📚 IB CCEA Physics: Calculation Practice Drills | IB CCEA 物理：计算题专项训练
Success in IB and CCEA Physics exams depends heavily on the ability to solve numerical problems with precision and speed. This article brings together key calculation topics from mechanics, waves, electricity, thermal physics, and modern physics, providing a structured set of drills that mirror the style of assessment questions. Each section focuses on the essential equations, common pitfalls, and step-by-step strategies. By working through these examples, you will sharpen your unit handling, algebraic manipulation, and critical reasoning skills, all of which are vital for achieving top marks.
在 IB 和 CCEA 物理考试中，能否精准、快速地完成计算题直接关系到最终成绩。本文汇集了力学、波、电学、热学和近代物理中最核心的计算专题，以贴近真题的方式组织训练。每节都围绕关键公式、常见易错点和分步解题策略展开。通过反复演练这些例题，你将强化单位处理、代数推理和批判性思维三项核心能力，为冲击高分打下扎实基础。
1. Kinematic Equations for Linear Motion | 直线运动的运动学方程
The four kinematic equations describe uniformly accelerated motion along a straight line. Remember that they apply only when acceleration is constant. The most common forms use u for initial velocity, v for final velocity, a for acceleration, s for displacement, and t for time: v = u + at, s = ut + ½at², v² = u² + 2as, and s = ½(u+v)t.
四个运动学方程描述匀变速直线运动，务必谨记它们只适用于加速度恒定的情形。最常用的形式涉及初速度 u、末速度 v、加速度 a、位移 s 和时间 t：v = u + at，s = ut + ½at²，v² = u² + 2as，以及 s = ½(u+v)t。
Always begin by writing down the five symbols and marking which values you know and which you need to find. Pay close attention to signs: take one direction as positive, typically the direction of initial motion, and then acceleration is negative if it opposes this direction. Converting all units to SI before substituting numbers prevents many errors.
开始解题时，先把五个符号列出来，标注已知量和待求量。特别注意正负号的选取：通常设初速度方向为正，若加速度与该方向相反则取负值。代入数值前将所有单位转换为国际单位制，可以避免大量计算失误。
Example: A car accelerates from rest at 2.5 m s⁻² for 8.0 s. Find the distance travelled.
We have u = 0, a = 2.5 m s⁻², t = 8.0 s, s = ?. Using s = ut + ½at² gives s = 0 + ½ × 2.5 × (8.0)² = 80 m.
例题：一辆汽车从静止开始以 2.5 m s⁻² 的加速度行驶 8.0 s，求通过的位移。
已知 u = 0，a = 2.5 m s⁻²，t = 8.0 s，s = ?。代入 s = ut + ½at² 得 s = 0 + ½ × 2.5 × (8.0)² = 80 m。
2. Newton’s Laws and Free-Body Force Calculations | 牛顿定律与受力分析计算
Newton’s second law, Fnet = ma, links the net force acting on an object to its acceleration. The net force is the vector sum of all forces, so drawing a clear free-body diagram is the first essential step. For objects on inclined planes, resolve weight into components parallel and perpendicular to the slope: mg sin θ down the plane and mg cos θ into the plane.
牛顿第二定律 Fnet = ma 将物体所受合外力与其加速度联系起来。合外力是所有力的矢量和，因此清晰绘制受力图是至关重要的第一步。对于斜面上的物体，需将重力分解为平行于斜面的分量 mg sin θ 和垂直于斜面的分量 mg cos θ。
When friction is involved, remember that kinetic friction fk = μk N, where N is the normal reaction. For static friction, fs ≤ μs N. Many students forget that the normal force is not always equal to mg; it changes on an incline or under an applied push or pull with a vertical component.
涉及摩擦力时，动摩擦 fk = μk N，其中 N 为支持力；静摩擦 fs ≤ μs N。很多同学常忘记支持力并不总等于 mg：在斜面上或当外力有竖直分量时，支持力的大小会改变。
Drill: A 5.0 kg block slides down a 30° incline with μk = 0.20. Determine its acceleration.
Resolve weight: mg sin30° = 5.0×9.8×0.5 = 24.5 N down the slope. N = mg cos30° = 5.0×9.8×0.866 = 42.4 N. Friction = μk N = 0.20×42.4 = 8.48 N. Net force down slope = 24.5 – 8.48 = 16.02 N. a = F/m = 16.02/5.0 = 3.2 m s⁻².
练习：5.0 kg 的滑块沿 30° 斜面下滑，动摩擦因数 μk = 0.20，求加速度。
重力分解：mg sin30° = 5.0×9.8×0.5 = 24.5 N 沿斜面向下。N = mg cos30° = 5.0×9.8×0.866 = 42.4 N。摩擦力 = μk N = 0.20×42.4 = 8.48 N。沿斜面合力 = 24.5 – 8.48 = 16.02 N。a = F/m = 16.02/5.0 = 3.2 m s⁻²。
3. Work, Energy and Power | 功、能与功率
The work done by a constant force is W = F d cos θ, where θ is the angle between the force and displacement. Kinetic energy is Ek = ½mv², and gravitational potential energy near the Earth’s surface is Ep = mgh. The work–energy principle states that the net work done on an object equals its change in kinetic energy.
恒力做功的公式为 W = F d cos θ，其中 θ 是力与位移的夹角。动能为 Ek = ½mv²，地表附近的重力势能为 Ep = mgh。功能原理指出，合力对物体做的功等于其动能的变化量。
Power is the rate of doing work: P = W/t. For an object moving at speed v under a constant force F in the same direction, the instantaneous power is also given by P = F v. Always check units: energy in joules (J) and power in watts (W).
功率是做功的速率：P = W/t。当物体在恒力 F 同方向下以速度 v 运动时，瞬时功率也可用 P = F v 计算。解题时务必检查单位：能量用焦耳 (J)，功率用瓦特 (W)。
Example: A 1200 kg car accelerates from 10 m s⁻¹ to 25 m s⁻¹ in 8.0 s. Calculate the average power delivered by the engine.
ΔEk = ½ × 1200 × (25² – 10²) = 600 × (625 – 100) = 600 × 525 = 315 000 J. Average power = ΔEk / t = 315 000 / 8.0 = 39 375 W ≈ 39 kW.
例题：一辆 1200 kg 的汽车在 8.0 s 内从 10 m s⁻¹ 加速到 25 m s⁻¹，求发动机的平均输出功率。
ΔEk = ½ × 1200 × (25² – 10²) = 600 × (625 – 100) = 600 × 525 = 315 000 J。平均功率 = ΔEk / t = 315 000 / 8.0 = 39 375 W ≈ 39 kW。
4. Momentum and Impulse | 动量与冲量
Momentum is a vector defined as p = mv. Impulse J equals the change in momentum: J = Δp = Favg Δt. In collision problems, use the principle of conservation of momentum for an isolated system: total momentum before collision equals total momentum after collision.
动量是矢量，定义为 p = mv。冲量 J 等于动量的变化：J = Δp = Favg Δt。在处理碰撞问题时，对于孤立系统应用动量守恒定律：碰撞前总动量等于碰撞后总动量。
For perfectly elastic collisions, kinetic energy is also conserved. In inelastic collisions, kinetic energy is not conserved, but momentum is always conserved in the absence of external forces. When two objects stick together, use m₁v₁ + m₂v₂ = (m₁+m₂)vfinal.
在完全弹性碰撞中，动能也守恒。非弹性碰撞中动能不守恒，但只要没有外力，动量依然守恒。当两个物体粘在一起运动时，使用 m₁v₁ + m₂v₂ = (m₁+m₂)vfinal。
Drill: A 0.50 kg trolley moving at 4.0 m s⁻¹ collides with a stationary 1.0 kg trolley. They stick together. Find the final speed.
Initial momentum = 0.50×4.0 + 1.0×0 = 2.0 kg m s⁻¹. Final momentum = (0.50+1.0)v = 1.5 v. Equate: 1.5 v = 2.0 → v = 1.33 m s⁻¹.
练习：质量 0.50 kg 的小车以 4.0 m s⁻¹ 的速度撞上静止的 1.0 kg 小车，两车粘在一起运动，求最终速度。
初始动量 = 0.50×4.0 + 1.0×0 = 2.0 kg m s⁻¹。末动量 = (0.50+1.0)v = 1.5 v。由动量守恒 1.5 v = 2.0 → v = 1.33 m s⁻¹。
5. Circular Motion and Gravitation | 圆周运动与引力
For an object moving in a circle of radius r at constant speed v, centripetal acceleration is a = v²/r and centripetal force is F = mv²/r. These point toward the centre. The force can be provided by tension, friction, or gravity. In vertical circles, energy conservation often links speed at different points with height changes.
物体以恒定速率 v 在半径为 r 的圆上运动时，向心加速度 a = v²/r，向心力 F = mv²/r，两者均指向圆心。这个力可以由拉力、摩擦力或引力提供。在竖直面内的圆周运动中，常需结合能量守恒将不同位置的速度与高度变化联系起来。
Newton’s law of gravitation: F = Gm₁m₂ / r². Near a planet’s surface, g = GM/R². For orbital motion, equate gravitational force to centripetal force: GmM/r² = mv²/r, leading to v = √(GM/r) and orbital period T² ∝ r³.
万有引力定律：F = Gm₁m₂ / r²。在行星表面附近，g = GM/R²。对于轨道运动，将引力与向心力等置：GmM/r² = mv²/r，由此导出 v = √(GM/r) 和周期关系 T² ∝ r³。
Example: A satellite orbits Earth at an altitude where g = 2.5 m s⁻². The radius of its orbit is 8.0 × 10⁶ m. Find its orbital speed.
g = v²/r → v = √(g r) = √(2.5 × 8.0×10⁶) = √(20×10⁶) = 4.47 × 10³ m s⁻¹.
例题：一卫星在 g = 2.5 m s⁻² 的高度上绕地球运行，轨道半径 8.0 × 10⁶ m，求轨道速率。
由 g = v²/r，得 v = √(g r) = √(2.5 × 8.0×10⁶) = √(20×10⁶) = 4.47 × 10³ m s⁻¹。
6. Simple Harmonic Motion | 简谐运动
For SHM, acceleration a = –ω²x, where ω = 2πf = 2π/T. Displacement can be written as x = A sin(ωt) or x = A cos(ωt). Maximum speed vmax = ωA, and maximum acceleration amax = ω²A. The period of a mass–spring system is T = 2π√(m/k), and for a simple pendulum T = 2π√(L/g).
在简谐运动中，加速度 a = –ω²x，其中 ω = 2πf = 2π/T。位移可写作 x = A sin(ωt) 或 x = A cos(ωt)。最大速度 vmax = ωA，最大加速度 amax = ω²A。弹簧振子的周期 T = 2π√(m/k)，单摆周期 T = 2π√(L/g)。
Energy in SHM is continuously exchanged between kinetic and potential forms, with total energy E = ½mω²A². At any displacement, v = ω√(A² – x²). This relationship is invaluable for finding speed at specific positions.
简谐运动中的能量在动能和势能之间不断转化，总能量 E = ½mω²A²。在任一位置，v = ω√(A² – x²)。这个关系在计算特定位置的速度时非常有用。
Drill: A pendulum has period 2.0 s on Earth where g = 9.8 m s⁻². Find its length.
T = 2π√(L/g) → L = gT²/(4π²) = 9.8×4.0 / (4×9.87) = 39.2 / 39.48 ≈ 0.99 m.
