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Astronaut momentum problem

  1. Mar 4, 2007 #1
    A 60kg astronaut floating in space simultaneously tosses away a 14-kg oxygen tank and a 5.8kg camera. The tank moves in the x direction at 1.6 m/s, and the astronaut recoils at .85m/s in a direction 200degrees counterclockwise from the x axis. Find the velocity of the camera.

    I can't find a way to start off this problem. I drew a diagram and everthing.
     
  2. jcsd
  3. Mar 4, 2007 #2

    hage567

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    Homework Helper

    Well, you know momentum must be conserved. So what is the total momentum before the toss and after? Find the expressions for each direction.
     
  4. Mar 4, 2007 #3


    The total momentum before the toss = 0 because there is no velocity.
    After the toss,
    Ptotal=Pastro+Ptank+Pcam
     
  5. Mar 4, 2007 #4
    Resolve the momenta into components.
     
  6. Mar 4, 2007 #5
    mtotvtot=mastrovastro+mtankvtank+mcamvcam

    like that?
     
  7. Mar 4, 2007 #6
    Nope, along the x and y axes.

    Since the total momentum is conserved, they must be conserved along the axes, too. So
    initial momentum along x axis = final momentum along x axis. Similarly for y.
     
  8. Mar 4, 2007 #7
    x: mtvtx=mavax+mcvcx+mtankvtankx
    y: mtvty=mavay+mcvcy

    i think this is what you mean right?
     
  9. Mar 4, 2007 #8
    Yes, that's right.

    Make sure you get the signs right when putting in the values of the velocities.
     
  10. Mar 4, 2007 #9

    ok, so i'm solving for the velocity of the camera but i don't have the total momentum, so how can i solve for the camera's velocity with two unknowns?
     
  11. Mar 4, 2007 #10
    What two unknowns? Read the question again. There's only one unknown per equation, and you can solve for them with the given info.
     
  12. Mar 4, 2007 #11
    Total momentum, and then the final velocity for the camera.

    Can i say

    -mavax=mcvcamx+mtankvtankx ???
     
  13. Mar 4, 2007 #12
    Sure you can. The initial momenta along both directions is zero, remember?
     
  14. Mar 4, 2007 #13

    right, but i thought i was solving for the final velocity
     
  15. Mar 4, 2007 #14
    Yes you are. It is because of the zero initial momentum (mtvtx = mtvty = 0) that you are able to write the equation as your previous post.
     
  16. Mar 3, 2008 #15
    can u say ( also have this problem, and this is how i was working it out)

    m_ast*v_ast*cos(200) + m_tank*v_tank + m_cam*v_cam*cos(20) = 0

    these two equation actualy equal each other (produce same answer)

    -mavax=mcvcamx+mtankvtankx

    all withrespect to the x axis

    can someone explain...
    why do you have to use the y axis or even worry about the x axis in this problem... you can relate teh whole thing to the x component w/ trig?

    x: mtvtx=mavax+mcvcx+mtankvtankx
    y: mtvty=mavay+mcvcy
     
    Last edited: Mar 3, 2008
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