AP Physics C: Mechanics

Unit 4: Linear Momentum第 4 单元：线动量

Impulse, collisions, and the conservation law that governs every interaction. Master the vector nature of momentum and learn when energy methods fail but momentum methods succeed.

冲量（impulse）、碰撞（collision），以及支配一切相互作用的守恒律。掌握动量的矢量本性，弄清在哪些情形下能量方法失效、而动量方法依然奏效。

10–20% Exam Weight考试占分 10–20% ~11–15 Class Periods约 11–15 课时 4 Topics4 个专题
Topic 4.1专题 4.1

Linear Momentum线动量

Momentum is the quantity that captures how much "motion" an object has — and in what direction. Unlike kinetic energy, momentum is a vector. Two identical cars traveling at the same speed in opposite directions have the same kinetic energy but opposite momenta. This distinction is essential for collision analysis — it's why momentum, not energy, captures the full picture of a system's state of motion.

动量（momentum）刻画了物体"运动得有多多"——以及朝哪个方向。与动能不同，动量是一个矢量。两辆完全相同、以同样速率反向行驶的汽车，动能相同但动量相反。这一区分对碰撞分析至关重要——正因如此，要完整描述系统的运动状态，靠的是动量而不是能量。

Definition of Linear Momentum线动量的定义
$$\vec{p} = m\vec{v}$$

Units: kg·m/s. Direction: same as $\vec{v}$.

单位：kg·m/s。方向与 $\vec{v}$ 一致。

Vector Nature Because momentum is a vector, you must assign a sign convention (or use components) when adding momenta. In one dimension, choosing rightward as positive, an object moving left has negative momentum. This is where many students first make errors in collision problems — always define your positive direction before you start.
矢量本性 动量是矢量，因此在求和时必须先定正负号约定（或拆成分量）。在一维情形下，取向右为正，则向左运动的物体动量为负。这正是学生最早在碰撞题里栽跟头的地方——动笔前一定先标好正方向。

Collisions and Explosions碰撞与爆炸

Momentum provides the natural language for describing two important classes of interactions. A collision is a model for an interaction in which the forces between the colliding objects are much larger than any external forces during the brief interaction time. Because of this, external forces can be neglected and momentum is approximately conserved. An explosion is the reverse: internal forces push objects within a system apart. The same conservation principle applies.

动量是描述两类重要相互作用的天然语言。碰撞（collision）的模型是：两物体间的内力在极短作用时间内远大于任何外力，因此外力可忽略，动量近似守恒。爆炸（explosion）则相反——系统内部的力把物体推开。但守恒原理同样适用。

Why Momentum Works for Collisions When we analyze a collision, we only compare the "immediately before" and "immediately after" states. During the collision itself, the forces can be wildly complicated, but we don't need to know the details — conservation of momentum handles the bookkeeping. This is what makes momentum methods so powerful: they bypass the need to know the force during the interaction.
为什么动量方法适合碰撞 分析碰撞时，我们只需要比较"刚碰撞前"和"刚碰撞后"两个状态。碰撞过程中的内力可能极其复杂，但我们不需要知道细节——动量守恒会自动把账算清楚。这正是动量方法的威力所在：它绕开了对碰撞过程中力的具体形式的依赖。
Worked Example — Explosion from Rest例题 —— 静止状态下的"爆炸"

A 3 kg cannon sits on a frictionless surface and fires
a 0.5 kg ball at 20 m/s to the right. Find the cannon's recoil velocity.

一门 3 kg 的炮置于无摩擦地面上，向右发射一颗 0.5 kg 的炮弹（速率 20 m/s）。求炮身的反冲速度（recoil velocity）。

System: cannon + ball. No external horizontal forces, so $p_\text{initial} = 0$ (both at rest).取系统为炮 + 炮弹。水平方向无外力，故 $p_\text{initial} = 0$（两者均静止）。
Conservation of momentum动量守恒
$$0 = m_\text{ball}\,v_\text{ball} + m_\text{cannon}\,v_\text{cannon}$$
$$0 = (0.5)(20) + (3)\,v_\text{cannon}$$
$$v_\text{cannon} = -\tfrac{10}{3} \approx -3.33\;\text{m/s}$$
The cannon recoils at 3.33 m/s to the left.炮身以 3.33 m/s 向左反冲。
A 4 kg object moves at 3 m/s to the right and a 2 kg object moves at 6 m/s to the left. What is the total momentum of the system? (Take rightward as positive.)4 kg 的物体向右以 3 m/s 运动，2 kg 的物体向左以 6 m/s 运动。系统的总动量是多少？（取向右为正。）
$24$ kg·m/s
$0$ kg·m/s
$−12$ kg·m/s
$12$ kg·m/s
Correct! $p = (4)(3) + (2)(-6) = 12 - 12 = 0$ kg·m/s. The momenta cancel exactly.正确！$p = (4)(3) + (2)(-6) = 12 - 12 = 0$ kg·m/s，两侧动量恰好抵消。
Use $\vec{p}_\text{total} = m_1 v_1 + m_2 v_2 = (4)(3) + (2)(-6) = 12 - 12 = 0$ kg·m/s. Answer: (B).由 $\vec{p}_\text{total} = m_1 v_1 + m_2 v_2 = (4)(3) + (2)(-6) = 12 - 12 = 0$ kg·m/s，选 (B)。
Topic 4.2专题 4.2

Change in Momentum and Impulse动量变化与冲量

Newton's Second Law: The Momentum FormNewton 第二定律：动量形式

Newton's second law is often written as $\vec{F} = m\vec{a}$, but this is actually a special case that assumes constant mass. Newton himself stated the law in terms of momentum: the net external force on a system equals the rate of change of that system's momentum. This more general formulation handles situations like rockets (where mass changes) and collisions (where force varies rapidly).

Newton 第二定律常被写成 $\vec{F} = m\vec{a}$，但这其实是质量恒定时的特例。Newton 本人是用动量来表述这条定律的：作用于系统的合外力等于系统动量的变化率。这种更一般的写法能处理质量变化的情形（如火箭）以及力随时间剧烈变化的情形（如碰撞）。

Newton's Second Law — General FormNewton 第二定律 —— 一般形式
$$\vec{F}_\text{net} = \frac{d\vec{p}}{dt}$$

The Impulse–Momentum Theorem动量定理

Integrating both sides of Newton's second law over a time interval gives the impulse–momentum theorem: the impulse delivered to an object equals its change in momentum. Impulse is defined as the integral of force over time — it's a vector quantity with the same direction as the net force.

将 Newton 第二定律两边对一段时间积分，得到动量定理（impulse–momentum theorem）：物体获得的冲量（impulse）等于它动量的变化。冲量定义为力对时间的积分，是一个与净力方向相同的矢量。

Impulse Definition冲量的定义
$$\vec{J} = \int_{t_1}^{t_2} \vec{F}_\text{net}(t)\,dt$$
Impulse–Momentum Theorem动量定理
$$\vec{J} = \int_{t_1}^{t_2} \vec{F}_\text{net}(t)\,dt = \Delta\vec{p} = \vec{p}_f - \vec{p}_0$$
Change in Momentum动量的变化
$$\Delta\vec{p} = \vec{p}_f - \vec{p}_0$$
Calculus Derivation — Two Lines to Memorize

Impulse–momentum drops straight out of Newton's second law. If the FRQ asks you to derive it, these two lines earn full credit.

