Edexcel Further Mechanics 1 Complete Guide — Edexcel 进阶数学力学1 完全指南

一、动量与冲量：冲量-动量原理及其应用 | Momentum and Impulse: The Impulse-Momentum Principle and Its Applications

Further Mechanics 1 的第一个核心概念是动量（momentum）。动量的定义是物体的质量与速度的乘积，记作 p = mv，单位是 kg m/s。与速度一样，动量是矢量，既有大小也有方向，因此在解题时必须先规定正方向。例如，一辆质量 1200 kg 的汽车以 25 m/s 向东行驶，它的动量就是 1200 × 25 = 30000 kg m/s，方向向东。

The first core concept in Further Mechanics 1 is momentum. Momentum is defined as the product of an object’s mass and its velocity, written p = mv, with units of kg m/s. Like velocity, momentum is a vector quantity: it has both magnitude and direction, so you must choose a positive direction before solving any problem. For example, a car of mass 1200 kg travelling east at 25 m/s has momentum 1200 × 25 = 30000 kg m/s directed eastwards.

与动量紧密相关的是冲量（impulse）。冲量定义为力与力作用时间的乘积，即 I = Ft，单位是 N s。冲量-动量原理指出：作用在物体上的冲量等于物体动量的变化量，即 Ft = mv – mu，其中 u 是初速度，v 是末速度。这个方程把力、时间和速度变化联系在一起，是 FM1 中最高频使用的工具之一。

Closely linked to momentum is impulse. Impulse is defined as the product of a force and the time for which it acts, I = Ft, with units of N s. The impulse-momentum principle states that the impulse acting on an object equals the change in its momentum: Ft = mv – mu, where u is the initial velocity and v is the final velocity. This equation connects force, time and velocity change, and it is one of the most frequently used tools in FM1.

例1：冲量改变运动方向 | Example 1: An Impulse Reversing the Direction of Motion

题目：一个质量为 0.5 kg 的球以 4 m/s 向右运动。一个向左的水平冲量 6 N s 作用在球上，求球的末速度。

Problem: A ball of mass 0.5 kg moves to the right at 4 m/s. A horizontal impulse of 6 N s acts on the ball towards the left. Find the final velocity of the ball.

解答：取向右为正方向，则初速度 u = 4 m/s，冲量 I = -6 N s。由冲量-动量原理：-6 = 0.5v – 0.5 × 4，整理得 0.5v = -6 + 2 = -4，所以 v = -8 m/s。负号表示球以 8 m/s 向左运动。

Solution: Take rightwards as positive, so the initial velocity is u = 4 m/s and the impulse is I = -6 N s. By the impulse-momentum principle: -6 = 0.5v – 0.5 × 4, which rearranges to 0.5v = -6 + 2 = -4, so v = -8 m/s. The negative sign means the ball moves leftwards at 8 m/s.

这个例子提醒我们两个要点：第一，冲量和速度都是矢量，正负号决定方向，弄错方向是考试中最常见的失分点；第二，物体可以先减速、停下、再反向加速，动量变化量 = 末动量 – 初动量这个式子本身就自动包含了方向的改变。现实中的应用包括汽车安全气囊和安全带：它们通过延长力的作用时间来减小冲击力，这正是 I = Ft 的直接体现。

This example highlights two key points. First, impulse and velocity are both vectors, and the signs determine direction; getting the direction wrong is one of the most common mark-losing errors in the exam. Second, an object can slow down, stop, and then accelerate in the opposite direction, and the equation change in momentum = final momentum – initial momentum automatically includes the direction change. Real-world applications include airbags and seat belts in cars: they increase the time over which the force acts, which reduces the impact force, a direct consequence of I = Ft.

