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

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

Mastering IGCSE Physics involves more than memorising formulas — it requires the ability to apply concepts to numerical and conceptual problems. This article presents a set of carefully chosen worked examples covering the core topics of the syllabus, from kinematics and forces to waves and radioactivity. Each example is solved step by step, with clear reasoning and correct units, so you can build confidence in your problem-solving skills.

掌握 IGCSE 物理不仅需要记住公式，更要求将概念应用于计算和概念性问题。本文精选了一系列涵盖课程核心主题的典型例题，从运动学和力到波和放射性。每个例题都按步骤详细求解，给出清晰的推理和正确的单位，帮助你建立解题信心。

1. Motion: Calculating Acceleration | 运动：计算加速度

A car accelerates from rest to a velocity of 24 m/s in 8.0 seconds. Calculate its acceleration.

一辆汽车从静止加速到 24 m/s，用时 8.0 秒。计算其加速度。

Known quantities: initial velocity u = 0 m/s, final velocity v = 24 m/s, time t = 8.0 s.

已知量：初速度 u = 0 m/s，末速度 v = 24 m/s，时间 t = 8.0 s。

The formula for uniform acceleration: a = (v – u) / t

匀加速度公式：a = (v – u) / t

a = (24 – 0) / 8.0 = 3.0 m/s²

The car’s acceleration is 3.0 metres per second squared.

汽车的加速度为 3.0 米每二次方秒。

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

A trolley of mass 2.5 kg is pulled with a resultant force of 10 N. Find its acceleration.

一辆质量为 2.5 kg 的小车受到 10 N 的合力。求其加速度。

Newton’s second law: F = m × a, so a = F / m.

牛顿第二定律：F = m × a，所以 a = F / m。

a = 10 N / 2.5 kg = 4.0 m/s²

The acceleration of the trolley is 4.0 m/s² in the direction of the resultant force.

小车的加速度为 4.0 m/s²，方向与合力方向相同。

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3. Energy: Conservation of Mechanical Energy | 能量：机械能守恒

A stone of mass 0.20 kg is dropped from a height of 15 m. Ignoring air resistance, calculate its speed just before hitting the ground. Take g = 10 m/s².

一颗质量为 0.20 kg 的石子从 15 m 高处落下。忽略空气阻力，计算它落地瞬间的速度。取 g = 10 m/s²。

Loss in gravitational potential energy = m × g × h

减少的重力势能 = m × g × h

Gain in kinetic energy = ½ × m × v²

增加的动能 = ½ × m × v²

By conservation of energy: m × g × h = ½ × m × v². The mass cancels out.

根据能量守恒：m × g × h = ½ × m × v²。质量可以约掉。

v² = 2 × g × h = 2 × 10 × 15 = 300

v = √300 ≈ 17.3 m/s

The stone hits the ground at approximately 17.3 m/s.

石子以大约 17.3 m/s 的速度撞击地面。

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4. Density: Mass, Volume and Density | 密度：质量、体积和密度

A metal block has a mass of 850 g and a volume of 340 cm³. Determine its density in both g/cm³ and kg/m³.

一个金属块的质量为 850 g，体积为 340 cm³。分别以 g/cm³ 和 kg/m³ 为单位计算其密度。

Formula: ρ = m / V

公式：ρ = m / V

ρ = 850 g / 340 cm³ = 2.5 g/cm³

To convert to kg/m³: 1 g/cm³ = 1000 kg/m³.

转换为 kg/m³：1 g/cm³ = 1000 kg/m³。

ρ = 2.5 × 1000 = 2500 kg/m³

The density of the metal is 2.5 g/cm³ or 2500 kg/m³.

该金属的密度为 2.5 g/cm³ 或 2500 kg/m³。

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5. Electricity: Ohm’s Law and Resistance | 电学：欧姆定律与电阻

A component has a potential difference of 6.0 V across it and carries a current of 0.40 A. Calculate its resistance.

一个元件两端的电势差为 6.0 V，通过的电流为 0.40 A。计算其电阻。

Ohm’s law states: V = I × R, which can be rearranged to R = V / I.

欧姆定律为：V = I × R，可变形为 R = V / I。

R = 6.0 V / 0.40 A = 15 Ω

The resistance of the component is 15 ohms.

该元件的电阻为 15 欧姆。

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6. Waves: Speed, Frequency and Wavelength | 波：波速、频率和波长

A water wave has a frequency of 5.0 Hz and a wavelength of 0.12 m. Calculate its speed.

一个水波的频率为 5.0 Hz，波长为 0.12 m。计算其波速。

The wave equation: v = f × λ

波速公式：v = f × λ

v = 5.0 × 0.12 = 0.60 m/s

The wave travels at 0.60 metres per second.

