AcademyMomentum Systems
Academy
Momentum and Impulse
Level 1 - Physics topic page in Momentum Systems.
Principle
Impulse is the time-integral of force, and it equals the change in momentum.
Notation
\(\vec{p}\)
linear momentum
\(\mathrm{kg\,m\,s^{-1}}\)
\(m\)
mass
\(\mathrm{kg}\)
\(\vec{v}\)
velocity
\(\mathrm{m\,s^{-1}}\)
\(\vec{J}\)
impulse
\(\mathrm{N\,s}\)
\(\vec{F}\)
force
\(\mathrm{N}\)
\(\Delta t\)
time interval
\(\mathrm{s}\)
Method
For a fixed-mass particle, Newton's second law can be written as a momentum-rate equation.
Define momentum
\[\vec{p}=m\vec{v}\]
Force rate
\[\sum\vec{F}=\frac{d\vec{p}}{dt}\]
This is equivalent to \(\sum\vec{F}=m\vec{a}\) when mass is fixed.
Integrate force
\[\int_{t_1}^{t_2}\sum\vec{F}\,dt=\vec{p}_f-\vec{p}_i\]
Average force
\[\vec{F}_{\mathrm{avg}}\Delta t=\Delta\vec{p}\]
The force-time graph shows impulse as area under the curve, not as the peak force alone.
The same impulse can come from a high force over a short time or a lower force over a longer time.
Rules
These are the compact impulse-momentum results from the integration above.
Momentum
\[\vec{p}=m\vec{v}\]
Impulse
\[\vec{J}=\int_{t_1}^{t_2}\vec{F}\,dt\]
Impulse momentum
\[\vec{J}=\Delta\vec{p}\]
Average force
\[\vec{J}=\vec{F}_{\mathrm{avg}}\Delta t\]
Examples
Question
A
\[0.30\,\mathrm{kg}\]
puck moving right at \[9.0\,\mathrm{m\,s^{-1}}\]
leaves left at \[3.0\,\mathrm{m\,s^{-1}}\]
Find the impulse on the puck.Answer
Take right as positive.
\[J_x=m(v_f-v_i)=0.30(-3.0-9.0)=-3.6\,\mathrm{N\,s}\]
The impulse is \[3.6\,\mathrm{N\,s}\]
left.Checks
- Momentum and impulse are vectors.
- \(\mathrm\{N\,s\}\) and \(\mathrm\{kg\,m\,s^\{-1\}}\) are equivalent units.
- Longer stopping time means smaller average force for the same momentum change.
- Use signed velocity, not speed alone.