Ice skaters collision with angle

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SUMMARY

The discussion focuses on a physics problem involving two ice skaters colliding at different velocities and angles. The first skater, weighing 50 kg, moves at 5 m/s, while the second skater, weighing 40 kg, moves at 4 m/s at a right angle to the first skater. After the collision, the first skater moves at 4 m/s at an angle of 25 degrees. Participants confirm the use of momentum equations to calculate the final velocities of both skaters and discuss the necessity of determining both magnitude and direction of the resulting velocity vector.

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  • Knowledge of trigonometric functions (sine, cosine, tangent)
  • Ability to perform calculations involving angles and magnitudes
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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.
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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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