📚 IGCSE CCEA Physics: Worked Examples Explained | IGCSE CCEA 物理：典型例题详解
Welcome to this comprehensive revision guide covering a selection of worked examples for IGCSE CCEA Physics. Each section tackles a typical exam-style question, providing clear step-by-step solutions and explanations. Mastering these examples will strengthen your problem-solving skills and deepen your understanding of key physics concepts.
欢迎阅读这本全面的复习指南，其中涵盖了 IGCSE CCEA 物理的精选典型例题。每个小节都解决一道典型的考试风格题目，提供清晰的逐步解答和解释。掌握这些例题将增强你的问题解决能力，加深你对关键物理概念的理解。
1. Motion with Constant Acceleration (SUVAT) | 匀加速直线运动
A car accelerates from rest at 2 m/s² for 8 seconds. Calculate the distance travelled and its final velocity.
一辆汽车从静止开始，以 2 m/s² 的加速度行驶 8 秒。计算其行驶距离和最终速度。
Step 1: List the known quantities. u = 0 m/s, a = 2 m/s², t = 8 s, s = ?, v = ?.
步骤1：列出已知量。 初始速度 u = 0 m/s，加速度 a = 2 m/s²，时间 t = 8 s，距离 s = ?，末速度 v = ?。
Step 2: Choose the equation for distance. Since time, acceleration, and initial velocity are known, use s = ut + ½at².
步骤2：选择距离公式。 已知时间、加速度和初速度，使用 s = ut + ½at²。
s = (0)(8) + ½ × 2 × (8)² = 0 + ½ × 2 × 64 = 64 m
Distance travelled is 64 m.
行驶距离为 64 m。
Step 3: Find the final velocity using v = u + at.
步骤3：用 v = u + at 求末速度。
v = 0 + 2 × 8 = 16 m/s
The car’s final velocity is 16 m/s.
汽车最终速度为 16 m/s。
2. Forces and Newton’s Second Law | 力与牛顿第二定律
A 5 kg box is pulled on a frictionless surface with a horizontal force of 20 N. Calculate its acceleration. Then, explain how the acceleration changes if a friction force of 5 N opposes the motion.
一个 5 kg 的箱子在无摩擦表面上受到 20 N 的水平拉力。计算其加速度。然后解释如果存在 5 N 的摩擦力，加速度将如何变化。
Part 1: No friction. Net force F = 20 N. Using F = ma, a = F / m.
第一部分：无摩擦。 净力 F = 20 N。由 F = ma 得 a = F / m。
a = 20 N / 5 kg = 4 m/s²
Acceleration is 4 m/s² in the direction of the pull.
加速度为 4 m/s²，方向与拉力相同。
Part 2: With friction. The frictional force opposes motion, so net force = applied force – friction = 20 N – 5 N = 15 N.
第二部分：有摩擦。 摩擦力阻碍运动，因此净力 = 外力 – 摩擦力 = 20 N – 5 N = 15 N。
a = 15 N / 5 kg = 3 m/s²
With friction, the acceleration decreases to 3 m/s².
加入摩擦力后，加速度降至 3 m/s²。
3. Energy, Work and Power | 能量、功和功率
A crane lifts a 200 kg mass vertically through a height of 15 m in 10 seconds. Calculate the work done by the crane, the gravitational potential energy gained by the mass, and the power output of the crane. (Take g = 10 N/kg)
一台起重机在 10 秒内将 200 kg 的重物垂直提升 15 m。计算起重机做的功、重物增加的重力势能以及起重机的输出功率。（取 g = 10 N/kg）
Work done and GPE gained: Work done = force × distance. The force required is equal to the weight of the mass, W = mg = 200 kg × 10 N/kg = 2000 N.
功和重力势能增量： 功 = 力 × 距离。所需的力等于物重，W = mg = 200 kg × 10 N/kg = 2000 N。
Work done = 2000 N × 15 m = 30 000 J
The gravitational potential energy (GPE) gained is equal to the work done against gravity, so GPE = mgh = 30 000 J.
重力势能 (GPE) 的增量等于克服重力所做的功，因此 GPE = mgh = 30 000 J。
Power output: Power = work done / time taken.
输出功率： 功率 = 做功 / 所用时间。
P = 30 000 J / 10 s = 3000 W
The crane’s power output is 3000 W (or 3 kW).
