📚 IGCSE AQA Physics: Worked Examples Explained in Detail | IGCSE AQA 物理：典型例题详解
Welcome to this comprehensive guide on tackling IGCSE AQA Physics problems. In this article, we walk through a selection of carefully chosen worked examples, breaking each one down into clear, logical steps. Whether you are aiming to consolidate your understanding of key formulas or sharpen your exam technique, these detailed solutions will help you build confidence. We cover topics from forces and motion to electricity, waves, and energy, always emphasising the method behind the answer.
欢迎阅读这篇关于攻克 IGCSE AQA 物理难题的全面指南。在本文中，我们将逐步解析一系列精心挑选的典型例题，把每一道题分解为清晰、合乎逻辑的步骤。无论你是想巩固对关键公式的理解，还是想打磨应试技巧，这些详细的解题过程都将帮助你建立信心。我们涵盖的内容包括力与运动、电学、波动和能量，始终强调答案背后的解题方法。
1. SUVAT Equations: Constant Acceleration | 匀加速运动方程 (SUVAT)
SUVAT equations are the foundation of kinematics in IGCSE Physics. They apply whenever an object moves with uniform acceleration in a straight line. The five variables are s (displacement), u (initial velocity), v (final velocity), a (acceleration), and t (time). You must identify three known quantities to find a fourth, then select the appropriate equation. A common mistake is using the wrong sign for acceleration when an object is decelerating.
SUVAT 方程是 IGCSE 物理中运动学的基础。只要物体沿直线做匀加速运动，这些方程就适用。五个变量分别是 s（位移）、u（初速度）、v（末速度）、a（加速度）和 t（时间）。你必须确定三个已知量才能求出第四个量，然后选择合适的方程。常见的错误是物体减速时加速度的符号用错。
Worked Example: A car accelerates from rest at 3 m/s² for 8 seconds. Calculate the distance travelled.
例题：一辆汽车从静止开始以 3 m/s² 的加速度行驶了 8 秒。计算行驶的距离。
Solution: Known values: u = 0 m/s, a = 3 m/s², t = 8 s. We need s. The best SUVAT equation is s = ut + ½at². Substituting gives s = (0 × 8) + ½ × 3 × 8² = 0 + ½ × 3 × 64 = 96 m. Always check your units and ensure the answer is reasonable — 96 m in 8 s makes sense for rapid acceleration.
解答：已知量：u = 0 m/s，a = 3 m/s²，t = 8 s。我们需要求 s。最合适的 SUVAT 方程是 s = ut + ½at²。代入得 s = (0 × 8) + ½ × 3 × 8² = 0 + ½ × 3 × 64 = 96 m。务必检查单位，并确保答案合理——快速加速下 8 秒行驶 96 m 是合理的。
2. Newton’s Second Law: F = ma | 牛顿第二定律：F = ma
Newton’s Second Law states that the resultant force acting on an object is equal to its mass multiplied by its acceleration. This relationship is fundamental to dynamics problems, especially when multiple forces are present. Always begin by drawing a free-body diagram showing all forces. Remember that the resultant force is the vector sum of all forces, so opposing forces must be subtracted. The unit of force is the newton (N), where 1 N = 1 kg m/s².
牛顿第二定律指出，作用在物体上的合力等于其质量乘以加速度。这一关系是动力学问题的基础，尤其是在存在多个力的情况下。始终从绘制显示所有力的自由体图开始。记住，合力是所有力的矢量和，因此方向相反的力必须相减。力的单位是牛顿 (N)，其中 1 N = 1 kg m/s²。
Worked Example: A 12 kg box is pulled along a frictionless surface by a horizontal force of 48 N. A resistive force of 12 N opposes the motion. Find the acceleration and the distance travelled in 5 seconds from rest.
例题：一个 12 kg 的箱子在光滑水平面上被 48 N 的水平力拉动。有一个 12 N 的阻力阻碍其运动。求加速度以及从静止开始 5 秒内行驶的距离。
Solution: Resultant force = 48 N − 12 N = 36 N. Using F = ma, we get 36 = 12 × a, so a = 3 m/s². Now apply s = ut + ½at² with u = 0: s = 0 + ½ × 3 × 5² = 37.5 m. This two-step problem shows how to combine force and kinematics concepts seamlessly.
