Year 9 CAIE Physics Unit Test Mock Paper Walkthrough | Year 9 CAIE 物理单元测试模拟卷解析

📚 Year 9 CAIE Physics Unit Test Mock Paper Walkthrough | Year 9 CAIE 物理单元测试模拟卷解析

This comprehensive walkthrough tackles a full Year 9 CAIE Physics unit test mock paper, covering key topics from mechanics, energy, waves, electricity and materials. Each question is broken down with clear, step-by-step solutions to reinforce your understanding and exam technique. Use this guide to identify common pitfalls and master the essential skills required by the CAIE curriculum.

这篇详尽的解析文章完整剖析了一套 Year 9 CAIE 物理单元测试模拟卷，覆盖力学、能量、波、电学和材料等重点主题。每道题目都配有清晰的、按步骤呈现的解答，帮助你巩固理解并掌握考试技巧。借助本指南，你可以识别常见错误，牢固掌握 CAIE 课程所要求的关键能力。

1. Distance–Time Graph Analysis | 距离–时间图分析

Question 1: A toy car moves along a straight track. The distance– 5 m from the start. (a) Calculate the average speed of the car between 2 s and 6 s. (b) Describe the motion of the car if the graph becomes a horizontal straight line after 6 s.

题目 1：一辆玩具小车在直轨道上运动。距离–时间图显示，t = 2 s 时小车距起点 15 m，t = 6 s 时距起点 45 m。(a) 计算小车在 2 s 到 6 s 之间的平均速度。(b) 若 6 s 后图线成为一条水平直线，请描述小车的运动状态。

For part (a), the average speed is the gradient of the distance– e formula is v = Δd / Δt, where Δd is the change in distance and Δt is the change in time.

第 (a) 部分中，平均速度是这段时间内距离–时间图线的斜率。所用公式为 v = Δd / Δt，其中 Δd 是距离的变化量，Δt 是时间的变化量。

Here, Δd = 45 m – 15 m = 30 m, and Δt = 6 s – the unit and check that your answer is a speed, not a velocity unless direction is given.

本题中，Δd = 45 m – 15 m = 30 m，Δt = 6 s – 2 s = 4 s。因此速度 v = 30 m / 4 s = 7.5 m/s。务必带上单位，并注意题目只要求速率，除非指明方向，否则不求速度。

For part (b), a horizontal line on a distance–time graph means the distance does not change as time increases. Therefore the car is stationary, i.e. its speed is zero.

第 (b) 部分，距离–时间图中的水平线表示距离不随时间变化。因此小车处于静止状态，即其速率为零。

2. Density Calculation | 密度计算

Question 2: A metal cube has a mass of 150 g and a volume of 30 cm³. (a) Calculate its density in g/cm³. (b) Express this density in kg/m³.

题目 2：一个金属立方体的质量为 150 g，体积为 30 cm³。(a) 计算它的密度，单位用 g/cm³。(b) 用 kg/m³ 表示该密度。

g/cm³. Memorising the formula and rearranging it will help you tackle any density problem.

密度 ρ 由公式 ρ = 质量 / 体积求得。代入已知数据，ρ = 150 g / 30 cm³ = 5 g/cm³。牢记该公式并会变形，能帮助你解决各种密度问题。

³ contains 100 × 100 × 100 = 1,000,000 cm³, while 1 kg = 1000 g, hence the factor of 1000.

要从 g/cm³ 转换为 kg/m³，记住 1 g/cm³ = 1000 kg/m³。乘以 1000 即可：5 g/cm³ × 1000 = 5000 kg/m³。这一结果合理，因为 1 m³ 含有 1,000,000 cm³，而 1 kg = 1000 g，因此换算因子为 1000。

3. Pressure and Force | 压强与力

Question 3: A rectangular block weighing 120 N rests on a table with a base area of 0.03 m². (a) Calculate the pressure exerted on the table. (b) Explain why the same block exerts a greater pressure when it stands on a smaller face.

题目 3：一个重 120 N 的长方体木块平放在桌面上，底座面积为 0.03 m²。(a) 计算它对桌面施加的压强。(b) 解释为何同一木块用更小的一面站立时，产生的压强更大。

is 0.03 m². So P = 120 N / 0.03 m² = 4000 Pa. The unit pascal (Pa) is equivalent to N/m².

压强 P = 力 / 面积。此处力为重量 120 N，面积为 0.03 m²。因此 P = 120 N / 0.03 m² = 4000 Pa。单位帕斯卡 (Pa) 等同于 N/m²。

For part (b), pressure is inversely proportional to area for a constant force. A smaller contact area concentrates the same force into a smaller region, leading to a larger pressure. This is why a sharp knife cuts more easily than a blunt one.

第 (b) 部分中，当力不变时，压强与面积成反比。更小的接触面积会把相同的力集中到更小的区域，从而产生更大的压强。这就是为什么锋利的刀比钝刀更容易切割。

4. Kinetic Energy and Gravitational Potential Energy | 动能与重力势能

Question 4: etic energy of the cyclist. (b) The cyclist then coasts up a 20 m high slope. Assuming 100% efficiency, how much gravitational potential energy is gained? (Use g = 10 N/kg.)

题目 4：一名总质量为 60 kg 的自行车手正以 8 m/s 的速度行驶。(a) 计算车手的动能。(b) 随后车手靠惯性滑上一个 20 m 高的斜坡。假设能量转换效率为 100%，求获得的重力势能为多少？（取 g = 10 N/kg）

= ½ × 60 kg × (8 m/s)² = ½ × 60 × 64 = 30 × 64 = 1920 J. Always square the velocity first.

