IB AQA Maths: Mechanics Key Points Explained | IB AQA 数学：力学考点精讲

📚 IB AQA Maths: Mechanics Key Points Explained | IB AQA 数学：力学考点精讲

Mechanics is the branch of mathematics concerned with the motion of objects and the forces that cause them. In the IB and AQA specifications, mechanics appears within the calculus and applied mathematics components, covering kinematics, Newton’s laws, energy, momentum, and more. Understanding these concepts is crucial for problem-solving in physics-related contexts.

力学是数学中研究物体运动及其受力的分支。在IB和AQA课程中，力学出现在微积分和应用数学部分，涵盖运动学、牛顿定律、能量、动量等。掌握这些概念对于解决物理情境中的问题至关重要。

1. Kinematics: Displacement, Velocity, Acceleration | 运动学：位移、速度、加速度

Kinematics describes motion without considering the forces causing it. Displacement (s) is the vector distance from a reference point, measured in metres (m). Velocity (v) is the rate of change of displacement with respect to time, and acceleration (a) is the rate of change of velocity. Both are vector quantities and can be expressed as derivatives:

运动学描述不考虑力的运动。位移（s）是相对于参考点的矢量距离，单位为米（m）。速度（v）是位移随时间的变化率，加速度（a）是速度随时间的变化率。两者均为矢量，并可表示为导数：

v = ds/dt, a = dv/dt = d²s/dt²

In one-dimensional motion along a straight line, the sign indicates direction. If the acceleration is constant, we can use SUVAT equations to model the motion. For variable acceleration, calculus becomes essential, differentiating or integrating between displacement, velocity, and acceleration functions.

在直线一维运动中，符号表示方向。若加速度恒定，我们可以使用SUVAT方程来建模运动。对于变加速度，微积分变得至关重要，需要在位移、速度和加速度函数之间进行微分或积分。

2. Constant Acceleration Equations (SUVAT) | 匀加速运动方程 (SUVAT)

For motion with constant acceleration a along a straight line, five key variables interrelate: initial velocity u, final velocity v, displacement s, time t, and acceleration a. The four SUVAT equations are:

对于加速度a恒定的直线运动，五个关键变量相互关联：初速度u、末速度v、位移s、时间t和加速度a。四个SUVAT方程为：

v = u + a t

s = u t + ½ a t²

s = (u + v)/2 × t

v² = u² + 2 a s

These equations only apply when acceleration is constant. When solving, list known quantities and identify the missing one, then select the appropriate equation. Take care with signs: choose a positive direction and maintain consistency, especially for vertical motion where gravity (g = 9.8 m/s²) acts downwards.

这些方程仅在加速度恒定时适用。解题时列出已知量并找出未知量，然后选择合适的方程。注意符号：选择正方向并保持一致性，尤其在垂直运动中重力（g = 9.8 m/s²）向下作用时。

3. Using Calculus in Kinematics | 运动学中的微积分应用

When acceleration varies with time, displacement, velocity, and acceleration are linked by differentiation and integration. Given a displacement function s(t), velocity v(t) = s'(t) and acceleration a(t) = v'(t) = s”(t). Conversely, if acceleration is given as a(t), velocity is v(t) = ∫ a(t) dt, and displacement is s(t) = ∫ v(t) dt.

当加速度随时间变化时，位移、速度和加速度通过微分和积分联系起来。给定位移函数s(t)，速度v(t) = s'(t)，加速度a(t) = v'(t) = s”(t)。反之，若给定加速度a(t)，则速度v(t) = ∫ a(t) dt，位移s(t) = ∫ v(t) dt。

Initial conditions must be used to find constants of integration. For a particle starting from rest, v(0) = 0; if it passes a certain point, s(0) may be known. In IB and AQA exams, problems often involve finding maximum displacement (when v = 0) or calculating distance travelled by considering changes in direction (where v changes sign).

必须使用初始条件来求出积分常数。对于从静止开始的粒子，v(0) = 0；若经过某点，s(0)可能已知。在IB和AQA考试中，问题通常涉及求最大位移（当v = 0时），或通过考虑方向变化（v改变符号处）来计算路程。

4. Newton’s Laws of Motion | 牛顿运动定律

Newton’s First Law: A body remains at rest or in uniform motion unless acted upon by a resultant external force. Second Law: The net force F acting on a body of mass m produces acceleration a such that F = ma. Third Law: Every action has an equal and opposite reaction.

