📚 Year 7 AQA Physics: In-depth Analysis of Past Exam Questions | Year 7 AQA 物理：历年真题深度解析
This article provides a thorough walkthrough of typical past paper questions for AQA Year 7 Physics. By examining real exam-style problems, we uncover the key skills, common errors and marking points you need to succeed. Whether you are revising energy stores, circuit diagrams or graph interpretation, this guide will build your confidence and exam technique.
本文全面梳理 AQA 七年级物理历年真题中的典型题目。通过分析真实考试风格的题目，我们发掘关键技能、常见错误和得分要点。无论你是在复习能量储存、电路图还是图表解读，本指南都将帮助你增强信心、提升应试技巧。
1. Exam Format and Expectations | 考试形式与要求
The AQA Year 7 Physics exam typically lasts 45 minutes and carries 50 marks. It mixes multiple-choice questions, short structured answers and longer 4–6 mark responses.
AQA 七年级物理考试通常持续 45 分钟，满分 50 分。题目包含选择题、简答题以及 4–6 分的长答题。
Questions are drawn from the full Key Stage 3 content: Energy, Forces, Waves, Matter, Electricity and Space. Many items test practical skills, so you must be able to describe methods, identify variables and process results.
试题涵盖完整的 KS3 内容：能量、力、波、物质、电和空间。许多题目考查实验技能，因此你必须能描述方法、识别变量并处理结果。
Command words like 'describe', 'explain' and 'compare' signal exactly what the examiner wants. Mastering these words is as important as knowing the science.
指令词如 'describe'、'explain' 和 'compare' 准确提示了考官的要求。掌握这些词汇与掌握科学知识同样重要。
2. Energy Stores and Transfers | 能量储存与转移
A classic past paper question asks: 'A ball is dropped from a height of 2 m. Describe the energy transfers that occur before it hits the ground.' (4 marks) Many students only mention gravitational potential energy (GPE) changing to kinetic energy but forget thermal energy transfers.
一道经典真题问道："从 2 米高处释放一个球。描述它撞击地面前的能量转移。" (4 分) 许多学生只提到重力势能转换为动能，却忘记了热能转移。
The full answer must state that the ball has maximum GPE at the start. As it falls, the GPE store decreases and is transferred to the kinetic energy store of the ball. Work is done against air resistance, so the thermal energy store of the surroundings increases. This shows the principle of conservation of energy — energy is never lost, only transferred.
完整答案必须说明球开始时重力势能最大。下落时，重力势能储存减少，转移到球的动能储存。克服空气阻力做功，因此周围环境的热能储存增加。这体现了能量守恒原理 —— 能量永不消失，只会转移。
total energy before = total energy after
转移前的总能量 = 转移后的总能量
In another question, you might be shown a Sankey diagram and asked to calculate useful output energy. Always subtract wasted energy from total input energy. A common mistake is reading the width of arrows incorrectly.
在另一类题目中，你可能会看到桑基图并被要求计算有用的输出能量。总是从总输入能量中减去浪费的能量。常见错误是错误地读取箭头的宽度。
3. Electrical Circuits and Current | 电路与电流
A typical exam task: 'Use standard circuit symbols to draw a series circuit containing a battery, a closed switch, a lamp and a voltmeter connected across the lamp.' (3 marks) The voltmeter must be drawn in parallel — drawing it in series is a frequent error that loses a mark.
典型考试任务："使用标准电路符号画出一个串联电路，包含电池、闭合开关、一盏灯泡和一个并联在灯泡两端的电压表。" (3 分) 电压表必须并联绘制 —— 画成串联是常见错误，会损失 1 分。
Voltmeters measure potential difference (in volts) and have very high resistance, so they need a shunt connection. Ammeters, in contrast, are placed in series to measure current (in amps).
电压表测量电势差（单位为伏特），电阻非常高，因此需要并联连接。而安培表则串联在电路中以测量电流（单位为安培）。
Another past paper asks: 'In a series circuit, the ammeter reads 0.4 A. A second identical bulb is added in series. Explain what happens to the ammeter reading.' The current decreases because total resistance increases while the battery voltage remains constant.
