IB Math: Mechanics Key Points | IB 数学：力学考点精讲

📚 IB Math: Mechanics Key Points | IB 数学：力学考点精讲

Mechanics forms a crucial part of the IB Mathematics: Applications and Interpretation HL course, bridging the gap between pure mathematics and physical reality. In this article, we will systematically break down the essential concepts of kinematics, forces, momentum, energy, and more, providing clear explanations and worked examples to help you excel in your IB exams.

力学是IB数学“应用与解释”高级水平课程的重要组成部分，它连接了纯数学与物理现实。本文将系统拆解运动学、力、动量、能量等核心概念，提供清晰的解释与示例，助你在IB考试中脱颖而出。

1. Kinematics in One Dimension | 一维运动学

Kinematics describes the motion of objects without considering the forces that cause the motion. In one dimension, we use displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t). The four SUVAT equations are fundamental: v = u + at, s = (u+v)t/2, s = ut + ½at², and v² = u² + 2as.

运动学描述物体的运动而不考虑导致运动的力。在一维中，我们使用位移(s)、初速度(u)、末速度(v)、加速度(a)和时间(t)。四个SUVAT方程是基础：v = u + at, s = (u+v)t/2, s = ut + ½at², v² = u² + 2as。

Always check that acceleration is constant before applying these equations. If acceleration varies with time, you must use calculus: velocity is the derivative of displacement with respect to time, and acceleration is the derivative of velocity.

使用这些方程前务必确认加速度恒定。若加速度随时间变化，则必须使用微积分：速度是位移对时间的导数，加速度是速度对时间的导数。

SUVAT equations: v = u + at, s = ut + ½at², v² = u² + 2as, s = (u+v)t/2.
SUVAT方程：v = u + at, s = ut + ½at², v² = u² + 2as, s = (u+v)t/2.
Calculus links: v = ds/dt, a = dv/dt = d²s/dt², and s = ∫ v dt, v = ∫ a dt.
微积分关系：v = ds/dt, a = dv/dt = d²s/dt², 以及 s = ∫ v dt, v = ∫ a dt.

2. Motion with Variable Acceleration | 变加速运动

When acceleration is not constant, you must express it as a function of time, a(t), or sometimes as a function of displacement, a(s). To find velocity from acceleration, integrate: v(t) = ∫ a(t) dt + C₁. The constant is determined by initial conditions. Similarly, displacement is found by integrating velocity.

当加速度不恒定时，需将其表示为时间函数a(t)，有时也为位移函数a(s)。由加速度求速度需积分：v(t) = ∫ a(t) dt + C₁。常数由初始条件确定。类似地，位移通过积分速度求得。

A common exam problem gives a(t) = 6t – 2, with initial velocity u = 3 m/s at t=0. Then v(t) = ∫(6t-2)dt = 3t² – 2t + C; using v(0)=3 gives C=3, so v(t) = 3t² – 2t + 3. Displacement s(t) = ∫ v(t) dt = t³ – t² + 3t + D; if s(0)=0, D=0.

常见考题：已知a(t) = 6t – 2，t=0时初速度u=3 m/s。则v(t) = ∫(6t-2)dt = 3t² – 2t + C；由v(0)=3得C=3，故v(t) = 3t² – 2t + 3。位移s(t) = ∫ v(t) dt = t³ – t² + 3t + D；若s(0)=0，则D=0。

You may also need to find distance traveled, not just displacement. Distance is the integral of the absolute value of velocity over the time interval, so you must identify when velocity changes sign.

可能还需要求路程，而不仅是位移。路程是速度绝对值在时间区间上的积分，因此必须确定速度何时变号。

3. Projectile Motion | 抛体运动

Projectile motion is two-dimensional motion under constant gravitational acceleration (g = 9.8 m/s² downwards). Resolve the initial velocity u into horizontal and vertical components: uₓ = u cos θ, uᵧ = u sin θ. The horizontal motion has constant velocity (aₓ = 0). The vertical motion has constant acceleration aᵧ = -g.

抛体运动是在恒定重力加速度(g = 9.8 m/s² 向下)下的二维运动。将初速度u分解为水平和竖直分量：uₓ = u cos θ, uᵧ = u sin θ。水平方向速度恒定(aₓ = 0)。竖直方向加速度恒定aᵧ = -g。

Apply SUVAT separately in each direction. Horizontal displacement: x = uₓ t. Vertical displacement: y = uᵧ t – ½ g t². The trajectory is parabolic: y = x tan θ – (g x²) / (2 u² cos² θ). Time of flight T = 2u sin θ / g, maximum height H = u² sin² θ / (2g), range R = u² sin 2θ / g.

