Year 10 CCEA Physics: Unit Test Mock Paper Walkthrough | Year 10 CCEA 物理：单元测试模拟卷解析

📚 Year 10 CCEA Physics: Unit Test Mock Paper Walkthrough | Year 10 CCEA 物理：单元测试模拟卷解析

This walkthrough guides you through a mock unit test for Year 10 CCEA Physics, covering motion, forces, energy, waves, electricity and radioactivity. Each question is broken down step by step so you can see exactly how to pick up every mark. Use it to diagnose weak spots and sharpen your exam technique.

这份解析带你逐题攻克 Year 10 CCEA 物理单元模拟卷，涵盖运动、力、能量、波、电学和放射性。每道题都拆解为详细的答题步骤，让你看清如何在考试中拿到满分。用它来检验薄弱环节、打磨答题规范。

1. Calculating Speed | 速度计算

Question: A cyclist travels 450 m in 30 seconds. Calculate her average speed.

题目：一名自行车骑手在30秒内骑行了450米，计算她的平均速度。

Use the formula average speed = total distance / total time. Substitute the numbers: speed = 450 m / 30 s = 15 m/s. Always include the unit; leaving it out loses a mark.

使用公式：平均速度 = 总路程 / 总时间。代入数据：速度 = 450 m ÷ 30 s = 15 m/s。一定别忘写单位，缺单位会丢分。

2. Interpreting a Distance–Time Graph | 距离—时间图解读

Question: The graph shows a car’s journey. Describe its motion between 0–10 s, 10–30 s, and 30–50 s.

题目：图形展示了一辆汽车的行程，分别描述它在0–10秒、10–30秒、30–50秒的运动状态。

From 0 to 10 s the line is straight and sloping upward, so the car moves at a constant speed. Between 10 and 30 s the line is horizontal, meaning the car is stationary. From 30 to 50 s the line slopes downward, showing the car returns to the start at constant speed. The gradient of a distance–time graph gives speed; zero gradient means zero speed.

0–10秒线段是向上倾斜的直线，表示汽车匀速行驶；10–30秒线段水平，汽车静止；30–50秒线段向下倾斜，汽车以恒定速度返回起点。距离—时间图的斜率代表速度，斜率为零则速度为零。

3. Newton’s First Law and Inertia | 牛顿第一定律与惯性

Question: Explain why a passenger lurches forward when a bus brakes suddenly.

题目：解释为何公共汽车急刹车时乘客会向前倾。

Newton’s First Law says an object continues in its state of rest or uniform motion unless acted on by a resultant force. The passenger’s body was moving forward with the bus. When the bus stops suddenly, the lower body is stopped by the seatbelt or friction, but the upper body continues moving forward due to inertia. This makes the passenger lurch forward.

牛顿第一定律指出，物体在不受外力作用时将保持静止或匀速直线运动状态。乘客的身体原本与公交车一起向前运动。公交车突然停下时，下半身被安全带或摩擦力拉住，但上半身因惯性继续向前运动，所以乘客会向前倾。

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

Question: A 1200 kg car experiences a driving force of 3000 N and a resistive force of 600 N. Calculate its acceleration.

题目：一辆质量为1200 kg的汽车受到3000 N的驱动力和600 N的阻力，计算它的加速度。

First find the resultant force: 3000 N – 600 N = 2400 N forwards. Then apply F = m a, rearranged to a = F / m: a = 2400 N / 1200 kg = 2 m/s². Remember that acceleration is a vector; state its direction (forwards) if asked.

