written papers. Paper 1 covers sections 1-5 (topics such as measurements, mechanics, materials, waves and electricity). Paper 2 covers sections 6-8 (thermal physics, nuclear physics and option topics). Paper 3 focuses on practical skills and data analysis. The supplementary “Paper 5” we discuss combines these areas into a single revision paper.
AQA A-level物理考纲通过三份笔试试卷进行评估。试卷1涵盖第1-5部分（测量、力学、材料、波和电学等主题）。试卷2涵盖第6-8部分（热物理、核物理和选修主题）。试卷3重点考查实验技能和数据分析。我们讨论的补充“试卷5”将这些领域合并到一份复习卷中。
Each paper uses a mix of multiple-choice, short-answer and extended-response questions. The mark scheme rewards correct physics, appropriate use of units, and sign conventions. For calculation questions, a correct final answer without working may still receive some credit, but full marks require clear substitution and rearrangement.
每份试卷混合使用选择题、简答题和扩展回答题。评分标准鼓励正确的物理表达、适当的单位和符号惯例使用。对于计算题，没有步骤的正确最终答案可能仍能得到部分分数，但满分需要清晰的代入和变形过程。
Understanding command words is essential: “state” means no explanation needed; “calculate” requires a numerical answer with working; “explain” demands a reason based on physics; “derive” asks for a mathematical proof from principles. Familiarity with these terms helps you target the required depth.
理解指令词至关重要：“state”表示不需要解释；“calculate”需要带有步骤的数值答案；“explain”需要基于物理原理的推理；“derive”要求从基本原理出发进行数学推导。熟悉这些术语有助于掌握答题深度。
2. Measurements and Uncertainties | 测量与不确定度
In the June 2018 examination, questions on measurements often tested absolute and percentage uncertainties, and the combination of uncertainties in final results. For example, a student measures a wire diameter as 0.52 mm using a micrometer with an uncertainty of ±0.01 mm. The percentage uncertainty in the diameter is (0.01 / 0.52) × 100 = 1.9%.
在2018年6月的考试中，关于测量的问题经常测试绝对不确定度和百分比不确定度，以及最终结果中不确定度的合成。例如，学生用千分尺测量一根金属丝的直径为0.52 mm，不确定度为±0.01 mm。直径的百分比不确定度为(0.01 / 0.52) × 100 = 1.9%。
When quantities are multiplied or divided, percentage uncertainties are added. If the diameter is used to compute the cross-sectional area A = π (d/2)², the percentage uncertainty in A is twice that in d, because the diameter is squared. Thus, the uncertainty in A becomes 2 × 1.9% = 3.8%.
当量相乘或相除时，百分比不确定度相加。如果用直径计算横截面积 A = π (d/2)²，A的百分比不确定度是d的两倍，因为直径被平方。因此A的不确定度变为2 × 1.9% = 3.8%。
A common mistake is to forget that a square root halves the percentage uncertainty, while a square doubles it. For the radius, r = d/2, the percentage uncertainty remains the same as d because dividing by a constant does not affect the percentage uncertainty.
常见错误是忘记平方根使百分比不确定度减半，而平方使百分比不确定度加倍。对于半径 r = d/2，其百分比不确定度与d相同，因为除以常数不影响百分比不确定度。
To reduce uncertainties, use a larger sample: measure the time for 20 oscillations instead of one, and use a digital vernier scale for repetitive measurements. Always record the instrument resolution and calculate the mean absolute deviation for repeated readings.
为减少不确定度，应使用更大的样本：测量20次振荡的时间而不是1次，并使用数字游标尺进行重复测量。始终记录仪器分辨率，并计算多次读数的平均绝对偏差。
3. Mechanics – Forces and Projectile Motion | 力学——力与抛体运动
Projectile motion is a staple of the AQA mechanics questions. A typical June 2018-style problem: a ball is thrown from a cliff with an initial speed of 20 m/s at 30° above the horizontal. The height of the cliff is 15 m. Calculate the maximum height reached above the ground and the horizontal range.
抛体运动是AQA力学问题的常见题型。一个典型的2018年6月风格问题：一个小球以20 m/s的初速度从悬崖上抛出，方向与水平方向成30°角。悬崖高度为15 m。计算小球达到的最大高度和水平射程。
The vertical component of the initial velocity is u_y = 20 sin 30° = 10 m/s. At maximum height, v_y = 0. Using v² = u² + 2as, we find s = (0 – 10²) / (2 × (-9.81)) = 5.10 m. Adding the cliff height, the maximum height above the ground is 15 + 5.10 = 20.10 m.
