The total momentum is the vector sum of all momenta. Assume that y-axis points northward, y-axis looks eastward:
"\\vec{p}_x=\\sum^n_{i=1} m_i\\vec{v}_{xi}=12\\cdot6=72\\text{ kg}\\cdot\\text{m\/s},\\\\\n\\space\\\\\n\\vec{p}_y=\\sum^n_{i=1} m_i\\vec{v}_{yi}=9\\cdot(-8.5)=-76.5\\text{ kg}\\cdot\\text{m\/s}."
We can find the magnitude of the total momentum:
"p=\\sqrt{p_x^2+p_y^2}=\\sqrt{72^2+(-76.5)^2}=105\\text{ kg}\\cdot\\text{m\/s}." The angle above x-axis (north of east) is
"\\theta=\\text{atan}\\frac{\\vec{p_y}}{\\vec{p_x}}=-46.7^\\circ," or 46.7° below x-axis (below east axis), or 46.7° south of east.
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