先求合力：3000 N – 600 N = 2400 N，方向向前。再使用 F = m a，变形为 a = F / m：a = 2400 N ÷ 1200 kg = 2 m/s²。加速度是矢量，题目若要求说明方向，记得写上“向前”。

5. Work Done and Gravitational Potential Energy | 做功与重力势能

Question: A 50 kg student climbs a flight of stairs 12 m high. Calculate the work done and the gain in gravitational potential energy. (g = 10 m/s²)

题目：一名50 kg的学生爬上一段12 m高的楼梯。计算他做的功和增加的重力势能。（取 g = 10 m/s²）

Work done = force × distance = weight × height. Weight = 50 kg × 10 m/s² = 500 N. Work done = 500 N × 12 m = 6000 J. The gain in gravitational potential energy (GPE) equals the work done against gravity, so GPE = 6000 J. Energy and work are both measured in joules.

做功 = 力 × 距离 = 重力 × 高度。重力 = 50 kg × 10 m/s² = 500 N。做功 = 500 N × 12 m = 6000 J。增加的重力势能等于克服重力所做的功，所以 GPE = 6000 J。功和能量单位都是焦耳。

6. Kinetic Energy and Velocity | 动能与速度

Question: An arrow of mass 0.15 kg is fired with kinetic energy 75 J. Find its speed.

题目：一支质量为0.15 kg的箭射出时具有75 J动能，求它的速度。

Kinetic energy formula: KE = ½ m v². Rearranging gives v = √(2 × KE / m). Substitute: v = √(2 × 75 J / 0.15 kg) = √(150 / 0.15) = √1000 ≈ 31.6 m/s. Always square the velocity first when using the formula forwards, and remember to take the square root when solving for v.

动能公式：KE = ½ m v²。变形得 v = √(2 × KE / m)。代入：v = √(2 × 75 J ÷ 0.15 kg) = √(150 ÷ 0.15) = √1000 ≈ 31.6 m/s。正向使用公式时先平方速度，求速度时记得开平方根。

7. Wave Properties: Frequency and Period | 波的性质：频率与周期

Question: A wave generator produces 20 complete waves in 5 seconds. Find the frequency and the period of the wave.

题目：一台造波机在5秒内产生了20个完整波形，求波的频率和周期。

Frequency = number of waves / time = 20 / 5 s = 4 Hz. Period is the time for one complete wave: T = 1 / f = 1 / 4 Hz = 0.25 s. The period can also be found directly: 5 s / 20 waves = 0.25 s per wave. These two quantities are reciprocals of each other.

频率 = 波数 / 时间 = 20 ÷ 5 s = 4 Hz。周期是一个完整波所用的时间：T = 1 / f = 1 / 4 Hz = 0.25 s。也可直接由 5 s ÷ 20 波 = 0.25 s/波得到。频率与周期互为倒数。

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

Question: Two resistors, 4 Ω and 6 Ω, are connected first in series, then in parallel across a 12 V battery. Compare the total resistance and the current from the battery in each case.

题目：两个电阻，4 Ω 和 6 Ω，先串联后并联，分别接在12 V电池两端。比较每种情况的总电阻和电池输出的总电流。

Connection	Total resistance	Current (I = V/R)
Series	R = 4 + 6 = 10 Ω	I = 12 / 10 = 1.2 A
Parallel	1/R = 1/4 + 1/6 = 5/12 → R = 12/5 = 2.4 Ω	I = 12 / 2.4 = 5 A

In series the total resistance is larger, so current is smaller. In parallel the total resistance is smaller, drawing a larger current. This matches the fact that parallel branches provide extra paths for current.

串联时总电阻更大，电流更小；并联时总电阻更小，电流更大。这符合并联支路为电流提供额外通路的事实。

9. Mains Electricity and Power | 市电与电功率

Question: An electric heater labelled 230 V, 2000 W is connected to the mains. Calculate the current it draws and the energy transferred if it runs for 2 hours. Give the energy in kilowatt-hours.

题目：标有230 V、2000 W的电暖器接入市电。计算它取用的电流，以及连续工作2小时传递的能量，并用千瓦时表示。

Use P = I V, so I = P / V = 2000 W / 230 V ≈ 8.7 A. Energy transferred = power × time = 2000 W × (2 × 3600 s) = 14 400 000 J. In kilowatt-hours: power in kW = 2 kW, time = 2 h, so energy = 2 kW × 2 h = 4 kWh. The kWh is a more convenient unit for large electrical energies.

