📚 Further Mechanics 1: Momentum, Energy and Collisions | 进阶力学1：动量、能量与碰撞
Further Mechanics 1 builds on the core mechanics you have already studied, introducing momentum, impulse, work, energy, power, collisions and centres of mass in a more rigorous way. This revision guide summarises the key concepts, formulas and common exam techniques needed for Further Maths modules such as Edexcel FM1 or equivalent.
进阶力学1是在你已学的核心力学基础上进一步深化，引入动量、冲量、功、能量、功率、碰撞和质心等概念，并以更严格的方式处理。本复习指南总结了 Further Maths 模块（如 Edexcel FM1 或同等课程）所需的关键概念、公式和常见考试技巧。
1. Momentum | 动量
ven by the equation p = mv. The SI unit is kilogram metre per second, written as kg m s⁻¹.
动量是一个矢量，定义为质量与速度的乘积。对于质量为 m、速度为 v 的质点，动量 p 由方程 p = mv 给出。其国际单位是千克米每秒，写作 kg m s⁻¹。
Because momentum is a vector, direction matters. In one-dimensional problems, you should choose a positive direction and replace velocities in the opposite direction with negative values. This is especially important in collision problems.
由于动量是矢量，方向非常重要。在一维问题中，你应选择一个正方向，并将反方向的速度用负值代入。这在碰撞问题中尤其重要。
2. Impulse | 冲量
by I = Ft. The SI unit is newton second, written as N s, which is equivalent to kg m s⁻¹.
冲量度量了力在一段时间内作用所引起的动量变化。对于恒力 F 作用时间 t，冲量 I 由 I = Ft 给出。其国际单位是牛秒，写作 N s，等价于 kg m s⁻¹。
Impulse is also a vector quantity. If a force acts in the negative direction, the impulse is negative relative to your chosen positive direction. In exam questions, impulses are often given as a single value with a direction.
冲量也是一个矢量。如果力的方向为负，冲量相对于你所选择的正方向就是负值。在考试题目中，冲量通常给出一个数值并标明方向。
3. Impulse-Momentum Principle | 冲量-动量原理
ty u and final velocity v, then the impulse I is given by the vector equation I = mv - mu.
冲量-动量原理指出，作用在质点上的冲量等于其动量的变化。如果质点的初速度为 u，末速度为 v，则冲量 I 由矢量方程 I = mv - mu 给出。
rmined by the chosen positive direction. This principle is extremely useful for finding unknown velocities or forces when collision times are known.
在一维问题中，该原理可写成 I = m(v - u)，其中符号由所选正方向决定。当已知碰撞时间时，该原理在求解未知速度或力时极为有用。
4. Conservation of Momentum | 动量守恒
and final velocities v₁ and v₂, conservation of momentum gives m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
当两个质点发生碰撞且系统不受外力作用时，碰撞前的总动量等于碰撞后的总动量。对于质量 m₁ 和 m₂、初速度 u₁ 和 u₂、末速度 v₁ 和 v₂，动量守恒给出 m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。
This equation is valid for all types of collisions, including elastic and inelastic collisions. In two-dimensional problems, momentum is conserved separately in the horizontal and vertical components.
该方程对所有类型的碰撞均成立，包括弹性碰撞和非弹性碰撞。在二维问题中，动量在水平方向和竖直方向分别守恒。
- Always identify the system and check that no external resultant force acts during the collision.
- 总是先确定系统，并检查碰撞过程中是否没有外力合力作用。
- When particles join together, use the combined mass in the after-collision term.
- 当质点粘在一起时，在碰撞后的项中使用合并后的质量。
5. Work Done by a Force | 力所做的功
the displacement d, the work done W is given by W = Fd cos θ. The SI unit is the joule, J.
当力的作用点沿力的方向发生位移时，力对物体做功。对于与位移 d 成角 θ 的恒力 F，所做的功 W 由 W = Fd cos θ 给出。其国际单位是焦耳，J。
he force is perpendicular to the displacement, no work is done. Work can be positive or negative depending on the direction of the force relative to the motion.
如果力与位移方向相同，则 θ = 0 且 W = Fd。如果力与位移垂直，则不做功。功可以为正也可以为负，取决于力相对于运动的方向。
6. Kinetic and Potential Energy | 动能与势能
ional field, given by GPE = mgh, where h is the vertical height above a chosen zero level.
动能是质点由于运动而具有的能量。对于质量为 m、速度为 v 的质点，动能 KE 由 KE = ½mv² 给出。重力势能是由于在重力场中的位置而具有的能量，由 GPE = mgh 给出，其中 h 是相对于所选零势能面的竖直高度。
Energy is a scalar quantity, so direction is not relevant. However, changes in height must be measured vertically, not along the slope. In mechanics problems, the zero level for GPE can be chosen to simplify calculations.
能量是标量，因此方向无关紧要。但是，高度的变化必须沿竖直方向测量，而不是沿斜面。在力学问题中，可以选择重力势能零势能面以简化计算。
7. Work-Energy Principle | 功-能原理
e with initial speed u and final speed v, the net work done W is given by W = ½mv² - ½mu².
功-能原理指出，作用在质点上的总功等于其动能的变化。对于初速度为 u、末速度为 v 的质点，净功 W 由 W = ½mv² - ½mu² 给出。
This principle is especially useful when forces vary or when acceleration is not constant. It can also be combined with gravitational potential energy changes: work done against gravity equals mgh, and this appears as a loss or gain in mechanical energy.