二、动量守恒：碰撞前后系统总动量不变 | Conservation of Momentum: Total Momentum Before a Collision Equals Total Momentum After

动量守恒定律是 FM1 的基石之一：在没有外力作用（或外力可以忽略）的封闭系统中，碰撞前后系统的总动量保持不变。对两个物体碰撞的情形，可以写成 m1u1 + m2u2 = m1v1 + m2v2，其中 u 表示碰撞前的速度，v 表示碰撞后的速度。注意：守恒的是”系统总动量”，单个物体的动量在碰撞中一定会改变。

The principle of conservation of momentum is one of the cornerstones of FM1: in a closed system with no external forces (or where external forces are negligible), the total momentum of the system is the same before and after a collision. For a collision between two objects, this can be written as m1u1 + m2u2 = m1v1 + m2v2, where u denotes velocities before the collision and v denotes velocities after. Note that it is the total momentum of the system that is conserved; the momentum of each individual object always changes during a collision.

使用动量守恒时必须注意：方程中的速度都是矢量，需要统一正方向；如果两个物体碰撞后粘在一起，则 v1 = v2 = v，方程简化为 m1u1 + m2u2 = (m1 + m2)v。此外，反冲（recoil）问题也可以看成动量守恒：例如枪发射子弹时，枪与子弹组成的系统初始总动量为零，子弹向前飞出的同时枪必然向后反冲。

When using conservation of momentum, remember that all velocities in the equation are vectors and a common positive direction must be chosen. If the two objects stick together after the collision, then v1 = v2 = v and the equation simplifies to m1u1 + m2u2 = (m1 + m2)v. Recoil problems can also be treated with momentum conservation: when a gun fires a bullet, the total momentum of the gun and bullet system is initially zero, so the gun must recoil backwards while the bullet flies forwards.

例2：两辆玩具车碰撞 | Example 2: Two Toy Trolleys Colliding

题目：质量分别为 2 kg 和 3 kg 的两辆玩具车沿同一直线相向而行，速度分别为 5 m/s 和 2 m/s。碰撞后两车粘在一起，求碰撞后共同速度的大小和方向。

Problem: Two toy trolleys of masses 2 kg and 3 kg move towards each other along the same straight line with speeds 5 m/s and 2 m/s respectively. After the collision they stick together. Find the magnitude and direction of their common velocity after the collision.

解答：取 2 kg 车的运动方向为正。碰撞前总动量 = 2 × 5 + 3 × (-2) = 10 – 6 = 4 kg m/s。碰撞后总动量 = (2 + 3)v = 5v。由守恒：5v = 4，v = 0.8 m/s，方向与 2 kg 车原来的运动方向相同。

Solution: Take the direction of the 2 kg trolley as positive. Total momentum before = 2 × 5 + 3 × (-2) = 10 – 6 = 4 kg m/s. Total momentum after = (2 + 3)v = 5v. By conservation: 5v = 4, so v = 0.8 m/s, in the same direction as the 2 kg trolley’s original motion.

三、功与功率：定义、计算公式与常见陷阱 | Work and Power: Definitions, Formulas and Common Pitfalls

功（work）的定义是：力在物体位移方向上的分量与位移大小的乘积，W = Fs cos θ，其中 θ 是力与位移方向之间的夹角，单位是焦耳（J）。当力的方向与位移方向一致时，W = Fs；当力与位移垂直时，做功为零。例如，人提着重物在水平地面上匀速行走，手提力竖直向上，与水平位移垂直，因此手提力不做功。

Work is defined as the product of the component of the force in the direction of displacement and the magnitude of the displacement: W = Fs cos θ, where θ is the angle between the force and the displacement, measured in joules (J). When the force acts in the same direction as the displacement, W = Fs; when the force is perpendicular to the displacement, no work is done. For example, a person carrying a heavy bag walks at constant speed along level ground; the upward lifting force is perpendicular to the horizontal displacement, so the lifting force does no work.

功率（power）是做功的快慢，定义为单位时间内所做的功，P = W/t，单位是瓦特（W）。对于恒力牵引问题，还有更实用的公式 P = Fv，即功率等于力与速度的乘积。功率分为输入功率（发动机产生的总功率）和输出功率（用于驱动运动的功率），两者之差对应能量的损耗。效率 = 输出功率 / 输入功率 × 100%，是考试中经常要求计算的量。

Power is the rate of doing work, defined as work done per unit time, P = W/t, measured in watts (W). For problems involving a constant driving force, the more practical formula P = Fv applies: power equals force multiplied by velocity. Power can be divided into input power (the total power produced by the engine) and output power (the power available to drive the motion), and the difference between the two corresponds to energy losses. Efficiency = output power / input power × 100%, a quantity frequently requested in exams.

