Conservation of momentum · 动量守恒
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| external force/ekˈstɜːnl fɔːs/ | 外力 | wài lì |
| conservation of momentum/ˌkɒnsəˈveɪʃn ɒv məʊˈmentəm/ | 动量守恒 | dòng liàng shǒu héng |
| collision/kəˈlɪʒn/ | 碰撞 | pèng zhuàng |
| explosion/ekˈspləʊʒn/ | 爆炸 | bào zhà |
| recoil/rɪˈkɔɪl/ | 反冲 | fǎn chōng |
| elastic/ɪˈlæstɪk/ | 弹性 | tán xìng |
| inelastic/ɪnɪˈlæstɪk/ | 非弹性 | fēi tán xìng |
| thrust/θrʌst/ | 推力 | tuī lì |
Push off and drift apart
- Two ice skaters stand still, then push on each other and glide apart.
- One goes left, one goes right — the total momentum stays zero.
- Nothing pushed from outside, so momentum was conserved.
推开后各自滑走
- 两个滑冰者静止站着,然后互相推一把,各自滑开。
- 一个向左,一个向右——总 动量保持为零。
- 没有外部的推力,所以动量 守恒。
The principle
- If there is no resultant external force 外力, the total momentum of a system stays constant.
- This is conservation of momentum 动量守恒.
Free-body diagram of a book being pulled on a table
原理
- 如果 没有合外力,系统的总动量保持不变。
- 这就是 动量守恒(conservation of momentum)。
书在桌上被拉动的自由体受力图
Conservation of momentum · 动量守恒
m₁u₁ + m₂u₂ = (m₁+m₂)v
In a collision the total momentum is conserved. · 在碰撞中,总 动量 守恒。
Total momentum stays constant when the resultant ____ force on the system is zero. · 当系统所受的合 ____ 力为零时,总动量保持不变。
With no resultant external force, the system's total momentum is conserved (internal forces come in third-law pairs and cancel). · 没有合外力时,系统的总动量守恒(内力成对出现,即第三定律力对,互相抵消)。
When it holds
- It works for collisions 碰撞, explosions 爆炸 and recoil 反冲 — any time outside forces cancel.
- In two dimensions, momentum is conserved along each direction on its own.
Forces on a falling object in a fluid
什么时候成立
- 它适用于 碰撞、爆炸和反冲——只要外力互相抵消。
- 在二维中,动量沿 每个方向 各自守恒。
流体中下落物体所受的力
Elastic 弹性 vs inelastic 非弹性
- In every collision, momentum is conserved.
- Elastic: kinetic energy is also conserved (relative speed of approach = of separation).
- Inelastic: some KE becomes heat/sound/deformation. If they stick, it is perfectly inelastic.
Newton's third law in an isolated two-particle system: equal and opposite forces
弹性碰撞与非弹性碰撞
- 在 每一次 碰撞中,动量都守恒。
- 弹性碰撞: 动能也守恒(接近的相对速度 = 分离的相对速度)。
- 非弹性碰撞: 一部分动能变成热、声或形变。如果它们粘在一起,就是完全非弹性。
孤立的两粒子系统中的牛顿第三定律:大小相等、方向相反的力
In an inelastic collision, momentum is still conserved. · 在非弹性碰撞中,动量仍然守恒。
Momentum is conserved in every collision (no external force). In an inelastic one it is the kinetic energy · 动能 that is not conserved. · 动量在 每一次 碰撞中都守恒(没有外力)。在非弹性碰撞中,不守恒的是动能。
Compared with collisions in general, what is special about an elastic · 弹性 collision? · 与一般的碰撞相比,弹性 碰撞有什么特别之处?
All collisions conserve momentum; an elastic collision also conserves kinetic energy. · 所有碰撞都守恒动量;弹性碰撞还守恒动能。
Solving a head-on collision
- Use signed velocities: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$.
- If it is elastic, add $u_1 - u_2 = -(v_1 - v_2)$ to get a second equation.
