Collisions homework problem

  1. A 1.0 kg magnetized air puck moving across a level table at 0.24 m/s approaches head-on a stationary, similarly magnetized air puck of mass 0.50 kg. If the "magnetic collision" is repulsive and perfectly elastic, determine:
    (a) the velocity of each puck after the collision
    (b) the velocity of both pucks at minimum separation
    (c) the total kinetic energy at minimum separation
    (d) the maximum potential energy stored in the magnetic force field during the collision


    For A) i simply used the conservation of momentum and KE and found the final velocities (1-dimensional)

    For B) I am not sure how to find th miniimum distance
     
  2. jcsd
  3. Doc Al

    Staff: Mentor

    Hint: When the pucks are as close to each other as they get, what's the relationship of their velocities?
     
  4. uhh..i am not sure abt this..velocities become the same..?
     
  5. Doc Al

    Staff: Mentor

    Exactly! (If the velocities weren't the same, then they'd keep getting closer.) So figure out what that velocity must be.
     
  6. so you are saying that the collision becomes the case of a perfectly inelastic collision...?
     
  7. if thats the case, then M1v= (M1+M2)V
    hence V= M1v/M1+M2
     
  8. Doc Al

    Staff: Mentor

    Exactly.
     
  9. ahh k..so for part c its quite staight foreward. we have to use0.5(M1+M2)V2
    For part d potential enegy will be max when the h is max. How do we find the h..?
     
  10. and also this is all 1-dimensional yes?
     
  11. Doc Al

    Staff: Mentor

    Right.
    What do you mean by "h is max"? Just use the fact that total energy is constant.

    Yes. (It's a "head-on" collision.)
     
  12. so then the potential energy will be equal to the kE in part C)..?
     
  13. Doc Al

    Staff: Mentor

    No, but the sum of PE + KE must be constant. (What's the initial total energy?)
     
  14. initial total energy is M1v1^2+ M2v2^2
     
  15. Doc Al

    Staff: Mentor

    Not exactly. Initially, only one puck is moving. (And that's not the correct formula for KE!)
     
  16. yes so M2v2^2 will 0..i know that hence the initial total energy will be 0.5*M1v1^2= 0.0288J. so then we can subtract the KE (from part c)frm this energy to find the PE
     
  17. Doc Al

    Staff: Mentor

    Now you've got it.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?