Year 7 CCEA Physics Unit Test Mock Paper Analysis | 七年级 CCEA 物理单元测试模拟卷解析

📚 Year 7 CCEA Physics Unit Test Mock Paper Analysis | 七年级 CCEA 物理单元测试模拟卷解析

This mock paper walkthrough is a guided revision resource for Year 7 CCEA Physics. It unpacks a typical unit test, covering forces, energy, electricity, waves, and matter. Each section presents one exam-style question, followed by a clear step‑by‑step solution and helpful tips. Use it to check understanding, spot common mistakes, and build confidence before the real assessment.

这份模拟卷解析是专为七年级 CCEA 物理设计的引导式复习材料。它拆解了一份典型的单元测试，涵盖力、能量、电、波和物质等主题。每一节都呈现一道考试风格的问题，然后给出清晰的分步解答和实用提示。用它来检验理解、发现常见错误，并在实际测评前树立信心。

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1. Forces and Motion Basics | 力与运动基础

Question: A student pushes a crate across the floor. The push force is 25 N to the right. While the crate moves at a steady speed, friction acts. Draw labelled arrows to show the two horizontal forces, and explain why the arrows must be equal in length.

题目：一名学生在地板上推动一个板条箱。推力为 25 N 向右。当板条箱以恒定速度移动时，摩擦力同时作用。请画出带标记的箭头表示这两个水平力，并解释为什么两个箭头的长度必须相等。

Solution: When an object moves at a constant speed in a straight line, the forces on it are balanced. The push force points right → with an arrow labelled ’25 N →’. Friction points left ← with an arrow also labelled ’25 N ←’. The arrows must be the same length because the forces have the same magnitude, giving a net force of zero.

解析：当物体沿直线以恒定速度运动时，作用在它上面的力互相平衡。推力指向右 →，标有“25 N →”。摩擦力指向左 ←，也标有“25 N ←”。两个箭头长度必须相同，因为这两个力大小相等，合力为零。

Key Tip: In CCEA physics, always check for balanced forces when motion is steady. Remember, forces are measured in newtons (N).

关键提示：在 CCEA 物理中，只要运动是稳定的，一定要检查力是否平衡。记住，力的单位是牛顿 (N)。

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2. Measuring Speed and Interpreting Graphs | 测量速度与图表解读

Question: A cyclist travels 300 metres in 25 seconds. Calculate the average speed. Then sketch a simple distance-time graph for a journey where the cyclist rides at constant speed for 10 s, stops for 5 s, then rides again at a slower speed for another 10 s.

题目：一名自行车手在 25 秒内行驶了 300 米。计算平均速度。然后勾勒一张简单的距离‑时间图，表示该车手先以恒定速度骑行 10 秒，然后停顿 5 秒，再以较慢的速度骑行 10 秒。

Solution: Average speed is total distance divided by total time.

speed = distance ÷ time

So speed = 300 m ÷ 25 s = 12 m/s. For the graph: the line rises steadily for the first 10 s (constant speed), becomes horizontal during the 5 s stop, then rises again but with a shallower slope for the last 10 s (slower speed).

解析：平均速度等于总距离除以总时间。

速度 = 距离 ÷ 时间

因此速度 = 300 m ÷ 25 s = 12 m/s。图表方面：一开始线条在 10 秒内匀速上升（恒定速度），在停顿的 5 秒内变为水平线，然后在最后 10 秒内再次上升，但斜率更平缓（较慢的速度）。

Common Mistake: Many students forget to include the stopping time when calculating average speed. In this question, the first part uses total time 25 s, but the graph sketch must clearly show the 5‑second horizontal section.

常见错误：很多学生在计算平均速度时会忘记把停顿时间算进去。本题第一问的总时间是 25 秒，但在画图时必须明确标出 5 秒的水平段。

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3. Mass, Weight, and Gravity | 质量、重量与重力

Question: A rock has a mass of 4 kg. On Earth, the gravitational field strength is approximately 10 N/kg. Calculate the weight of the rock. Explain what would happen to the weight if the rock were taken to the Moon, where gravity is weaker, but its mass stays the same.

