📚 Year 9 Edexcel Physics: Unit Test Mock Paper Breakdown | Year 9 Edexcel 物理：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test for Year 9 Edexcel Physics. Every question is broken down with model answers, key concepts, and common mistakes to avoid. Use this revision resource to boost your confidence and sharpen your exam technique.
本文为 Year 9 Edexcel 物理单元测试模拟卷提供详细解析。每道题都配有标准答案、关键概念和需要避免的常见错误。使用这份复习资料来增强信心，提升应试技巧。
1. Resultant Force and Balanced Forces | 合力与平衡力
Question: A wooden block is pushed to the right with a force of 12 N. Friction acts to the left with a force of 7 N. Calculate the resultant force on the block and state its direction. (2 marks)
题目：一个木块受到向右12 N的推力，同时受到向左7 N的摩擦力。计算木块所受的合力，并说明方向。（2分）
right. When forces act in opposite directions, the resultant acts in the direction of the larger force. This is an unbalanced force, so the block will accelerate to the right according to Newton's second law.
合力通过相减反向的力求得。此处净力 = 12 N（右）− 7 N（左） = 5 N，方向向右。当力方向相反时，合力的方向与较大的力一致。这是一个非平衡力，因此根据牛顿第二定律，木块会向右加速。
Common mistake: adding the forces instead of subtracting, or forgetting to state the direction. Always include both magnitude and direction for a vector quantity.
常见错误：将两个力相加而不是相减，或者忘记指出方向。对于矢量，务必同时给出大小和方向。
2. Calculating Average Speed | 计算平均速度
Question: A cyclist travels 36 km in 2 hours. Calculate her average speed in km/h. Then convert your answer to m/s. (3 marks)
题目：一位自行车骑行者在2小时内行驶了36 km。计算她的平均速度（单位 km/h），再将答案换算成 m/s。（3分）
m/s, multiply by 1000/3600 (or divide by 3.6): 18 ÷ 3.6 = 5 m/s. So average speed = 5 m/s.
使用公式：速度 = 距离 ÷ 时间。以 km/h 为单位：v = 36 km ÷ 2 h = 18 km/h。换算为 m/s 时，乘以 1000/3600（或除以 3.6）：18 ÷ 3.6 = 5 m/s。因此平均速度为 5 m/s。
ber 1 m/s = 3.6 km/h. Many students forget to divide by 3.6 and give 64.8 m/s incorrectly.
避免使用错误的换算系数；记住 1 m/s = 3.6 km/h。许多学生忘记除以 3.6，错误地给出 64.8 m/s。
3. Energy Stores and Transfers | 能量储存与转移
Question: A ball is held 2 m above the ground and then dropped. State the main energy store when the ball is held, just before hitting the ground, and describe the energy transfer. (3 marks)
题目：一个球被举在离地2 m高处然后释放。分别说出球在被举着时、即将撞击地面前的主要能量储存形式，并描述能量的转移过程。（3分）
d, the energy has been mostly converted to kinetic energy (ignoring air resistance). The transfer is: GPE → kinetic energy.
当球被举着时，它具有重力势能。下落过程中，重力势能转化为动能。在即将撞击地面前，能量几乎全部转化为动能（不计空气阻力）。转移过程为：重力势能 → 动能。
Some students may confuse 'energy store' with 'energy transfer'. In Edexcel, stores are kinetic, thermal, GPE, elastic, chemical, etc. Transfers happen mechanically, electrically, by heating, or by radiation.
一些学生可能混淆“能量储存”与“能量转移”。Edexcel 中，储存形式包括动能、热能、重力势能、弹性势能、化学能等。转移方式有机地、电地、加热或辐射。
4. Series Circuit Rules | 串联电路规则
Question: Two identical lamps are connected in series with a 12 V battery. The current through the first lamp is 0.5 A. State the current through the second lamp and the voltage across each lamp. (3 marks)
题目：两盏相同的灯泡与一个12 V 电池串联。通过第一盏灯的电流为0.5 A。说出通过第二盏灯的电流以及每盏灯两端的电压。（3分）
ical, each gets half of the total 12 V, so 6 V per lamp. Total voltage = 6 V + 6 V = 12 V.
在串联电路中，各处电流相等。因此通过第二盏灯的电流也是 0.5 A。电压被分摊：由于灯泡相同，每盏分得总电压的一半，即每盏灯 6 V。总电压 = 6 V + 6 V = 12 V。
s in series, or that both lamps get 12 V. Remember: series = same current, shared voltage.
常见错误是认为串联电路中电流会分流，或两盏灯都得到12 V。请记住：串联 = 电流相同，电压分摊。
5. Transverse and Longitudinal Waves | 横波与纵波
Question: Give two differences between transverse and longitudinal waves. Give one example of each. (4 marks)
题目：给出横波与纵波的两个区别，并各举一个例子。（4分）
In transverse waves, the oscillations are perpendicular to the direction of energy transfer; in longitudinal waves, oscillations are parallel. Transverse waves can travel in a vacuum (e.g., light), whereas longitudinal waves cannot. Examples: light is a transverse wave; sound is a longitudinal wave.
横波中，振动方向与能量传递方向垂直；纵波中，振动方向与能量传递方向平行。横波可以在真空中传播（如光），而纵波不能。例子：光属于横波；声音属于纵波。
Use a simple table to compare:
| Feature | Transverse | Longitudinal |
|---|---|---|
| Oscillation vs direction | Perpendicular | Parallel |
| Can travel through vacuum? | Yes | No |
| Example | Light, water waves | Sound, seismic P-waves |
6. Density of a Solid | 固体的密度
Question: A metal cube has a mass of 540 g and a volume of 200 cm³. Calculate its density in g/cm³ and kg/m³. (3 marks)
题目：一个金属立方体的质量为 540 g，体积为 200 cm³。计算其密度，分别以 g/cm³ 和 kg/m³ 表示。（3分）
1 g/cm³ = 1000 kg/m³). Thus 2.7 × 1000 = 2700 kg/m³. This density is typical of aluminium.
