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Homework Help: Physics question answer needed by 2-19-06

  1. Feb 17, 2006 #1
    I have a few questions that I have to have answered by 2-19-05 for a class on the following day. I have tried to solve the problems and have looked in several physics books for help but really have no idea as to the answers. If you can help, I would greatly appreciate it. Again, I am basically physics illiterate (hence the name) so you will need to put your reply in basic terms. Thanks .

    There is a rocket. The total momentum is zero relative to the station where the rocket is taking off from. The pilot fires the engine for just a few seconds, converting a mass "Mfuel" from liquid to a gas that shoots off into space behind the ship at a speed "Vfuel." A person at the station sees a flash of gas shoot off to the left and the ship moves to the right.

    1. If we know "Mship", "Mfuel", and "Vgas" we can figure out what "Vship" is. let "Mship"=1000kg, "Mfuel"=5kg, and "Vgas"=-5000m/s. What was the ship's speed (V) to the right when it left the dock?

    2. A big rocket at launch has a huge ball of flaming gas that is deflected by the ground at the beginning. Does the gas hitting the ground have anything to do with the rocket taking off? Think carefully here--is this really any different than the ship leaving the station--was there anything to "push against" there? Explain.

    3. We develop this as a two object system (ship and fuel-gas). How would you explain a rocket ship to another student who has seen a model rocket, bottle rocket or fireworks shot into the air and wonders how that works. Did you ever blow up a balloon and just let it go?

    Again, I really need these answers and explinations by 2-19-06. Thanks for any and all help on this.
  2. jcsd
  3. Feb 17, 2006 #2
    Do you understand the concepts of momentum?
  4. Feb 17, 2006 #3
    momentum concepts

    Yes, I have the basics. My physics book goes over that part well. My teacher said that these problems have something to do with Newtons 3rd Law....I know that this law has to do with 2 objects (when one object exerts a force on a second object, the second object exerts an equal force int he opposite direction on the first object). Basically, action force and reaction force. I also get the part where the ship is moving in one direction and the gas is moving in the opposite direction. It is the actual setting up the math problem that I am having difficulty with in #1.
  5. Feb 17, 2006 #4
    For #1 question I have: 1000kg X ? + 5kg X -5000m/s = 0

    This is what I get from the problem. I am not sure if this is correct or what to do with it even if it is correct.
  6. Feb 17, 2006 #5

    Well it sounds like you understand the concepts well enough so it shouldn't be too hard of a problem, you just need to remember that the momentum of the system is always conserved, so if you can figure out what the initial momentum is you can find what the velocity of the rocket afterwards must be in order for the momentum to remain constant.
  7. Feb 17, 2006 #6
    What is the mathematical definition of momentum? What two things make it up?
  8. Feb 17, 2006 #7
    momentum = mass X velocity

    The problem stated that "at the dock the rocket includes the vehicle and the fuel that will be burned sitting at rest, so the total momentum is zero relative to the station.

    Not too sure what this means but I think it is important info otherwise it wouldn't be there???
  9. Feb 17, 2006 #8
    the momentum before = the momentum afterward soooo,

    0 = 0

    But I don't know what to do with the other numbers in #1
  10. Feb 17, 2006 #9
    Ok and what is always true about momentum, it's conserved right? So can you come up with expressions for the total initial and final momentum in terms of the initial and final velocities of the rocket and its fuel?
  11. Feb 17, 2006 #10
    Conserved, yes. If the object interact with only one another, each object can have its momentum changed in the interaction, provided that the total momentum after it occurs is the same as it was before. The expressions part is where I am lost. I have the basics but the actual setting up of the math problem is where it all goes bad.
  12. Feb 17, 2006 #11
    Well in the first case since they are one object you can consider the expression

    (m_ship + m_fuel)v_shipandfuel = 0 because they are just sitting on the platform.

    Can you figure out what the case has to be after they start moving in oppsoite directions?
  13. Feb 17, 2006 #12
    Okay...I might have something.

    m_fuel = 5kg
    v_gas= -5000m/s
    m_ship= 1000kg

    5kg x -5000m/s divided by 1000kg

    = -25000 divided by 1000

    = -25

    Am I totally off base or at least in the ballpark??
  14. Feb 17, 2006 #13
    m_ship = 1000kg + m_fuel = 5kg = 1005kg

    I dont know how to get the v_ship and fuel number

    When they begin to move in opposite directions one is a positive number and the other is a negative number so they cancell eachother out to equal zero.
  15. Feb 17, 2006 #14
    You're really close you just need to remember that momentum is a vector and direction matters they both can't have negative velocity otherwise the momentum wouldn't be conserved.

    you have something along the lines of

    let m = mass of fuel
    let M = mass of rocket

    let v = velocity of fuel
    and let V = velocity of the rocket.

    Now since momentum is conserved and we know that the initial momentum was zero we know that

    0 = mv + MV this is just a statement that momentum is conserved in this system.

    You know m, M, and v so you can find V.
  16. Feb 17, 2006 #15
    That was just an example of an expression for the initial momentum, which is obviously zero since neither object is moving thus the velocity must have been 0.
  17. Feb 17, 2006 #16
    5kg (-5000m/s) + 1000kg (V)

    -25000 + ??

    I cannot figure out how to find Velocity or rocket. I keep thinking to divide the -25000 by 1000 to get an answer of -25 but this still seems wrong
  18. Feb 17, 2006 #17
    sorry...Velocity of rocket
  19. Feb 17, 2006 #18
    I already gave you the equation which you have to use, you can't solve for anything the way you wrote it, you need to have an equation.
  20. Feb 17, 2006 #19
    how about 5kg(-5000m/s) + 1000kg(25) = 0

    ???????????????????? Maybe??????????????
  21. Feb 17, 2006 #20
    Yes, you see you can't get anything from saying

    mv + MV

    you have to say

    mv + MV = 0

    Yes the answer should be 25m/s.
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