📚 Edexcel Further Mechanics 2 Core Knowledge Points | Edexcel 进阶力学 2 核心知识点精讲
This revision article distils the essential definitions, formulas and problem‑solving strategies for Edexcel AS and A Level Further Mathematics – Further Mechanics 2. Every topic, from vector momentum to vertical circular motion, is presented in parallel English and Chinese to support deep understanding and exam success.
本文梳理了 Edexcel 进阶数学（进阶力学 2）模块的核心定义、公式与解题思路。每个知识点均以中英双语对照呈现，帮助你在复习中精准把握考点，提升解题能力。
1. Momentum and Impulse in Two Dimensions | 二维动量与冲量
The impulse‑momentum principle in vector form states that the impulse I applied to a particle equals the change in its momentum: I = m(v – u). In two dimensions, you must treat the i and j components separately, applying conservation of momentum vectorially for an isolated system.
矢量形式的冲量–动量定理为：作用在质点上的冲量 I 等于其动量的变化量 I = m(v – u)。在二维问题中，需要将 i、j 分量独立处理，并对孤立系统运用动量守恒的矢量形式。
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Impulse can also be expressed as I = F t, where F is a constant force vector. When a particle receives an impulse, its velocity changes instantaneously, but its position remains unchanged at that instant.
冲量还可以表示为恒力矢量 F 与时间 t 的乘积：I = F t。当质点受到冲量作用时，其速度瞬时改变，但位置在该时刻不发生变化。
2. Oblique Impact of Smooth Spheres | 光滑球体的斜碰
For a smooth sphere hitting another smooth sphere obliquely, resolve the velocities along the line of centres (the normal) and perpendicular to it. Smoothness guarantees no impulsive friction, so the velocity components perpendicular to the line of centres remain unchanged for both spheres.
处理两光滑球体的斜碰时，应将速度沿连心线（法线）和垂直连心线分解。因为光滑接触面无脉冲摩擦力，所以两球在垂直连心线方向的速度分量保持不变。
Newton’s experimental law of restitution applies along the line of centres: e = (v₂ₙ – v₁ₙ) / (u₁ₙ – u₂ₙ), where uₙ and vₙ denote normal velocity components before and after impact, and 0 ≤ e ≤ 1.
牛顿恢复系数定律沿连心线方向成立：e = (v₂ₙ – v₁ₙ) / (u₁ₙ – u₂ₙ)，其中 uₙ、vₙ 表示碰撞前后沿法向的速度分量，且 0 ≤ e ≤ 1。
Conservation of linear momentum along the line of centres provides a second equation. The loss in kinetic energy can be expressed as ΔKE = ½ m₁(m₂/(m₁+m₂))(1-e²)(u₁ₙ–u₂ₙ)² in one‑dimension analogy, but in oblique impact the loss reduces to only that associated with the normal components.
沿连心线方向的动量守恒提供第二个方程。动能损失可参照一维碰撞的表达式，但斜碰中只有法向分量贡献能量损失。
3. Work, Energy and Power in Vector Form | 矢量形式的功、能与功率
When a constant force F moves its point of application through a displacement s, the work done is the scalar product W = F·s = |F||s|cos θ. For a variable force, integrate: W = ∫ F·dr. The work–energy principle states that the total work done by all forces equals the change in kinetic energy.
恒力 F 作用点在位移 s 上做的功为标量积 W = F·s = |F||s|cos θ；变力则需积分 W = ∫ F·dr。动能定理指出：所有力做的总功等于质点动能的变化量。
Kinetic energy (KE) is ½ m v², and gravitational potential energy (GPE) is mgh near Earth’s surface. Power is the rate of doing work, P = dW/dt = F·v, and it measures the energy transferred per unit time.