题目：一块岩石的质量为 4 kg。在地球上，引力场强度约为 10 N/kg。计算这块岩石的重量。解释如果将岩石带到引力较弱的月球上，重量会发生什么变化，而其质量保持不变。

Solution: Weight is calculated using the equation:

W = m × g

Here, W = 4 kg × 10 N/kg = 40 N. On the Moon, the value of g is smaller (about 1.6 N/kg), so the weight would become much less (around 6.4 N). Mass is the amount of matter and does not depend on gravity, so mass remains 4 kg.

解析：重量用下式计算：

重量 = 质量 × 引力场强度

这里重量 = 4 kg × 10 N/kg = 40 N。在月球上，g 值较小（约为 1.6 N/kg），所以重量会小得多（约为 6.4 N）。质量是物体所含物质的多少，与引力无关，因此质量仍然是 4 kg。

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4. Energy Forms and Transfers | 能量形式与转移

Question: A battery‑powered torch is switched on. Describe the main energy transfer that occurs, naming the input energy and the useful output energies. Also mention one way energy is wasted in this system.

题目：打开一个电池供电的手电筒。描述发生的主要能量转移，并说出输入能量和有用的输出能量。同时指出该系统浪费能量的一种方式。

Solution: The energy transfer is: chemical potential energy stored in the battery → electrical energy in the wires → light energy (useful) + thermal energy (wasted heat) emitted from the bulb. The torch also wastes energy as sound if there is a faint buzzing, but the main wasted energy is heat that does not produce light.

解析：能量转移为：电池中储存的化学势能 → 导线中的电能 → 灯泡发出的光能（有用）+ 热能（浪费的热量）。如果手电筒有微弱的嗡嗡声，也会有声音形式的能量浪费，但最主要的浪费是那些不产生光的热量。

CCEA Note: Always use the term ‘chemical potential energy’ for stored energy in batteries and food. Useful output is energy transferred in the way we want, while wasted energy spreads into the surroundings.

CCEA 提示：对于电池和食物中储存的能量，始终使用“化学势能”这一术语。有用的输出是我们希望获得的能量转移方式，而浪费的能量会散逸到周围环境中。

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5. Simple Electric Circuits | 简单电路

Question: Draw a simple series circuit containing one cell, a switch, and two buzzers. Explain what happens to the buzzers when the switch is closed. A student adds a third buzzer in series – predict how the sound of each buzzer will change, and give a reason.

题目：画一个简单的串联电路，包含一节电池、一个开关和两个蜂鸣器。解释当开关闭合时蜂鸣器会发生什么。一个学生又在电路中串联了第三个蜂鸣器——预测每个蜂鸣器的声音会如何变化，并说明理由。

Solution: In the circuit, electricity flows from the positive terminal of the cell, through the closed switch, and then through both buzzers back to the negative terminal. All buzzers will sound at the same loudness because the current is the same everywhere in a series circuit. Adding a third buzzer increases the total resistance, so the current from the cell decreases. All three buzzers will be quieter, because a smaller current means each buzzer converts less electrical energy into sound.

解析：电路中，电流从电池正极流出，经过闭合的开关，然后依次通过两个蜂鸣器回到负极。所有蜂鸣器发出相同音量，因为串联电路中各处的电流相等。添加第三个蜂鸣器会增加总电阻，因此电池提供的电流变小。三个蜂鸣器都会变得小声，因为较小的电流意味着每个蜂鸣器将电能转化为声音的能量更少。

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6. Series and Parallel Circuits | 串联与并联电路

Question: Two bulbs are connected in parallel with a single cell. Bulb A is removed from its socket. Explain what happens to Bulb B. Compare this with what would happen if the two bulbs were connected in series and one bulb were removed.

题目：两个灯泡以并联方式与一节电池连接。灯泡 A 从灯座上取下。解释灯泡 B 会发生什么。将其与两个灯泡串联且取下一个灯泡的情况进行对比。

Solution: In a parallel circuit, each bulb has its own separate loop connected directly to the cell. Removing Bulb A does not break the loop for Bulb B, so Bulb B stays lit. In a series circuit, however, all components share the same single loop. Removing one bulb creates a gap, so the circuit is broken and both bulbs go out.

解析：在并联电路中，每个灯泡都有自己独立的回路直接与电池相连。取下灯泡 A 不会断开灯泡 B 的回路，所以灯泡 B 仍然亮着。而在串联电路中，所有元件共享同一个回路。取下任意一个灯泡都会造成断路，电路中断，两个灯泡都会熄灭。

Think of it like: parallel = separate motorways; series = one single track. This comparison often appears in CCEA multiple‑choice questions.