Step 1 — Start from the general form of Newton's second law:
$$\vec{F}_\text{net} = \frac{d\vec{p}}{dt}$$
Step 2 — Integrate both sides over the collision time interval:
$$\int_{t_1}^{t_2}\vec{F}_\text{net}\,dt = \int_{\vec{p}_0}^{\vec{p}_f} d\vec{p} = \vec{p}_f - \vec{p}_0 = \Delta\vec{p}$$
The left side is the impulse by definition, giving the theorem:
$$\vec{J} = \Delta\vec{p}$$
微积分推导 —— 把这两步背下来

动量定理直接由 Newton 第二定律导出。若 FRQ 要求推导，这两行就能拿到满分。

第 1 步 —— 由 Newton 第二定律的一般形式出发：
$$\vec{F}_\text{net} = \frac{d\vec{p}}{dt}$$
第 2 步 —— 两边对碰撞时间区间积分：
$$\int_{t_1}^{t_2}\vec{F}_\text{net}\,dt = \int_{\vec{p}_0}^{\vec{p}_f} d\vec{p} = \vec{p}_f - \vec{p}_0 = \Delta\vec{p}$$
左边按定义即为冲量，于是得到定理：
$$\vec{J} = \Delta\vec{p}$$
When to Use Integral Form vs. $\bar F\,\Delta t$

Use $\vec{J} = \int \vec{F}(t)\,dt$ when the problem gives you $F(t)$ as a function (algebraic or graphical) — the integral is just the area under the $F$–$t$ curve.

Use $\vec{J} = \bar{\vec{F}}\,\Delta t$ when the problem tells you the collision lasts $\Delta t$ and asks for the average force. Rearrange to $\bar F = \Delta p / \Delta t$.

The two forms are consistent — $\bar F \equiv \tfrac{1}{\Delta t}\int F\,dt$ — but never mix them in the same equation.

何时用积分形式，何时用 $\bar F\,\Delta t$

当题目给出 $F(t)$（无论是代数式还是图像）时，用 $\vec{J} = \int \vec{F}(t)\,dt$——积分就是 $F$–$t$ 曲线下方的面积。

当题目告知碰撞持续时间 $\Delta t$ 并问平均力时，用 $\vec{J} = \bar{\vec{F}}\,\Delta t$，整理为 $\bar F = \Delta p / \Delta t$。

两种形式在概念上是自洽的——$\bar F \equiv \tfrac{1}{\Delta t}\int F\,dt$——但同一个方程里千万别混用。

Real-World Application — Safety Engineering The impulse–momentum theorem is the reason airbags and crumple zones work. In a car crash, the change in momentum (and hence the impulse) is fixed — the car must come to rest. But by increasing the collision time, the average force is reduced: $\vec{F}_\text{avg} = \Delta\vec{p} / \Delta t$. A longer stopping time means a smaller peak force on the passengers.
实际应用 —— 安全工程 动量定理正是安全气囊（airbag）和溃缩区（crumple zone）能保命的原因。在车祸中，动量的变化（也即冲量）是定值——车必须停下来。但若延长碰撞时间，平均力就会变小：$\vec{F}_\text{avg} = \Delta\vec{p} / \Delta t$。停止时间越长，乘客受到的峰值力就越小。

Graphical Interpretation图像解读

On a force vs. time graph, the impulse delivered to a system is the area under the curve. On a momentum vs. time graph, the net external force at any instant is the slope of the curve. These graphical connections are heavily tested on the AP exam.

在力—时间图上，冲量等于曲线下方的面积。在动量—时间图上，任意瞬时的净外力等于该曲线的斜率。这些图像关系在 AP 考试中频繁出现。

Graph ↔ Calculus Connections for Momentum动量的图像 ↔ 微积分关系
From由To求Operation运算On the Graph图像意义
$F(t)$$J$ (impulse)（冲量）$J = \int F\,dt$Area under $F$-$t$ curve$F$–$t$ 曲线下方的面积
$p(t)$$F_\text{net}$$F = dp/dt$Slope of $p$-$t$ graph$p$–$t$ 图的斜率
$J$$\Delta p$$J = \Delta p$Area = change in $p$面积 = 动量变化
Impulse Explorer — F(t) Graph冲量探究器 —— F(t) 图
Adjust the force profile. The shaded area equals the impulse = change in momentum. Axes remain fixed for comparison.调整力的形状。阴影面积即为冲量，也就是动量变化。坐标轴范围保持固定，便于对比。
Impulse J冲量 J
—
N·s
Δv
—
m/s
F avg平均力 F̄
—
N
Exam Tip — Bouncing vs. Stopping A ball that bounces off a wall receives a larger impulse than one that sticks. The bouncing ball reverses direction, so $\Delta p = m v_f - m(-v_i)$ — the magnitudes add. This is a classic AP question: a bouncing ball can deliver nearly twice the impulse of a ball that stops.
应试提醒 —— 反弹 vs. 黏住 撞墙后弹回的球比黏在墙上的球受到更大的冲量。弹回的球速度方向反转，故 $\Delta p = m v_f - m(-v_i)$——两个方向的大小相加。这是 AP 反复考的经典题：反弹球能传递接近"完全停下"情形两倍的冲量。

Variable-Mass Systems变质量系统

When an object's mass changes with time (such as a rocket expelling fuel), the impulse–momentum theorem in its general form still applies. If velocity is constant but mass changes:

当物体的质量随时间变化（例如喷射燃料的火箭），动量定理的一般形式依然成立。若速度恒定而质量变化：

Variable-Mass Form变质量形式
$$\vec{F}_\text{net} = \frac{d\vec{p}}{dt} = \frac{dm}{dt}\vec{v}$$

Valid when velocity is constant but mass changes (e.g., sand falling onto a conveyor belt).

当速度恒定但质量变化时成立（例如沙子落到匀速传送带上）。

For a rocket, the general form is $\vec{F}_\text{net} = m\frac{d\vec{v}}{dt} + \frac{dm}{dt}\vec{v}_\text{exhaust}$. The AP exam occasionally tests this — know the concept even if you aren't asked to derive the full rocket equation.

对于火箭，一般形式为 $\vec{F}_\text{net} = m\frac{d\vec{v}}{dt} + \frac{dm}{dt}\vec{v}_\text{exhaust}$。AP 考试偶尔会考——即便不要求你完整推出火箭方程，也要懂这个概念。

Worked Example — Impulse from a Variable Force例题 —— 由变力求冲量

A force F(t) = 6t N acts on a 3 kg object from rest
for t = 0 to t = 4 s. Find the impulse and final speed.

力 F(t) = 6t N 作用于一个 3 kg 的物体（初始静止），
持续 t = 0 到 t = 4 s。求冲量和末速率。

Step 1第 1 步
Impulse冲量
$$J = \int _0^4 6t dt = 3t^2 |_0^4 = 3(16) − 0 = 48\;\text{N}\cdot s$$
Step 2第 2 步
Apply impulse–momentum theorem应用动量定理
$$J = \Delta p = m \times \Delta v$$
$$48 = 3 \times (v_f − 0)$$
$$v_f = 16\;\text{m/s}$$
Worked Example — Impulse from F(t) Graph (Triangle)例题 —— 由 F(t) 图（三角形）求冲量

A triangular F(t) profile: F rises linearly from 0 to 20 N
over 0.5 s, then drops back to 0 over the next 0.5 s.
The object has mass 4 kg and is initially at rest.

三角形 F(t) 图：F 在 0.5 s 内由 0 线性升至 20 N，
再在随后 0.5 s 内降回 0。
物体质量 4 kg，初始静止。

Impulse = area under the $F$–$t$ curve. The triangle has base 1.0 s and height 20 N:冲量 = $F$–$t$ 曲线下方的面积。三角形底为 1.0 s、高为 20 N：
$$J = \tfrac{1}{2}(1.0\;\text{s})(20\;\text{N}) = 10\;\text{N}\cdot\text{s}$$
Final velocity末速度
$$v_f = \frac{J}{m} = \frac{10}{4} = 2.5\;\text{m/s}$$
A 0.15 kg ball moving at 10 m/s to the right bounces off a wall and returns at 8 m/s to the left. What is the impulse delivered by the wall? (Rightward = positive.)一个 0.15 kg 的球以 10 m/s 向右运动，撞墙后以 8 m/s 向左反弹。墙施加的冲量是多少？（取向右为正。）
$0.3$ N·s to the left向左
$1.5$ N·s to the left向左
$1.5$ N·s to the right向右
$2.7$ N·s to the left向左
Correct! $J = \Delta p = m v_f - m v_i = 0.15(-8) - 0.15(10) = -1.2 - 1.5 = -2.7$ N·s. The impulse is 2.7 N·s to the left (negative direction).正确！$J = \Delta p = m v_f - m v_i = 0.15(-8) - 0.15(10) = -1.2 - 1.5 = -2.7$ N·s。冲量大小为 2.7 N·s，方向向左（负方向）。
$J = \Delta p = 0.15(-8) - 0.15(+10) = -1.2 - 1.5 = -2.7$ N·s. The direction reversal means the magnitudes add. Answer: (D).$J = \Delta p = 0.15(-8) - 0.15(+10) = -1.2 - 1.5 = -2.7$ N·s。方向反转使两边数值相加。选 (D)。
Topic 4.3专题 4.3

Conservation of Linear Momentum线动量守恒

Conservation of momentum is one of the most fundamental principles in all of physics. During the brief instant of a collision, external forces like gravity and friction are dwarfed by the enormous contact forces between the colliding objects. So we model the system as having zero net external force during the collision, and momentum is conserved.

