1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the direction of the momentum (by drawing?)

  1. Jan 27, 2010 #1
    1. The problem statement, all variables and given/known data
    http://img191.imageshack.us/img191/5851/lolz32.png [Broken]

    A missle explodes in the air, forming 3 pieces equal in mass. Their masses are 0,03kg.
    First piece started moving with a velocity of 60m/s, which makes its momentum
    The second piece started moving with a velocity of 55m/s, making its momentum

    The velocity directions are shown in the picture.

    Find the momentum of the third piece by constructing/drawing.
    The angle of the third piece's movement relative to vertical diretion.
    The direction of the momentum of the third piece.

    2. Relevant equations


    3. The attempt at a solution

    I have no idea how to solve this.
    I'd bet the last 2 things are practically the same thing, though.

    Thanks in advance,
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 27, 2010 #2
    Apparently you are supposed to assume that the initial momentum of the missile was 0.
    There's no way to solve this otherwise. If the inital momentum was 0 then the momemta of the 3 pieces still have to add to 0, both in the x and in the y direction.
  4. Jan 27, 2010 #3
    Oh sorry, I forgot to mention the missle stopped, and then exploded.
    I thought of that solution for a second, but wasnt sure at all. I will give it a try.
  5. Jan 27, 2010 #4
    I guess it worked. Hope I got the right answer.

    Thanks you very much good sir !
  6. Jan 28, 2010 #5
    Ok, I think I actually reached a point where I dont get it anymore.

    I added the 2 first momentas, getting the third one to be -3,45.
    But that doesnt seem to be the right answer. The right answer is supposed to be ~2,441, which you can get by Pythagoras.

    But I dont really get it. If its ~2,44, the momentums wont add up to 0, which was the missles initial momentum. I mean, arent I supposed to add the 2 vectors in order to get the third one (opposite direction ofcourse)?

    Please help.
  7. Jan 28, 2010 #6


    User Avatar

    Staff: Mentor

    Momentum (not moments) has a magnitude and direction. So, as stated in the initial help, the sum of the momentum in the x direction nas to be zero, and the sum in the y direction also has to be zero. Given the way you've drawn the figure, the 3rd momentum vector has to point up to the left, in order to cancel out the other two vectors.

    Does that help?
  8. Jan 28, 2010 #7
    Yes, I know it has to point up to the left, but why is its momentum -2,44, instead of -3,45?

    Because p1+p2+p3=0
    But in this case
  9. Jan 28, 2010 #8


    User Avatar

    Staff: Mentor

    What are the x and y components of p1?

    What are the x and y components of p2?

    What does that mean the x and y components of p3 need to be?

    What is the resulting magnitude and direction of p3? Give the answer and show it on your drawing (showing the x and y components of p3, and the resulting magnitude and direction).
  10. Jan 28, 2010 #9
    I understand now, Im a moron.

    Thanks for the help.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook