IB WJEC Maths: Newton's Laws Exam Focus | IB WJEC 数学：牛顿定律考点精讲

📚 IB WJEC Maths: Newton’s Laws Exam Focus | IB WJEC 数学：牛顿定律考点精讲

Newton’s laws form the backbone of classical mechanics, and in IB and WJEC Mathematics courses they are applied through calculus, vector resolution, and differential equations. This article unpacks the essential exam techniques and common pitfalls so you can confidently tackle any mechanics problem involving forces and motion.

牛顿定律是经典力学的支柱，在 IB 和 WJEC 数学课程中，它们通过微积分、矢量分解和微分方程进行应用。本文将拆解核心考试技巧与常见陷阱，让你自信应对任何涉及力和运动的力学题目。

1. Newton’s First Law and Equilibrium | 牛顿第一定律与平衡

Newton’s First Law states that an object remains at rest or in uniform motion in a straight line unless acted upon by a resultant external force. In mathematical modelling, this means the vector sum of all forces equals zero when the system is in equilibrium.

牛顿第一定律指出，除非受到合外力的作用，物体将保持静止或匀速直线运动状态。在数学建模中，这意味着当系统处于平衡状态时，所有力的矢量和为零。

Always start by drawing a free-body diagram showing weight, normal reaction, tension, friction, and any applied forces. Resolve forces in perpendicular directions (usually horizontally and vertically, or parallel and perpendicular to an inclined plane) and set the net force in each direction to zero. This yields simultaneous equations that can be solved for unknown forces or angles.

始终从画受力分析图开始，标出重力、法向反力、张力、摩擦力和所有外加力。在互相垂直的方向上分解力（通常沿水平和竖直方向，或沿斜面平行与垂直方向），并令每个方向的合力为零。这将产生可求解未知力或角度的联立方程组。

For a particle on a smooth inclined plane of angle θ, the component of weight down the slope is mg sin θ, and the normal reaction is mg cos θ. Equilibrium often involves friction: F ≤ μR, where μ is the coefficient of friction and R is the normal reaction.

对于在倾角为 θ 的光滑斜面上的质点，重力沿斜面的分力为 mg sin θ，法向反力为 mg cos θ。平衡常涉及摩擦力：F ≤ μR，其中 μ 为摩擦系数，R 为法向反力。

2. Newton’s Second Law and Equations of Motion | 牛顿第二定律与运动方程

The iconic formulation F = ma is a differential equation in disguise. For systems with constant mass, the resultant force F (in newtons) equals mass m (kg) multiplied by acceleration a (ms⁻²). When forces vary with time, position, or velocity, you must write m(dv/dt) = F(t, x, v) and integrate.

标志性的公式 F = ma 实际上是一个微分方程。对于质量不变的系统，合力 F（牛）等于质量 m（千克）乘以加速度 a（米每二次方秒）。当力随时间、位置或速度变化时，必须写出 m(dv/dt) = F(t, x, v) 并进行积分。

F = m a ⇌ m (dv/dt) = F

Exam questions frequently ask you to find velocity or displacement by integrating acceleration with respect to time. Remember initial conditions: at t = 0, u = initial velocity, s = 0. Also note that a = v(dv/dx) can be used when acceleration is given as a function of displacement.

考题常要求通过对加速度关于时间积分来求速度或位移。务必记住初始条件：t = 0 时，初速为 u，位移 s = 0。还要注意，当加速度表达为位移的函数时，可用 a = v(dv/dx)。

3. Newton’s Third Law and Connected Particles | 牛顿第三定律与连接体

Newton’s Third Law: when two bodies interact, they exert forces on each other that are equal in magnitude and opposite in direction. In connected particle problems (pulleys, tow-bars), the tension in a light inextensible string is the same throughout, and the accelerations of the connected bodies have the same magnitude.

牛顿第三定律：当两个物体相互作用时，它们彼此施加的力大小相等、方向相反。在连接体问题中（滑轮、拖杆），轻质且不可伸长的绳子两端张力大小相同，相连物体的加速度大小相等。

Treat each particle separately: draw its free-body diagram, apply F = ma in the direction of motion. Set up equations for each mass and solve for acceleration and tension. If friction or a rough surface is involved, include the frictional force acting opposite to motion on the appropriate body.

对每个质点分别处理：画出受力图，沿运动方向应用 F = ma。对每个质量建立方程，联立求解加速度和张力。若涉及摩擦或粗糙表面，要在相应物体上加入与运动方向相反的摩擦力。

A classic WJEC-style problem is two masses on a horizontal table connected by a string passing over a pulley at the edge, with one mass hanging vertically. The tension and acceleration are found by solving: T = m₁ a and m₂ g − T = m₂ a.