分别对每个方向应用SUVAT。水平位移：x = uₓ t。竖直位移：y = uᵧ t – ½ g t²。轨迹呈抛物线：y = x tan θ – (g x²) / (2 u² cos² θ)。飞行时间T = 2u sin θ / g，最大高度H = u² sin² θ / (2g)，射程R = u² sin 2θ / g。

Quantity (量)	Formula (公式)
Time of flight (飞行时间)	T = 2u sin θ / g
Maximum height (最大高度)	H = u² sin² θ / (2g)
Range (射程)	R = u² sin 2θ / g

4. Forces and Newton’s Laws | 力与牛顿定律

Newton’s three laws govern dynamics. First law: an object remains at rest or in uniform motion unless acted upon by a net external force. Second law: F = ma, where F is the net force, m is mass, a is acceleration. Third law: for every action there is an equal and opposite reaction.

牛顿三定律支配动力学。第一定律：物体会保持静止或匀速直线运动，除非受到净外力作用。第二定律：F = ma，其中F为净力，m为质量，a为加速度。第三定律：每一个作用力都有一个大小相等、方向相反的反作用力。

Always draw a free-body diagram showing all forces: weight (mg downwards), normal reaction (perpendicular to surface), tension (along a string or rod), friction (opposing motion), and external applied forces. Resolve forces into components along perpendicular axes, then apply ΣF = ma in each direction.

务必画出受力分析图，标示所有力：重力(mg向下)、法向反力(垂直于表面)、张力(沿绳或杆)、摩擦力(阻碍运动)和外力。将力沿垂直轴分解，然后对每个方向应用ΣF = ma。

For objects on an inclined plane at angle θ to the horizontal, the weight component down the slope is mg sin θ, and the normal reaction is mg cos θ. Friction f ≤ μR, where μ is the coefficient of friction and R is the normal reaction. Limiting friction f = μR occurs when motion is about to start.

对于与水平面成θ角的斜面上的物体，沿斜面的重力分量为mg sin θ，法向反力为mg cos θ。摩擦力f ≤ μR，μ为摩擦系数，R为法向反力。当运动即将开始时，极限摩擦力f = μR。

5. Connected Particles and Pulleys | 连接体与滑轮

For systems of connected particles, treat each particle separately or consider the whole system. If a string is inextensible, accelerations of connected bodies have the same magnitude. If the string is light, tension is constant throughout its length. Pulleys are often assumed smooth and light, so tension is unchanged across them.

对于连接体系统，可分别处理每个物体或考虑整体。若绳子不可伸长，连接体的加速度大小相同。若绳子轻质，张力在整个绳长上恒定。滑轮通常假设光滑轻质，因此张力在滑轮两侧不变。

Example: two masses m₁ and m₂ connected by a light inextensible string over a smooth pulley. Equation for m₁: T – m₁g = m₁a (if m₂ > m₁, m₁ accelerates up). For m₂: m₂g – T = m₂a. Solve simultaneously to get a = (m₂ – m₁)g/(m₁ + m₂) and T = 2m₁m₂g/(m₁ + m₂).

例：两质量m₁和m₂由轻质不可伸长绳连接，跨过光滑滑轮。对m₁：T – m₁g = m₁a (若m₂ > m₁，m₁向上加速)。对m₂：m₂g – T = m₂a。联立求解得a = (m₂ – m₁)g/(m₁ + m₂)，T = 2m₁m₂g/(m₁ + m₂)。

If one particle rests on a table, include friction. Suppose m₁ on a rough horizontal table, coefficient of friction μ, connected by string over pulley to hanging m₂. Then T – μm₁g = m₁a and m₂g – T = m₂a. Solving gives a = (m₂ – μm₁)g/(m₁ + m₂).

若一物体置于桌面，需考虑摩擦。设m₁在粗糙水平桌上，摩擦系数μ，通过绳跨滑轮连接悬挂的m₂。则T – μm₁g = m₁a，m₂g – T = m₂a。解得a = (m₂ – μm₁)g/(m₁ + m₂)。

6. Momentum and Impulse | 动量与冲量

Linear momentum p = mv is a vector quantity. The principle of conservation of momentum states that in the absence of external forces, the total momentum of a system remains constant. For collisions or explosions: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.

线动量p = mv是矢量。动量守恒定律指出，在无外力作用下，系统总动量保持不变。对于碰撞或爆炸：m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。

Impulse J = FΔt = Δp = m(v – u). Impulse equals the change in momentum. In force-time graphs, the area under the curve represents impulse. Impulse is a vector and has units N s or kg m/s.