解答：合力 = 48 N − 12 N = 36 N。利用 F = ma，得 36 = 12 × a，所以 a = 3 m/s²。现在应用 s = ut + ½at²，其中 u = 0：s = 0 + ½ × 3 × 5² = 37.5 m。这道两步题展示了如何无缝地结合力和运动学的概念。
3. Momentum and Conservation | 动量与动量守恒
Momentum is a vector quantity defined as the product of mass and velocity, p = mv. The principle of conservation of momentum states that in a closed system with no external forces, the total momentum before a collision or explosion equals the total momentum after. This is immensely useful for solving problems involving collisions, recoil, and explosions. Always assign a positive direction to one side and treat velocities in the opposite direction as negative.
动量是一个矢量，定义为质量与速度的乘积，p = mv。动量守恒定律指出，在没有外力的封闭系统中，碰撞或爆炸前的总动量等于碰撞或爆炸后的总动量。这对于解决涉及碰撞、反冲和爆炸的问题非常有用。始终规定一个方向为正，并将相反方向的速度视为负值。
Worked Example: A 2 kg trolley moving at 6 m/s to the right collides with a stationary 4 kg trolley. They stick together. Find their common velocity after the collision.
例题：一辆 2 kg 的小车以 6 m/s 的速度向右运动，与一辆静止的 4 kg 小车碰撞。它们粘在一起。求碰撞后它们的共同速度。
Solution: Take right as positive. Total momentum before = (2 × 6) + (4 × 0) = 12 kg m/s. After collision, combined mass = 2 + 4 = 6 kg. Let v be the common velocity. Conservation gives 12 = 6 × v, so v = 2 m/s to the right. Note how the stuck-together condition simplifies the final momentum calculation.
解答：设向右为正。碰撞前总动量 = (2 × 6) + (4 × 0) = 12 kg m/s。碰撞后，总质量 = 2 + 4 = 6 kg。设 v 为共同速度。根据守恒定律，12 = 6 × v，所以 v = 2 m/s 向右。注意粘在一起的条件简化了最终动量的计算。
4. Work Done and Gravitational Potential Energy | 做功与重力势能
Work done is the energy transferred when a force moves an object through a distance in the direction of the force. The equation is W = Fd. Gravitational potential energy (GPE) gained by an object lifted in a uniform gravitational field is ΔEₚ = mgΔh. These two concepts are linked: the work done lifting an object at constant speed equals the gain in GPE, assuming no energy losses. Understanding this equivalence helps solve many lifting and ramp problems.
做功是指力使物体沿力的方向移动一段距离时所传递的能量。方程为 W = Fd。在均匀重力场中，物体被提升所获得的重力势能 (GPE) 为 ΔEₚ = mgΔh。这两个概念相互关联：以恒定速度提升物体所做的功等于重力势能的增加量，假设没有能量损失。理解这种等价关系有助于解决许多关于提升和斜面的问题。
Worked Example: A 25 kg sack is lifted vertically through 3 m at constant speed. Calculate the work done and the gain in GPE. (g = 9.8 N/kg)
例题：一个 25 kg 的袋子以恒定速度被垂直提升 3 m。计算所做的功和增加的重力势能。(g = 9.8 N/kg)
Solution: Weight = mg = 25 × 9.8 = 245 N. At constant speed, lifting force = weight = 245 N. Work done = Fd = 245 × 3 = 735 J. Gain in GPE = mgΔh = 25 × 9.8 × 3 = 735 J. The equality confirms energy conservation. Always show both calculations when a question asks for them.
解答：重量 = mg = 25 × 9.8 = 245 N。以恒定速度提升，提升力 = 重量 = 245 N。做功 = Fd = 245 × 3 = 735 J。增加的重力势能 = mgΔh = 25 × 9.8 × 3 = 735 J。两者相等，验证了能量守恒。当题目要求两项计算时，务必两者都展示。
5. Kinetic Energy and Velocity | 动能与速度
Kinetic energy (KE) is the energy possessed by a moving object, given by Eₖ = ½mv². This formula reveals a squared relationship with velocity, meaning doubling the speed quadruples the kinetic energy. This is a crucial concept for understanding braking distances and impact forces. Rearranging the equation to find velocity from KE is a common exam skill: v = √(2Eₖ/m).
动能 (KE) 是运动物体所具有的能量，由 Eₖ = ½mv² 给出。该公式揭示了动能与速度的平方关系，这意味着速度加倍会使动能变为原来的四倍。这是理解刹车距离和撞击力的关键概念。根据动能求速度时，需要重新整理公式，这是一项常见的考试技能：v = √(2Eₖ/m)。
Worked Example: A 0.5 kg ball has 64 J of kinetic energy. Find its speed.
