📚 Year 7 OCR Physics: Unit Test Mock Paper Analysis | 单元测试模拟卷解析
This article walks you through a typical Year 7 OCR Physics unit test with full explanations for every question. You will see how concepts such as energy stores, forces, circuits, waves and particle theory are assessed, and learn how to avoid common mistakes. Use it as a revision guide and a confidence booster before your real test.
本文带你全面解析一套典型的Year 7 OCR物理单元测试模拟卷，每题都有详细讲解。你将看到能量储存、力、电路、波和粒子理论等概念是如何考查的，并学会避开常见错误。它既是复习指南，也是考前提升信心的利器。
1. Mock Paper Overview | 模拟卷概览
The mock paper is designed to reflect OCR's Year 7 syllabus for physics. It contains 10 questions: 4 multiple-choice, 3 short-answer calculations, 1 circuit diagram task, 1 graph interpretation task and 1 extended explanation using particle theory. Topics covered include energy stores and transfers, forces, speed, electrical circuits, light, sound, and the particle model of matter. The total marks add up to 30, and you should aim to complete it in 40 minutes.
本模拟卷依据OCR七年级物理大纲设计，包含10道题目：4道选择题、3道简答计算题、1道电路作图题、1道图表解读题和1道运用粒子理论的扩展解释题。涵盖能量储存与转移、力、速度、电路、光、声音和物质的粒子模型。总分30分，建议在40分钟内完成。
Working through this analysis will help you identify which topics need extra practice. Each section below takes a cluster of questions, presents the question, shows the correct approach, and explains the underlying physics.
通过本次解析，你能发现哪些主题还需要额外练习。下面每个小节集中一类题目，先呈现题目，再展示正确解法，并解释背后的物理原理。
2. Energy Stores and Transfers (Q1–Q2) | 能量储存与转移（选择题1–2）
Question 1: Which one of the following is an example of gravitational potential energy? A. A moving car B. A stretched spring C. A book on a high shelf D. A glowing light bulb.
第1题：下列哪一项是重力势能的例子？A. 行驶的汽车 B. 拉伸的弹簧 C. 放在高处书架上的书 D. 发光的灯泡。
Correct answer: C. Gravitational potential energy is energy stored in an object because of its height above the ground. A moving car has kinetic energy, a stretched spring stores elastic potential energy, and a glowing bulb emits light and heat. Only the book on a high shelf has gravitational potential energy.
正确答案：C。重力势能是物体因为离地高度而储存的能量。行驶的汽车具有动能，拉伸的弹簧储存弹性势能，发光的灯泡释放光和热。只有高处书架上的书具有重力势能。
Question 2: A lamp transfers 30 J of energy from the mains. 15 J is transferred as light, and the rest is transferred as heat. How much energy is wasted as heat? Show your calculation.
第2题：一盏灯从电源中转移了30焦耳的能量。其中15焦耳转化为光，剩余的能量转化为热。有多少能量以热的形式浪费？写出计算过程。
y in = 30 J, useful light energy = 15 J. By the conservation of energy, wasted heat = 30 J – that energy is never lost, only transferred and sometimes dissipated in less useful forms.
解答：输入总能量=30 J，有用的光能=15 J。根据能量守恒，浪费的热能=30 J – 15 J = 15 J。这类题考查你是否理解能量不会消失，只是被转移，有时以不太有用的形式耗散。
d energy. Always write down the conservation equation: Total in = useful out + wasted out.
常见错误：忘记做减法，或混淆有用能量和浪费能量。务必写出守恒方程：输入总能量=有用输出+浪费输出。
3. Forces and Motion (Q3–Q4) | 力与运动（选择题3和计算题4）
Question 3: A student pushes a box with a force of 10 N to the right. Friction acts on the box with a force of 4 N to the left. What is the size and direction of the resultant force? A. 14 N to the right B. 14 N to the left C. 6 N to the right D. 6 N to the left.
第3题：一个学生用10 N的力向右推一个箱子，摩擦力以4 N的力向左作用在箱子上。合力的大小和方向是什么？A. 14 N向右 B. 14 N向左 C. 6 N向右 D. 6 N向左。
wer: C. The forces act in opposite directions along a straight line, so you subtract: 10 N – e right, the resultant force is 6 N to the right. Balanced forces would give zero resultant; here they are unbalanced, so the box will accelerate to the right.
