📚 Typical Example Problems Explained | GCSE CIE 物理：典型例题详解
In CIE IGCSE Physics (0625), mastering the application of physical formulas and principles through worked examples is essential for success. This article provides a series of carefully selected typical problems, each solved step by step with clear explanations in both English and Chinese. By studying these examples, you will reinforce your understanding of key topics such as kinematics, dynamics, energy, waves, electricity, and radioactivity. All working follows CIE conventions and uses standard units.
在 CIE IGCSE 物理 (0625) 中，通过典型例题掌握物理公式和原理的运用是取得成功的关键。本文精选了一系列典型题目，逐一用中英双语分步详解。学习这些例题将帮助你巩固运动学、动力学、能量、波、电学和放射性等重点内容。所有解题过程均遵循 CIE 规范并使用标准单位。
1. Kinematics: Acceleration and Distance | 运动学：加速度与距离
A car accelerates uniformly from rest to a velocity of 20 m/s in 10 seconds. Calculate (a) the acceleration, and (b) the distance travelled during this time.
一辆汽车从静止匀加速到 20 m/s，用时 10 秒。计算 (a) 加速度；(b) 这段时间内行驶的距离。
The acceleration a is defined as the rate of change of velocity: a = (v – u) / t, where u is initial velocity, v is final velocity and t is time.
加速度 a 定义为速度的变化率：a = (v – u) / t，其中 u 是初速度，v 是末速度，t 是时间。
Substituting the given values: a = (20 m/s – 0 m/s) / 10 s = 2 m/s².
代入给定数值：a = (20 m/s – 0 m/s) / 10 s = 2 m/s²。
The distance s can be found using the average velocity method for uniform acceleration: s = (u + v) × t / 2 = (0 + 20) × 10 / 2 = 100 m. Alternatively, the equation s = ut + ½at² gives the same result: s = 0 + ½ × 2 m/s² × (10 s)² = 100 m.
距离 s 可用匀加速运动的平均速度法求得：s = (u + v) × t / 2 = (0 + 20) × 10 / 2 = 100 m。或者用公式 s = ut + ½at² 同样得到：s = 0 + ½ × 2 m/s² × (10 s)² = 100 m。
Remember: the equation s = (u + v)t/2 is only valid when acceleration is constant. A common mistake is using average speed (v/2) when acceleration is not uniform.
注意：s = (u + v)t/2 仅适用于加速度恒定的情况。常见错误是在加速度不均衡时仍然使用平均速度 (v/2)。
2. Motion Graphs: Distance–Time | 运动图像：距离–时间图
Sketch a distance–time graph for a car that travels at constant speed for 5 seconds, then stops for 3 seconds, and finally returns to the starting point at a constant speed, taking another 6 seconds. Describe how to obtain speed from the graph.
画出一辆汽车的距离–时间图：它先以恒定速度行驶 5 秒，然后停止 3 秒，最后以恒定速度返回起点，耗时 6 秒。说明如何从图中获取速度。
A distance–time graph has time on the horizontal axis and distance from the starting point on the vertical axis. The slope (gradient) of the graph represents speed.
距离–时间图以时间为横轴，以离开起点的距离为纵轴。图线的斜率（梯度）表示速度。
For the first 5 s, the line is a straight sloping upward, indicating constant speed. From 5 s to 8 s, the line is horizontal (zero slope), meaning the car is stationary. From 8 s to 14 s, the line slopes downward back to the time axis, showing constant speed in the return direction.
前 5 秒，图线是一条向上倾斜的直线，表示恒定速度。5 秒到 8 秒，图线水平（斜率为零），表示汽车静止。8 秒到 14 秒，图线向下倾斜回到时间轴，表示汽车以恒定速度返回。
To find the speed during the first stage, calculate the gradient: rise / run = distance travelled / time taken. For the return stage, the gradient will be negative, indicating motion towards the start. The magnitude of the gradient gives the speed.
求第一阶段的速度，计算斜率：上升量 / 时间跨度 = 行驶距离 / 所用时间。返回阶段斜率为负，表明物体朝向起点运动。梯度的绝对值即为速度。
3. Force and Motion: Newton’s Second Law | 力与运动：牛顿第二定律
A block of mass 5 kg is pulled along a frictionless surface by a resultant force of 15 N. Calculate the acceleration of the block.
一个质量为 5 kg 的木块在无摩擦表面上受到 15 N 的合力拉动。计算木块的加速度。
Newton’s second law states that resultant force F = m a, where m is mass and a is acceleration. Rearranging gives a = F / m.
牛顿第二定律指出，合力 F = m a，其中 m 是质量，a 是加速度。移项得 a = F / m。
F = 15 N, m = 5 kg, so a = 15 N / 5 kg = 3 m/s². The direction of acceleration is the same as the direction of the resultant force.
