📚 IB Mathematics: Newton’s Laws – Key Concepts and Problem Solving | IB 数学：牛顿定律考点精讲
In IB Mathematics, especially in the Applications and Interpretation (AI) course and certain optional topics in Analysis and Approaches (AA), Newton’s laws of motion offer a powerful context for modelling with differential equations. The central task is to translate a physical description of forces into a differential equation, solve it using calculus techniques, and interpret the resulting functions for velocity, displacement, or time. This article covers the key problem types, essential methods, and common pitfalls, helping you master this cross-topic area where pure mathematics meets applied physics.
在 IB 数学中，特别是在应用与解释（AI）课程以及分析与方法（AA）的某些选修内容里，牛顿运动定律为使用微分方程建模提供了一个强有力的背景。核心任务是将力的物理描述转化为微分方程，用微积分技巧求解，并解释得到的关于速度、位移或时间的函数。本文涵盖主要问题类型、必备方法和常见陷阱，助你掌握这一纯数学与应用物理交汇的主题。
1. Newton’s Second Law as a Differential Equation | 牛顿第二定律作为微分方程
Newton’s second law in its most useful form for constant mass is F = m a, where acceleration a is the derivative of velocity v with respect to time t, i.e. a = dv/dt. Since velocity itself is the derivative of displacement x, we can also write a = d²x/dt². Therefore, when a resultant force F acts on a particle of mass m, the law instantly becomes a differential equation: m dv/dt = F. The force F may be constant, a function of time F(t), a function of velocity F(v), or even a function of position F(x). IB problems predominantly feature F(v) such as air resistance models.
对于恒定质量，牛顿第二定律最常用的形式是 F = m a，其中加速度 a 是速度 v 对时间 t 的导数，即 a = dv/dt。因为速度本身是位移 x 的导数，我们也可写 a = d²x/dt²。因此，当合力 F 作用于质量为 m 的质点时，定律立即转化为一个微分方程：m dv/dt = F。力 F 可以是常数、时间的函数 F(t)、速度的函数 F(v)，甚至是位置的函数 F(x)。IB 考题中主要涉及如空气阻力模型中的 F(v)。
Once the differential equation is set up, solving it usually requires separation of variables if F depends only on v, or direct integration if F is purely a function of t. Initial conditions, such as v(0)=v₀ or x(0)=x₀, are then used to determine the constants of integration. Being fluent with these first-order ODE techniques is the mathematical core of the topic.
一旦建立起微分方程，求解通常需要当 F 仅依赖于 v 时使用分离变量法，或当 F 仅是 t 的函数时直接积分。然后利用初始条件，如 v(0)=v₀ 或 x(0)=x₀，确定积分常数。熟练运用这些一阶常微分方程技巧是该主题的数学核心。
2. Constant Force and Uniform Acceleration | 恒力与匀加速运动
When the resultant force is constant, F = constant, the acceleration a = F/m is also constant. Starting from dv/dt = a and integrating with respect to time yields v = u + a t, where u is the initial velocity. Integrating v = dx/dt = u + a t gives the displacement function x = x₀ + u t + ½ a t². These are the familiar suvat equations. In an IB exam, you might be asked to derive them from the differential equation rather than quoting them, to demonstrate the link with calculus.
当合力恒定时，F = constant，加速度 a = F/m 也是常数。由 dv/dt = a 出发对时间积分得到 v = u + a t，其中 u 为初速度。再对 v = dx/dt = u + a t 积分，得到位移函数 x = x₀ + u t + ½ a t²。这就是熟悉的 suvat 方程。在 IB 考试中，你可能会被要求从微分方程推导这些公式，而不是直接引用，以体现与微积分的联系。
This simplest case is the foundation for analysing motion under gravity near the Earth’s surface, where the acceleration is g ≈ 9.8 m s⁻² downwards. Be careful with sign conventions: taking upward as positive makes a = −g. Mastering constant acceleration problems builds confidence for the more challenging resistive force scenarios ahead.
这一最简单情形是分析地球表面附近重力下运动的基础，加速度为向下的 g ≈ 9.8 m s⁻²。注意符号约定：取向上为正方向会使 a = −g。掌握匀加速问题能为应对更具挑战性的阻力情形树立信心。
3. Resistance Proportional to Velocity | 阻力与速度成正比
A frequently occurring model assumes that a resistive force acts opposite to motion with magnitude proportional to the speed: FR = −k v, where k > 0. For a body driven by a constant force P (e.g., an engine thrust) and subject to such resistance, the equation of motion is m dv/dt = P − k v. This is a first-order linear ODE that can be solved by separation of variables: dv/(P − k v) = dt/m. Integrating both sides gives (−1/k) ln|P − k v| = t/m + C. Using the initial condition v(0)=0 leads to v(t) = (P/k)(1 − e^{-kt/m}).
