Conservation of Angular Momentum of particle

[Versaiteis]

Homework Statement

A 2.80 kg particle that is moving horizontally over a floor with velocity (-3.00 m/s) \widehat{j}undergoes a completely inelastic collision with a 4.00 kg particle that is moving horizontally over the floor with velocity (4.55 m/s)\widehat{i}. 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\widehat{k} 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.

 

[Andrew Mason]

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 \vec{r} \text{ and } \vec{v}.

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

AM

 

[Versaiteis]

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

L = L_{1} + L_{2}

L_{1} = 2.8 * (-0.5 * -3.00)
L_{2} = 4.0 * ( 0 - (-0.1 * 4.55))

Sure enough

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

Thank you for your help Andrew

 

[Andrew Mason]

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

L = L_{1} + L_{2}

L_{1} = 2.8 * (-0.5 * -3.00)
L_{2} = 4.0 * ( 0 - (-0.1 * 4.55))

Sure enough

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

Thank you for your help Andrew

No problemo.

AM