📚 Year 9 OCR Physics: Unit Test Mock Paper Walkthrough | Year 9 OCR物理：单元测试模拟卷解析
This walkthrough covers a typical Year 9 OCR Physics unit test, breaking down each question with model answers, key revision points and common pitfalls. Use it to check your understanding and sharpen your exam technique.
本解析涵盖了一份典型的 Year 9 OCR 物理单元测试卷，逐题给出标准答案、关键复习点与常见错误分析。用它来检验你的理解并提升应试技巧。
1. Question 1: Kinetic Energy | 第1题：动能
A car of mass 1200 kg is travelling at a speed of 15 m/s. Calculate its kinetic energy. [2 marks]
一辆质量为1200 kg的汽车以15 m/s的速度行驶。计算它的动能。[2分]
Write the kinetic energy formula and substitute the values carefully. Remember that speed must be squared before multiplying.
写出动能公式并仔细代入数据。记住速度要先平方再与质量相乘。
KE = ½mv²
KE = ½ × 1200 × (15)² = ½ × 1200 × 225 = 600 × 225 = 135 000 J (135 kJ)
The kinetic energy is 135 000 J, which can also be written as 135 kJ. Always include the correct unit; missing units often lose a mark.
动能是135 000 J，也可写成135 kJ。务必写上正确的单位；漏写单位通常会扣分。
2. Question 2: Resistance Calculation | 第2题：电阻计算
A student connects a resistor to a power supply. The potential difference across the resistor is 6 V and the current flowing through it is 0.4 A. Calculate the resistance. [2 marks]
一名学生将电阻连接到电源上。电阻两端的电势差为6 V，流过电阻的电流为0.4 A。计算电阻值。[2分]
Apply Ohm's law in its rearranged form. Make sure you use the correct equation and units.
应用欧姆定律的变形公式。确保使用正确的公式和单位。
R = V ÷ I
R = 6 V ÷ 0.4 A = 15 Ω
The resistance is 15 Ω. In a 'show that' or calculation question, always write the formula, show the substitution and give the answer with the unit ohm (Ω).
电阻为15 Ω。在“证明”或计算题中，始终写出公式、代入步骤并给出带单位欧姆(Ω)的答案。
3. Question 3: Density | 第3题：密度
A metal block has a mass of 540 g and a volume of 200 cm³. Calculate its density in g/cm³. [2 marks]
一金属块的质量为540 g，体积为200 cm³。计算其密度，单位为g/cm³。[2分]
Use the density equation. The units g/cm³ are already consistent, so no conversion is needed.
使用密度公式。单位g/cm³已是一致的，无需换算。
density = mass ÷ volume
density = 540 g ÷ 200 cm³ = 2.7 g/cm³
The density is 2.7 g/cm³. This value is typical of aluminium. Students often confuse mass and weight – remember density depends on mass, not weight.
密度为2.7 g/cm³。该值对应典型的铝。同学们常混淆质量与重量——请记住密度取决于质量而非重量。
4. Question 4: Wave Speed | 第4题：波速
A sound wave has a frequency of 50 Hz and a wavelength of 0.6 m. Calculate the wave speed. [2 marks]
一个声波的频率为50 Hz，波长为0.6 m。计算波速。[2分]
Apply the wave equation. Both frequency and wavelength are given in standard units.
应用波动方程。频率和波长都已用标准单位给出。
v = f × λ
v = 50 Hz × 0.6 m = 30 m/s
The wave speed is 30 m/s. Always check that frequency is in hertz (Hz) and wavelength in metres (m) to get speed in m/s.
波速为30 m/s。务必检查频率的单位为赫兹(Hz)、波长为米(m)，这样得到的波速单位才是m/s。
5. Question 5: Gas Pressure and Particle Motion | 第5题：气体压强与粒子运动
Explain, in terms of particles, why the pressure of a fixed mass of gas increases when its temperature is raised while the volume remains constant. [3 marks]
用粒子观点解释，为什么一定质量的气体在体积不变时，温度升高会导致压强增大。[3分]
Raising the temperature gives the gas particles more kinetic energy. This influences both the frequency and the force of collisions.
升高温度使气体粒子获得更多的动能。这会影响碰撞的频率和力度。
The particles move faster on average. They collide with the walls of the container more often. Each collision also exerts a larger force because the particles are moving with greater momentum.
粒子平均运动速度加快。它们更频繁地撞击容器壁。由于粒子动量更大，每次碰撞施加的力也更大。
Since pressure is caused by the total force of particle impacts per unit area, a higher collision rate and greater impact force cause the pressure to rise.
