\subsection{Linear Momentum}

This subsection introduces linear momentum as the vector state variable for translational motion. In AP mechanics, momentum is the quantity that naturally leads into impulse and conservation ideas.

\dfn{Particle and system momentum}{Let $m$ denote the mass of a particle, let $\vec{v}$ denote its velocity measured in a chosen inertial reference frame, and let $\vec{p}$ denote its linear momentum. The \emph{linear momentum} of the particle is
\[
\vec{p}=m\vec{v}.
\]
Because $m$ is a scalar and $\vec{v}$ is a vector, $\vec{p}$ is a vector in the same direction as $\vec{v}$. Its SI unit is
\[
1\,\mathrm{kg\cdot m/s}.
\]

For a system of $N$ particles labeled by $i=1,2,\dots,N$, let $m_i$ denote the mass of particle $i$, let $\vec{v}_i$ denote its velocity, and let $\vec{p}_i=m_i\vec{v}_i$ denote its momentum. The total linear momentum of the system is
\[
\vec{p}_{\mathrm{sys}}=\sum_{i=1}^N \vec{p}_i=\sum_{i=1}^N m_i\vec{v}_i.
\]}

\nt{Momentum is not the same thing as speed. Speed is the scalar $v=|\vec{v}|$, while momentum is the vector $\vec{p}=m\vec{v}$. Two objects can have the same speed but different momenta if their masses are different or if they move in different directions. Momentum also depends on the chosen reference frame because velocity does: an object at rest in one frame has $\vec{p}=\vec{0}$ in that frame, but it can have nonzero momentum in another frame. In one-dimensional motion, a negative momentum component means motion in the negative coordinate direction; it does not mean negative mass or negative speed.}

\mprop{Component and system relations}{Let
\[
\vec{v}=v_x\hat{\imath}+v_y\hat{\jmath}+v_z\hat{k}
\]
be the velocity of a particle of mass $m$, and let
\[
\vec{p}=p_x\hat{\imath}+p_y\hat{\jmath}+p_z\hat{k}
\]
be its momentum. For a system of particles, let
\[
M=\sum_{i=1}^N m_i
\]
denote the total mass, and let $\vec{v}_{\mathrm{cm}}$ denote the center-of-mass velocity.

\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\item Momentum components are found component-by-component from the velocity:
\[
p_x=mv_x,
\qquad
p_y=mv_y,
\qquad
p_z=mv_z.
\]
In one-dimensional motion along the $x$-axis,
\[
p_x=mv_x.
\]

\item For a single particle, the momentum magnitude is
\[
p=|\vec{p}|=m|\vec{v}|=mv.
\]
In two dimensions,
\[
p=\sqrt{p_x^2+p_y^2},
\]
and in three dimensions,
\[
p=\sqrt{p_x^2+p_y^2+p_z^2}.
\]

\item System momentum is the vector sum of the particle momenta:
\[
\vec{p}_{\mathrm{sys}}=\sum_{i=1}^N \vec{p}_i.
\]
Therefore its components also add:
\[
p_{\mathrm{sys},x}=\sum_{i=1}^N p_{i,x},
\qquad
p_{\mathrm{sys},y}=\sum_{i=1}^N p_{i,y},
\qquad
p_{\mathrm{sys},z}=\sum_{i=1}^N p_{i,z}.
\]
Equivalently,
\[
\vec{p}_{\mathrm{sys}}=M\vec{v}_{\mathrm{cm}}.
\]
\end{enumerate}}

\qs{Worked example}{In a laboratory frame, two pucks slide on nearly frictionless ice. Puck 1 has mass $m_1=2.0\,\mathrm{kg}$ and velocity
\[
\vec{v}_1=(1.5\hat{\imath}+0.50\hat{\jmath})\,\mathrm{m/s}.
\]
Puck 2 has mass $m_2=1.0\,\mathrm{kg}$ and velocity
\[
\vec{v}_2=(-1.0\hat{\imath}+4.0\hat{\jmath})\,\mathrm{m/s}.
\]
Let $\vec{p}_1$ and $\vec{p}_2$ denote the individual momenta, let $\vec{p}_{\mathrm{sys}}$ denote the total momentum, let $M$ denote the total mass, and let $\theta$ denote the direction of $\vec{p}_{\mathrm{sys}}$ measured counterclockwise from the positive $x$-axis.

