📚 IGCSE OCR Physics: Detailed Worked Examples | IGCSE OCR 物理典型例题详解
IGCSE OCR Physics exams reward clear, step-by-step working just as much as the correct final answer. This article walks you through ten carefully chosen worked examples spanning the entire syllabus, from mechanics to radioactivity. Each example is broken down into the data you are given, the equation you need, the substitution, and the final result. Reading through these will sharpen your problem‑solving skills and help you avoid common pitfalls.
IGCSE OCR 物理考试不仅看重最终答案，更看重清晰、分步的解题过程。本文为你精选了十道覆盖力学、电学、波动、热学和核物理等全考纲的典型例题，逐步讲解每一题的已知数据、所用公式、代入过程与最终结果。认真研读这些范例，能有效提升你的解题能力，并帮助你避开常见失分点。
1. Speed and Acceleration | 速度与加速度
A car accelerates uniformly from rest at 2.0 m/s² for 5.0 s. Calculate its final velocity and the distance travelled in this time.
一辆汽车从静止开始以 2.0 m/s² 的加速度匀加速行驶 5.0 秒。计算该时间段内的末速度和行驶距离。
Given data: initial velocity u = 0 m/s, acceleration a = 2.0 m/s², time t = 5.0 s. To find final velocity v, use v = u + a t. Substituting: v = 0 + (2.0 × 5.0) = 10 m/s. For distance s, use s = u t + ½ a t². s = 0 × 5.0 + ½ × 2.0 × (5.0)² = 0.5 × 2.0 × 25 = 25 m.
已知：初速度 u = 0 m/s，加速度 a = 2.0 m/s²，时间 t = 5.0 s。求末速度 v 用公式 v = u + a t，代入得 v = 0 + (2.0 × 5.0) = 10 m/s。求距离 s 用 s = u t + ½ a t²，s = 0 × 5.0 + ½ × 2.0 × (5.0)² = 0.5 × 2.0 × 25 = 25 m。
Always check units: m/s for velocity, m for distance. The term ½ at² handles the uniformly changing speed correctly.
始终核查单位：速度单位为 m/s，距离单位为 m。½ a t² 项正确地处理了均匀变化的速度。
2. Newton’s Second Law | 牛顿第二定律
A resultant force of 450 N acts on a car of mass 1200 kg. Find the acceleration of the car. The car then hits a wall and comes to rest in 0.25 s from an initial speed of 8.0 m/s. Calculate the average force exerted by the wall on the car.
一辆质量为 1200 kg 的汽车受到 450 N 的合力。求汽车的加速度。随后该车以 8.0 m/s 的速度撞上墙壁并在 0.25 s 内停下，计算墙壁对汽车的平均作用力。
First part: F = m a → a = F / m = 450 / 1200 = 0.375 m/s². Second part: find deceleration a using a = (v − u) / t. Here u = 8.0 m/s, v = 0, t = 0.25 s. So a = (0 − 8.0) / 0.25 = −32 m/s². The negative sign indicates deceleration. Magnitude of average force F = m |a| = 1200 × 32 = 38 400 N (or 3.84 × 10⁴ N). The force is opposite to the original motion.
第一部分：F = m a → a = F / m = 450 / 1200 = 0.375 m/s²。第二部分：计算减速度 a = (v − u) / t，u = 8.0 m/s，v = 0，t = 0.25 s。a = (0 − 8.0) / 0.25 = −32 m/s²。负号表示减速。平均力的大小 F = m |a| = 1200 × 32 = 38 400 N（即 3.84 × 10⁴ N），方向与运动方向相反。
3. Energy Transfers and Work Done | 能量转换与做功
A crane lifts a 500 kg steel beam vertically through a height of 15 m at constant speed. The gravitational field strength g is 9.8 N/kg. Calculate the work done by the crane and the gain in gravitational potential energy of the beam.
一台起重机以恒定速度将 500 kg 的钢梁竖直提升 15 m。重力场强度 g = 9.8 N/kg。计算起重机所做的功及钢梁增加的重力势能。
The force needed to lift the beam at constant speed equals its weight W = m g = 500 × 9.8 = 4900 N. Work done = force × distance moved in direction of force = 4900 × 15 = 73 500 J (73.5 kJ). By the conservation of energy, the gain in gravitational potential energy (GPE) is also 73 500 J (∆GPE = m g h).
匀速提升所需的力等于钢梁的重力 W = m g = 500 × 9.8 = 4900 N。做功 = 力 × 沿力方向移动的距离 = 4900 × 15 = 73 500 J (73.5 kJ)。根据能量守恒，增加的重力势能同样为 73 500 J (∆GPE = m g h)。
In an exam, always write the formula, substitute numbers with units, and give the answer to an appropriate number of significant figures.