一道经典的 WJEC 风格题是水平桌面上两个物体通过绕过桌边滑轮的绳子相连，一物体竖直悬挂。通过求解 T = m₁ a 和 m₂ g − T = m₂ a 来得到张力和加速度。

4. Constant Force Motion and SUVAT Equations | 恒力作用下的运动与 SUVAT 方程

When resultant force is constant, acceleration is constant, and you can use the SUVAT equations: v = u + at, s = ut + ½ at², v² = u² + 2as, s = (u+v)t/2. These are derived from integrating constant acceleration but are indispensable for quick calculations.

合力恒定时加速度恒定，可使用 SUVAT 方程：v = u + at, s = ut + ½ at², v² = u² + 2as, s = (u+v)t/2。它们由恒定加速度积分导出，但能实现快速计算。

Be sure to assign a positive direction. Displacement, velocity, and acceleration are vectors; if an object decelerates, a is negative. Write down the five variables: s, u, v, a, t. Identify which three are known, which one you need, and choose the appropriate equation without missing required conversions (e.g. km/h to m/s).

务必规定正方向。位移、速度和加速度都是矢量；若物体减速，则 a 为负值。列出五个变量：s, u, v, a, t。确定已知哪三个、需要求哪个，选择合适的方程，不要遗漏必要的单位换算（如 km/h 转 m/s）。

5. Variable Force and Calculus: Relating v, a, s, t | 变力与微积分：v, a, s, t 的关联

In many IB HL and WJEC Mechanics papers, force depends on time (F = f(t)), velocity (air resistance proportional to v or v²), or position (spring force). To find velocity, integrate acceleration a = F/m with respect to time: v = ∫ a dt. For displacement, s = ∫ v dt.

在许多 IB HL 和 WJEC 力学试卷中，力依赖于时间（F = f(t)）、速度（空气阻力与 v 或 v² 成正比）或位置（弹力）。要计算速度，对加速度 a = F/m 关于时间积分：v = ∫ a dt。求位移时，s = ∫ v dt。

When acceleration is a function of displacement, use the relation a = v(dv/dx). Separate variables: ∫ v dv = ∫ a(x) dx. Apply boundary conditions carefully—often initial velocity and initial position are given. This method is essential for problems involving springs obeying Hooke’s law.

当加速度是位移的函数时，使用关系式 a = v(dv/dx)。分离变量：∫ v dv = ∫ a(x) dx。小心应用边界条件——通常会给出初速度和初始位置。对于涉及胡克定律的弹簧问题，该方法十分关键。

Example: A particle of mass 2 kg moves along a line under a force (6 − 2t) N. At t = 0, v = 1 ms⁻¹. Acceleration a = (6 − 2t)/2 = 3 − t. Integrating, v = ∫(3−t) dt = 3t − ½ t² + C; using v(0)=1 gives C=1, so v = 3t − ½ t² + 1.

示例：质量 2 kg 的质点在力 (6 − 2t) N 作用下沿直线运动。t = 0 时 v = 1 ms⁻¹。加速度 a = (6 − 2t)/2 = 3 − t。积分得 v = ∫(3−t) dt = 3t − ½ t² + C；代入 v(0)=1 得 C=1，故 v = 3t − ½ t² + 1。

6. Resistance Forces and Terminal Velocity | 阻力与终极速度

When an object falls through a fluid, it experiences resistance that increases with speed. Newton’s second law then becomes: m(dv/dt) = mg − kv (for linear resistance) or mg − kv² (for quadratic). Terminal velocity occurs when net force is zero, i.e. mg = kv_term or mg = kv_term².

当物体在流体中下落时，会受到随速度增大的阻力。牛顿第二定律此时变为：m(dv/dt) = mg − kv（线性阻力）或 mg − kv²（平方阻力）。当合力为零时达到终极速度，即 mg = kv_term 或 mg = kv_term²。

To find v(t), treat the differential equation as separable. For linear resistance: dv / (g − (k/m)v) = dt. Integrating gives an expression for velocity that asymptotically approaches the terminal value. On the exam, you may be given a differential equation and asked to verify a solution or find a particular integral.

