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Center of mass and distance

  1. Nov 1, 2009 #1
    A 500-kg cannon and a supply of 59 cannon balls, each with a mass of 19.0 kg, are inside a sealed railroad car with a mass of 50000 kg and a length of 41 m. The cannon fires to the right; the car recoils to the left. The cannon balls remain in the car after hitting the wall. After all the cannon balls have been fired, what is the greatest distance the car can have moved from its original position?What is the speed of the car after all the cannon balls have come to rest on the right side?




    xcm= (m1x1 + m2x2) / (m1+m2)



    I tried to solve for the left part of the equation and then the right but I don't know how to set it up, please help!
     
  2. jcsd
  3. Nov 1, 2009 #2

    Doc Al

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    If you measure the center of mass of the railroad car and its contents from the left side of the car, what's the most it can change?

    If you measure the center of mass of the railroad car and its contents with respect to the tracks, how much does it change?
     
  4. Nov 1, 2009 #3
    i set up the equations as (5000(x1) + 1121(41)) = xcm but i dont know what to use for the x's and also what to use for the next equation to find the difference between the two
     
  5. Nov 1, 2009 #4
    i mean 1621 instead of 1121
     
  6. Nov 1, 2009 #5

    Doc Al

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    Measure everything from the left edge of the car. Assume the car is uniform (although it doesn't matter). Assume the cannon is on the left side of the car. The cannon balls move from the left side to the right side. Compute the center of mass for each case and compare. (Don't forget to divide by the total mass.)
     
  7. Nov 1, 2009 #6
    so xcm= total mass of everything + the right side of the car divided by the total mass

    xcm= 2121kg (x1) + 19kg (41m) / 2121 + 19

    and then continue this for each time a cannon ball is fired

    I am still confused I don't comprehend how this equation works
     
  8. Nov 1, 2009 #7

    Doc Al

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    xcm = (xcar*mcar + xballs*mballs + xcannon*mcannon)/mtotal
     
  9. Nov 1, 2009 #8
    ok i got that, but still I have a problem figuring out what x's to use for each case

    I would say that the movement of the cannon balls would be 41 m but we don't know that for the car or the cannon
     
  10. Nov 1, 2009 #9

    Doc Al

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    The cannon balls are the only things that move, so it really doesn't matter what you put for the others. (But I gave you some suggestions in post #5.) All you really care about is how the COM changes when the balls move.
     
  11. Nov 1, 2009 #10
    well i set up the equation (5500 (x1) + 0) /6621 for the left side when the cannon balls haven't not moved and for the right i set up the equation 5500(x1) +1121(41) / 6621 but then I don't have x1 and also when i figuring that out I don't come up with the right answer
     
  12. Nov 1, 2009 #11

    Doc Al

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    You don't need x1 to find how the center of mass changes. What did you get when you subtracted those two? (You'll see that the terms with x1 cancel out.)
     
  13. Nov 1, 2009 #12
    I got 6.942 m but that is not correct
     
  14. Nov 1, 2009 #13

    Doc Al

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    Oops. It looks like you have the mass of the car (and thus the total mass) incorrect:
    Is the mass 50 000 kg or 5 000 kg?
     
  15. Nov 1, 2009 #14
    Thanks so much!
     
  16. Nov 7, 2009 #15
    So what did you end up doing that worked? I have the same problem and I just can't get it.
     
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