可以这样想：并联好比分开的高速公路；串联好比一条单轨。这种对比经常出现在 CCEA 的选择题中。

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7. Magnets and Magnetic Fields | 磁铁与磁场

Question: Describe how you would use iron filings and a plotting compass to map the magnetic field pattern around a bar magnet. Then state the rule for the interaction when two north poles are brought close together.

题目：描述你将如何使用铁屑和指南针来描绘一根条形磁铁周围的磁场图样。然后说明当两个北极相互靠近时的相互作用规律。

Solution: Place a sheet of paper over the bar magnet and gently sprinkle iron filings on top. Tap the paper lightly – the filings align along the magnetic field lines, showing curves from the north pole to the south pole. A plotting compass can be placed at different points; its needle always points along the field line direction. When two north poles are brought close, they repel each other (like poles repel).

解析：将一张纸盖在条形磁铁上，然后轻轻撒上铁屑。轻敲纸张——铁屑会沿着磁感线排列，显示出从北极指向南极的曲线。将指南针放在不同点，其指针始终指向磁感线的方向。当两个北极靠近时，它们相互排斥（同极相斥）。

CCEA detail: Field lines are drawn with arrows pointing from N to S. Ensure you never draw lines crossing. Also, the magnetic force is strongest where the lines are closest together (at the poles).

CCEA 细节：磁感线要画上箭头，由 N 指向 S。注意线条绝不能交叉。另外，磁感线最密集的地方磁力最强（即磁极处）。

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8. States of Matter and Particle Arrangement | 物态与粒子排列

Question: Use the particle model to explain why a solid has a fixed shape and volume, while a liquid has a fixed volume but takes the shape of its container. In your answer, refer to the arrangement, movement, and forces between particles.

题目：利用粒子模型解释为什么固体具有固定的形状和体积，而液体具有固定的体积却会呈现容器的形状。回答时要提到粒子的排列、运动以及粒子间的力。

Solution: In a solid, particles are packed closely together in a regular pattern. They vibrate about fixed positions and are held by strong forces, so the solid keeps its own shape. In a liquid, particles are still close but arranged randomly and can slide past each other. The forces are weaker, allowing the liquid to flow and take the shape of its container, but the particles stay close enough to keep the volume constant.

解析：在固体中，粒子紧密排列成规则的图案。它们在固定的位置上振动，并被强大的作用力束缚住，因此固体能保持自身的形状。在液体中，粒子仍然紧密但排列无规则，且能相互滑动。粒子间作用力较弱，使得液体可以流动并呈现容器的形状，但粒子间距离仍足够近，从而体积保持不变。

Memory aid: Solids – shape and volume fixed; liquids – volume fixed, shape not; gases – neither fixed. This links easily to density and changes of state.

记忆窍门：固体——形状和体积都固定；液体——体积固定，形状不固定；气体——两者都不固定。这与密度和状态变化的知识点直接关联。

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9. Sound Waves | 声波

Question: An alarm clock is placed inside a glass jar. When the air is pumped out, the sound becomes much quieter. Explain this observation using the idea of longitudinal waves and a medium. Include the term ‘vibration’ in your answer.

题目：把一个闹钟放在玻璃罐内。当抽出空气后，声音变得非常小。运用纵波和介质的概念来解释这一现象。回答中要包括“振动”一词。

Solution: Sound is produced by vibrating objects and travels as longitudinal waves. These waves need a medium (solid, liquid, or gas) to transmit vibrations from particle to particle. When air is removed from the jar, there are very few particles left, so the vibrations cannot be passed on efficiently. The sound almost disappears because sound cannot travel through a vacuum.

解析：声音是由振动的物体产生的，并以纵波的形式传播。这些波需要介质（固体、液体或气体）将振动从一个粒子传递到下一个粒子。当罐内的空气被抽走时，几乎没有粒子剩下，因此振动无法有效传递。声音几乎消失，因为声音不能在真空中传播。

CCEA often links this to the bell‑jar experiment. Note: light can travel through a vacuum, but sound cannot – a classic comparison question.

CCEA 常把这点与钟罩实验联系起来。注意：光可以在真空中传播，但声音不能——这是一个经典的比较题。

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10. Light and the Law of Reflection | 光与反射定律

Question: A ray of light strikes a plane mirror. The angle of incidence is 35°. State the Law of Reflection and calculate the angle of reflection. Draw a labelled diagram showing the incident ray, reflected ray, normal line, and both angles.