求解正面碰撞
- 使用带符号的速度:$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$。
- 如果是弹性碰撞,再加上 $u_1 - u_2 = -(v_1 - v_2)$ 得到第二个方程。
A $2.0\ \text{kg}$ trolley at $3.0\ \dfrac{\text{m}}{\text{s}}$ hits a stationary $1.0\ \text{kg}$ trolley and they stick. Find their common velocity. · 一辆 $2.0\ \text{kg}$ 的小车以 $3.0\ \dfrac{\text{m}}{\text{s}}$ 撞上一辆静止的 $1.0\ \text{kg}$ 小车并粘在一起。求它们共同的速度。
Momentum: $2.0 \times 3.0 = (2.0 + 1.0)\,v$, so $v = \dfrac{6.0}{3.0} = 2.0\ \dfrac{\text{m}}{\text{s}}$. · 动量:$2.0 \times 3.0 = (2.0 + 1.0)\,v$,所以 $v = \dfrac{6.0}{3.0} = 2.0\ \dfrac{\text{m}}{\text{s}}$。
Collisions in two dimensions
- Split each velocity into perpendicular components.
- Apply conservation of momentum along each axis separately.
In two dimensions, resolve along two axes and conserve momentum on each one separately
二维碰撞
- 把每个速度分解成 互相垂直的分量。
- 沿每个坐标轴分别应用动量守恒。
在二维中,沿两条轴分解,并在每条轴上分别守恒动量
In two dimensions, momentum is conserved separately along each perpendicular direction. · 在二维中,动量沿每个互相垂直的方向各自守恒。
Yes — resolve into components and apply conservation of momentum to each axis on its own. · 是的——分解成分量,并对每个坐标轴分别应用动量守恒。
Rockets and recoil
- A rocket throws gas one way and is pushed the other way (Newton's third law).
- Thrust 推力 $F = \dot{m}\,u$ — mass thrown per second times its speed.
火箭与反冲
- 火箭把气体往一个方向喷,自己被推向另一个方向(牛顿第三定律)。
- 推力 $F = \dot{m}\,u$——每秒喷出的质量乘以它的速度。
A rocket ejects gas at $100\ \dfrac{\text{m}}{\text{s}}$ with a mass-flow rate of $2.0\ \dfrac{\text{kg}}{\text{s}}$. What thrust does it feel? · 一枚火箭以 $100\ \dfrac{\text{m}}{\text{s}}$ 喷出气体,质量流率为 $2.0\ \dfrac{\text{kg}}{\text{s}}$。它受到的推力是多少?
Thrust $F = \dot{m}\,u = 2.0 \times 100 = 200\ \text{N}$. · 推力 $F = \dot{m}\,u = 2.0 \times 100 = 200\ \text{N}$。
Worked example: a head-on collision
- A $1200\ \text{kg}$ car moving at $15\ \text{m/s}$ east collides head-on with an $800\ \text{kg}$ car moving at $10\ \text{m/s}$ west. They lock together. Find their common velocity.
- Choose east as positive. The second car's velocity is then $-10\ \text{m/s}$.
- Momentum before: $(1200)(15) + (800)(-10) = 18000 - 8000 = 10000\ \text{kg m/s}$.
- After: $(1200 + 800)v = 10000$, so $v = 5.0\ \text{m/s}$, still east.
- The commonest error is adding both speeds because the total is wanted. Choose a positive direction, give the opposing velocity a minus sign, and then add.
例题:迎面相撞
- 一辆 $1200\ \text{kg}$ 的车以 $15\ \text{m/s}$ 向东行驶,与一辆 $800\ \text{kg}$、以 $10\ \text{m/s}$ 向西行驶的车迎面相撞。两车锁在一起。求它们共同的速度。
- **取向东为正。**于是第二辆车的速度是 $-10\ \text{m/s}$。
- 碰前动量:$(1200)(15) + (800)(-10) = 18000 - 8000 = 10000\ \text{kg m/s}$。
- 碰后:$(1200 + 800)v = 10000$,所以 $v = 5.0\ \text{m/s}$,仍然向东。
- 最常见的错误是因为要求"总量"就把两个速率相加。要选定一个正方向,给反向的速度加负号,然后再相加。
Elastic against inelastic
- An elastic collision conserves total kinetic energy, and equivalently the relative speed of approach equals the relative speed of separation.
- An inelastic collision conserves momentum but transfers some kinetic energy to other forms, usually internal energy and sound.
- Momentum is conserved in both. That is what makes it the useful quantity: it survives a collision that kinetic energy does not.
