IB & WJEC Physics: Calculation Question Drill | IB与WJEC物理：计算题专项训练

📚 IB & WJEC Physics: Calculation Question Drill | IB与WJEC物理：计算题专项训练

Physics calculations form the backbone of exam success in both IB and WJEC specifications. This guide provides a structured approach to mastering quantitative problem‑solving, from fundamental unit handling to advanced mechanics, fields, and thermodynamics. Each section pairs a key technique with examples that mirror the style and depth required by these boards.

物理计算是IB和WJEC考试成功的关键。本指南提供了一套系统量化解题方法，涵盖从基本单位处理到进阶力学、场和热力学的内容。每一节将关键技巧与贴近这两个考试局题型风格和深度的例题配对，帮助你在计算题中稳操胜券。

1. Extracting Data and Visualising the Problem | 提取数据与可视化问题

Begin by highlighting numerical values and their units directly on the question paper. Sketch a clear diagram labelling all forces, velocities, and field directions. Convert every quantity into SI base units before plugging into equations.

首先直接在题目纸上标出所有数值及其单位。绘制清晰的示意图，标出所有的力、速度以及场的方向。在代入方程之前，将每一个物理量转换成国际单位制基本单位。

For example, if a car accelerates from 54 km/h to 90 km/h over 120 m, rewrite 54 km/h = 15 m/s and 90 km/h = 25 m/s. The sketch then shows an arrow for initial velocity, final velocity, and displacement, making the kinematics choice obvious.

例如，一辆汽车从54 km/h加速到90 km/h，行驶了120 m，则先把54 km/h改写为15 m/s，90 km/h为25 m/s。示意图上再画出初速度、末速度和位移的箭头，运动学公式的选择就变得一目了然。

2. Units, Significant Figures and Dimensional Checking | 单位、有效数字与量纲检查

Always write the unit next to each substituted value. Use dimensional analysis as a quick error detector: if a force calculation gives kg·m/s, you have missed a factor of s⁻¹. IB and WJEC mark schemes consistently penalise missing or mismatched units and incorrect significant figures.

始终在每一个代入值旁写下单位。利用量纲分析快速查错：如果力的计算结果得到kg·m/s，就说明漏掉了s⁻¹因子。IB和WJEC的评分标准一向会扣掉缺少或单位错误以及有效数字不正确的分数。

In final answers, match the least number of significant figures from the given data. For intermediate steps, keep one extra figure to avoid rounding errors. A mistake often seen is reporting g = 9.8 N/kg as 10 N/kg and then losing precision in energy sums.

最终答案的有效数字应与已知数据中最少的保持一致。中间步骤多保留一位数字以避免舍入误差。一个常见错误是把g=9.8 N/kg当成10 N/kg，结果在能量求和中丢失精度。

Length: m, mm → 10⁻³ m, km → 10³ m
Mass: g → 10⁻³ kg, tonne → 10³ kg
Time: minutes → 60 s, hours → 3600 s
Force: N = kg·m/s²
Energy: J = N·m = kg·m²/s²

长度：m，mm → 10⁻³ m，km → 10³ m
质量：g → 10⁻³ kg，吨 → 10³ kg
时间：分钟 → 60 s，小时 → 3600 s
力：N = kg·m/s²
能量：J = N·m = kg·m²/s²

3. Selecting the Right Equation & Rearranging | 选择正确方程并移项

Write the relevant formula from the data booklet before substituting. Rearrange the equation symbolically first, then insert numbers. This reduces substitution slips and shows the examiner your logic even if arithmetic fails.

代入前先将数据手册中的相关公式写下来。先用符号移项，再代入数字。这样可以减少代换错误，即使计算失误，考官也能看清你的逻辑。

For example, to find final temperature in a thermal energy transfer where Q = mcΔθ, rearrange to Δθ = Q/(mc) before putting in Q = 12 500 J, m = 0.45 kg, c = 4200 J/(kg·°C).