动量守恒（conservation of momentum）是整个物理学最根本的原理之一。在碰撞发生的极短瞬间，重力、摩擦等外力远小于物体间巨大的接触力，因此我们把系统模型为"碰撞期间净外力为零"，于是动量守恒。

Conservation of Linear Momentum线动量守恒
$$\sum \vec{p}_\text{before} = \sum \vec{p}_\text{after}$$

When $\vec{F}_\text{ext,net} = 0$ on the system.

当系统所受净外力 $\vec{F}_\text{ext,net} = 0$ 时成立。

Key Principle Momentum is conserved in all interactions, provided you choose the right system. If external forces act, expand your system to include the source of those forces, or recognize that momentum is transferred between the system and its environment via impulse: $\vec{J}_\text{ext} = \Delta\vec{p}_\text{system}$.
核心原则 只要选对系统，任何相互作用中动量都是守恒的。如果有外力存在，要么把外力来源也并入系统，要么承认动量通过冲量在系统与环境之间转移：$\vec{J}_\text{ext} = \Delta\vec{p}_\text{system}$。

Center-of-Mass Velocity质心速度

A collection of objects with individual momenta can be described as one system with one center-of-mass velocity. This velocity is constant whenever the net external force on the system is zero — even if objects within the system are bouncing off each other chaotically.

由若干各自具有动量的物体组成的整体，可用一个系统、一种质心速度（center-of-mass velocity）来描述。当系统所受净外力为零时，该速度保持恒定——即使内部物体之间正在杂乱无章地碰撞。

Center-of-Mass Velocity质心速度
$$\vec{v}_\text{cm} = \frac{\sum m_i \vec{v}_i}{\sum m_i} = \frac{\vec{p}_\text{total}}{M_\text{total}}$$
Powerful Shortcut In any collision with no external forces, $\vec{v}_\text{cm}$ is the same before and after. For a perfectly inelastic collision (objects stick together), the final velocity of the combined object is $\vec{v}_\text{cm}$. This gives you the answer instantly without setting up equations.
超快捷径 在任何无外力作用的碰撞中，$\vec{v}_\text{cm}$ 在碰撞前后保持不变。对于完全非弹性碰撞（perfectly inelastic collision，物体粘在一起），合并后整体的末速度就是 $\vec{v}_\text{cm}$。这样不用列任何方程就能直接得出答案。

System Selection系统选择

Whether momentum is conserved depends on what you choose as your system. A bullet hitting a block has external forces if you analyze the bullet alone (the block pushes on it), but if you analyze the bullet + block system, the collision forces become internal and momentum is conserved. Choosing the right system is a critical skill.

动量是否守恒取决于你把什么当作系统。如果只分析撞向木块的子弹，由于木块对子弹施力，存在外力；而如果把子弹和木块视为一个系统，碰撞力就变成内力，动量守恒。选对系统是一项关键技能。

System Selection Rules If $\vec{F}_\text{ext,net} = 0$ on the system → total momentum is constant. If $\vec{F}_\text{ext,net} \neq 0$ → momentum transfers between the system and the environment via $\vec{J} = \Delta\vec{p}$. Newton's third law guarantees that internal forces always cancel. The impulse exerted by object A on B is equal and opposite to the impulse exerted by B on A.
系统选择准则 若系统满足 $\vec{F}_\text{ext,net} = 0$ → 总动量保持不变。 若 $\vec{F}_\text{ext,net} \neq 0$ → 动量通过冲量在系统与环境之间转移：$\vec{J} = \Delta\vec{p}$。 Newton 第三定律保证所有内力都成对抵消。A 对 B 施加的冲量与 B 对 A 的冲量大小相等、方向相反。
Worked Example — Two-Cart Explosion例题 —— 两小车"爆炸"

Two carts are on a frictionless track, held together by a compressed spring.
Cart A (2 kg) and Cart B (3 kg). The spring releases.
After the explosion, Cart A moves at 6 m/s to the left. Find Cart B's velocity.

两辆小车置于无摩擦轨道上，由压缩的弹簧夹在中间。
A 车 2 kg、B 车 3 kg，弹簧释放后
A 车以 6 m/s 向左运动。求 B 车的速度。

System: Cart A + Cart B. No external horizontal forces, so $p_\text{initial} = 0$ (both at rest).系统：A 车 + B 车。水平方向无外力，故 $p_\text{initial} = 0$（两车均静止）。
Conservation of momentum动量守恒
$$0 = m_A v_A + m_B v_B$$
$$0 = (2)(-6) + (3)\,v_B$$
$$v_B = \tfrac{12}{3} = +4\;\text{m/s (to the right)}$$
Check with the center-of-mass velocity (zero before → must stay zero after):用质心速度校验（碰撞前为零 → 碰撞后必须仍为零）：
$$v_\text{cm} = \frac{(2)(-6) + (3)(4)}{5} = \frac{-12 + 12}{5} = 0\;\checkmark$$
A 5 kg block at rest on frictionless ice is hit by a 1 kg ball moving at 12 m/s. The ball sticks to the block. What is the velocity of the center of mass before and after the collision?无摩擦冰面上静止的 5 kg 木块被一个以 12 m/s 飞来的 1 kg 球击中并粘住。碰撞前后系统的质心速度分别是多少？
$12$ m/s before, $2$ m/s after碰撞前 $12$ m/s，碰撞后 $2$ m/s
$6$ m/s before, $6$ m/s after碰撞前 $6$ m/s，碰撞后 $6$ m/s
$2$ m/s before, $2$ m/s after碰撞前 $2$ m/s，碰撞后 $2$ m/s
$2.4$ m/s before, $2$ m/s after碰撞前 $2.4$ m/s，碰撞后 $2$ m/s
Correct! $v_\text{cm} = (1)(12) / (1+5) = 12/6 = 2$ m/s, both before and after. The center-of-mass velocity is always conserved when external forces are zero.正确！$v_\text{cm} = (1)(12) / (1+5) = 12/6 = 2$ m/s，碰撞前后都是 2 m/s。只要没有外力，质心速度始终守恒。
$v_\text{cm} = \sum m_i v_i / \sum m_i = (1)(12)/(1+5) = 2$ m/s. This doesn't change because no external forces act. Answer: (C).$v_\text{cm} = \sum m_i v_i / \sum m_i = (1)(12)/(1+5) = 2$ m/s。因为没有外力，这一速度不会变。选 (C)。
Topic 4.4专题 4.4

Elastic and Inelastic Collisions弹性碰撞与非弹性碰撞

Momentum is conserved in all collisions (provided no external forces act). But kinetic energy is not necessarily conserved — and the amount of kinetic energy retained is what distinguishes different types of collisions.

所有碰撞（在无外力的前提下）都满足动量守恒，但动能未必守恒——保留下来的动能多少正是区分不同碰撞类型的关键。

Collision Taxonomy Elastic: Both momentum and kinetic energy are conserved. $K_i = K_f$. Inelastic: Momentum is conserved, but kinetic energy decreases. $K_f < K_i$. Some KE is transformed into heat, sound, or deformation. Perfectly inelastic: Objects stick together. Maximum KE loss (consistent with momentum conservation). $v_f = (m_1 v_1 + m_2 v_2) / (m_1 + m_2)$.
碰撞分类 弹性碰撞（elastic collision）：动量和动能都守恒，$K_i = K_f$。 非弹性碰撞（inelastic collision）：动量守恒，但动能下降，$K_f < K_i$；部分动能转化为热、声或形变能。 完全非弹性碰撞（perfectly inelastic collision）：物体粘在一起，在满足动量守恒的前提下动能损失最大：$v_f = (m_1 v_1 + m_2 v_2) / (m_1 + m_2)$。

Perfectly Inelastic Collisions完全非弹性碰撞

In a perfectly inelastic collision, the objects stick together and move with a common final velocity. This gives the simplest algebra — one equation, one unknown.