例题：一个 0.5 kg 的球具有 64 J 的动能。求它的速度。
Solution: Using Eₖ = ½mv², substitute: 64 = ½ × 0.5 × v². Simplify: 64 = 0.25 × v². So v² = 64 ÷ 0.25 = 256. Taking the square root gives v = 16 m/s. Always double-check that you have halved the mass correctly before rearranging, as this step often trips students up.
解答：使用 Eₖ = ½mv²，代入：64 = ½ × 0.5 × v²。简化：64 = 0.25 × v²。所以 v² = 64 ÷ 0.25 = 256。取平方根得 v = 16 m/s。务必仔细检查在重新整理前是否正确地将质量减半，因为这一步常常让学生出错。
6. Springs and Hooke’s Law | 弹簧与胡克定律
Hooke’s Law states that the extension of a spring is directly proportional to the applied force, provided the elastic limit is not exceeded. This is expressed as F = kx, where k is the spring constant (N/m) and x is the extension (m). The spring constant indicates the stiffness of the spring. Elastic potential energy stored is Eₑ = ½Fx or Eₑ = ½kx². Graphically, the area under a force-extension graph gives the work done.
胡克定律指出，只要不超过弹性限度，弹簧的伸长量与施加的力成正比。这表示为 F = kx，其中 k 是弹簧常数（N/m），x 是伸长量（m）。弹簧常数表示弹簧的刚度。储存的弹性势能为 Eₑ = ½Fx 或 Eₑ = ½kx²。在图像上，力-伸长量图下方的面积表示所做的功。
Worked Example: A spring stretches 4 cm when a 6 N load is hung from it. Find the spring constant in N/m, and the energy stored at this extension.
例题：一个弹簧在悬挂 6 N 的负载时伸长了 4 cm。求以 N/m 为单位的弹簧常数，以及在此伸长量下储存的能量。
Solution: Convert extension to metres: x = 4 cm = 0.04 m. Using F = kx, we get 6 = k × 0.04, so k = 6 ÷ 0.04 = 150 N/m. Energy stored Eₑ = ½Fx = ½ × 6 × 0.04 = 0.12 J. Always convert cm to m in physics calculations to maintain SI consistency and avoid factor-of-100 errors.
解答：将伸长量转换为米：x = 4 cm = 0.04 m。使用 F = kx，得 6 = k × 0.04，所以 k = 6 ÷ 0.04 = 150 N/m。储存的能量 Eₑ = ½Fx = ½ × 6 × 0.04 = 0.12 J。在物理计算中务必将 cm 转换为 m，以保持国际单位制的一致性，避免因差 100 倍而出错。
7. Moments and Equilibrium | 力矩与平衡
A moment is the turning effect of a force about a pivot, calculated as moment = force × perpendicular distance from the pivot. The principle of moments states that for a system in rotational equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments. This principle is essential for solving problems involving levers, beams, and balances. Always identify the pivot clearly and measure perpendicular distances accurately.
力矩是力绕支点产生的转动效应，计算公式为：力矩 = 力 × 到支点的垂直距离。力矩原理指出，对于处于转动平衡的系统，顺时针力矩之和等于逆时针力矩之和。这一原理对于解决涉及杠杆、横梁和天平的问题至关重要。务必清楚地确定支点，并准确测量垂直距离。
Worked Example: A uniform metre rule pivoted at its 50 cm mark has a 2 N weight hanging at the 20 cm mark. What weight must hang at the 80 cm mark to balance the rule horizontally? (Ignore the mass of the rule.)
例题：一把均匀的米尺在其 50 cm 刻度处作为支点，一个 2 N 的重物悬挂在 20 cm 刻度处。要使米尺水平平衡，必须在 80 cm 刻度处悬挂多大的重量？（忽略米尺的质量。）
Solution: Pivot is at 50 cm. The 2 N weight at 20 cm is 30 cm from the pivot (50 − 20). Its moment is anticlockwise: 2 × 0.30 = 0.60 N m. For balance, the clockwise moment must equal 0.60 N m. The weight W at 80 cm is 30 cm from the pivot (80 − 50). So W × 0.30 = 0.60, giving W = 2 N. Symmetry gives the same weight because the distances are equal on both sides.
