📚 Year 9 WJEC Physics: Unit Test Mock Paper Walkthrough | Year 9 WJEC 物理：单元测试模拟卷解析
Welcome to our detailed walkthrough of a Year 9 WJEC Physics unit test mock paper. In this article, we break down twelve key questions covering energy, electricity, waves, forces, matter and practical skills. Each solution is presented step by step with paired English and Chinese explanations to help you master core concepts and exam technique.
欢迎阅读我们针对 Year 9 WJEC 物理单元测试模拟卷的详细解析。本文分解了十二道核心试题，涵盖能量、电学、波、力、物质以及实验技能。每道题的解答都逐步呈现，配有英中对照的解释，帮助你掌握核心概念和考试技巧。
1. Energy – Gravitational Potential Energy | 第1题：重力势能
m. Calculate the gravitational potential energy (GPE) gained. (Take g = 10 N/kg) [2 marks]
题目：一块质量为5 kg的岩石被垂直提升4 m。计算获得的重力势能。（取 g = 10 N/kg）[2分]
Ep = m × g × h
Ep = m × g × h
Step 1: Write the GPE formula, where m is mass in kg, g is the gravitational field strength in N/kg, and h is the vertical height in metres.
步骤1：写出重力势能公式，其中 m 为质量（kg），g 为重力场强度（N/kg），h 为垂直高度（m）。
Step 2: Substitute the known values: m = 5 kg, g = 10 N/kg, h = 4 m → Ep = 5 × 10 × 4.
步骤2：代入已知数值：m = 5 kg，g = 10 N/kg，h = 4 m → Ep = 5 × 10 × 4。
oule (J). Final answer: 200 J. Always check that your answer is given in the correct unit.
步骤3：计算：5 × 10 = 50，然后 50 × 4 = 200。能量的单位是焦耳 (J)。最终答案：200 J。务必检查答案的单位是否正确。
2. Energy – Kinetic Energy Calculation | 第2题：动能计算
The question: A car of mass 800 kg moves at a steady speed of 15 m/s. Calculate its kinetic energy (KE). [2 marks]
题目：一辆质量为800 kg的汽车以15 m/s的恒定速度行驶。计算其动能。[2分]
KE = ½ × m × v²
KE = ½ × m × v²
Step 1: Recall the kinetic energy formula: KE = ½ × m × v², with m in kg and v in m/s.
步骤1：回顾动能公式：KE = ½ × m × v²，质量 m 单位为 kg，速度 v 单位为 m/s。
tep 2: Insert the values: m = 800 kg, v = 15 m/s. The term v² means v × v = 15 × 15 = 225.
步骤2：代入数值：m = 800 kg，v = 15 m/s。v² 表示 v × v = 15 × 15 = 225。
90,000. Energy is in joules, so the answer is 90,000 J. This can also be written as 90 kJ.
步骤3：计算：½ × 800 × 225。先算 ½ × 800 = 400，再算 400 × 225 = 90,000。能量单位是焦耳，答案为 90,000 J，也可写为 90 kJ。
3. Electricity – Total Resistance in Series | 第3题：串联总电阻
The question: Two resistors, 4 Ω and 6 Ω, are connected end-to-end in a circuit. Calculate the total resistance. [1 mark]
题目：两个电阻，4 Ω 和 6 Ω，在电路中首尾相连。计算总电阻。[1分]
Rₜ = R₁ + R₂
Rₜ = R₁ + R₂
Step 1: For resistors in series, the total resistance Rₜ is simply the sum of the individual resistances.
步骤1：对于串联电阻，总电阻 Rₜ 就是每个电阻值的直接相加。
Step 2: Add the two values: R₁ = 4 Ω, R₂ = 6 Ω, so Rₜ = 4 Ω + 6 Ω = 10 Ω.
步骤2：将两个值相加：R₁ = 4 Ω，R₂ = 6 Ω，因此 Rₜ = 4 Ω + 6 Ω = 10 Ω。
No further calculation is needed; the ohm (Ω) is the unit of resistance. Always remember that adding more resistors in series increases the total resistance.
无需进一步计算；欧姆 (Ω) 是电阻的单位。请记住，串联更多电阻会增大总电阻。
4. Electricity – Current in a Series Circuit | 第4题：串联电路中的电流
The question: In the same series circuit, the current through the 4 Ω resistor is 0.5 A. What is the current through the 6 Ω resistor? Explain your reasoning. [2 marks]
题目：在同一个串联电路中，通过 4 Ω 电阻的电流为 0.5 A。通过 6 Ω 电阻的电流是多少？请解释原因。[2分]
Step 1: In a series circuit, the current has only one path. Therefore the current is the same at every point in the circuit.