另一份真题问："在一个串联电路中，电流表读数为 0.4 A。串联接入一个相同的灯泡。解释电流表读数会如何变化。" 电流会减小，因为在电池电压不变的情况下总电阻增大了。
4. Forces and Motion: Speed and Graphs | 力与运动：速度与图表
Calculation questions are a staple. For example: 'A cyclist travels 300 metres in 60 seconds. Calculate her speed.' The equation needed is:
计算题是必考内容。例如："一位自行车手在 60 秒内行驶了 300 米。计算她的速度。" 所需公式为：
speed = distance ÷ time
速度 = 路程 ÷ 时间
s include the unit m/s; omitting it often loses a mark. Check that your answer is sensible — walking pace is about 1.5 m/s, so 5 m/s is a brisk run.
代入数字：速度 = 300 m ÷ 60 s = 5 m/s。始终牢记带上单位 m/s；遗漏单位通常会丢分。检查答案是否合理 —— 步行速度约 1.5 m/s，5 m/s 大概是快跑的速度。
Distance–time graphs are equally common. A horizontal line means the object is stationary. A straight sloping line indicates constant speed, and the gradient gives the speed. If asked to 'describe the motion', use precise phrases like 'constant speed of 4 m/s for the first 10 s, then stationary for 5 s'.
路程–时间图同样常见。水平线表示物体静止。倾斜直线表示匀速运动，斜率就是速度。如果被要求"描述运动"，使用精确表述，如 "前 10 秒以 4 m/s 匀速运动，然后静止 5 秒"。
5. Sound and Light Waves | 声波与光波
An exam favourite: 'Explain why sound cannot travel through a vacuum, but light can.' This tests the particle model. Sound is produced by vibrations that travel through a medium by passing from particle to particle. In a vacuum there are no particles, so no transmission.
一道考试热门题："解释为什么声音不能在真空中传播，而光可以。"这考查粒子模型。声音由振动产生，通过介质中粒子之间的传递而传播。真空中没有粒子，因此无法传播。
Light, on the other hand, is an electromagnetic wave that can travel through empty space. This is a key distinction: sound is a mechanical wave; light is not.
另一方面，光是一种电磁波，可以在真空中传播。这是一个关键区别：声音是机械波，光则不是。
When a ray of light reflects off a plane mirror, the law of reflection states:
当光线经平面镜反射时，反射定律为：
angle of incidence = angle of reflection
入射角 = 反射角
Both angles are measured from the normal, an imaginary line at 90° to the surface. In diagrams, always draw the normal as a dashed line and mark the angles with arrows.
两个角度都是从法线量起的，法线是一条垂直于表面的假想线。在作图时，始终用法线为虚线，并用箭头标出角度。
6. The Particle Model and States | 粒子模型与状态
The particle model explains the properties of solids, liquids and gases. A past paper might ask: 'Using the particle model, explain why a gas can be compressed but a solid cannot.' The key is spacing: in a gas, particles are far apart with empty space between them, so they can be pushed closer together. In a solid, particles are already closely packed with very little space.
粒子模型解释固体、液体和气体的性质。一份真题可能会问："用粒子模型解释为什么气体可被压缩而固体不能。"关键在于间距：气体中粒子相距很远，之间存在空隙，因此可以被挤压得更紧密；固体中粒子已紧密排列，空间极小。
For a 6-mark 'describe the differences' question, structure your answer with arrangement, movement and energy. For solids: regular pattern, vibrate in fixed positions, lowest energy. Liquids: particles close but disordered, can slide past each other, more energy. Gases: widely spaced, move rapidly in all directions, highest energy.
对于 6 分"描述区别"的题目，按排列、运动与能量来组织答案。固体：规则排列，在固定位置振动，能量最低。液体：粒子紧密但无序，可相互滑动，能量较高。气体：粒子间距大，向各个方向快速运动，能量最高。
A common mistake is to say particles 'melt' or 'evaporate'. Particles themselves do not change state; the arrangement changes due to added or removed thermal energy.
常见错误是说粒子自己"熔化"或"蒸发"。粒子本身不会改变状态；变化的是由于热能的增加或移除而导致的排列方式。
7. Space Physics: Gravity and Orbits | 空间物理：重力与轨道
A typical Year 7 exam question: 'Explain why the Moon stays in orbit around the Earth.' The answer requires linking gravity to circular motion. There is a gravitational force of attraction between Earth and the Moon. This force acts towards the centre of the Earth, providing the centripetal force that changes the Moon's direction constantly. Without this force, the Moon would move off in a straight line.