解答：支点在 50 cm 处。2 N 的重物在 20 cm 处，距离支点 30 cm (50 − 20)。它的力矩是逆时针方向：2 × 0.30 = 0.60 N m。为了平衡，顺时针力矩必须等于 0.60 N m。在 80 cm 处的重量 W 距离支点也是 30 cm (80 − 50)。所以 W × 0.30 = 0.60，得出 W = 2 N。由于两侧距离相等，对称性给出了相同的重量。
8. Pressure in Fluids | 流体的压强
Pressure is defined as the force acting per unit area, P = F/A. In a fluid, pressure increases with depth due to the weight of the fluid above, as given by P = hρg, where h is the depth, ρ is the fluid density, and g is the gravitational field strength. This equation assumes the fluid is incompressible and at rest. Atmospheric pressure at sea level is approximately 100,000 Pa (or 100 kPa).
压强定义为单位面积上作用的力，P = F/A。在流体中，由于上方流体的重量，压强随着深度的增加而增加，由 P = hρg 给出，其中 h 是深度，ρ 是流体密度，g 是重力场强度。该方程假设流体不可压缩且处于静止状态。海平面的大气压强大约为 100,000 Pa（或 100 kPa）。
Worked Example: A diving bell is submerged in seawater of density 1030 kg/m³. Calculate the pressure at a depth of 25 m, given g = 9.8 N/kg. Express your answer in kPa and state the total pressure including atmospheric pressure (101 kPa).
例题：一个潜水钟浸没在密度为 1030 kg/m³ 的海水中。计算在 25 m 深度处的压强，已知 g = 9.8 N/kg。以 kPa 为单位表示你的答案，并说明包括大气压 (101 kPa) 在内的总压强。
Solution: Pressure due to water: P = hρg = 25 × 1030 × 9.8 = 252,350 Pa = 252.35 kPa. Total pressure = water pressure + atmospheric pressure = 252.35 + 101 = 353.35 kPa. Notice how the depth significantly increases the total pressure — nearly 3.5 times atmospheric pressure.
解答：由水产生的压强：P = hρg = 25 × 1030 × 9.8 = 252,350 Pa = 252.35 kPa。总压强 = 水压 + 大气压 = 252.35 + 101 = 353.35 kPa。注意深度是如何显著增加总压强的——接近大气压的 3.5 倍。
9. Electrical Power and Energy Transfer | 电功率与能量传递
Electrical power is the rate at which electrical energy is transferred, calculated using P = IV, P = I²R, or P = V²/R. The choice of formula depends on the known quantities. Energy transferred is E = Pt = IVt, typically measured in joules (J) or kilowatt-hours (kWh) for larger domestic consumption. Understanding which formula is most efficient saves time in multi-step problems.
电功率是电能传递的速率，可使用 P = IV、P = I²R 或 P = V²/R 计算。公式的选择取决于已知量。传递的能量为 E = Pt = IVt，通常以焦耳 (J) 计量，对于较大的家庭用电量则以千瓦时 (kWh) 计量。理解哪个公式最有效率可以节省多步骤问题的时间。
Worked Example: A 230 V electric heater draws a current of 4 A. Calculate its power rating and the energy transferred in 10 minutes in joules.
例题：一个 230 V 的电热器消耗 4 A 的电流。计算其额定功率以及在 10 分钟内以焦耳为单位的能量传递量。
Solution: Power P = IV = 4 × 230 = 920 W. Time t = 10 min = 600 s. Energy E = Pt = 920 × 600 = 552,000 J (or 552 kJ). If the question asked for kWh, we would use hours: E = 0.92 kW × (10/60) h = 0.153 kWh. Showing both conversions demonstrates full understanding.
解答：功率 P = IV = 4 × 230 = 920 W。时间 t = 10 min = 600 s。能量 E = Pt = 920 × 600 = 552,000 J（或 552 kJ）。如果题目要求以 kWh 为单位，我们会使用小时：E = 0.92 kW × (10/60) h = 0.153 kWh。展示两种转换方法能体现对概念的全面理解。
10. Wave Speed and the Wave Equation | 波速与波动方程
The wave equation v = fλ links wave speed (v), frequency (f), and wavelength (λ). This relationship holds for all types of waves, including sound, light, and water waves. Frequency is measured in hertz (Hz), wavelength in metres (m), and speed in metres per second (m/s). When solving problems, check whether you are given frequency or period (T = 1/f), and adjust accordingly.
波动方程 v = fλ 将波速 (v)、频率 (f) 和波长 (λ) 联系起来。这一关系适用于所有类型的波，包括声波、光波和水波。频率以赫兹 (Hz) 为单位，波长以米 (m) 为单位，速度以米每秒 (m/s) 为单位。在解题时，检查给定的是频率还是周期 (T = 1/f)，并相应地调整。
Worked Example: A sound wave travels at 340 m/s in air. If its wavelength is 0.85 m, calculate the frequency and the time period of the wave.
