📚 IGCSE Physics: Calculation Practice Masterclass | IGCSE 物理计算题专项训练
Welcome to the IGCSE Physics calculation masterclass. This article focuses on the essential quantitative skills required for the IGCSE Physics examination. You will work through major calculation topics, each supported by key formulas, step-by-step strategies and worked examples. Consistent practice with these patterns will boost your confidence and speed when solving numerical problems under exam conditions.
欢迎来到 IGCSE 物理计算专项训练。本文聚焦于 IGCSE 物理考试所必需的核心定量技能。你将逐步攻克各大计算专题，每个专题都配有核心公式、分步解题策略和典型例题。针对这些题型进行持续练习，将显著提升你在考场上解决计算问题的信心与速度。
1. Speed, Velocity and Acceleration | 速度与加速度计算
Average speed is defined as the total distance travelled divided by the total time taken: v = s / t. When motion is in a straight line with uniform acceleration, the following equations of motion apply, where u is initial velocity, v is final velocity, a is acceleration, t is time and s is displacement.
平均速率定义为总路程除以总时间：v = s / t。当物体沿直线做匀加速运动时，可以使用以下运动学方程，其中 u 为初速度，v 为末速度，a 为加速度，t 为时间，s 为位移。
v = u + at
s = ut + ½ at²
v² = u² + 2as
Example: A car accelerates uniformly from 10 m/s to 30 m/s in 5 seconds. Calculate (a) the acceleration, and (b) the distance travelled during this time.
例题：一辆汽车从 10 m/s 匀加速至 30 m/s，用时 5 秒。计算 (a) 加速度，(b) 这段时间内行驶的距离。
Solution part (a): Use v = u + at, rearrange to a = (v – u)/t = (30 – 10)/5 = 4 m/s².
解答 (a)：使用 v = u + at，变形得 a = (v – u)/t = (30 – 10)/5 = 4 m/s²。
Solution part (b): Use s = ut + ½ at² = 10×5 + 0.5×4×25 = 50 + 50 = 100 m. Alternatively, v² = u² + 2as gives 900 = 100 + 2×4×s, so s = 100 m.
解答 (b)：使用 s = ut + ½ at² = 10×5 + 0.5×4×25 = 50 + 50 = 100 m。也可用 v² = u² + 2as，得 900 = 100 + 2×4×s，解得 s = 100 m。
2. Forces and Newton’s Second Law | 力与牛顿第二定律
Newton’s second law states that the resultant force acting on an object is equal to the product of its mass and acceleration: F = m a. Weight is the force due to gravity: W = m g, where g = 9.8 m/s² (often approximated as 10 m/s² in IGCSE). Remember to resolve forces into components when they act at an angle.
牛顿第二定律指出，作用在物体上的合力等于物体质量与加速度的乘积：F = m a。重力是由于引力产生的力：W = m g，其中 g = 9.8 m/s²（IGCSE 中常近似为 10 m/s²）。当力作用有角度时，记得对力进行分解。
Example: A block of mass 8 kg is pulled along a smooth horizontal surface by a force of 40 N. Calculate the acceleration. If the same block is now pulled with the same force but there is a friction force of 10 N opposing the motion, find the new acceleration.
例题：一个质量为 8 kg 的木块在光滑水平面上受到 40 N 的拉力。求加速度。若同一木块在受到相同拉力的同时受到 10 N 的摩擦阻力，求新的加速度。
Solution without friction: F = m a → a = F/m = 40/8 = 5 m/s². With friction: Resultant force = 40 – 10 = 30 N, a = 30/8 = 3.75 m/s².
解答（无摩擦）：a = F/m = 40/8 = 5 m/s²。有摩擦时：合力 = 40 – 10 = 30 N，a = 30/8 = 3.75 m/s²。
Weight example: Calculate the weight of a 5 kg object on Earth (g = 10 m/s²). W = m g = 5 × 10 = 50 N.
重力示例：计算 5 kg 物体在地球上的重量（g = 10 m/s²）。W = m g = 5 × 10 = 50 N。
3. Momentum and Impulse | 动量与冲量
Momentum p is the product of an object’s mass and velocity: p = m v. The principle of conservation of momentum states that in a closed system, total momentum before a collision equals total momentum after, provided no external forces act. Impulse is the change in momentum and equals force multiplied by time: Ft = Δp = m(v – u).