练习：一个单摆在地球表面 (g = 9.8 m s⁻²) 的周期为 2.0 s，求摆长。
T = 2π√(L/g) → L = gT²/(4π²) = 9.8×4.0 / (4×9.87) = 39.2 / 39.48 ≈ 0.99 m。
7. Electric Fields and Potential | 电场与电势
Coulomb’s law gives the force between two point charges: F = kQq/r², where k = 8.99×10⁹ N m² C⁻². Electric field strength E = F/q, and for a point charge E = kQ/r². The force on a charge in a uniform field is F = qE, and the work done moving a charge through a potential difference V is W = qV.
库仑定律给出两点电荷间的作用力：F = kQq/r²，其中 k = 8.99×10⁹ N m² C⁻²。电场强度 E = F/q，对点电荷 E = kQ/r²。电荷在匀强电场中受力 F = qE，将电荷移动经过电势差 V 所做的功 W = qV。
In a uniform electric field between parallel plates separated by distance d with potential difference V, E = V/d. The electronvolt (eV) is a convenient energy unit: 1 eV = 1.60×10⁻¹⁹ J. When an electron accelerates through 500 V, its kinetic energy gain is 500 eV = 8.0×10⁻¹⁷ J.
在间距为 d、电势差为 V 的平行板间的匀强电场中，E = V/d。电子伏特 (eV) 是常用的能量单位：1 eV = 1.60×10⁻¹⁹ J。一个电子经 500 V 电压加速后，动能增加 500 eV = 8.0×10⁻¹⁷ J。
Example: Two parallel plates are 0.020 m apart with 200 V across them. Find the electric field strength and the force on an electron placed between them.
E = V/d = 200 / 0.020 = 10000 V m⁻¹ = 1.0×10⁴ N C⁻¹. Force F = eE = 1.6×10⁻¹⁹ × 1.0×10⁴ = 1.6×10⁻¹⁵ N.
例题：两平行板相距 0.020 m，电势差为 200 V，求电场强度以及置于其中电子所受的力。
E = V/d = 200 / 0.020 = 10000 V m⁻¹ = 1.0×10⁴ N C⁻¹。力 F = eE = 1.6×10⁻¹⁹ × 1.0×10⁴ = 1.6×10⁻¹⁵ N。
8. DC Circuits and Internal Resistance | 直流电路与内阻
Ohm’s law V = IR applies to ohmic conductors at constant temperature. For a circuit with emf ε, internal resistance r, and external load R, the terminal potential difference is V = ε – Ir. The current in the circuit is I = ε / (R + r). Power delivered to the external circuit is P = I²R, and maximum power transfer occurs when R = r.
欧姆定律 V = IR 适用于恒温下的欧姆导体。对于电动势为 ε、内阻为 r、外接负载 R 的电路，端电压为 V = ε – Ir。电路中电流 I = ε / (R + r)。外电路获得的功率 P = I²R，且当 R = r 时输出功率最大。
For resistors in series, Rtotal = R₁ + R₂ + …; in parallel, 1/Rtotal = 1/R₁ + 1/R₂ + … Kirchhoff’s current law states that the sum of currents entering a junction equals the sum leaving. Kirchhoff’s voltage law states that the sum of emfs equals the sum of IR products around any closed loop.
电阻串联时，Rtotal = R₁ + R₂ + …；并联时，1/Rtotal = 1/R₁ + 1/R₂ + …。基尔霍夫电流定律指出，流入节点的电流之和等于流出电流之和；电压定律指出，绕任意闭合回路的电动势之和等于各电阻上 IR 之和。
Drill: A battery of emf 12.0 V and internal resistance 0.50 Ω is connected to a 5.5 Ω resistor. Find the current and terminal voltage.
I = ε / (R+r) = 12.0 / (5.5+0.5) = 12.0/6.0 = 2.0 A. V = ε – Ir = 12.0 – 2.0×0.5 = 11.0 V.