例3：斜向拉力做功 | Example 3: Work Done by an Oblique Pulling Force

题目：一个人用与水平方向成 30° 的力 50 N 拉着雪橇在水平地面上前进 20 m，求拉力做的功。

Problem: A person pulls a sledge along level ground with a force of 50 N at 30° to the horizontal over a distance of 20 m. Find the work done by the pulling force.

解答：W = Fs cos θ = 50 × 20 × cos 30° = 1000 × 0.866 ≈ 866 J。注意不能直接写成 50 × 20 = 1000 J，因为拉力并不完全沿位移方向；只有水平分量 50 cos 30° 在做功。

Solution: W = Fs cos θ = 50 × 20 × cos 30° = 1000 × 0.866 ≈ 866 J. Note that you must not simply write 50 × 20 = 1000 J, because the pulling force is not entirely along the direction of displacement; only its horizontal component 50 cos 30° does work.

四、动能与重力势能：能量守恒的起点 | Kinetic and Gravitational Potential Energy: The Starting Point of Energy Conservation

动能（kinetic energy）是物体由于运动而具有的能量，KE = ½mv²，单位是焦耳。注意动能是标量，永远非负，且与速度的平方成正比：速度加倍，动能变为原来的四倍。重力势能（gravitational potential energy）是物体由于位置升高而储存的能量，GPE = mgh，其中 h 是相对参考面的高度差。

Kinetic energy is the energy an object possesses due to its motion: KE = ½mv², measured in joules. Note that kinetic energy is a scalar, always non-negative, and proportional to the square of the speed: doubling the speed quadruples the kinetic energy. Gravitational potential energy is the energy stored in an object due to its height: GPE = mgh, where h is the height above a chosen reference level.

做功-能量原理（work-energy principle）把两者联系起来：合力所做的净功等于物体动能的变化量，即 W(净) = ½mv² – ½mu²。当只有重力做功时，机械能守恒：½mu² + mgh1 = ½mv² + mgh2。当存在摩擦力等非保守力时，机械能不守恒，损失的能量转化为热能，此时需要把摩擦力做的负功计入方程：初始机械能 + 外力做功 = 末机械能 + 摩擦力损耗。

The work-energy principle links the two: the net work done by the resultant force equals the change in kinetic energy, W(net) = ½mv² – ½mu². When only gravity does work, mechanical energy is conserved: ½mu² + mgh1 = ½mv² + mgh2. When non-conservative forces such as friction are present, mechanical energy is not conserved and the lost energy is converted to heat; the equation must then include the negative work done by friction: initial mechanical energy + work done by external forces = final mechanical energy + energy lost to friction.

例4：斜面滑下与摩擦损耗 | Example 4: Sliding Down a Slope with Friction

题目：一个质量 4 kg 的物块从倾角 30°、长 10 m 的粗糙斜面顶端由静止滑下，摩擦力恒为 8 N。求物块到达斜面底端时的速度。

Problem: A block of mass 4 kg slides from rest down a rough slope of length 10 m inclined at 30°. The friction force is constant at 8 N. Find the speed of the block when it reaches the bottom of the slope.

解答：斜面高度 h = 10 sin 30° = 5 m。初始机械能 = mgh = 4 × 9.8 × 5 = 196 J。摩擦力做功损耗 = 8 × 10 = 80 J。到达底端时机械能 = 196 – 80 = 116 J，全部为动能：½ × 4 × v² = 116，v² = 58，v ≈ 7.6 m/s。

Solution: The vertical height of the slope is h = 10 sin 30° = 5 m. Initial mechanical energy = mgh = 4 × 9.8 × 5 = 196 J. Energy lost to friction = 8 × 10 = 80 J. At the bottom, mechanical energy = 196 – 80 = 116 J, all in the form of kinetic energy: ½ × 4 × v² = 116, so v² = 58 and v ≈ 7.6 m/s.