正确答案：C。两个力沿直线方向相反，因此要相减：10 N – 4 N = 6 N。因为较大的力朝右，合力为6 N方向向右。若二力平衡，合力为零；这里不平衡，所以箱子会向右加速。
onds. Calculate the average speed. Use the formula speed = distance ÷ time. Give the unit.
第4题：一个骑自行车的人用25秒行驶了150米。计算平均速度。使用公式速度=距离÷时间，并写出单位。
es per second). Always show the formula, substitute the numbers and include the correct unit. If distance is in metres and time in seconds, speed is in m/s.
答案：速度=150 m ÷ 25 s = 6 m/s（米每秒）。务必先写公式，代入数字，并带上正确的单位。如果距离是米、时间是秒，速度单位就是m/s。
= v × t) or time (t = d ÷ v). Practise using the triangle method if it helps you remember.
技巧：你可以变形公式求距离（d = v × t）或时间（t = d ÷ v）。若有助于记忆，可以用“公式三角形”辅助练习。
4. Electrical Circuits (Q5–Q6) | 电路（选择题5与作图题6）
Question 5: In a series circuit containing two identical bulbs and one cell, one bulb is unscrewed and removed from its holder. What happens to the other bulb? A. It stays on at the same brightness B. It stays on but becomes brighter C. It goes out D. It flickers.
第5题：在一个包含两个相同灯泡和一个电池的串联电路中，其中一个灯泡被拧下并从灯座中取出。另一个灯泡会怎样？A. 保持相同亮度发光 B. 保持发光但变亮 C. 熄灭 D. 闪烁。
Correct answer: C. A series circuit provides only one loop for current. If a gap appears (the removed bulb), the circuit is incomplete and current stops everywhere, so the other bulb goes out. This reflects that current is the same in all parts of a series circuit, and a break stops all current.
正确答案：C。串联电路只有一个电流回路。如果出现断开（取下的灯泡），电路不再完整，各处电流停止，所以另一个灯泡熄灭。这反映了串联电路中各处电流相等，一旦断开，所有电流消失。
Question 6: Draw a circuit diagram using standard symbols for a circuit that contains a single cell, a switch (open in the diagram) and one lamp. Then explain why the lamp is off in your diagram.
第6题：用标准符号画出一个包含一个电池、一个开关（在图中为断开状态）和一个灯泡的电路图。然后解释图中灯泡为何不亮。
Diagram guidance: Use the correct symbol for a cell (long line positive, short line negative), a switch with a gap, and a lamp as a circle with a cross or a bulb symbol. The lamp is off because the open switch creates a break in the circuit, so there is no complete path for current to flow.
作图指导：使用正确的电池符号（长线为正极、短线为负极），断开状态的开关（有空隙），灯泡可用一个圆圈内含叉号或专用灯泡符号。灯泡不亮是因为断开的开关导致电路中存在断路，电流没有完整的通路。
Many students forget that the switch must be drawn open to explain the lamp being off. Practise drawing common circuit symbols and labelling them clearly.
很多学生忘记为了解释灯泡不亮，开关必须画成断开状态。要练习绘制常见电路符号并进行清晰的标注。
5. Waves and Light (Q7) | 波与光（简答题7）
Question 7: A ray of light strikes a plane mirror. State the law of reflection. Then, if the angle between the incident ray and the mirror is 35°, calculate the angle of reflection.
第7题：一束光线射向平面镜。陈述反射定律。然后，如果入射光线与镜面的夹角为35°，计算反射角。
Law of reflection: The angle of incidence equals the angle of reflection. This is always measured between the ray and the normal – a dotted line drawn perpendicular to the mirror at the point of incidence.
反射定律：入射角等于反射角。角度总是在光线与法线之间测量——法线是在入射点处垂直于镜面的虚线。
the incident ray and the mirror is 35°, so the angle between the ray and the normal is 90° – dence is 55°, and by the law, the angle of reflection is also 55°. Draw a clear diagram with the normal and label all angles.