典型的七年级考题："解释月球为何能保持在绕地球的轨道上。"答案需要将引力与圆周运动联系起来。地球和月球之间存在引力。这个力指向地球中心，提供了不断改变月球运动方向的向心力。没有这个力，月球将沿直线飞出。
Students often confuse mass and weight. Mass is the amount of matter (kg) and stays the same everywhere. Weight is the force due to gravity (N) and changes with location. On Earth, weight is calculated by:
学生经常混淆质量与重量。质量是物质的量（千克），在任何地方都保持不变。重量是由重力产生的力（牛顿），会随位置变化。在地球上，重量计算公式为：
weight (N) = mass (kg) × gravitational field strength (N/kg)
重量（牛）= 质量（千克）× 引力场强度（牛/千克）
The gravitational field strength on Earth is approximately 10 N/kg. So a 5 kg object weighs 50 N. On the Moon it would weigh only about 8.3 N because the gravitational field strength is smaller.
地球表面的引力场强度约为 10 N/kg。因此 5 kg 的物体重 50 N。在月球上它仅重约 8.3 N，因为引力场强度更小。
8. Practical Investigations and Variables | 实验探究与变量
A past paper scenario: 'A student investigates how the mass hung on a spring affects its extension. She adds 100 g masses one at a time and measures the new length.' The independent variable is the mass (or force) applied. The dependent variable is the extension of the spring. Controlled variables include the type of spring, the starting point (original length) and the room temperature.
真题情景："一名学生探究悬挂在弹簧上的质量如何影响弹簧的伸长量。她逐一添加 100 g 的砝码，并测量新的长度。" 自变量是所加的质量（或力）。因变量是弹簧的伸长量。控制变量包括弹簧种类、起始点（原长）和室温。
When asked to improve the investigation, candidates should suggest using a ruler with finer divisions, repeating measurements and calculating an average, and checking that the spring has not exceeded its elastic limit. The term 'elastic limit' often appears in mark schemes.
当被要求改进实验时，考生应建议使用刻度更细的尺子、重复测量并计算平均值，以及检查弹簧是否未超过弹性极限。"弹性极限"一词经常出现在评分方案中。
Always remember: 'extension' is the increase in length, not the total length. Many students lose marks by misreading the table heading.
务必记住："伸长量"是长度增加量，而非总长度。许多学生因看错表格标题而失分。
9. Interpreting Data and Graphs | 解读数据与图表
Graph questions are worth high marks. You may be given a force–extension graph or a distance–time graph and asked to describe the trend. Use the language of proportionality: 'As force increases, extension increases linearly' or 'The graph shows a straight line through the origin, indicating a directly proportional relationship.'
图表题分值较高。可能会给出力–伸长图或路程–时间图，要求描述变化趋势。使用比例关系的表述："力增大，伸长量成线性增大"或"图像为过原点的直线，表明成正比关系"。
When plotting your own graph, always label the axes with quantity and unit, e.g. 'Force (N)' and 'Extension (cm)'. Choose a sensible scale that uses more than half the grid. Draw a single clean best-fit line; never connect dots with a zigzag ruler.
自己绘制图表时，坐标轴务必标注物理量和单位，如"力 (N)"和"伸长量 (cm)"。选择合理的比例，使图形占据坐标纸一半以上区域。用直尺画一条平滑的最佳拟合线；切勿将点连成锯齿形。
In table analysis, look for numerical patterns. If the extension doubles when the force doubles, you can conclude that extension is proportional to force — this is Hooke's Law in its simplest form.
在表格分析中，寻找数字规律。如果力加倍时伸长量也加倍，可以得出结论：伸长量与力成正比 —— 这就是最简单形式的胡克定律。
10. Using Equations in Physics | 使用物理公式
Year 7 students are expected to recall and apply three key equations:
七年级学生需要记忆并应用三个关键公式：
speed = distance ÷ time
速度 = 路程 ÷ 时间
weight = mass × gravitational field strength
重量 = 质量 × 引力场强度
pressure = force ÷ area
压强 = 压力 ÷ 受力面积
In a typical pressure calculation: ' t of 40 N and rests on a face of area 0.5 m². Calculate the pressure exerted on the floor.' Pressure = 40 N ÷ 0.5 m² = 80 N/m² (or 80 Pa). Always show the substitution step clearly.
典型的压强计算题："一块木块重 40 N，以 0.5 m² 的面放置在地板上。计算它对地板施加的压强。" 压强 = 40 N ÷ 0.5 m² = 80 N/m²（或 80 Pa）。始终清晰地展示代入步骤。
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