密度 ρ = 质量 ÷ 体积。以 g/cm³ 为单位：ρ = 540 g ÷ 200 cm³ = 2.7 g/cm³。换算为 kg/m³ 时，乘以 1000（因为 1 g/cm³ = 1000 kg/m³），得到 2.7 × 1000 = 2700 kg/m³。该密度作为铝的特征值。
Ensure you use the correct formula and units. Many students forget the conversion or misinterpret the volume.
确保使用正确的公式和单位。许多学生忘记单位换算或误读体积。
7. Hooke's Law and Spring Extension | 胡克定律与弹簧伸长
Question: A spring stretches by 4 cm when a 10 N weight is hung from it. Calculate the spring constant k. If the weight is doubled, what is the new extension? (Assume the limit of proportionality is not exceeded.) (4 marks)
题目：在弹簧下悬挂重 10 N 的物体时，弹簧伸长 4 cm。计算弹簧的劲度系数 k。如果重物重量翻倍，新的伸长量是多少？（假定不超过弹性限度）（4分）
Hooke' n e = F / k = 20 N / 250 N/m = 0.08 m = 8 cm. Extension is directly proportional to force.
胡克定律：F = k × e。先将伸长量换算为米：4 cm = 0.04 m。然后 k = F / e = 10 N / 0.04 m = 250 N/m。如果力翻倍为 20 N，新的伸长量 e = F / k = 20 N / 250 N/m = 0.08 m = 8 cm。伸长量与力成正比。
Common pitfalls: forgetting to convert cm to m, or miscalculating k. Always show unit conversion clearly.
常见陷阱：忘记将 cm 换算为 m，或计算 k 时出错。务必清晰展示单位转换。
8. Sound Waves: Frequency and Amplitude | 声波：频率与振幅
Question: A sound wave produced by a tuning fork has a frequency of 440 Hz. Explain what frequency and amplitude mean in terms of sound. How would the sound change if the frequency is increased? (3 marks)
题目：一支音叉产生 440 Hz 的声波。解释频率和振幅在声音中的含义。如果频率增加，声音会如何变化？（3分）
Frequency is the number of complete oscillations per second, measured in hertz (Hz). In sound, frequency determines pitch: a higher frequency gives a higher pitch. Amplitude is the maximum displacement from the rest position and determines loudness. If frequency is increased, the pitch becomes higher, but loudness remains unchanged if amplitude stays the same.
频率是每秒完整振动的次数，单位为赫兹（Hz）。在声音中，频率决定音调：频率越高，音调越高。振幅是偏离平衡位置的最大位移，决定响度。如果频率增加，音调变高，但如果振幅不变，响度不变。
Students often confuse pitch with loudness. Remember: frequency ↔ pitch, amplitude ↔ loudness.
学生常混淆音调和响度。请记住：频率 ↔ 音调，振幅 ↔ 响度。
9. Work Done by a Force | 力做的功
Question: A person pushes a trolley with a constant force of 50 N over a distance of 8 m. Calculate the work done. If this work is done in 10 seconds, what is the power? (4 marks)
题目：一个人用 50 N 的恒力推动手推车移动了 8 m。计算所做的功。如果这些功在 10 秒内完成，功率是多少？（4分）
me taken = 400 J ÷ 10 s = 40 W. The unit of power is the watt (W), meaning 1 J per second.
功 W = 力 × 力的方向上移动的距离。W = 50 N × 8 m = 400 J。功率 P = 功 ÷ 时间 = 400 J ÷ 10 s = 40 W。功率的单位是瓦特（W），表示每秒 1 焦耳。
Be careful: the distance must be in the direction of the force. If the force is at an angle, only the component along the distance does work, but this is Year 9, so forces are parallel to motion.
注意：距离必须是在力的方向上。如果力与运动方向有夹角，只有平行分量做功，但是 Year 9 阶段力的方向与运动方向一致。
10. Law of Reflection | 光的反射定律
Question: A ray of light strikes a plane mirror at an angle of incidence of 30°. State the angle of reflection. Draw a simple diagram and label the incident ray, reflected ray, and normal. Explain the law of reflection. (4 marks)
题目：一条光线以 30° 的入射角射向平面镜。说出反射角。画简图并标出入射光线、反射光线和法线。解释反射定律。（4分）
The law of reflection states that the angle of incidence equals the angle of reflection, both measured from the normal. Therefore, if the angle of incidence is 30°, the angle of reflection is also 30°. The normal is an imaginary line perpendicular to the mirror surface at the point of incidence.
光的反射定律指出，入射角等于反射角，两者均从法线量起。因此，入射角为 30° 时，反射角也是 30°。法线是在入射点处垂直于镜面的一条假想线。
Many students measure angles from the mirror surface instead of the normal, giving 60°. Always measure from the normal, not the mirror.
许多学生从镜面而不是法线量角，得出 60°。务必从法线量起，而不是镜面。
In a diagram: the incident ray approaches the mirror, the reflected ray leaves, and the normal is drawn as a dashed line. Angles are marked with arcs.
在图中：入射光线射向镜面，反射光线离开，法线用虚线画出。角度用弧线标记。
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