F = 15 N，m = 5 kg，所以 a = 15 N / 5 kg = 3 m/s²。加速度方向与合力方向相同。
Always ensure that force is the net (resultant) force. If friction or other forces are present, they must be subtracted to find the resultant force before applying F = m a.
务必确保所用的力为净力（合力）。如果存在摩擦力或其他力，必须先相减求出合力，再应用 F = m a。
4. Energy: Kinetic and Potential Energy Conversions | 能量：动能与势能的转化
A ball of mass 2 kg is dropped from rest at a height of 5 m above the ground. Assuming air resistance is negligible, calculate (a) the kinetic energy just before hitting the ground, and (b) the speed at that instant. (Take g = 9.8 m/s²)
一个质量为 2 kg 的球从离地 5 m 高处静止释放。假设空气阻力不计，计算 (a) 着地前的动能；(b) 该瞬间的速度。（取 g = 9.8 m/s²）
As the ball falls, gravitational potential energy (PE) is converted to kinetic energy (KE). At the top, PE = m g h and KE = 0. Just before impact, all initial PE has become KE (assuming no energy losses).
球下落时，重力势能 (PE) 转化为动能 (KE)。最高点时，PE = m g h，KE = 0。着地前瞬间，初始势能全部转化为动能（假设无能量损失）。
Initial PE = 2 kg × 9.8 m/s² × 5 m = 98 J. Therefore, KE just before impact = 98 J.
初始 PE = 2 kg × 9.8 m/s² × 5 m = 98 J。因此着地前动能 = 98 J。
Using KE = ½ m v², we solve for v: v = √(2 KE / m) = √(2 × 98 J / 2 kg) = √(98) ≈ 9.90 m/s. Note that v = √(2 g h) gives the same result directly.
利用 KE = ½ m v² 求解 v：v = √(2 KE / m) = √(2 × 98 J / 2 kg) = √(98) ≈ 9.90 m/s。注意直接用 v = √(2 g h) 可得同样结果。
5. Pressure: Force and Area | 压强：力与面积
A rectangular crate weighing 400 N has a base area of 0.5 m². Calculate the pressure exerted on the floor when the crate is placed flat on its base.
一个重 400 N 的长方体箱子底面积为 0.5 m²。计算箱子平放在地面时对地面施加的压强。
Pressure P is defined as the normal force per unit area: P = F / A, where F is the force acting perpendicular to the surface and A is the area of contact.
压强 P 定义为单位面积上所受的法向力：P = F / A，其中 F 是垂直于表面的作用力，A 是接触面积。
Here, the force is the weight of the crate, F = 400 N, and the area A = 0.5 m². So P = 400 N / 0.5 m² = 800 Pa (or N/m²).
这里，力为箱子重量 F = 400 N，面积 A = 0.5 m²。所以 P = 400 N / 0.5 m² = 800 Pa（或 N/m²）。
If the crate were placed on a smaller face, the pressure would be greater because the area would be smaller. Pressure is inversely proportional to area for a constant force.
如果箱子用较小的面放置，由于面积更小，压强会更大。在力恒定时，压强与面积成反比。
6. Waves: Frequency, Wavelength and Wave Speed | 波：频率、波长与波速
A water wave has a frequency of 5 Hz and a wavelength of 0.2 m. Calculate the speed of the wave. If the frequency is doubled while the speed remains constant, what happens to the wavelength?
一水波的频率为 5 Hz，波长为 0.2 m。计算波速。如果波速不变而频率加倍，波长会怎样？
The wave equation is v = f λ, where v is wave speed, f is frequency and λ is wavelength.
波动方程为 v = f λ，其中 v 是波速，f 是频率，λ 是波长。
Substituting: v = 5 Hz × 0.2 m = 1.0 m/s.
代入得：v = 5 Hz × 0.2 m = 1.0 m/s。
If speed stays constant and frequency doubles, λ must halve, because v = f λ = (2f) × (λ/2). Therefore, the new wavelength would be 0.1 m.
如果波速不变，频率加倍，则由 v = f λ 可知波长必须减半。因此新波长为 0.1 m。
7. Electricity: Ohm’s Law | 电学：欧姆定律
A resistor of 6 Ω is connected to a power supply, and the current through it is measured to be 3 A. Calculate the voltage across the resistor. What power is dissipated in the resistor?
一个 6 Ω 的电阻连接到电源上，测得通过的电流为 3 A。计算电阻两端的电压和电阻消耗的功率。
Ohm’s law states that V = I R, where V is voltage, I is current and R is resistance. It applies when temperature remains constant.
欧姆定律指出 V = I R，其中 V 是电压，I 是电流，R 是电阻。在温度恒定时适用。
Thus, V = 3 A × 6 Ω = 18 V.