题目：一束光线照射到平面镜上，入射角为 35°。陈述光的反射定律并计算反射角。画一个带标记的示意图，标出入射光线、反射光线、法线以及两个角度。

Solution: The Law of Reflection states: the angle of incidence equals the angle of reflection. Both angles are measured from the normal (the line perpendicular to the mirror surface). Therefore, if the angle of incidence = 35°, the angle of reflection = 35°. The diagram must show a dashed normal line at 90° to the mirror, with the incident and reflected rays on opposite sides of the normal and the angles clearly labelled.

解析：反射定律指出：入射角等于反射角。两个角都从法线（垂直于镜面的线）量起。因此，如果入射角 = 35°，那么反射角 = 35°。示意图中必须有一条垂直于镜面的虚线法线，入射光线和反射光线分居法线两侧，并清楚标出两个角度。

Common error: Many students measure angles from the mirror surface rather than the normal. Remember, the normal is always at 90° to the mirror.

常见错误：很多学生从镜面而不是法线量起。要记住，法线总是与镜面成 90°。

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11. Data Handling and Graph Skills | 数据处理与图表技能

Question: An experiment measures the extension of a spring as weights are added. The results are shown in the table. Plot a graph of extension (cm) against weight (N) and use it to find the weight needed for an extension of 6.0 cm.

Weight (N)	0	2	4	6	8
Extension (cm)	0	1.5	3.0	4.5	6.0

题目：一项实验测量了弹簧在增加砝码时的伸长量。结果如下表所示。画出伸长量（cm）随重量（N）变化的图表，并用它求出产生 6.0 cm 伸长量所需的重量。

Solution: The table shows a clear pattern: extension increases by 1.5 cm for every 2 N added. This is a directly proportional relationship. Plotting the points gives a straight line through the origin. From the table, an extension of 6.0 cm corresponds to a weight of 8 N. Checking the graph confirms this value. If asked for a weight beyond the data, you would need to extend the line and read the value, assuming the spring does not exceed its elastic limit.

解析：表格呈现清晰的规律：每增加 2 N 重量，伸长量增加 1.5 cm。这是一种正比关系。将点描出后得到一条过原点的直线。从表格中可以看出，伸长量 6.0 cm 对应重量为 8 N。查阅图表可确认该数值。如果要求数据点之外的重量，则需要延长直线并读取数值，前提是弹簧未超过弹性极限。

CCEA Assessment Objective: You must be able to plot points with small crosses, draw a best‑fit straight line, and read interpolation data accurately.

CCEA 评估目标：你必须能够用小叉号描点，画出最佳拟合直线，并准确读取内插数据。

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12. Revision Summary and Top Tips | 复习总结与得分技巧

The mock paper shows that CCEA Year 7 Physics tests a mix of factual recall, graphical interpretation, and application of equations. Strengthen your method by always writing the formula, substituting numbers with units, and showing the final answer with the correct unit. For long‑answer questions, use scientific keywords like ‘balanced forces’, ‘longitudinal wave’, and ‘angle of incidence’ that examiners expect. Practise drawing circuit symbols and magnetic field lines, as these manual skills are often tested in the written paper.

这份模拟卷表明，CCEA 七年级物理测试融合了对事实性记忆、图表解读和公式应用的综合考查。通过始终坚持写公式、代入带单位的数值并给出带正确单位的最终答案，可以强化解题方法。对于长篇回答题，要使用考官期待的科学关键词，如“平衡力”“纵波”“入射角”等。同时要练习绘制电路符号和磁感线，因为这些动手技能常常在书面考试中考查。

Quick recall checklist: v = d ÷ t, W = m × g, angle of incidence = angle of reflection. Sound needs a medium; light doesn’t. Like poles repel; unlike poles attract. Solids have fixed shape; gases have neither fixed shape nor volume. Knowing these fundamentals will help you tackle any unit test with confidence.

快速回忆清单：v = d ÷ t，W = m × g，入射角 = 反射角。声音需要介质；光不需要。同极相斥，异极相吸。固体有固定形状；气体形状和体积都不固定。掌握这些基础知识，你将能自信地应对任何单元测试。

Published by TutorHao | Physics Revision Series | aleveler.com

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