为求 v(t)，将该微分方程视作可分离变量的。对线性阻力：dv / (g − (k/m)v) = dt。积分得到速度表达式，随时间趋近于终极值。在考试中，可能会给出微分方程并要求验证解或求特解。

7. Motion on an Inclined Plane | 斜面上的运动

Resolving weight mg into components parallel and perpendicular to the slope is a foundational skill. For a plane inclined at angle θ to the horizontal, the component down the slope is mg sin θ, and the normal reaction R equals mg cos θ unless other forces act perpendicularly.

将重力 mg 沿斜面平行和垂直方向分解是一项基本技能。对于与水平面成 θ 角的斜面，沿斜面向下的分力为 mg sin θ，法向反力 R 等于 mg cos θ，除非有其他垂直力作用。

If friction acts, the frictional force = μR when motion occurs or is impending. Apply F = ma along the slope: for a particle sliding down a rough plane, mg sin θ − μ mg cos θ = ma. Thus acceleration a = g (sin θ − μ cos θ). This expression is frequently examined.

若有摩擦力，当发生运动或即将运动时，摩擦力 = μR。沿斜面应用 F = ma：对于沿粗糙斜面下滑的质点，mg sin θ − μ mg cos θ = ma。因此加速度 a = g (sin θ − μ cos θ)。该表达式常考。

When an additional force pulls the particle up the incline, include it in the resolution. Always check whether the frictional force reverses direction depending on the attempted motion. Draw a clear, labelled diagram to avoid sign errors.

当有额外力沿斜面拉动物体时，把它计入力的分解。始终检查摩擦力方向是否随运动趋势而反转。画清晰且带标注的受力图可避免符号错误。

8. Impulse and Momentum | 冲量与动量

Newton’s second law originally stated in terms of momentum: F = dp/dt where p = mv. Impulse is the change in momentum: J = ∫ F dt = mv − mu. In problems with constant force, impulse = F × t. For variable forces, the area under a force–time graph gives impulse.

牛顿第二定律最初用动量表述：F = dp/dt，其中 p = mv。冲量为动量的变化量：J = ∫ F dt = mv − mu。在恒力问题中，冲量 = F × t。对于变力，力–时间图下的面积即为冲量。

Conservation of momentum is used for collisions and explosions in isolated systems. In one dimension, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. Coupled with Newton’s experimental law for direct impact (coefficient of restitution e = (v₂ − v₁) / (u₁ − u₂)), you can solve for final velocities.

动量守恒用于孤立系统中的碰撞和爆炸。在一维情形下，m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。结合牛顿碰撞实验定律（恢复系数 e = (v₂ − v₁) / (u₁ − u₂)），可求解末速度。

9. Circular Motion and Newton’s Law | 圆周运动与牛顿定律

For a particle moving in a circle of radius r at constant speed v, there is a centripetal acceleration a = v²/r directed towards the centre. Newton’s second law gives the resultant force towards the centre: F = mv²/r = mrω² (where ω = v/r is angular speed).

对于在半径为 r 的圆上做匀速运动的质点，存在指向圆心的向心加速度 a = v²/r。牛顿第二定律给出指向圆心的合力：F = mv²/r = mrω²（其中 ω = v/r 为角速度）。

Common scenarios include a car rounding a banked curve, a conical pendulum, or a mass on a string swung in a horizontal circle. Resolve forces vertically and radially. For a conical pendulum: T cos θ = mg and T sin θ = m v²/r. Eliminate T to find relationships among θ, v, and r.

常见情形包括汽车在斜弯道上行驶、圆锥摆以及绳子一端系着质量在水平面内旋转。沿竖直和径向分解力。对圆锥摆：T cos θ = mg，T sin θ = m v²/r。消去 T 求出 θ、v 和 r 间的关系。

In vertical circular motion, speed varies; energy conservation often combines with radials Newton’s law. At the top position, the resultant force toward the centre is T + mg = mv²/r, while at the bottom it is T − mg = mv²/r.

在竖直圆周运动中，速度变化；常用能量守恒结合径向牛顿定律。在最高点，指向圆心的合力为 T + mg = mv²/r；在最低点则为 T − mg = mv²/r。

10. Differential Equations from Newton’s Laws | 由牛顿定律建立微分方程

IB Analysis & Approaches and WJEC Mechanics both require you to formulate first-order differential equations from verbal descriptions of forces. A classic example: “The rate of decrease of velocity is proportional to the square of velocity.” This translates to dv/dt = −kv².

IB 分析与方法以及 WJEC 力学都要求根据力的文字描述列出一阶微分方程。经典示例：“速度减小的速率与速度的平方成正比。” 这转化为 dv/dt = −kv²。

Solve by separating variables: ∫ v⁻² dv = ∫ −k dt → −1/v = −kt + C. Using initial velocity v(0) = u, find C = −1/u, thus v = u/(1 + kut). Such solutions are often plotted and examined for long-term behaviour like terminal velocity or stopping distance.

