\subsection{Impulse and Momentum Transfer}

This subsection connects the local momentum law $\vec{F}_{\text{net}}=d\vec{p}/dt$ to finite-time interactions such as hits, kicks, and collisions, where a large force acts for a short time and changes momentum by a measurable amount.

\dfn{Momentum change and impulse}{Let a body of constant mass $m$ have velocity $\vec{v}$. Its linear momentum is
\[
\vec{p}=m\vec{v}.
\]
Let $\vec{p}_i$ and $\vec{p}_f$ denote the initial and final momenta over some time interval from $t_i$ to $t_f$. The change in momentum is
\[
\Delta \vec{p}=\vec{p}_f-\vec{p}_i.
\]
Let $\vec{F}_{\text{net}}(t)$ denote the net external force on the body during that interval. The impulse delivered to the body is
\[
\vec{J}=\int_{t_i}^{t_f} \vec{F}_{\text{net}}\,dt.
\]
If the net force is constant, then this reduces to
\[
\vec{J}=\vec{F}_{\text{net}}\Delta t,
\qquad
\Delta t=t_f-t_i.
\]
The SI unit of impulse is $\mathrm{N\,s}$, which is equivalent to $\mathrm{kg\,m/s}$.}

\thm{Impulse-momentum theorem}{Let $\vec{p}(t)$ denote the momentum of a body and let $\vec{F}_{\text{net}}(t)$ denote the net external force on it. Over any time interval from $t_i$ to $t_f$,
\[
\vec{J}=\int_{t_i}^{t_f} \vec{F}_{\text{net}}\,dt=\Delta \vec{p}=\vec{p}_f-\vec{p}_i.
\]
Thus the net impulse on a body equals its change in momentum. In one dimension along the $x$-axis, let $J_x$, $F_{\text{net},x}$, and $p_x$ denote the corresponding $x$-components. Then
\[
J_x=\int_{t_i}^{t_f} F_{\text{net},x}\,dt=\Delta p_x.
\]}

\nt{Impulse is a vector, so its direction is the direction of $\Delta \vec{p}$. A body can be moving in one direction while the impulse points in the opposite direction if the interaction slows or reverses the motion. Impulse depends on both force and time: a large force acting briefly can produce the same impulse as a smaller force acting longer. If an average net force $\vec{F}_{\text{avg}}$ over a time interval $\Delta t$ is defined so that it has the same effect as the actual time-varying force, then
\[
\vec{J}=\vec{F}_{\text{avg}}\Delta t.
\]
On a force-versus-time graph, the signed area under the curve gives impulse. In component form, the signed area under an $F_x(t)$ graph gives $J_x=\Delta p_x$.}

\pf{Short derivation from $d\vec{p}/dt$}{Start with the local momentum law
\[
\vec{F}_{\text{net}}=\frac{d\vec{p}}{dt}.
\]
Multiply by $dt$:
\[
\vec{F}_{\text{net}}\,dt=d\vec{p}.
\]
Now integrate from $t_i$ to $t_f$:
\[
\int_{t_i}^{t_f} \vec{F}_{\text{net}}\,dt=\int_{t_i}^{t_f} d\vec{p}.
\]
The left side is the impulse $\vec{J}$, and the right side is the total change in momentum:
\[
\vec{J}=\vec{p}_f-\vec{p}_i=\Delta \vec{p}.
\]
This is the impulse-momentum theorem.}

\qs{Worked example}{Choose the positive $x$-axis to the right. A tennis ball of mass $m=0.150\,\mathrm{kg}$ moves horizontally with initial velocity
\[
\vec{v}_i=-18.0\,\hat{\imath}\,\mathrm{m/s}.
\]
Let $F_x(t)$ denote the net horizontal force on the ball during contact. The force is directed to the right and has the following force-versus-time graph: it increases linearly from $0$ at $t=0$ to $900\,\mathrm{N}$ at $t=4.0\,\mathrm{ms}$, then decreases linearly back to $0$ at $t=8.0\,\mathrm{ms}$.

Find (a) the impulse delivered to the ball, (b) the ball's final momentum, (c) the ball's final velocity, and (d) the average net force during contact.}

\sol Let $\Delta t=8.0\times 10^{-3}\,\mathrm{s}$ denote the contact time. Because the force is entirely in the $+x$ direction, the impulse is the area under the triangular $F_x(t)$ graph in the positive direction:
\[
\vec{J}=\left(\tfrac12 \right)(\Delta t)(900\,\mathrm{N})\,\hat{\imath}.
\]
Substitute $\Delta t=8.0\times 10^{-3}\,\mathrm{s}$:
\[
\vec{J}=\left(\tfrac12 \right)(8.0\times 10^{-3}\,\mathrm{s})(900\,\mathrm{N})\,\hat{\imath}=3.6\,\hat{\imath}\,\mathrm{N\,s}.
\]
Using $1\,\mathrm{N\,s}=1\,\mathrm{kg\,m/s}$,
\[
\vec{J}=3.6\,\hat{\imath}\,\mathrm{kg\,m/s}.
\]

Now compute the initial momentum:
\[
\vec{p}_i=m\vec{v}_i=(0.150\,\mathrm{kg})(-18.0\,\hat{\imath}\,\mathrm{m/s})=-2.70\,\hat{\imath}\,\mathrm{kg\,m/s}.
\]
From the impulse-momentum theorem,
\[
\vec{J}=\Delta \vec{p}=\vec{p}_f-\vec{p}_i,
\]
so
\[
\vec{p}_f=\vec{p}_i+\vec{J}.
\]
Substitute the values:
\[
\vec{p}_f=(-2.70+3.60)\,\hat{\imath}\,\mathrm{kg\,m/s}=0.90\,\hat{\imath}\,\mathrm{kg\,m/s}.
\]

Then the final velocity is
\[
\vec{v}_f=\frac{\vec{p}_f}{m}=\frac{0.90\,\hat{\imath}\,\mathrm{kg\,m/s}}{0.150\,\mathrm{kg}}=6.0\,\hat{\imath}\,\mathrm{m/s}.
\]
The positive sign shows that the ball leaves moving to the right.

For the average net force, use
\[
\vec{J}=\vec{F}_{\text{avg}}\Delta t.
\]
Therefore,
\[
\vec{F}_{\text{avg}}=\frac{\vec{J}}{\Delta t}=\frac{3.6\,\hat{\imath}\,\mathrm{N\,s}}{8.0\times 10^{-3}\,\mathrm{s}}=4.5\times 10^2\,\hat{\imath}\,\mathrm{N}.
\]

So the results are
\[
\vec{J}=3.6\,\hat{\imath}\,\mathrm{N\,s},
\qquad
\vec{p}_f=0.90\,\hat{\imath}\,\mathrm{kg\,m/s},
\]
\[
\vec{v}_f=6.0\,\hat{\imath}\,\mathrm{m/s},
\qquad
\vec{F}_{\text{avg}}=4.5\times 10^2\,\hat{\imath}\,\mathrm{N}.
\]
The graph interpretation is essential here: the triangular area under $F_x(t)$ gives the impulse directly, and that impulse determines the momentum transfer.
