IGCSE AQA Physics: Calculation Practice | IGCSE AQA 物理：计算题专项训练

📚 IGCSE AQA Physics: Calculation Practice | IGCSE AQA 物理：计算题专项训练

Calculation questions carry significant weight in IGCSE AQA Physics exams. Success depends on systematic problem-solving: identifying known quantities, selecting the correct formula, converting units, and presenting the final answer with proper units and significant figures. This article provides targeted training with worked examples for the most common calculation types.

计算题在 IGCSE AQA 物理考试中占有重要分值。成功取决于系统的解题方法：识别已知量、选择正确的公式、换算单位，并以恰当的单位和有效数字呈现最终答案。本文针对最常见的计算类型提供专项训练和例题解析。

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1. Kinematics Equations (SUVAT) | 运动学方程 (SUVAT)

The five SUVAT equations describe motion with uniform acceleration in a straight line: v = u + at, s = ½(u+v)t, s = ut + ½at², v² = u² + 2as, and s = vt − ½at². Always define a positive direction; quantities acting opposite become negative.

五个 SUVAT 方程描述匀加速直线运动：v = u + at、s = ½(u+v)t、s = ut + ½at²、v² = u² + 2as 和 s = vt − ½at²。始终定义正方向；反方向的矢量取负值。

Before choosing an equation, list every known quantity with its symbol and value, and note the unknown you need to find. Select the equation that contains only that unknown.

选择方程之前，列出每个已知量的符号和数值，并标明需要求的未知量。选择只含有该未知量的方程。

Example: A stone falls from rest off a cliff and takes 3.0 s to hit the ground. Take g = 10 m/s². Find the height of the cliff.

例题：一块石头从悬崖边静止下落，3.0 秒后触地。取 g = 10 m/s²。求悬崖高度。

Step 1: Known values – u = 0, t = 3.0 s, a = 10 m/s² (downward), unknown displacement s.

第1步：已知量 – u = 0，t = 3.0 s，a = 10 m/s²（向下），未知量位移 s。

Step 2: Choose the equation without v: s = ut + ½at² = 0 + ½ × 10 × (3.0)² = 45 m.

第2步：选择不含 v 的方程：s = ut + ½at² = 0 + ½ × 10 × (3.0)² = 45 m。

The cliff is 45 m high. Always include units and check that the answer is sensible.

悬崖高度为 45 m。务必包含单位并检查答案是否合理。

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2. Newton’s Second Law and Resultant Force | 牛顿第二定律与合力

Newton’s second law states that the resultant force F = ma, where m is mass (kg), a is acceleration (m/s²), and F is in newtons (N). F must be the net force; if multiple forces act, resolve or subtract them first.

牛顿第二定律指出合力 F = ma，其中 m 为质量（kg），a 为加速度（m/s²），F 单位为牛顿（N）。F 必须是净合力；如果多个力作用，需先分解或相减。

Example: A 12 kg crate is pulled along a horizontal floor by a 40 N force. A friction force of 10 N opposes the motion. Calculate the acceleration.

例题：一个 12 kg 箱子受到 40 N 水平拉力在地板上运动，摩擦力为 10 N 与运动方向相反。求加速度。

Step 1: Resultant force = 40 N − 10 N = 30 N (in the direction of the pull).

第1步：合力 = 40 N − 10 N = 30 N（沿拉力方向）。

Step 2: a = F / m = 30 N / 12 kg = 2.5 m/s².

第2步：a = F / m = 30 N / 12 kg = 2.5 m/s²。

Always state the direction of acceleration when relevant.

在相关时，始终说明加速度的方向。

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3. Momentum and Impulse | 动量与冲量

Momentum p = mv is a vector, with unit kg m/s. In collisions or explosions, total momentum is conserved if no external resultant force acts.

动量 p = mv 是矢量，单位 kg m/s。在碰撞或爆炸中，若无外部合力作用，总动量守恒。

Impulse = change in momentum = FΔt = Δp. It equals the area under a force-time graph.

冲量 = 动量变化量 = FΔt = Δp。它等于力-时间图线下的面积。

Example: A 1500 kg car travelling at 20 m/s collides with a stationary 1000 kg car. They stick together. Find their common velocity just after impact.

例题：一辆 1500 kg 汽车以 20 m/s 的速度行驶，与一辆静止的 1000 kg 汽车碰撞并连在一起。求撞击后瞬间的共同速度。

Conservation of momentum: (1500 × 20) + (1000 × 0) = (1500 + 1000) × v → 30000 = 2500v → v = 12 m/s.

动量守恒：(1500 × 20) + (1000 × 0) = (1500 + 1000) × v → 30000 = 2500v → v = 12 m/s。

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4. Work, Energy and Power | 功、能和功率

Work done = W = Fd cosθ; when force and displacement are in the same direction, W = Fd. Kinetic energy Ek = ½mv², gravitational potential energy Ep = mgh. Power P = W/t or P = Fv for constant speed.

做功 = W = Fd cosθ；当力与位移方向相同时，W = Fd。动能 Ek = ½mv²，重力势能 Ep = mgh。功率 P = W/t 或匀速时 P = Fv。

Example: A crane lifts a 500 kg load vertically through 20 m at constant speed in 10 s. Take g = 10 N/kg. Calculate the power output.

例题：一台起重机将 500 kg 重物匀速垂直提升 20 m，用时 10 s。取 g = 10 N/kg。求输出功率。

Lifting force equals weight = mg = 500 × 10 = 5000 N. Work done = 5000 N × 20 m = 100000 J. Power = 100000 J / 10 s = 10000 W = 10 kW.

提升力等于重力 = mg = 500 × 10 = 5000 N。做功 = 5000 N × 20 m = 100000 J。功率 = 100000 J / 10 s = 10000 W = 10 kW。

Alternatively, speed v = 20 m / 10 s = 2 m/s; P = Fv = 5000 × 2 = 10000 W.

或者，速度 v = 20 m / 10 s = 2 m/s；P = Fv = 5000 × 2 = 10000 W。

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5. Density and Pressure | 密度与压强

Density ρ = m / V (kg/m³). Pressure p = F / A (Pa). In a liquid, pressure at depth p = hρg, where h is the vertical depth in metres.

密度 ρ = m / V（kg/m³）。压强 p = F / A（Pa）。在液体中，深度处的压强 p = hρg，其中 h 为竖直深度（米）。

Example: A cylindrical tank of base area 0.02 m² contains water of density 1000 kg/m³ to a depth of 0.50 m. Calculate the water pressure on the base and the total force on the base.

例题：一个底面积 0.02 m² 的圆柱形水箱，装有密度 1000 kg/m³ 的水深 0.50 m。求水对箱底的压强和总压力。

Pressure p = hρg = 0.50 × 1000 × 10 = 5000 Pa. Force F = pA = 5000 × 0.02 = 100 N.

压强 p = hρg = 0.50 × 1000 × 10 = 5000 Pa。压力 F = pA = 5000 × 0.02 = 100 N。

Remember that liquid pressure depends only on vertical depth and density, not on the shape of the container.

记住，液体压强只取决于竖直深度和密度，与容器形状无关。

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6. Ohm’s Law and Circuit Calculations | 欧姆定律与电路计算

Ohm’s law: V = IR, where V is potential difference (V), I is current (A), and R is resistance (Ω).

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