deletions | additions       diff --git a/Analysis and Interpretation.tex b/Analysis and Interpretation.tex  index 4449817..86976a0 100644  --- a/Analysis and Interpretation.tex  +++ b/Analysis and Interpretation.tex 
...
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.