一个经常出现的模型假设阻力与速度大小成正比且方向与运动相反：FR = −k v，其中 k > 0。对于受恒定驱动力 P（如发动机推力）和此类阻力作用的物体，运动方程为 m dv/dt = P − k v。这是一个一阶线性常微分方程，可通过分离变量法求解：dv/(P − k v) = dt/m。两边积分得 (−1/k) ln|P − k v| = t/m + C。利用初始条件 v(0)=0 得出 v(t) = (P/k)(1 − e^{-kt/m})。
As t grows large, the exponential term decays to zero, and the velocity approaches the terminal velocity vterm = P/k. The solution reveals a characteristic exponential approach to a steady speed, a pattern that IB examiners love. A similar analysis applies to a falling object with linear air resistance, where mg replaces the driving force. Always clearly state the limiting value and relate it to the physical balance of forces.
随着 t 增大，指数项衰减至零，速度趋近终端速度 vterm = P/k。该解揭示了以指数方式趋近稳定速度的特征模式，这一模式深受 IB 考官喜爱。类似的分析也适用于含线性空气阻力的落体，此时驱动力为 mg。务必清晰说明极限值并将其与力的物理平衡联系起来。
4. Resistance Proportional to Square of Velocity | 阻力与速度平方成正比
There are situations where resistance is better modelled by a quadratic dependence: FR = −k v² (or −k v|v| to keep direction correct). For a free-fall drop under gravity with such resistance, the equation becomes m dv/dt = mg − k v², assuming downward positive. Separating variables leads to ∫ dv/(mg/k − v²) = (k/m) ∫ dt. Using the standard integral ∫ dv/(a² − v²) = (1/(2a)) ln| (a+v)/(a−v) | + C, one obtains a logarithmic expression for velocity, or an equivalent hyperbolic tangent form: v = a tanh( (a k/m) t ), where a = √(mg/k) is the terminal velocity.
有些情况下阻力更适合用二次方模型：FR = −k v²（或 −k v|v| 以保持方向正确）。对于具有此类阻力的自由落体，设向下为正，方程变为 m dv/dt = mg − k v²。分离变量需要计算 ∫ dv/(mg/k − v²) = (k/m) ∫ dt。利用标准积分 ∫ dv/(a² − v²) = (1/(2a)) ln| (a+v)/(a−v) | + C，可得到速度的对数表达式，或与之等价的双曲正切形式：v = a tanh( (a k/m) t )，其中 a = √(mg/k) 是终端速度。
From an IB Maths perspective, the key skill is handling the partial fractions decomposition of 1/(a² − v²) to carry out the integration step by step. You should also be prepared to find the time taken to reach a given fraction of terminal velocity. This model tests your integration chops and your ability to manipulate logarithms and exponentials without losing physical meaning.
从 IB 数学角度看，关键技能是先将 1/(a² − v²) 分解为部分分式，再逐步完成积分。你还需要能够求出达到终端速度特定比例所需的时间。这一模型既检验积分功底，也考察在不丢失物理意义的前提下处理对数和指数的能力。
5. Projectile Motion: Horizontal and Vertical DEs | 抛体运动：水平与垂直微分方程
Projectile problems under uniform gravity neglect air resistance, turning the motion into two independent second-order differential equations. Horizontally there is no force, so m d²x/dt² = 0, which integrates to dx/dt = u cosθ and x = (u cosθ) t. Vertically, m d²y/dt² = −mg, giving dy/dt = u sinθ − g t and y = (u sinθ) t − ½ g t². These parametric equations are the standard projectile formulas; the IB Mathematics twist is that you must derive them from Newton’s law using integration and initial conditions.
在均匀重力且忽略空气阻力的抛体问题中，运动被分解为两个独立的二阶微分方程。水平方向无力，故 m d²x/dt² = 0，积分得 dx/dt = u cosθ 及 x = (u cosθ) t。竖直方向，m d²y/dt² = −mg，得 dy/dt = u sinθ − g t 及 y = (u sinθ) t − ½ g t²。这些参数方程就是标准抛体公式；IB 数学的不同之处在于你必须通过牛顿定律和积分以及初始条件将它们推导出来。
More advanced IB questions may introduce a horizontal or vertical drag that couples the equations, e.g., a resistance force proportional to the speed vector. In that case the component equations are no longer independent, and you might need to use an integrating factor or a numerical method. Such questions stretch your ability to set up realistic differential equation systems from Newton’s vector law F = m a.