五、胡克定律与弹性绳：张力与伸长量的线性关系 | Hooke’s Law and Elastic Strings: The Linear Relation between Tension and Extension

FM1 中处理弹性绳（elastic string）和弹簧（spring）时使用胡克定律。对弹性绳，张力 T = λx/l，其中 λ 是绳的弹性模量（modulus of elasticity，单位 N），l 是自然长度，x 是伸长量；对弹簧，张力 T = kx，其中 k 是劲度系数（单位 N/m）。两者都是线性关系：伸长量越大，张力越大，且张力始终指向恢复原长的方向。

Hooke’s law is used for elastic strings and springs in FM1. For an elastic string, the tension is T = λx/l, where λ is the modulus of elasticity (measured in N), l is the natural length and x is the extension; for a spring, the tension is T = kx, where k is the stiffness constant (measured in N/m). Both are linear relations: the greater the extension, the greater the tension, and the tension always acts towards restoring the natural length.

使用胡克定律时有几个关键细节。第一，λ 的单位是牛顿而不是 N/m，这与 k 不同，很多同学在这里写错单位而丢分。第二，弹性绳只能承受张力，不能承受压力，一旦松驰（x = 0），绳中张力立即为零；弹簧则可以拉伸也可以压缩。第三，弹性极限（elastic limit）之内胡克定律才成立，超过极限绳或弹簧会永久变形甚至断裂。

Several details matter when using Hooke’s law. First, the units of λ are newtons, not N/m, which is different from k, and many students lose marks by writing the wrong units here. Second, an elastic string can only sustain tension and cannot take compression; as soon as it becomes slack (x = 0), the tension drops to zero immediately. A spring, by contrast, can be stretched or compressed. Third, Hooke’s law only holds within the elastic limit; beyond it, the string or spring becomes permanently deformed or even breaks.

例5：悬挂重物求伸长量 | Example 5: Finding the Extension of a Hanging String

题目：一条自然长度 2 m、弹性模量 49 N 的弹性绳，上端固定，下端挂一个质量 3 kg 的物块，物块静止悬挂。取 g = 9.8 m/s²，求绳的伸长量。

Problem: An elastic string of natural length 2 m and modulus of elasticity 49 N has one end fixed and supports a 3 kg block hanging at rest from the other end. Taking g = 9.8 m/s², find the extension of the string.

解答：物块静止，绳中张力等于重力：T = 3 × 9.8 = 29.4 N。由胡克定律 T = λx/l：29.4 = 49x/2，解得 x = 29.4 × 2 / 49 = 1.2 m。此时绳的总长度为 2 + 1.2 = 3.2 m。

Solution: Since the block is at rest, the tension equals the weight: T = 3 × 9.8 = 29.4 N. By Hooke’s law T = λx/l: 29.4 = 49x/2, giving x = 29.4 × 2 / 49 = 1.2 m. The total length of the string is then 2 + 1.2 = 3.2 m.

六、弹性势能：拉伸储存的能量如何计算 | Elastic Potential Energy: How to Calculate Energy Stored in a Stretched String

拉伸弹性绳或弹簧时，我们对它做功，能量以弹性势能（elastic potential energy, EPE）的形式储存起来。弹性势能的公式为 EPE = λx²/(2l)（弹性绳）或 EPE = ½kx²（弹簧）。注意弹性势能永远为正，且与伸长量的平方成正比：伸长量翻倍，储存的能量变为原来的四倍。

When we stretch an elastic string or spring, we do work on it and the energy is stored as elastic potential energy (EPE). The formula is EPE = λx²/(2l) for an elastic string, or EPE = ½kx² for a spring. Note that EPE is always positive and proportional to the square of the extension: doubling the extension quadruples the stored energy.

含弹性绳的能量守恒问题在考试中非常常见，典型场景是：物块从某高度自由下落，撞到自然悬挂的弹性绳下端，把绳拉伸到最大伸长量后瞬时停下。此时能量方程为：损失的动能 + 损失的重力势能 = 储存的弹性势能。这类题目的关键是把”下降的总距离”和”绳的伸长量”区分清楚：如果物块从绳的自然长度位置开始下落，则下降距离 = 伸长量；如果从绳上方更高处下落，则下降距离 = 额外高度 + 伸长量。

Energy conservation problems involving elastic strings are very common in exams. A typical scenario: a block falls freely from a height and hits the lower end of a hanging elastic string, stretching it until it momentarily stops at maximum extension. The energy equation is then: kinetic energy lost + gravitational potential energy lost = elastic potential energy stored. The key to these problems is distinguishing between the total distance fallen and the extension of the string. If the block starts falling from the natural-length position, the distance fallen equals the extension; if it starts higher above the string, the distance fallen equals the extra height plus the extension.