牛顿第一定律：若不受合外力作用，物体将保持静止或匀速运动状态。第二定律：作用在质量为m的物体上的净力F产生加速度a，满足F = ma。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。

In mechanics problems, always draw a clear force diagram, resolve forces into components, and apply F = ma in the direction of motion. The unit of force is the newton (N). Remember that weight is mg and acts downwards. Normal reaction and tension forces are common; treat them as vectors.

在力学问题中，一定要画出清晰的受力图，将力分解为分量，并沿运动方向应用F = ma。力的单位是牛顿（N）。记住重力是mg，向下作用。法向反力和张力很常见；将它们视为矢量处理。

5. Connected Particles | 连接体问题

When two or more particles are connected by a light inextensible string over a smooth pulley, or placed on surfaces, they share the same magnitude of acceleration (if the string remains taut). Apply Newton’s second law to each particle separately to form simultaneous equations, then solve for acceleration and tension.

当两个或多个物体通过轻质不可伸长的绳子跨过光滑滑轮相连，或放置在表面上时，它们具有大小相同的加速度（若绳子保持拉紧）。分别对每个物体应用牛顿第二定律，建立联立方程，然后求解加速度和张力。

Consider the direction of motion for each particle. If one particle moves downwards, the other moves along the table or upwards. Assign positive direction consistently. Pulley problems often neglect mass and friction of the pulley unless stated otherwise. The tension T is the same on both sides of a smooth pulley.

考虑每个物体的运动方向。如果一个物体向下运动，另一个则沿桌面或向上运动。一致地分配正方向。除非特别说明，滑轮问题通常忽略滑轮的质量和摩擦。对于光滑滑轮，张力T在滑轮两侧大小相等。

6. Projectile Motion | 抛体运动

A projectile is an object moving under the influence of gravity alone, with no air resistance. Its motion can be analysed horizontally and vertically independently. Horizontal velocity u cos θ remains constant; vertical motion uses constant acceleration a = –g (if upwards is positive). Key equations:

抛体是仅受重力影响、无空气阻力的运动物体。其运动可分解为水平方向和竖直方向独立分析。水平速度u cos θ 保持不变；竖直运动使用恒定加速度a = –g（若向上为正）。关键方程：

Horizontal: x = (u cos θ) t

Vertical: y = (u sin θ) t – ½ g t², v_y = u sin θ – g t

Time of flight, maximum height, and range are derived by setting y = 0 for total flight time, v_y = 0 for max height, and substituting back. Always resolve initial velocity into components, and remember that the trajectory is a parabola.

飞行时间、最大高度和射程通过设y = 0求总飞行时间，设v_y = 0求最大高度，并代入求得。始终将初速度分解为分量，并记住轨迹为抛物线。

7. Forces and Equilibrium | 力与平衡

A particle is in equilibrium when the resultant force acting on it is zero. This means the vector sum of all forces equals zero. For coplanar forces, this gives two independent equations: sum of horizontal components = 0, sum of vertical components = 0.

当作用在一个质点上的合外力为零时，质点处于平衡状态。这意味着所有力的矢量和为零。对于共面力，可得到两个独立方程：水平分量之和 = 0，竖直分量之和 = 0。

Use resolving or triangle of forces methods. If three forces act at a point in equilibrium, they can form a closed triangle. Lami’s theorem can be applied: for three concurrent forces P, Q, R with opposite angles α, β, γ, P/sin α = Q/sin β = R/sin γ.

使用力的分解法或力三角形法。若在一点处有三个力平衡，它们可构成闭合三角形。可应用拉密定理：对于三个共点力P、Q、R，其对边夹角分别为α、β、γ，有P/sin α = Q/sin β = R/sin γ。

8. Friction | 摩擦力

Friction opposes relative motion or tendency of motion between surfaces. Static friction F ≤ μ_s R, where μ_s is the coefficient of static friction and R is the normal reaction. Kinetic (dynamic) friction F_k = μ_k R, where μ_k is the coefficient of kinetic friction and is usually less than μ_s.