步骤1：在串联电路中，电流只有一条路径。因此，电路中各点的电流处处相等。
Step 2: Since the current through the 4 Ω resistor is given as 0.5 A, the current through the 6 Ω resistor must also be 0.5 A.
步骤2：已知通过 4 Ω 电阻的电流为 0.5 A，所以通过 6 Ω 电阻的电流也必定是 0.5 A。
Key concept: Components in series share the same current, but the voltage across each may differ. This is a very common exam fact that many students overlook.
关键概念：串联元件电流相同，但各元件两端的电压可能不同。这是考试中非常常见且许多学生容易忽略的事实。
5. Waves – Frequency and Period | 第5题：频率与周期
The question: A water wave has a period of 0.02 s. Calculate the frequency of the wave. [2 marks]
题目：一个水波的周期为 0.02 s。计算该波的频率。[2分]
f = 1 ÷ T
f = 1 ÷ T
her. The formula is f = 1 / T, where T is the time for one complete wave cycle in seconds.
步骤1：频率 (f) 和周期 (T) 互为倒数。公式为 f = 1 / T，T 是完成一个完整波周期所需的时间，单位为秒。
Step 2: Substitute T = 0.02 s into the formula: f = 1 / 0.02.
步骤2：将 T = 0.02 s 代入公式：f = 1 / 0.02。
Step 3: 1 divided by 0.02 equals 50. Therefore the frequency is 50 hertz (Hz). A high frequency like 50 Hz means the wave oscillates 50 times per second.
步骤3：1 除以 0.02 等于 50。因此频率为 50 赫兹 (Hz)。像 50 Hz 这样较高的频率意味着该波每秒振荡 50 次。
6. Waves – Wave Speed Calculation | 第6题：波速计算
The question: A sound wave travelling through air has a frequency of 500 Hz and a wavelength of 0.68 m. Calculate the speed of the sound wave. [2 marks]
题目：在空气中传播的声波频率为 500 Hz，波长为 0.68 m。计算该声波的速度。[2分]
v = f × λ
v = f × λ
e equation links speed (v), frequency (f) and wavelength (λ). The standard form is v = fλ.
步骤1：波动方程将波速 (v)、频率 (f) 和波长 (λ) 联系起来。标准形式为 v = fλ。
Step 2: Insert the known values: f = 500 Hz, λ = 0.68 m. So, v = 500 × 0.68.
步骤2：代入已知数值：f = 500 Hz，λ = 0.68 m。因此 v = 500 × 0.68。
is metres per second (m/s). Answer: 340 m/s, which matches the typical speed of sound in air. Always show the multiplication step clearly.
步骤3：进行乘法：500 × 0.68 = 340。波速的单位是米每秒 (m/s)。答案：340 m/s，这与空气中声音的典型速度一致。务必清晰地展示乘法步骤。
7. Forces – Pressure Calculation | 第7题：压强计算
The question: A block exerts a force of 40 N on a table surface of area 0.2 m². Calculate the pressure acting on the table. [2 marks]
题目：一个块体对桌面施加 40 N 的力，受力面积为 0.2 m²。计算作用在桌面上的压强。[2分]
P = F / A
P = F / A
unit area (A). The formula is P = F ÷ A, with force in newtons and area in square metres.
步骤1：压强 (P) 定义为单位面积 (A) 上的力 (F)。公式为 P = F ÷ A，力的单位为牛顿，面积单位为平方米。
Step 2: Substitute F = 40 N and A = 0.2 m² into the equation: P = 40 ÷ 0.2.
步骤2：将 F = 40 N 和 A = 0.2 m² 代入公式：P = 40 ÷ 0.2。
Step 3: 40 divided by 0.2 equals 200. The unit of pressure is the pascal (Pa), which is equivalent to N/m². Final answer: 200 Pa. Remember that a smaller area produces a larger pressure for the same force.
步骤3：40 除以 0.2 等于 200。压强的单位是帕斯卡 (Pa)，等同于 N/m²。最终答案：200 Pa。记住，力相同时，面积越小压强越大。
8. Forces – Moment of a Force | 第8题：力矩
The question: A spanner is used to tighten a nut. A force of 15 N is applied at a perpendicular distance of 0.25 m from the centre of the nut. Calculate the moment exerted. [2 marks]
题目：一把扳手用于拧紧螺母。在垂直于螺母中心 0.25 m 处施加了 15 N 的力。计算产生的力矩。[2分]
M = F × d
M = F × d
, where d is the perpendicular distance from the pivot to the line of action of the force.
步骤1：力的力矩（转动效应）由 M = F × d 给出，其中 d 是从支点到力作用线的垂直距离。
Step 2: Insert the values: F = 15 N, d = 0.25 m. So M = 15 × 0.25.