练习：电动势 12.0 V、内阻 0.50 Ω 的电池外接 5.5 Ω 电阻，求电流和端电压。
I = ε / (R+r) = 12.0 / (5.5+0.5) = 12.0/6.0 = 2.0 A。V = ε – Ir = 12.0 – 2.0×0.5 = 11.0 V。
9. Magnetic Force and Electromagnetic Induction | 磁场力与电磁感应
A charged particle moving with velocity v perpendicular to magnetic field B experiences a force F = qvB (or F = qvB sin θ for any angle). This force provides the centripetal force for circular motion: qvB = mv²/r, yielding radius r = mv/(qB). For a current-carrying wire of length L in a uniform field, F = BIL sin θ.
电荷以速度 v 垂直穿过磁场 B 时受力 F = qvB (一般情况为 F = qvB sin θ)。该力提供圆周运动的向心力：qvB = mv²/r，由此得出半径 r = mv/(qB)。对处于匀强磁场中的载流直导线，安培力 F = BIL sin θ。
Faraday’s law states that induced emf equals the rate of change of magnetic flux linkage: ε = –N ΔΦ/Δt. Magnetic flux Φ = BA cos θ. For a conductor of length L moving perpendicularly through a field at speed v, the motional emf is ε = BLv. Lenz’s law determines the direction of induced current.
法拉第电磁感应定律指出，感应电动势等于磁通链变化率的负值：ε = –N ΔΦ/Δt。磁通量 Φ = BA cos θ。长为 L 的导体以速度 v 垂直切割磁感线时，动生电动势 ε = BLv。楞次定律用于判断感应电流的方向。
Example: A proton (q = 1.6×10⁻¹⁹ C, m = 1.67×10⁻²⁷ kg) enters a 0.30 T field at 2.0×10⁶ m s⁻¹. Determine the radius of its path.
r = mv/(qB) = (1.67×10⁻²⁷ × 2.0×10⁶) / (1.6×10⁻¹⁹ × 0.30) = (3.34×10⁻²¹) / (4.8×10⁻²⁰) = 0.070 m = 7.0 cm.
例题：一个质子 (q = 1.6×10⁻¹⁹ C, m = 1.67×10⁻²⁷ kg) 以 2.0×10⁶ m s⁻¹ 的速度垂直进入 0.30 T 的磁场，求轨道半径。
r = mv/(qB) = (1.67×10⁻²⁷ × 2.0×10⁶) / (1.6×10⁻¹⁹ × 0.30) = (3.34×10⁻²¹) / (4.8×10⁻²⁰) = 0.070 m = 7.0 cm。
10. Thermal Physics and Ideal Gases | 热物理与理想气体
The ideal gas equation is pV = nRT, where n is the number of moles and R = 8.31 J K⁻¹ mol⁻¹. Alternatively, pV = NkT, where N is the number of molecules and k = 1.38×10⁻²³ J K⁻¹. The average translational kinetic energy of a molecule is (3/2) kT. Always convert temperature to kelvin: T(K) = T(°C) + 273.
理想气体状态方程为 pV = nRT，其中 n 为摩尔数，R = 8.31 J K⁻¹ mol⁻¹。也可写成 pV = NkT，N 为分子数，k = 1.38×10⁻²³ J K⁻¹。分子的平均平动动能为 (3/2) kT。解题时必须将温度换算为开尔文：T(K) = T(°C) + 273。
For specific heat capacity, Q = mcΔθ, and for latent heat, Q = mL. In calorimetry problems, energy lost by hotter objects equals energy gained by cooler ones, assuming no external heat loss. Power input in electrical heating is P = VI, and total energy supplied is P × t.
比热容公式 Q = mcΔθ；潜热公式 Q = mL。在量热学问题中，若无热损失，高温物体失去的热量等于低温物体获得的热量。电加热时输入功率 P = VI，提供的总能量为 P × t。
Drill: A gas cylinder of volume 0.030 m³ contains helium at 27 °C and pressure 4.0×10⁵ Pa. How many moles of gas are present?
T = 27+273 = 300 K. n = pV/(RT) = (4.0×10⁵ × 0.030) / (8.31 × 300) = 12000 / 2493 = 4.81 mol.
