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Spring and conservation of energy

  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data
    a spring is pulled back .2 meters and a ball is fired. the ball travels 5 meters before stopping 1.5 meters bellow it's firing position which was on a table. there is a picture that I have drawn and attached in a pdf. I have also listed all the givens except the mass. i forgot to do that one. the mass is given and is 1kg.


    2. Relevant equations
    conservation of energy.
    1/2kx^2 = 1/mv^2

    3. The attempt at a solution
    also attached in the pdf. i figured it would be easier since there is a picture and lots of equations.

    only allowed to use conservation of energy and momentum. also straight line motion stuff. no calc.
     

    Attached Files:

  2. jcsd
  3. Oct 27, 2009 #2

    Doc Al

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    Staff: Mentor

    Looks OK to me. You started out with: 1/2kx2 + mgh = 1/2mv2, which is true but not helpful. You then switched to 1/2kx2 = 1/2mvx2, which is better.
     
  4. Oct 28, 2009 #3
    but the big assumption i made was that tx=ty here. the ball travels on the surface of the table and then falls off.... so only then does it start traveling in the y direction. so wouldn't the ty timer start ticking when this happens? which is long after the tx timer has started... or do both timers start when the projectile starts and the y direction just breaks into two problems with different accelerations?
     
  5. Oct 28, 2009 #4

    Doc Al

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    Staff: Mentor

    I think I see what you're asking. When you compute the speed using Vx = Δx/ty, the Δx should be measured from the table edge (where the ball leaves the table), not the starting point back at the spring. And ty is the time it takes to fall, also measured from the moment the ball leaves the table.
     
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