📚 IB Physics: Calculation Intensive Practice | IB 物理：计算题专项训练
Success in IB Physics hinges on the ability to translate physical principles into precise, well-structured numerical solutions. This guide walks you through the most demanding calculation topics, from SUVAT to quantum phenomena, reinforcing technique, common pitfalls, and exam-ready strategies.
在IB物理中，能否取得高分取决于你是否能将物理原理转化为准确、条理清晰的计算过程。本文梳理了从匀加速运动到量子现象等最高频的计算专题，通过中英对照的方式强化解题技巧、规避常见错误、打磨应考策略。
1. Kinematics and SUVAT Mastery | 运动学与匀加速方程精通
Kinematics problems demand fluency with the five SUVAT equations. Always list the known quantities (s, u, v, a, t) and identify the target variable before selecting an equation. Be mindful of sign conventions when vectors point in opposite directions.
运动学问题要求你熟练运用五个匀加速运动方程。先列出已知量(s、u、v、a、t)，确定待求量，再选择合适的公式。当矢量方向相反时，务必注意正负号的设定。
A ball is thrown vertically upward with speed 15.0 m s⁻¹. Using g = 9.81 m s⁻², the time to reach maximum height is found via v = u + at with v = 0:
一球以15.0 m s⁻¹ 的初速度竖直上抛，取 g = 9.81 m s⁻²。到达最高点时有 v = 0，由 v = u + at 可得上升时间：
0 = 15.0 − 9.81 t ⇒ t = 1.53 s
Notice that the acceleration due to gravity is negative relative to the upward positive direction. The same approach works for projection at an angle: resolve the initial velocity into horizontal and vertical components, then treat the two motions independently.
注意重力加速度与设定的正方向相反，因此取负值。同样的思路也适用于斜抛运动：将初速度分解为水平和竖直分量，然后独立处理两个方向的运动。
| Equation | Missing variable |
|---|---|
| v = u + at | s |
| s = ut + ½at² | v |
| v² = u² + 2as | t |
| s = ½(u + v)t | a |
| s = vt − ½at² | u |
Memorise the table above; during exams, write the five equations on your scrap paper immediately to reduce retrieval errors.
牢记上表；考试时立刻将五个方程默写在草稿纸上，可以有效减少因记忆混淆导致的错误。
2. Forces, Free-Body Diagrams and Newton’s Laws | 受力分析与牛顿定律
Every dynamics calculation starts with a correctly drawn free-body diagram. Identify all forces, resolve weight along an incline if present, and apply ΣF = ma in each perpendicular direction.
每道动力学计算题都应从正确的受力图开始。画出所有受力，若有斜面则分解重力，并在正交的两个方向上分别应用 ΣF = ma。
A block of mass 4.0 kg rests on a 30° slope with negligible friction. The acceleration down the slope is g sin θ = 9.81 sin30° = 4.91 m s⁻². When friction μ = 0.25 is added, the net force becomes mg sin θ − μ mg cos θ, and acceleration is reduced.
质量为 4.0 kg 的滑块静置于 30° 光滑斜面上，下滑加速度为 g sin θ = 9.81 sin30° = 4.91 m s⁻²。若摩擦系数 μ = 0.25，则合力变为 mg sin θ − μ mg cos θ，加速度减小。
Always check if the system moves with constant velocity (a = 0) or if a string tension links two masses. In connected-body problems, write separate equations for each mass and solve simultaneously.
务必确认系统是匀速运动(a = 0)还是通过细绳连接着两个物体。在连接体问题中，分别对每个物体列出方程，然后联立求解。
3. Work, Energy and Power | 功、能与功率
The work-energy theorem ΔEₖ = Wₙₑₜ is your fastest route in many calculation problems. Kinetic energy ½mv², gravitational potential mgΔh, and elastic potential ½kx² must be evaluated with consistent units.
在许多计算题中，动能定理 ΔEₖ = Wₙₑₜ 是最快捷的路径。动能 ½mv²、重力势能 mgΔh 和弹性势能 ½kx² 的计算必须使用一致的单位。
A 2.0 kg mass dropped from rest falls 10.0 m. Neglecting air resistance, the speed upon impact is derived from energy conservation: mgh = ½mv² ⇒ v = √(2gh) = √(2×9.81×10.0) ≈ 14.0 m s⁻¹.
