deletions | additions       diff --git a/Analysis and Interpretation.tex b/Analysis and Interpretation.tex  index 706fecb..e254325 100644  --- a/Analysis and Interpretation.tex  +++ b/Analysis and Interpretation.tex 
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The time it takes for the ball to travel the length of each arrow is approximately $0.456\units s$. 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* m  \end{equation}  Adding the initial momentums of each ball, i.e. from their starting position to the collision, will result in the initial momentum of the system. This should be approximately equal to the final momentum of the system. This is due to the principle of conservation of momentum. 
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\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}\\  &= \langle 0.024,0.015\rangle\units m * 0.146\units kg  / 0.456 \units s* 0.146\units kg  + \vec r\sub{2i} / .456 \units s * 0.057 \units kg\\ \end{align}