📚 IGCSE Edexcel Physics: Worked Examples Explained | IGCSE Edexcel 物理：典型例题详解
Working through past-paper style problems is one of the most effective ways to master key concepts and boost exam confidence. In this article, we present twelve carefully chosen worked examples covering the major topics of the Edexcel IGCSE Physics specification. Each solution is broken down into logical steps, with clear reasoning in both English and Chinese, helping you to build problem-solving skills while reinforcing essential theory.
通过历年真题风格的例题进行练习是掌握核心概念、提升考试信心的最有效方法之一。本文精选了十二道涵盖 Edexcel IGCSE 物理大纲主要知识点的典型例题，逐一详解。每道题的解答都按逻辑步骤分解，并用中英双语清晰解释，帮助你培养解题能力，同时巩固基本理论。
1. Kinematics: Acceleration and Distance | 运动学：加速度和距离
A car accelerates uniformly from rest to a speed of 20 m/s in 5.0 s. Calculate (a) the acceleration, and (b) the distance travelled during this time.
一辆汽车从静止开始均匀加速，5.0 秒内速度达到 20 m/s。计算 (a) 加速度，(b) 此段时间内行驶的距离。
Step 1: Write down the known quantities. Initial speed u = 0 m/s, final speed v = 20 m/s, time t = 5.0 s.
步骤1：写出已知量。初速度 u = 0 m/s，末速度 v = 20 m/s，时间 t = 5.0 s。
Step 2: For part (a), use the definition of acceleration a = (v – u) / t.
步骤2：(a) 小题使用加速度的定义 a = (v – u) / t。
a = (20 – 0) / 5.0 = 4.0 m/s²
Step 3: For part (b), the distance s can be found using s = ut + ½ a t². Since u = 0, this simplifies to s = ½ a t² = ½ × 4.0 × (5.0)² = ½ × 4.0 × 25 = 50 m. Alternatively, average speed (0+20)/2 = 10 m/s × 5 s = 50 m.
步骤3：(b) 小题求距离 s，可使用 s = ut + ½ a t²。由于 u = 0，化简为 s = ½ a t² = ½ × 4.0 × (5.0)² = ½ × 4.0 × 25 = 50 m。也可用平均速度法：(0+20)/2 = 10 m/s × 5 s = 50 m。
2. Newton’s Second Law: Resultant Force | 牛顿第二定律：合力
A 1200 kg car experiences a forward driving force of 3000 N. The total resistive force (air resistance and friction) is 600 N. Determine the acceleration of the car.
一辆质量为 1200 kg 的汽车受到 3000 N 的前进驱动力，总阻力（空气阻力和摩擦力）为 600 N。求汽车的加速度。
Step 1: Find the resultant (net) force acting on the car. Resultant force = driving force – resistive force = 3000 N – 600 N = 2400 N forward.
步骤1：求出作用在汽车上的合力。合力 = 驱动力 – 阻力 = 3000 N – 600 N = 2400 N，方向向前。
Step 2: Apply Newton’s second law Fnet = m a. Rearrange to a = Fnet / m.
步骤2：应用牛顿第二定律 Fnet = m a。变形得 a = Fnet / m。
a = 2400 N / 1200 kg = 2.0 m/s²
3. Momentum and Impulse: Collision | 动量和冲量：碰撞
A trolley of mass 2.0 kg moving at 3.0 m/s collides with and sticks to a stationary trolley of mass 1.0 kg. Calculate the velocity of the combined trolleys immediately after the collision.
一辆质量为 2.0 kg 的小车以 3.0 m/s 的速度运动，与另一辆质量为 1.0 kg 的静止小车碰撞后粘在一起。求碰撞后瞬间两小车共同的速度。
Step 1: This is an inelastic collision where momentum is conserved. Write the conservation equation: total momentum before = total momentum after.
步骤1：这是一个完全非弹性碰撞，动量守恒。写出守恒方程：碰前总动量 = 碰后总动量。
Step 2: Before collision, momentum of moving trolley = 2.0 kg × 3.0 m/s = 6.0 kg m/s; stationary trolley momentum = 0. So total p before = 6.0 kg m/s.
步骤2：碰撞前，运动小车的动量 = 2.0 kg × 3.0 m/s = 6.0 kg m/s；静止小车动量 = 0。所以碰前总动量 p = 6.0 kg m/s。
Step 3: After collision, combined mass = 2.0 + 1.0 = 3.0 kg. Let v be the common velocity. Then 3.0 × v = 6.0, so v = 2.0 m/s in the original direction.
步骤3：碰撞后，总质量 = 2.0 + 1.0 = 3.0 kg。设共同速度为 v。则 3.0 × v = 6.0，所以 v = 2.0 m/s，方向与原方向相同。
4. Work, Energy and Power: Kinetic and Potential Energy | 功、能和功率：动能与势能
A ball of mass 0.50 kg is thrown vertically upwards with a speed of 8.0 m/s. Ignoring air resistance, calculate the maximum height it reaches. (g = 10 m/s²)
一个质量为 0.50 kg 的小球以 8.0 m/s 的初速度竖直上抛。忽略空气阻力，计算它能到达的最大高度。（g = 10 m/s²）
Step 1: At the maximum height, the kinetic energy is completely converted to gravitational potential energy. So ½ m v² = m g h. Note that mass m cancels.