例题：一个声波在空气中的传播速度为 340 m/s。如果它的波长是 0.85 m，计算该波的频率和周期。
Solution: Using v = fλ, rearrange to f = v/λ = 340 ÷ 0.85 = 400 Hz. Time period T = 1/f = 1/400 = 0.0025 s (or 2.5 ms). This frequency lies within the range of human hearing (20 Hz to 20 kHz), which is a useful real-world check for sound problems.
解答：使用 v = fλ，重新整理得 f = v/λ = 340 ÷ 0.85 = 400 Hz。周期 T = 1/f = 1/400 = 0.0025 s（或 2.5 ms）。这个频率在人耳可听范围内 (20 Hz 到 20 kHz)，这对于声波问题是一个有用的实际检验。
11. Resistance and Ohm’s Law | 电阻与欧姆定律
Ohm’s Law states that the current through a conductor is directly proportional to the potential difference across it, provided temperature remains constant. This is expressed as V = IR. Resistance combines with circuit rules: in series, total resistance Rₜ = R₁ + R₂ + …; in parallel, 1/Rₜ = 1/R₁ + 1/R₂ + … . Recognising series and parallel arrangements quickly is a vital exam skill that saves precious minutes.
欧姆定律指出，在温度保持不变的条件下，通过导体的电流与导体两端的电势差成正比。这表示为 V = IR。电阻与电路规则结合：在串联电路中，总电阻 Rₜ = R₁ + R₂ + …；在并联电路中，1/Rₜ = 1/R₁ + 1/R₂ + … 。快速识别串联和并联结构是一项重要的考试技能，可以节省宝贵的时间。
Worked Example: Two resistors of 6 Ω and 3 Ω are connected in parallel, and this combination is connected in series with a 4 Ω resistor. Calculate the total resistance across the whole arrangement.
例题：两个电阻，6 Ω 和 3 Ω，并联连接，然后这个并联组合与一个 4 Ω 的电阻串联。计算整个电路的总电阻。
Solution: For the parallel pair: 1/Rp = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, so Rp = 6/3 = 2 Ω. Then total resistance Rₜ = Rp + 4 = 2 + 4 = 6 Ω. Always work stepwise — simplify parallel branches first, then add series resistances. A common error is adding all resistances directly as if they were in series.
解答：对于并联部分：1/Rp = 1/6 + 1/3 = 1/6 + 2/6 = 3/6，所以 Rp = 6/3 = 2 Ω。然后总电阻 Rₜ = Rp + 4 = 2 + 4 = 6 Ω。始终按步骤进行——先简化并联支路，再加上串联电阻。一个常见错误是像对待串联电路那样直接相加所有电阻。
12. Specific Heat Capacity | 比热容
Specific heat capacity (c) is the energy required to raise the temperature of 1 kg of a substance by 1°C. The thermal energy transferred is Q = mcΔθ, where m is mass, c is specific heat capacity, and Δθ is the temperature change. Water has a particularly high specific heat capacity of 4200 J/(kg °C), which makes it excellent for thermal storage. Be precise with temperature changes — Δθ is always final temperature minus initial temperature.
比热容 (c) 是使 1 kg 物质的温度升高 1°C 所需的能量。传递的热能为 Q = mcΔθ，其中 m 是质量，c 是比热容，Δθ 是温度变化。水的比热容特别高，为 4200 J/(kg °C)，这使得它非常适合热储存。要精确处理温度变化——Δθ 总是末温度减去初温度。
Worked Example: How much energy is needed to heat 2.5 kg of water from 15°C to 85°C? (cw = 4200 J/(kg °C)). If a 2000 W heater is used, how long will this take, assuming no heat losses?
例题：将 2.5 kg 的水从 15°C 加热到 85°C 需要多少能量？(cw = 4200 J/(kg °C))。如果使用 2000 W 的加热器，假设没有热量损失，需要多长时间？
Solution: Δθ = 85 − 15 = 70°C. Q = mcΔθ = 2.5 × 4200 × 70 = 735,000 J. Using P = E/t, rearrange to t = E/P = 735,000 ÷ 2000 = 367.5 s (about 6.13 minutes). This type of multi-concept problem tests your ability to link thermal physics with electrical power seamlessly.
解答：Δθ = 85 − 15 = 70°C。Q = mcΔθ = 2.5 × 4200 × 70 = 735,000 J。使用 P = E/t，重新整理得 t = E/P = 735,000 ÷ 2000 = 367.5 s（约 6.13 分钟）。这类多概念问题考查你将热物理学与电功率无缝联系的能力。
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