动量 p 是物体质量与速度的乘积：p = m v。动量守恒定律指出，在没有外力作用的封闭系统中，碰撞前的总动量等于碰撞后的总动量。冲量等于动量的变化量，也等于力乘以时间：Ft = Δp = m(v – u)。
Example: A 0.5 kg ball moves at 20 m/s and strikes a wall, rebounding at 12 m/s in the opposite direction. Calculate (a) the initial momentum, (b) the final momentum, and (c) the impulse exerted by the wall.
例题：一个 0.5 kg 的球以 20 m/s 的速度撞击墙壁，并以 12 m/s 的速度反向弹回。计算 (a) 初动量，(b) 末动量，(c) 墙壁施加的冲量。
Solution: Take the initial direction as positive. Initial momentum = 0.5 × 20 = 10 kg m/s. Final velocity = -12 m/s, so final momentum = 0.5 × (-12) = -6 kg m/s. Impulse = final momentum – initial momentum = -6 – 10 = -16 N s. The magnitude is 16 N s.
解答：取初速度方向为正。初动量 = 0.5 × 20 = 10 kg m/s。末速度 = -12 m/s，末动量 = 0.5 × (-12) = -6 kg m/s。冲量 = -6 – 10 = -16 N s，其大小为 16 N s。
4. Pressure | 压强计算
Pressure is defined as force per unit area: P = F / A, measured in pascals (Pa) where 1 Pa = 1 N/m². For a liquid, pressure at a depth h is given by P = ρ g h, where ρ is the density of the liquid and g is the gravitational field strength.
压强定义为单位面积所受的力：P = F / A，单位为帕斯卡 (Pa)，1 Pa = 1 N/m²。液体中深度 h 处的压强由 P = ρ g h 给出，其中 ρ 为液体密度，g 为重力场强度。
Example 1: A force of 600 N acts on an area of 0.3 m². Find the pressure. P = 600 / 0.3 = 2000 Pa.
示例 1：600 N 的力作用在 0.3 m² 的面积上。求压强。P = 600 / 0.3 = 2000 Pa。
Example 2: Calculate the pressure exerted by water (ρ = 1000 kg/m³) at a depth of 15 m. Take g = 10 m/s². P = 1000 × 10 × 15 = 150 000 Pa (1.5 × 10⁵ Pa).
示例 2：计算水深 15 m 处水（ρ = 1000 kg/m³）产生的压强，取 g = 10 m/s²。P = 1000 × 10 × 15 = 150 000 Pa (1.5 × 10⁵ Pa)。
5. Energy, Work and Power | 能量、功与功率
Work done W is the product of force and distance moved in the direction of the force: W = F d. Kinetic energy KE = ½ m v², gravitational potential energy GPE = m g h. Power is the rate of doing work: P = W / t, measured in watts (W).
功 W 等于力与物体在力的方向上移动距离的乘积：W = F d。动能 KE = ½ m v²，重力势能 GPE = m g h。功率是做功的速率：P = W / t，单位为瓦特 (W)。
Example: A 2 kg object is lifted vertically through a height of 5 m at constant speed. (a) Calculate the increase in GPE (g = 10 m/s²). (b) If this lifting takes 4 seconds, what is the power developed?
例题：一个 2 kg 的物体以恒定速度被竖直提升 5 m。(a) 计算重力势能的增加量 (g = 10 m/s²)。(b) 若提升过程耗时 4 秒，求产生的功率。
Solution: (a) GPE = m g h = 2 × 10 × 5 = 100 J. (b) Work done = 100 J, power = W/t = 100/4 = 25 W.
解答：(a) GPE = 2 × 10 × 5 = 100 J。(b) 做功 = 100 J，功率 = 100/4 = 25 W。
Energy conversion example: The same object falls freely from 5 m. Find its speed just before impact, assuming all GPE converts to KE. ½ m v² = m g h, cancel m: ½ v² = 10×5 → v² = 100 → v = 10 m/s.
能量转化示例：同一物体从 5 m 自由下落，假设所有重力势能转化为动能，求落地前的速度。½ m v² = m g h，消去 m 得 ½ v² = 50，v² = 100，v = 10 m/s。
6. Specific Heat Capacity and Latent Heat | 比热容与潜热
When an object is heated, the thermal energy transferred is Q = m c Δθ, where c is the specific heat capacity and Δθ is the temperature change. When a substance changes state, the energy transferred is Q = m L, where L is the specific latent heat (of fusion or vaporisation).
