Conservation of Angular Momentum of particle

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Versaiteis
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Homework Statement



A 2.80 kg particle that is moving horizontally over a floor with velocity (-3.00 m/s) [tex]\widehat{j}[/tex]undergoes a completely inelastic collision with a 4.00 kg particle that is moving horizontally over the floor with velocity (4.55 m/s)[tex]\widehat{i}[/tex]. The collision occurs at xy coordinates (-0.50 m, -0.10 m). After the collision, what is the angular momentum of the stuck-together particles with respect to the origin?

answer = 6.02[tex]\widehat{k}[/tex] kg*m^2/s

Homework Equations



L = m( r x v )

The Attempt at a Solution



I can't seem to understand exactly what I'm supposed to do, what keeps throwing me off is the two different coordinate systems, ij and xy. No matter what combinations of vectors I use, nothing comes close, so I guess what I'm looking for is a place to start.
 
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Versaiteis said:
L = m( r x v )

The Attempt at a Solution



I can't seem to understand exactly what I'm supposed to do, what keeps throwing me off is the two different coordinate systems, ij and xy. No matter what combinations of vectors I use, nothing comes close, so I guess what I'm looking for is a place to start.
Keep in mind that the angular momentum vector is always perpendicular to the plane of [itex]\vec{r} \text{ and } \vec{v}[/itex].

The i, j, and k axes are all perpendicular to each other, of course. [tex]\hat i[/tex] is the unit vector along the x-axis in the direction of + x. [tex]\hat j[/tex] is the unit vector along the y-axis in the direction of +y. The [tex]\hat k[/tex] direction is perpendicular to both the i and j unit vectors.

AM
 
Oh I see, the two systems are directly related so I can simply say

L = L[tex]_{1}[/tex] + L[tex]_{2}[/tex]

L[tex]_{1}[/tex] = 2.8 * (-0.5 * -3.00)
L[tex]_{2}[/tex] = 4.0 * ( 0 - (-0.1 * 4.55))

Sure enough

L = 6.02[tex]\widehat{k}[/tex] kg*m^2/s

Thank you for your help Andrew
 
Versaiteis said:
Oh I see, the two systems are directly related so I can simply say

L = L[tex]_{1}[/tex] + L[tex]_{2}[/tex]

L[tex]_{1}[/tex] = 2.8 * (-0.5 * -3.00)
L[tex]_{2}[/tex] = 4.0 * ( 0 - (-0.1 * 4.55))

Sure enough

L = 6.02[tex]\widehat{k}[/tex] kg*m^2/s

Thank you for your help Andrew
No problemo.

AM