📚 Year 8 WJEC Physics: Unit Test Mock Paper Walkthrough | 八年级 WJEC 物理：单元测试模拟卷解析
This article provides a step-by-step walkthrough of a mock unit test for Year 8 WJEC Physics. The paper covers core topics: energy stores, forces, electricity, waves, density, pressure, particle theory, and space. Each question is broken down with model answers and clear explanations to strengthen your understanding and exam technique.
本文详细解析了一份为八年级 WJEC 物理设计的单元测试模拟卷。试卷涵盖能量储存、力、电、波、密度、压强、粒子理论和太空等核心主题。每道题都配有标准答案和清晰的解析，帮助你加深理解并提升应试技巧。
1. Mock Paper Overview | 模拟卷概览
The mock paper contains 10 questions worth a total of 50 marks. Question types include multiple-choice, short-structured answers, calculations, and a diagram-based task. The topics are selected to reflect the WJEC Year 8 scheme of work, ensuring you revise the most important concepts before your test.
模拟卷包含 10 道题，总分 50 分。题型包括选择题、简答题、计算题以及一道图表任务。选题紧扣 WJEC 八年级教学大纲，确保你在考前复习到最重要的概念。
You will need to recall definitions, apply formulas (density, pressure, gravitational potential energy), interpret circuit diagrams, and compare wave types. Showing all working and using correct units is essential for calculation questions.
你需要记忆定义、运用公式（密度、压强、重力势能）、解读电路图并比较不同类型的波。在计算题中，写出完整的演算步骤并使用正确单位至关重要。
2. Question 1: Energy Stores | 题一：能量储存
Question 1 (Multiple choice): A pupil lifts a chair from the floor onto a desk. Which energy store increases the most in the chair? A. Kinetic energy store B. Gravitational potential energy store C. Thermal energy store D. Chemical energy store
第 1 题（选择题）：一名学生将椅子从地板举到课桌上。椅子中哪种能量储存增加最多？ A. 动能储存 B. 重力势能储存 C. 热能储存 D. 化学能储存
The correct choice is B, gravitational potential energy store. Lifting an object against gravity transfers energy into the gravitational store of that object, raising its position in a gravitational field.
正确答案是 B，重力势能储存。将物体克服重力举起时，能量会转移到该物体的重力储能中，使其在引力场中的位置升高。
Gravitational potential energy depends on mass m, height h and the strength of gravity g. The energy change can be calculated using the formula below. Since the chair's height increases while mass and g stay constant, the store increases.
重力势能取决于质量 m、高度 h 和引力强度 g。其变化可用以下公式计算。在椅子质量与 g 不变的情况下，高度增加，因此重力势能储存增加。
ΔE = mgΔh
3. Question 2: Forces and Motion | 题二：力与运动
Question 2: A bicycle is travelling along a straight, flat road at a constant speed of 6 m/s. Describe the sizes of the driving force and the resistive forces acting on the bicycle. Explain your answer.
第 2 题：一辆自行车在平直道路上以 6 m/s 的恒定速度行驶。描述作用于自行车的驱动力与阻力的大小关系，并解释原因。
The driving force and the total resistive forces (air resistance and friction) are equal in size. When an object moves at constant speed in a straight line, the resultant force is zero.
驱动力与总阻力（空气阻力和摩擦力）大小相等。当物体沿直线做匀速运动时，合外力为零。
This is a direct application of Newton's First Law. A zero resultant force means the forces are balanced, so there is no change in velocity. If the driving force were larger, the bicycle would accelerate.
这是牛顿第一定律的直接应用。合外力为零意味着力相互平衡，因此速度不发生变化。如果驱动力更大，自行车就会加速。
4. Question 3: Circuit Symbols | 题三：电路符号
Question 3: Draw the standard circuit symbols for a cell and a filament lamp. Then draw a simple series circuit containing one cell, one open switch, and one filament lamp.
第 3 题：画出电池和灯丝灯泡的标准电路符号。然后画出一个由一个电池、一个打开的开关和一个灯丝灯泡组成的简单串联电路。
For a cell, draw a long thin line and a short thick line parallel to each other. The long line represents the positive terminal, the short thick line the negative terminal. A filament lamp is shown as a circle enclosing a cross (or a looped wire symbol inside a circle).