起重机的输出功率为 3000 W（或 3 kW）。
4. Momentum and Impulse | 动量与冲量
A 0.5 kg ball travels horizontally at 10 m/s, hits a wall, and rebounds at 6 m/s in the opposite direction. The contact time with the wall is 0.2 s. Calculate the change in momentum of the ball and the average force exerted by the wall on the ball.
一个 0.5 kg 的球以 10 m/s 的水平速度运动，撞击墙壁后以 6 m/s 的速度反向弹回。与墙壁的接触时间为 0.2 s。计算球的动量变化以及墙壁对球施加的平均力。
Change in momentum: Take the initial direction as positive. Initial momentum = m × u = 0.5 kg × 10 m/s = 5 kg m/s. Final momentum = m × v = 0.5 kg × (-6 m/s) = -3 kg m/s (negative due to opposite direction).
动量变化： 取初速度方向为正。初动量 = m × u = 0.5 kg × 10 m/s = 5 kg m/s。末动量 = m × v = 0.5 kg × (-6 m/s) = -3 kg m/s（负号表示方向相反）。
Δp = final momentum – initial momentum = -3 – 5 = -8 kg m/s
The change in momentum is 8 kg m/s in the direction of the rebound (the magnitude is 8 kg m/s).
动量变化大小为 8 kg m/s，方向与反弹方向相同。
Average force: Impulse = change in momentum = F × Δt.
平均力： 冲量 = 动量变化 = F × Δt。
F = Δp / Δt = -8 kg m/s / 0.2 s = -40 N
The negative sign indicates the force is opposite to the initial direction. The average force exerted by the wall is 40 N.
负号表示力的方向与初速度方向相反。墙壁施加的平均力为 40 N。
5. Waves – Refraction in a Ripple Tank | 波 – 波纹槽中的折射
Water waves travel from deep water into shallow water. In deep water, the speed is 12 cm/s and the wavelength is 4 cm. In shallow water, the speed reduces to 9 cm/s. Calculate the wavelength in shallow water. If the incident wavefront makes an angle of 30° to the normal, determine the angle of refraction.
水波从深水区进入浅水区。在深水区，波速为 12 cm/s，波长为 4 cm。在浅水区，波速降至 9 cm/s。计算浅水区的波长。若入射波阵面与法线的夹角为 30°，求折射角。
Finding the wavelength: Frequency remains constant. f = v / λ. In deep water, f = 12 cm/s / 4 cm = 3 Hz.
求波长： 频率保持不变。f = v / λ。在深水区，f = 12 cm/s / 4 cm = 3 Hz。
In shallow water, λ’ = v’ / f = 9 cm/s / 3 Hz = 3 cm.
在浅水区，λ’ = v’ / f = 9 cm/s / 3 Hz = 3 cm。
Angle of refraction: For wave refraction, Snell’s law is: sin i / sin r = v₁ / v₂ (where i is the incident angle, 30°).
折射角： 对于波的折射，斯涅尔定律为：sin i / sin r = v₁ / v₂（其中 i 为入射角，30°）。
sin 30° / sin r = 12 / 9 = 4/3
Thus, sin r = sin 30° × 3/4 = 0.5 × 0.75 = 0.375.
因此，sin r = sin 30° × 3/4 = 0.5 × 0.75 = 0.375。
r = sin⁻¹(0.375) ≈ 22.0°
The wavelength in shallow water is 3 cm and the refracted angle is approximately 22°.
浅水区波长为 3 cm，折射角约为 22°。
6. Electrical Circuits – Ohm’s Law and Resistance | 电路 – 欧姆定律与电阻
A resistor of unknown value is connected to a 6 V battery, and a current of 0.3 A flows through it. Calculate its resistance. Then, two such identical resistors are connected first in series and then in parallel across the same 6 V battery; find the total current drawn from the battery in each case.
一个未知阻值的电阻器连接到 6 V 电池上，流过 0.3 A 的电流。计算其电阻。然后将两个相同的这种电阻器先串联、再并联在同一个 6 V 电池上；分别求出每种情况下从电池流出的总电流。
Single resistor: Using Ohm’s law, R = V / I = 6 V / 0.3 A = 20 Ω.
单个电阻： 使用欧姆定律，R = V / I =
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