初速度的竖直分量为 u_y = 20 sin 30° = 10 m/s。在最高点处 v_y = 0。使用 v² = u² + 2as，求得 s = (0 – 10²) / (2 × (-9.81)) = 5.10 m。加上悬崖高度，小球距地面的最大高度为 15 + 5.10 = 20.10 m。
To find the range, determine the total flight time from the vertical motion. The displacement from the cliff top to the ground is y = -15 m. Using y = u_y t + ½ a t², we substitute: -15 = 10t – 4.905 t². Rearranging gives 4.905 t² – 10t – 15 = 0. Solving this quadratic: t = [10 + √(100 + 4 × 4.905 × 15)] / (2 × 4.905) = [10 + √(394.3)] / 9.81 = [10 + 19.86] / 9.81 = 3.04 s.
为了求射程，需要从竖直运动确定总飞行时间。小球从悬崖顶到地面的位移为 y = -15 m。使用 y = u_y t + ½ a t²，代入：-15 = 10t – 4.905 t²。整理得 4.905 t² – 10t – 15 = 0。解这个二次方程：t = [10 + √(100 + 4 × 4.905 × 15)] / (2 × 4.905) = [10 + √(394.3)] / 9.81 = [10 + 19.86] / 9.81 = 3.04 s。
The horizontal velocity is constant, u_x = 20 cos 30° = 17.32 m/s. Therefore, range = u_x × t = 17.32 × 3.04 = 52.6 m. When answering, always show the quadratic equation and reject the negative root for time.
水平速度恒定，u_x = 20 cos 30° = 17.32 m/s。因此射程 = u_x × t = 17.32 × 3.04 = 52.6 m。回答时，务必写出二次方程并舍去负的时间根。
For forces, remember that Newton’s second law is F = ma, and when multiple forces act on an incline, resolve the weight into components parallel and perpendicular to the slope. Friction is μR, where R is the normal reaction. In the June 2018 paper, a question asked to calculate the coefficient of friction given a constant velocity down a slope of 20°.
对于力的问题，记住牛顿第二定律为 F = ma。当物体在斜面上受到多个力作用时，要分解质量为平行和垂直于斜面的分量。摩擦力为 μR，其中R为法向反作用力。在2018年6月试卷中，有一道题要求根据物体以恒定速度沿20°斜面下滑来计算摩擦系数。
At constant velocity, the resultant force is zero. The component of weight down the slope is mg sin θ, and friction opposes motion up the slope. Hence mg sin θ = μ mg cos θ, which simplifies to μ = tan θ = tan 20° = 0.364. This is a classic result that saves calculation time.
在恒定速度下，合力为零。重量沿斜面方向的分量为 mg sin θ，摩擦力沿斜面向上。因此 mg sin θ = μ mg cos θ，化简为 μ = tan θ = tan 20° = 0.364。这是一个经典的结论，可以节省计算时间。
4. Materials – Young Modulus and Energy | 材料——杨氏模量与能量
The Young modulus is defined as the ratio of tensile stress to tensile strain. A June 2018 question presented a steel wire of length 2.00 m and diameter 0.50 mm. When a 15 N load is hung from the wire, it stretches by 1.20 mm. The cross-sectional area is A = π (0.25 × 10⁻³)² = 1.96 × 10⁻⁷ m².
杨氏模量定义为拉伸应力与拉伸应变之比。2018年6月的一道题给出一根长度为2.00 m、直径为0.50 mm的钢丝。当悬挂15 N负载时，钢丝伸长1.20 mm。横截面积为 A = π (0.25 × 10⁻³)² = 1.96 × 10⁻⁷ m²。
Stress is σ = F/A = 15 / 1.96 × 10⁻⁷ = 7.65 × 10⁷ Pa. Strain is ε = ΔL / L = 1.2 × 10⁻³ / 2.00 = 6.00 × 10⁻⁴. Young modulus E = σ/ε = 7.65 × 10⁷ / 6.00 × 10⁻⁴ = 1.28 × 10¹¹ Pa.
应力为 σ = F/A = 15 / 1.96 × 10⁻⁷ = 7.65 × 10⁷ Pa。应变为 ε = ΔL / L = 1.2 × 10⁻³ / 2.00 = 6.00 × 10⁻⁴。杨氏模量 E = σ/ε = 7.65 × 10⁷ / 6.00 × 10⁻⁴ = 1.28 × 10¹¹ Pa。
The area under the stress-strain graph up to the elastic limit represents the strain energy per unit volume. In a multiple-choice question, students were asked to identify the correct unit of strain energy; the answer is J/m³, which is equivalent to Pa.