更高级的 IB 问题可能引入水平或垂直方向的阻力，使方程耦合，例如阻力与速度矢量成正比。此时分量方程不再独立，你可能需要用到积分因子或数值方法。这类题目能拓展你从牛顿矢量定律 F = m a 建立现实微分方程系统的能力。
6. Connected Particles and Pulley Systems | 连接体与滑轮系统
Typical IB mechanics problems involve two masses connected by a light inextensible string passing over a smooth pulley. By applying Newton’s second law to each mass and using the constraint that the accelerations have equal magnitude, you obtain a system of equations. For masses m₁ and m₂ with m₁ > m₂, the equations are m₁g − T = m₁a and T − m₂g = m₂a. Solving simultaneously yields the constant acceleration a = (m₁ − m₂)g/(m₁ + m₂) and the tension T = (2m₁m₂g)/(m₁ + m₂).
典型的 IB 力学问题涉及两个通过轻质不可伸长绳跨过光滑滑轮连接的物体。对每个物体应用牛顿第二定律，并利用加速度大小相等这一约束条件，便可得到一个方程组。对于质量 m₁ 和 m₂ 且 m₁ > m₂，方程为 m₁g − T = m₁a 和 T − m₂g = m₂a。联立求解得到恒定加速度 a = (m₁ − m₂)g/(m₁ + m₂) 和张力 T = (2m₁m₂g)/(m₁ + m₂)。
In a maths exam the follow-up often asks to find the velocity or displacement by integrating a(t) = a. The differential equations here are trivial because the acceleration is constant, but the process of deriving that constant from the force equations is what’s being assessed. Always draw a clear free-body diagram and assign consistent positive direction before writing equations.
在数学考试中，后续常要求通过对 a(t)=a 积分来求速度或位移。这里的微分方程很简单，因为加速度恒定，但评估的正是从受力方程导出该常数的过程。写方程前务必画出清晰的受力图并指定一致的正方向。
7. Momentum, Impulse and Integral Forms | 动量、冲量与积分形式
Newton’s original second law is expressed in terms of momentum p = m v: F = dp/dt. For a constant mass this reduces to F = m a, but the momentum form is essential when mass changes (e.g., rocket propulsion) or when force acts over a short time interval. Integrating dp/dt = F(t) with respect to time gives the impulse-momentum theorem: ∫ F(t) dt = Δp = m(v_f − v_i). In IB Mathematics, you may be asked to compute impulse from a known force function or to solve the differential equation dp/dt = F(t) for p(t) and v(t).
牛顿原始的力学第二定律用动量 p = m v 表示：F = dp/dt。对于恒定质量，它简化为 F = m a，但当质量变化（如火箭推进）或力在短时间内作用时，动量形式不可或缺。对 dp/dt = F(t) 关于时间积分，得到冲量-动量定理：∫ F(t) dt = Δp = m(v_f − v_i)。在 IB 数学中，你可能需要根据已知力函数计算冲量，或求解微分方程 dp/dt = F(t) 得出 p(t) 和 v(t)。
The momentum approach also appears in collision problems where the force-time graph is given, and the area under the curve equals the impulse. The differential equation framework ties neatly into integral calculus, emphasising that the change in momentum is the accumulation of force over time. Always pay attention to vector directions when dealing with impulses.
动量方法也出现在碰撞问题中，此时给出力-时间图，曲线下方面积等于冲量。微分方程框架与积分学紧密相连，强调动量的变化是力对时间的累积。处理冲量时务必留意矢量的方向。
8. Slope Fields for Velocity Equations | 速度微分方程的斜率场
In the IB AI course, differential equations that cannot be solved analytically are explored using slope fields. For a first-order ODE of the form dv/dt = f(v,t) derived from Newton’s laws, the slope field gives a graphical representation of how velocity changes with time. By plotting short line segments with slope f(v,t) at grid points, students can sketch solution curves that follow the field. This is particularly useful for qualitative analysis, such as determining terminal velocity behaviour or stability of equilibrium solutions.
在 IB AI 课程中，无法解析求解的微分方程通过斜率场来探索。对于从牛顿定律导出的形如 dv/dt = f(v,t) 的一阶常微分方程，斜率场给出了速度如何随时间变化的图形表示。通过在网格点绘制斜率为 f(v,t) 的短线段，学生可以勾勒出沿场方向的解曲线。这在定性分析中尤其有用，例如判断终端速度行为或平衡解的稳定性。
An example setup might be dv/dt = g − (k/m) v² for a falling object. Drawing the slope field reveals that all solutions approach the terminal velocity v = √(mg/k), regardless of the starting velocity. IB exam questions might ask you to match a slope field to a given physical scenario or to sketch a particular solution curve given an initial condition.
