Conservation of Linear Momentum · 线性动量守恒
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| conservation/ˌkɒnsəˈveɪʃn/ | 守恒 | shǒu héng |
| isolated system/ˈaɪsəleɪtɪd ˈsɪstəm/ | 孤立系统 | gū lì xì tǒng |
A cannon kicks back — and the books balance
- Fire a cannon and it lurches backward as the ball flies forward.
- The forward momentum of the ball is matched by the backward momentum of the cannon.
- Before firing, nothing moved: total momentum was zero. After firing, it is still zero.
- Momentum was not created — it was shared out, and the total was conserved 守恒.
大炮向后一顿——账却是平的
- 开炮时,炮身向后一顿,炮弹向前飞出。
- 炮弹向前的动量,恰好被炮身向后的动量抵消。
- 开炮前什么都不动:总动量为零。开炮后,它仍然是零。
- 动量没有被创造——它被分摊出去,而总量被守恒。
The conservation law
- In an isolated system 孤立系统 (no external net force), total momentum stays constant.
- Total momentum before an interaction equals total momentum after.
- $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ (using $u$ for before, $v$ for after).
- It follows directly from Newton's third law: the internal forces cancel in pairs.
守恒定律
- 在孤立系统(没有外部合力)中,总动量保持恒定。
- 相互作用之前的总动量等于之后的总动量。
- $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$(用 $u$ 表示之前,$v$ 表示之后)。
- 它直接来自牛顿第三定律:内力成对相消。
Momentum is conserved · 动量守恒
Set the masses and speeds, run the collision, and check the total momentum before equals after. · 设置质量和速度，运行碰撞，检查碰撞前后的总动量是否相等。
A $1000\ \text{kg}$ cannon fires a $5\ \text{kg}$ ball at $200\ \tfrac{\text{m}}{\text{s}}$. What is the cannon's recoil speed, in $\tfrac{\text{m}}{\text{s}}$? (Total momentum starts at zero.) · 一门质量为 $1000\ \text{kg}$ 的大炮发射一颗质量为 $5\ \text{kg}$ 的炮弹，速度为 $200\ \tfrac{\text{m}}{\text{s}}$。大炮的反冲速度是多少，单位为 $\tfrac{\text{m}}{\text{s}}$？（总动量从零开始。）
$0 = 1000 v + 5(200)$, so $v = -1000/1000 = -1\ \tfrac{\text{m}}{\text{s}}$ — speed $1\ \tfrac{\text{m}}{\text{s}}$ backward. · $0 = 1000 v + 5(200)$，所以$v = -1000/1000 = -1\ \tfrac{\text{m}}{\text{s}}$ — 向后速度$1\ \tfrac{\text{m}}{\text{s}}$。
In an isolated collision, the total momentum after equals the total momentum before. · 在孤立碰撞中，碰撞后的总动量等于碰撞前的总动量。
That is exactly conservation of momentum: $\sum p_{\text{before}} = \sum p_{\text{after}}$. · 这正是动量守恒：$\sum p_{\text{before}} = \sum p_{\text{after}}$。
Why the internal forces cancel
- During a collision, the two objects push on each other equally and oppositely.
- These internal impulses are equal and opposite, so they cancel in the total.
- With no external force, nothing changes the system's total momentum.
- Each object's momentum can change wildly — but their sum does not.
内力为何相消
- 碰撞时,两个物体等大反向地互相推。
- 这些内部冲量等大反向,所以在总和中相消。
- 没有外力,就没有东西改变系统的总动量。
- 每个物体的动量可以剧烈变化——但它们的和不变。
Conservation of momentum follows directly from Newton's ____ law. · 动量守恒直接遵循牛顿的____定律。
The equal-and-opposite internal forces (third law) cancel, leaving the total momentum unchanged. · 等大的内部相互作用力（第三定律）相互抵消，使总动量保持不变。
Solving with conservation
- Write down the total momentum before, set it equal to the total after, and solve.
- Remember to include signs — momentum is a vector.
- Recoil (guns, rockets), explosions, and every collision obey this law.
- It works even when you have no idea about the messy forces during contact.
用守恒解题
- 写下之前的总动量,令它等于之后的总动量,然后求解。
- 记得带上符号——动量是矢量。
- 反冲(枪、火箭)、爆炸,以及每一次碰撞都遵守这条定律。
- 即使你对接触时那些杂乱的力一无所知,它也照样成立。
A $2\ \text{kg}$ trolley at $3\ \tfrac{\text{m}}{\text{s}}$ hits a stationary $1\ \text{kg}$ trolley and they stick. What is their common speed after, in $\tfrac{\text{m}}{\text{s}}$? · 一辆 $2\ \text{kg}$ 的推车在 $3\ \tfrac{\text{m}}{\text{s}}$ 下撞击一辆静止的 $1\ \text{kg}$ 推车，两者粘在一起。碰撞后它们的共同速度是多少，单位为 $\tfrac{\text{m}}{\text{s}}$？
Before: $p = 2(3) = 6$. After: $3v = 6$, so $v = 2\ \tfrac{\text{m}}{\text{s}}$. · 之前：$p = 2(3) = 6$。之后：$3v = 6$，所以$v = 2\ \tfrac{\text{m}}{\text{s}}$。
Select all · 所有 situations where total momentum is conserved. · 选择所有总动量守恒的情况。
The first three have no external net force. Strong external friction adds outside impulse, so momentum is not conserved there. · 前三种情况无外部净力。强外部摩擦力增加了外部冲量，因此那里的动量不守恒。
Momentum is only conserved for an isolated system — one with no external net force. Friction, gravity or a wall can add outside impulse and change the total. Choose your system so the big forces are internal, then conservation applies.
动量只对孤立系统守恒——即没有外部合力的系统。摩擦、重力或墙都能加入外部冲量,改变总量。选择你的系统,让大的力都是内部的,守恒就适用。
Momentum is conserved only when the system is: · 仅当系统是：时，动量才守恒
With no external net force, the internal forces cancel and total momentum stays constant. · 在没有外部净力的情况下，内部力相互抵消，总动量保持不变。
A $2\ \text{kg}$ trolley at $3\ \tfrac{\text{m}}{\text{s}}$ hits a stationary $1\ \text{kg}$ trolley and they stick together.
- Before: $p = 2(3) + 1(0) = 6\ \text{kg}\cdot\tfrac{\text{m}}{\text{s}}$.
- After: combined mass $3\ \text{kg}$ moves at $v$, so $3v = 6 \Rightarrow v = 2\ \tfrac{\text{m}}{\text{s}}$.
一辆 $2\ \text{kg}$ 的小车以 $3\ \tfrac{\text{m}}{\text{s}}$ 撞上一辆静止的 $1\ \text{kg}$ 小车,两者粘在一起。
- 之前:$p = 2(3) + 1(0) = 6\ \text{kg}\cdot\tfrac{\text{m}}{\text{s}}$。
- 之后:合并质量 $3\ \text{kg}$ 以 $v$ 运动,所以 $3v = 6 \Rightarrow v = 2\ \tfrac{\text{m}}{\text{s}}$。
Conservation of momentum: in an isolated system, total momentum is unchanged, so $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$. It comes from Newton's third law (internal forces cancel). Keep track of signs, and it solves recoil, explosions and collisions.
动量守恒:在孤立系统中,总动量不变,所以 $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$。它来自牛顿第三定律(内力相消)。带上符号,它就能解反冲、爆炸和碰撞。