Change in Momentum and Impulse
Introduced| English |
|---|
| impulse/ˈɪmpʌls/ |
| impulse-momentum theorem/ˈɪmpʌls məʊˈmentəm ˈθɪərəm/ |
| contact time/ˈkɒntækt taɪm/ |
Why airbags save lives
- In a crash, your body must stop -- there is no avoiding that.
- An airbag cannot change how much your motion changes, only how long it takes.
- By stretching the stop from a hundredth of a second to a tenth, it cuts the force tenfold.
- The physics behind this is impulse.
Impulse
- Impulse 冲量 is force applied over time:
- It is the area under a force-versus-time graph.
- A big force for a short time can equal a small force for a long time.
A force of $50\ \text{N}$ acts for $0.4\ \text{s}$. What impulse does it deliver (in N·s)?
$J = F\Delta t = 50 \times 0.4 = 20\ \text{N}\cdot\text{s}$.
The impulse delivered by a varying force equals the area under its force-versus-time graph.
$J = \int F\,dt$ is exactly that area.
The impulse-momentum theorem
- The impulse-momentum theorem 冲量动量定理 says impulse equals the change in momentum:
- To change an object's momentum, apply a force for some time.
- The bigger the impulse, the bigger the change in motion.
An impulse of $12\ \text{N}\cdot\text{s}$ acts on a $3\ \text{kg}$ cart at rest. Its final speed (in m/s)?
$J = \Delta p = m\Delta v$, so $\Delta v = 12/3 = 4\ \tfrac{\text{m}}{\text{s}}$ from rest.
Newton's law, deeper
- Differentiating gives Newton's second law in its truest form:
- Force is the rate of change of momentum.
- For constant mass this reduces to the familiar $F = ma$.
Impulse and momentum change
Impulse is force times time, and it equals the change in momentum. Sort each case.
In its most general form, the net force equals the rate of change of ____.
$\vec{F}_{net} = d\vec{p}/dt$ -- force is the time rate of change of momentum.
Stretch the time, soften the blow
- For a fixed change in momentum, a longer contact time 接触时间 means a smaller average force ($F = \Delta p / \Delta t$).
- Airbags, crumple zones, bent knees on landing, and catching an egg gently all use this.
- Same momentum change, gentler force.
An airbag reduces injury by...
The momentum change is fixed; the airbag stretches the time, so the average force $F = \Delta p/\Delta t$ drops.
A ball hitting a wall and bouncing back experiences a momentum change that is... a ball that hits and stops.
A bounce reverses the velocity (e.g. $+20$ to $-20$), a change of $40$; stopping is only a change of $20$.
A $0.15\ \text{kg}$ ball hits a wall at $20\ \tfrac{\text{m}}{\text{s}}$ and bounces back at $20\ \tfrac{\text{m}}{\text{s}}$.
- Change in momentum: $\Delta p = m(v_f - v_i) = 0.15(-20 - 20) = -6\ \tfrac{\text{kg}\cdot\text{m}}{\text{s}}$.
- If the contact lasts $0.01\ \text{s}$, the average force is $|\Delta p|/\Delta t = 6/0.01 = 600\ \text{N}$.
A bounce changes momentum more than a stop. Coming in at $20$ and leaving at $-20$ is a change of $40$ (in speed units), not $20$. Forgetting the sign flip on a bounce is a classic impulse mistake.
Impulse $\vec{J} = \int \vec{F}\,dt$ (the area under an $F$-$t$ graph) equals the change in momentum: $\vec{J} = \Delta\vec{p}$. Since $\vec{F} = d\vec{p}/dt$, stretching the contact time for a fixed momentum change lowers the average force -- the secret of every airbag.