动能为 ½ m v²，重力势能（近地表）为 mgh。功率是做功的快慢，P = dW/dt = F·v，表示单位时间内传递的能量。
4. Elastic Potential Energy and Energy Methods | 弹性势能与能量法
An elastic string or spring obeys Hooke’s law: the tension (or thrust) T is proportional to the extension x from natural length l, T = λx / l, where λ is the modulus of elasticity. The elastic potential energy (EPE) stored is EPE = λ x² / (2l).
弹性绳或弹簧遵循胡克定律：拉力（或推力）T 与伸长量 x 满足 T = λx / l，其中 λ 为弹性模量，l 为自然长度。储存的弹性势能为 EPE = λ x² / (2l)。
Problems involving vertical oscillation or release from rest are best tackled by writing a conservation of energy equation: initial energy (GPE + KE + EPE) = final energy. Remember that a string goes slack if extension becomes zero, after which its tension and EPE drop instantly to zero.
处理垂直振动或从静止释放的问题时，常用能量守恒：初态总能量（重力势能+动能+弹性势能）= 末态总能量。注意当绳恢复原长（伸长量为零）后会变松弛，此后张力和弹性势能立即消失。
5. Centres of Mass of Uniform Laminas | 匀质薄板的重心
The centre of mass (CoM) of a uniform lamina is the point where its entire weight can be considered to act. For standard shapes, use the results below. In all cases, the lamina is of uniform density and negligible thickness.
匀质薄板的重心（质心）可视为其全部重力作用的位置。下列表格给出了标准形状的重心位置，所有薄板均假设密度均匀且厚度可忽略。
| Shape / 形状 | Centre of mass position / 重心位置 |
|---|---|
| Uniform rod / 匀质细杆 | Midpoint |
| Rectangular lamina / 矩形薄板 | Intersection of diagonals |
| Triangular lamina / 三角形薄板 | Intersection of medians (2/3 along median from vertex) |
| Sector of a circle (radius r, angle 2α at centre, α in radians) / 扇形 (半径r，圆心角2α) | Distance from centre: (2r sin α) / (3α) |
To find the CoM of a composite lamina, treat it as a collection of simple shapes. Take moments about an axis: (total mass) × x̄ₜₒₜₐₗ = Σ (massᵢ × x̄ᵢ). For missing parts, treat the cut‑out as negative mass.
求组合薄板重心时，将其分解为简单图形，对参考轴取矩：(总质量) × x̄ₜₒₜₐₗ = Σ (质量ᵢ × x̄ᵢ)。若图形带有挖空部分，可将挖空区域视为负质量。
6. Centres of Mass of Solids of Revolution | 旋转体的重心
A solid of revolution is generated by rotating a curve y = f(x) about the x‑axis (or a similar axis). The centre of mass lies on the axis of symmetry. The x‑coordinate of the CoM is given by x̄ = (∫ x dV) / (∫ dV), where the volume element is dV = π y² dx.
旋转体由曲线 y = f(x) 绕 x 轴（或类似轴）旋转而成，其重心位于对称轴上。重心的 x 坐标可通过公式 x̄ = (∫ x dV) / (∫ dV) 求出，其中体积元 dV = π y² dx。
When the solid is made of a uniform material, its mass M is proportional to its volume V; thus the same integrals apply. Problems may require finding the position of the CoM of a frustum or a hemisphere using calculus.
对于匀质材料，质量 M 与体积 V 成正比，因此上述积分公式可以直接使用。考题中常要求通过微积分求出圆锥台或半球体的重心位置。
7. Simple Harmonic Motion (SHM) – Definitions and Equations | 简谐运动（SHM）– 定义与方程
A particle moves with simple harmonic motion when its acceleration is directed towards a fixed point and is proportional to the displacement from that point: a = –ω² x, where ω is a positive constant called the angular frequency. The negative sign indicates that acceleration always opposes displacement.
若质点的加速度始终指向一个固定点，且大小与位移成正比，则它做简谐运动：a = –ω² x，其中 ω 为正常数（角频率）。负号表示加速度方向始终与位移方向相反。
x = A sin(ω t + ε) or x = A cos(ω t + δ)
The amplitude A is the maximum displacement. The period T = 2π/ω does not depend on the amplitude. An extremely useful relation linking speed and displacement is v² = ω²(A² – x²), which is derived by integrating v dv/dx = –ω² x.