冲量J = FΔt = Δp = m(v – u)。冲量等于动量的变化。在力-时间图中，曲线下面积代表冲量。冲量是矢量，单位为N·s或kg·m/s。

Collisions can be elastic (kinetic energy conserved) or inelastic (some kinetic energy lost). In a perfectly inelastic collision, bodies stick together. Coefficient of restitution e = (speed of separation)/(speed of approach). e = 1 for elastic, 0 ≤ e < 1 for inelastic.

碰撞可分为弹性(动能守恒)或非弹性(部分动能损失)。在完全非弹性碰撞中，物体粘在一起。恢复系数e = (分离速度)/(接近速度)。弹性碰撞e = 1，非弹性0 ≤ e < 1。

7. Work, Energy and Power | 功、能量与功率

Work done by a constant force: W = F s cos θ, where θ is the angle between force and displacement. Work is a scalar measured in joules (J). The work-energy principle: net work done on an object equals its change in kinetic energy.

恒力做功：W = F s cos θ，其中θ为力与位移的夹角。功是标量，单位为焦耳(J)。功能原理：对物体做的净功等于其动能的变化。

Kinetic energy KE = ½ mv². Gravitational potential energy GPE = mgh (near Earth’s surface). Mechanical energy = KE + GPE. Conservation of mechanical energy holds when only conservative forces (like gravity) do work. In presence of friction, mechanical energy is not conserved; work done against friction = ΔME.

动能KE = ½ mv²。重力势能GPE = mgh (地表附近)。机械能 = KE + GPE。当只有保守力(如重力)做功时，机械能守恒。存在摩擦时，机械能不守恒；克服摩擦做功 = 机械能的变化。

Power P = ΔW/Δt = Fv (for constant force in direction of motion). Unit: watt (W). Average power = total work / time. Instantaneous power = Fv. Be careful with units when speed is given in km/h, convert to m/s.

功率P = ΔW/Δt = Fv (力与运动方向一致时)。单位：瓦特(W)。平均功率 = 总功 / 时间。瞬时功率 = Fv。注意单位转换，若速度单位是km/h，需转换为m/s。

8. Moments and Equilibrium | 力矩与平衡

The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action: M = F × d. Moments can cause rotation; clockwise moments are often taken as negative, anticlockwise as positive.

力对某点的力矩等于力的大小乘以该点到力作用线的垂直距离：M = F × d。力矩会引起转动；通常取顺时针力矩为负，逆时针为正。

For a body in static equilibrium, two conditions must be satisfied: (1) the vector sum of all forces is zero (ΣF = 0), and (2) the sum of moments about any point is zero (ΣM = 0). These conditions allow solving for unknown forces and reactions.

物体处于静力平衡必须满足两个条件：(1) 合力为零(ΣF = 0)，(2) 对任意点的合力矩为零(ΣM = 0)。利用这些条件可求解未知力和反力。

Uniform rods have weight acting at their centre. For a non-uniform rod, the centre of mass is not necessarily at the geometric centre. Pivots and supports: normal reactions at supports can be found by taking moments about a suitable point to eliminate an unknown force.

均匀杆的重力作用在其几何中心。非均匀杆的质心不一定在几何中心。支点与支撑：通过选取合适点取矩以消去某个未知力，可求出支撑处的法向反力。

9. Centre of Mass | 质心

The centre of mass is the point where the entire mass of a body can be considered concentrated for purposes of translational motion. For a system of particles: r_cm = (Σ m_i r_i) / Σ m_i. In two dimensions, coordinates (x̄, ȳ) = (Σ m_i x_i / M, Σ m_i y_i / M).

质心是为了平动可以将物体全部质量视为集中于此的点。对于质点系：r_cm = (Σ m_i r_i) / Σ m_i。二维坐标 (x̄, ȳ) = (Σ m_i x_i / M, Σ m_i y_i / M)。

Standard uniform shapes: rod (midpoint), rectangular lamina (intersection of diagonals), circular disc (centre), triangular lamina (intersection of medians, centroid at one-third height from base). For composite bodies, treat as collection of point masses at the centres of mass of component parts.

标准均匀形状：杆(中点)，矩形薄片(对角线交点)，圆盘(圆心)，三角形薄片(中线交点，形心距底边三分之一高)。对于组合体，可视为各部件质心处的质点集合处理。

If a part is removed (a hole), treat the removed portion as negative mass. For example, a circular lamina with a smaller circle removed: treat as whole positive mass at centre plus negative mass at the hole’s centre.

若移除某部分(如孔洞)，将被移除部分视为负质量。例如，一个圆形薄片挖去一个小圆：视为完整正质量位于其中心，加上孔洞中心处的负质量。

10. Coefficient of Friction and Inclined Planes | 摩擦系数与斜面

Friction is a resistive force that opposes relative motion. Static friction prevents motion up to a limit f_s ≤ μ_s R. Kinetic friction acts during motion: f_k = μ_k R. Usually μ_k < μ_s. The friction force always acts parallel to the contact surface.

