AS Physics Unit 2 (June 2022) — Question Paper Walkthrough | AS物理Unit 2（2022年6月）真题解析

📚 AS Physics Unit 2 (June 2022) — Question Paper Walkthrough | AS物理Unit 2（2022年6月）真题解析

This article provides a structured walkthrough of the AQA AS Physics Unit 2 (June 2022) examination paper, focusing on the core topics of mechanics and materials. We analyse question types, common examiner expectations, and the key equations you must command to secure top marks.

本文系统解析 AQA AS 物理 Unit 2（2022年6月）试卷，聚焦力学与材料两大核心板块。我们分析题型特征、考官评分期望以及你必须熟练掌握的关键方程，帮助你在考试中稳拿高分。

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1. Exam Format and Weighting | 试卷结构与分值分布

The AQA AS Physics Unit 2 paper is a written examination lasting 1 hour 30 minutes, carrying 70 marks and contributing 50% of the AS qualification. The paper contains a mixture of short-answer questions, calculations, and extended response items, with Section A focusing on multiple-choice and structured questions, while Section B tests deeper problem-solving skills.

AQA AS 物理 Unit 2 试卷为笔试，时长 1 小时 30 分钟，满分 70 分，占 AS 总成绩的 50%。试卷题型包括简答题、计算题与拓展回答题；其中 A 部分以选择题和结构性问题为主，B 部分侧重考查深层问题解决能力。

• Total marks: 70 | 总分：70 分

• Duration: 1 hour 30 minutes | 时长：1 小时 30 分钟

• Materials allowed: calculator, ruler, protractor | 允许使用：计算器、直尺、量角器

One crucial strategy is time management: with 70 marks in 90 minutes, you have just over one minute per mark. Calculation-heavy questions on moments or Young modulus often demand more time, so complete shorter definition questions first.

一个关键策略是时间管理：70 分对应 90 分钟，每题每分钟略多于一分钟。涉及力矩或杨氏模量的计算题往往耗时更多，因此应先完成较短的定义为题。

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2. Defining the Key Quantities | 核心物理量定义辨析

In June 2022, the first questions tested precise definitions of scalar and vector quantities. A scalar has magnitude only, whereas a vector has both magnitude and direction. You must also know examples: speed, mass, and energy are scalars; velocity, force, and displacement are vectors.

2022年6月试卷的开篇题目考查了标量与矢量的精确定义。标量只有大小，矢量同时具有大小和方向。你必须熟记示例：速率、质量、能量为标量；速度、力、位移为矢量。

A second common definition is the moment of a force: the product of the force and the perpendicular distance from the line of action of the force to the pivot. The unit is newton metre (N m).

另一个常见定义是力的力矩：力与力的作用线到转轴垂直距离的乘积，单位为牛顿米（N·m）。

Moment = F × d (perpendicular distance) | 力矩 = 力 × 垂直距离

Examiners penalise students who use the direct distance instead of the perpendicular distance. Always draw the perpendicular from the pivot to the line of action.

考官会惩罚使用直线距离而非垂直距离的考生。务必从转轴画垂线至力的作用线。

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3. Motion Graphs and SUVAT Equations | 运动图像与SUVAT方程

Unit 2 frequently presents velocity-time or displacement-time graphs. In the June 2022 paper, a velocity-time graph for a vehicle undergoing uniform acceleration required you to calculate acceleration from the gradient and displacement from the area under the graph.

Unit 2 常考速度-时间图或位移-时间图。2022年6月试卷中，一辆汽车做匀加速运动的 v-t 图要求你通过斜率计算加速度，通过图线与横轴围成的面积计算位移。

a = (v − u) / t | s = ut + ½at² | v² = u² + 2as

You should also recall that the area under an acceleration-time graph gives the change in velocity. A common exam mistake is confusing the gradient of a displacement-time graph (velocity) with the gradient of a velocity-time graph (acceleration).

