IGCSE CIE Physics: Typical Example Questions Explained | IGCSE CIE 物理：典型例题详解

📚 IGCSE CIE Physics: Typical Example Questions Explained | IGCSE CIE 物理：典型例题详解

This revision article provides detailed, step-by-step solutions to ten carefully selected IGCSE CIE Physics questions. Each worked example reinforces core concepts, common calculations, and typical exam techniques. The bilingual format helps you grasp both the English terminology and the Chinese explanations simultaneously.

本文通过精选的十道 IGCSE CIE 物理典型例题，提供详细的分步解析。每道题都强化核心概念、常见计算和典型考试技巧。中英双语格式帮助你同步掌握英文术语和中文解释。

1. Kinematics – Uniform Acceleration | 运动学 – 匀加速直线运动

A car accelerates uniformly from 10 m/s to 30 m/s in 5 seconds. Calculate the acceleration and the distance travelled.

一辆汽车从 10 m/s 均匀加速到 30 m/s，用时 5 秒。计算加速度和行驶距离。

First, find the acceleration using the definition formula.

首先，利用定义式求加速度。

a = (v – u) / t

Substitute the given values: a = (30 – 10) / 5 = 20 / 5 = 4 m/s².

代入已知数值：a = (30 – 10) / 5 = 20 / 5 = 4 m/s²。

Now calculate the distance. You can use either s = ut + ½at² or average velocity × time.

现在计算距离。可以用 s = ut + ½at² 或平均速度 × 时间。

s = ut + ½at²

s = (10 × 5) + ½ × 4 × (5)² = 50 + 2 × 25 = 100 m.

s = (10 × 5) + ½ × 4 × (5)² = 50 + 2 × 25 = 100 m。

Alternative method: average velocity = (u + v)/2 = 20 m/s, so s = 20 m/s × 5 s = 100 m. Both give the same answer.

另一种解法：平均速度 = (u + v)/2 = 20 m/s，因此 s = 20 m/s × 5 s = 100 m。两种方法结果一致。

2. Newton’s Second Law | 牛顿第二定律

A block of mass 4 kg is pulled on a smooth surface with a net force of 20 N. Determine the acceleration.

一个 4 kg 的木块在光滑水平面上受到 20 N 的净力作用。求加速度。

F = m a

Rearranging gives a = F / m = 20 N / 4 kg = 5 m/s².

整理得 a = F / m = 20 N / 4 kg = 5 m/s²。

If friction of 4 N were present, the net force would be 20 N – 4 N = 16 N, and the acceleration would drop to 4 m/s². Always identify the resultant force first.

若存在 4 N 的摩擦力，净力为 20 N – 4 N = 16 N，加速度将变为 4 m/s²。务必先确定合外力。

3. Work Done and Gravitational Potential Energy | 功与重力势能

A worker lifts a box of weight 50 N vertically upwards through a height of 2 m. Calculate the work done and the gain in gravitational potential energy (g.p.e.).

一位工人将重 50 N 的箱子竖直向上提升 2 m。计算做功和增加的重力势能。

Work Done = Force × distance moved in direction of force

Work = 50 N × 2 m = 100 J.

功 = 50 N × 2 m = 100 J。

ΔEₚ = weight × Δh = 50 N × 2 m = 100 J

Since the box is lifted at constant speed, all the work done is converted to gravitational potential energy, so the g.p.e. gained equals 100 J.

因箱子匀速提升，所有功转化为重力势能，因此势能增加 100 J。

4. Power – Rate of Doing Work | 功率 – 做功的快慢

A motor lifts a 100 kg load through a vertical height of 5 m in 10 s. Take g = 10 N/kg. Find the power of the motor.

一台电动机在 10 s 内将 100 kg 的重物竖直提升 5 m。取 g = 10 N/kg。求电动机的功率。

First find the weight of the load: W = m g = 100 × 10 = 1000 N.

先求重物的重量：W = m g = 100 × 10 = 1000 N。

Work done = force × distance = 1000 N × 5 m = 5000 J.

做功 = 力 × 距离 = 1000 N × 5 m = 5000 J。

P = Work / time = 5000 J / 10 s = 500 W

If the load moves up at a constant speed, you can also use P = F v, where v = distance / time = 5/10 = 0.5 m/s, giving P = 1000 N × 0.5 m/s = 500 W.

若重物匀速上升，也可用 P = F v，v = 5/10 = 0.5 m/s，得到 P = 1000 N × 0.5 m/s = 500 W。

5. Series Circuit – Resistors | 串联电路 – 电阻

Two resistors, 4 Ω and 6 Ω, are connected in series across a 12 V battery. Calculate the total resistance, the current in the circuit, and the potential difference across each resistor.

两个电阻值分别为 4 Ω 和 6 Ω 的电阻串联后接在 12 V 电池两端。求总电阻、电路中的电流以及每个电阻两端的电压。

R_total = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω

Using Ohm’s law, I = V / R = 12 V / 10 Ω = 1.2 A.

运用欧姆定律，I = V / R = 12 V / 10 Ω = 1.2 A。

Now V₁ = I × R₁ = 1.2 A × 4 Ω = 4.8 V, and V₂ = 1.2 A × 6 Ω = 7.2 V. Check: 4.8 + 7.2 = 12 V.

接着 V₁ = I × R₁ = 1.2 A × 4 Ω = 4.8 V，V₂ = 1.2 A × 6 Ω = 7.2 V。验证：4.8 + 7.2 = 12 V。

6. Resistance of a Wire – Length and Area | 导线的电阻 – 长度与截面积

A uniform metal wire has a resistance of 10 Ω. It is stretched so that its length becomes twice the original length, without any loss of metal (volume remains constant). Find its new resistance.

