Ice skaters collision with angle

AI Thread Summary
Two ice skaters collide, with one weighing 50 kg moving at 5 m/s and the other 40 kg at 4 m/s perpendicular to the first. After the collision, the 50 kg skater moves at 4 m/s at a 25-degree angle from their original direction. To find the 40 kg skater's velocity post-collision, the components of momentum must be calculated using the angles and magnitudes of the velocities. The final velocity will be a vector requiring both magnitude and direction, with the angle determined by arctan of the velocity components. The discussion emphasizes the importance of correctly applying momentum conservation equations and understanding vector components.
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.
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You have v1fx and v1fy in the equations for components of momentum. How do you get them, knowing magnitude and angle of velocity ?
 
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°))?
 
fantisism said:
Is v1fx=(4 m/s)(cos(25°)) and v1fy=(4 m/s)(sin(25°))?
Yes.
 
ehild said:
Yes.
And then with that information, solve for v2fx and v2fy?
 
fantisism said:
And then with that information, solve for v2fx and v2fy?
Yes.
 
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)?
 
  • #10
fantisism said:
To find the angle, would it be arctan(v2fy/v2fx)?
Yes.
 
  • #11
Alright. I can solve the second subquestion myself. Thank you again!
 
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