例如，要在热传递中求末温，Q = mcΔθ，先移项得到Δθ = Q/(mc)，再代入Q = 12 500 J, m = 0.45 kg, c = 4200 J/(kg·°C)。

v = u + at

v² = u² + 2as

s = ut + ½at²

These three kinematic equations suffice for constant linear acceleration. Check whether the question involves time; if not, the second equation is fastest.

以上三个运动学方程足以处理匀变速直线运动。判断问题是否涉及时间；若不涉及，第二个方程最快。

4. Kinematics in One and Two Dimensions | 一维与二维运动学

Treat horizontal and vertical motions independently. Projectile problems require splitting initial velocity into vₓ = v cos θ and vᵧ = v sin θ. Vertical motion uses a = −g, while horizontal motion has constant velocity.

水平与竖直运动要分开处理。抛体问题需要将初速度分解为vₓ = v cos θ和vᵧ = v sin θ。竖直方向加速度a = −g，水平方向则匀速。

A ball kicked at 22 m/s at 35° above the ground: vₓ = 22 cos 35° ≈ 18.0 m/s, vᵧ = 22 sin 35° ≈ 12.6 m/s. Time of flight t = 2vᵧ/g = 2×12.6/9.81 ≈ 2.57 s. Range = vₓ × t = 18.0 × 2.57 ≈ 46.3 m.

一只足球以22 m/s的速度、仰角35°踢出：vₓ = 22 cos 35° ≈ 18.0 m/s，vᵧ = 22 sin 35° ≈ 12.6 m/s。飞行时间t = 2vᵧ/g = 2×12.6/9.81 ≈ 2.57 s。射程 = vₓ × t = 18.0 × 2.57 ≈ 46.3 m。

When an object is projected horizontally from a height, uᵧ = 0, and the time to hit the ground depends only on height: t = √(2h/g).

物体从高处水平抛出时，uᵧ = 0，落地时间仅取决于高度：t = √(2h/g)。

5. Newton’s Laws and Free‑Body Force Calculations | 牛顿定律与受力分析计算

Draw a free‑body diagram showing all forces as arrows from the centre of mass. Resolve inclined plane weight: component down the slope mg sin θ, normal reaction mg cos θ. Apply ΣF = ma along the direction of acceleration.

画出受力图，以质心为起点用箭头表示所有力。斜面上重力分解：沿斜面分量mg sin θ，法向反作用力mg cos θ。沿加速度方向运用ΣF = ma。

For a 5.0 kg block sliding down a 30° incline with friction μ = 0.25: weight component = 5×9.81×sin30° = 24.5 N. Normal R = 5×9.81×cos30° = 42.5 N. Friction f = μR = 0.25×42.5 = 10.6 N. Net force = 24.5 − 10.6 = 13.9 N. Acceleration a = 13.9/5.0 = 2.78 m/s².

一个5.0 kg的滑块沿30°斜面下滑，摩擦系数μ = 0.25：重力下滑分量 = 5×9.81×sin30° = 24.5 N。法向支持力R = 5×9.81×cos30° = 42.5 N。摩擦力f = μR = 0.25×42.5 = 10.6 N。合力 = 24.5 − 10.6 = 13.9 N。加速度a = 13.9/5.0 = 2.78 m/s²。

Connected‑body problems (pulleys) require writing ΣF = ma for each mass and solving simultaneous equations. Tension is the same throughout a light inextensible string.

连接体问题（滑轮）需要分别对每个物体写ΣF = ma，解联立方程组。轻绳不可伸长时绳中张力处处相等。

6. Work, Energy Conservation and Power | 功、能量守恒与功率

Work done = F × d × cos θ, where θ is the angle between force and displacement. Kinetic energy Eₖ = ½mv², gravitational potential energy Eₚ = mgΔh. When non‑conservative forces like friction do work, use Wₙ.ₑₜ = ΔEₖ + ΔEₚ.

做功 = F × d × cos θ，其中θ为力与位移的夹角。动能Eₖ = ½mv²，重力势能Eₚ = mgΔh。当存在摩擦力等非保守力做功时，用Wₙₑₜ = ΔEₖ + ΔEₚ。

Example: a 1200 kg car travels at 20 m/s. Brakes apply a constant 6000 N force. Using energy, ½mv² = F × d → ½×1200×20² = 6000 × d, so d = 40 m. This avoids kinematic multiple steps.

