📚 Year 10 CAIE Physics: Unit Test Mock Paper Walkthrough | 十年级 CAIE 物理：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test for Year 10 CAIE Physics. It covers key topics such as kinematics, forces, energy, thermal physics, waves, electricity and radioactivity. Each question is accompanied by step-by-step solutions in both English and Chinese to reinforce understanding and exam technique.
本文提供了十年级 CAIE 物理单元测试模拟卷的详细解析。内容涵盖运动学、力、能量、热物理学、波、电学和放射性等核心主题。每个问题都配有中英双语的逐步解答，以加深理解并提升应试技巧。
1. Kinematics – Speed and Acceleration | 运动学 – 速度和加速度
A car accelerates uniformly from rest. Its velocity-time graph is a straight line reaching 20 m s⁻¹ in 5 s.
一辆汽车从静止开始匀加速。它的速度-时间图像是一条直线，在5秒内达到20 m s⁻¹。
Calculate the acceleration. Acceleration a = (v – u) / t = (20 – 0) / 5 = 4 m s⁻².
计算加速度。加速度 a = (v – u) / t = (20 – 0) / 5 = 4 m s⁻²。
Calculate the distance travelled in 5 s. Distance = area under velocity-time graph = ½ × base × height = ½ × 5 s × 20 m s⁻¹ = 50 m.
计算5秒内行驶的距离。距离 = 速度-时间图下的面积 = ½ × 底 × 高 = ½ × 5 s × 20 m s⁻¹ = 50 m。
a = 4 m s⁻², s = 50 m
2. Forces and Newton’s Second Law | 力与牛顿第二定律
A box of mass 10 kg rests on a smooth horizontal surface. A horizontal force of 30 N acts on it.
一个质量为10 kg的箱子静止在光滑水平面上，受到30 N的水平推力。
Find the acceleration. Using F = m a, a = F / m = 30 N / 10 kg = 3 m s⁻².
求加速度。由 F = m a 得 a = F / m = 30 N / 10 kg = 3 m s⁻²。
If a frictional force of 10 N opposes the motion, the net force becomes (30 – 10) = 20 N. New acceleration a’ = 20 N / 10 kg = 2 m s⁻².
如果运动受到10 N的摩擦力阻碍，合外力变为 (30 – 10) = 20 N。新加速度 a’ = 20 N / 10 kg = 2 m s⁻²。
Without friction: a = 3 m s⁻²; With friction: a = 2 m s⁻²
3. Energy, Work and Power | 能量、功和功率
A object of mass 2 kg is released from rest at a height of 6 m above the ground. Take g = 10 m s⁻².
一个质量为2 kg的物体从离地6 m高处由静止释放。取 g = 10 m s⁻²。
Calculate the initial gravitational potential energy. Eₚ = m g h = 2 kg × 10 m s⁻² × 6 m = 120 J.
计算初始重力势能。Eₚ = m g h = 2 kg × 10 m s⁻² × 6 m = 120 J。
Neglecting air resistance, all this energy converts to kinetic energy just before impact. Eₖ = 120 J. Using Eₖ = ½ m v², v = √(2Eₖ / m) = √(240 / 2) = √120 ≈ 10.95 m s⁻¹.
忽略空气阻力，这些能量在撞击前全部转化为动能。Eₖ = 120 J。由 Eₖ = ½ m v²，v = √(2Eₖ / m) = √(240 / 2) = √120 ≈ 10.95 m s⁻¹。
If the fall lasts exactly 2 s, the average power delivered by gravity is P = work done / time = 120 J / 2 s = 60 W.
若下落过程恰好持续2 s，重力做功的平均功率为 P = 做功 / 时间 = 120 J / 2 s = 60 W。
Eₚ = 120 J, v ≈ 10.95 m s⁻¹, P = 60 W
4. Density and Pressure | 密度和压强
A metal block has a mass of 540 g and a volume of 200 cm³. It is placed on a horizontal table with a contact area of 0.02 m².
一金属块的质量为540 g，体积为200 cm³。它平放在水平桌面上，接触面积为0.02 m²。
Calculate its density in kg m⁻³. First convert: mass m = 0.54 kg, volume V = 200 cm³ = 200 × 10⁻⁶ m³ = 2.0 × 10⁻⁴ m³. Density ρ = m / V = 0.54 kg / (2.0 × 10⁻⁴ m³) = 2700 kg m⁻³.
计算其密度，单位 kg m⁻³。单位换算：质量 m = 0.54 kg，体积 V = 200 cm³ = 200 × 10⁻⁶ m³ = 2.0 × 10⁻⁴ m³。密度 ρ = m / V = 0.54 kg / (2.0 × 10⁻⁴ m³) = 2700 kg m⁻³。
The force exerted on the table is its weight: F = m g = 0.54 kg × 10 m s⁻² = 5.4 N. Pressure p = F / A = 5.4 N / 0.02 m² = 270 Pa.
