GCSE CCEA Physics: Worked Examples Explained | GCSE CCEA 物理：典型例题详解

📚 GCSE CCEA Physics: Worked Examples Explained | GCSE CCEA 物理：典型例题详解

This article walks you through common GCSE CCEA Physics exam questions, demonstrating step-by-step solutions and key concepts. Mastering worked examples is one of the most effective ways to prepare for your examination.

本文将通过逐步解析 GCSE CCEA 物理常见考题，帮助你掌握解题方法和核心概念。攻克典型例题是备考最有效的方式之一。

1. Motion: Acceleration & Distance from a v-t Graph | 运动：从速度-时间图求加速度与距离

A car starts from rest and accelerates uniformly to 20 m/s in 10 seconds. It then travels at a constant speed of 20 m/s for 20 seconds, before decelerating uniformly to rest in a further 10 seconds. (a) Calculate the acceleration during the first 10 s. (b) Calculate the total distance travelled by the car.

一辆汽车从静止开始匀加速，在10秒内达到20 m/s。然后以20 m/s的恒定速度行驶20秒，最后在10秒内匀减速至静止。(a) 计算前10秒的加速度。(b) 计算汽车行驶的总距离。

(a) Acceleration = change in velocity / time taken. a = (v – u) / t = (20 – 0) / 10 = 2 m/s².

(a) 加速度 = 速度变化量 ÷ 时间。a = (v – u) / t = (20 – 0) / 10 = 2 m/s²。

(b) The distance travelled is the area under the velocity-time graph. The area can be split into three shapes: a triangle (0–10 s), a rectangle (10–30 s) and a triangle (30–40 s). Total distance = (½ × 10 × 20) + (20 × 20) + (½ × 10 × 20) = 100 + 400 + 100 = 600 m.

(b) 行驶距离等于速度-时间图下的面积。该面积可分成三个图形：一个三角形（0–10秒）、一个矩形（10–30秒）和一个三角形（30–40秒）。总距离 = (½ × 10 × 20) + (20 × 20) + (½ × 10 × 20) = 100 + 400 + 100 = 600 m。

You can also use the trapezium area formula: ½ × (sum of parallel sides) × height. The parallel sides are the velocities at 0 s and 40 s (both 0 m/s) and the total time is 40 s, but for non-zero velocities it is easier to use the area method shown.

你也可以用梯形面积公式：½ × (平行边之和) × 高。平行边即0秒和40秒时的速度（均为零），总时长为40秒，但对于非零速度段，还是用上述面积法更直观。

2. Forces: Newton’s Second Law | 力：牛顿第二定律

A car of mass 1200 kg accelerates at 2.5 m/s². (a) Calculate the resultant force on the car. (b) If a resistive force of 500 N acts against the motion, what driving force must the engine provide?

一辆质量为1200 kg的汽车以2.5 m/s²加速。(a) 计算作用在车上的合力。(b) 如果运动过程中存在500 N的阻力，发动机需要提供多大的驱动力？

(a) Resultant force F = m × a = 1200 kg × 2.5 m/s² = 3000 N.

(a) 合力 F = m × a = 1200 kg × 2.5 m/s² = 3000 N。

(b) Driving force – resistive force = resultant force. Therefore, driving force = resultant force + resistive force = 3000 N + 500 N = 3500 N.

(b) 驱动力 – 阻力 = 合力。因此，驱动力 = 合力 + 阻力 = 3000 N + 500 N = 3500 N。

3. Energy: Kinetic Energy and Work Done | 能量：动能与做功

A cyclist and bicycle have a combined mass of 80 kg. The cyclist accelerates from 5 m/s to 15 m/s. Calculate the increase in kinetic energy. State the work done by the cyclist.

一位骑行者与自行车的总质量为80 kg。他从5 m/s加速到15 m/s。计算动能的增加量，并说出骑行者所做的功。

Kinetic energy KE = ½ m v². Initial KE = ½ × 80 × 5² = 40 × 25 = 1000 J. Final KE = ½ × 80 × 15² = 40 × 225 = 9000 J. Increase in KE = 9000 – 1000 = 8000 J.

动能 KE = ½ m v²。初始动能 = ½ × 80 × 5² = 40 × 25 = 1000 J。末动能 = ½ × 80 × 15² = 40 × 225 = 9000 J。动能增加量 = 9000 – 1000 = 8000 J。

Assuming no energy losses, the work done by the cyclist equals the increase in kinetic energy, so work done = 8000 J.

假设没有能量损失，骑行者所做的功等于动能的增加量，因此做功 = 8000 J。

4. Electricity: Series Resistance and Ohm’s Law | 电学：串联电阻与欧姆定律

A 4 Ω resistor and a 6 Ω resistor are connected in series to a 12 V battery. Calculate: (a) the total resistance, (b) the current in the circuit, (c) the voltage across each resistor.