使用 P = I V，得 I = P / V = 2000 W ÷ 230 V ≈ 8.7 A。传递的能量 = 功率 × 时间 = 2000 W × (2 × 3600 s) = 14 400 000 J。以千瓦时表示：功率为2 kW，时间2 h，能量 = 2 kW × 2 h = 4 kWh。千瓦时是度量大量电能的更方便单位。

10. Radioactive Decay and Half-Life | 放射性衰变与半衰期

Question: A sample of a radioactive isotope has an initial count rate of 800 counts per minute. Its half-life is 3 hours. What will the count rate be after 9 hours? Explain what half-life means.

题目：某种放射性同位素样品的初始计数率为每分钟800次，半衰期为3小时。求9小时后的计数率，并解释半衰期的含义。

Half-life is the time taken for half of the radioactive nuclei in a sample to decay, or for the activity to halve. After one half-life (3 h): 800 → 400 cpm. After two half-lives (6 h): 400 → 200 cpm. After three half-lives (9 h): 200 → 100 cpm. The count rate drops to 100 counts per minute. Background count rate is usually subtracted first in such questions; here it is assumed negligible.

半衰期是指样品中一半的放射性原子核发生衰变，或者活度减半所需的时间。经过一个半衰期（3 h）：800 → 400 次/分；两个半衰期（6 h）：400 → 200 次/分；三个半衰期（9 h）：200 → 100 次/分。计数率降至每分钟100次。此类题目通常需先扣除本底计数率，此处假设可忽略。

11. Electromagnetic Spectrum Order | 电磁波谱排序

Question: List the following in order of increasing frequency: microwaves, visible light, X-rays, radio waves, ultraviolet.

题目：将以下电磁波按频率递增的顺序排列：微波、可见光、X射线、无线电波、紫外线。

The electromagnetic spectrum in order of increasing frequency (and decreasing wavelength) is: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. So the correct order from the given list is: radio waves, microwaves, visible light, ultraviolet, X-rays. Remember that frequency and energy are proportional: higher frequency means higher photon energy.

电磁波谱按频率递增（波长递减）的顺序为：无线电波、微波、红外线、可见光、紫外线、X射线、伽马射线。因此题目所列的正确顺序是：无线电波、微波、可见光、紫外线、X射线。记住频率与能量成正比：频率越高，光子能量越大。

12. Using Circuit Symbols to Draw Diagrams | 用电路符号绘制电路图

Question: Draw a circuit containing a cell, a switch, an ammeter in series with a lamp, and a voltmeter measuring the potential difference across the lamp.

题目：画一个包含电池、开关、与灯泡串联的电流表以及测量灯泡两端电压的电压表的电路图。

Key rules: the ammeter must be in series so all current passes through it; the voltmeter must be in parallel with the lamp. Start by drawing the cell, then add the switch in series. Connect the ammeter in series with the lamp, then complete the series loop back to the cell. Finally, add the voltmeter across the lamp, making sure it does not break the series circuit. Use standard symbols: a single cell (long and short line), a switch (open or closed), a circle with ‘A’ for ammeter, a circle with ‘V’ for voltmeter, and a circle with an ‘X’ for lamp.

关键规则：电流表必须串联，让全部电流通过它；电压表必须并联在灯泡两端。先画出电池，接着串联开关，再将电流表与灯泡串联，然后闭合回路回到电池。最后将电压表并联在灯泡两端，确保其不切断串联回路。使用标准符号：单节电池（一长一短竖线）、开关（断开或闭合）、圆圈内写 A 是电流表、V 是电压表、打叉的圆圈表示灯泡。

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Year 10 CCEA Physics: Unit Test Mock Paper Walkthrough | Year 10 CCEA 物理：单元测试模拟卷解析