考试中务必写出公式，代入带单位的数字，并给出合理有效位数的答案。
4. Density and Pressure | 密度与压强
A solid aluminium cube has a side length of 0.040 m and a mass of 0.173 kg. The cube rests on a flat surface. Calculate (a) the density of aluminium, (b) the pressure exerted on the surface by the cube.
一块实心的铝质立方体边长为 0.040 m，质量为 0.173 kg。该立方体静置于水平面上。计算 (a) 铝的密度，(b) 立方体对支承面产生的压强。
Volume of cube V = (0.040)³ = 6.4 × 10⁻⁵ m³. Density ρ = mass / volume = 0.173 / (6.4 × 10⁻⁵) = 2703 kg/m³ (≈ 2700 kg/m³ to 2 s.f.). Area of one face A = (0.040)² = 1.6 × 10⁻³ m². Weight W = m g = 0.173 × 9.8 ≈ 1.6954 N. Pressure p = force / area = 1.6954 / (1.6 × 10⁻³) ≈ 1060 Pa (or 1.06 × 10³ Pa).
立方体体积 V = (0.040)³ = 6.4 × 10⁻⁵ m³。密度 ρ = 质量 / 体积 = 0.173 / (6.4 × 10⁻⁵) = 2703 kg/m³ (取两位有效数字约 2700 kg/m³)。一个面的面积 A = (0.040)² = 1.6 × 10⁻³ m²。重力 W = m g = 0.173 × 9.8 ≈ 1.6954 N。压强 p = 力 / 面积 = 1.6954 / (1.6 × 10⁻³) ≈ 1060 Pa (或 1.06 × 10³ Pa)。
5. Ohm’s Law and Series Circuits | 欧姆定律与串联电路
A 6.0 Ω resistor and a 12 Ω resistor are connected in series across a 9.0 V battery. Calculate the total resistance, the current flowing from the battery, and the potential difference across the 12 Ω resistor.
一只 6.0 Ω 电阻与一只 12 Ω 电阻串联后接在 9.0 V 电池两端。计算总电阻、电池输出的电流以及 12 Ω 电阻两端的电压。
In series, R_total = R₁ + R₂ = 6.0 + 12 = 18 Ω. Using Ohm’s law V = I R → I = V / R_total = 9.0 / 18 = 0.50 A. The current is the same everywhere in a series circuit. Potential difference across the 12 Ω resistor: V₁₂ = I R₁₂ = 0.50 × 12 = 6.0 V. (The remaining 3.0 V appears across the 6.0 Ω resistor.)
串联电路中 R_total = R₁ + R₂ = 6.0 + 12 = 18 Ω。由欧姆定律 V = I R 得 I = V / R_total = 9.0 / 18 = 0.50 A。串联电路中各处电流相等。12 Ω 电阻两端的电压 V₁₂ = I R₁₂ = 0.50 × 12 = 6.0 V。（剩余 3.0 V 加在 6.0 Ω 电阻上。）
6. Resistance and Power | 电阻与电功率
A 24 V heater has a power rating of 60 W when connected to the correct supply. Calculate the current through the heater and its resistance. The heater is used for 3.0 hours; determine the energy transferred in kilowatt‑hours.
一台额定功率为 60 W 的电暖器接在 24 V 电源上正常工作。计算通过暖器的电流及其电阻。若暖器工作 3.0 小时，求以千瓦时为单位计量的转移能量。
Power P = V I, so I = P / V = 60 / 24 = 2.5 A. Resistance R = V / I = 24 / 2.5 = 9.6 Ω. Alternatively, R = V² / P = (24)² / 60 = 576 / 60 = 9.6 Ω. Energy E = P t. Convert power to kW: 60 W = 0.060 kW. Time = 3.0 h. E = 0.060 kW × 3.0 h = 0.18 kWh. (In joules, E = 60 × 3.0 × 3600 = 648 000 J.)
功率 P = V I，因此 I = P / V = 60 / 24 = 2.5 A。电阻 R = V / I = 24 / 2.5 = 9.6 Ω；也可用 R = V² / P = (24)² / 60 = 576 / 60 = 9.6 Ω。能量 E = P t，功率单位转换为千瓦：60 W = 0.060 kW，时间 3.0 h，E = 0.060 kW × 3.0 h = 0.18 kWh。（若用焦耳，E = 60 × 3.0 × 3600 = 648 000 J。）
7. Wave Speed and Frequency | 波速与频率
A water wave has a wavelength of 0.80 m and a frequency of 5.0 Hz. Calculate its speed. If the same wave enters a shallower region where its speed drops to 3.2 m/s, what is its new wavelength? (Assume frequency remains constant.)