一个 2.0 kg 的物体从静止下落 10.0 m，不计空气阻力，落地速度由能量守恒求得：mgh = ½mv² ⇒ v = √(2gh) = √(2×9.81×10.0) ≈ 14.0 m s⁻¹。
Power calculations often involve P = Fv for a force acting on a moving object. For an engine exerting constant thrust on a vehicle, ensure you use instantaneous speed when power is specified.
功率计算常涉及 P = Fv，即作用于运动物体上的力与速度的乘积。若发动机提供恒定推力，要注意当功率给定时需使用瞬时速度。
4. Momentum and Collisions | 动量与碰撞
Momentum p = mv is a vector; its conservation applies in isolated systems. For collisions, classify as elastic (kinetic energy conserved) or inelastic, and always write before-and-after momentum components.
动量 p = mv 是矢量，在孤立系统中动量守恒。处理碰撞时，先判断是弹性碰撞(动能守恒)还是非弹性碰撞，并写出碰撞前后的动量分量。
Two carts, 1.0 kg moving at 4.0 m s⁻¹ and 2.0 kg at rest, stick together after collision. By momentum conservation: (1.0)(4.0) = (3.0)v ⇒ v = 1.33 m s⁻¹. Kinetic energy is NOT conserved; the lost energy becomes heat or deformation.
两辆小车，1.0 kg 以 4.0 m s⁻¹ 运动，2.0 kg 静止，碰撞后粘在一起。动量守恒：(1.0)(4.0) = (3.0)v ⇒ v = 1.33 m s⁻¹。动能并不守恒；损失的能量转化为内能或形变能。
Impulse J = FΔt = Δp is especially useful when the force varies with time; a graph of F vs. t gives impulse as the area under the curve.
冲量 J = FΔt = Δp 在力随时间变化时尤其有用；F-t 图线下面积即为冲量。
5. Circular Motion and Gravitation | 圆周运动与万有引力
Centripetal acceleration a = v²/r = ω²r always points toward the centre. Combine it with Newton’s second law to find the net radial force. Do not invent a “centrifugal force” in inertial reference frames.
向心加速度 a = v²/r = ω²r 始终指向圆心，与牛顿第二定律结合可求径向合力。在惯性参考系中不要虚构”离心力”。
For a 1200 kg car rounding a curve of radius 50 m at 12 m s⁻¹, the required frictional force is F = mv²/r = 1200×144/50 = 3456 N. If friction cannot supply this, the car skids.
一辆 1200 kg 的汽车以 12 m s⁻¹ 驶过半径 50 m 的弯道，所需摩擦力为 F = mv²/r = 1200×144/50 = 3456 N。若实际摩擦力不足，汽车将打滑。
Newton’s law of gravitation F = GMm/r² yields orbital velocity v = √(GM/r) for a satellite. Equate gravitational force to centripetal requirement, and remember to convert all distances to metres.
万有引力定律 F = GMm/r² 可推导出卫星的轨道速度 v = √(GM/r)。令万有引力等于所需向心力，并注意将所有距离统一为米。
6. Thermal Physics and Ideal Gases | 热学与理想气体
Specific heat Q = mcΔT and latent heat Q = mL are fundamental. Use absolute temperatures in kelvin for gas law calculations: pV = nRT and pV = NkₘT, where T must be in K.
比热容 Q = mcΔT 和潜热 Q = mL 是计算的基础。运用气体定律时，温度必须使用开尔文：pV = nRT 与 pV = NkₘT。
A sample of 0.50 mol of an ideal gas occupies 0.020 m³ at 300 K. The pressure is p = nRT/V = (0.50×8.31×300)/(0.020) = 6.23×10⁴ Pa. If the gas expands isothermally to double the volume, pressure halves.
0.50 mol 理想气体在 300 K 时占据 0.020 m³，压强 p = nRT/V = (0.50×8.31×300)/(0.020) = 6.23×10⁴ Pa。若气体等温膨胀至两倍体积，压强减半。
Average kinetic energy of a molecule ⟨Eₖ⟩ = (3/2)kₘT. Use this to link microscopic motion to macroscopic temperature, and mind that kₘ is Boltzmann’s constant 1.38×10⁻²³ J K⁻¹.