由于压强是由单位面积上粒子碰撞的总力产生的，更高的碰撞率和更大的冲击力就会导致压强增大。
6. Question 6: Work Done and Power | 第6题：做功与功率
onds. Use g = 10 N/kg. Calculate (a) the work done, and (b) the power developed. [4 marks]
一台起重机在20秒内将500 kg的重物竖直吊起10 m。取g = 10 N/kg。计算 (a) 所做的功，以及 (b) 产生的功率。[4分]
First find the weight (force) needed to lift the load. Work done is this force multiplied by the distance moved in the direction of the force.
先求出吊起重物所需的力（重量）。做功就是这个力乘以沿力的方向移动的距离。
weight = m × g = 500 × 10 = 5 000 N
work done = force × distance = 5 000 N × 10 m = 50 000 J (50 kJ)
Power is the rate of doing work. Divide the work done by the time taken.
功率是做功的快慢。用所做的功除以所用时间。
power = work ÷ time = 50 000 J ÷ 20 s = 2 500 W (2.5 kW)
Therefore, the work done is 50 000 J and the power is 2 500 W. Many students forget to convert mass into weight first – a common mistake.
因此，做功为50 000 J，功率为2 500 W。很多同学会忘记先将质量换算为重量——这是一个常见错误。
7. Question 7: Energy Transfers in a Torch | 第7题：手电筒中的能量转移
Describe the useful and wasted energy transfers that take place when a battery-powered torch is switched on. [3 marks]
描述当使用电池的手电筒打开时，发生的有用能量转移和浪费的能量转移。[3分]
Energy is transferred from the chemical energy store of the battery to the electrical pathway in the wires.
能量从电池的化学能储存转移到导线中的电路径。
In the bulb, most of the electrical energy is converted into light (useful output) but some is always transferred to the thermal energy store of the surroundings (wasted).
在灯泡中，大部分电能转化为光（有用输出），但总有一部分转移到周围环境的热能储存（浪费）。
questions, always name the energy stores at the start and end, and the pathway in between.
因此能量转移路径为：化学能 → 电能 → 光能 + 热能。在OCR的考题中，一定要说出起始与最终的能量储存以及中间的能量转移方式。
8. Question 8: Refraction of Light | 第8题：光的折射
A ray of light travels from air into glass. State what happens to its speed and its direction. [2 marks]
一束光从空气斜射入玻璃。说明它的速度和方向会发生什么变化。[2分]
When light enters a denser medium, it slows down. The change in speed causes the ray to change direction unless it strikes the boundary along the normal.
当光进入光密介质时，速度减慢。速度的变化导致光线改变方向，除非是沿法线入射。
The speed of light decreases in glass compared with air. The ray bends towards the normal (the imaginary line perpendicular to the surface).
与空气相比，光在玻璃中的速度减小。光线会向法线（垂直于界面的假想线）偏折。
Always use the terms 'towards the normal' and 'slows down' – these are the exact phrases OCR examiners expect to see.
一定要使用“向法线偏折”和“减慢”——这正是OCR考官期望看到的关键词。
9. Question 9: Filament Lamp I-V Characteristic | 第9题：白炽灯丝的 I-V 特性
A student investigates a filament lamp. Sketch the current-voltage (I-V) graph and explain why it is not a straight line. [3 marks]
一名学生研究了一个白炽灯丝灯泡。画出电流-电压（I-V）特性曲线的示意图，并解释为什么它不是一条直线。[3分]
The graph starts steep and then curves, becoming less steep as the voltage increases. This shape shows that the resistance of the lamp increases with current.
图像开始较陡，然后随着电压增大而弯曲，变得越来越平缓。这个形状表明灯泡的电阻随电流增大而增大。
As the current flows through the filament, it heats up. The metal filament's resistance increases with temperature. Therefore, for the same increase in voltage, the current increases by a smaller amount at higher voltages.
当电流通过灯丝时，灯丝温度升高。金属灯丝的电阻随温度升高而增大。因此，在较高电压下，相同电压增量所引起的电流增加量较小。
The resulting curve is not a straight line through the origin, making the filament lamp a non-ohmic conductor. In your sketch, label axes (current vs voltage) and show the curved shape clearly.
所得到的曲线不是一条过原点的直线，因此白炽灯丝属于非线性电阻。在你的示意图中，标出坐标轴（电流-电压），并清晰画出弯曲的形状。
Published by TutorHao | Physics Revision Series | aleveler.com
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