一个4 Ω和一个6 Ω的电阻串联后接在12 V电池上。计算：(a) 总电阻，(b) 电路中的电流，(c) 每个电阻两端的电压。

(a) In series, total resistance R_total = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω.

(a) 串联时，总电阻 R_total = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω。

(b) Using Ohm’s Law, current I = V / R_total = 12 V / 10 Ω = 1.2 A.

(b) 根据欧姆定律，电流 I = V / R_total = 12 V / 10 Ω = 1.2 A。

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

A water wave has a frequency of 5 Hz and a wavelength of 0.4 m. Calculate the wave speed and the period of the wave.

一个水波的频率为5 Hz，波长为0.4 m。计算波速和波的周期。

Wave speed v = f × λ = 5 Hz × 0.4 m = 2 m/s.

波速 v = f × λ = 5 Hz × 0.4 m = 2 m/s。

Period T = 1 / f = 1 / 5 Hz = 0.2 s. The wave completes one full oscillation every 0.2 seconds.

周期 T = 1 / f = 1 / 5 Hz = 0.2 s。波每0.2秒完成一次全振动。

6. Radioactivity: Half-life Calculations | 放射性：半衰期计算

A radioactive source has an initial count rate of 800 counts per minute. Its half-life is 2 hours. What will the count rate be after 6 hours?

某放射源初始计数率为每分钟800次，半衰期为2小时。求6小时后的计数率。

Number of half-lives = total time / half-life = 6 hours / 2 hours = 3 half-lives.

半衰期个数 = 总时间 ÷ 半衰期 = 6小时 ÷ 2小时 = 3个半衰期。

After each half-life, the count rate halves. Count rate = 800 × (½)³ = 800 × 1/8 = 100 counts per minute.

每经过一个半衰期，计数率减半。计数率 = 800 × (½)³ = 800 × 1/8 = 100 次/分钟。

Alternative table method:

也可用表格法：

Half-lives elapsed	0	1	2	3
Count rate (min⁻¹)	800	400	200	100

7. Moments: Principle of Moments | 力矩：力矩原理

A uniform metre rule of weight 1.0 N is pivoted at its centre (50 cm mark). A 2.0 N weight is hung at the 20 cm mark. Determine at which mark a 3.0 N weight must be hung to balance the rule horizontally.

一根重量为1.0 N的均匀米尺在其中心（50 cm刻度处）支起。在20 cm刻度处悬挂2.0 N的重物。若要使米尺水平平衡，一个3.0 N的重物应悬挂在哪个刻度处？

The weight of the rule acts at the pivot, so it produces zero moment about the pivot. Anticlockwise moment = 2.0 N × (50 – 20) cm = 2.0 N × 30 cm = 60 N cm.

米尺自身的重力作用在支点上，因此对支点不产生力矩。逆时针力矩 = 2.0 N × (50 – 20) cm = 2.0 N × 30 cm = 60 N cm。

For balance, clockwise moment = anticlockwise moment. Let d be the distance from the pivot to the 3.0 N weight (on the right side). 3.0 N × d = 60 N cm → d = 20 cm. So the 3.0 N weight must be placed at the (50 + 20) cm = 70 cm mark.

平衡时，顺时针力矩 = 逆时针力矩。设3.0 N重物到支点的距离为d（在右侧），3.0 N × d = 60 N cm → d = 20 cm。因此3.0 N重物应挂在(50 + 20) cm = 70 cm刻度处。

8. Density and Pressure | 密度与压强

A metal block has a mass of 0.5 kg and a volume of 2.0 × 10⁻⁴ m³. (a) Calculate its density. (b) The block rests on a flat surface with a base area of 10 cm². Calculate the pressure it exerts on the surface. (Take g = 10 N/kg).

一个金属块质量为0.5 kg，体积为2.0 × 10⁻⁴ m³。(a) 计算其密度。(b) 该金属块平放在面积为10 cm²的表面上，求它对表面产生的压强。（取g = 10 N/kg）

(a) Density ρ = mass / volume = 0.5 kg / (2.0 × 10⁻⁴ m³) = 2500 kg/m³.

(a) 密度 ρ = 质量 / 体积 = 0.5 kg / (2.0 × 10⁻⁴ m³) = 2500 kg/m³。

(b) Weight = m × g = 0.5 kg × 10 N/kg = 5 N. Area in m²: 10 cm² = 10 × 10⁻⁴ m² = 10⁻³ m². Pressure p = Force / Area = 5 N / 10⁻³ m² = 5000 Pa.