你还应记得：加速度-时间图下方的面积等于速度变化量。常见易错点是混淆位移-时间图的斜率（速度）与速度-时间图的斜率（加速度）。

For the June 2022 question, a car accelerated from rest at 2.0 m s⁻² for 6.0 s, then travelled at constant velocity for 4.0 s, then braked to rest in 3.0 s. You would compute the total distance by splitting the graph into a triangle, a rectangle, and a triangle:

对于2022年6月那道题，汽车从静止以 2.0 m s⁻² 加速 6.0 s，再匀速 4.0 s，最后在 3.0 s 内刹停。你应该将图像分割为一个三角形、一个矩形和另一个三角形来计算总距离：

• Phase 1 distance: ½ × 6 × 12 = 36 m | 第一阶段距离：½ × 6 × 12 = 36 m

• Phase 2 distance: 12 × 4 = 48 m | 第二阶段距离：12 × 4 = 48 m

• Phase 3 distance: ½ × 3 × 12 = 18 m | 第三阶段距离：½ × 3 × 12 = 18 m

• Total distance: 36 + 48 + 18 = 102 m | 总距离：36 + 48 + 18 = 102 m

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4. Newton’s Laws of Motion | 牛顿运动定律

The June 2022 paper included a structured question on Newton’s three laws. Newton’s first law states that an object remains at rest or in uniform motion unless acted on by a resultant force. Newton’s second law gives F = ma (resultant force equals mass times acceleration). Newton’s third law states that if object A exerts a force on object B, then object B exerts an equal and opposite force on object A.

2022年6月试卷包含一道关于牛顿三大定律的结构化问题。牛顿第一定律：物体在不受合外力时保持静止或匀速直线运动。牛顿第二定律：F = ma（合外力等于质量乘以加速度）。牛顿第三定律：若物体 A 对物体 B 施力，则物体 B 对物体 A 施加大小相等、方向相反的力。

F = ma | P = mv | F = Δp / Δt

A typical calculation: a 1200 kg car experiences a driving force of 3600 N and a resistive force of 900 N. The resultant force is 3600 − 900 = 2700 N, so the acceleration is 2700 / 1200 = 2.25 m s⁻².

一道典型计算：质量为 1200 kg 的汽车受到 3600 N 的驱动力和 900 N 的阻力。合外力为 3600 − 900 = 2700 N，因此加速度为 2700 / 1200 = 2.25 m s⁻²。

For Newton’s third law, examiners often test the distinction between a balanced pair (acting on the same object) and a Newton’s third law pair (acting on different objects). In the June paper, you had to identify the pair of forces representing action and reaction, for example the weight of a book on a table and the contact force of the table on the book.

关于牛顿第三定律，考官常考平衡力对（作用在同一物体上）与作用力反作用力对（作用在不同物体上）的区分。在6月试卷中，你需要识别作用力与反作用力对，例如书的重力与桌面对书的支持力。

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5. Work, Energy, and Power | 功、能与功率

This section is a high-yield topic in Unit 2. Work done is defined as the product of force and displacement in the direction of the force. Energy is the capacity to do work. Power is the rate of doing work or transferring energy.

该板块是 Unit 2 的高频考点。功的定义是力与力的方向上位移的乘积。能是做功的能力。功率是做功或能量转移的速率。

W = Fs cos θ | KE = ½mv² | GPE = mgh | P = W / t = Fv

In the June 2022 paper, a question asked you to calculate the power of a motor that lifts a 40 kg mass through a height of 15 m in 20 s. The work done against gravity is mgh = 40 × 9.81 × 15 = 5886 J, and the power output is 5886 / 20 ≈ 294 W.

在2022年6月试卷中，一道题要求计算电动机的功率：将 40 kg 的重物提升 15 m，用时 20 s。克服重力做的功为 mgh = 40 × 9.81 × 15 = 5886 J，输出功率为 5886 / 20 ≈ 294 W。

Examiner feedback from June 2022 noted that candidates frequently lost marks by omitting the angle θ in W = Fs cos θ. When a force is applied at an angle to the displacement, you must resolve the force into the component parallel to the displacement. For example, if the force is 50 N at 30° to the horizontal and the displacement is 10 m, then W = 50 × 10 × cos 30° ≈ 433 J.