在完全非弹性碰撞中，物体碰后粘在一起以共同速度运动。这种情形的代数最简——一个方程一个未知数。

Perfectly Inelastic Collision完全非弹性碰撞
$$m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2)\,v_f$$
Worked Example — Ballistic Pendulum例题 —— 弹道摆（ballistic pendulum）

A 10 g bullet embeds in a 2 kg pendulum bob. The combined mass rises to a height of 0.45 m. Find the bullet's initial speed.

10 g 的子弹嵌入一个 2 kg 的摆球中。合并后的整体上升到 0.45 m 高。求子弹的初速。

Stage 1 — Energy conservation (after collision, swinging upward)第 1 阶段 —— 能量守恒（碰撞后向上摆动）
$$\tfrac{1}{2}(m+M)v_f^2 = (m+M)gh$$
$$v_f = \sqrt{2gh} = \sqrt{2(10)(0.45)} = 3\;\text{m/s}$$
Stage 2 — Momentum conservation (the collision itself)第 2 阶段 —— 动量守恒（碰撞本身）
$$m\,v_{\text{bullet}} = (m+M)\,v_f$$
$$(0.010)\,v_{\text{bullet}} = (2.010)(3)$$
$$v_{\text{bullet}} = 603\;\text{m/s}$$

Common trap: do not apply energy conservation to the collision — it is perfectly inelastic, so kinetic energy is NOT conserved there. Use momentum for the collision and energy for the swing.

常见陷阱：不要对碰撞段套用能量守恒——它是完全非弹性碰撞，动能不守恒。碰撞段用动量，上升段才用能量。

Worked Example — Perfectly Inelastic Collision with KE Tracking (FRQ-style)例题 —— 完全非弹性碰撞 + 动能追踪（FRQ 风格）

A 2.0 kg cart moving at 6.0 m/s collides with and sticks to a 4.0 kg cart at rest on a frictionless track. (a) Find the common final velocity. (b) Find $KE_i$, $KE_f$, and $\Delta KE$. (c) What fraction of the initial kinetic energy was lost? (d) Where did that energy go?

一辆 2.0 kg 的小车以 6.0 m/s 撞向无摩擦轨道上静止的 4.0 kg 小车，撞后粘在一起。(a) 求共同末速度；(b) 求 $KE_i$、$KE_f$、$\Delta KE$；(c) 损失的动能占初动能的几分之几？(d) 这部分能量到哪里去了？

(a) Conservation of momentum(a) 动量守恒
$$m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2)\,v_f$$
$$(2.0)(6.0) + (4.0)(0) = (6.0)\,v_f \;\Rightarrow\; v_f = 2.0\;\text{m/s}$$
(b) Kinetic energies — side-by-side(b) 动能 —— 并排核对
$$KE_i = \tfrac{1}{2}(2.0)(6.0)^2 + 0 = 36\;\text{J}$$
$$KE_f = \tfrac{1}{2}(6.0)(2.0)^2 = 12\;\text{J}$$
$$\Delta KE = KE_f - KE_i = 12 - 36 = -24\;\text{J}$$
(c) Fraction of KE lost(c) 损失动能占比
$$\frac{|\Delta KE|}{KE_i} = \frac{24}{36} = \tfrac{2}{3} \approx 67\%$$
(d) Where the energy went(d) 能量去哪了
The missing 24 J is converted into heat, sound, and deformation of the coupling — not gone, just no longer mechanical. Momentum is conserved ($p_i = p_f = 12$ kg·m/s) but KE is not, which is the defining property of an inelastic collision.缺失的 24 J 转化为热能、声能和挂钩的形变能——并非消失，只是不再以机械能的形式存在。动量守恒（$p_i = p_f = 12$ kg·m/s），但动能不守恒——这正是非弹性碰撞的定义性特征。
Shortcut For a perfectly inelastic collision with target at rest, the fraction of KE retained is $\tfrac{m_1}{m_1+m_2}$ — here $\tfrac{2}{6} = \tfrac{1}{3}$, matching the $12/36$ above. The fraction lost is $\tfrac{m_2}{m_1+m_2}$.
捷径 对于靶物体静止的完全非弹性碰撞，保留下来的动能占比为 $\tfrac{m_1}{m_1+m_2}$——此例为 $\tfrac{2}{6} = \tfrac{1}{3}$，正与上面的 $12/36$ 吻合。损失的占比为 $\tfrac{m_2}{m_1+m_2}$。
Worked Example — Elastic Collision: Verify Both Conservation Laws例题 —— 弹性碰撞：同时核验两条守恒律

A 3.0 kg ball moving at 4.0 m/s collides elastically head-on with a 1.0 kg ball at rest. (a) Find both final velocities. (b) Verify that both momentum AND kinetic energy are conserved.

3.0 kg 的球以 4.0 m/s 与静止的 1.0 kg 球发生正向弹性碰撞。(a) 求两球的末速；(b) 同时核验动量与动能均守恒。

(a) Target-at-rest elastic formulas(a) 静止靶弹性碰撞公式
$$v_{1f} = \frac{m_1-m_2}{m_1+m_2}\,v_{1i} = \frac{3-1}{4}(4) = 2.0\;\text{m/s}$$
$$v_{2f} = \frac{2m_1}{m_1+m_2}\,v_{1i} = \frac{6}{4}(4) = 6.0\;\text{m/s}$$
(b) Momentum check(b) 动量核验
$$p_i = (3.0)(4.0) + 0 = 12\;\text{kg}\cdot\text{m/s}$$
$$p_f = (3.0)(2.0) + (1.0)(6.0) = 6 + 6 = 12\;\text{kg}\cdot\text{m/s}\;\checkmark$$
Kinetic energy check动能核验
$$KE_i = \tfrac{1}{2}(3.0)(4.0)^2 = 24\;\text{J}$$
$$KE_f = \tfrac{1}{2}(3.0)(2.0)^2 + \tfrac{1}{2}(1.0)(6.0)^2 = 6 + 18 = 24\;\text{J}\;\checkmark$$
$\Delta KE = 0$, which is the definition of elastic.$\Delta KE = 0$，这正是弹性碰撞的定义。

On the FRQ, write both conservation lines explicitly — graders look for $p_i = p_f$ AND $KE_i = KE_f$ as the two defining statements for an elastic collision.

在 FRQ 中，请把两条守恒方程都明确写出——批卷者会找 $p_i = p_f$ 与 $KE_i = KE_f$ 这两条作为弹性碰撞的定义性陈述。

Elastic Collisions弹性碰撞

Elastic collisions conserve both momentum and kinetic energy, giving you two equations. For the special case of a 1D elastic collision where object 2 is initially at rest:

