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

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

This article walks through ten typical GCSE WJEC Physics questions, showing detailed step-by-step solutions. Each worked example targets a key skill from the specification, such as motion graphs, forces, energy, circuits, waves, nuclear decay, moments and specific heat capacity. By following the reasoning and calculations, you can strengthen your problem-solving approach and be fully prepared for the exam.

本文精选十道 WJEC GCSE 物理典型例题，逐步展示详细解答过程。每道题覆盖考纲核心技能，包括运动图像、力、能量、电路、波、核衰变、力矩和比热容。通过跟随解题思路与计算，你可以强化解题方法，为考试做好充分准备。

1. Finding Acceleration and Distance from a Velocity-Time Graph | 从速度-时间图求加速度和距离

Question: A car accelerates uniformly from rest to 20 m/s in 5 seconds. It then travels at constant velocity for a further 10 seconds. Draw the velocity-time graph and use it to calculate the acceleration in the first 5 s and the total distance travelled.

题目：一辆汽车从静止开始匀加速，5 秒内速度达到 20 m/s，然后匀速行驶 10 秒。画出速度-时间图，并利用图像计算前 5 秒的加速度和总共行驶的距离。

The graph has two sections: a straight line from (0,0) to (5,20) and a horizontal line from (5,20) to (15,20).

图像分为两段：一条从 (0,0) 到 (5,20) 的直线，和一条从 (5,20) 到 (15,20) 的水平线。

Acceleration a = gradient of the velocity-time graph = (change in velocity) / time taken.

加速度 a = 速度-时间图的斜率 = 速度变化量 / 所用时间。

a = (v – u) / t = (20 – 0) / 5 = 4 m/s²

a = (20 – 0) / 5 = 4 m/s²

Distance travelled = area under the graph. Split into two areas: a triangle for the acceleration phase and a rectangle for constant velocity.

行驶距离 = 图像下的面积。分成两个区域：加速阶段的三角形和匀速阶段的矩形。

Triangle area = ½ × base × height = ½ × 5 s × 20 m/s = 50 m.

三角形面积 = ½ × 底 × 高 = ½ × 5 s × 20 m/s = 50 m。

Rectangle area = length × height = 10 s × 20 m/s = 200 m.

矩形面积 = 长 × 高 = 10 s × 20 m/s = 200 m。

Total distance = 50 + 200 = 250 m.

总距离 = 50 + 200 = 250 m。

2. Applying Newton’s Second Law | 应用牛顿第二定律

Question: A resultant force of 300 N acts on a trolley of mass 60 kg. Calculate its acceleration. A frictional force of 50 N opposes the motion. What is the new acceleration if the driving force remains unchanged?

题目：合力 300 N 作用在一辆质量为 60 kg 的小车上，计算其加速度。如果存在 50 N 的摩擦力阻碍运动，且驱动力保持不变，新的加速度是多少？

First, use F = m a with the resultant force.

首先，使用 F = m a 代入合力。

a = F / m = 300 N / 60 kg = 5 m/s²

When friction acts, the resultant force = driving force – friction. The driving force is the original 300 N, so resultant = 300 – 50 = 250 N.

当存在摩擦力时，合力 = 驱动力 – 摩擦力。驱动力仍是原来的 300 N，所以合力 = 300 – 50 = 250 N。

New acceleration a = 250 N / 60 kg = 4.17 m/s² (approx 4.2 m/s²).

新加速度 a = 250 N / 60 kg = 4.17 m/s²（约 4.2 m/s²）。

3. Calculating Kinetic Energy and Gravitational Potential Energy | 计算动能和重力势能

Question: A ball of mass 0.5 kg is dropped from a height of 8 m. Ignoring air resistance, calculate its gravitational potential energy at the top, its kinetic energy just before hitting the ground, and its speed at that moment. (g = 10 N/kg)

题目：一个质量为 0.5 kg 的小球从 8 m 高处下落。忽略空气阻力，计算它在最高点的重力势能、即将撞击地面前的动能，以及那一刻的速度。（g = 10 N/kg）

Gravitational potential energy E_p = m g h = 0.5 × 10 × 8 = 40 J.