振幅 A 为最大位移，周期 T = 2π/ω 与振幅无关。一个十分重要的速度–位移关系式为 v² = ω²(A² – x²)，由积分加速度表达式 v dv/dx = –ω² x 得出。
8. SHM of Springs and Energy Considerations | 弹簧振子的 SHM 与能量分析
A light spring of stiffness k exhibits SHM for a mass m attached horizontally. Here ω² = k/m and the period is T = 2π √(m/k). For a vertical spring, the equilibrium extension e balances mg, and the motion about this equilibrium is still SHM with the same ω, independent of gravity.
劲度系数为 k 的轻弹簧连接质量 m 可产生简谐运动：水平振子满足 ω² = k/m，周期 T = 2π √(m/k)。对于竖直弹簧，平衡伸长量 e 平衡重力 mg，以平衡位置为中心的振动仍是 SHM，且 ω 与重力无关。
The total mechanical energy in SHM is constant and can be written as E = ½ m v² + ½ k x² (for a spring) or generally E = ½ m ω² A². This conservation law helps solve problems where the spring may go slack when extension ceases.
简谐运动中的总机械能守恒：对于弹簧振子，E = ½ m v² + ½ k x²；一般情形下 E = ½ m ω² A²。运用能量守恒可以方便地处理弹簧松弛等非持续简谐运动段的问题。
9. Circular Motion in a Horizontal Plane (Conical Pendulum) | 水平面内的圆周运动（圆锥摆）
In a conical pendulum, a particle is attached to a light string and moves in a horizontal circle at constant angular speed ω. The string traces a cone, and the particle’s path has radius r = L sin θ, where L is the string length and θ is the half‑angle from the vertical.
在圆锥摆中，质点系于轻绳一端，以恒定角速度 ω 在水平面内做圆周运动，绳描绘出一个圆锥面。轨道半径 r = L sin θ，L 为绳长，θ 为绳与竖直方向的夹角。
Resolving forces vertically gives T cos θ = mg; radially, T sin θ = m r ω². Eliminating T leads to ω² = g tan θ / r and the period Tₚ = 2π √(L cos θ / g). The period depends only on the vertical depth L cos θ.
竖直方向力平衡：T cos θ = mg；径向向心力：T sin θ = m r ω²。消去张力可得 ω² = g tan θ / r，周期 Tₚ = 2π √(L cos θ / g)。周期仅与竖直高度 L cos θ 有关。
10. Motion in a Vertical Circle | 竖直面内的圆周运动
When a particle moves on the inside of a smooth circular track or is whirled on a string in a vertical circle, the speed varies with height. Conservation of energy between the lowest point A and a general point P gives ½ m u² = ½ m v² + mgh, where u is the speed at A and h is the vertical height gained.
当质点沿光滑圆形轨道内侧运动，或用绳牵动物体在竖直面内做圆周运动时，速率随高度变化。由最低点 A 到任意点 P 的能量守恒给出：½ m u² = ½ m v² + mgh，其中 u 为 A 点速率，h 为上升的竖直高度。
The required condition to complete a full circle differs for a string (can go slack) and a rod (remains taut). For a string, the tension must be non‑negative at the highest point: T ≥ 0 implies v² ≥ gR, so the minimum speed at the bottom is √(5gR). For a light rod, the rod can push, so the minimal speed at the bottom can be as low as √(4gR), provided v ≥ 0 at the top.
能否完成完整圆周运动的条件因约束不同而异：轻绳在最高点张力不可小于零，即 v² ≥ gR，由此可推出最低点最小速率为 √(5gR)；轻杆在最高点可提供支撑，故最低点最小速率可低至 √(4gR)，只需保证最高点速率 v ≥ 0 即可。
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