动能 KE = ½ m v²。代入 m = 60 kg，v = 8 m/s：KE = ½ × 60 kg × (8 m/s)² = ½ × 60 × 64 = 30 × 64 = 1920 J。务必先计算速度的平方。

2 000 J. Since all kinetic energy is assumed to convert to GPE, this is the gained energy.

重力势能 GPE = m g h。本题中 m = 60 kg，g = 10 N/kg，h = 20 m。因此 GPE = 60 × 10 × 20 = 12 000 J。由于假定所有动能都转化为重力势能，这就是获得的重力势能。

5. Wave Speed, Frequency and Wavelength | 波速、频率与波长

Question 5: A water wave has a frequency of 25 Hz and a wavelength of 0.4 m. (a) Calculate the speed of the wave. (b) Determine the period of the wave.

题目 5：一个水波的频率为 25 Hz，波长为 0.4 m。(a) 计算该波的波速。(b) 求波的周期。

λ by v = f × λ. Plugging in the values: v = 25 Hz × 0.4 m = 10 m/s. Note that 1 Hz is 1/s.

波速 v 与频率 f、波长 λ 之间的关系为 v = f × λ。代入数值：v = 25 Hz × 0.4 m = 10 m/s。注意 1 Hz 即为 1/s。

cy: T = 1 / f = 1 / 25 s = 0.04 s. This means each complete wave cycle takes 0.04 seconds.

周期 T 是频率的倒数：T = 1 / f = 1 / 25 s = 0.04 s。这意味着每个完整的波动周期耗时 0.04 秒。

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6. Series Circuit Analysis | 串联电路分析

Question 6: A 12 V battery is connected in series with a 3 Ω resistor and a 6 Ω resistor. (a) Calculate the total resistance. (b) Find the current in the circuit. (c) Determine the voltage across each resistor.

题目 6：一个 12 V 电池与一个 3 Ω 电阻和一个 6 Ω 电阻串联。(a) 计算总电阻。(b) 求电路中的电流。(c) 确定每个电阻两端的电压。

For resistors in series, total resistance R_total = R₁ + R₂ = 3 Ω + 6 Ω = 9 Ω. In a series circuit the current is the same everywhere.

对于串联电阻，总电阻 R_总 = R₁ + R₂ = 3 Ω + 6 Ω = 9 Ω。串联电路中各处电流均相等。

Using Ohm’s law, I = V / R_total = 12 V / 9 Ω ≈ 1.33 A. The current through both resistors is 1.33 A.

根据欧姆定律，I = V / R_总 = 12 V / 9 Ω ≈ 1.33 A。流过两个电阻的电流均为 1.33 A。

₂ = 1.33 A × 6 Ω = 8.0 V. Notice that 4 V + 8 V = 12 V, which confirms the supply voltage.

现在计算 3 Ω 电阻两端的电压 V₁ = I × R₁ = 1.33 A × 3 Ω = 4.0 V；6 Ω 电阻两端的电压 V₂ = 1.33 A × 6 Ω = 8.0 V。注意 4 V + 8 V = 12 V，与电源电压一致，验证了计算结果。

7. Hooke’s Law and Spring Constant | 胡克定律与弹簧常数

Question 7: A spring stretches by 0.12 m when a force of 6 N is applied. (a) Calculate the spring constant k. (b) Predict the extension when a 9 N force is applied, assuming the elastic limit is not exceeded.

题目 7：某弹簧在 6 N 拉力作用下伸长了 0.12 m。(a) 计算弹簧常数 k。(b) 假设未超过弹性限度，预计施加 9 N 的力时弹簧的伸长量是多少。

e spring constant, and x is the extension. Rearranging, k = F / x = 6 N / 0.12 m = 50 N/m.

胡克定律表述为 F = k × x，其中 F 是力，k 是弹簧常数，x 是伸长量。变形公式得 k = F / x = 6 N / 0.12 m = 50 N/m。

18 m. The extension is directly proportional to the applied force only within the elastic limit.

对于 9 N 的力，x = F / k = 9 N / 50 N/m = 0.18 m。伸长量与所施加的力成正比，但这只在弹性限度内成立。

8. Resultant Force and Acceleration | 合力与加速度

Question 8: A car of mass 800 kg has an engine thrust of 4000 N acting forward. Friction and air resistance total 600 N acting backwards. (a) Calculate the resultant force on the car. (b) Determine the car’s acceleration.

题目 8：一辆质量为 800 kg 的汽车，发动机施加向前的推力 4000 N，而摩擦和空气阻力总和为向后的 600 N。(a) 计算汽车所受的合力。(b) 求汽车的加速度。

Resultant force F_net = forward force – resistive force = 4000 N – 00 N = 3400 N in the forward direction. Always assign a direction to your resultant force.

合力 F_net = 向前力 – 阻力 = 4000 N – 600 N = 3400 N，方向向前。一定要为合力标出方向。

Apply Newton’s second law: F_net = m × a, so a = F_net / m = 3400 N / 800 kg = 4.25 m/s². The car accelerates forward at 4.25 m/s².

应用牛顿第二定律：F_net = m × a，因此 a = F_net / m = 3400 N / 800 kg = 4.25 m/s²。汽车以 4.25 m/s² 的加速度向前加速。

9. Refractive Index from Speeds | 用速度求折射率

Question 9: ls from air into a glass block. The speed of light in air is 3.0 × 10⁸ m/s, and inside the

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