弹性碰撞同时满足动量和动能守恒，因而给出两个方程。一维、靶物 2 初始静止的特殊情形：

1D Elastic Collision (Target at Rest)一维弹性碰撞（靶静止）
$$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}\,v_{1i} \qquad v_{2f} = \frac{2m_1}{m_1 + m_2}\,v_{1i}$$
Elegant Shortcut — Relative Velocity For elastic collisions, the relative speed of approach equals the relative speed of separation: $v_{1i} - v_{2i} = -(v_{1f} - v_{2f})$. This replaces the kinetic energy equation and is much easier to use because it's linear rather than quadratic.
优雅捷径 —— 相对速度 对于弹性碰撞，相对接近速率等于相对分离速率：$v_{1i} - v_{2i} = -(v_{1f} - v_{2f})$。这条线性关系可以替代二次的动能方程，使用起来要容易得多。
Special Cases — Elastic Collisions (Target at Rest)特殊情形 —— 弹性碰撞（靶静止）
Case情形After Collision碰撞后Example举例
$m_1 = m_2$Ball 1 stops; Ball 2 takes its velocity球 1 停下，球 2 取得原来的速度Newton's cradleNewton 摆（牛顿摆）
$m_1 \gg m_2$Ball 1 barely slows; Ball 2 flies at $\approx 2v_{1i}$球 1 几乎不减速，球 2 以约 $2v_{1i}$ 飞出Bowling ball → tennis ball保龄球撞网球
$m_1 \ll m_2$Ball 1 bounces back at $\approx -v_{1i}$; Ball 2 barely moves球 1 以约 $-v_{1i}$ 反弹，球 2 几乎不动Tennis ball → wall网球撞墙
Collision Explorer碰撞探究器
Compare elastic, perfectly inelastic, and custom collisions. Axes stay fixed so you can compare scenarios.比较弹性、完全非弹性以及自定义参数下的碰撞结果。坐标轴范围保持固定，便于对比。
v₁ finalv₁ 末
—
m/s
v₂ finalv₂ 末
—
m/s
KE retained动能保留
—
%
v_cm
—
m/s

Two-Dimensional Collisions二维碰撞

In 2D collisions, momentum is conserved independently in each direction — this gives you two equations (one for $x$, one for $y$). The AP exam expects you to handle these by decomposing all velocity vectors into components.

在二维碰撞中，动量在每个方向上都独立守恒——这给出两个方程（$x$ 方向一个，$y$ 方向一个）。AP 考试要求你将所有速度矢量分解为分量来处理。

2D Collision Strategy Always set up a coordinate system where one axis aligns with the initial direction of motion. Write $\sum p_x^\text{before} = \sum p_x^\text{after}$ and $\sum p_y^\text{before} = \sum p_y^\text{after}$ as two separate equations. If one object is initially at rest, the initial $y$-momentum is zero, which means the $y$-momenta of the two objects after the collision must cancel.
二维碰撞策略 建坐标系时，永远让一条轴沿初始运动方向。把 $\sum p_x^\text{before} = \sum p_x^\text{after}$ 和 $\sum p_y^\text{before} = \sum p_y^\text{after}$ 写成两个独立方程。若有一个物体初始静止，则初始 $y$ 方向动量为零，碰撞后两物体的 $y$ 分量动量必须相互抵消。
Worked Example — 2D Collision例题 —— 二维碰撞

A 2 kg puck at 5 m/s in +x hits a stationary 3 kg puck.
After: the 2 kg puck moves at 3 m/s at 30° above x-axis.
Find the 3 kg puck's velocity.

2 kg 的冰球沿 +x 方向以 5 m/s 撞向静止的 3 kg 冰球。
碰后 2 kg 冰球以 3 m/s 沿 x 轴上方 30° 方向运动。
求 3 kg 冰球的速度。

$x$-momentum:$x$ 方向动量：
$$(2)(5) + 0 = (2)(3\cos 30°) + (3)\,v_{2x}$$
$$10 = 5.196 + 3\,v_{2x} \;\Rightarrow\; v_{2x} = 1.601\;\text{m/s}$$
$y$-momentum:$y$ 方向动量：
$$0 = (2)(3\sin 30°) + (3)\,v_{2y}$$
$$0 = 3 + 3\,v_{2y} \;\Rightarrow\; v_{2y} = -1\;\text{m/s}$$
Magnitude and direction大小与方向
$$|\vec{v}_2| = \sqrt{1.601^2 + 1^2} \approx 1.89\;\text{m/s}$$
$$\theta = \arctan\!\left(\tfrac{-1}{1.601}\right) \approx -32°$$
Measured below the $x$-axis.从 $x$ 轴向下量起。
A 3 kg ball at 4 m/s hits a stationary 1 kg ball elastically. Using the elastic collision formulas, what is the speed of the 1 kg ball after the collision?3 kg 的球以 4 m/s 与静止的 1 kg 球发生弹性碰撞。用弹性碰撞公式，1 kg 球碰后的速率是多少？
$6$ m/s
$4$ m/s
$3$ m/s
$8$ m/s
Correct! $v_{2f} = \frac{2m_1}{m_1+m_2}v_{1i} = \frac{2(3)}{3+1}(4) = \frac{6}{4}(4) = 6$ m/s.正确！$v_{2f} = \frac{2m_1}{m_1+m_2}v_{1i} = \frac{2(3)}{3+1}(4) = \frac{6}{4}(4) = 6$ m/s。
$v_{2f} = \frac{2m_1}{m_1+m_2}v_{1i} = \frac{6}{4}(4) = 6$ m/s. Answer: (A).$v_{2f} = \frac{2m_1}{m_1+m_2}v_{1i} = \frac{6}{4}(4) = 6$ m/s，选 (A)。
In a head-on elastic collision, a 1 kg ball moving at 4 m/s hits a stationary 1 kg ball. What are the final velocities?在正向弹性碰撞中，1 kg 球以 4 m/s 撞向静止的 1 kg 球。两球末速度分别为？
(A) Both at 2 m/s(A) 两球都为 2 m/s
(B) Ball 1: 0 m/s, Ball 2: 4 m/s(B) 球 1：0 m/s；球 2：4 m/s
(C) Ball 1: 4 m/s, Ball 2: 0 m/s(C) 球 1：4 m/s；球 2：0 m/s
(D) Ball 1: −2 m/s, Ball 2: 6 m/s(D) 球 1：−2 m/s；球 2：6 m/s
Correct. For equal masses in an elastic collision with the target at rest, the projectile stops and the target takes on the projectile's original velocity. $v_{1f} = \frac{m_1-m_2}{m_1+m_2}v_{1i} = 0$ and $v_{2f} = \frac{2m_1}{m_1+m_2}v_{1i} = v_{1i} = 4$ m/s.正确。当两质量相等、靶静止的弹性碰撞中，入射物体停下，靶取得入射物体原本的速度。$v_{1f} = \frac{m_1-m_2}{m_1+m_2}v_{1i} = 0$，$v_{2f} = \frac{2m_1}{m_1+m_2}v_{1i} = v_{1i} = 4$ m/s。
With $m_1 = m_2$, the elastic formulas give $v_{1f} = 0$ and $v_{2f} = v_{1i} = 4$ m/s. The first ball stops; the second moves at 4 m/s. Answer: (B).当 $m_1 = m_2$ 时，弹性碰撞公式给出 $v_{1f} = 0$、$v_{2f} = v_{1i} = 4$ m/s。第一球停下，第二球以 4 m/s 运动。选 (B)。
Energy Loss in Collisions碰撞中的动能损失
See how the mass ratio affects the fraction of kinetic energy lost in a perfectly inelastic collision (target at rest).观察在完全非弹性碰撞（靶静止）中，质量比如何决定动能损失的比例。
KE Retained保留动能
—
%
KE Lost损失动能
—
%
v_f / v₁ᵢ
—
Exam Preparation考试备考

Exam Strategy考试策略

✎ Multiple-Choice Questions✎ 多选题（Multiple Choice）

MCQs in this unit test your ability to compute impulse from $F(t)$ graphs (area under the curve), compare momenta and impulses in different scenarios, determine whether a collision is elastic or inelastic from given data, and apply momentum conservation in 1D and 2D. Always assign a sign convention first, and watch for questions where a bouncing object delivers more impulse than a stopping one. Many questions can be answered quickly using the special-case elastic collision results (equal masses, heavy-on-light, light-on-heavy).

本单元的 Multiple Choice 部分考查：从 $F(t)$ 图（曲线下面积）计算冲量；在不同情境下比较动量与冲量；根据已知数据判断碰撞是弹性还是非弹性；以及在一维和二维下应用动量守恒。永远先定正负号约定，并留意"反弹比停下传递的冲量更大"这种陷阱。许多题用弹性碰撞的特殊情形（等质量、大撞小、小撞大）能一眼解决。

✎ Free-Response — Experimental Design (FRQ 3)✎ 自由作答 —— 实验设计（FRQ 第 3 题）

The third FRQ on the AP exam is an Experimental Design and Analysis question, which frequently draws from this unit. You may be asked to design an experiment to verify momentum conservation using carts on a track, or to determine an unknown mass using a collision. Key skills: describe what measurements you'd take (masses, velocities before and after), explain how you'd use them to test conservation, discuss sources of error (friction, imprecise timing, rotational effects), and linearize your data for graphing.