金属块对桌面的作用力为其重力：F = m g = 0.54 kg × 10 m s⁻² = 5.4 N。压强 p = F / A = 5.4 N / 0.02 m² = 270 Pa。
ρ = 2700 kg m⁻³, p = 270 Pa
5. Specific Heat Capacity | 比热容
A 0.8 kg aluminium block is heated by an electric heater that supplies 9600 J of energy. The temperature rises from 22 °C to 42 °C.
一个0.8 kg的铝块被电加热器提供了9600 J能量，温度从22 °C升高到42 °C。
Calculate the specific heat capacity of aluminium. Δθ = 42 – 22 = 20 °C. Using E = m c Δθ, c = E / (m Δθ) = 9600 J / (0.8 kg × 20 °C) = 600 J kg⁻¹ °C⁻¹.
计算铝的比热容。Δθ = 42 – 22 = 20 °C。由 E = m c Δθ 得 c = E / (m Δθ) = 9600 J / (0.8 kg × 20 °C) = 600 J kg⁻¹ °C⁻¹。
Explain why a real experiment might give a slightly higher value for the specific heat capacity. This is because some energy is lost to the surroundings as heat, so more energy is recorded to achieve the same temperature rise, making the calculated c appear larger.
解释为什么实际实验测得的比热容值可能偏大。这是因为部分能量以热的形式散失到周围环境中，因此要获得相同的温升需记录更多的能量，使得计算出的 c 值偏大。
c = 600 J kg⁻¹ °C⁻¹
6. Waves – Frequency, Wavelength and Speed | 波 – 频率、波长和波速
A sound wave travels at 340 m s⁻¹ with a frequency of 250 Hz.
一列声波以340 m s⁻¹的速度传播，频率为250 Hz。
Calculate the wavelength. Using the wave equation v = f λ, λ = v / f = 340 m s⁻¹ / 250 Hz = 1.36 m.
计算波长。由波速公式 v = f λ，λ = v / f = 340 m s⁻¹ / 250 Hz = 1.36 m。
If the frequency is doubled to 500 Hz while the speed remains the same, the new wavelength becomes λ’ = 340 / 500 = 0.68 m. The wavelength is halved.
若频率加倍到500 Hz而波速不变，新波长 λ’ = 340 / 500 = 0.68 m。波长减半。
λ = 1.36 m, when f doubles, λ halves
7. Electricity – Ohm’s Law and Circuits | 电学 – 欧姆定律与电路
A resistor has a potential difference of 6.0 V across it and a current of 0.5 A flowing through it.
一个电阻两端的电压为6.0 V，通过的电流为0.5 A。
Calculate its resistance. R = V / I = 6.0 V / 0.5 A = 12 Ω.
计算其电阻。R = V / I = 6.0 V / 0.5 A = 12 Ω。
Two identical such resistors are connected in parallel across a 6.0 V supply. The total resistance R_total: 1/R_total = 1/12 + 1/12 = 2/12, so R_total = 6 Ω. The total current I_total = V / R_total = 6.0 V / 6 Ω = 1 A. Alternatively, since each branch carries 0.5 A, total current = 1 A.
两个相同的这种电阻并联后接在6.0 V电源上。总电阻 R_total：1/R_total = 1/12 + 1/12 = 2/12，故 R_total = 6 Ω。总电流 I_total = V / R_total = 6.0 V / 6 Ω = 1 A。或者，由于每条支路通过0.5 A，总电流为1 A。
Single R = 12 Ω; parallel: R_total = 6 Ω, I_total = 1 A
8. Radioactive Decay and Half-life | 放射性衰变和半衰期
A radioactive isotope has a half-life of 4 minutes. Its initial activity is 800 Bq.
一种放射性同位素的半衰期为4分钟，初始活度为800 Bq。
How many minutes pass until the activity falls to 100 Bq? Every half-life the activity halves: 800 → 400 (after 4 min) → 200 (8 min) → 100 (12 min). So 12 minutes are required.
活度降到100 Bq需要多少分钟？每经过一个半衰期活度减半：800 → 400 (4分钟后) → 200 (8分钟) → 100 (12分钟)。所以需要12分钟。
Determine the activity after 8 minutes. After 8 minutes, two half-lives have elapsed, so activity = 800 Bq / 2² = 200 Bq.
求8分钟后的活度。8分钟后，经历了两个半衰期，活度 = 800 Bq / 2² = 200 Bq。
12 min to reach 100 Bq; activity at 8 min = 200 Bq
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