该原理在力变化或加速度不恒定时尤其有用。它也可以与重力势能的变化相结合：克服重力所做的功等于 mgh，这表现为机械能的减少或增加。
Work-energy equation: Fd = ½mv² - ½mu² + mgh
功-能方程：Fd = ½mv² - ½mu² + mgh
8. Power | 功率
P developed is given by P = Fv. The SI unit of power is the watt, W, where 1 W = 1 J s⁻¹.
功率是做功的速率。如果恒力 F 使物体以速度 v 运动，则产生的功率 P 由 P = Fv 给出。功率的国际单位是瓦特，W，其中 1 W = 1 J s⁻¹。
e vehicle is moving at constant speed, the driving force equals the total resistance force.
对于沿道路行驶的车辆，发动机产生的驱动力通常使用 P = Fv 计算。如果车辆匀速行驶，驱动力等于总阻力。
- Power is a scalar quantity.
- 功率是标量。
- When speed changes, use the instantaneous relationship P = Fv at the relevant instant.
- 当速度变化时，在相关时刻使用瞬时关系 P = Fv。
9. Coefficient of Restitution | 恢复系数
after impact, then e = (v₂ - v₁)/(u₁ - u₂), provided the positive direction is consistent.
恢复系数，记作 e，用于度量碰撞的弹性程度。它定义为沿碰撞线分离相对速度与接近相对速度之比。对于两个质点，如果碰撞前速度为 u₁ 和 u₂，碰撞后速度为 v₁ 和 v₂，则在正方向一致的前提下 e = (v₂ - v₁)/(u₁ - u₂)。
s e = 1, while a perfectly inelastic collision, where particles stick together, has e = 0.
对于大多数碰撞，e 的值介于 0 和 1 之间。完全弹性碰撞的 e = 1，而完全非弹性碰撞（质点粘在一起）的 e = 0。
10. Direct Collisions | 一维直接碰撞
In a direct collision, the velocities of the two particles act along the same straight line. Two equations are used to solve for the two unknown final velocities: conservation of momentum and the coefficient of restitution equation.
在一维直接碰撞中，两个质点的速度沿同一直线作用。使用两个方程来求解两个未知的末速度：动量守恒方程和恢复系数方程。
- (-2)), so v₂ - v₁ = 3. Solving simultaneously gives v₁ = -1.4 m s⁻¹ and v₂ = 1.6 m s⁻¹.
例如，若 m₁ = 2 kg，m₂ = 3 kg，u₁ = 4 m s⁻¹，u₂ = -2 m s⁻¹，且 e = 0.5，首先写出动量方程 2(4) + 3(-2) = 2v₁ + 3v₂，化简得 2 = 2v₁ + 3v₂。恢复系数方程给出 0.5 = (v₂ - v₁)/(4 - (-2))，所以 v₂ - v₁ = 3。联立求解得 v₁ = -1.4 m s⁻¹ 和 v₂ = 1.6 m s⁻¹。
| Equation | Use |
| m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ | Momentum conservation in all collisions |
| e = (v₂ - v₁)/(u₁ - u₂) | Restitution along line of impact |
中文对照：动量守恒 m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ 用于所有碰撞；恢复系数 e = (v₂ - v₁)/(u₁ - u₂) 用于沿碰撞线方向。
11. Oblique Collisions with a Smooth Plane | 与光滑平面的斜碰
When a particle collides obliquely with a smooth fixed plane, the velocity component parallel to the plane remains unchanged because the plane is smooth and exerts no friction. The component perpendicular to the plane is reversed and multiplied by the coefficient of restitution e.
当质点与光滑固定平面发生斜碰时，平行于平面的速度分量保持不变，因为平面光滑且不产生摩擦。垂直于平面的速度分量则反向并乘以恢复系数 e。
posite direction. The rebound angle β satisfies tan β = (u cos α)/(e u sin α) = cot α / e.
如果质点以速度 u 与水平面成角 α 撞击平面，则平行分量为 u cos α，垂直分量为 u sin α。碰撞后，平行分量仍为 u cos α，而垂直分量变为反方向的 e u sin α。反弹角 β 满足 tan β = (u cos α)/(e u sin α) = cot α / e。
This component method avoids vector rotation and is the standard approach for oblique collisions in Further Mechanics 1.
这种分量方法避免了矢量旋转，是进阶力学1中处理斜碰的标准方法。
12. Centre of Mass | 质心
ion calculations. For particles with masses mᵢ and position coordinates xᵢ along a line, the centre of mass position x̄ is given by x̄ = Σmᵢxᵢ / Σmᵢ.
质点系的质心是这样一点：在平动计算中，系统的全部质量可视为集中于此。对于质量 mᵢ 且沿直线坐标 xᵢ 的质点，质心位置 x̄ 由 x̄ = Σmᵢxᵢ / Σmᵢ 给出。
For two-dimensional systems, the x and y coordinates are calculated separately using the same weighted average idea. Uniform rods, laminae and composite shapes can be treated by splitting them into simple parts.
对于二维系统，x 坐标和 y 坐标分别使用相同的加权平均思想计算。均匀杆、薄片和组合形状可以通过拆分为简单部分来处理。
- The centre of mass of a uniform rod lies at its midpoint.
- 均匀杆的质心位于其中点。
- For a uniform triangular lamina, the centre of mass is at the intersection of the medians.
- 对于均匀三角形薄片，质心位于中线的交点。
- Composite bodies: use moments of mass about a chosen axis.
- 组合体：对所选轴使用质量矩。
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