一道水波波长为 0.80 m，频率为 5.0 Hz。计算波速。若同一列波进入较浅水域后波速降为 3.2 m/s，假定频率不变，求新的波长。
The wave equation: v = f λ. In deep water: v = 5.0 × 0.80 = 4.0 m/s. When the wave enters shallower water, frequency stays at 5.0 Hz. Using v = f λ again, λ = v / f = 3.2 / 5.0 = 0.64 m. The wavelength shortens as the wave slows down.
波动方程：v = f λ。在深水区：v = 5.0 × 0.80 = 4.0 m/s。进入浅水后频率仍为 5.0 Hz，由 v = f λ 得 λ = v / f = 3.2 / 5.0 = 0.64 m。波速减慢时波长变短。
8. Specific Heat Capacity | 比热容
An electric immersion heater supplies 12 000 J of energy to 0.50 kg of water in an insulated container. The temperature of the water rises from 22 °C to 32 °C. Calculate the specific heat capacity of water from this data. The heater is rated 50 W; how long was it switched on?
一个浸没式电加热器向绝热容器中 0.50 kg 的水提供了 12 000 J 的能量，水温从 22 °C 升高到 32 °C。据此计算水的比热容。若加热器功率为 50 W，加热了多长时间？
Temperature change ∆θ = 32 − 22 = 10 °C. Energy supplied Q = m c ∆θ → c = Q / (m ∆θ) = 12 000 / (0.50 × 10) = 12 000 / 5.0 = 2400 J/(kg °C). This is close to the accepted value. Time t = Energy / Power = 12 000 J / 50 W = 240 s (4.0 minutes). Always check significant figures: the data here give 2400 J/(kg °C) to 2 s.f. and 240 s to 2 s.f.
温度变化 ∆θ = 32 − 22 = 10 °C。能量 Q = m c ∆θ → c = Q / (m ∆θ) = 12 000 / (0.50 × 10) = 12 000 / 5.0 = 2400 J/(kg °C)，与标准值接近。时间 t = 能量 / 功率 = 12 000 J / 50 W = 240 s（4.0 分钟）。注意有效数字：数据给出比热容 2400 J/(kg °C)（两位有效数字），时间 240 s（两位有效数字）。
9. Half‑life and Radioactive Decay | 半衰期与放射性衰变
A sample of radioactive material has an initial count rate of 640 counts per minute after background has been subtracted. The count rate drops to 80 counts per minute in 36 minutes. Determine the half‑life of the material.
某放射性样品经扣除本底后初始计数率为每分钟 640 次。36 分钟后计数率降为每分钟 80 次。求该物质的半衰期。
Find how many half‑lives have elapsed: 640 → 320 → 160 → 80. That is three half‑lives. Total time = 36 minutes, so one half‑life = 36 / 3 = 12 minutes. Alternatively, use the ratio: (½)ⁿ = 80/640 = 1/8 → 2⁻ⁿ = 2⁻³ → n = 3. So t₁/₂ = 12 min.
确定经过了几个半衰期：640 → 320 → 160 → 80，共 3 个半衰期。总时间 36 分钟，因此一个半衰期 = 36 / 3 = 12 分钟。也可用比值法：(½)ⁿ = 80/640 = 1/8 → 2⁻ⁿ = 2⁻³ → n = 3，故半衰期为 12 分钟。
10. Motor Effect and Magnetic Force | 电动机效应与磁场力
A straight wire of length 0.15 m carries a current of 4.0 A perpendicular to a uniform magnetic field of flux density 0.60 T. Calculate the force on the wire. If the current is reversed, what happens to the direction of the force?
一根长 0.15 m 的直导线中通有 4.0 A 电流，且电流方向与 0.60 T 的匀强磁场方向垂直。计算导线受力。若电流反向，力的方向如何变化？
When current is perpendicular to the field, F = B I L. F = 0.60 × 4.0 × 0.15 = 0.36 N. The direction is given by Fleming’s left‑hand rule. If the current is reversed, the direction of the force reverses as well (provided the magnetic field direction stays the same).
当电流与磁场垂直时，F = B I L。F = 0.60 × 4.0 × 0.15 = 0.36 N。方向由弗莱明左手定则确定。若电流反向，力的方向也随之反向（磁场方向不变的前提下）。
In an exam you may be asked to sketch the arrangement and label the force direction. Always show the angle between I and B is 90° for maximum force.
考试中可能要求画出装置简图并标出力的方向。记得当电流与磁场夹角为 90° 时安培力最大。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导