分子平均动能 ⟨Eₖ⟩ = (3/2)kₘT。用它可以将微观运动与宏观温度联系起来，注意 kₘ 为玻尔兹曼常量 1.38×10⁻²³ J K⁻¹。
7. Waves and Oscillations | 波与振动
The wave equation v = fλ is simple but must be applied with consistent units. For periodic waves, period T = 1/f, and angular frequency ω = 2πf. In standing wave problems, node-to-node distance = λ/2.
波动方程 v = fλ 虽简单，但单位必须一致。对周期波，周期 T = 1/f，角频率 ω = 2πf。在驻波问题中，相邻波节间距为 λ/2。
A guitar string of length 0.65 m vibrates in its fundamental mode at 440 Hz. The wavelength of the standing wave is 2L = 1.30 m, so wave speed v = fλ = 440×1.30 = 572 m s⁻¹.
一根长 0.65 m 的吉他弦以基频 440 Hz 振动，驻波波长为 2L = 1.30 m，因此波速 v = fλ = 440×1.30 = 572 m s⁻¹。
Intensity I = P/A follows an inverse-square law for a point source: I ∝ 1/r². Combine with amplitude-squared relation to compare power or displacement.
对点波源，强度 I = P/A 遵循平方反比定律 I ∝ 1/r²。结合强度与振幅平方的关系，可以比较功率或位移。
8. Electric Fields and DC Circuits | 电场与直流电路
Coulomb’s law F = kₑq₁q₂/r² and electric field E = F/q = kₑQ/r² share the same inverse-square form. Always convert charge to coulombs and distance to metres; kₑ = 8.99×10⁹ N m² C⁻².
库仑定律 F = kₑq₁q₂/r² 和电场强度 E = F/q = kₑQ/r² 同为平方反比形式。务必把电荷换算为库仑，距离换算为米；kₑ = 8.99×10⁹ N m² C⁻²。
In circuits, Ohm’s law V = IR is the starting point. For series resistors, Rₜₒₜ = R₁ + R₂ + …; for parallel, 1/Rₜₒₜ = 1/R₁ + 1/R₂. Power dissipated P = IV = I²R = V²/R must be calculated with the correct potential difference.
在电路中，欧姆定律 V = IR 是基础。串联电阻 Rₜₒₜ = R₁ + R₂ + …；并联电阻 1/Rₜₒₜ = 1/R₁ + 1/R₂。计算消耗功率 P = IV = I²R = V²/R 时，务必使用正确的电压值。
A 12.0 V battery powers two 6.0 Ω resistors in parallel. The equivalent resistance is 3.0 Ω, so total current I = 12.0/3.0 = 4.0 A. Each branch carries 2.0 A, and each resistor dissipates P = I²R = (2.0)²×6.0 = 24 W.
12.0 V 电池为两个 6.0 Ω 并联电阻供电。等效电阻为 3.0 Ω，总电流 I = 12.0/3.0 = 4.0 A。每个支路电流 2.0 A，各电阻消耗功率 P = I²R = (2.0)²×6.0 = 24 W。
9. Magnetism and Electromagnetic Induction | 磁场与电磁感应
Force on a moving charge F = qvB sinθ, and on a current-carrying wire F = BIL sinθ. The right-hand rule determines direction; angle θ is between velocity (or current) and the magnetic field.
运动电荷受力 F = qvB sinθ，载流导线受力 F = BIL sinθ。用右手定则判断方向；角 θ 是速度(或电流)与磁场方向之间的夹角。
Magnetic flux Φ = BA cosθ links geometry to induction. Faraday’s law ε = −N ΔΦ/Δt gives the induced emf; the negative sign emphasises Lenz’s law. For a rod moving perpendicularly in a uniform field, ε = BLv.
磁通量 Φ = BA cosθ 把几何与感应联系起来。法拉第定律 ε = −N ΔΦ/Δt 给出感应电动势；负号体现了楞次定律。对于在匀强磁场中垂直移动的导体棒，ε = BLv。
Transformers operate on the principle χ = Vₚ/Vₛ = Nₚ/Nₛ and, for ideal transformers, Pₚ = Pₛ. Always check whether the question assumes 100% efficiency.