物体受热时传递的热能为 Q = m c Δθ，其中 c 为比热容，Δθ 为温度变化量。当物质发生物态变化时，传递的能量为 Q = m L，其中 L 为比潜热（熔化潜热或汽化潜热）。
Example: How much energy is required to heat 3 kg of water from 20°C to 100°C? (c = 4200 J/(kg °C)). Q = 3 × 4200 × (100 – 20) = 3 × 4200 × 80 = 1 008 000 J (≈ 1.01 MJ).
例题：将 3 kg 水从 20°C 加热至 100°C 需要多少能量？(c = 4200 J/(kg °C))。Q = 3 × 4200 × 80 = 1 008 000 J (≈ 1.01 MJ)。
Latent heat example: Calculate the energy required to melt 0.5 kg of ice at 0°C without a temperature change. (specific latent heat of fusion of ice = 334 000 J/kg). Q = m L = 0.5 × 334 000 = 167 000 J.
潜热示例：计算使 0.5 kg 0°C 的冰完全熔化而不升温所需的能量（冰的比熔化潜热 = 334 000 J/kg）。Q = 0.5 × 334 000 = 167 000 J。
7. Waves: Speed, Refraction and Critical Angle | 波：波速、折射与临界角
The wave equation relates speed v, frequency f and wavelength λ: v = f λ. For light, refraction at a boundary is described by Snell’s law: n = sin i / sin r, where i is the angle of incidence and r is the angle of refraction. The refractive index n can also be linked to the critical angle c by n = 1 / sin c.
波动方程关联波速 v、频率 f 和波长 λ：v = f λ。对于光，界面上的折射由斯涅尔定律描述：n = sin i / sin r，其中 i 为入射角，r 为折射角。折射率 n 与临界角 c 的关系为 n = 1 / sin c。
Example 1: A water wave has a frequency of 5 Hz and a wavelength of 0.4 m. Find the wave speed. v = f λ = 5 × 0.4 = 2 m/s.
示例 1：水波的频率为 5 Hz，波长为 0.4 m，求波速。v = 5 × 0.4 = 2 m/s。
Example 2: Light passes from air into glass with an angle of incidence of 45° and an angle of refraction of 28°. Calculate the refractive index of the glass. n = sin 45° / sin 28° ≈ 0.7071 / 0.4695 ≈ 1.51.
示例 2：光从空气射入玻璃，入射角为 45°，折射角为 28°，求玻璃的折射率。n = sin 45° / sin 28° ≈ 0.7071 / 0.4695 ≈ 1.51。
Example 3: If the refractive index of a medium is 1.5, find its critical angle. sin c = 1/n = 1/1.5 ≈ 0.6667, so c = sin⁻¹(0.6667) ≈ 41.8°.
示例 3：某介质的折射率为 1.5，求其临界角。sin c = 1/1.5 ≈ 0.6667，故 c ≈ 41.8°。
8. Ohm’s Law and Resistance | 欧姆定律与电阻
Ohm’s law states that the current I through a resistor is directly proportional to the potential difference V across it, provided temperature remains constant: V = I R. For resistors in series, Rtotal = R₁ + R₂ + … For two resistors in parallel, the combined resistance is given by 1/Rtotal = 1/R₁ + 1/R₂, or Rtotal = (R₁ × R₂) / (R₁ + R₂).
欧姆定律指出，在温度恒定的条件下，流过电阻的电流 I 与其两端的电压 V 成正比：V = I R。电阻串联时，总电阻 Rtotal = R₁ + R₂ + … 两个电阻并联时，总电阻由 1/Rtotal = 1/R₁ + 1/R₂ 计算，或 Rtotal = (R₁ × R₂) / (R₁ + R₂)。
Example: A 12 V battery is connected to a 6 Ω resistor in series with a 3 Ω resistor. Calculate (a) the total resistance, (b) the current in the circuit, and (c) the potential difference across the 6 Ω resistor.
例题：一个 12 V 的电池与 6 Ω 和 3 Ω 的电阻串联。计算 (a) 总电阻，(b) 电路中的电流，(c) 6 Ω 电阻两端的电压。
Solution: (a) Rtotal = 6 + 3 = 9 Ω. (b) I = V / Rtotal = 12 / 9 = 1.33 A. (c) V across 6 Ω = I × 6 = 1.33 × 6 = 8 V. Check: V across 3 Ω = 4 V; sum = 12 V.
解答：(a) Rtotal = 9 Ω。(b) I = 12 / 9 ≈ 1.33 A。(c) 6 Ω 两端电压 = 1.33 × 6 = 8 V。验证：3 Ω 电阻电压为 4 V，总和为 12 V。
9. Electrical Power
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