1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Use Conservation of Momentum or Conservation of KE for spring problem

  1. Dec 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Block A with mass 13kg is moving right on a frictionless table. Block B with mass 1kg is moving left. Block B has a spring on its left side, the side that will hit Block A. THe question is: Which is the same for both after the collision?
    1. speed
    2. velocity
    3. acceleration
    4. KE
    5. magnitude of momentum


    2. Relevant equations
    Conservation of momentum
    Conservation of energy


    3. The attempt at a solution
    I eliminated choices 1-3, but I don't know if you use the conservation of momentum or the conservation of energy for this problem. I know that the collision has to be elastic to use CoE but I know you can always use conservation of momentum. So, which one is the right answer?
     
  2. jcsd
  3. Dec 14, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    An ideal spring will provide a perfectly elastic collision. Sure, the collision will "occur" over some small period of time and distance that is longer than it would be for a pair of infinitely "stiff" blocks, but it will still be elastic.

    So you should ask yourself the same question assuming that a perfectly elastic collision just occurred.
     
  4. Dec 14, 2011 #3
    But I was pretty sure it was elastic. But wouln't that mean that both KE and momentum are the same?
     
  5. Dec 14, 2011 #4

    gneill

    User Avatar

    Staff: Mentor

    The totals, yes (conservation laws). But they are asking what is the same for both.
     
  6. Dec 14, 2011 #5
    Exactly, so wouldn't both KE and momentum be the same ? But there's only one right answer.
     
  7. Dec 14, 2011 #6
    Here's the problem.
     

    Attached Files:

  8. Dec 14, 2011 #7

    gneill

    User Avatar

    Staff: Mentor

    No. The blocks will have different KE and momentum. 0.5m1v12 will not equal 0.5 m2v22, and m1v1 will not equal m2v2.
     
  9. Dec 14, 2011 #8

    gneill

    User Avatar

    Staff: Mentor

    That is a different problem! The spring connection is permanent in this one. In the problem you posed, the spring was attached only to the second block.
     
  10. Dec 14, 2011 #9
    I'm sorry, I mixed up two problems. But in the problem I gave, wouldn't KE and momentum both be the same?
     
  11. Dec 14, 2011 #10

    gneill

    User Avatar

    Staff: Mentor

    No, the blocks have different masses and different velocities (both before and after the collision).

    Conservation laws apply to TOTALS for a given system, not individual elements of the system.
     
  12. Dec 14, 2011 #11
    But then velocity and acceleration can't be the same.
     
  13. Dec 14, 2011 #12

    gneill

    User Avatar

    Staff: Mentor

    Well the velocities probably won't be the same (you can calculate the final velocities if you wish to do the collision math). But why do you say the accelerations can't be the same? What forces are acting on the blocks AFTER the collision?
     
  14. Dec 14, 2011 #13
    isn't this the same principle as two people on a frictionless surface pushing each other apart? I still think it's momentum. Can you please explain using formulas why it's acceleration? I would think that acceleration for the larger mass would be less.
     
  15. Dec 14, 2011 #14

    gneill

    User Avatar

    Staff: Mentor

    AFTER the collision is complete, what are the forces acting on each block?
     
  16. Dec 14, 2011 #15
    just the force of the spring. I'm now wondering if this has to do with Center of Mass. We haven't covered that yet though, so I don't know.
     
  17. Dec 14, 2011 #16

    gneill

    User Avatar

    Staff: Mentor

    :confused: I thought that the spring was only connected to block B? The problem didn't say anything about the spring sticking to block A during the collision.

    Are you mixing up two different problems again?
     
  18. Dec 14, 2011 #17
    Oh, I'm sorry. To clear it all up, it's all in the image. it's stuck to both blocks and when the blocks are pulled apart, they contract
     
  19. Dec 14, 2011 #18

    gneill

    User Avatar

    Staff: Mentor

    You have two separate threads (with the same title) with two different problem statements. The problems statements are not the same. The original post in this thread stated that the spring was connected only to block B. Is this a separate problem or are you running two threads for the same problem?

    We cannot help you solve a problem unless the problem statement is clear and consistent. Please choose one scenario per thread and stick to it, or choose one thread and abandon the other.
     
  20. Dec 14, 2011 #19
    Very well, go to this link for the problem. https://www.physicsforums.com/showthread.php?p=3669539#post3669539
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Use Conservation of Momentum or Conservation of KE for spring problem
Loading...