摩擦力是阻碍相对运动的阻力。静摩擦力在极限内阻止运动：f_s ≤ μ_s R。动摩擦力在运动中作用：f_k = μ_k R。通常μ_k < μ_s。摩擦力始终平行于接触面。

On an inclined plane, resolve weight: component down slope = mg sin θ, perpendicular = mg cos θ. If the object is in equilibrium, friction must balance the component of weight down slope: f = mg sin θ. For limiting equilibrium (just about to slip), μ_s R = mg sin θ, and R = mg cos θ, giving μ_s = tan θ. This relation is often used to find the coefficient of static friction experimentally.

在斜面上，分解重力：沿斜面分量 = mg sin θ，垂直分量 = mg cos θ。若物体平衡，摩擦力必须平衡重力沿斜面分量：f = mg sin θ。对于极限平衡(即将滑动)，μ_s R = mg sin θ，且R = mg cos θ，得μ_s = tan θ。该关系常用于实验测定静摩擦系数。

For acceleration down a rough incline: ma = mg sin θ – μ_k mg cos θ, so a = g(sin θ – μ_k cos θ). If the object is projected up the slope, both weight component and friction oppose motion, giving a = -g(sin θ + μ_k cos θ).

粗糙斜面上的向下加速度：ma = mg sin θ – μ_k mg cos θ，故a = g(sin θ – μ_k cos θ)。若物体沿斜面上抛，重力分量和摩擦均阻碍运动，得a = -g(sin θ + μ_k cos θ)。

11. Variable Forces and Simple Harmonic Motion | 变力与简谐运动

In many mechanics problems, force varies with position, such as a spring obeying Hooke’s Law: F = -kx. The equation of motion leads to simple harmonic motion (SHM): ma = -kx → a = -(k/m)x. The solution is x = A sin(ωt + φ) or x = A cos(ωt + φ), with angular frequency ω = √(k/m).

在许多力学问题中，力随位置变化，例如服从胡克定律的弹簧：F = -kx。运动方程引出简谐运动：ma = -kx → a = -(k/m)x。解为x = A sin(ωt + φ)或x = A cos(ωt + φ)，角频率ω = √(k/m)。

Key SHM relationships: v = ω √(A² – x²), maximum speed v_max = ωA at equilibrium; maximum acceleration a_max = ω²A at extremes. Period T = 2π/ω = 2π √(m/k) for a mass-spring system. For a simple pendulum, T = 2π √(L/g) for small amplitudes.

简谐运动关键关系：v = ω √(A² – x²)，平衡位置最大速度v_max = ωA；端点最大加速度a_max = ω²A。周期T = 2π/ω，对于弹簧振子T = 2π √(m/k)。对于单摆，小角度周期T = 2π √(L/g)。

Energy in SHM: total mechanical energy E = ½ mω²A², kinetic energy KE = ½ mω²(A² – x²), potential energy PE = ½ mω²x². Energy continuously interchanges between kinetic and potential forms.

简谐运动中的能量：总机械能E = ½ mω²A²，动能KE = ½ mω²(A² – x²)，势能PE = ½ mω²x²。能量在动能和势能之间连续转换。

12. Exam Techniques and Common Pitfalls | 应试技巧与常见误区

Always read carefully whether vector quantities like velocity or acceleration are required, or just their magnitudes. Pay attention to units: convert to SI units (metres, seconds, kilograms) before substituting into formulas. g = 9.8 m/s² unless otherwise specified.

务必仔细阅读题目，明确要求的是矢量（如速度、加速度）还是仅大小。注意单位：代入公式前统一转换为国际单位制（米、秒、千克）。g 未特别说明时取 9.8 m/s²。

When applying conservation of momentum or energy, clearly define the system and check for external forces. In collisions, direction matters: assign positive direction and stick with it consistently. Show clear working and free-body diagrams; marks are awarded for method.

应用动量或能量守恒时，明确系统并检查外力。碰撞问题中方向至关重要：设定正方向并一致使用。展示清晰的解题步骤和受力图；该方法可获得过程分。

Common mistakes: forgetting to resolve forces correctly on inclined planes, using v² = u² + 2as with a negative displacement incorrectly, mixing up degrees and radians in trigonometric functions, and neglecting the vertical component of initial velocity in projectile motion. Practice past paper questions extensively.

常见错误：斜面上力分解不当，错误使用v² = u² + 2as时的负位移，三角函数中混淆角度与弧度，抛体运动中忽略初速度的竖直分量。务必大量练习历年真题。

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IB Math: Mechanics Key Points | IB 数学：力学 考点精讲