📚 Edexcel IAL FM02 Further Mechanics 2 Exam Paper Key Concepts (May 2023) | Edexcel IAL FM02 进阶力学2试卷核心知识点精讲（2023年5月）
In the Edexcel IAL Further Mathematics AS examination, the FM02 paper (Further Mechanics 2) tests advanced mechanics topics that build on basic kinematics and dynamics. The 15 May 2023 sitting covered a range of concepts including momentum in two dimensions, oblique impacts, centre of mass, energy methods, elastic materials and simple harmonic motion. This article explains the core ideas you need to master for this unit, with worked examples and key formulas.
在爱德思IAL进阶数学AS考试中，FM02试卷（进阶力学2）考察建立在基础运动学和动力学之上的高级力学主题。2023年5月15日的考试涵盖了一系列概念，包括二维动量、斜碰撞、质心、能量法、弹性材料以及简谐运动。本文将讲解你需要掌握的核心要点，并提供示例和关键公式。
1. Momentum and Impulse in Two Dimensions | 二维动量与冲量
In FM02, momentum and impulse are treated as vector quantities, meaning you must handle both horizontal and vertical components. The impulse-momentum principle states that the impulse I applied to a particle equals the change in its momentum: I = m(v – u). When dealing with two particles, the total momentum of the system is conserved in the absence of external impulses, which gives you a pair of component equations.
在FM02中，动量和冲量都是矢量，因此必须同时处理水平分量和垂直分量。冲量-动量原理指出，作用在质点上的冲量I等于其动量的变化量：I = m(v – u)。对于两个质点组成的系统，若没有外力冲量，系统的总动量守恒，由此可写出两个分量方向的方程。
Iₓ = m(vₓ – uₓ), Iy = m(vy – uy)
Always begin by sketching a clear diagram, labelling velocity vectors before and after any impulse, and then resolve parallel and perpendicular to a chosen direction, such as the line of centres during a collision. For a single particle experiencing an impulse, the magnitude of the impulse is |I| = √Iₓ² + Iy².
一定要先画出清晰的示意图，标出冲量作用前后各质点的速度矢量，然后沿选定的方向（例如碰撞时的连心线方向）进行分解。对于单个质点受到冲量，冲量的大小为 |I| = √Iₓ² + Iy²。
- Key tip: In oblique impact problems, write momentum conservation separately along the line of impact and perpendicular to it.
- 重要提示：在斜碰撞问题中，要分别沿碰撞线方向和垂直于碰撞线方向写出动量守恒方程。
2. Oblique Impacts and Coefficient of Restitution | 斜碰撞与恢复系数
For a smooth head-on impact, Newton's law of restitution relates the relative speeds before and after collision: e = (v₂ – v₁)/(u₁ – smooth spheres, this law applies only to velocity components along the line of impact. If n is a unit vector along this line, the restitution equation is (v₁·n – v₂·n) = –e (u₁·n – u₂·n). The tangential components of velocity remain unchanged because the spheres are smooth.
对于光滑的正碰，牛顿恢复定律给出了碰撞前后相对速度的关系：e = (v₂ – v₁)/(u₁ – u₂)。在光滑球体的斜碰撞中，该定律仅适用于碰撞线方向的速度分量。若n是沿碰撞线的单位矢量，恢复系数方程为 (v₁·n – v₂·n) = –e (u₁·n – u₂·n)。因为球面光滑，切向速度分量保持不变。
e = (speed of separation) / (speed of approach) along the line of impact
ponds to a completely inelastic collision where the particles coalesce. Some FM02 questions ask you to find the angle of deflection or the loss in kinetic energy, which can be calculated from the velocity components.
注意 0 ≤ e ≤ 1。e = 1 对应完全弹性碰撞（动能守恒），而 e = 0 对应于完全非弹性碰撞，此时质点会粘在一起。部分FM02题目会要求计算偏转角或动能损失，这些都可以通过速度分量求得。
| e = 1 | Perfectly elastic; kinetic energy is conserved. | 完全弹性；动能守恒。 |
| 0 < e < 1 | Partially elastic; some kinetic energy is lost. | 部分弹性；部分动能损失。 |
| e = 0 | Completely inelastic; particles stick together. | 完全非弹性；质点粘在一起。 |
3. Statics of Rigid Bodies: Equilibrium and Toppling | 刚体静力学：平衡与倾倒
A rigid body is in equilibrium when the resultant force is zero and the resultant moment about any point is zero. In FM02, problems often involve a lamina on an inclined plane or a body resting on a rough surface. You need to resolve forces parallel and perpendicular to the plane, and take moments about a suitable point—often the point where the normal reaction acts, to eliminate unknown forces.
