📚 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。
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– 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。
ight. 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 – e. 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。
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。
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 Rtotal = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω.
(a) 串联时，总电阻 Rtotal = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω。
(b) Using Ohm's Law, current I = V / Rtotal = 12 V / 10 Ω = 1.2 A.
(b) 根据欧姆定律，电流 I = V / Rtotal = 12 V / 10 Ω = 1.2 A。
1.2 A × 4 Ω = 4.8 V; V₆ = I × 6 Ω = 1.2 A × 6 Ω = 7.2 V. (Check: 4.8 V + 7.2 V = 12.0 V).
(c) V₄ = I × 4 Ω = 1.2 A × 4 Ω = 4.8 V；V₆ = I × 6 Ω = 1.2 A × 6 Ω = 7.2 V。（检验：4.8 V + 7.2 V = 12.0 V）
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。
d 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。
-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 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。
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
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。
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' 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。
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。
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。
econds 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）。
ilowatts: 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
llowing 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
Published by TutorHao | GCSE Physics Revision Series | aleveler.com
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