例题：一辆1200 kg的汽车以20 m/s行驶，刹车施加恒力6000 N。用能量法：½mv² = F × d → ½×1200×20² = 6000 × d，解得d = 40 m。这省去了运动学多步计算。

Power P = work done/time = F × v for constant velocity. A lift raising 800 kg at 1.5 m/s needs power P = mg × v = 800×9.81×1.5 ≈ 11 800 W.

功率P = 做功/时间 = F × v （匀速时）。电梯以1.5 m/s提升800 kg，所需功率P = mg × v = 800×9.81×1.5 ≈ 11 800 W。

7. Momentum, Impulse and Collisions | 动量、冲量与碰撞

Momentum p = mv is a vector. Impulse J = F Δt = Δp. In collisions, use conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. For elastic collisions, kinetic energy is also conserved; otherwise it is inelastic.

动量p = mv是矢量。冲量J = F Δt = Δp。碰撞中使用动量守恒：m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。弹性碰撞动能同时守恒；非弹性碰撞只动量守恒。

A 0.15 kg tennis ball arrives at 35 m/s and is struck back at 45 m/s. Contact time 0.030 s. Impulse = Δp = 0.15×(45 − (−35)) = 0.15×80 = 12 kg·m/s. Average force = 12/0.030 = 400 N.

一个0.15 kg的网球以35 m/s飞来，以45 m/s被击回，接触时间0.030 s。冲量 = Δp = 0.15×(45 − (−35)) = 0.15×80 = 12 kg·m/s。平均力 = 12/0.030 = 400 N。

Remember to assign a positive direction and treat reversing velocities as negative.

要设定正方向，反向速度取负值。

8. Circular Motion and Gravitational Fields | 圆周运动与引力场

Centripetal force is F = mv²/r = mω²r, where v = ωr. Always identify the physical origin of this force: tension, friction, gravity, or normal reaction.

向心力F = mv²/r = mω²r，且v = ωr。务必明确向心力的实际来源：张力、摩擦力、引力或法向反作用力。

For a satellite, gravitational force provides centripetal force: GMm/r² = mv²/r, giving v = √(GM/r). For Earth, GM = gR², so near the surface vₒᵣᵦᵢₜ = √(gR).

对卫星而言，万有引力提供向心力：GMm/r² = mv²/r，得v = √(GM/r)。对地球，GM = gR²，所以近地轨道速度vₒᵣᵦᵢₜ = √(gR)。

Gravitational field strength g = F/m = GM/r². Weightlessness in orbit occurs because the astronaut is in free fall, not because gravity is zero.

引力场强度g = F/m = GM/r²。轨道中的失重并非因为引力为零，而是因为宇航员处于自由落体状态。

9. Electric Fields, Potential and Circuits | 电场、电势与电路

Electric field strength E = F/q = V/d for uniform fields. Force on a charge F = qE; on an electron F = eE. Work done W = qV, electronvolt 1 eV = 1.60×10⁻¹⁹ J.

电场强度E = F/q = V/d（匀强电场）。电荷受力F = qE；电子受力F = eE。做功W = qV，电子伏特1 eV = 1.60×10⁻¹⁹ J。

Circuit analysis relies on Kirchhoff’s laws: ΣI(in) = ΣI(out), ΣV(loop) = 0. For series resistors Rₜₑₛₑᵣᵢₑ = R₁ + R₂. For parallel, 1/Rₜₑₛₑ = 1/R₁ + 1/R₂. Internal resistance r causes terminal voltage V = ε − Ir.

电路分析依靠基尔霍夫定律：ΣI(入) = ΣI(出)，ΣV(回路) = 0。串联电阻Rₜₑₛₑᵣᵢₑ = R₁ + R₂；并联电阻1/Rₜₑₛₑ = 1/R₁ + 1/R₂。内阻r使端电压V = ε − Ir。

Power in a resistor P = IV = I²R = V²/R. Calculate total energy using E = P t.