Find $\vec{p}_1$, $\vec{p}_2$, $\vec{p}_{\mathrm{sys}}$, the magnitude $|\vec{p}_{\mathrm{sys}}|$, the direction $\theta$, and the center-of-mass velocity $\vec{v}_{\mathrm{cm}}$.}

\sol Use $\vec{p}=m\vec{v}$ for each puck.

For puck 1,
\[
\vec{p}_1=m_1\vec{v}_1=(2.0\,\mathrm{kg})(1.5\hat{\imath}+0.50\hat{\jmath})\,\mathrm{m/s}.
\]
Therefore,
\[
\vec{p}_1=(3.0\hat{\imath}+1.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
\]

For puck 2,
\[
\vec{p}_2=m_2\vec{v}_2=(1.0\,\mathrm{kg})(-1.0\hat{\imath}+4.0\hat{\jmath})\,\mathrm{m/s}.
\]
So,
\[
\vec{p}_2=(-1.0\hat{\imath}+4.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
\]

Now add the momenta component-by-component:
\[
\vec{p}_{\mathrm{sys}}=\vec{p}_1+\vec{p}_2.
\]
Hence,
\[
\vec{p}_{\mathrm{sys}}=(3.0-1.0)\hat{\imath}+(1.0+4.0)\hat{\jmath}.
\]
Therefore,
\[
\vec{p}_{\mathrm{sys}}=(2.0\hat{\imath}+5.0\hat{\jmath})\,\mathrm{kg\cdot m/s}.
\]

Its magnitude is
\[
|\vec{p}_{\mathrm{sys}}|=\sqrt{(2.0)^2+(5.0)^2}\,\mathrm{kg\cdot m/s}=\sqrt{29}\,\mathrm{kg\cdot m/s}.
\]
Numerically,
\[
|\vec{p}_{\mathrm{sys}}|\approx 5.39\,\mathrm{kg\cdot m/s}.
\]

To find the direction, use the component ratio. Since both components of $\vec{p}_{\mathrm{sys}}$ are positive, the vector lies in the first quadrant. Thus,
\[
\tan\theta=\frac{p_{\mathrm{sys},y}}{p_{\mathrm{sys},x}}=\frac{5.0}{2.0}=2.5.
\]
So,
\[
\theta=\tan^{-1}(2.5)\approx 68.2^\circ.
\]

Now find the center-of-mass velocity. The total mass is
\[
M=m_1+m_2=2.0\,\mathrm{kg}+1.0\,\mathrm{kg}=3.0\,\mathrm{kg}.
\]
Using
\[
\vec{p}_{\mathrm{sys}}=M\vec{v}_{\mathrm{cm}},
\]
we get
\[
\vec{v}_{\mathrm{cm}}=\frac{\vec{p}_{\mathrm{sys}}}{M}=\frac{(2.0\hat{\imath}+5.0\hat{\jmath})\,\mathrm{kg\cdot m/s}}{3.0\,\mathrm{kg}}.
\]
Therefore,
\[
\vec{v}_{\mathrm{cm}}=\left(\frac{2.0}{3.0}\hat{\imath}+\frac{5.0}{3.0}\hat{\jmath}\right)\,\mathrm{m/s}
\]
or numerically,
\[
\vec{v}_{\mathrm{cm}}\approx (0.667\hat{\imath}+1.67\hat{\jmath})\,\mathrm{m/s}.
\]

So the individual and system momenta are
\[
\vec{p}_1=(3.0\hat{\imath}+1.0\hat{\jmath})\,\mathrm{kg\cdot m/s},
\qquad
\vec{p}_2=(-1.0\hat{\imath}+4.0\hat{\jmath})\,\mathrm{kg\cdot m/s},
\]
\[
\vec{p}_{\mathrm{sys}}=(2.0\hat{\imath}+5.0\hat{\jmath})\,\mathrm{kg\cdot m/s},
\]
with magnitude
\[
|\vec{p}_{\mathrm{sys}}|\approx 5.39\,\mathrm{kg\cdot m/s},
\]
direction
\[
\theta\approx 68.2^\circ,
\]
and center-of-mass velocity
\[
\vec{v}_{\mathrm{cm}}\approx (0.667\hat{\imath}+1.67\hat{\jmath})\,\mathrm{m/s}.
\]
This example shows why momentum must be handled as a vector: the total momentum is found by adding components, not by adding speeds.