- Check whether a collision is elastic: compute the total kinetic energy before and after using the full speed of each body, not a component. If they match, it is elastic.
- A "perfectly inelastic" collision is the one where the bodies stick together, which loses the most kinetic energy possible while still conserving momentum.
弹性碰撞与非弹性碰撞
- 弹性碰撞守恒总动能,等价地说,接近的相对速率等于分离的相对速率。
- 非弹性碰撞守恒动量,但把一部分动能转移为其他形式,通常是内能和声。
- 两者都守恒动量。这正是动量之所以有用的地方:它能挺过一场动能挺不过的碰撞。
- *判断碰撞是否弹性:*用每个物体的完整速率(不是某个分量)算出碰前碰后的总动能。若相等,就是弹性的。
- "完全非弹性"碰撞就是物体粘在一起的那种,它在仍然守恒动量的前提下损失了最多的动能。
A 1200 kg car at 15 m/s east hits an 800 kg car at 10 m/s west head-on and they lock together. What is their common velocity, in m/s east? · 一辆 1200 kg 的车以 15 m/s 向东与一辆 800 kg、以 10 m/s 向西的车迎面相撞并锁在一起。它们共同的速度是多少 m/s(向东)?
East positive: 18000 - 8000 = 10000 kg m/s, shared by 2000 kg, so 5.0 m/s east. Adding both speeds without the minus sign gives 13 m/s, which is the classic wrong answer. · 向东为正:18000 - 8000 = 10000 kg m/s,由 2000 kg 共同分担,所以是向东 5.0 m/s。不加负号直接相加会得到 13 m/s,那是经典的错误答案。
Marks that slip away
- State the condition: momentum is conserved provided no resultant external force acts on the system. Without it the statement is incomplete.
- Choose a positive direction and give opposing velocities a minus sign. Never add two speeds in a head-on collision.
- Kinetic energy uses the full speed of each body, never one component.
- Momentum is a vector, so in two dimensions it is conserved separately in each of two perpendicular directions.
- An elastic collision conserves kinetic energy; an inelastic one does not. Both conserve momentum.
容易丢掉的分
- 要说出条件:动量守恒的前提是系统不受合外力。没有它,这句话就不完整。
- 选定一个正方向,给反向的速度加负号。迎面相撞时绝不能把两个速率直接相加。
- 动能用每个物体的完整速率,绝不用某一个分量。
- 动量是矢量,所以在二维中它在两个互相垂直的方向上分别守恒。
- 弹性碰撞守恒动能;非弹性碰撞不守恒。但两者都守恒动量。
Match each collision type to what it conserves. · 把每种碰撞与它守恒的量配对。
Momentum survives every collision, which is exactly why it is the useful quantity. Kinetic energy does not. · 动量在每一种碰撞中都存活下来,这正是它有用的原因。动能则不然。
A complete statement of the principle of conservation of momentum includes which of these? Select all · 所有 that apply. · 动量守恒原理的完整表述包含下列哪些?选出所有适用的。
Elasticity is irrelevant: momentum is conserved in every collision. Leaving out the external-force condition is what loses the mark. · 是否弹性无关紧要:每一种碰撞都守恒动量。丢分的是漏掉了外力那个条件。
Put the test for whether a collision is elastic in order. · 把判断碰撞是否弹性的检验按顺序排列。
Use the full speed, not a component: kinetic energy is a scalar built from the whole velocity. · 要用完整速率而不是某个分量:动能是由整个速度构成的标量。
In a two-dimensional collision, momentum is conserved separately in each of two perpendicular directions. · 在二维碰撞中,动量在两个互相垂直的方向上分别守恒。
Momentum is a vector, so its conservation is a vector equation, and a vector equation is two scalar equations in a plane. That is what makes 2-D collision questions solvable. · 动量是矢量,所以它的守恒是一个矢量方程,而平面内的矢量方程就是两个标量方程。这正是二维碰撞题可解的原因。
You've got it
- no external resultant force → total momentum is conserved
- momentum is conserved in all collisions; elastic ones also conserve KE
- solve 1-D collisions with $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ (use signs)
你掌握了
- 没有合外力 → 总动量守恒
- 动量在 所有 碰撞中守恒;弹性 碰撞还守恒动能
- 用 $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ 求解一维碰撞(注意符号)