AP 考试的第 3 道 FRQ 是"实验设计与分析（Experimental Design）"题，常考本单元内容。你可能被要求：设计一个用轨道小车验证动量守恒的实验，或借助一次碰撞测量未知质量。关键技能：清楚说明要测哪些量（质量、碰前与碰后速度）；解释如何用这些量检验守恒；讨论至少一种误差来源（摩擦、计时不准、转动效应）；并把数据线性化便于作图。

FRQ Step-by-Step 1. Identify the system and state that you assume external forces are negligible. 2. Write conservation of momentum in component form if 2D. 3. Check elastic vs. inelastic — if the problem says "sticks together," it's perfectly inelastic; use the simpler equation. 4. Show units and box your final answer. 5. For experimental design, describe the procedure sequentially, list all measurements, explain how to use them, and discuss at least one source of error.
FRQ 步骤清单 1. 选定系统，并声明假设外力可忽略。 2. 写出动量守恒；二维情形要按分量写。 3. 区分弹性 vs. 非弹性——若题目说"粘在一起"，就是完全非弹性，用更简单的方程。 4. 写出单位，并把最终答案画框圈起。 5. 实验设计题：按顺序写出操作流程、列出所有需要测量的量、说明如何利用它们，并讨论至少一种误差来源。
Pitfalls陷阱

Common Mistakes常见错误

Top Point-Losing Errors

1. Forgetting that momentum is a vector. You must assign signs or use components. Simply adding magnitudes is wrong whenever objects move in different directions.

2. Confusing elastic and perfectly inelastic. "Elastic" conserves KE. "Perfectly inelastic" means they stick together. Many students swap these under time pressure.

3. Assuming KE is always conserved. Kinetic energy is conserved only in elastic collisions. Momentum is conserved in all collisions (when external forces are negligible). Don't conflate the two.

4. Using the wrong equation. $v_f = (m_1 v_1 + m_2 v_2)/(m_1 + m_2)$ is only valid for perfectly inelastic collisions. Don't use it for elastic collisions.

5. Ignoring direction change in impulse. A ball bouncing back has a larger $|\Delta p|$ than a ball that stops. The reversal of direction doubles the momentum change.

6. Applying momentum conservation when external forces are significant. If a collision occurs on an incline or with friction over a long time, external forces matter.

7. Mixing up impulse and momentum. Impulse is the change in momentum ($\vec{J} = \Delta\vec{p}$), not the momentum itself.

最容易丢分的错误

1. 忘了动量是矢量。必须先定正负号或拆成分量。只要物体方向不同，直接加大小就是错的。

2. 弄混"弹性"和"完全非弹性"。"elastic" 守恒动能；"perfectly inelastic" 指碰后粘在一起。许多学生在时间紧迫时把这两个互换。

3. 误以为动能总是守恒。只有弹性碰撞守恒动能；任何忽略外力的碰撞都守恒动量。两者不要混为一谈。

4. 用错公式。$v_f = (m_1 v_1 + m_2 v_2)/(m_1 + m_2)$ 只对完全非弹性碰撞成立，别拿它去算弹性碰撞。

5. 忽视方向变化对冲量的影响。反弹回去的球 $|\Delta p|$ 更大。方向反转使动量变化"翻倍"。

6. 在外力不可忽略的情形下乱用动量守恒。如果碰撞发生在斜面上，或长时间含摩擦，外力就不能忽略。

7. 把冲量和动量当一回事。冲量是动量的变化（$\vec{J} = \Delta\vec{p}$），不是动量本身。

Review复习

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点击卡片可翻面。

Linear Momentum线动量
$$\vec{p} = m\vec{v}$$
Vector, same direction as $\vec{v}$. Units: kg·m/s.矢量，方向与 $\vec{v}$ 一致。单位：kg·m/s。
Newton's 2nd Law
(Momentum Form)Newton 第二定律
（动量形式）
$$\vec{F}_\text{net} = \frac{d\vec{p}}{dt}$$
Force = rate of change of momentum; handles variable-mass systems too.力 = 动量变化率；同样适用于变质量系统。
Impulse–Momentum
Theorem冲量—动量
定理
$$\vec{J} = \int \vec{F}\,dt = \Delta\vec{p}$$
Area under the $F$–$t$ curve equals $\Delta p$.$F$–$t$ 曲线下方的面积等于 $\Delta p$。
Conservation of
Momentum动量守恒
$$\sum \vec{p}_\text{before} = \sum \vec{p}_\text{after}$$
Valid when $\vec{F}_\text{ext,net} = 0$ on the chosen system.当所选系统满足 $\vec{F}_\text{ext,net} = 0$ 时成立。
Perfectly Inelastic
Collision完全非弹性
碰撞
$$v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$$
Objects stick together → maximum KE loss; momentum still conserved.物体粘合 → 动能损失最大；动量依然守恒。
Elastic Collision
(Equal Masses)弹性碰撞
（等质量）
$$v_{1f}=0,\; v_{2f}=v_{1i}$$
Equal-mass elastic collision with target at rest — projectile stops, target takes its speed.靶静止、等质量弹性碰撞：入射体停下，靶取得入射速度。
Relative Velocity Rule
(Elastic Only)相对速度规则
（仅适用弹性碰撞）
$$v_{1i} - v_{2i} = -(v_{1f} - v_{2f})$$
Elastic collisions: approach speed = separation speed.弹性碰撞：接近速率 = 分离速率。
Center-of-Mass
Velocity质心速度
$$\vec{v}_\text{cm} = \frac{\sum m_i \vec{v}_i}{\sum m_i}$$
Constant when $\vec{F}_\text{ext,net} = 0$.当 $\vec{F}_\text{ext,net} = 0$ 时保持不变。
Assessment测评