画电池时，画一长一短两条平行线，长线代表正极，短粗线代表负极。灯丝灯泡的符号是一个圆圈内画一个交叉（或一个环状灯丝的符号）。
To construct the series circuit, connect the components in a single loop: the cell's positive terminal to one side of the switch, the other side of the switch to the lamp, and then back to the cell's negative terminal. Since the switch is open, no current will flow.
构建串联电路时，需将各元件连接成单一回路：电池正极接开关一端，开关另一端接灯泡，再回到电池负极。由于开关是断开的，电路中没有电流。
5. Question 4: Series vs Parallel Circuits | 题四：串联与并联电路
Question 4: Two identical bulbs are connected to a battery. In circuit A they are in series; in circuit B they are in parallel. Which circuit will have brighter bulbs? Use ideas about voltage and current to justify your answer.
第 4 题：两个相同的灯泡接入电池。电路 A 为串联，电路 B 为并联。哪个电路中的灯泡更亮？请运用电压和电流的概念加以解释。
The bulbs in circuit B (parallel) will be brighter. In a parallel circuit, each bulb gets the full battery voltage. In a series circuit, the voltage is shared between the bulbs, so each receives only half the supply voltage.
电路 B（并联）中的灯泡更亮。在并联电路中，每个灯泡都能获得电池的全部电压。而在串联电路中，电压被两个灯泡分摊，每个灯泡仅获得电源电压的一半。
With identical bulbs, a lower voltage across each bulb in series results in a smaller current through them, and thus less power dissipated as light and heat. In parallel, full voltage drives full current through each bulb, making them shine at full brightness.
对于相同的灯泡，串联时每个灯泡两端的电压较低，导致通过的电流较小，因此转化为光和热的功率较少。并联时，每个灯泡在满电压下驱动满电流，因而达到正常亮度。
| Circuit A (Series) | Circuit B (Parallel) |
| Voltage shared → dimmer | Full voltage across each → brighter |
6. Question 5: Waves | 题五：波
Question 5: State one example of a transverse wave and one example of a longitudinal wave. For each, describe the direction of particle vibration relative to the direction of energy transfer.
第 5 题：分别举出一个横波和一个纵波的例子。针对每种波，描述质点振动方向与能量传递方向的关系。
Light and all electromagnetic waves are transverse. In a transverse wave, particles vibrate perpendicular (at right angles) to the direction the wave travels. Sound waves are longitudinal. Here, particles vibrate parallel to the direction of wave travel.
光及所有电磁波都是横波。在横波中，质点振动的方向与波的传播方向垂直。声波是纵波，质点振动的方向与波的传播方向平行。
A water ripple is another good transverse example: the surface moves up and down while the ripple travels horizontally. A slinky pushed and pulled along its length creates compressions and rarefactions, demonstrating a longitudinal wave.
水波也是一个很好的横波例子：水面上下起伏，而波纹沿水平方向传播。沿水平方向推拉弹簧会产生疏密相间的区域，这正演示了纵波。
7. Question 6: Density | 题六：密度
Question 6: A solid aluminium block has a mass of 135 g and a volume of 50 cm³. Calculate its density. State the formula, show your working and give the correct unit.
第 6 题：一块实心铝块的质量为 135 g，体积为 50 cm³。计算它的密度。写出公式，展示演算过程并注明正确单位。
. Always check that the unit matches the quantities used; here the unit is grams per cubic centimetre.
密度等于质量除以体积。公式如下。代入数值：密度 = 135 g ÷ 50 cm³ = 2.7 g/cm³。务必检查单位是否与所用物理量匹配；这里单位是克每立方厘米。
ρ = m / V
If the question asked for density in kg/m³, you would convert mass to kg and volume to m³. A density of 2.7 g/cm³ is equal to 2700 kg/m³.
若题目要求以 kg/m³ 为单位，你需要将质量换算为千克，体积换算为立方米。2.7 g/cm³ 等于 2700 kg/m³。
8. Question 7: Pressure | 题七：压强
Question 7: A student exerts a force of 450 N on the floor while standing on one foot. The area of the shoe sole is 0.03 m². Calculate the pressure on the floor.