一根均匀金属丝的电阻为 10 Ω。将其均匀拉长至原长的两倍，且金属没有损失（体积保持不变）。求新电阻。

R = ρ L / A

When stretched, volume V = A × L stays constant. If new length L’ = 2L, then new cross‑sectional area A’ = V / (2L) = A / 2.

拉伸时，体积 V = A × L 不变。若新长度 L’ = 2L，则新截面积 A’ = V/(2L) = A / 2。

So new resistance R’ = ρ (2L) / (A/2) = 4 ρ L / A = 4R. Hence R’ = 4 × 10 Ω = 40 Ω.

因此新电阻 R’ = ρ (2L) / (A/2) = 4 ρ L / A = 4R。故 R’ = 4 × 10 Ω = 40 Ω。

A quick rule: for a wire of constant volume, resistance is proportional to the square of its length (R ∝ L²).

简明规律：体积不变时，导线的电阻与长度的平方成正比 (R ∝ L²)。

7. Wave Speed, Frequency and Wavelength | 波速、频率与波长

A water wave has a frequency of 5 Hz and a wavelength of 2 m. Calculate the speed of the wave. Also, if the periodic time shown on an oscilloscope is 0.2 s, determine its frequency.

一个水波的频率为 5 Hz，波长为 2 m。计算波速。另外，若示波器显示周期为 0.2 s，求该波的频率。

v = f λ

v = 5 Hz × 2 m = 10 m/s.

v = 5 Hz × 2 m = 10 m/s。

For the oscilloscope trace: frequency f = 1 / T, where T is the periodic time. f = 1 / 0.2 s = 5 Hz.

对于示波器波形：频率 f = 1 / T，T 为周期。f = 1 / 0.2 s = 5 Hz。

This same relationship applies to all waves, including sound and light, as long as you work in consistent SI units.

这一关系适用于所有波，包括声波和光波，只要使用一致的国际单位即可。

8. Specific Heat Capacity – Thermal Energy | 比热容 – 热能

A 0.5 kg aluminium block is heated from 20 °C to 80 °C. The specific heat capacity of aluminium is 900 J/(kg °C). Find the thermal energy supplied. If a 50 W electric heater is used and all energy goes to the block, how long will it take?

一个 0.5 kg 的铝块从 20 °C 加热到 80 °C。铝的比热容为 900 J/(kg °C)。求供给的热能。如果使用 50 W 的电热器且能量全部传给铝块，需要多长时间？

Q = m c Δθ

Δθ = 80 – 20 = 60 °C. Q = 0.5 × 900 × 60 = 27 000 J.

Δθ = 80 – 20 = 60 °C。Q = 0.5 × 900 × 60 = 27 000 J。

Power P = energy / time ⇒ time t = E / P = 27 000 J / 50 W = 540 s (9 minutes).

功率 P = 能量 / 时间 ⇒ 时间 t = E / P = 27 000 J / 50 W = 540 s（9 分钟）。

In practice, some energy is always lost to the surroundings, so the actual heating time would be longer.

实际上总会有部分能量散失到环境中，因此实际加热时间会更长。

9. Radioactive Decay – Half-life | 放射性衰变 – 半衰期

A sample of radioactive isotope gives a corrected count rate of 800 counts per minute. The half-life of the isotope is 3 hours. Estimate the count rate after 9 hours.

某放射性同位素样品的修正计数率为 800 次/分。该同位素的半衰期为 3 小时。估算 9 小时后的计数率。

Number of half-lives elapsed = total time / half-life = 9 h / 3 h = 3 half-lives.

经过的半衰期个数 = 总时间 / 半衰期 = 9 h / 3 h = 3 个半衰期。

After each half-life the count rate halves:

每经过一个半衰期计数率减半：

Start: 800

初始：800
After 1 half‑life: 400

1 个半衰期后：400
After 2 half‑lives: 200

2 个半衰期后：200
After 3 half‑lives: 100 counts per minute.

3 个半衰期后：100 次/分。

Always remember to subtract background radiation if it was included in the initial reading. Here the count rate is already corrected.

记住，如果初始读数包含本底辐射，必须先减去。这里已是修正后的计数率。

10. Moments – Equilibrium of a Beam | 力矩 – 杠杆平衡

A uniform metre rule is pivoted at its 50 cm mark. A 20 N weight is suspended at the 20 cm mark. At which mark on the other side should a 25 N weight be placed to keep the rule balanced?

一根均匀的米尺在 50 cm 刻度处支起。在 20 cm 处悬挂 20 N 的砝码。另一个 25 N 的砝码应悬挂在支点另一侧的哪个刻度位置，才能使尺保持水平平衡？

The distance of the 20 N force from the pivot is 50 cm – 20 cm = 30 cm (0.30 m).

20 N 力到支点的距离为 50 cm – 20 cm = 30 cm (0.30 m)。

Clockwise moment = Anticlockwise moment

20 N × 0.30 m = 25 N × d, where d is the distance from the pivot on the opposite side.

20 N × 0.30 m = 25 N × d，d 为支点另一侧的距离。

6 = 25 × d ⇒ d = 6 / 25 = 0.24 m = 24 cm.

6 = 25 × d ⇒ d = 6 / 25 = 0.24 m = 24 cm。

Therefore the 25 N weight should be hung at the 50 cm + 24 cm = 74 cm mark.

因此 25 N 砝码应挂在 50 cm + 24 cm = 74 cm 刻度处。

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IGCSE CIE Physics: Typical Example Questions Explained | IGCSE CIE 物理：典型例题详解