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Momentum in Two Dimensions

  1. Jul 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Object 1:
    m1=1.9x10^4 kg
    v1'= 972 m/s [E5.1*N]

    Object 2:
    m2=1.7x10^4 kg
    v2'=944 m/s [E5.9*S]

    Determine their original speed when they were linked together.

    2. Relevant equations

    Pti = Ptf
    m1v1 + m2v2 = m1v1' + m2v2'
     
    Last edited: Jul 22, 2009
  2. jcsd
  3. Jul 21, 2009 #2

    Dick

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    You want to do a problem like this by expressing the velocities as vectors like (x,y), where x is the east component and y is the north component. Add the two final momentum vectors and equate that to the initial momentum vector and solve for the vector v. The to find the speed find the magnitude of v. I have can only guess what your direction information (like [E5.1*N]) means. 5.1 degrees east of north?
     
  4. Jul 21, 2009 #3
    By E5.1N, I mean 5.1 degrees north of east; the two craft were traveling together in the east direction initially.

    So, for a trigonometric solution, I would add p1' and p2' vectors, apply the cosine law to find p'. Then I would equate p' with v(m1+m2) and solve for v?

    Is my algebraic solution in the original post fine? I will try the trig solution right now.
     
  5. Jul 21, 2009 #4
    Like I said, im not very good at trig so my attempt for a diagram got me even more confused (i dont think im doing it right). This is what I came up with:

    http://img199.yfrog.com/img199/8992/vectoraddition.jpg [Broken]

    where

    p1' = m1v1'
    =(1.9x10^4)(972)
    =1.8x10^7

    p2' = m2v2'
    =(1.7x10^4)(944)
    =1.6x10^7

    p1' is at an angle of 5.1, p2' is at an angle of 5.9 (which is in the triangle because of the alternate pattern) and 180-5.1-5.9=169.

    I know im wrong, but I hope that by showing my work, you can help me get things right.
     
    Last edited by a moderator: May 4, 2017
  6. Jul 21, 2009 #5

    Dick

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    You don't need a cosine law to add vectors. You just add them. But I may have been misjudging the level of your course. Have you done vectors? If you know the initial direction of both craft is east then you are doing the right thing. You are equating the eastward components of momentum before and after the collision. But I get an answer of more like 954m/s. You might not be keeping enough decimal places. You have the velocities to three decimal places. Try keeping three decimal places. But to two decimal places, your first answer is correct.
     
  7. Jul 21, 2009 #6

    Dick

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    Your diagram is sort of correct. But you can't equate the 5.9 angles unless you know the base is horizontal. And you aren't using vectors in your solution. But like I said before, I don't know that you are doing anything wrong. Did they tell you that both craft were heading exactly east before the separation?
     
    Last edited by a moderator: May 4, 2017
  8. Jul 21, 2009 #7
    Yes, we have covered vectors before. I also get 954, since before I rounded off the sum of m1v1' and m2v2'. I'm glad that I at least have the answer right.

    Yes, the picture (of the problem) I have shows it. I'm happy that my triangle ediagram is correct.

    Oh. I guess when you said "Add the two final momentum vectors and equate that to the initial momentum vector and solve for the vector v" that was before you knew they were traveling east? If that's the case, how would I approach a graphical trig solution?
     
  9. Jul 21, 2009 #8

    Dick

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    Yes. You didn't tell us that. If they hadn't you should work it with vectors.
     
  10. Jul 21, 2009 #9

    Dick

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    You would work out the two components of the vectors you labelled p1' and p2', add them and figure out the two components of the vector you labelled p'. Then find it's length.
     
  11. Jul 21, 2009 #10
    Oh, so I would need 3 vector diagrams (one for x dimension, one for y, then the last one combining them)? It was my understanding that i added p1' and p2' without separating them into their components.
     
  12. Jul 21, 2009 #11

    Dick

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    Your diagram has two components, horizontal and vertical distance. A two dimensional vector is written like (x,y). Are you sure you studied vectors?
     
  13. Jul 21, 2009 #12
    Yes, though a bit rusty (been a long time). But what I mean about the 3 diagrams is:

    First Diagram: p1x' + p2x' = px'
    Second Diagram: p1y' + p2y' = py'
    Third Diagram: px' + py' = p'

    I hope that makes sense and I hope I follow you.
     
  14. Jul 21, 2009 #13

    Dick

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    No, no. They are all the same diagram. You ARE rusty. Can you split the momentum p1' for Spacecraft 1:
    m1=1.9x10^4 kg
    v1'= 972 m/s [E5.1*N]

    into (E,N) components?
     
  15. Jul 22, 2009 #14
    Ohhh.

    So it would be:
    First: p1' = p1x' + p1y
    Second: p2' = p2x' + p2y'
    Third: p' = (p1x' + p2x') + (p1y' + p2y')

    Splitting them up into the components would be:
    x= (1.9x10^4)(972cos5.1)
    y= (1.9x10^4)(972sin5.1)

    I hope I didnt' screw anything up here, otherwise i really need to brush up.
     
  16. Jul 22, 2009 #15

    Dick

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    Right. The momentum vector for p1' is ((1.9x10^4)(972cos5.1),(1.9x10^4)(972sin5.1))*kg*m/s. But I'm still not sure you need to solve this problem that way.
     
  17. Jul 22, 2009 #16
    Thanks for the help. I just wanted to know how to do the trig solution, I don't really need to solve it that way for the assignment.
     
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