因此 V = 3 A × 6 Ω = 18 V。
Power P can be calculated using P = I V or P = I² R. Using P = I² R: P = (3 A)² × 6 Ω = 9 × 6 = 54 W. Alternatively, P = 3 A × 18 V = 54 W.
功率 P 可以用 P = I V 或 P = I² R 计算。用 P = I² R：P = (3 A)² × 6 Ω = 9 × 6 = 54 W。或用 P = 3 A × 18 V = 54 W。
8. Series and Parallel Circuits | 串联和并联电路
Two resistors of 4 Ω and 6 Ω are connected in series across a 12 V battery. Calculate (a) the total resistance, (b) the current in the circuit, and (c) the voltage across the 4 Ω resistor.
两个阻值分别为 4 Ω 和 6 Ω 的电阻串联后接在 12 V 电池两端。计算 (a) 总电阻；(b) 电路中的电流；(c) 4 Ω 电阻两端的电压。
For resistors in series, the total resistance R_total = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω.
串联电阻的总电阻 R_total = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω。
The current I is the same through all components in a series circuit. Using Ohm’s Law: I = V / R_total = 12 V / 10 Ω = 1.2 A.
串联电路中电流处处相等。用欧姆定律：I = V / R_total = 12 V / 10 Ω = 1.2 A。
The voltage across the 4 Ω resistor is V_4 = I × R = 1.2 A × 4 Ω = 4.8 V. The remaining voltage (12 V – 4.8 V = 7.2 V) drops across the 6 Ω resistor, which can also be verified.
4 Ω 电阻两端的电压为 V_4 = I × R = 1.2 A × 4 Ω = 4.8 V。剩余电压（12 V – 4.8 V = 7.2 V）降落在 6 Ω 电阻上，可以自行验证。
9. Electromagnetism: The Motor Effect | 电磁学：电动机效应
A straight wire carries a current from left to right and is placed in a uniform magnetic field directed vertically downwards. In which direction does the force on the wire act?
一根直导线通有从左向右的电流，置于方向竖直向下的匀强磁场中。导线所受的力方向如何？
The direction of the force on a current-carrying conductor in a magnetic field is given by Fleming’s left-hand rule. Extend the thumb, first finger and second finger of the left hand mutually at right angles. The first finger points in the direction of the magnetic field (North to South), the second finger points in the direction of conventional current (positive to negative), and the thumb gives the direction of the force (motion).
通电导体在磁场中所受力的方向由弗莱明左手定则确定。将左手的拇指、食指和中指互相垂直伸出。食指指向磁场方向（北到南），中指指向常规电流方向（正到负），拇指即指向力的方向（运动方向）。
Here, magnetic field is downwards (first finger pointing down), current is to the right (second finger pointing right). Aligning the fingers, the thumb points out of the page (towards the observer). So the force acts out of the page.
此题中，磁场向下（食指朝下），电流向右（中指朝右）。将手指排好，拇指指向纸面外（指向观察者）。因此力垂直于纸面向外。
If the current or field direction is reversed, the force direction will also reverse. The magnitude of the force can be increased by increasing current, increasing magnetic field strength, or using a longer length of wire in the field.
如果电流或磁场方向反向，力的方向也随之反向。增加电流、增强磁场或增加导线在磁场中的有效长度都可增大作用力的大小。
10. Radioactivity: Half-Life | 放射性：半衰期
A radioactive isotope has a half-life of 2 days. A sample initially contains 80 g of the isotope. How much of the isotope remains after 8 days?
某放射性同位素的半衰期为 2 天。一个样品最初含有 80 g 该同位素。8 天后还剩下多少克？
Half-life is the time taken for half of the radioactive nuclei in a sample to decay. After each half-life, the remaining mass is halved.
半衰期是指样品中一半的放射性原子核发生衰变所需要的时间。每经过一个半衰期，剩余质量减半。
Number of half-lives elapsed = total time / half-life = 8 days / 2 days = 4 half-lives.
经历的半衰期数 = 总时间 / 半衰期 = 8 天 / 2 天 = 4 个半衰期。
Remaining mass = initial mass × (½)^number of half-lives = 80 g × (½)⁴ = 80 g × (1/16) = 5 g.
剩余质量 = 初始质量 × (½)^(半衰期数) = 80 g × (½)⁴ = 80 g × (1/16) = 5 g。
Thus, after 8 days, only 5 g of the original isotope remains. The rest has decayed into other elements. Note that the mass does not disappear; it is transformed according to the decay series.
因此，8 天后只剩下 5 g 原始同位素。其余已衰变成其他元素。注意质量并没有消失，而是按照衰变链发生了转化。
Published by TutorHao | Physics Revision Series | aleveler.com
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