一个示例设置可以是落体的 dv/dt = g − (k/m) v²。绘制斜率场可揭示所有解都趋近终端速度 v = √(mg/k)，无论初速度如何。IB 考题可能会要求你将某个斜率场匹配到给定的物理情景，或根据初始条件画出特定的解曲线。
9. Euler’s Method for Numerical Solutions | 欧拉法数值求解
When a Newtonian differential equation resists analytic integration, Euler’s method provides a simple numerical scheme. Starting from the ODE dv/dt = f(v,t) with initial condition v(t₀)=v₀, the method iterates v_{n+1} = v_n + h × f(v_n, t_n), where h = Δt is the step size. To find displacement simultaneously, you can apply x_{n+1} = x_n + h × v_n. In IB AI examinations, you are often given a table layout and asked to carry out two or three steps, demonstrating an understanding of the approximation that uses the slope at the current point to step forward.
当牛顿力学微分方程难以解析积分时，欧拉法提供了一种简单的数值方案。从常微分方程 dv/dt = f(v,t) 和初始条件 v(t₀)=v₀ 出发，该方法迭代计算 v_{n+1} = v_n + h × f(v_n, t_n)，其中 h = Δt 为步长。如需同时求位移，可应用 x_{n+1} = x_n + h × v_n。在 IB AI 考试中，常会给出一个表格布局，要求你执行两到三步计算，以展示对该近似法的理解——它利用当前点的斜率向前推进。
Consider a boat of mass m with engine force P and resistance kv, where the ODE is dv/dt = (P − kv)/m. Choosing a suitable step size, you can approximate the velocity at successive times. The accuracy improves with smaller h, but the IB typically tests whether you can apply the formula correctly rather than discussing error. Newton’s law continuous model thus connects directly to the discrete world of numerical calculus.
考虑一艘质量为 m、推力为 P 且阻力为 kv 的船，其常微分方程为 dv/dt = (P − kv)/m。选择合适的步长，可以近似各连续时刻的速度。步长 h 越小精度越高，但 IB 通常检验你是否能正确应用公式，而非讨论误差。牛顿定律的连续模型由此直接进入数值微积分的离散世界。
10. Common Pitfalls and Exam Tips | 常见错误与应试技巧
IB examiners frequently observe mistakes that can be avoided with careful practice. Ensure you respect the direction of forces: when a resistive force acts opposite to motion, its sign must be negative relative to the positive direction. Confusing v and |v| in quadratic drag leads to incorrect differential equations. Always define your positive direction explicitly at the start and check that initial conditions match this convention.
IB 考官经常见到一些通过细心练习即可避免的错误。要确保尊重力的方向：当阻力与运动方向相反时，其符号必须与正方向相反。在二次阻力中混淆 v 与 |v| 会导致错误的微分方程。务必在开始时明确指定正方向，并检查初始条件是否与该约定一致。
Other errors include forgetting the constant of integration, misapplying the separation of variables, or mishandling the absolute value when integrating 1/(a − kv). A sketch of the slope field or a quick dimensional analysis often catches algebraic slips. Finally, when a question gives a force function, translate it directly to ma, not to some memorised formula, to stay rigorous and flexible.
其他错误包括忘记积分常数、错误应用分离变量法，或在积分 1/(a − kv) 时对绝对值处理不当。画一下斜率场草稿或快速量纲分析常能发现代数笔误。最后，当题目给出力函数时，应直接将其转化为 ma，而非套用某个记忆公式，以保持严谨和灵活。
11. Worked Example: Car With Linear Resistance | 示例精讲：带线性阻力的汽车
Problem: A car of mass 800 kg produces a constant driving force of 2000 N and experiences air resistance modelled by 40v N, where v is the speed in m s⁻¹. The car starts from rest. Find an expression for its speed as a function of time, and determine the time taken to reach 90% of the terminal speed.
问题：一辆质量 800 kg 的汽车产生 2000 N 的恒定驱动力，并受到 40v N 的空气阻力（v 为以 m s⁻¹ 为单位的速度）。汽车从静止起步。求速度关于时间的表达式，并确定达到终端速度的 90% 所需的时间。
Solution: Let the direction of motion be positive. Newton’s second law gives 800 dv/dt = 2000 − 40v, which simplifies to dv/dt = 2.5 − 0.05v. Separate variables: dv/(2.5 − 0.05v) = dt. Integrating yields −20 ln|2.5 − 0.05v| = t + C. Using v(0)=0 gives C = −20 ln 2.5, so −20 ln|2.5 − 0.05v| = t − 20 ln 2.5. Rearranging, v(t) = 50(1 −
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