Ice skaters collision with angle

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fantisism
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Homework Statement


Two ice skaters crash into each other. Before they collide, one of them (50 kg) is skating in a straight line at 5 m/s, the other (40 kg) is skating at 4 m/s in a straight line at 90 degrees to the 50 kg skater’s direction. After the collision, the 50 kg skater is moving at 4 m/s at an angle of 25 degrees relative to their original direction. Calculate the velocity of the 40 kg skater after the collision. Is the total kinetic energy constant?

Homework Equations


m1v1ix+m2v2ix=m1v1fx+m2v2fx
m1v1iy+m2v2iy=m1v1fy+m2v2fy

The Attempt at a Solution


I drew a diagram before collision and after collision. Really, all I need to know is where in the formula(s) would I need to include the angle and I SHOULD be able to figure it out from there.
FullSizeRender.jpg
 
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ehild said:
You have v1fx and v1fy in the equations for components of momentum. How do you get them, knowing magnitude and angle of velocity ?
Is v1fx=(4 m/s)(cos(25°)) and v1fy=(4 m/s)(sin(25°))?
 
ehild said:
Yes.
And then with that information, solve for v2fx and v2fy?
 
ehild said:
Yes.
And then once I figure out what those two velocities are, I would square them individually and take the square root of the sums? So, sqrt((v2fx)2+(v2fy)2? And that would be the answer for the first question?
 
fantisism said:
And then once I figure out what those two velocities are, I would square them individually and take the square root of the sums? So, sqrt((v2fx)2+(v2fy)2? And that would be the answer for the first question?
The final velocity is asked. It is a vector with magnitude and direction. I think the problem wants both the magnitude and angle.
 
ehild said:
The final velocity is asked. It is a vector with magnitude and direction. I think the problem wants both the magnitude and angle.
To find the angle, would it be arctan(v2fy/v2fx)?
 
Alright. I can solve the second subquestion myself. Thank you again!