重力势能 E_p = m g h = 0.5 × 10 × 8 = 40 J。

By conservation of energy, kinetic energy at the bottom E_k = E_p = 40 J.

根据能量守恒，底部的动能 E_k = 重力势能 = 40 J。

Kinetic energy formula: E_k = ½ m v².

动能公式：E_k = ½ m v²。

40 = ½ × 0.5 × v² → 40 = 0.25 v² → v² = 160 → v = √160 ≈ 12.6 m/s

4. Series Circuit Analysis | 串联电路分析

Question: Two resistors, 4 Ω and 6 Ω, are connected in series to a 12 V battery. Calculate the total resistance, the current in the circuit, and the voltage across the 6 Ω resistor.

题目：两个电阻，4 Ω 和 6 Ω，串联后接到 12 V 电池上。计算总电阻、电路中的电流以及 6 Ω 电阻两端的电压。

Total resistance R_total = R₁ + R₂ = 4 + 6 = 10 Ω.

总电阻 R_total = R₁ + R₂ = 4 + 6 = 10 Ω。

Current I = V / R_total = 12 V / 10 Ω = 1.2 A.

电流 I = V / R_total = 12 V / 10 Ω = 1.2 A。

Voltage across 6 Ω resistor V = I × R = 1.2 A × 6 Ω = 7.2 V.

6 Ω 电阻两端的电压 V = I × R = 1.2 A × 6 Ω = 7.2 V。

5. Electrical Power and Energy Transferred | 电功率与能量转移

Question: A 230 V mains kettle draws a current of 8 A. Calculate its power and the energy transferred in 5 minutes. Give the energy in joules and kilowatt-hours.

题目：一个 230 V 的家用电水壶工作电流为 8 A。计算其功率以及在 5 分钟内转移的能量。能量分别用焦耳和千瓦时表示。

Power P = V × I = 230 V × 8 A = 1840 W (1.84 kW).

功率 P = V × I = 230 V × 8 A = 1840 W（1.84 kW）。

Time t = 5 min = 5 × 60 = 300 s.

时间 t = 5 分钟 = 5 × 60 = 300 秒。

Energy E = P × t = 1840 W × 300 s = 552 000 J (or 552 kJ).

能量 E = P × t = 1840 W × 300 s = 552 000 J（或 552 kJ）。

In kilowatt-hours: E = 1.84 kW × (5/60) h = 1.84 × 0.0833 = 0.153 kWh (about 0.15 kWh).

用千瓦时表示：E = 1.84 kW × (5/60) h = 1.84 × 0.0833 = 0.153 kWh（约 0.15 kWh）。

6. Using the Wave Equation | 运用波动方程

Question: A water wave has a frequency of 5 Hz and a wavelength of 0.4 m. Calculate its speed. If the frequency is doubled but the speed stays the same, what happens to the wavelength?

题目：水波的频率为 5 Hz，波长为 0.4 m。计算波速。如果频率加倍而波速不变，波长会如何变化？

Wave equation: v = f λ.

波动方程：v = f λ。

v = 5 Hz × 0.4 m = 2 m/s

If frequency doubles (f = 10 Hz) and v remains 2 m/s, then λ = v / f = 2 / 10 = 0.2 m. The wavelength halves.

如果频率加倍 (f = 10 Hz) 而 v 仍然为 2 m/s，则 λ = v / f = 2 / 10 = 0.2 m。波长减半。

7. Balancing Nuclear Decay Equations | 配平核衰变方程

Question: Uranium-238 emits an alpha particle to become thorium. Write the balanced nuclear equation. Also, carbon-14 decays by beta emission to nitrogen. Write the balanced equation.

题目：铀-238 释放一个 α 粒子变成钍。写出配平的核方程。碳-14 通过 β 衰变变成氮，写出配平的核方程。

Alpha particle is ⁴₂He, so for uranium-238 (²³⁸₉₂U):

α 粒子是 ⁴₂He，铀-238 为 ²³⁸₉₂U：

²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He

Mass numbers: 238 = 234 + 4. Atomic numbers: 92 = 90 + 2. Balanced.