电阻功率P = IV = I²R = V²/R。用E = P t计算总能量。

Quantity	Series	Parallel
Current I	Same	Splits
Voltage V	Splits	Same
Resistance R	Adds	Reciprocal sum

10. Magnetic Fields and Electromagnetic Induction | 磁场与电磁感应

Force on a current‑carrying wire F = BIL sin θ, where θ is angle between B and current. For a moving charge F = Bqv sin θ. Use Fleming’s left‑hand rule for direction.

通电导线所受安培力F = BIL sin θ，θ为B与电流的夹角。运动电荷受力F = Bqv sin θ。用弗莱明左手定则判断方向。

Magnetic flux Φ = BA cos φ, flux linkage = NΦ. Faraday’s law: ε = −N ΔΦ/Δt. Lenz’s law gives the induced current direction opposing the change in flux.

磁通量Φ = BA cos φ，磁链 = NΦ。法拉第定律：ε = −N ΔΦ/Δt。楞次定律指出感应电流方向总是阻碍磁通量的变化。

For a conductor of length L moving perpendicularly through field B at speed v, induced emf ε = BLv. Calculate power dissipated in a load connected to the generator using P = ε²/R.

长度为L的导线垂直穿过磁场B、以速度v运动时，感应电动势ε = BLv。外接负载时用P = ε²/R计算耗散功率。

11. Thermal Physics and Ideal Gases | 热物理与理想气体

Specific heat capacity Q = mcΔθ, latent heat Q = mL. When substance changes phase, temperature remains constant. Use the absolute temperature T in Kelvin for all gas calculations.

比热容Q = mcΔθ，潜热Q = mL。物态变化过程中温度保持不变。气体计算中一律使用开尔文绝对温标T。

Ideal gas equation pV = nRT, where n = mass/molar mass. Boltzmann constant k = R/Nₐ. The average kinetic energy of a molecule Eₖ = 3/2 kT.

理想气体方程pV = nRT，其中n = 质量/摩尔质量。玻尔兹曼常数k = R/Nₐ。分子的平均动能Eₖ = 3/2 kT。

When combining gas laws, hold the constant variables: Boyle’s law p₁V₁ = p₂V₂ at constant T; Charles’ law V₁/T₁ = V₂/T₂ at constant p; pressure law p₁/T₁ = p₂/T₂ at constant V.

使用气体定律时要注意不变条件：波义耳定律等温p₁V₁ = p₂V₂；查理定律等压V₁/T₁ = V₂/T₂；压强定律等容p₁/T₁ = p₂/T₂。

12. Simple Harmonic Motion and Wave Calculations | 简谐运动与波动计算

SHM defining equation: a = −ω²x. Displacement x = A sin(ωt) or x = A cos(ωt). Maximum speed vₘₐₓ = ωA, maximum acceleration aₘₐₓ = ω²A. Period of a mass‑spring T = 2π√(m/k), pendulum T = 2π√(L/g).

简谐运动特征方程：a = −ω²x。位移x = A sin(ωt)或x = A cos(ωt)。最大速度vₘₐₓ = ωA，最大加速度aₘₐₓ = ω²A。弹簧振子周期T = 2π√(m/k)，单摆T = 2π√(L/g)。

Wave speed v = fλ. Intensity I ∝ A² and I = P/(4πr²) for a spherical wave. Refractive index n = c/v. Snell’s law n₁ sin θ₁ = n₂ sin θ₂. Critical angle sin C = 1/n.

波速v = fλ。强度I ∝ A²，球面波I = P/(4πr²)。折射率n = c/v。斯涅耳定律n₁ sin θ₁ = n₂ sin θ₂。临界角sin C = 1/n。

Double‑slit fringe spacing Δx = λD/d. Be careful with units: D and d in metres, λ in metres. Diffraction grating d sin θ = nλ gives sharp maxima.

双缝干涉条纹间距Δx = λD/d。注意单位：D和d用米，λ用米。衍射光栅d sin θ = nλ给出尖锐极大值。

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IB & WJEC Physics: Calculation Question Drill | IB与WJEC物理：计算题专项训练