变压器基于 χ = Vₚ/Vₛ = Nₚ/Nₛ 工作，对理想变压器有 Pₚ = Pₛ。解题时先确认题目是否假定100%效率。
10. Atomic, Nuclear and Quantum Physics | 原子、核与量子物理
Photoelectric effect: hf = φ + Eₖₘₐₓ. Frequency below the threshold f₀ = φ/h produces no emission regardless of intensity. Kinetic energy maximum is found from stopping potential eVs = Eₖₘₐₓ.
光电效应：hf = φ + Eₖₘₐₓ。频率低于截止频率 f₀ = φ/h 时，无论光强多大均无电子逸出。最大动能由遏止电压求得：eVs = Eₖₘₐₓ。
De Broglie wavelength λ = h/p = h/(mv) connects wave and particle properties. For an electron accelerated through 100 V, first find v from ½mv² = eV, then calculate λ.
德布罗意波长 λ = h/p = h/(mv) 将波动性与粒子性联系起来。对于被 100 V 电压加速的电子，先由 ½mv² = eV 求得速度，再计算波长。
Radioactive decay follows A = λN and half-life T₁/₂ = ln2/λ. The exponential decay law N = N₀ e⁻λt appears frequently; use natural logarithms to solve for time when N/N₀ is given.
放射性衰变遵循 A = λN，半衰期 T₁/₂ = ln2/λ。指数衰减律 N = N₀ e⁻λt 经常出现；已知 N/N₀ 时可用自然对数求出时间。
11. Uncertainties and Data Analysis | 不确定度与数据分析
Every measured value in IB Physics carries an absolute uncertainty ±Δx. The fractional uncertainty is Δx/x, and percentage uncertainty is (Δx/x)×100%. In calculations, propagate uncertainties using the appropriate rules.
IB 物理中每一个测量值都带有绝对不确定度 ±Δx。相对不确定度为 Δx/x，百分不确定度为 (Δx/x)×100%。计算时，应按规则传递不确定度。
For addition/subtraction, add absolute uncertainties. For multiplication/division, add fractional (or percentage) uncertainties. When a quantity is raised to a power, multiply the fractional uncertainty by that power.
加减运算时，要将绝对不确定度相加；乘除运算时，要将相对不确定度(或百分不确定度)相加。当物理量被乘方时，相对不确定度应乘以该指数。
Example: A rectangle’s sides are measured as (5.0±0.2) cm and (10.0±0.3) cm. Area = 50 cm². Fractional uncertainties: 0.2/5.0 = 0.04, 0.3/10.0=0.03, sum = 0.07. Absolute uncertainty in area = 0.07×50 ≈ 4 cm², so Area = (50±4) cm².
示例：矩形边长测量值为 (5.0±0.2) cm 和 (10.0±0.3) cm。面积 = 50 cm²。相对不确定度：0.2/5.0 = 0.04，0.3/10.0 = 0.03，总和 0.07。面积绝对不确定度 = 0.07×50 ≈ 4 cm²，故面积 = (50±4) cm²。
In graphs, error bars represent uncertainties. The line of best fit and the worst-fit lines help determine uncertainty in gradient and intercept—an essential skill for Internal Assessment.
在图表中，误差棒表示不确定度。最佳拟合线和最差拟合线用于确定斜率与截距的不确定度——这是内部评估（IA）中的关键技能。
12. Exam Calculation Tactics and Error Prevention | 考场计算策略与防错要诀
Write “GIVEN” quantities with their symbols and values immediately. Convert all units to SI before plugging into equations. Scientific notation (e.g., 6.02×10²³) keeps numbers manageable.
拿到题目立即用符号写出 “已知” 量及其数值。代入方程前将所有单位转换为国际单位制。科学记数法(如 6.02×10²³)能让数字保持简洁。
Check dimensional consistency: if you are solving for force but your expression yields kg m s⁻¹, you have found momentum, not force. Always box your final answer with correct significant figures.
检查量纲是否匹配：如果你要求出力，但表达式得出的单位是 kg m s⁻¹，那实际求的是动量而不是力。最终答案要用方框标出，并取正确的有效数字。
Practice paced calculation: IB Paper 1 allows ~1.5 minutes per question; Paper 2 demands extended multi-step solutions. Trust your raw algebraic manipulation before touching the calculator, and re-read the question to ensure you have answered exactly what was asked.
进行限时训练：IB 试卷一每题约 1.5 分钟，试卷二则需要呈现多步骤的详细解答。先用代数推导，再动用计算器；最后重读题目，确认你的答案完整回应了所问。
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