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Need quick help! - Ball swings down, hits block - Speed of the block?

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Ok I cant seem to figure this out. Maybe im tired, or strung out on coffee :), but I need some help!

    The image shows the question with diagram:

    30hqag4.jpg



    2. Relevant equations



    3. The attempt at a solution

    Iv done this a few different ways, but all WRONG. The CORRECT answer is V = 0.767 m/s

    First I found the speed of the ball by going mgh (.300)(9.8)(.500) which gave me 1.47 m/s

    Then I found the speed as it left the block after hitting it by the same method. Which would be 0.588 m/s

    So then I simply used: m1v1i + m2v2i = m1v1f + m2v2f

    (.300)(1.47) + (.200)(0) = (.300)(.588) + (.200)(x) and solve for x gave me: 1.32 m/s

    WHAT am I doing WRONG?

    Very appreciate any help on this one guys. Thanks!
     
  2. jcsd
  3. Nov 22, 2011 #2
    Personally, I would determine the potential energy that the object has at hi, and then determine the potential energy at hf. The difference between these values will equal the gain in kinetic energy of the box.
     
  4. Nov 22, 2011 #3
    Can you possibly elaborate on that for me? :)
     
  5. Nov 22, 2011 #4
    Have you explored any energy theorems in Physics yet, pertaining to potential energy and kinetic energy?
     
  6. Nov 22, 2011 #5
    Yes I have.

    So, you are saying determine the potential energy at Hi and Potential energy at Hf

    Once I have those to peices of data, what do I do with them?
     
  7. Nov 22, 2011 #6
    You will notice that at h2 the object has a smaller potential energy than it did when it was at hi; that is, energy was transferred from the ball to the block, which can be seen as kinetic energy.
     
  8. Nov 22, 2011 #7
    so is it like....

    So to calculate the potential energy at Height initial: its (m)(g)(h) ?

    same with height final? (m)(g)(h) ?
     
  9. Nov 22, 2011 #8
    Yes, assuming constant mass, we have,
    [tex]m_{ball}g(h_1-h_2)=\frac{1}{2}mv_{block}^2[/tex]
     
  10. Nov 22, 2011 #9
    Is that expression saying that the difference in potential energy = the same ammount in Kinetic energy?
     
  11. Nov 22, 2011 #10
    In words, the loss of energy depicted by the swinging ball can be accounted for by the gain in kinetic energy of the box. The potential energy of the ball changed did it not? Law of of conservation of mass implies the energy of the system remains constant, and hence is merely a transfer from the ball to the box.
     
  12. Nov 22, 2011 #11
    So I have then kinetic energy of .882 applied to the block.

    How do I use that to find the blocks speed?
     
  13. Nov 22, 2011 #12
    What is the formula for kinetic energy? I already provided it to you.
     
  14. Nov 22, 2011 #13

    gneill

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    Staff: Mentor

    mgh yields potential energy in Joules, not speed. You need to use the kinetic energy equation relating velocity to KE in order to find the speed that comes from converting this PE to KE.
     
  15. Nov 22, 2011 #14
    KE = 1/2mv^2
     
  16. Nov 23, 2011 #15
    How do you do this? :( Relating Velocity to KE?
     
  17. Nov 23, 2011 #16

    gneill

    User Avatar

    Staff: Mentor

    See your post immediately above! KE = (1/2) MV2.

    If some potential energy PE = Mgh is converted to kinetic energy KE = (1/2) MV2, then that means KE = PE.

    Mgh = (1/2) MV2. Solve for V.
     
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