📚 GCSE OCR Physics: Worked Examples Step-by-Step | GCSE OCR 物理：典型例题详解
Mastering GCSE OCR Physics requires a solid grasp of key principles and the ability to apply them confidently to a variety of problems. This article presents detailed, step-by-step worked examples drawn from core topics such as mechanics, electricity, energy, waves and nuclear physics. Each example is paired with clear explanations in both English and Chinese to support bilingual revision and deepen your problem-solving skills.
掌握 GCSE OCR 物理要求牢固掌握核心原理，并能够自信地将它们应用于各类问题。本文提供逐步详解的典型例题，涵盖力学、电学、能量、波动与核物理等核心主题。每个例题都配有中英文双语解释，以支持双语复习并深化你的解题能力。
1. Newton’s Second Law and Motion | 牛顿第二定律与运动学
Newton’s second law states that the acceleration of an object is proportional to the net force acting on it and inversely proportional to its mass. The relationship is written as:
F = m a
牛顿第二定律指出，物体的加速度与作用在其上的净力成正比，与质量成反比。该关系写作：
F = m a
where F is the net force in newtons (N), m is mass in kilograms (kg) and a is acceleration in metres per second squared (m/s²).
其中 F 是净力（牛顿，N），m 是质量（千克，kg），a 是加速度（米每平方秒，m/s²）。
Example: A car of mass 1200 kg experiences a constant net forward force of 3000 N. Calculate its acceleration and determine how long it takes to reach 15 m/s from rest.
例题：一辆质量为 1200 kg 的汽车受到 3000 N 的恒定净力。计算其加速度，并求从静止加速到 15 m/s 所需的时间。
Solution: Rearranging F = m a gives a = F / m = 3000 N / 1200 kg = 2.5 m/s².
解答：由 F = m a 得 a = F / m = 3000 N / 1200 kg = 2.5 m/s²。
Using the kinematic equation v = u + a t, with initial velocity u = 0, final velocity v = 15 m/s and a = 2.5 m/s²:
t = v / a = (15 m/s) / (2.5 m/s²) = 6 s
使用运动学方程 v = u + a t，初速度 u = 0，末速度 v = 15 m/s，a = 2.5 m/s²：
t = v / a = (15 m/s) / (2.5 m/s²) = 6 s
The car accelerates at 2.5 m/s² and reaches the target speed in 6 seconds.
汽车以 2.5 m/s² 加速，用 6 秒达到目标速度。
2. Work, Energy and Efficiency | 功、能量与效率
When a force moves an object through a distance, work is done. For lifting an object, the work done against gravity is equal to the gain in gravitational potential energy:
W = m g h
当力使物体移动一段距离时，就做了功。对于提升物体，克服重力做的功等于增加的重力势能：
W = m g h
where m is mass, g is gravitational field strength (10 N/kg on Earth) and h is the height raised.
其中 m 为质量，g 为引力场强（地球取 10 N/kg），h 为提升高度。
Example: An electric motor lifts a 50 kg load vertically through 12 m. The motor receives 8000 J of electrical energy. Calculate the useful work done on the load and the efficiency of the motor.
例题：一台电动机将 50 kg 的重物竖直提升 12 m。电动机接收 8000 J 的电能。计算对重物做的有用功和电动机的效率。
Solution: Useful work done W = m g h = 50 kg × 10 N/kg × 12 m = 6000 J.
解答：有用功 W = m g h = 50 kg × 10 N/kg × 12 m = 6000 J。
Efficiency = (useful output energy / total input energy) × 100% = (6000 J / 8000 J) × 100% = 75%.
Efficiency = 75%
效率 = (有用输出能量 / 总输入能量) × 100% = (6000 J / 8000 J) × 100% = 75%。
效率 = 75%
The motor is 75% efficient, meaning 25% of the input energy is dissipated as heat or sound.
电动机效率为 75%，意味着 25% 的输入能量以热或声音的形式散失。
3. Circuit Rules and Ohm’s Law | 电路规律与欧姆定律
In a series circuit, current is the same everywhere, and the total potential difference (voltage) from the battery is shared across components. Ohm’s law links voltage, current and resistance:
V = I R
在串联电路中，各处电流相同，电池提供的总电压被各元件分担。欧姆定律将电压、电流与电阻联系起来：
V = I R
Example: A 12 V battery is connected in series with a 4 Ω resistor and a 6 Ω resistor. Determine the total current and the voltage across each resistor.
例题：一个 12 V 电池与一个 4 Ω 和一个 6 Ω 的电阻器串联。求总电流及各电阻两端的电压。
Solution: Total resistance Rtotal = 4 Ω + 6 Ω = 10 Ω.
解答：总电阻 R总 = 4 Ω + 6 Ω = 10 Ω。
Using V = I R, total current I = V / Rtotal = 12 V / 10 Ω = 1.2 A.
