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Conservation of Momentum in two dimensions

  1. Mar 1, 2007 #1

    Mitchtwitchita

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    Hey Everyone,

    I've just signed up for this forum because I'm currently taking a correspondence course and there's one problem in particular that I'm having problems with and there isn't really anybody I can ask. Can anybody here help me with this?

    Two spacecrafts from different nations have linked in space and are coasting with their engines off, heading directly towards Mars. The spacecrafts are thrust apart by large springs. Spacecraft 1, with a mass of 1.9 x 10^4 Kg, then has a velocity of 3.5 x 10^3 Km/h at 5.1 degrees to its original direction. Spacecraft 2, whose mass is 1.7 x 10^4 Kg, has a velocity of 3.4 x 10^3 Km/h at 5.9 degrees to its original direction. Determine the original speed of the two crafts when they were linked together.

    I arrived at an answer of 125 m/s. But I know that's way off! Again, can anybody help me out please?
     
    Mitchtwitchita, Mar 1, 2007
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  3. Mar 1, 2007 #2

    Doc Al

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    Staff: Mentor

    I presume that the spacecraft move on opposite sides of the original direction. (For example: if one angles north of the original direction, the other angles south.)

    Show how you arrived at your answer.
     
    Doc Al, Mar 1, 2007
  4. Mar 1, 2007 #3

    Mitchtwitchita

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    yes you are correct that they move on opposite sides of the origin.

    This is how I arrived at that answer:

    Given: m1=1.9 x 10^4 Kg
    m2=1.7 x10^4 Kg
    v1f=3.5 x 10^3 Km/h = 972 m/s
    v2f=3.4 x 10^3 Km/h = 944 m/s

    Also: v1 = v2 = Pti

    Initial Momentum:

    P1i =m1v1
    =unknown

    P2 =m2v2i
    =unknown

    Final Momentum:

    P1f =m1v1f
    =1.9 x 10^4 Kg * 972 m/s [5.1 degrees Up]
    =1.8 x 10^7 Kg*m/s

    P2f =m2v2f
    =1.7 x 10^4 * 944 m/s [5.9 degrees down]
    =1.6 x 10^7 Kg*m/s

    Since P1i + P2i =Pti and Pti = Ptf

    Therefore, Pti = P1f + P2f

    (Pti)^2 = (P2f)^2 + (P1f)^2 - 2(P1f)*(P2f)Cos11degrees
    =(1.6 x 10^7 Kg*m/s)^2 + (1.8 x 10^7 Kg*m/s)^2 - 2(1.8 x 10^7 Kg*m/s) * (1.6 x 10^7 Kg*m/s)Cos11degrees
    =2.6 x 10^14 + 3.4 x 10^14 - 5.9 x 10^14 Cos11degrees
    =2.6 x 10^14 + 3.4 x 10^14 - 5.8 x 10^14
    =2 x 10^13 (Kg*m/s)^2

    Pti = the square root of 2 x 10^13 Kg*m/s
    = 4.5 x 10^6 Kg*m/s

    Therefore, the magnitude of the total initial momentum of the rocket was 4.5 x 10^6 Kg*m/s

    P=mv
    Therefore, 4.5 x 10^6 Kg*m/s = 36000 Kg(v)
    v=4.5 x 10^6 Kg*m/s / 36000 Kg
    = 125 m/s

    Somehow I don't think I had the right approach.
     
    Mitchtwitchita, Mar 1, 2007
  5. Mar 2, 2007 #4

    Doc Al

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    Staff: Mentor

    vector addition

    Nothing wrong with using the law of cosines to find the resultant, but you have the wrong triangle and thus the wrong angle.

    Draw a diagram of the vector addition you are trying to model: Draw an arrow representing P1i (draw it making an angle of 5.1 degrees below the x-axis), then add to it an arrow representing P2i (draw it with an angle of 5.9 degrees above the x-axis). Make sure to put the tail of P2i at the head of P1i to correctly represent vector addition.

    Once you've drawn the triangle properly, then you can apply the law of cosines. What's the angle between those two sides of the new triangle?
     
    Doc Al, Mar 2, 2007
  6. Mar 2, 2007 #5

    apelling

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    Perhaps its worth pointing out that a solution can also be found by finding the components of P1f and P2f parallel to the original direction of motion. Add these components and you have the initial momentum Pti.

    The perpendicular components should be equal and opposite so do not contribute to Pti.
    It gets the same answer of course but is slightly quicker I think.
     
    apelling, Mar 2, 2007
  7. Mar 2, 2007 #6

    Doc Al

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    Staff: Mentor

    That's exactly how I would solve this problem. But there are several ways to skin this cat. To the OP: Do it both ways and compare! :wink:
     
    Doc Al, Mar 2, 2007
  8. Mar 2, 2007 #7

    Mitchtwitchita

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    Thanks Doc The new vector diagram worked!!! I was always crappy with Trigonometry! I appreciate it Doc.
     
    Mitchtwitchita, Mar 2, 2007
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