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Conservation of Momentum of a bomb shell

  1. May 2, 2013 #1
    I must be thinking way to hard about this problem. Can someone please give me a nudge in the right direction?
    1. The problem statement, all variables and given/known data
    A shell traveling with speed v0 exactly horizontally and due north explodes into two equal mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed v0. What is the velocity of the other fragment?

    2. Relevant equations

    Let M=m1+m2 (the mass of the shell before explosion equals the sum of the masses of the fragments)

    Conservation of momentum: v0*M=v0*m1+V2*m2 (where V2 is the vector pointing in the x direction and v0*m1 is the momentum of the fragment traveling directly upwards)

    3. The attempt at a solution

    When I work this out, everything pretty much cancels out and V2 = v0. Except I think this is too easy. When I try to shift this problem so that the initial velocity of the shell is traveling in the x direction, and the fragments both exit with an x and y component with magnitudes at a 90 degree angle to each other, I get a completely different answer (V2=(2*sqrt(2)-1)*v0). And this doesnt seem right either.

    Please give me a hand
     
  2. jcsd
  3. May 2, 2013 #2

    ehild

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    Momentum is vector, and conservation of momentum holds for each component. Write an equation for both the horizontal and vertical components of the shell and its pieces.


    ehild
     
  4. May 2, 2013 #3
    Your first attempt was correct. It does cancel out each other...what is your doubt?
     
  5. May 2, 2013 #4

    ehild

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    Once vo is the horizontal velocity the shell travels before explosion. After explosion, one piece travels vertically up with speed vo. You can not add horizontal and vertical components of vectors.

    ehild
     
  6. May 2, 2013 #5
    Ah..now I get you...ya, echild is right...gotta use vector or cartesian way to add them...
     
  7. May 2, 2013 #6
    ok so I got V2=vo*cos(45) in the x direction. Does that sound reasonable?
     
  8. May 2, 2013 #7

    SammyS

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    What do you mean by the x direction?

    Usually, x is horizontal and y is vertical. For this problem that would mean that the xy-plane is vertical with the x-axis pointing North (or South).

    Please define your coordinate system.
     
  9. May 2, 2013 #8
    Yes, I am using the general cartesian coordinate system where Y is north and X is east. My original velocity vector (before explosion) is pointing 45 degrees from horizontal. The mass m1 has a velocity exactly north, and I have extrapolated that the mass m2 has a velocity pointing exactly east.
     
  10. May 2, 2013 #9

    SammyS

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    If the y-axis points North and the x-axis points East, then what is up, the vertical direction? That must be the z-axis.

    The initial velocity vector, v0, is due North and horizontal, according to the given information.For the set of coordinate axes you have chosen here, that means that v0 is purely in the y direction.

    If v1 is the velocity of the fragment that goes vertically up, (magnitude of vector, v1, is v0, the magnitude of vector, v0) then v1 is purely in the z direction.

    Neither of these two velocities has any component Eastward, i.e. neither has an x component.


    You may want to rethink your choice of coordinates.

    The coordinate system usually used to describe this situation has the x axis pointing North, and the y-axis pointing vertically.
     
  11. May 2, 2013 #10
    forget the coordinate system. I can work in any conventional or non conventional set of coordinates. But that is not the issue right now since V1 and V2 are exactly perpendicular to each other. All I want to do is find the value of V2 if the vector V1 is 90 degrees perpendicular to V2 and the original (Pre explosion) velocity vector has these components (which are perpendicular to each other).
     
  12. May 2, 2013 #11
    This is a 2 dimensional problem not 3 dimensional.
     
  13. May 2, 2013 #12

    SammyS

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    Yes. That's is consistent with what I said in post #7, part of which I "Quote" below.

    So, as I suggested, choosing the x axis to point North, and the y-axis to point up (vertically) will express the problem in two dimensions. Everything occurs in the xy plane.
     
  14. May 2, 2013 #13
    Ok. Now that that's out of the way, I have already solved the problem using vector notation in the x-y-plane, as I stated in my original post. If you know weather or not V2=(2*sqrt(2)-1)*v0, is correct or not, it would actually help me.
     
  15. May 2, 2013 #14

    SammyS

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    I can't comment on that answer until I see how you got that answer.


    Up to this point, your choice of axes has been unclear at best, and perhaps inconsistent. At some point in the above discussion, you have a fragment heading east and the x-axis pointing east. (None of the velocities or momenta have any components in the eastward direction.)

    I'm not trying to be a pain or be condescending, it's just that until your coordinate system is well-defined and you give components of the velocities and/or momenta in terms of that coordinate system, it makes no sense to deal with any other details of the problem. Once a relevant coordinate system is established, the problem you are trying to solve is not too difficult.
     
  16. May 2, 2013 #15
    i will post a picture. I re-did it this morning and it should be more clear.
     
  17. May 2, 2013 #16
    here it is. Please ignore the center of mass problem on the bottom (problem 3.15). As you can see, I chose the original velocity vector to be in the x-y-plane, with x and y components. The resulting fragments each have one of the components. [Broken]
     
    Last edited by a moderator: May 6, 2017
  18. May 2, 2013 #17
    I guess that didnt work. The picture is here: http://imgur.com/0WNwbtA [Broken] .
    -Thx
     
    Last edited by a moderator: May 6, 2017
  19. May 2, 2013 #18

    SammyS

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    Yes, that image is way to large to display in a post.

    It turns out that the coordinate system is the issue here.

    You show that the initial velocity vector, v0, has equal x and y coordinates.

    According to the problem statement
    The "... shell [is] traveling with speed v0 exactly horizontally and due north ... ".​
    Therefore, v0 will have only one non-zero component if the x-axis points due North.

    Please define your coordinate system.
     
    Last edited by a moderator: May 6, 2017
  20. May 2, 2013 #19
    I shouldn't have to. I can just work with vector components.
     
  21. May 2, 2013 #20

    SammyS

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    It's impossible to give components of v0 without specifying a coordinate system.

    If you would like, I can specify a coordinate system.



    ... Come to think of it I've already given two systems, one of which I said I preferred.
     
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