Looking for verification on a momentum & collision problem

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The discussion focuses on calculating the momentum and final velocity of two runners colliding in a perfectly inelastic manner. Runner1, with a mass of 60 kg and a velocity of 5 m/s at 20 degrees North of East, has a momentum of 300 kg·m/s. Runner2, with a mass of 80 kg and a velocity of 4 m/s at 20 degrees North of West, has a momentum of -320 kg·m/s. The final velocity after the collision is calculated using the conservation of momentum, resulting in a velocity of approximately -0.049 m/s in the x-direction and 4.16 m/s in the y-direction, leading to an angle of -89 degrees North of East.

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PHYclueless
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Here is the problem:
Runner1 is going 20 degrees NofE with a mass of 60kg and v=5m/s. Runner2 is going 20 degrees NofW with a mass of 80kg and v=4m/s when they collide. What are their respective momentums prior to the collision? What is their final velocity and direction after collision if it was perfectly inelastic collision?

For question1 I got L=300kgm/s for runner1 and L=-320kgm/s for runner2.

For question2 I'm a little perplexed. This is what I did.
(x) m1v1+m2v2=(m1+m2)vf
-4sin20(80kg)+5sin20(60kg)=(60kg+80kg)vf
-109+103=140vf
-6=140vfx
vf(x)=-.049m/s

(y) m1v1+m2v2=(m1+m2)vf
4cos20(80kg)+5cos20(60kg)=(60kg+80kg)vf
301+282=140vf
583=140vf
vf(y)=4.16m/s

Then I took the TAN-1(4.16/-.049) to find theta and it gave me -89degrees North of East.

Does this look right?

Thank you to anyone who can give me some insight. I appreciate your help.
 
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Your method is right.
But your vector algebra is slightly flawed.
For (x) take cosine and not sine.
Viceversa for (y) .
 

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