Conservation of Linear Momentum · 线性动量守恒
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| isolated system/ˈaɪsəleɪtɪd ˈsɪstəm/ | 孤立系统 | gū lì xì tǒng |
| conservation of linear momentum/ˌkɒnsəˈveɪʃn ɒv ˈlɪnɪə məʊˈmentəm/ | 动量守恒 | dòng liàng shǒu héng |
| internal forces/ɪnˈtɜːnl ˈfɔːsɪz/ | 内力 | nèi lì |
Two skaters push apart
- Two skaters stand still on ice, then shove off each other.
- They glide away in opposite directions -- with no outside help at all.
- Somehow their motions stay perfectly balanced.
- Behind it is one of the most powerful rules in physics.
两名滑冰者互相推开
- 两名滑冰者静止地站在冰上,然后互相一蹬。
- 他们朝相反方向滑开——完全没有外界帮助。
- 不知怎地,他们的运动始终保持完美平衡。
- 这背后是物理学中最强大的规律之一。
Conservation of momentum
- If the net external force on a system is zero -- an isolated system 孤立系统 -- its total momentum stays constant.
- Whatever total momentum it starts with, it keeps.
- This is the conservation of linear momentum 动量守恒.
动量守恒
- 如果系统所受的净外力为零——即孤立系统——它的总动量保持不变。
- 无论它以多大的总动量开始,它都保持不变。
- 这就是动量守恒。
Momentum is conserved only when... · 只有当...时，动量才守恒
An isolated system -- zero net external force -- conserves momentum. Internal forces are fine. · 孤立系统——零净外力——动量守恒。内力是可以的。
Internal forces cannot change it
- Internal forces 内力 come in third-law pairs that cancel across the system.
- So a push between the parts (like the skaters' shove) can never change the total.
- Only an external force can.
内力无法改变它
- 内力成对出现(第三定律),在系统内部抵消。
- 所以各部分之间的推力(如滑冰者的一蹬)永远无法改变总量。
- 只有外力才能。
An internal explosion inside a floating system changes the system's total momentum. · 漂浮系统内部的内部爆炸改变了系统的总动量。
Explosion forces are internal (third-law pairs), so the total momentum is unchanged. · 爆炸力是内力（第三定律对），因此总动量保持不变。
Forces between the parts of a system are called ____ forces and cannot change the total momentum. · 系统各部分之间的力被称为____力，不能改变总动量。
Internal forces come in canceling third-law pairs, so they leave the total momentum unchanged. · 内力成对出现并相互抵消（第三定律对），因此它们不改变总动量。
Collisions and explosions
- The law applies to collisions (objects hitting) and explosions (one thing flying apart) alike.
- In an explosion, the pieces fly off so their momenta still sum to the original total.
- Pick your system so the forces you care about are internal.
碰撞与爆炸
- 这条定律对碰撞(物体相撞)和爆炸(一个东西飞散)都适用。
- 在爆炸中,碎片飞出,它们的动量之和仍等于原来的总量。
- 选择你的系统,使你关心的力都是内力。
Conservation of momentum · 动量守恒
In a collision with no outside forces, the total momentum before equals the total momentum after. · 在没有外力的碰撞中，碰撞前的总动量等于碰撞后的总动量。
A $50\ \text{kg}$ skater at rest throws a $5\ \text{kg}$ ball at $8\ \tfrac{\text{m}}{\text{s}}$. The skater's recoil speed (in m/s)? · 一个静止的 $50\ \text{kg}$ 滑冰者将一个 $5\ \text{kg}$ 的球以 $8\ \tfrac{\text{m}}{\text{s}}$ 抛出。滑冰者的反冲速度是多少（单位：m/s）？
$0 = 5(8) + 50 v$, so $v = -40/50 = -0.8\ \tfrac{\text{m}}{\text{s}}$ -- the skater recoils at $0.8\ \tfrac{\text{m}}{\text{s}}$. · $0 = 5(8) + 50 v$，因此 $v = -40/50 = -0.8\ \tfrac{\text{m}}{\text{s}}$ ——滑冰者以 $0.8\ \tfrac{\text{m}}{\text{s}}$ 反冲。
A $4\ \text{kg}$ object at rest explodes into a $1\ \text{kg}$ piece moving right at $6\ \tfrac{\text{m}}{\text{s}}$ and a $3\ \text{kg}$ piece. The $3\ \text{kg}$ piece's speed (in m/s)? · 一个静止的 $4\ \text{kg}$ 物体爆炸成一个 $1\ \text{kg}$ 的碎片向右以 $6\ \tfrac{\text{m}}{\text{s}}$ 运动，以及一个 $3\ \text{kg}$ 的碎片。$3\ \text{kg}$ 碎片的速度是多少（单位：m/s）？
$0 = 1(6) + 3v$, so $v = -2\ \tfrac{\text{m}}{\text{s}}$ -- the $3\ \text{kg}$ piece moves left at $2\ \tfrac{\text{m}}{\text{s}}$. · $0 = 1(6) + 3v$，因此 $v = -2\ \tfrac{\text{m}}{\text{s}}$ ——$3\ \text{kg}$ 碎片向左以 $2\ \tfrac{\text{m}}{\text{s}}$ 运动。
Two dimensions
- Momentum is a vector, so it is conserved separately along each axis.
- Conserve the $x$-momenta and the $y$-momenta independently.
- Two equations let you solve two-dimensional collisions.
二维情形
- 动量是矢量,所以它沿每个轴分别守恒。
- 分别独立地守恒 $x$ 方向动量和 $y$ 方向动量。
- 两个方程就能解二维碰撞。
In a two-dimensional collision, momentum is conserved separately along each axis. · 在二维碰撞中，动量沿每个轴分别守恒。
Momentum is a vector, so its x- and y-components are each conserved. · 动量是一个矢量，因此其 x 分量和 y 分量各自守恒。
A $60\ \text{kg}$ astronaut at rest throws a $2\ \text{kg}$ wrench at $10\ \tfrac{\text{m}}{\text{s}}$.
- Total momentum starts at zero, so it must stay zero: $0 = (2)(10) + (60)v$.
- The astronaut recoils at $v = -20/60 \approx -0.33\ \tfrac{\text{m}}{\text{s}}$ backward.
一名 $60\ \text{kg}$ 的宇航员静止,抛出一个 $2\ \text{kg}$ 的扳手,速度 $10\ \tfrac{\text{m}}{\text{s}}$。
- 总动量从零开始,所以必须保持为零:$0 = (2)(10) + (60)v$。
- 宇航员以 $v = -20/60 \approx -0.33\ \tfrac{\text{m}}{\text{s}}$ 向后反冲。
Momentum is conserved only when the net external force is zero. If a strong outside force acts (a wall, the ground, a rocket engine's exhaust leaving), you must either include it or choose a bigger system that makes it internal.
只有在净外力为零时动量才守恒。如果有强的外力作用(墙、地面、火箭喷出的气体离开),你必须要么把它包括进来,要么选一个更大的系统使它变成内力。
In an isolated system (zero net external force), total momentum is conserved -- conservation of linear momentum. Internal forces cancel in pairs and cannot change the total; only external forces can. It holds for collisions and explosions, and in 2D you conserve each axis separately.
在孤立系统(净外力为零)中,总动量守恒——即动量守恒。内力成对抵消,无法改变总量;只有外力才能。它对碰撞和爆炸都成立,二维时沿每个轴分别守恒。