质量数：238 = 234 + 4。原子序数：92 = 90 + 2。已配平。

Beta particle is ⁰₋₁e. Carbon-14 (¹⁴₆C) beta decays:

β 粒子为 ⁰₋₁e。碳-14 (¹⁴₆C) 发生 β 衰变：

¹⁴₆C → ¹⁴₇N + ⁰₋₁e

Mass number unchanged (14), atomic number increases by 1 (6 → 7).

质量数不变 (14)，原子序数增加 1 (6 → 7)。

8. Working with Half-Life | 半衰期计算

Question: A radioactive sample has a half-life of 3 days. Its initial activity is 800 counts per second. Calculate the activity after 12 days and determine the fraction of the sample remaining after 9 days.

题目：一个放射性样品的半衰期为 3 天，初始活度为每秒 800 次计数。计算 12 天后的活度，并确定 9 天后样品剩余的比例。

Number of half-lives in 12 days = 12 / 3 = 4 half-lives.

12 天内的半衰期个数 = 12 / 3 = 4 个半衰期。

Activity halves each half-life: after 1: 800/2=400; 2: 200; 3: 100; 4: 50 counts/s.

每经过一个半衰期活度减半：1 个后 800/2=400；2 个后 200；3 个后 100；4 个后 50 次/秒。

Alternatively, activity = 800 × (½)⁴ = 800 × 1/16 = 50 counts/s.

或用公式，活度 = 800 × (½)⁴ = 800 × 1/16 = 50 次/秒。

After 9 days (3 half-lives), fraction remaining = (½)³ = 1/8.

9 天（3 个半衰期）后，剩余比例 = (½)³ = 1/8。

9. Applying the Principle of Moments | 应用力矩原理

Question: A uniform metre ruler is pivoted at its 50 cm mark. A 2 N weight is hung at the 10 cm mark. At which mark must a 4 N weight be hung to balance the ruler?

题目：一根均匀的米尺在 50 cm 刻度处支起。在 10 cm 刻度处挂一个 2 N 的重物。要将米尺平衡，应在哪个刻度处挂一个 4 N 的重物？

Distance of 2 N force from pivot = 50 cm – 10 cm = 40 cm (anticlockwise moment).

2 N 的力到支点的距离 = 50 cm – 10 cm = 40 cm（逆时针力矩）。

Let the 4 N weight be hung at distance d from pivot on the opposite side (clockwise moment). For balance, anticlockwise moment = clockwise moment.

设 4 N 的重物挂在支点另一侧，距离为 d（顺时针力矩）。平衡时，逆时针力矩 = 顺时针力矩。

2 N × 0.40 m = 4 N × d → d = 0.20 m = 20 cm from pivot.

2 N × 0.40 m = 4 N × d → d = 0.20 m = 20 cm（距支点 20 cm）。

Position on ruler = 50 cm + 20 cm = 70 cm mark.

在米尺上的位置 = 50 cm + 20 cm = 70 cm 刻度处。

10. Specific Heat Capacity and Thermal Energy | 比热容与热能

Question: An aluminium block of mass 2 kg is heated by a 50 W electric heater for 5 minutes. The temperature rises from 20 °C to 32 °C. Calculate the specific heat capacity of aluminium and the energy supplied. Assume no heat losses.

题目：一个质量为 2 kg 的铝块用 50 W 电热器加热 5 分钟。温度从 20 °C 升高到 32 °C。计算铝的比热容和提供的能量。假设没有热量损失。

Energy supplied E = P × t = 50 W × (5 × 60) s = 50 × 300 = 15 000 J.

提供的能量 E = P × t = 50 W × (5 × 60) s = 50 × 300 = 15 000 J。

Temperature change Δθ = 32 – 20 = 12 °C (or 12 K).

温度变化 Δθ = 32 – 20 = 12 °C（或 12 K）。

Using ΔE = m c Δθ → c = ΔE / (m Δθ) = 15 000 / (2 × 12) = 15 000 / 24 = 625 J/(kg °C).

使用 ΔE = m c Δθ → c = ΔE / (m Δθ) = 15 000 / (2 × 12) = 15 000 / 24 = 625 J/(kg °C)。

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