例6：下落拉伸弹性绳 | Example 6: A Falling Block Stretching an Elastic String

题目：一条自然长度 1.5 m、弹性模量 60 N 的弹性绳上端固定。一个质量 2 kg 的物块系在绳下端，从绳自然长度位置由静止释放，求最大伸长量。取 g = 10 m/s²。

Problem: An elastic string of natural length 1.5 m and modulus 60 N has its upper end fixed. A block of mass 2 kg attached to the lower end is released from rest at the natural-length position. Find the maximum extension. Take g = 10 m/s².

解答：设最大伸长量为 x，此时物块下落的距离等于 x。重力势能损失 = mgx = 2 × 10 × x = 20x；弹性势能储存 = λx²/(2l) = 60x²/(2 × 1.5) = 20x²。能量守恒：20x = 20x²，即 x² = x，x = 0 或 x = 1。最大伸长量为 1 m。

Solution: Let the maximum extension be x; the distance fallen by the block is then also x. Gravitational potential energy lost = mgx = 2 × 10 × x = 20x; elastic potential energy stored = λx²/(2l) = 60x²/(2 × 1.5) = 20x². By conservation of energy: 20x = 20x², so x² = x, giving x = 0 or x = 1. The maximum extension is 1 m.

七、一维弹性碰撞：恢复系数与牛顿实验定律 | Elastic Collisions in One Dimension: Coefficient of Restitution and Newton’s Experimental Law

恢复系数（coefficient of restitution）e 是描述碰撞”弹性程度”的量，由牛顿实验定律（Newton’s experimental law）定义：e = (v2 – v1)/(u1 – u2)，其中 u1、u2 是碰撞前两物体沿碰撞方向的速度，v1、v2 是碰撞后的速度，所有速度都取同一正方向。e 的取值范围是 0 ≤ e ≤ 1：e = 1 表示完全弹性碰撞（动能无损失），e = 0 表示完全非弹性碰撞（两物体粘在一起），0 < e < 1 表示部分弹性碰撞。

The coefficient of restitution e measures how elastic a collision is, defined by Newton’s experimental law: e = (v2 – v1)/(u1 – u2), where u1 and u2 are the velocities of the two objects before the collision, v1 and v2 are the velocities after, all taken along the line of impact with a common positive direction. The coefficient lies in the range 0 ≤ e ≤ 1: e = 1 means a perfectly elastic collision (no kinetic energy lost), e = 0 means a perfectly inelastic collision (the objects stick together), and 0 < e < 1 means a partially elastic collision.

求解一维弹性碰撞的标准方法是联立两个方程：动量守恒方程 m1u1 + m2u2 = m1v1 + m2v2 和恢复系数方程 v2 – v1 = e(u1 – u2)。把第二个方程写成 v2 = v1 + e(u1 – u2) 代入第一个方程，即可解出 v1 和 v2。如果题目中给出”碰撞后动能损失了百分之几”，则需要额外利用动能公式列出第三个方程。

The standard method for solving one-dimensional elastic collisions is to solve two simultaneous equations: conservation of momentum m1u1 + m2u2 = m1v1 + m2v2 and the restitution equation v2 – v1 = e(u1 – u2). Writing the second as v2 = v1 + e(u1 – u2) and substituting into the first gives v1 and v2 directly. If the question states that a certain percentage of kinetic energy is lost in the collision, a third equation based on the kinetic energy formula must be added.

例7：部分弹性碰撞求解 | Example 7: Solving a Partially Elastic Collision

题目：质量 3 kg 的物体 A 以 6 m/s 向右运动，与静止的质量 1 kg 的物体 B 发生碰撞，恢复系数 e = 0.5。求碰撞后 A 和 B 的速度。

Problem: Object A of mass 3 kg moves to the right at 6 m/s and collides with object B of mass 1 kg at rest. The coefficient of restitution is e = 0.5. Find the velocities of A and B after the collision.