刚体平衡的条件是合力为零，且对任意点的合力矩为零。在FM02中，题目常涉及斜面上的薄板或置于粗糙面上的物体。你需要沿平行和垂直于平面分解力，并对合适的点（通常是法向反力作用点）取矩，以消去未知力。
e weight falls outside the base area, toppling occurs. For a uniform rectangular lamina on a rough incline, the critical angle for toppling is given by tan θ = (base width)/(2×height of centre of mass).
判断物体是倾倒还是滑动，需要比较引起转动的力矩与最大可能的摩擦力。若重力作用线超出支撑面，就会发生倾倒。对于放在粗糙斜面上的均匀矩形薄板，倾倒的临界角满足 tan θ = (底边宽)/(2×质心高度)。
- Sliding: requires the component of weight down the slope to exceed limiting friction.
- 滑动：需要重力沿斜面的分量超过极限摩擦力。
- Toppling: occurs if the moment of the weight about the pivot edge exceeds any restraining moment.
- 倾倒：若重力对支点棱边的力矩超过任何约束力矩，即发生倾倒。
4. Centre of Mass of Plane Laminas | 平面薄板的质心
The centre of mass of a system of particles is the point (x̄, ȳ) where the entire mass can be considered to act. For a plane lamina, the coordinates are given by
质点系的质心是可将全部质量视为集中作用的那一点 (x̄, ȳ)。对于平面薄板，质心坐标由下式给出
x̄ = (∑ mᵢ xᵢ) / ∑ mᵢ, ȳ = (∑ mᵢ yᵢ) / ∑ mᵢ
In FM02, you will work with uniform laminas whose mass is proportional to area. Standard shapes (rectangles, triangles, sectors, compound shapes) have known centres of mass that can be combined using area‑weighted averages. For a composite lamina, treat each part as a point mass at its own centre of mass, then apply the above formulas.
在FM02中，你将处理均匀薄板，其质量与面积成正比。标准图形（矩形、三角形、扇形、组合图形）的质心已知，可以通过面积加权平均来组合。对于组合薄板，可将每一部分视作集中在各自质心的质点，然后代入上述公式。
When a lamina is suspended freely from a point, it will hang with its centre of mass vertically below the point of suspension. This fact is frequently used to find coordinates of the centre of mass, or to determine the angle a side makes with the vertical.
当薄板从某一点自由悬挂时，质心一定位于悬挂点正下方。这一性质常被用来求解质心坐标，或确定某边与竖直线之间的角度。
- Uniform rectangular lamina: centre at intersection of diagonals.
- 均匀矩形薄板：质心在对角线交点。
- Uniform triangular lamina: centre of mass at the intersection of medians, ⅓ of the way from the base to the opposite vertex.
- 均匀三角形薄板：质心位于三条中线的交点，距底边为对应顶点到底边距离的⅓。
5. Work, Energy and Power with Variable Forces | 变力作用下的功、能量与功率
Work done by a force is the integral of the force with respect to displacement in the direction of the force. For a one‑dimensional variable force F(x),
力所做的功是力沿其作用方向的位移积分。对于一维变力 F(x)，
W = ∫ F(x) dx (from x₁ to x₂)
This idea is often combined with the work–energy principle: the net work done on a particle equals its change in kinetic energy. If a resistance force depends on speed or displacement, you may need to integrate a v dv/dx relationship.
这一思想常与功能原理结合：合力对质点所做的净功等于质点动能的变化量。若阻力依赖于速度或位移，你可能需要积分 v dv/dx 的关系式。
where v is the instantaneous speed in the direction of the force. When a vehicle moves against resistance, the driving force is often found via P/v. Be prepared to set up a differential equation linking acceleration, velocity and power.
功率是做功的快慢，P = Fv，其中 v 是沿力方向的瞬时速率。当车辆克服阻力运动时，驱动力常通过 P/v 求出。要准备好建立加速度、速度和功率之间的微分方程。
P = Fv, F – R = m a
where R is
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