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David Gronlund edited Analysis and Interpretation.tex
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\section{Analysis and Interpretation}
The time it takes for the ball to travel the length of each arrow is approximately
$0.456\units replace_content.456\units s$. Divide the components of each arrow by $.456\units s$ to get the average velocity of each ball at that moment. Then multiply that by the ball's mass to get its average momentum. For example, the equation for momentum vector $\vec p$ with a distance vector $\vec r$, a time of $\Delta t$, and a mass of $m$ would be:
\begin{equation}
\vec p = \vec r * m / \Delta t
...
Plugging our data in these equations results in the following initial total momentum
\begin{align}
\vec p\sub{total,i} &= \vec p\sub{1i} + \vec p\sub{2i}\\ &= \vec r\sub{1i} / \Delta t * m\sub{1} + \vec r\sub{2i} / \Delta t * m\sub{2}\\ &= \langle2.4,1.5\rangle \sci{-2} \units m * 0.146 \units kg / 0.456 \units s + \langle-1.5,1.8\rangle \sci{-2} \units m * 0.057 \units kg / 0.456 \units s\\ &=
\langle7.68,4.80\rangle \sci{-3} \langle0.030,0.019\rangle \units kgm/s +
\langle-4.80,5.76\rangle \sci{-3} \langle-0.048,0.058\rangle \units kgm/s\\ &=
\langle2.88,10.56\rangle \sci{-3} \langle-0.018,0.077\rangle \units kgm/s
\end{align}
and again for the final momentum
\begin{align} \vec p\sub{total,f} &= \vec p\sub{1f} + \vec p\sub{2f}\\ &= \vec r\sub{1f} / \Delta t * m\sub{1} + \vec r\sub{2f} / \Delta t * m\sub{2}\\ &= \langle-2.5,1.6\rangle \sci{-2} \units m * 0.146 \units kg / 0.456 \units s + \langle0.4,1.6\rangle \sci{-2} \units m * 0.057 \units kg / 0.456 \units s\\ &= \langle-0.03,0.02\rangle \units kgm/s + \langle0.013,0.051\rangle \units kgm/s\\ &= \langle-0.017,0.071\rangle \units kgm/s
\end{align} and according to conservation of momentum both final and initial momentum must be equal. Accounting for measurement errors our results really closely match this initial assumption.