该波的传播速度为 0.60 米每秒。

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7. Thermal Physics: Specific Heat Capacity | 热物理学：比热容

An electric heater supplies 48 000 J of energy to 1.5 kg of water. The temperature of the water rises from 22 °C to 42 °C. Calculate a value for the specific heat capacity of water.

一个电加热器为 1.5 kg 的水提供了 48 000 J 的能量。水的温度从 22 °C 上升到 42 °C。计算水的比热容的值。

Thermal energy equation: Q = m × c × Δθ

热量公式：Q = m × c × Δθ

Temperature change Δθ = 42 – 22 = 20 °C.

温度变化 Δθ = 42 – 22 = 20 °C。

c = Q / (m × Δθ) = 48 000 / (1.5 × 20) = 48 000 / 30 = 1600 J/(kg °C)

The specific heat capacity is 1600 J/(kg °C). (Note: the accepted value for water is about 4200 J/(kg °C); this example uses a smaller value for practice.)

比热容为 1600 J/(kg °C)。（注意：水的公认值约为 4200 J/(kg °C)；本例题使用较小值进行练习。）

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8. Radioactivity: Half-Life Calculation | 放射性：半衰期计算

A radioactive sample has an initial count rate of 800 counts per second. Its half-life is 3.0 days. Predict the count rate after 9.0 days.

一个放射性样品的初始计数率为每秒 800 次。其半衰期为 3.0 天。预测 9.0 天后的计数率。

Number of half-lives elapsed = total time / half-life = 9.0 days / 3.0 days = 3 half-lives.

经过的半衰期次数 = 总时间 / 半衰期 = 9.0 天 / 3.0 天 = 3 个半衰期。

After each half-life, the count rate halves. After 3 half-lives:

每经过一个半衰期，计数率减半。经过 3 个半衰期后：

Count rate = 800 / 2³ = 800 / 8 = 100 counts per second

The predicted count rate after 9.0 days is 100 counts/s.

9.0 天后的预测计数率为 100 次/秒。

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

A uniform metre ruler of weight 1.0 N is pivoted at the 40 cm mark. A 0.50 N weight is hung at the 10 cm mark. Where should a 2.0 N weight be placed to balance the ruler?

一根重 1.0 N 的均匀米尺支点在 40 cm 刻度处。一个 0.50 N 的重物挂在 10 cm 刻度处。一个 2.0 N 的重物应挂在何处才能使米尺平衡？

The weight of the ruler acts at its centre of mass, the 50 cm mark. Taking moments about the pivot (40 cm):

米尺的重力作用在其质心（50 cm 刻度处）。以支点（40 cm）为转轴计算力矩：

Clockwise moment (ruler) = 1.0 N × (50 cm – 40 cm) = 1.0 N × 10 cm = 10 N cm.

顺时针力矩（米尺）= 1.0 N × (50 cm – 40 cm) = 1.0 N × 10 cm = 10 N cm。

Anticlockwise moment (0.50 N weight) = 0.50 N × (40 cm – 10 cm) = 0.50 N × 30 cm = 15 N cm.

逆时针力矩（0.50 N 重物）= 0.50 N × (40 cm – 10 cm) = 0.50 N × 30 cm = 15 N cm。

Total anticlockwise moment = 15 N cm; total clockwise moment = 10 N cm. To balance, an extra clockwise moment of 5 N cm is needed from the 2.0 N weight.

总逆时针力矩 = 15 N cm；总顺时针力矩 = 10 N cm。为平衡，2.0 N 重物需提供 5 N cm 的额外顺时针力矩。

Let the distance from the pivot to the 2.0 N weight be d on the clockwise side. Then 2.0 N × d = 5 N cm, so d = 2.5 cm. The position on the ruler = 40 cm + 2.5 cm = 42.5 cm mark.

设 2.0 N 重物距支点的距离为 d（在顺时针一侧）。则 2.0 N × d = 5 N cm，所以 d = 2.5 cm。重物应挂在 40 cm + 2.5 cm = 42.5 cm 刻度处。

The 2.0 N weight must be hung at the 42.5 cm mark.

2.0 N 重物必须挂在 42.5 cm 刻度处。

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10. Waves: Refractive Index and Critical Angle | 波：折射率与临界角

Light travels from glass into air. The refractive index of the glass is 1.5. Calculate the critical angle for the glass–air boundary.

光从玻璃射入空气。玻璃的折射率为 1.5。计算玻璃–空气界面的临界角。

The relationship for critical angle c: sin c = 1 / n (when the second medium is air, n = 1.0).

临界角 c 的关系式：sin c = 1 / n（当第二种介质为空气时，n = 1.0）。

sin c = 1 / 1.5 = 0.6667

c = sin⁻¹(0.6667) ≈ 41.8°

The critical angle is approximately 41.8 degrees. For angles of incidence greater than this in the glass, total internal reflection occurs.

临界角约为 41.8 度。当光在玻璃中入射角大于此角度时，会发生全内反射。

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