2022年6月的考官反馈指出，考生常在 W = Fs cos θ 中遗漏角度 θ。当力与位移方向成一定角度时，必须将力分解为平行于位移的分量。例如，若力为 50 N，与水平方向成 30°，位移为 10 m，则 W = 50 × 10 × cos 30° ≈ 433 J。

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6. Momentum and Collisions | 动量与碰撞

Momentum is the product of mass and velocity, p = mv. The principle of conservation of momentum states that in a closed system, the total momentum before an interaction equals the total momentum after the interaction. This is essential for solving collision problems.

动量为质量与速度的乘积，p = mv。动量守恒定律表明：在封闭系统中，相互作用前总动量等于相互作用后总动量。这对解决碰撞问题至关重要。

p = mv | m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

In the June 2022 paper, a trolley A of mass 0.8 kg moving at 3.0 m s⁻¹ collides head-on with a stationary trolley B of mass 1.2 kg. After the collision they stick together. Calculate the common velocity:

在2022年6月试卷中，质量 0.8 kg、速度 3.0 m s⁻¹ 的小车 A 与静止的 1.2 kg 小车 B 发生正碰，碰后粘在一起。计算共同速度：

0.8 × 3.0 + 0 = (0.8 + 1.2) × v ⇒ v = 2.4 / 2.0 = 1.2 m s⁻¹

This type of collision is perfectly inelastic because kinetic energy is not conserved. Before the collision, KE = ½ × 0.8 × 3.0² = 3.6 J. After, KE = ½ × 2.0 × 1.2² = 1.44 J. The missing 2.16 J is transferred to internal (thermal) energy, sound, and deformation of the trolleys.

这种碰撞属于完全非弹性碰撞，因为动能不守恒。碰撞前动能 KE = ½ × 0.8 × 3.0² = 3.6 J。碰撞后 KE = ½ × 2.0 × 1.2² = 1.44 J。损失的 2.16 J 转化为内能（热能）、声能以及小车的形变。

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7. Moments, Couples, and Equilibrium | 力矩、力偶与平衡

The principle of moments states that for a body in equilibrium, the sum of the clockwise moments about any pivot equals the sum of the anticlockwise moments. In the June 2022 paper, a uniform beam of weight 60 N and length 4.0 m was pivoted at one end. A force F applied at the other end held it horizontal.

力矩原理：处于平衡状态的物体，绕任意转轴的顺时针力矩之和等于逆时针力矩之和。在2022年6月试卷中，一根重 60 N、长 4.0 m 的均匀梁在一端铰接，另一端施加力 F 使其保持水平。

Σ clockwise moments = Σ anticlockwise moments | 顺时针力矩之和 = 逆时针力矩之和

For a uniform beam, the weight acts at the centre, 2.0 m from the pivot. Therefore F × 4.0 = 60 × 2.0, giving F = 30 N. A couple is a pair of equal and opposite parallel forces whose lines of action do not coincide; the moment of a couple is force × perpendicular distance between the forces.

对于均匀梁，重力作用于重心，距转轴 2.0 m。因此 F × 4.0 = 60 × 2.0，解得 F = 30 N。力偶是一对大小相等、方向相反且不共线的平行力；力偶矩等于力乘以两力之间的垂直距离。

Examiners often require you to state the condition for equilibrium additionally as the resultant force being zero. This means both translational equilibrium (ΣF = 0) and rotational equilibrium (Σ moment = 0) must hold simultaneously.