摩擦力阻碍两表面之间的相对运动或运动趋势。静摩擦力F ≤ μ_s R，其中μ_s是静摩擦系数，R是法向反力。动摩擦力F_k = μ_k R，其中μ_k是动摩擦系数，通常小于μ_s。

When an object is on the point of sliding, friction reaches its limiting value F_max = μR (where μ can be μ_s if not specified). In inclined plane problems, resolve weight into components parallel and perpendicular to the plane, and apply F = ma along the slope including friction opposing motion.

当物体处于即将滑动的临界点时，摩擦力达到极限值F_max = μR（若未指定，μ可为μ_s）。在斜面问题中，将重力分解为平行和垂直于斜面的分量，并沿斜面应用F = ma，摩擦力方向与运动趋势相反。

9. Work, Energy and Power | 功、能量和功率

Work done by a constant force F moving an object a distance d in the direction of the force is W = F d. When force is at angle θ to the displacement, W = F d cos θ. Energy is measured in joules (J). Kinetic energy = ½ m v², gravitational potential energy = m g h.

恒力F沿力的方向移动物体距离d所做的功为W = F d。当力与位移夹角为θ时，W = F d cos θ。能量单位为焦耳（J）。动能 = ½ m v²，重力势能 = m g h。

The work–energy principle states that the total work done by all forces equals the change in kinetic energy: W_total = ΔKE. In the absence of non-conservative forces, mechanical energy is conserved. Power is the rate of doing work: P = W / t or P = F v (for constant force and velocity).

功能原理指出，所有力所做的总功等于动能的变化：W_total = ΔKE。在没有非保守力的情况下，机械能守恒。功率是做功的速率：P = W / t 或 P = F v（对于恒力与速度）。

10. Momentum and Impulse | 动量与冲量

Linear momentum p = m v is a vector quantity measured in kg m/s. Impulse J is the change in momentum caused by a force acting over time: J = F Δt = Δp = m v – m u. Impulse equals the area under a force–time graph.

线性动量p = m v是一个矢量，单位为kg m/s。冲量J是由力在一段时间内作用引起的动量变化：J = F Δt = Δp = m v – m u。冲量等于力–时间图下的面积。

The principle of conservation of momentum states that for a system with no external forces, total momentum before an interaction equals total momentum after. This is essential for collision and explosion problems. Always define a positive direction and handle vector signs carefully.

动量守恒定律指出，没有外力作用的系统，相互作用前的总动量等于相互作用后的总动量。这对碰撞与爆炸问题至关重要。务必定义正方向并小心处理矢量符号。

11. Moments | 力矩

The moment of a force about a point is the product of the force magnitude and the perpendicular distance from the point to the line of action of the force: M = F d. Moments are measured in N m. A moment tends to cause rotation; clockwise and anticlockwise directions are distinguished.

力对一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。力矩单位为N m。力矩趋向于引起转动；需区分顺时针和逆时针方向。

For a rigid body in equilibrium, both the resultant force and the resultant moment must be zero. Take moments about a pivot or any convenient point to eliminate unknown forces. The principle of moments: sum of clockwise moments = sum of anticlockwise moments about the same point.

对于刚体平衡，合外力与合力矩都必须为零。对支点或任一方便的点取力矩，以消去未知力。力矩原理：对同一点的顺时针力矩之和 = 逆时针力矩之和。

12. Collisions and Coefficient of Restitution | 碰撞与恢复系数

When two particles collide along a straight line, they exert equal and opposite impulses. The coefficient of restitution e is defined as the ratio of relative speed after collision to relative speed before: e = (v₂ – v₁) / (u₁ – u₂). For perfectly elastic collisions e = 1, for perfectly inelastic (coalescing) e = 0.

当两个质点沿直线碰撞时，它们施加大小相等、方向相反的冲量。恢复系数e定义为碰撞后相对速度与碰撞前相对速度之比：e = (v₂ – v₁) / (u₁ – u₂)。对于完全弹性碰撞e = 1，完全非弹性（结合）碰撞e = 0。

Using conservation of momentum together with Newton’s law of restitution allows us to find final velocities. In IB and AQA exams, these problems often ask for the loss in kinetic energy as well, which is ΔKE = KE_initial – KE_final, useful for assessing collision type.

结合动量守恒与牛顿恢复定律，可求出最终速度。在IB和AQA考试中，这类问题还常要求计算动能损失 ΔKE = KE_initial – KE_final，这有助于判断碰撞类型。

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IB AQA Maths: Mechanics Key Points Explained | IB AQA 数学：力学 考点精讲