步骤2：代入数值：F = 15 N, d = 0.25 m。因此 M = 15 × 0.25。
Step 3: Multiply to get 3.75. The unit is newton-metre (Nm). Answer: 3.75 Nm. To increase the moment, you can increase the force or use a longer spanner to increase the distance.
步骤3：相乘得 3.75。力矩的单位是牛顿·米 (Nm)。答案：3.75 Nm。要增大力矩，可以加大力或使用更长的扳手以增加距离。
9. Matter – Density Calculation | 第9题：密度计算
The question: A solid metal cube has a mass of 250 g and a volume of 50 cm³. Calculate its density. [2 marks]
题目：一个实心金属立方体质量为 250 g，体积为 50 cm³。计算其密度。[2分]
ρ = m / V
ρ = m / V
ume (V). The formula is ρ = m / V. Always be careful with the units given in the question.
步骤1：密度 (ρ) 等于质量 (m) 除以体积 (V)。公式为 ρ = m / V。务必注意题目给出的单位。
Step 2: Substitute m = 250 g and V = 50 cm³: ρ = 250 / 50.
步骤2：代入 m = 250 g 和 V = 50 cm³：ρ = 250 / 50。
typical density for metals like iron or tin. If you need to convert, 1 g/cm³ = 1000 kg/m³.
步骤3：250 ÷ 50 = 5。由于质量单位为克，体积单位为立方厘米，因此密度为 5 g/cm³。这是铁或锡等金属的典型密度。如需要转换，1 g/cm³ = 1000 kg/m³。
10. Matter – Particle Model and States | 第10题：粒子模型与物态
The question: Explain why a solid has a fixed shape but a liquid can flow. Use ideas about the arrangement and movement of particles. [3 marks]
题目：解释为什么固体有固定形状而液体可以流动。请利用粒子的排列和运动加以说明。[3分]
Marking point 1: In a solid, particles are packed closely together in a regular, fixed arrangement. They can only vibrate about fixed positions and cannot move past one another, so the solid keeps its shape.
得分点1：在固体中，粒子紧密堆积，排列规则且位置固定。它们只能在固定位置振动，无法彼此滑过，因此固体保持其形状。
Marking point 2: In a liquid, particles are still close together but arranged irregularly. They have more kinetic energy and can slide past each other, allowing the liquid to flow and take the shape of its container.
得分点2：在液体中，粒子仍然彼此靠近但排列不规则。它们具有更多的动能，可以相互滑过，从而使液体能够流动并呈现容器的形状。
Marking point 3: The forces of attraction between particles are stronger in a solid, holding them in neat rows, while in a liquid these forces are weaker, letting particles move around.
得分点3：固体中粒子间的吸引力更强，能将粒子固定在整齐的行列中；而液体中的吸引力较弱，允许粒子四处移动。
11. Energy Resources – Fossil Fuels and Renewables | 第11题：化石燃料与可再生能源
The question: State one advantage and one disadvantage of using wind turbines to generate electricity. [2 marks]
题目：说出使用风力涡轮机发电的一个优点和一个缺点。[2分]
Advantage: Wind energy is a renewable resource, meaning it will not run out. It also produces no greenhouse gases during operation, which helps reduce climate change.
优点：风能是一种可再生能源，这意味着它不会耗尽。发电过程中不产生温室气体，有助于减缓气候变化。
Disadvantage: Wind turbines depend on the weather; when the wind isn't blowing strongly, they generate little or no electricity. They can also be noisy and some people consider them an eyesore on the landscape.
缺点：风力涡轮机依赖天气；当风力不足时，它们发电量很少甚至不发电。它们也可能产生噪音，并有人觉得它们破坏了景观。
Exam tip: When answering comparison questions, always use linking words like 'whereas' or 'on the other hand'. A structured two-point answer scores more reliably than a long paragraph.
考试提示：回答比较类问题时，总是使用诸如“而”、“另一方面”之类的连接词。结构清晰的两点式回答比长段落更容易稳定得分。
12. Practical Skills – Precision of a Measurement | 第12题：测量精度
The question: A student uses a ruler marked in millimetres to measure the length of a spring. The reading is 12.3 cm. State the precision of this measurement. [1 mark]
题目：一位学生使用毫米刻度的尺子测量弹簧的长度。读数为 12.3 cm。说明此次测量的精度。[1分]
Step 1: Precision refers to the smallest change that can be detected with the instrument. On a millimetre ruler, the smallest division is 1 mm, which is 0.1 cm.
步骤1：精度是指仪器能检测到的最小变化。在毫米尺上，最小刻度
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