Unit Quiz单元测验

Q1. A 60 kg person jumps off a 200 kg boat (both initially at rest on frictionless water). If the person's velocity is 3 m/s relative to the dock, the boat's velocity is:一名 60 kg 的人从一艘 200 kg 的小船上跳下（两者初始在无摩擦水面上静止）。若该人相对岸边的速度为 3 m/s，则船的速度为：
(A)   3 m/s in the same direction3 m/s，同向
(B)   1.2 m/s opposite1.2 m/s，反向
(C)   0.9 m/s opposite0.9 m/s，反向
(D)   0.6 m/s opposite0.6 m/s，反向
$0 = 60(3) + 200(v_b)$, so $v_b = -180/200 = -0.9$ m/s. The boat moves at 0.9 m/s opposite to the person.$0 = 60(3) + 200(v_b)$，得 $v_b = -180/200 = -0.9$ m/s。船以 0.9 m/s 与人反向运动。
Conservation of momentum: $0 = m_p v_p + m_b v_b \Rightarrow v_b = -60(3)/200 = -0.9$ m/s. Answer: (C).动量守恒：$0 = m_p v_p + m_b v_b \Rightarrow v_b = -60(3)/200 = -0.9$ m/s。选 (C)。
Q2. A force $F(t) = 6t$ N acts on a 3 kg object from $t = 0$ to $t = 4$ s. If the object starts from rest, its final speed is:力 $F(t) = 6t$ N 从 $t = 0$ 作用到 $t = 4$ s 于一个 3 kg 物体（初始静止）。末速率为：
(A)   8 m/s
(B)   16 m/s
(C)   24 m/s
(D)   48 m/s
$J = \int_0^4 6t\,dt = 3t^2\big|_0^4 = 48$ N·s. $\Delta p = J = 48$ kg·m/s. $v_f = 48/3 = 16$ m/s.$J = \int_0^4 6t\,dt = 3t^2\big|_0^4 = 48$ N·s。$\Delta p = J = 48$ kg·m/s。$v_f = 48/3 = 16$ m/s。
Impulse $= \int_0^4 6t\,dt = 3(16) = 48$ N·s. By impulse–momentum theorem, $v_f = J/m = 48/3 = 16$ m/s. Answer: (B).冲量 $= \int_0^4 6t\,dt = 3(16) = 48$ N·s。由动量定理，$v_f = J/m = 48/3 = 16$ m/s。选 (B)。
Q3. A 2 kg ball moving at 5 m/s collides elastically with a 6 kg ball at rest. What is the velocity of the 2 kg ball after the collision?2 kg 球以 5 m/s 与静止的 6 kg 球发生弹性碰撞。碰后 2 kg 球的速度为：
(A)   +1.25 m/s
(B)   0 m/s
(C)   −1.25 m/s
(D)   −2.5 m/s
$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i} = \frac{2-6}{2+6}(5) = \frac{-4}{8}(5) = -2.5$ m/s. The ball bounces back.$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i} = \frac{2-6}{2+6}(5) = \frac{-4}{8}(5) = -2.5$ m/s。球被反弹。
$v_{1f} = \frac{m_1-m_2}{m_1+m_2}v_{1i} = \frac{-4}{8}(5) = -2.5$ m/s (bounces back). Answer: (D).$v_{1f} = \frac{m_1-m_2}{m_1+m_2}v_{1i} = \frac{-4}{8}(5) = -2.5$ m/s（反弹）。选 (D)。
Q4. Two identical objects undergo a perfectly inelastic collision. Object 1 moves at $v$ and Object 2 is at rest. What fraction of the initial kinetic energy remains?两个完全相同的物体发生完全非弹性碰撞。物体 1 以速率 $v$ 运动，物体 2 静止。初始动能保留了多少？
(A)   1/2
(B)   1/4
(C)   3/4
(D)   0
$v_f = mv/(m+m) = v/2$. $K_i = \frac{1}{2}mv^2$. $K_f = \frac{1}{2}(2m)(v/2)^2 = \frac{1}{4}mv^2$. Fraction: $K_f/K_i = 1/2$.$v_f = mv/(m+m) = v/2$。$K_i = \frac{1}{2}mv^2$，$K_f = \frac{1}{2}(2m)(v/2)^2 = \frac{1}{4}mv^2$。比值 $K_f/K_i = 1/2$。
$v_f = v/2$. $K_f = \frac{1}{2}(2m)(v/2)^2 = \frac{mv^2}{4}$. $K_i = \frac{mv^2}{2}$. Ratio = $1/2$. Answer: (A).$v_f = v/2$，$K_f = \frac{1}{2}(2m)(v/2)^2 = \frac{mv^2}{4}$，$K_i = \frac{mv^2}{2}$，比值为 $1/2$。选 (A)。
Q5. A 0.2 kg ball hits the floor at 8 m/s and bounces up at 6 m/s. The impulse delivered by the floor is (take upward as positive):0.2 kg 的球以 8 m/s 撞向地面，并以 6 m/s 反弹。地面给球的冲量是（取向上为正）：
(A)   0.4 kg·m/s upward0.4 kg·m/s，向上
(B)   1.2 kg·m/s downward1.2 kg·m/s，向下
(C)   2.8 kg·m/s upward2.8 kg·m/s，向上
(D)   2.8 kg·m/s downward2.8 kg·m/s，向下
$p_i = 0.2(-8) = -1.6$ kg·m/s (downward). $p_f = 0.2(+6) = +1.2$ kg·m/s (upward). $J = p_f - p_i = 1.2 - (-1.6) = +2.8$ kg·m/s (upward).$p_i = 0.2(-8) = -1.6$ kg·m/s（向下）；$p_f = 0.2(+6) = +1.2$ kg·m/s（向上）。$J = p_f - p_i = 1.2 - (-1.6) = +2.8$ kg·m/s（向上）。
With up positive: $\Delta p = 0.2(6) - 0.2(-8) = 1.2 + 1.6 = 2.8$ kg·m/s upward. Answer: (C).取向上为正：$\Delta p = 0.2(6) - 0.2(-8) = 1.2 + 1.6 = 2.8$ kg·m/s，向上。选 (C)。
Worked Example — 2-D Perfectly Inelastic Collision (FRQ Style)例题 —— 二维完全非弹性碰撞（FRQ 风格）

Puck A ($m_A = 2.0\;\text{kg}$) moves east at $4.0\;\text{m/s}$ on a frictionless air table. Puck B ($m_B = 3.0\;\text{kg}$) moves north at $3.0\;\text{m/s}$. They collide and stick. (a) Find the velocity (magnitude and direction, measured north of east) of the combined puck after the collision. (b) Compute the kinetic energy lost in the collision and express it as a fraction of the initial KE.

在无摩擦气垫桌上，冰球 A（$m_A = 2.0\;\text{kg}$）向东以 $4.0\;\text{m/s}$ 运动，冰球 B（$m_B = 3.0\;\text{kg}$）向北以 $3.0\;\text{m/s}$ 运动。两球相撞后粘在一起。(a) 求合并体碰后的速度大小与方向（以"北偏东"角表示）；(b) 求碰撞中损失的动能，并表达成初动能的几分之几。

Identify明辨

Principle: $\vec p$ is conserved component-by-component when no external impulsive forces act. Perfectly inelastic ⇒ a single combined mass $m_A + m_B$ at the final velocity. KE is not conserved in inelastic collisions; the lost KE goes to internal modes (deformation, sound, heat).

原理：在没有外冲量力时，$\vec p$ 按分量分别守恒。完全非弹性碰撞 ⇒ 合并体质量 $m_A + m_B$ 以共同末速度运动。非弹性碰撞中动能不守恒，损失的动能转入内部自由度（形变、声、热）。

$m_A = 2.0\;\text{kg}$, $\vec v_A = 4.0\,\hat\imath\;\text{m/s}$ $m_B = 3.0\;\text{kg}$, $\vec v_B = 3.0\,\hat\jmath\;\text{m/s}$ Stick on contact接触即粘合
Set Up建模

Conserve $p_x$ and $p_y$ separately:

$p_x$ 与 $p_y$ 分别守恒：

$$m_A v_{A,x} + m_B v_{B,x} = (m_A + m_B)\,v_{f,x}$$
$$m_A v_{A,y} + m_B v_{B,y} = (m_A + m_B)\,v_{f,y}$$
Execute求解
(a) Final velocity components(a) 末速度分量
$$v_{f,x} = \frac{(2.0)(4.0) + (3.0)(0)}{5.0} = \frac{8.0}{5.0} = 1.60\;\text{m/s}$$
$$v_{f,y} = \frac{(2.0)(0) + (3.0)(3.0)}{5.0} = \frac{9.0}{5.0} = 1.80\;\text{m/s}$$
$$|\vec v_f| = \sqrt{1.60^2 + 1.80^2} = \sqrt{2.56 + 3.24} = \sqrt{5.80} \approx 2.41\;\text{m/s}$$
$$\theta = \arctan\!\left(\frac{v_{f,y}}{v_{f,x}}\right) = \arctan\!\left(\frac{1.80}{1.60}\right) \approx 48.4^{\circ}\;\text{north of east}$$
(b) KE before, after, and lost(b) 碰前、碰后及损失的动能
$$K_i = \tfrac{1}{2}(2.0)(4.0)^2 + \tfrac{1}{2}(3.0)(3.0)^2 = 16.0 + 13.5 = 29.5\;\text{J}$$
$$K_f = \tfrac{1}{2}(5.0)(2.41)^2 = \tfrac{1}{2}(5.0)(5.80) = 14.5\;\text{J}$$
$$\Delta K = K_i - K_f = 29.5 - 14.5 = 15.0\;\text{J}\;\;(\text{lost})$$
$$\frac{\Delta K}{K_i} = \frac{15.0}{29.5} \approx 0.508 = 50.8\%$$
Evaluate校验
$\vec p$ check: initial $\vec p = (8.0\,\hat\imath + 9.0\,\hat\jmath)\;\text{kg}\cdot\text{m/s}$; final $\vec p = 5.0(1.60\,\hat\imath + 1.80\,\hat\jmath) = (8.0\,\hat\imath + 9.0\,\hat\jmath)\;\text{kg}\cdot\text{m/s}$. ✓$\vec p$ 校验：碰前 $\vec p = (8.0\,\hat\imath + 9.0\,\hat\jmath)\;\text{kg}\cdot\text{m/s}$；碰后 $\vec p = 5.0(1.60\,\hat\imath + 1.80\,\hat\jmath) = (8.0\,\hat\imath + 9.0\,\hat\jmath)\;\text{kg}\cdot\text{m/s}$。✓
A perfectly inelastic collision in the lab frame loses the maximum possible KE consistent with momentum conservation; it does not lose all the KE because the system still moves. ✓实验室系下的完全非弹性碰撞在动量守恒前提下损失的动能达到最大；但并非全部损失——系统整体仍在运动。✓
In the centre-of-mass frame the combined object is at rest after the collision, so all of the COM-frame KE is lost — the $14.5\;\text{J}$ that survives in the lab frame is precisely $\tfrac{1}{2}(m_A + m_B)|\vec v_\text{cm}|^2$, the bulk-translation kinetic energy. ✓在质心系下，合并体碰后静止，故质心系的动能全部损失——实验室系中保留的 $14.5\;\text{J}$ 恰好等于 $\tfrac{1}{2}(m_A + m_B)|\vec v_\text{cm}|^2$，即整体平动动能。✓
Worked Example — Ballistic Pendulum + Spring Recoil (FRQ Style)例题 —— 弹道摆 + 弹簧反冲（FRQ 风格）