计算：入射光线与镜面的夹角为35°，因此光线与法线的夹角为90° – 35° = 55°。所以入射角是55°，根据定律，反射角也是55°。画一个清晰的图，标出法线和所有角度。
Watch out: When a question gives the angle to the mirror surface, always convert to the angle to the normal before applying the law.
注意：如果题目给出的是与镜面的夹角，在使用定律前一定要先转换成与法线的夹角。
6. Sound and Vibrations (Q8) | 声音与振动（填空题8）
__________ of the vibration, and the loudness depends on the __________ of the vibration.
第8题：补全句子：声音是由__________产生的。音调取决于振动的__________，响度取决于振动的__________。
Answers: Sound is produced by vibrations. Pitch depends on the frequency (how many waves per second). Loudness depends on the amplitude (the size or energy of the vibration). Higher frequency gives a higher pitch; larger amplitude makes a louder sound.
答案：声音由振动产生。音调取决于频率（每秒振动次数），响度取决于振幅（振动的大小或能量）。频率越高，音调越高；振幅越大，声音越响。
In an exam, you might be asked to relate these to a waveform. A tall wave has high amplitude; a closely spaced wave has high frequency. Use the correct terms and avoid mixing up pitch and loudness.
考试中可能会让你在波形图上辨别这些。高的波形振幅大；密集的波形频率高。要使用正确的术语，不要混淆音调和响度。
7. Particle Model of Matter (Q9–Q10) | 物质的粒子模型（简答题9–10）
Question 9: Use the particle model to explain why a metal spoon left in a hot cup of tea gets warm all the way to the handle.
第9题：用粒子模型解释为什么放在热茶里的金属勺连勺柄都会变热。
Answer: The particles in the hot tea have lots of kinetic energy and vibrate vigorously. They bump into the particles in the metal spoon, passing on energy. This is conduction. The energy passes from particle to particle along the spoon, so eventually even the handle’s particles gain energy and the whole spoon feels warm.
答案：热茶中的粒子动能很大，振动剧烈。它们撞击金属勺中的粒子，把能量传递过去。这是热传导。能量沿着勺子从一个粒子传递到下一个，最终连勺柄的粒子也获得能量，整个勺子都变热了。
Question 10: Describe the arrangement and movement of particles in a solid and in a gas. Include a simple labelled sketch in your answer to show the differences.
第10题：描述固体和气体中粒子的排列与运动方式。在答案中包含一个简单的标注示意图，展示两者的差异。
Model answer: In a solid, particles are packed closely in a regular pattern. They vibrate about fixed positions but cannot move past each other. This gives solids a fixed shape and volume. In a gas, particles are far apart with no regular arrangement. They move quickly and randomly in all directions, filling any container. A sketch should show particles in a solid in an ordered grid, with small vibration arrows, and in a gas as widely spaced dots with random motion arrows. Use labels like ‘regular’, ‘close’, ‘vibrate’ for solid; ‘random’, ‘far apart’, ‘move freely’ for gas.
标准答案：固体中粒子紧密排列成规则的图案。它们在固定位置附近振动，但不能相互滑动，因此固体有固定的形状和体积。气体中粒子相距很远，没有规则的排列。它们快速而随机地向四面八方运动，充满整个容器。示意图中，固体粒子应画成有序网格状并附上小振动箭头；气体粒子画成离散的圆点并附上随机运动箭头。标签可以写“规则”、“紧密”、“振动”对应固体；“随机”、“间隔大”、“自由运动”对应气体。
Particle diagrams are quick wins: always include them to boost your mark if the question asks for a comparison of states.
粒子示意图能轻易得分：只要是要求比较物态的题目，别忘了画图并标注。
8. Graph Interpretation Skills (Distance–Time Graph) | 图表解读技能（距离–时间图）
Question from the mock: A distance–time graph shows three sections: 0–5 s line sloping up steadily, 5–10 s horizontal line, 10–15 s line sloping downwards back to the start. Describe the motion in each section and calculate the speed for the first section.
