Change in Momentum and Impulse · 动量变化与冲量
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| impulse/ˈɪmpʌls/ | 冲量 | chōng liàng |
Why an airbag saves you — the physics of a soft landing
- In a crash your momentum must drop to zero. That change is fixed; you cannot avoid it.
- But you can choose how long it takes to happen — and that changes the force.
- Stop in a split second and the force is huge; stop over half a second and it is gentle.
- Airbags, crumple zones and bent knees all work by stretching the stopping time.
安全气囊为何救你——柔和着陆的物理
- 碰撞中你的动量必须降到零。这个变化是定死的;你无法避免。
- 但你可以选择让它发生得多久——而这会改变力的大小。
- 一瞬间停下,力就巨大;用半秒停下,力就柔和。
- 安全气囊、溃缩区和弯曲的膝盖,都靠拉长停下的时间起作用。
Impulse changes momentum
- Impulse 冲量 is force multiplied by the time it acts: $\vec J = \vec F\,\Delta t$.
- The impulse–momentum theorem: impulse equals the change in momentum, $\vec J = \Delta \vec p$.
- So $\vec F\,\Delta t = m\vec v_f - m\vec v_i$.
- Impulse is measured in $\text{N}\cdot\text{s}$, which equals $\text{kg}\cdot\tfrac{\text{m}}{\text{s}}$.
冲量改变动量
- 冲量是力乘以它作用的时间:$\vec J = \vec F\,\Delta t$。
- 冲量–动量定理:冲量等于动量的变化,$\vec J = \Delta \vec p$。
- 所以 $\vec F\,\Delta t = m\vec v_f - m\vec v_i$。
- 冲量的单位是 $\text{N}\cdot\text{s}$,它等于 $\text{kg}\cdot\tfrac{\text{m}}{\text{s}}$。
The impulse–momentum theorem says impulse equals the change in ____. · 冲量-动量定理指出冲量等于____的变化。
$\vec J = \Delta \vec p$ — impulse equals the change in momentum. · $\vec J = \Delta \vec p$ — 冲量等于动量的变化。
A force of $20\ \text{N}$ acts for $0.5\ \text{s}$. What is the impulse, in $\text{N·s}$? · 一个$20\ \text{N}$的力作用了$0.5\ \text{s}$。冲量是多少，单位为$\text{N·s}$？
$J = F\Delta t = 20 \times 0.5 = 10\ \text{N·s}$.
Impulse is the area under a force–time graph
- When the force varies, the impulse is the area under the force–time curve.
- A short, sharp spike and a long, gentle push can deliver the same impulse.
- Same area ⇒ same change in momentum, whatever the shape.
- This is why we care about the whole interaction, not just the peak force.
冲量是力–时间图下的面积
- 当力变化时,冲量是力–时间曲线下的面积。
- 一个短促尖锐的峰和一个长而柔和的推,可以传递相同的冲量。
- 面积相同 ⇒ 动量变化相同,无论形状如何。
- 这就是我们关心整个相互作用、而不仅是峰值力的原因。
The impulse delivered by a varying force equals the area under its force–time graph. · 变力提供的冲量等于其力-时间图线下的面积。
Impulse is the accumulated $F\,\Delta t$ — exactly the area under the force–time curve. · 冲量是累积的$F\,\Delta t$ — 正好是力-时间曲线下的面积。
Trade force for time
- For a fixed change in momentum, $F$ and $\Delta t$ are inversely related: $F = \dfrac{\Delta p}{\Delta t}$.
- Longer contact time ⇒ smaller force. Shorter time ⇒ bigger force.
- A caught cricket ball hurts less if you move your hands back as you catch.
- Follow-through in sport does the opposite — a longer push builds more momentum.