通过分离变量求解：∫ v⁻² dv = ∫ −k dt → −1/v = −kt + C。代入初速 v(0) = u 得 C = −1/u，故 v = u/(1 + kut)。此类解常被画出图像并考查长期行为，如终极速度或制动距离。

Another exam favourite is Newton’s Law of Cooling: dT/dt = −k(T − T₀), where T₀ is ambient temperature. The solution T = T₀ + (T_initial − T₀)e⁻^kt appears in many past papers and tests understanding of exponential models.

另一个考试热门为牛顿冷却定律：dT/dt = −k(T − T₀)，其中 T₀ 为环境温度。解 T = T₀ + (T_初始 − T₀)e⁻^kt 在历年真题中反复出现，考查对指数模型的理解。

11. Worked Exam-Style Problem | 典型考题示例

Problem: A particle of mass 0.5 kg moves along a straight line. At time t seconds, the force acting is (3e⁻²^t) N in the positive direction. When t = 0, the particle is at rest at the origin. Find an expression for velocity v(t) and displacement s(t).

题目：质量 0.5 kg 的质点沿直线运动。在 t 秒时，作用力为沿正向的 (3e⁻²^t) N。t = 0 时质点静止于原点。求速度 v(t) 和位移 s(t) 的表达式。

Solution: a = F/m = 3e⁻²^t / 0.5 = 6e⁻²^t. Integrate: v = ∫ 6e⁻²^t dt = −3e⁻²^t + C. Using v(0) = 0, 0 = −3 + C → C = 3, so v = 3 − 3e⁻²^t. For displacement: s = ∫ (3 − 3e⁻²^t) dt = 3t + (3/2) e⁻²^t + D. With s(0) = 0, 0 = 0 + 3/2 + D → D = −3/2. Hence s = 3t + (3/2)e⁻²^t − 3/2.

解答：a = F/m = 3e⁻²^t / 0.5 = 6e⁻²^t。积分：v = ∫ 6e⁻²^t dt = −3e⁻²^t + C。代入 v(0) = 0 得 0 = −3 + C → C = 3，故 v = 3 − 3e⁻²^t。位移：s = ∫ (3 − 3e⁻²^t) dt = 3t + (3/2) e⁻²^t + D。由 s(0) = 0 得 0 = 0 + 3/2 + D → D = −3/2。因此 s = 3t + (3/2)e⁻²^t − 3/2。

Note how initial conditions are used stepwise, and the final expressions can be checked by differentiation. This exemplifies the rigorous approach expected in IB and WJEC mark schemes.

注意初始条件如何逐步使用，最终表达式可通过微分验证。这体现了 IB 和 WJEC 评分方案所期望的严谨做法。

12. Exam Strategy and Common Mistakes | 应试策略与常见错误

Strategy tips: Always list given data and convert units before plugging into equations. Write the statement of Newton’s second law in the direction of motion explicitly. For variable force problems, state which form of acceleration you are using: dv/dt or v dv/dx. Show integration limits or include the constant of integration and use initial conditions clearly.

策略提示：在代入方程之前始终列出已知数据并转换单位。沿运动方向明确写出牛顿第二定律的表达式。对于变力问题，写明你所用的加速度形式：dv/dt 或 v dv/dx。标出积分上下限（或包含积分常数）并清晰地使用初始条件。

Common mistakes include: forgetting to include all forces in the free-body diagram; sign errors when resolving on an inclined plane; using v = u + at for non-constant acceleration; failing to state the direction of vectors; and not checking the dimensions of final answers. In differential equations, be mindful that the constant of integration must be evaluated using the specific initial state, not generic values.

常见错误包括：受力图中遗漏某些力；在斜面上分解时符号出错；对非匀变速使用 v = u + at；未说明矢量的方向；未检查最终答案的量纲。在微分方程中，注意积分常数必须用特定初始状态求值，而非通用值。

Finally, practice past paper questions under timed conditions. Newton’s laws in maths are not just about memorising formulas but about building a logical sequence from force diagram to equation to integration. Mastering this flow will secure you top marks.

最后，在限时条件下练习历年真题。数学中的牛顿定律不仅关乎记忆公式，更需构建从受力图到方程再到积分的逻辑链条。掌握这一流程将确保你获得高分。

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IB WJEC Maths: Newton’s Laws Exam Focus | IB WJEC 数学：牛顿定律 考点精讲