应力-应变图到弹性极限之间的面积代表单位体积的应变能。在一道选择题中，学生被要求识别应变能的正确单位；答案是 J/m³，也等价于 Pa。
When a force F stretches a spring by extension x, the elastic potential energy is ½ F x. For a wire under tension, the internal energy stored is ½ stress × strain × volume. Do not confuse this with the work done, which is F × ΔL only when F is constant.
当力F将弹簧拉伸伸长量x时，弹性势能为 ½ F x。对于受拉力的金属丝，储存的内能为 ½ 应力 × 应变 × 体积。不要将其与功混淆：功为 F × ΔL 仅在F为常数时成立。
In extended-response questions, you may be asked to compare the behaviour of a ductile material and a brittle material. Ductile materials show a yield point and plastic region, while brittle materials fracture without plastic deformation. Always quote stress and strain values from the graph where possible.
在扩展回答题中，你可能被要求比较延性材料和脆性材料的行为。延性材料呈现屈服点和塑性区域，而脆性材料在没有塑性变形的情况下断裂。尽可能从图中引用应力和应变数值。
5. Waves and Optics | 波动与光学
Diffraction grating problems are common in the June 2018 paper. A grating has 600 lines per mm, so the slit spacing is d = 1 / (600 × 10³) = 1.67 × 10⁻⁶ m. Monochromatic light of wavelength 589 nm is incident normally. For the second-order maximum, n = 2, we use d sin θ = n λ.
衍射光栅问题在2018年6月试卷中常见。一个光栅每毫米有600条刻线，因此狭缝间距为 d = 1 / (600 × 10³) = 1.67 × 10⁻⁶ m。波长为589 nm的单色光垂直入射。对于二级极大值，n = 2，使用 d sin θ = n λ。
sin θ = (2 × 589 × 10⁻⁹) / (1.67 × 10⁻⁶) = 1.178 / 1.67 = 0.7054. Hence θ = 44.9°. The maximum order visible occurs when sin θ = 1: n_max = d / λ = 1.67 × 10⁻⁶ / 589 × 10⁻⁹ = 2.84, so the highest order is 2.
sin θ = (2 × 589 × 10⁻⁹) / (1.67 × 10⁻⁶) = 1.178 / 1.67 = 0.7054。因此 θ = 44.9°。可见的最大级数出现在 sin θ = 1 时：n_max = d / λ = 1.67 × 10⁻⁶ / 589 × 10⁻⁹ = 2.84，因此最高级数为2。
Be careful: if n_max is not an integer, you must round down to the nearest integer. In this case, 2.84 rounds down to 2, not up to 3, because sin θ cannot exceed 1. This is a favourite exam trap.
注意：如果n_max不是整数，必须向下取整到最近的整数。在这种情况下，2.84取整为2，而不是3，因为 sin θ 不能超过1。这是考试中常见的陷阱。
For thin-film interference or the double-slit experiment, the fringe spacing Δx = λ L / s, where L is the distance from the slits to the screen and s is the slit separation. A question gave L = 1.5 m, s = 0.30 mm and λ = 550 nm, giving Δx = (550 × 10⁻⁹ × 1.5) / (0.30 × 10⁻³) = 2.75 × 10⁻³ m.
对于薄膜干涉或双缝实验，条纹间距 Δx = λ L / s，其中L为狭缝到屏幕的距离，s为狭缝间距。一道题给出 L = 1.5 m，s = 0.30 mm，λ = 550 nm，得到 Δx = (550 × 10⁻⁹ × 1.5) / (0.30 × 10⁻³) = 2.75 × 10⁻³ m。
When asked to describe the appearance of interference fringes, mention that they are equally spaced, of the same wavelength (colour), and alternate bright and dark regions. If white light is used, the central fringe is white and higher-order fringes show spectral colours.
当被要求描述干涉条纹的外观时，要提到它们是等间距的、波长（颜色）相同，并且亮暗交替。如果使用白光，中央条纹为白色，高级别条纹呈现光谱色彩。
Refraction questions often require Snell’s law: n₁ sin θ₁ = n₂ sin θ₂. For total internal reflection, the critical angle satisfies sin θ_c = n₂ / n₁ when n₁ > n₂. In optical fibres, the acceptance angle is related to the numerical aperture; a typical calculation involves applying sin θ_c to find the maximum angle of incidence.
折射问题通常需要斯涅尔定律：n₁ sin θ₁ = n₂ sin θ₂。对于全反射，临界角满足 sin θ_c = n₂ / n₁，其中 n₁ > n₂。在光纤中，接收角与数值孔径有关；典型的计算包括应用 sin θ_c 来找到最大入射角。
6. Electricity – Circuits and Internal Resistance | 电学——电路与内阻
A classic internal resistance question from June 2018 uses a battery of e.m.f. 1.5 V and internal resistance 0.50 Ω connected to an external resistor of 2.50 Ω. The total resistance is 3.00 Ω, so the current is I = E / (R + r) = 1.5 / 3.0 = 0.50 A.