用力换时间
- 对固定的动量变化,$F$ 与 $\Delta t$ 成反比:$F = \dfrac{\Delta p}{\Delta t}$。
- 接触时间越长 ⇒ 力越小。时间越短 ⇒ 力越大。
- 接住板球时如果你把手往后移,就没那么疼。
- 运动中的"随挥"作用相反——更长的推积累更多动量。
Impulse and momentum change · 冲量与动量变化
Impulse is force times time and equals the change in momentum. Sort each case. · 冲量是力乘以时间，等于动量的变化。对每种情况进行排序。
How does an airbag reduce the force on a passenger in a crash? · 安全气囊如何在碰撞中减少对乘客的冲击力？
The change in momentum is fixed. A longer stopping time $\Delta t$ makes $F = \Delta p/\Delta t$ smaller. · 动量的变化是固定的。更长的停止时间$\Delta t$会使$F = \Delta p/\Delta t$更小。
For the same change in momentum, a longer contact time means the force is: · 对于相同的动量变化，更长的接触时间意味着力：
$F = \Delta p / \Delta t$: a bigger $\Delta t$ gives a smaller force. · $F = \Delta p / \Delta t$：更大的$\Delta t$导致更小的力。
The change in momentum in a crash is fixed by the speeds involved — an airbag does not reduce it. What the airbag reduces is the force, by making the stop take longer ($F = \Delta p / \Delta t$). Same $\Delta p$, longer $\Delta t$, smaller $F$.
碰撞中动量的变化由所涉及的速度定死——安全气囊不能减少它。气囊减少的是力,靠让停下过程更久($F = \Delta p / \Delta t$)。同样的 $\Delta p$、更长的 $\Delta t$、更小的 $F$。
A $0.5\ \text{kg}$ ball hits a wall at $6\ \tfrac{\text{m}}{\text{s}}$ and bounces back at $4\ \tfrac{\text{m}}{\text{s}}$. What is the size of the impulse from the wall, in $\text{N·s}$? · 一个$0.5\ \text{kg}$的球以$6\ \tfrac{\text{m}}{\text{s}}$撞击墙壁并以$4\ \tfrac{\text{m}}{\text{s}}$反弹。墙壁施加的冲量大小是多少，单位为$\text{N·s}$？
$\Delta p = 0.5(-4) - 0.5(6) = -5$, so the impulse has size $5\ \text{N·s}$ (away from the wall). · $\Delta p = 0.5(-4) - 0.5(6) = -5$，因此冲量大小为$5\ \text{N·s}$（远离墙壁）。
A $0.5\ \text{kg}$ ball hits a wall at $6\ \tfrac{\text{m}}{\text{s}}$ and bounces straight back at $4\ \tfrac{\text{m}}{\text{s}}$.
- Taking "toward the wall" as positive: $\Delta p = m v_f - m v_i = 0.5(-4) - 0.5(6) = -5\ \text{kg}\cdot\tfrac{\text{m}}{\text{s}}$.
- The impulse from the wall is $5\ \text{N}\cdot\text{s}$, directed away from the wall.
一个 $0.5\ \text{kg}$ 的球以 $6\ \tfrac{\text{m}}{\text{s}}$ 撞墙,并以 $4\ \tfrac{\text{m}}{\text{s}}$ 径直弹回。
- 取"朝墙"为正:$\Delta p = m v_f - m v_i = 0.5(-4) - 0.5(6) = -5\ \text{kg}\cdot\tfrac{\text{m}}{\text{s}}$。
- 墙给出的冲量是 $5\ \text{N}\cdot\text{s}$,方向背离墙。
Impulse $\vec J = \vec F\,\Delta t$ equals the change in momentum: $\vec F\,\Delta t = \Delta \vec p$ (the impulse–momentum theorem). It is the area under a force–time graph. For a fixed $\Delta p$, a longer time means a smaller force — the secret of airbags.
冲量 $\vec J = \vec F\,\Delta t$ 等于动量的变化:$\vec F\,\Delta t = \Delta \vec p$(冲量–动量定理)。它是力–时间图下的面积。对固定的 $\Delta p$,时间越长意味着力越小——这就是安全气囊的秘密。