第 7 题：某学生单脚站立时，对地面施加 450 N 的力。鞋底面积为 0.03 m²。计算地面受到的压强。
e P = 450 N / 0.03 m² = 15 000 Pa (or 15 kPa). The unit pascal (Pa) is equivalent to N/m².
压强等于力除以受力面积。应用公式 P = F/A，可得 P = 450 N / 0.03 m² = 15000 Pa（或 15 kPa）。帕斯卡（Pa）等价于 N/m²。
P = F / A
If the same student stood on both feet, the area would double, halving the pressure to 7500 Pa. This illustrates why sharp objects with small area produce high pressure with even modest forces.
若该学生双脚站立，受力面积加倍，压强将减半至 7500 Pa。这解释了为何尖锐物体因面积小而即使受力不大也能产生很大压强。
9. Question 8: Particle Model | 题八：粒子模型
Question 8: Use the particle model to explain why a solid has a fixed shape and cannot be compressed, whereas a gas fills its container and can be easily compressed.
第 8 题：运用粒子模型解释：为什么固体有固定的形状且不能被压缩，而气体会充满整个容器并容易被压缩。
In a solid, particles are packed closely in a regular arrangement. Strong forces hold them in fixed positions, so they can only vibrate. This gives the solid a definite shape and volume, and very little empty space means it is incompressible.
在固体中，粒子紧密排列成规则的结构。强大的作用力使它们固定在原位，只能做振动。因此固体拥有确定的形状和体积，且几乎没有空隙，所以不可压缩。
In a gas, particles are far apart with negligible forces between them. They move rapidly in all directions and fill any available volume. The large spaces between particles allow a gas to be compressed easily when pressure is applied.
在气体中，粒子相距很远，彼此间的作用力可忽略不计。它们快速向各个方向运动，充满任何可用空间。粒子间的大量空隙使气体在施加压力时容易被压缩。
10. Question 9: Space | 题九：太空
Question 9: List the eight planets of our Solar System in order of increasing distance from the Sun. State why Pluto is no longer classified as a planet.
第 9 题：按离太阳距离由近到远的顺序，列出太阳系的八大行星。并说明冥王星为何不再被归类为行星。
The correct order is: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. A mnemonic like 'My Very Easy Method Just Speeds Up Naming' can help you remember.
正确的顺序是：水星、金星、地球、火星、木星、土星、天王星、海王星。可以用口诀帮助记忆。
Pluto was reclassified as a dwarf planet in 2006 because it has not cleared its neighbouring region of other objects – one of the three criteria required by the IAU for a full planet. It shares its orbit with other icy bodies in the Kuiper Belt.
冥王星于 2006 年被重新分类为矮行星，因为它未能清除其轨道附近的其他天体——这是国际天文学联合会界定行星的三个条件之一。冥王星与柯伊伯带中的其他冰质小天体共享轨道。
11. Question 10: Energy Transfers and Sankey Diagrams | 题十：能量传递与桑基图
Question 10: An electric motor lifts a load. The motor receives 100 J of electrical energy and produces 60 J of useful gravitational potential energy. The rest is wasted as heat and sound. Draw a Sankey diagram for this process, and calculate the efficiency.
第 10 题：一台电动机提起重物。电动机获得 100 J 的电能，并输出 60 J 有用的重力势能，其余以热能和声能的形式浪费。画出这一过程的桑基图，并计算效率。
In a Sankey diagram, the main arrow represents the total input energy (100 J), drawn with a width proportional to the energy. It splits into a useful energy arrow (60 J) pointing forward and a wasted energy arrow (40 J) branching downwards.
在桑基图中，主箭头表示总的输入能量（100 J），其宽度与能量大小成正比。它分叉为向前的有用能量箭头（60 J）和向下分支的浪费能量箭头（40 J）。
Efficiency = (useful output energy / total input energy) × 100%. Here, efficiency = (60 J / 100 J) × 100% = 60%. This means 60% of the input energy is converted usefully, while 40% is dissipated to the thermal store of the surroundings.
效率 =（有用输出能量 / 总输入能量）× 100%。此处效率 =（60 J / 100 J）× 100% = 60%。这表明 60% 的输入能量被转化为有用功，其余 40% 散失到周围环境的热能储存中。
Efficiency = (Useful output ÷ Total input) × 100%
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