步骤1：在最高点，动能完全转化为重力势能。因此 ½ m v² = m g h。注意质量 m 可以消去。
Step 2: Rearrange: h = v² / (2g).
步骤2：变形得 h = v² / (2g)。
h = (8.0)² / (2 × 10) = 64 / 20 = 3.2 m
5. Density and Pressure: Liquid Pressure | 密度和压强：液体压强
A swimming pool has a depth of 2.5 m. Calculate the water pressure at the bottom of the pool. Density of water = 1000 kg/m³, g = 10 m/s².
一个游泳池水深 2.5 m。求池底的水压。水的密度 = 1000 kg/m³，g = 10 m/s²。
Step 1: Use the formula for pressure due to a liquid column: p = ρ g h, where ρ is density, g is gravitational field strength, and h is depth.
步骤1：使用液体压强公式：p = ρ g h，其中 ρ 是密度，g 是重力场强度，h 是深度。
Step 2: Substitute values: p = 1000 × 10 × 2.5.
步骤2：代入数值：p = 1000 × 10 × 2.5。
p = 25 000 Pa (or 25 kPa)
Note that atmospheric pressure also acts on the surface, but the question only asks for water pressure, so the gauge pressure is 25 kPa.
注意大气压强也作用于水面，但本题只要求水产生的压强，因此表压为 25 kPa。
6. Electricity: Ohm’s Law and Series Circuits | 电学：欧姆定律与串联电路
A 12 V battery is connected in series with two resistors: R₁ = 4.0 Ω and R₂ = 6.0 Ω. Calculate (a) the total resistance, (b) the current in the circuit, and (c) the potential difference across R₂.
一个 12 V 电池与两个电阻 R₁ = 4.0 Ω 和 R₂ = 6.0 Ω 串联。计算 (a) 总电阻，(b) 电路中的电流，(c) R₂ 两端的电压。
Step 1: For series resistors, total resistance Rtotal = R₁ + R₂ = 4.0 + 6.0 = 10.0 Ω.
步骤1：串联电阻的总电阻 Rtotal = R₁ + R₂ = 4.0 + 6.0 = 10.0 Ω。
Step 2: Use Ohm’s law V = I R to find the current. I = V / Rtotal = 12 V / 10.0 Ω = 1.2 A.
步骤2：用欧姆定律 V = I R 求电流。I = V / Rtotal = 12 V / 10.0 Ω = 1.2 A。
Step 3: The p.d. across R₂ is V₂ = I × R₂ = 1.2 A × 6.0 Ω = 7.2 V. Check: p.d. across R₁ = 1.2 × 4.0 = 4.8 V; total = 4.8 + 7.2 = 12 V.
步骤3：R₂ 两端的电压 V₂ = I × R₂ = 1.2 A × 6.0 Ω = 7.2 V。验证：R₁ 两端电压 = 1.2 × 4.0 = 4.8 V，总和为 4.8 + 7.2 = 12 V。
7. Electrical Power and Energy | 电功率与电能
An electric heater operates at 230 V and draws a current of 5.0 A. Calculate (a) the power rating of the heater, and (b) the energy transferred in 3.0 minutes in joules and kilowatt-hours.
一台电暖器工作电压为 230 V，电流为 5.0 A。计算 (a) 暖器的额定功率，(b) 3.0 分钟内传递的能量，分别用焦耳和千瓦时表示。
Step 1: Power P = I V = 5.0 A × 230 V = 1150 W (1.15 kW).
步骤1：功率 P = I V = 5.0 A × 230 V = 1150 W（即 1.15 kW）。
Step 2: Energy in joules: E = P t, where t must be in seconds. 3.0 min = 180 s. E = 1150 W × 180 s = 207 000 J ≈ 2.07 × 10⁵ J.
步骤2：以焦耳计的能量：E = P t，时间必须用秒。3.0 分钟 = 180 s。E = 1150 W × 180 s = 207 000 J ≈ 2.07 × 10⁵ J。
Step 3: Energy in kWh: E = 1.15 kW × (3.0 / 60) h = 1.15 × 0.05 = 0.0575 kWh. Alternatively, 207 000 J / (3.6 × 10⁶ J/kWh) gives the same.
步骤3：以千瓦时计：E = 1.15 kW × (3.0 / 60) h = 1.15 × 0.05 = 0.0575 kWh。或用 207 000 J / (3.6 × 10⁶ J/kWh) 得到相同结果。
8. Waves: The Wave Equation | 波：波动方程
A sound wave has a frequency of 440 Hz and a wavelength of 0.78 m. Calculate the speed of sound in air.
一列声波的频率为 440 Hz，波长为 0.78 m。计算声音在空气中的传播速度。
Step 1: Use the wave equation v = f λ.