考官常要求你额外说明平衡条件为合外力为零。这意味着平动平衡（ΣF = 0）和转动平衡（合力矩 = 0）必须同时成立。

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8. Stress, Strain, and the Young Modulus | 应力、应变与杨氏模量

This is the core materials topic in Unit 2. Stress is defined as the force per unit cross-sectional area, measured in pascals (Pa) or N m⁻². Strain is the extension per unit original length, which has no units.

这是 Unit 2 中材料部分的核心内容。应力定义为单位横截面积所受的力，单位为帕斯卡（Pa）或 N m⁻²。应变是单位原长的伸长量，无单位。

σ = F / A | ε = ΔL / L | E = σ / ε = FL / (AΔL)

In the June 2022 question, a steel wire of length 2.50 m and diameter 0.50 mm was stretched by a force of 300 N. The extension was measured as 1.9 mm. Calculate the Young modulus:

在2022年6月的题目中，一根长 2.50 m、直径 0.50 mm 的钢丝受到 300 N 的拉力，测得伸长为 1.9 mm。计算杨氏模量：

A = π(d/2)² = π(0.25 × 10⁻³)² = 1.96 × 10⁻⁷ m²

E = (300 × 2.50) / (1.96 × 10⁻⁷ × 1.9 × 10⁻³) = 2.0 × 10¹¹ Pa

This value is characteristic of steel. A common error in this calculation is failing to convert the diameter to metres before computing the area, or using the diameter instead of the radius. Always check your units.

该值符合钢的特征。此计算中的常见错误是：计算横截面积前未将直径换算为米，或误用直径而非半径。务必检查单位。

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9. Force-Extension Graphs and Elastic Strain Energy | 力-伸长图像与弹性应变能

The June 2022 paper included a force-extension graph for a spring. The gradient of the linear region gives the spring constant k. The area under the force-extension graph represents the elastic strain energy stored in the wire or spring.

2022年6月试卷包含弹簧的力-伸长图像。线性区域的斜率给出劲度系数 k。力-伸长图像下方的面积表示储存在金属丝或弹簧中的弹性应变能。

Eₑ = ½ F ΔL = ½ k ΔL²

For example, if a spring extends by 40 mm under a force of 12 N, the stored energy is ½ × 12 × 0.040 = 0.24 J. Note that at the elastic limit the graph no longer obeys Hooke’s law, and the energy calculation using the triangular area is only valid up to that limit.

例如，若弹簧在 12 N 力作用下伸长 40 mm，储存的弹性势能为 ½ × 12 × 0.040 = 0.24 J。注意在弹性极限处图像不再遵循胡克定律，三角形面积法计算能量仅在弹性极限内有效。

Examiner reports for June 2022 highlighted that many students drew a straight line through the plastic region or incorrectly labelled the elastic limit as the breaking point. Be sure to label four key points: the limit of proportionality, the elastic limit, the yield point, and the breaking point.

2022年6月的考官报告强调，许多学生在塑性区仍画直线，或错误地将弹性极限标为断裂点。请务必标注四个关键点：比例极限、弹性极限、屈服点与断裂点。

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10. Springs in Series and Parallel | 弹簧的串联与并联

The June 2022 paper extended the spring question by asking you to compare the effective spring constant when two identical springs are arranged in series versus in parallel. This is a classic AS-level requirement.

2022年6月试卷延伸了弹簧问题，要求比较两根相同弹簧串联与并联时的等效劲度系数，这是 AS 阶段的经典要求。

Arrangement | 连接方式	Effective k | 等效劲度系数
Series | 串联	1/k_total = 1/k + 1/k = 2/k, so k_total = k/2
Parallel | 并联	k_total = k + k = 2k

For two identical springs each of spring constant 50 N m⁻¹, the series arrangement gives 25 N m⁻¹ while the parallel arrangement gives 100 N m⁻¹. Intuitively, in parallel the springs share the load but extend by the same amount, requiring twice the force for the same extension.

