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Simple Harmonic Motion

  1. Jan 21, 2007 #1
    1. The problem statement, all variables and given/known data

    http://www.physics.umd.edu/courses/Phys122/Noyes/HW/hw1.pdf (pg 3)

    2. Relevant equations

    Im guessing that E = K + U has something to do with it


    3. The attempt at a solution

    Not really sure where to start..I tried a few approaches but none of them got me anywheres E = 1/2mv^2 + 1/2 kx^2..but im missing too many things..

    Any help?
     
  2. jcsd
  3. Jan 21, 2007 #2
    when the bullet colides, it sticks. This means that energy is not conserved but something else is.
     
  4. Jan 21, 2007 #3
    by the way, the fact that the clay sticks and that the collision is therefore perfectly inelastic is one of those "assumptions"
     
    Last edited: Jan 21, 2007
  5. Jan 22, 2007 #4
    conservation of linear momentum?
     
  6. Jan 22, 2007 #5
    yes. now you can use that in combination with energy.
     
  7. Jan 22, 2007 #6
    so I have [tex]\frac{mv +Mv_2}{m+M} = v_f[/tex] where [tex]Mv_2[/tex] is going to be 0 since it is at rest?

    and [tex]E=1/2mv^2 + 1/2kx^2[/tex]

    Do i let m = m+M? and will 1/2kx^2 be 0 since the block can be considered to be at point x=0?...set it to zero and solve for v?
     
  8. Jan 22, 2007 #7
    you actually dont need to use energy untill you are ready to find the amplitude. You already have enough information to find the velocity.
     
  9. Jan 22, 2007 #8
    So imediately after the collision, the spring has no effect on the velocity?
     
  10. Jan 22, 2007 #9
    that's correct. two ways to understand that
    1) imediately after colision, spring is not compressed and therefore applies no force.

    2) even if it were compressed, a force causes an acceleration which only changes velocity of some period of time.
     
  11. Jan 22, 2007 #10
    Solved it..Thanks for the help
     
  12. Jan 22, 2007 #11
    good job, np
     
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