解答：取向右为正。动量守恒：3 × 6 + 1 × 0 = 3v1 + v2，即 3v1 + v2 = 18。恢复系数：v2 – v1 = 0.5 × (6 – 0) = 3。由第二式 v2 = v1 + 3 代入第一式：3v1 + v1 + 3 = 18，4v1 = 15，v1 = 3.75 m/s，v2 = 6.75 m/s。碰撞后 A 以 3.75 m/s、B 以 6.75 m/s 均向右运动。

Solution: Take rightwards as positive. Conservation of momentum: 3 × 6 + 1 × 0 = 3v1 + v2, so 3v1 + v2 = 18. Restitution: v2 – v1 = 0.5 × (6 – 0) = 3. Substituting v2 = v1 + 3 into the first equation: 3v1 + v1 + 3 = 18, so 4v1 = 15, giving v1 = 3.75 m/s and v2 = 6.75 m/s. After the collision A moves right at 3.75 m/s and B moves right at 6.75 m/s.

八、二维弹性碰撞：分量法与斜碰分析 | Elastic Collisions in Two Dimensions: Component Method and Oblique Impacts

当碰撞不在同一直线上发生时，需要把速度分解到两个互相垂直的方向上：沿碰撞线方向（line of centres，即碰撞瞬间两球球心连线方向）和垂直于碰撞线方向。动量守恒和恢复系数方程只在碰撞线方向上成立；在垂直于碰撞线的方向上，对于光滑物体，速度分量保持不变。

When a collision does not occur along a single straight line, velocities must be resolved into two perpendicular directions: along the line of centres (the line joining the centres of the two spheres at the instant of impact) and perpendicular to it. Conservation of momentum and the restitution equation apply only along the line of centres; in the perpendicular direction, for smooth objects, the velocity component is unchanged.

最典型的二维问题是小球撞击光滑固定墙壁（smooth fixed wall）。设小球以速度 u 与墙面法线成角 α 撞向墙面，恢复系数为 e，则碰撞后：垂直于墙面的速度分量由 u cos α 变为 -e u cos α（方向反转、大小乘以 e），平行于墙面的速度分量 u sin α 保持不变。因此碰撞后速度 v 满足 v² = (e u cos α)² + (u sin α)²，速度与法线的夹角 β 满足 tan β = sin α / (e cos α)。注意：因为平行分量不变，碰撞后的速度与法线夹角通常大于碰撞前的夹角，即 β > α。

The most typical two-dimensional problem is a sphere hitting a smooth fixed wall. Suppose the sphere approaches the wall with speed u at angle α to the normal, with coefficient of restitution e. After impact: the component perpendicular to the wall changes from u cos α to -e u cos α (direction reversed, magnitude multiplied by e), while the component parallel to the wall, u sin α, is unchanged. Hence the speed after impact satisfies v² = (e u cos α)² + (u sin α)², and the angle β of the velocity to the normal satisfies tan β = sin α / (e cos α). Since the parallel component is unchanged, the angle to the normal after impact is usually larger than before, so β > α.

例8：斜碰光滑墙 | Example 8: Oblique Impact with a Smooth Wall

题目：一个小球以速度 10 m/s 与光滑墙面的法线成 60° 角撞向墙面，恢复系数 e = 0.6。求碰撞后小球的速度大小和与法线的夹角。

Problem: A small sphere strikes a smooth wall at 60° to the normal with speed 10 m/s. The coefficient of restitution is e = 0.6. Find the speed of the sphere after impact and its angle to the normal.

解答：垂直分量 = 10 cos 60° = 5 m/s，碰撞后变为 0.6 × 5 = 3 m/s（方向反转）；平行分量 = 10 sin 60° ≈ 8.66 m/s（不变）。v = √(3² + 8.66²) ≈ √84 ≈ 9.17 m/s。tan β = 8.66 / 3 ≈ 2.887，β ≈ 70.9°。可见 β > 60°，符合预期。

Solution: The normal component is 10 cos 60° = 5 m/s, which becomes 0.6 × 5 = 3 m/s after impact (reversed); the parallel component is 10 sin 60° ≈ 8.66 m/s (unchanged). Hence v = √(3² + 8.66²) ≈ √84 ≈ 9.17 m/s. tan β = 8.66 / 3 ≈ 2.887, so β ≈ 70.9°. Indeed β > 60°, as expected.