模拟卷题目：一张距离–时间图显示三个部分：0–5秒线段平稳上升；5–10秒为水平线；10–15秒线段向下倾斜回到起点。描述每段的运动状态，并计算第一段的速度。
Answer: 0– tance ÷ time. Suppose distance rises from 0 to 200 m, then speed = 200 m ÷ 5 s = 40 m/s. 5–10 s: The line is horizontal, so distance does not change – the object is stationary. 10– owards the start at constant speed. The gradient is negative. This shows a return journey.
答案：0–5秒：物体从起点出发，以恒定速度远离。速度=距离变化÷时间。假设距离从0上升到200米，则速度=200 m ÷ 5 s = 40 m/s。5–10秒：线段水平，距离不变——物体静止。10–15秒：线段向下倾斜，表示物体以恒定速度返回起点，梯度为负，说明这是一段返程运动。
Key skill: The steeper the line, the greater the speed. A flat section means zero speed. Always use the gradient to find speed on a distance–time graph.
核心技能：线段越陡，速度越快。水平部分意味着速度为零。在距离–时间图上，一定要用斜率求速度。
9. Common Mistakes and How to Avoid Them | 常见错误与避免方法
d in kilograms. On Earth, weight = mass × 10 N/kg, but mass does not change with location.
错误1：混淆质量和重量。重量是一种力，单位是牛顿；质量的单位是千克。在地球上，重量=质量×10 N/kg，但质量不随位置变化。
Mistake 2: Forgetting that in a series circuit, current is the same everywhere, but voltage is shared. In a parallel circuit, the p.d. across each branch is the same and currents add up. Drawing circuits incorrectly loses marks.
错误2：忘记串联电路中各处电流相等而电压分压；并联电路中各支路电压相等且干路电流等于各支路电流之和。电路图画错会丢分。
Mistake 3: Not reading the scale when interpreting graphs. Always check the axes units and intervals. On a distance–time graph, a straight line does not always mean stationary; check if it is sloping or flat.
错误3：解读图表时不看刻度。务必检查坐标轴的单位和分度值。在距离–时间图上，直线不一定表示静止，要区分是倾斜还是水平。
e r. Refraction is bending of light when it passes into a different medium, due to speed change.
错误4：混淆反射和折射。反射是光线从表面弹回，入射角等于反射角。折射是光线进入不同介质时因速度改变而弯曲。
To avoid these, practise past questions and build a glossary of key terms. When you finish a mock paper, check your answers against the mark scheme and note where you lost marks.
为避免这些错误，多练习历年真题，建立关键术语词汇表。做完模拟卷后，对照评分标准批改并记录丢分点。
10. Exam Technique Tips and Extra Practice | 考试技巧提示与额外练习
Always show your working in calculations, even if it is simple. This secures method marks. Use a ruler and pencil for diagrams, and label them clearly. Manage your time: if you are stuck, move on and come back later. Read questions twice to catch instruction words like ‘describe’, ‘explain’ or ‘calculate’.
计算题始终要写过程，即使再简单，这能保证方法分。画图用尺子和铅笔，标注清楚。合理分配时间：遇到卡住的题先跳过，之后再看。仔细读题两次，抓住“描述”、“解释”或“计算”等指令词。
Here are three extra practice questions:
1. A bus moves 300 m in 20 s. What is its speed? (15 m/s)
2. In a parallel circuit, one bulb blows and goes out. What happens to the other bulb? (It stays on at the same brightness.)
3. Explain why a puddle of water dries up faster on a hot day using the particle model. (Particles gain energy, move faster, escape from the liquid surface more quickly.)
额外练习三道题：1. 一辆公交车20秒行驶300米，速度是多少？（15 m/s）2. 在并联电路中，一个灯泡烧坏熄灭，另一个灯泡会怎样？（保持相同亮度发光）3. 用粒子模型解释为什么水坑在热天干得更快。（粒子获得能量，运动加快，更快地脱离液体表面。）
Try these under timed conditions and compare your reasoning with the explanations given in this article. Repetition is key to building confidence.
请在定时条件下尝试这些题目，并对照本文的分析进行比较。反复练习是建立信心的关键。
Published by TutorHao | Physics Revision Series | aleveler.com
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