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Block and the bullet

  1. Apr 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Which equation could adequately be used to determine how high the block goes after being hit by the bullet (a height h)? (see figure)


    (m+M)gh+ksh=1/2(m+M)V^2
    (m+M)gh+1/2ksh^2=1/2(m+M)V^2
    (m+M)v+ksh=(m+M)V

    2. Relevant equations
    KE=1/2mv^2
    Ugrav=Mgh
    PE of spring = 1/2ks(s^2 final - s^2 initial)

    3. The attempt at a solution

    Based off of the equations, I believe the answer should be the second equation. I don't understand what else it could be..

    I have only one attempt
     

    Attached Files:

  2. jcsd
  3. Apr 9, 2015 #2

    SammyS

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    Some of the variables in those equations need to be definrd for us.

    I see L1, m, v, and h in the figure.

    What are k, s, M, V, an s without k ? Is that ks rather than k⋅s ?
     
  4. Apr 9, 2015 #3
    Maybe you could explain why you think it couldn't be the other two equations you excluded?
     
  5. Apr 9, 2015 #4
    Because the other two equations are missing components to their equations
     
  6. Apr 9, 2015 #5
    1/2ks(s^2 final - s^2 initial)

    ks = spring constant (k_s_)
    (s^2 final - s^2 initial) = final stretch - initial stretch
    v= velocity
    m=mass1
    M=mass2
     
  7. Apr 9, 2015 #6
    And what components do these two equations have that are missing that the other one isn't?
     
  8. Apr 9, 2015 #7
    (m+M)gh+ksh=1/2(m+M)V^2 the spring constant should be 1/2ksh^2 not ksh
    (m+M)gh+1/2ksh^2=1/2(m+M)V^2
    (m+M)v+ksh=(m+M)V Ugrav should be (m+M)gh instead of velocity multiplying the masses (m+M)v, also (m+M)V is missing V^2
     
  9. Apr 9, 2015 #8
    I think you have your answer then.
     
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