九、能量损失与完全非弹性碰撞：粘在一起的问题 | Energy Loss and Perfectly Inelastic Collisions: When Objects Stick Together

任何 e < 1 的碰撞都会损失动能，损失的动能转化为热、声和形变能。计算动能损失的通用方法是：分别算出碰撞前后的总动能，然后相减，ΔKE = (½m1u1² + ½m2u2²) - (½m1v1² + ½m2v2²)。注意动能是标量，计算时直接使用速度的大小（速度的平方），不需要考虑方向符号。

Every collision with e < 1 loses kinetic energy, which is converted into heat, sound and deformation energy. The general method for calculating the energy loss is to find the total kinetic energy before and after the collision and subtract: ΔKE = (½m1u1² + ½m2u2²) - (½m1v1² + ½m2v2²). Note that kinetic energy is a scalar, so calculations use the speed (the square of the velocity) directly, with no direction signs involved.

完全非弹性碰撞（e = 0）是”粘在一起”的特殊情形，此时 v1 = v2 = v，动量守恒方程简化为 m1u1 + m2u2 = (m1 + m2)v。完全非弹性碰撞损失的能量是所有碰撞类型中最大的：对于给定的碰撞前动量，粘在一起意味着系统的末动能最小。这个结论可以这样理解：动量相同而质量越大，动能越小，因为 KE = p²/(2m)。

A perfectly inelastic collision (e = 0) is the special “sticking together” case, where v1 = v2 = v and the momentum equation simplifies to m1u1 + m2u2 = (m1 + m2)v. The energy lost in a perfectly inelastic collision is the largest possible for the given initial momentum: sticking together means the system ends up with the smallest possible kinetic energy. This can be understood through KE = p²/(2m): for a fixed momentum, greater mass means smaller kinetic energy.

例9：黏土块碰撞后的能量损失 | Example 9: Energy Lost When Two Lumps of Clay Collide

题目：质量 2 kg 的黏土块以 8 m/s 向右运动，与静止的质量 6 kg 的黏土块发生完全非弹性碰撞。求碰撞损失的动能。

Problem: A lump of clay of mass 2 kg moving right at 8 m/s collides perfectly inelastically with a stationary lump of clay of mass 6 kg. Find the kinetic energy lost in the collision.

解答：碰撞后共同速度 v = (2 × 8 + 6 × 0)/(2 + 6) = 16/8 = 2 m/s。碰撞前动能 = ½ × 2 × 8² = 64 J；碰撞后动能 = ½ × 8 × 2² = 16 J。损失动能 = 64 – 16 = 48 J，占初始动能的 75%。

Solution: The common velocity after the collision is v = (2 × 8 + 6 × 0)/(2 + 6) = 16/8 = 2 m/s. Kinetic energy before = ½ × 2 × 8² = 64 J; kinetic energy after = ½ × 8 × 2² = 16 J. Energy lost = 64 – 16 = 48 J, which is 75% of the initial kinetic energy.

十、FM1 考试技巧：常见题型与失分点 | FM1 Exam Techniques: Common Question Types and Where Students Lose Marks

FM1 的试卷题目虽然背景多样，但题型高度可预测。最常见的题型包括：冲量-动量问题（求冲量、末速度或平均作用力）、两物体碰撞（利用动量守恒 + 恢复系数联立求解）、斜碰光滑墙（分量法）、能量守恒问题（含或不含摩擦力、含弹性绳）、以及功率-牵引力问题（P = Fv 结合牛顿第二定律 F – R = ma）。

Although FM1 exam questions come in varied contexts, the question types are highly predictable. The most common types include: impulse-momentum problems (finding impulse, final velocity or average force), two-object collisions (solving conservation of momentum together with the restitution equation), oblique impacts with smooth walls (component method), energy conservation problems (with or without friction, and with elastic strings), and power-driving-force problems (P = Fv combined with Newton’s second law F – R = ma).