2018年6月的一道经典内阻问题使用一个电动势为1.5 V、内阻为0.50 Ω的电池，连接到2.50 Ω的外部电阻。总电阻为3.00 Ω，因此电流为 I = E / (R + r) = 1.5 / 3.0 = 0.50 A。
The terminal potential difference is V = E – I r = 1.5 – (0.50 × 0.50) = 1.25 V. This is the voltage measured across the external resistor. If a voltmeter of finite resistance is placed across the terminals, it draws current and lowers the terminal pd; ideal voltmeters have infinite resistance.
端电压为 V = E – I r = 1.5 – (0.50 × 0.50) = 1.25 V。这是测量外部电阻两端的电压。如果将一个有限内阻的电压表并联在电池两端，它会分走电流并降低端电压；理想电压表内阻为无穷大。
Another question involved a potential divider. Two resistors, R₁ = 2.0 kΩ and R₂ = 3.0 kΩ, are connected in series across a 5.0 V supply. The output voltage across R₂ is V_out = V_supply × R₂ / (R₁ + R₂) = 5.0 × 3.0 / 5.0 = 3.0 V.
另一道题涉及分压器。两个电阻 R₁ = 2.0 kΩ 和 R₂ = 3.0 kΩ 串联在一个5.0 V电源两端。R₂两端的输出电压为 V_out = V_supply × R₂ / (R₁ + R₂) = 5.0 × 3.0 / 5.0 = 3.0 V。
Kirchhoff’s laws were tested in an extended-response question. The first law (junction rule) states that the sum of currents entering a junction equals the sum leaving it. The second law (loop rule) states that the sum of e.m.f.s around a closed loop equals the sum of potential differences. When applying these to a circuit with two loops, define a sign convention for current directions and be consistent.
基尔霍夫定律在一道扩展题中被考查。第一定律（节点定律）指出，进入节点的电流之和等于离开节点的电流之和。第二定律（回路定律）指出，绕闭合回路的电动势之和等于电势差之和。在应用于有两个回路的电路时，定义电流方向的符号惯例并保持一致。
For example, in a circuit with two batteries, 12 V and 6 V, and resistors 4 Ω and 2 Ω in series, the net e.m.f. is 6 V if the batteries oppose each other. The current is 6 / 6 = 1.0 A. The direction of current is determined by the larger e.m.f. If you set up simultaneous equations, double-check the polarity of each resistor’s voltage drop.
例如，在包含两个电池（12 V和6 V）以及串联的4 Ω和2 Ω电阻的电路中，如果电池反向，净电动势为6 V。电流为 6 / 6 = 1.0 A。电流方向由较大的电动势决定。如果建立联立方程，请仔细检查每个电阻电压降的极性。
In practical questions, you may need to sketch a circuit to measure the internal resistance of a cell. The typical method is to vary the external resistance using a rheostat, record the terminal pd V and current I, and plot V against I. The y-intercept is the e.m.f. and the negative gradient is the internal resistance.
在实验题中，你可能需要绘制测量电池内阻的电路图。典型方法是使用变阻器改变外部电阻，记录端电压V和电流I，并绘制V随I变化的图像。y截距为电动势，斜率的绝对值为内阻。
7. Circular and Simple Harmonic Motion | 圆周运动与简谐运动
Uniform circular motion questions often calculate centripetal force. For an object of mass 0.80 kg moving at 2.0 m/s in a circle of radius 0.50 m, the centripetal acceleration is a = v² / r = 4.0 / 0.50 = 8.0 m/s². The centripetal force is F = ma = 0.80 × 8.0 = 6.4 N.
匀速圆周运动的问题通常计算向心力。一个质量为0.80 kg的物体以2.0 m/s的速度在半径为0.50 m的圆周上运动，向心加速度为 a = v² / r = 4.0 / 0.50 = 8.0 m/s²。向心力为 F = ma = 0.80 × 8.0 = 6.4 N。
In the context of a banked track or vertical circle, the resultant of the normal reaction and weight provides the centripetal force. A common error is to take the normal reaction alone as the centripetal force; instead, the component of the normal force toward the centre is the relevant one.
在斜面轨道或竖直圆周的语境中，法向反作用力与重力的合力提供向心力。一个常见错误是将法向反作用力单独当作向心力；实际上，是法向力指向中心的分量起作用。
For simple harmonic motion (SHM), the defining equation is a = -ω² x
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