由 V = I R，总电流 I = V / R总 = 12 V / 10 Ω = 1.2 A。
Voltage across 4 Ω resistor: V₁ = I × 4 Ω = 1.2 A × 4 Ω = 4.8 V.
Voltage across 6 Ω resistor: V₂ = I × 6 Ω = 1.2 A × 6 Ω = 7.2 V.
4 Ω 电阻上的电压：V₁ = I × 4 Ω = 1.2 A × 4 Ω = 4.8 V。
6 Ω 电阻上的电压：V₂ = I × 6 Ω = 1.2 A × 6 Ω = 7.2 V。
Summarised in a table:
汇总于表格：
| Resistor | Resistance (Ω) | Voltage (V) | Current (A) |
|---|---|---|---|
| R₁ | 4 | 4.8 | 1.2 |
| R₂ | 6 | 7.2 | 1.2 |
Notice that 4.8 V + 7.2 V = 12 V, satisfying the conservation of energy in the circuit.
注意 4.8 V + 7.2 V = 12 V，符合电路中能量守恒。
4. Specific Heat Capacity | 比热容
The thermal energy required to change an object’s temperature depends on its mass, its specific heat capacity and the temperature change:
Q = m c Δθ
改变物体温度所需的热能与物体的质量、比热容以及温度变化有关：
Q = m c Δθ
where Q is energy transferred (J), m is mass (kg), c is specific heat capacity (J/(kg °C)) and Δθ is temperature change (°C).
其中 Q 为转移的能量（J），m 为质量（kg），c 为比热容（J/(kg °C)），Δθ 为温度变化（°C）。
Example: How much energy is required to heat 2.0 kg of water from 20 °C to 80 °C? The specific heat capacity of water is 4200 J/(kg °C).
例题：将 2.0 kg 水从 20 °C 加热到 80 °C 需要多少能量？水的比热容为 4200 J/(kg °C)。
Solution: Δθ = 80 °C − 20 °C = 60 °C.
Q = m c Δθ = 2.0 kg × 4200 J/(kg °C) × 60 °C = 504,000 J (or 504 kJ).
解答：Δθ = 80 °C − 20 °C = 60 °C。
Q = m c Δθ = 2.0 kg × 4200 J/(kg °C) × 60 °C = 504,000 J（即 504 kJ）。
Thus, over half a million joules of energy are needed to achieve this temperature rise.
因此，需要超过 50 万焦耳的能量来实现这个温升。
5. Half-Life Calculations | 半衰期计算
The half-life of a radioactive isotope is the time taken for half of its unstable nuclei to decay, or for the count rate to fall by half. After n half-lives, the remaining quantity is (1/2)n of the original.
放射性同位素的半衰期是指一半的不稳定原子核发生衰变，或者计数率减半所需的时间。经过 n 个半衰期后，剩余量是原来的 (1/2)n。
Example: A radioactive sample has an initial count rate of 800 counts per minute and a half-life of 2 days. Predict the count rate after 6 days.
例题：某放射性样品的初始计数率为 800 次/分钟，半衰期为 2 天。预测 6 天后的计数率。
Solution: Number of half-lives elapsed = total time / half-life = 6 days / 2 days = 3.
解答：经过的半衰期次数 = 总时间 / 半衰期 = 6 days / 2 days = 3。
After 1 half-life: count rate = 800 / 2 = 400 counts/min.
After 2 half-lives: 400 / 2 = 200 counts/min.
After 3 half-lives: 200 / 2 = 100 counts/min.
经过 1 个半衰期：计数率 = 800 / 2 = 400 次/分钟。
经过 2 个半衰期：400 / 2 = 200 次/分钟。
经过 3 个半衰期：200 / 2 = 100 次/分钟。
Therefore, after 6 days the count rate will be 100 counts per minute.
因此，6 天后计数率将为 100 次/分钟。
6. Wave Equation Applications | 波动方程应用
For all waves, the speed v is related to frequency f and wavelength λ by the wave equation:
v = f λ
对所有波，波速 v 与频率 f 和波长 λ 的关系由波动方程给出：
v = f λ
Example: A water wave has a frequency of 5 Hz and a wavelength of 0.4 m. Calculate the wave speed. If the wave speed remains the same but the frequency is increased to 10 Hz, what is the new wavelength?
例题：水波的频率为 5 Hz，波长为 0.4 m。计算波速。如果波速保持不变，但频率增加到 10 Hz，新的波长是多少？
Solution: Using v = f λ, v = 5 Hz × 0.4 m = 2 m/s.
解答：由 v = f λ，v = 5 Hz × 0.4 m = 2 m/s。
When frequency becomes 10 Hz, rearrange to λ = v / f = 2 m/s / 10 Hz = 0.2 m.
当频率变为 10 Hz 时，变形得 λ = v / f = 2 m/s / 10 Hz = 0.2 m。
The new wavelength is 0.2 metres, showing that higher frequency means shorter wavelength for the same speed.
新的波长为 0.2 米，表明在相同波速下，
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