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Elastic and Inelastic Collisions in One Dimension - Need Help, due tonight!

  1. Apr 14, 2010 #1
    I have a lab report due tonight. We were working on these in lab and I'm not sure how to tell which one was elastic or inelastic. We had three cases. We had to use the conservation of momentum and kinetic energy formulas.


    A ) Cart A and B have equal masses. Car B is at rest. Car A strikes car B. They don't stick together. Photogate measures different final velocities for each. Conservation of momentum -> 26% difference, and conservation of kinetic energy was 49 % difference.

    B) Cart A has larger mass than Cart B. Cart B is at rest. Car A strikes Car B, and again they measure different velocities. Percent difference for momentum was 8.9 % and for kinetic energy it was 26 %.

    C.1) Cart A and Cart B have equal masses, but Cart A is inverted, so that the two magnets on each car attract each other, ( I think we are simulating an elastic collision here, but Im not sure). Cart B is at rest. Cart A strikes Cart B and they stick together, both presenting with same final velocities. Momentum % diff. was 8.7% and kinetic energy was 57 %.

    C.2)After Cart A, passes the photogate, it measures a different velocity, (slower).


    So, the questions are, which collision is inelastic/elastic and why? And what is the importance/use of the second velocity measure in part C?

    These are my assumptions, Please tell me if I am correct, and why or why not am I wrong.

    A ) The collision is elastic because the carts didnt stick together, and their final velocities were different.

    B) The collision is elastic because the carts didnt stick together, and their final velocities were different.

    C.1) The collision is inelastic because the carts stuck together, and their final velocities were the same.

    C.2) The velocity measure after Cart A passed the photogate is different than when the collision took place because the carts are going slower? I dont know what the importance is though.


    Please, any input is appreciated! I'm deperate here, thank you!
     
    Last edited: Apr 14, 2010
  2. jcsd
  3. Apr 14, 2010 #2
    Any help is appreciated! plz :)
     
  4. Apr 14, 2010 #3
    Look up the definition of (in)elastic collisions. Also, you should make clear if they're asking about it being a theoretical elastic collision or a practical one. No practical collision is ever elastic.

    Also, this belongs in the homework forum.
     
  5. Apr 14, 2010 #4
    I dont know, the prof. just said to specify if its elastic or inelastic. It's a lab experiment,

    Inelastic = kinetic energy is not conserved
    elastic = kinetic energy is conserved

    idk how to tell if its conserved or not due to possible experimental error.
     
    Last edited: Apr 14, 2010
  6. Apr 14, 2010 #5
    well perform some try-out calculations. If kinetic energy is conserved in a clash, does that mean they stick together or veer off independently? Then look at how your experiments are set and see which ones should theoretically make for an elastic clash and which not.
     
  7. Apr 14, 2010 #6
    only in the third collision did they stick together, only because they had magnets, but i guess thats supposed to look like an inelastic collision, but its where the kinetic enery was least conserved.


    also, does anyone know what the second velocity in part C means?
     
  8. Apr 14, 2010 #7
    Why "but"? It's exactly as you should expect from an inelastic collision, think about it

    Concerning the velocity in part C, again, think theoretically. What would you expect in an ideal (in)elastic situation in the second part of C. Compare that with what you got as a result.

    Good luck
     
  9. Apr 14, 2010 #8
    Okay, so in the part C, after the photogate measures the final velocity, 23 m/s, ( which is the same for both carts, since they are stuck together after all), it runs across a certain short distance, until the first cart, A, runs thru the photogate (While still stuck to the second car) and the velocity is measured at 18 m/s. So, it means it slowed down?

    If so, why would this second velocity be useful, the prof, said to write about its importance.

    Thanks for ur help!
     
  10. Apr 14, 2010 #9
    link the velocity with the definitions of elastic and inelastic. that's all I can say without literally saying it :p remember the carts are of equal mass
     
  11. Apr 14, 2010 #10
    So, I guess it would be useful (the second velocity in the third collision), because it would help determine the velocity as its slowing down. The 18 m/s measure in Car A, would also be the same for B, then? Am I close?
     
  12. Apr 14, 2010 #11
    no.

    it's not hard, think more about it, lay connections.

    Read the definitions of elastic and inelastic again... Think about speed... mass... :p
     
  13. Apr 14, 2010 #12
    Is it to verify that no kinetic energy has been lost?
     
  14. Apr 14, 2010 #13
    or in this case...
     
  15. Apr 14, 2010 #14
    not conserved? gahh. im stumped!
     
  16. Apr 14, 2010 #15
    btw these are my calculations

    I.
    Ps = MaVa = (0.519)(0.42) = 0.22 N*s
    Ps’ = MaV’a + MbV’b = (0.519)(0) + (0.511)(0.33) = 0.17 N*s

    % diff. = [|0.22 - 0.17| / (0.22 + 0.17 / (2))] x 100 = 26 %

    K = (.5)(MaV^2a) = (0.5)(0.519)(0.42)^2 = 0.046 J
    K’ = (.5)(MaV’^2a) + (0.5)(MbV’^2b) = (0.5)(0.519)(0) + (0.5)(0.511)(0.33)^2 = 0.028 J

    % diff. = [|0.046 - 0.028| / (0.046 + 0.028 / (2))] x 100 = 49 %

    II.
    Ps = MaVa = (1.013)(0.46) = 0.47 N*s
    Ps’ = MaV’a + MbV’b = (1.013)(0.16) + (0.511)(0.53) = 0.43

    % diff. = [|0.47 - 0.43| / (0.47 + 0.43 / (2))] x 100 = 8.9 %

    K = (.5)(MaV^2a) = (0.5) (1.013)(0.46)^2 =0.11 J
    K’ = (.5)(MaV’^2a) + (0.5)(MbV’^2b) = (0.5)(1.013)(0.16)^2 + (0.5)(0.511)(0.53)^2 = 0.085 J

    % diff. = [|0.11 - 0.085| / (0.11 + 0.085 / (2))] x 100 = 26 %

    III.
    Ps = MaVa = (0.519)(0.42) = 0.22 N*s
    Ps’ = MaV’a + MbV’b = (0.519)(0.23) + (0.511)(0.23) = 0.24 N*s

    % diff. = [|0.22 - 0.24| / (0.22 + 0.24 / (2))] x 100 = 8.7 %

    K = (.5)(MaV^2a) = (0.5)(0.519)(0.42)^2 = 0.046 J
    K’ = (.5)(MaV’^2a) + (0.5)(MbV’^2b) = (0.5)(0.519)(0.23)^2 + (0.5)(0.511)(0.23)^2 = 0.027 J

    % diff. = [|0.046 - 0.027| / (0.046 + 0.027 / (2))] x 100 = 57%

    V”a = 0.18 m/s
     
  17. Apr 14, 2010 #16
    I'm confused. Right after the collision it measures the speed and then another time a bit later?
     
  18. Apr 14, 2010 #17
    Yes, when CartB is struck by A, the photogate measures its speed (which is the same for both as they are stuck together), then Cart A passes under the photogate still stuck to B, and it measures a lower velocity. I dont know what this means.
     
  19. Apr 14, 2010 #18
    It seems that the only thing you can say, is that there is friction... huh weird :s sorry
     
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