步骤1：使用波动方程 v = f λ。
v = 440 Hz × 0.78 m = 343.2 m/s ≈ 340 m/s
Remember that frequency is measured in hertz (Hz), wavelength in metres (m), giving speed in metres per second (m/s).
记住频率的单位是赫兹（Hz），波长的单位是米（m），速度的单位为米每秒（m/s）。
9. Radioactivity: Half-life | 放射性：半衰期
A sample of a radioactive isotope has an activity of 800 Bq. After 6 days, the activity drops to 100 Bq. Calculate the half-life of the isotope.
某放射性同位素样品的初始活度为 800 Bq。6 天后活度降至 100 Bq。计算该同位素的半衰期。
Step 1: Determine how many times the activity has halved. Original activity 800 → 400 (1 half-life) → 200 (2) → 100 (3). So 3 half-lives have elapsed.
步骤1：确定活度减半的次数。初始活度 800 → 400（1 个半衰期） → 200（2 个） → 100（3 个）。因此经过了 3 个半衰期。
Step 2: Total time = 6 days, so one half-life t½ = 6 days / 3 = 2 days.
步骤2：总时间 = 6 天，因此一个半衰期 t½ = 6 天 / 3 = 2 天。
10. Gas Laws: Boyle’s Law | 气体定律：波义耳定律
A fixed mass of gas occupies a volume of 0.20 m³ at a pressure of 150 kPa. The temperature is kept constant while the gas is compressed to a volume of 0.12 m³. Find the new pressure.
一定质量的气体在 150 kPa 压强下占据 0.20 m³ 的体积。温度保持不变，气体被压缩至 0.12 m³。求新的压强。
Step 1: For a fixed mass of gas at constant temperature, Boyle’s law applies: p₁ V₁ = p₂ V₂.
步骤1：对于一定质量、温度恒定的气体，适用波义耳定律：p₁ V₁ = p₂ V₂。
Step 2: Rearrange to find p₂: p₂ = (p₁ V₁) / V₂ = (150 kPa × 0.20 m³) / 0.12 m³ = 30 / 0.12 = 250 kPa.
步骤2：变形求 p₂：p₂ = (p₁ V₁) / V₂ = (150 kPa × 0.20 m³) / 0.12 m³ = 30 / 0.12 = 250 kPa。
11. Electromagnetic Induction: Transformer Turns Ratio | 电磁感应：变压器匝数比
A step-down transformer is designed to convert a primary voltage of 230 V to a secondary voltage of 11.5 V. The primary coil has 2000 turns. Calculate the number of turns needed on the secondary coil.
一个降压变压器旨在将 230 V 的初级电压转换为 11.5 V 的次级电压。初级线圈有 2000 匝。计算次级线圈所需的匝数。
Step 1: For an ideal transformer, Vp / Vs = Np / Ns, where p denotes primary and s secondary.
步骤1：对于理想变压器，Vp / Vs = Np / Ns，其中 p 代表初级，s 代表次级。
Step 2: Rearrange: Ns = (Vs × Np) / Vp = (11.5 V × 2000) / 230 V = 23000 / 230 = 100 turns.
步骤2：变形得 Ns = (Vs × Np) / Vp = (11.5 V × 2000) / 230 V = 23000 / 230 = 100 匝。
12. Astrophysics: Redshift and Hubble’s Law | 天体物理：红移和哈勃定律
A distant galaxy shows a redshift corresponding to a recessional speed of 2.4 × 10⁷ m/s. Using the Hubble constant H₀ = 2.2 × 10⁻¹⁸ s⁻¹, estimate the distance to the galaxy in metres and in light-years. (1 ly ≈ 9.5 × 10¹⁵ m)
一个遥远星系的红移对应的退行速度为 2.4 × 10⁷ m/s。已知哈勃常数 H₀ = 2.2 × 10⁻¹⁸ s⁻¹，估算该星系的距离，并以米和光年表示。（1 光年 ≈ 9.5 × 10¹⁵ m）
Step 1: Hubble’s law states v = H₀ d, so d = v / H₀.
步骤1：哈勃定律为 v = H₀ d，因此 d = v / H₀。
Step 2: d = (2.4 × 10⁷ m/s) / (2.2 × 10⁻¹⁸ s⁻¹) = 1.09 × 10²⁵ m ≈ 1.1 × 10²⁵ m.
步骤2：d = (2.4 × 10⁷ m/s) / (2.2 × 10⁻¹⁸ s⁻¹) = 1.09 × 10²⁵ m ≈ 1.1 × 10²⁵ m。
Step 3: Convert to light-years: d = 1.09 × 10²⁵ m / (9.5 × 10¹⁵ m/ly) = 1.15 × 10⁹ ly ≈ 1.2 billion light-years.
步骤3：转换为光年：d = 1.09 × 10²⁵ m / (9.5 × 10¹⁵ m/ly) = 1.15 × 10⁹ ly ≈ 12 亿光年。
Published by TutorHao | Physics Revision Series | aleveler.com
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