以下是历届考生最常见的失分点，务必逐一检查。第一，方向符号：所有矢量必须统一正方向，未规定正方向直接列方程会被扣分。第二，单位混乱：λ 的单位是 N、k 的单位是 N/m、冲量的单位是 N s，三者容易写混。第三，恢复系数的方向：牛顿实验定律公式中的速度必须沿碰撞线方向取值，符号处理错误会导致 e 为负值或大于 1 的荒谬结果。第四，弹性绳松弛：弹性绳只能承受张力，计算时要注意 x ≥ 0 的约束。第五，有效数字：Edexcel 官方要求答案保留 3 位有效数字（除非题目另有说明），约 9.8 m/s² 时中间步骤可多保留几位。

Here are the most common mark-losing errors made by past candidates; check each one carefully. First, direction signs: all vectors need a common positive direction, and writing equations without stating a positive direction loses marks. Second, unit confusion: λ is measured in N, k in N/m, and impulse in N s; these are easily mixed up. Third, the direction in the restitution formula: the velocities in Newton’s experimental law must be taken along the line of impact, and sign errors can produce absurd results such as a negative e or e > 1. Fourth, slack elastic strings: an elastic string can only sustain tension, so remember the constraint x ≥ 0. Fifth, significant figures: Edexcel requires answers to 3 significant figures unless stated otherwise, and when using g = 9.8 m/s² keep extra digits in intermediate steps.

最后，建立规范的解题流程：先画受力图，标明正方向；再列出已知量和未知量；然后选择适用的原理（动量守恒、能量守恒、冲量-动量、胡克定律、恢复系数）；最后代入数值求解并检查结果的合理性，比如速度方向是否与直觉相符、能量损失是否为正值。这个流程能帮助你在考场上稳定发挥，把 FM1 的分数稳稳拿下。

Finally, develop a disciplined problem-solving routine: draw a force diagram and mark the positive direction; list the known and unknown quantities; choose the applicable principle (conservation of momentum, conservation of energy, impulse-momentum, Hooke’s law, or the restitution equation); then substitute values, solve, and check the reasonableness of the answer, such as whether velocity directions match intuition and whether the energy loss is positive. This routine will help you perform consistently in the exam and secure full marks in FM1.

Summary | 总结

本文系统梳理了 Edexcel A-Level 进阶数学 Further Mechanics 1 的核心内容：动量与冲量（p = mv，Ft = mv – mu）、动量守恒（m1u1 + m2u2 = m1v1 + m2v2）、功与功率（W = Fs cos θ，P = Fv）、动能与重力势能（½mv²，mgh）、胡克定律与弹性势能（T = λx/l，EPE = λx²/(2l)）、一维与二维弹性碰撞（恢复系数 e 与分量法）。

This article has systematically covered the core content of Edexcel A-Level Further Mathematics Further Mechanics 1: momentum and impulse (p = mv, Ft = mv – mu), conservation of momentum (m1u1 + m2u2 = m1v1 + m2v2), work and power (W = Fs cos θ, P = Fv), kinetic and gravitational potential energy (½mv², mgh), Hooke’s law and elastic potential energy (T = λx/l, EPE = λx²/(2l)), and elastic collisions in one and two dimensions (the coefficient of restitution e and the component method).

掌握 FM1 的关键在于三点：一是矢量的方向意识，所有动量、冲量、速度问题都必须统一正方向；二是能量视角，把动能、势能、弹性势能和摩擦损耗放在同一个能量方程中统筹考虑；三是公式的适用条件，恢复系数只沿碰撞线方向成立，弹性绳不能承受压力。把这三点落实到每一道题的规范流程中，FM1 的高分自然水到渠成。

The key to mastering FM1 lies in three things: first, vector direction awareness, since every momentum, impulse and velocity problem requires a common positive direction; second, the energy perspective, balancing kinetic energy, potential energy, elastic potential energy and friction losses in a single energy equation; third, the conditions under which each formula applies, since the restitution equation only holds along the line of impact and elastic strings cannot take compression. Apply these three points within a disciplined routine for every problem, and top marks in FM1 will follow naturally.

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