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Rocket maximum velocity

  1. Jan 7, 2015 #1
    1. The problem statement, all variables and given/known data
    The whole problem is in picture, but simply it says to find the max velocity of rocket, with those given units.
    The whole problem is worked out with pounds, but when I try to do it for meter and newtons I get like 50 times bigger answer.
    The units :
    Total Impulse of motor: 2.5 Newtons-second or 0.5056 pound-second
    Duration of thrust: 0.42s
    Burnout weight of model: 25.7 grams or 0.0787 pound
    2. Relevant equations
    the equation to find max is
    vmax = ( total impulse of motor x g ) / burnout weight of model

    3. The attempt at a solution
    I tried it like ten times but I get around 600 meters per second, and I have no idea why.
    I did converted the grams to kilograms as are the newtons, and I bet that there is small thing that I overlooked so I came here. HELP
    Thanks
     

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  3. Jan 7, 2015 #2

    Bystander

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    Okay, that's the burnout mass --- now, what's the burnout weight?
     
  4. Jan 7, 2015 #3
    well, they got this by subtracting the propellant weight( 3.5g) from liftoff weight of model (39.2g)
    And in the example which they did with pounds they just use those values mentioned above.
     
  5. Jan 7, 2015 #4

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    "Weight," a unit of force, not to be confused with mass as so easily occurs when working with English units.
     
  6. Jan 7, 2015 #5
    The book is "Handbook of model rocketry" and it is into, so I don't think the author even considered it and made it simple, therefore it does not really mater as he uses those values and plugs them into the equation. My problem is that I don't get the right answer when pluging the IS units
     
  7. Jan 7, 2015 #6

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    SI mass x 9.8m/s2, at earth's surface gives you an SI force in Newtons.
     
  8. Jan 7, 2015 #7
    here is what i tried
    (2.5 N per seconds)( 9.8 m/s2 that is square) / 0.0357g
    and the answer I got is 686.97 meter per second,
    but the right answer should be around 62 meter per second ( I know that as I converted the 206.86 feet per second(answer that was in the book) to meter per second)
     
  9. Jan 7, 2015 #8

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    The 25.7 g is a mass. It has to be multiplied by g to give you a weight.
     
  10. Jan 7, 2015 #9
    EDIT: Typo the it is supposed to be 35.7 not 25.7. Sorry for mistake.
    so...
    (2.5 newtons per second x 9.8 meter/s2) / ( .0357 x 9.8)
    so i get 70
     
  11. Jan 7, 2015 #10

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    Happier with the numbers now?
     
  12. Jan 7, 2015 #11
    not realy, but thx
     
  13. Jan 7, 2015 #12
    But why in the example the author did not multiplied by g too? why did he kep it same?
     
  14. Jan 7, 2015 #13

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    It could be a typesetting error, it could be something he thought he'd explained somewhere else in the associated text, or a proofreading error. Haven't read the book, and can't really say.
     
  15. Jan 7, 2015 #14

    SteamKing

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    There's a couple of things to clear up.

    1 Newton = 1 kg-m/s2 = 0.22481 pounds force

    so an impulse of 2.5 Newton-seconds = 2.5 * 0.22481 = 0.5620 lbs-s, not 0.5056 lbs-s as printed in the text.

    The burned out mass of the rocket = 35.7 grams ( = 0.0357 kg), which is equivalent to a weight of 0.0787 lbs (there are 453.6 grams in 1 pound)

    The formula for the max. velocity of the rocket is vmax = Impulse / (burned out mass), which using SI units would be
    vmax = 2.5 N-s / 0.0357 kg = 70 m/s [Notice that the gravitational acceleration g is not needed here because you are using the burned out mass of the rocket]

    If you are not using SI units, then the formula for vmax = Impulse / (burned out mass) = Impulse / (burned out weight / g) = (Impulse * g) / (burned out weight)

    So, in imperial units, vmax = 0.5620 lbs-s * 32.2 ft/s2/0.0787 lbs = 230 feet/s = 70.1 m/s
     
  16. Jan 7, 2015 #15
    THANK YOU!!
     
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