(b) 重量 = m × g = 0.5 kg × 10 N/kg = 5 N。面积以平方米计：10 cm² = 10 × 10⁻⁴ m² = 10⁻³ m²。压强 p = 力 / 面积 = 5 N / 10⁻³ m² = 5000 Pa。

9. Hooke’s Law | 胡克定律

A spring extends by 2.0 cm when a force of 5.0 N is applied. (a) Calculate the spring constant k. (b) How much would the spring extend if an 8.0 N load is hung from it?

一个弹簧在受到5.0 N的力时伸长了2.0 cm。(a) 计算弹簧劲度系数k。(b) 如果挂上8.0 N的负载，该弹簧会伸长多少？

(a) Hooke’s Law: F = k × x, so k = F / x. Convert extension to metres: x = 2.0 cm = 0.02 m. k = 5.0 N / 0.02 m = 250 N/m.

(a) 胡克定律：F = k × x，因此 k = F / x。将伸长量转换为米：x = 2.0 cm = 0.02 m。k = 5.0 N / 0.02 m = 250 N/m。

(b) Using the same spring constant, x = F / k = 8.0 N / 250 N/m = 0.032 m = 3.2 cm. (Assuming the elastic limit is not exceeded.)

(b) 使用同样的劲度系数，x = F / k = 8.0 N / 250 N/m = 0.032 m = 3.2 cm。（假设未超出弹性限度）

10. Specific Heat Capacity | 比热容

An electric heater supplies 10 000 J of energy to a 2.0 kg aluminium block, raising its temperature from 20 °C to 30 °C. Calculate the specific heat capacity of aluminium.

一个电加热器向2.0 kg的铝块提供了10 000 J 的能量，使其温度从20 °C升高到30 °C。计算铝的比热容。

Energy transferred ΔQ = m × c × Δθ. Rearranging: c = ΔQ / (m × Δθ). Temperature change Δθ = 30 °C – 20 °C = 10 °C. c = 10 000 J / (2.0 kg × 10 °C) = 500 J/(kg °C).

能量转移 ΔQ = m × c × Δθ。整理得：c = ΔQ / (m × Δθ)。温度变化 Δθ = 30 °C – 20 °C = 10 °C。c = 10 000 J / (2.0 kg × 10 °C) = 500 J/(kg °C)。

This means 500 joules of energy are needed to raise the temperature of 1 kilogram of aluminium by 1 degree Celsius.

这意味着将1千克铝的温度升高1摄氏度需要500焦耳的能量。

11. Electrical Power and Energy | 电功率与电能

A 60 W filament lamp is left on for 5 hours. Calculate the energy transferred in (a) joules, and (b) kilowatt-hours.

一个60 W的白炽灯持续亮了5小时。计算所消耗的能量，分别以(a)焦耳，(b)千瓦时为单位。

(a) Energy E = Power × time. Seconds in 5 hours = 5 × 3600 s = 18 000 s. E = 60 W × 18 000 s = 1 080 000 J (or 1.08 MJ).

(a) 能量 E = 功率 × 时间。5小时的秒数 = 5 × 3600 s = 18 000 s。E = 60 W × 18 000 s = 1 080 000 J（或1.08 MJ）。

(b) Convert power to kilowatts: 60 W = 0.06 kW. Energy in kWh = power (kW) × time (h) = 0.06 kW × 5 h = 0.3 kWh.

(b) 将功率转换为千瓦：60 W = 0.06 kW。以kWh计的能量 = 功率(kW) × 时间(h) = 0.06 kW × 5 h = 0.3 kWh。

12. Nuclear Equations | 核方程

Complete the following nuclear decay equations. (a) Thorium-234 undergoes beta decay: ²³⁴₉₀Th → ²³⁴₉₁Pa + ? (b) Radium-226 undergoes alpha decay: ²²⁶₈₈Ra → ? + ⁴₂He.

完成下列核衰变方程。(a) 钍-234发生β衰变：²³⁴₉₀Th → ²³⁴₉₁Pa + ? (b) 镭-226发生α衰变：²²⁶₈₈Ra → ? + ⁴₂He。

(a) In beta decay, a neutron turns into a proton and emits an electron (beta particle). The atomic number increases by 1, mass number remains the same. The missing particle is an electron: ⁰₋₁e.

(a) β衰变中，一个中子转化为一个质子并放出一个电子（β粒子）。原子序数增加1，质量数不变。缺失的粒子是电子：⁰₋₁e。

Complete equation: ²³⁴₉₀Th → ²³⁴₉₁Pa + ⁰₋₁e.

完整方程：²³⁴₉₀Th → ²³⁴₉

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GCSE CCEA Physics: Worked Examples Explained | GCSE CCEA 物理：典型例题详解