A bullet of mass $m = 0.020\;\text{kg}$ is fired horizontally with speed $v_0$ into a wooden block of mass $M = 1.98\;\text{kg}$ that hangs at rest from a long massless cord. The bullet embeds, and the bullet+block swings to a maximum height $h = 0.20\;\text{m}$ above its starting level. (a) Find the speed of the bullet+block just after impact. (b) Find the bullet's initial speed $v_0$. (c) The bullet+block then falls back, separates from the cord at the lowest point, and slides along a horizontal surface into a spring of constant $k = 500\;\text{N/m}$. Find the maximum spring compression, ignoring friction. Use $g = 9.8\;\text{m/s}^2$.

质量 $m = 0.020\;\text{kg}$ 的子弹以水平速度 $v_0$ 射入由长而轻的绳悬挂、初始静止的木块（$M = 1.98\;\text{kg}$）。子弹嵌入后，"子弹 + 木块"摆动到比初始位置高 $h = 0.20\;\text{m}$ 处。(a) 求碰后瞬间"子弹 + 木块"的速度；(b) 求子弹的初速 $v_0$；(c) 此后"子弹 + 木块"摆回最低点，与绳脱离，在水平面上滑入弹簧常数 $k = 500\;\text{N/m}$ 的弹簧。忽略摩擦，求弹簧的最大压缩量。取 $g = 9.8\;\text{m/s}^2$。

Identify明辨

Principle: three physically distinct steps strung together — (i) embedding is a perfectly inelastic collision (momentum conserved, KE not), (ii) the swing up is energy-conserving (KE → gravitational PE on a pendulum cord that does no work), (iii) the spring compression is energy-conserving (KE → elastic PE). The key is to apply the right conservation law in each step and never carry KE conservation through the collision.

原理：把三个物理上不同的阶段串起来——(i) 嵌入是完全非弹性碰撞（动量守恒、动能不守恒）；(ii) 摆上升能量守恒（动能 → 重力势能，绳张力不做功）；(iii) 压缩弹簧能量守恒（动能 → 弹性势能）。关键是每一步只用对应的守恒律，绝不把动能守恒带过碰撞那一步。

Bullet $m$, block $M$, total $m + M$子弹 $m$、木块 $M$、合计 $m + M$ Swing height $h$摆动高度 $h$ Spring $k$, gravity $g$弹簧 $k$、重力 $g$
Set Up建模

Three equations, one per step:

每一阶段一个方程，共三个：

$$\text{(i) Collision: }\quad m v_0 = (m + M)\,v_1$$
$$\text{(ii) Swing: }\quad \tfrac{1}{2}(m + M)\,v_1^2 = (m + M)\,g h$$
$$\text{(iii) Spring: }\quad \tfrac{1}{2}(m + M)\,v_1^2 = \tfrac{1}{2}k x_\text{max}^2$$
Execute求解
(a) Speed just after embedding(a) 嵌入瞬间合并体的速度
From step (ii) — the swing tells us $v_1$ directly:由步骤 (ii) ——摆上升直接给出 $v_1$：
$$v_1 = \sqrt{2gh} = \sqrt{2(9.8)(0.20)} = \sqrt{3.92} \approx 1.98\;\text{m/s}$$
(b) Bullet's initial speed(b) 子弹初速
From step (i):由步骤 (i)：
$$v_0 = \frac{m + M}{m}\,v_1 = \frac{2.00}{0.020}(1.98) = 100 \times 1.98 = 198\;\text{m/s}$$
(c) Spring compression after the swing back(c) 回摆后压缩弹簧的最大量
When the bullet+block returns to the bottom of its swing it has the same speed $v_1$ (by energy conservation on the pendulum). Then step (iii):合并体回到摆的最低点时速率仍为 $v_1$（摆上升与回落能量守恒）。再用步骤 (iii)：
$$x_\text{max} = v_1 \sqrt{\frac{m + M}{k}} = 1.98 \sqrt{\frac{2.00}{500}} = 1.98 \times 0.0632 \approx 0.125\;\text{m}$$
Evaluate校验
Energy lost in the embedding: $K_i = \tfrac{1}{2}(0.020)(198)^2 = 392\;\text{J}$; $K_1 = \tfrac{1}{2}(2.00)(1.98)^2 \approx 3.92\;\text{J}$. Almost 99% of the bullet's KE is dissipated in the wood — characteristic of a low-mass projectile striking a much heavier target. ✓嵌入阶段损失的动能：$K_i = \tfrac{1}{2}(0.020)(198)^2 = 392\;\text{J}$；$K_1 = \tfrac{1}{2}(2.00)(1.98)^2 \approx 3.92\;\text{J}$。约 99% 的子弹动能在木块中耗散——轻入射重靶的典型结果。✓
The single crucial trap on this kind of FRQ: never write $\tfrac{1}{2}m v_0^2 = (m + M)gh$. KE is not conserved across the embedding step. Momentum is. ✓这类 FRQ 最关键的陷阱：千万别写 $\tfrac{1}{2}m v_0^2 = (m + M)gh$。嵌入过程中动能不守恒，守恒的是动量。✓
Spring step: $\tfrac{1}{2}k x_\text{max}^2 = \tfrac{1}{2}(500)(0.125)^2 \approx 3.91\;\text{J}$ matches $K_1$ to round-off — energy is conserved across the swing+spring sequence because no friction or non-conservative work was assumed there. ✓弹簧段：$\tfrac{1}{2}k x_\text{max}^2 = \tfrac{1}{2}(500)(0.125)^2 \approx 3.91\;\text{J}$，与 $K_1$ 在取整精度内吻合——摆+弹簧整个过程没有摩擦或其它非保守功，能量守恒。✓
Unit 4 Formula Quick Reference第 4 单元公式速查
$\vec{p} = m\vec{v}$
$\vec{F}_\text{net} = d\vec{p}/dt$
$\vec{J} = \int \vec{F}\,dt = \Delta\vec{p}$
$\vec{F}_\text{avg} = \Delta\vec{p}/\Delta t$
$\sum \vec{p}_i = \sum \vec{p}_f$
$\vec{v}_\text{cm} = \sum m_i\vec{v}_i / \sum m_i$
$v_f = \frac{m_1 v_1 + m_2 v_2}{m_1+m_2}$ (perf. inelastic)（完全非弹性）
$v_{1f} = \frac{m_1-m_2}{m_1+m_2}v_{1i}$ (elastic, $v_{2i}=0$)（弹性，$v_{2i}=0$）
$v_{2f} = \frac{2m_1}{m_1+m_2}v_{1i}$ (elastic, $v_{2i}=0$)（弹性，$v_{2i}=0$）
$F_\text{net} = \frac{dm}{dt}v$ (variable mass)（变质量）