对于两根劲度系数均为 50 N m⁻¹ 的相同弹簧，串联等效为 25 N m⁻¹，并联等效为 100 N m⁻¹。直觉上，并联时两根弹簧分担负载但伸长量相同，需要两倍的力才能产生相同的伸长量。

When the springs are in series, each spring experiences the full load but extends only half as much as the single spring under the same force, so the total extension is doubled, and the effective stiffness halves.

串联时，每根弹簧承受全部负载，但在相同拉力下伸长量仅为单根弹簧的一半，因此总伸长量加倍，等效刚度减半。

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11. Experimental Techniques and Errors | 实验方法与误差分析

Unit 2 always contains at least one question on experimental procedure. For the Young modulus experiment, you should know how to measure the diameter of a wire accurately using a micrometer screw gauge at several points along the wire, and why the original length should be measured with a metre ruler while the wire is under a small initial tension to remove any kinks.

Unit 2 至少包含一道实验方法题。对于杨氏模量实验，你应当知道如何用千分尺沿金属丝多个位置精确测量直径，以及为何应在金属丝施加轻微初张力后用米尺测量原始长度，以消除任何弯折。

Typical sources of error include: the wire not being exactly uniform, parallax error when reading the extension scale, and thermal expansion of the wire. To improve accuracy, use a travelling microscope to measure the extension and average multiple diameter measurements.

典型误差来源包括：金属丝并非完全均匀、读取伸长量刻度时产生视差误差、金属丝的热膨胀。为提高准确度，应使用移测显微镜测量伸长量，并对多个直径测量值取平均。

The June 2022 paper asked why a small initial load is applied to the wire before taking measurements. The answer: it ensures the wire is straight and taut so the measured extension corresponds to the additional load only, and any slack is eliminated before the experiment begins.

2022年6月试卷询问为何在实验测量前要给金属丝施加一个小初始载荷。答案是：确保金属丝绷直拉紧，使测得的伸长量仅对应额外载荷，消除初始松弛。

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12. Exam Strategy and Common Pitfalls | 考试策略与常见陷阱

Based on the June 2022 mark scheme, the most frequently lost marks were due to: missing units, incorrect significant figures, and not showing working in multi-step calculations. Follow these guidelines to maximise your score:

根据2022年6月的评分标准，最常见的失分原因包括：遗漏单位、有效数字错误以及多步计算中未展示过程。遵循以下建议可以最大化得分：

• Always show your substitution step clearly before giving the final answer | 在给出最终答案前，清晰展示数值代入步骤

• Quote answers to two or three significant figures, consistent with the data provided | 答案使用 2 至 3 位有效数字，与题干数据保持一致

• Draw a diagram for equilibrium and moment problems, marking distances from the pivot | 对平衡和力矩问题画图，标出到转轴的距离

• Check whether the angle in W = Fs cos θ is measured between the force and the displacement | 检查 W = Fs cos θ 中角度 θ 是否为力与位移方向之间的夹角

• Convert all units to base SI units (mm to m, km h⁻¹ to m s⁻¹) before substituting into equations | 代入方程前将所有单位转换为国际单位制基本单位（mm 转 m，km h⁻¹ 转 m s⁻¹）

For the June 2022 specific paper, the extended writing question required you to compare the behaviour of a ductile material and a brittle material on the same stress-strain graph. A ductile material (such as copper) shows plastic deformation before fracture with a large strain to failure, while a brittle material (such as glass) fractures at the elastic limit with little or no plastic deformation.

针对2022年6月的具体试卷，拓展写作题要求你在同一应力-应变图中比较延性材料与脆性材料的力学行为。延性材料（如铜）在断裂前经历明显塑性变形，破坏应变大；脆性材料（如玻璃）在弹性极限处即脆断，几乎没有塑性变形。

You must also describe the key features of the graph: the linear Hooke’s law region for both materials, the steeper gradient for the brittle material (higher Young modulus), and the absence of a yield point for the brittle material.

你还必须描述图像的关键特征：两种材料都存在线性胡克定律区，脆性材料斜率更大（杨氏模量更高），且脆性材料无屈服点。

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