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Conservation of energy, how far must a spring be compressed

  1. Nov 10, 2012 #1
    Two children are playing a game in which the try to hit a small box on the floor with a marble fired from a spring loaded gun mounted on a table. The target box is horizontal distance D=2.20 m from the edge of the table. Bobby compresses the spring 0.011m but the center of the marble falls 0.27 m short of the center of the box How far should Rhoda compress the spring to score a direct hit? Assume that neither the spring nor ball encounters friction in the gun.

    So I know that the ball will leave the gun with initial kinetic energy of 1/2mv^2 = the 1/2 k Δx^2 of the springs potential energy. I tried isolating velocity and using that in my kinematic equation d=v(t)+1/2at^2 and solved for time thinking that maybe I could find the height of the table and find the right velocity from there then work backwards to spring potential energy, but that didn't work because I don't know k or m and am not sure if I as n the right track
     
  2. jcsd
  3. Nov 10, 2012 #2

    Doc Al

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    Staff: Mentor

    By what factor must the initial velocity increase to reach the box?

    Hint: Make use of the fact that initial velocity is horizontal in all cases.
     
  4. Nov 10, 2012 #3
    New velocity= 1.14(d)/t I can plug that back into my kinetic=spring potential but I still have too may unknowns
     
  5. Nov 10, 2012 #4

    Doc Al

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    Think in terms of ratios. To increase the velocity by a factor of 1.14, by what factor must Δx increase?
     
  6. Nov 10, 2012 #5
    1.14

    So are you saying that all I had to do to solve this was make a ratio of it all? That sounds far too easy
     
  7. Nov 10, 2012 #6

    Doc Al

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    Yep.
    Sometimes a problem looks harder than it is. :smile:
     
  8. Nov 10, 2012 #7
    Wow...thank you
     
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