练习：容积 0.030 m³ 的氦气瓶在 27 °C 时压强为 4.0×10⁵ Pa，求气体的摩尔数。
T = 27+273 = 300 K。n = pV/(RT) = (4.0×10⁵ × 0.030) / (8.31 × 300) = 12000 / 2493 = 4.81 mol。
11. Photoelectric Effect and Quantum Calculations | 光电效应与量子计算
Photon energy E = hf = hc/λ, where h = 6.63×10⁻³⁴ J s and c = 3.00×10⁸ m s⁻¹. The photoelectric equation is hf = Φ + Ek max, where Φ is the work function (minimum energy to eject an electron). The threshold frequency f₀ is given by hf₀ = Φ. Kinetic energy Ek max can be measured as the stopping potential Vs: Ek max = e Vs.
光子能量 E = hf = hc/λ，其中 h = 6.63×10⁻³⁴ J s，c = 3.00×10⁸ m s⁻¹。光电效应方程为 hf = Φ + Ek max，其中 Φ 是逸出功（使电子逸出的最小能量）。极限频率 f₀ 满足 hf₀ = Φ。最大动能可以用遏止电势 Vs 来测量：Ek max = e Vs。
Calculations often involve converting between electronvolts and joules. Be comfortable with both unit systems: 1 eV = 1.60×10⁻¹⁹ J. When light intensity increases, the number of photons per second increases, but photon energy remains unchanged if frequency is constant.
计算时常需在电子伏特和焦耳之间转换。要熟练使用两种单位制：1 eV = 1.60×10⁻¹⁹ J。光强增加时，每秒光子数增加，但若频率不变，单个光子的能量保持不变。
Example: Light of wavelength 250 nm falls on a metal with work function 3.0 eV. Find the maximum kinetic energy of photoelectrons in eV.
Photon energy E = hc/λ = (6.63×10⁻³⁴ × 3.00×10⁸) / (250×10⁻⁹) = 7.96×10⁻¹⁹ J. Convert to eV: 7.96×10⁻¹⁹ / 1.60×10⁻¹⁹ = 4.975 eV. Ek max = E – Φ = 4.975 – 3.0 = 1.98 eV.
例题：波长为 250 nm 的光照射在逸出功为 3.0 eV 的金属上，求光电子的最大动能（以 eV 为单位）。
光子能量 E = hc/λ = (6.63×10⁻³⁴ × 3.00×10⁸) / (250×10⁻⁹) = 7.96×10⁻¹⁹ J。换算为 eV：7.96×10⁻¹⁹ / 1.60×10⁻¹⁹ = 4.975 eV。Ek max = E – Φ = 4.975 – 3.0 = 1.98 eV。
12. Nuclear Physics and Decay Calculations | 核物理与衰变计算
The activity A of a radioactive sample is the number of decays per second, measured in becquerels (Bq). Activity decreases exponentially: A = A₀ e^(–λt), where λ is the decay constant. The half-life t₁/₂ is related to λ by λ = ln2 / t₁/₂. The number of undecayed nuclei N follows the same law: N = N₀ e^(–λt).
放射性样品的活度 A 是每秒衰变次数，单位为贝克勒尔 (Bq)。活度按指数衰减：A = A₀ e^(–λt)，λ 为衰变常量。半衰期 t₁/₂ 与 λ 的关系是 λ = ln2 / t₁/₂。未衰变的原子核数 N 也遵循同样的规律：N = N₀ e^(–λt)。
Mass–energy equivalence is given by ΔE = Δm c². In nuclear reactions, the mass defect corresponds to the binding energy. Common calculations require converting atomic mass units (u) to energy: 1 u = 931.5 MeV. Always balance mass numbers and atomic numbers in nuclear equations.
质能方程 ΔE = Δm c² 将质量与能量联系起来。核反应中的质量亏损对应结合能。常见计算需将原子质量单位 (u) 转换为能量：1 u = 931.5 MeV。在写核反应方程时，必须配平质量数和电荷数。
Drill: A radioactive isotope has a half-life of 8.0
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