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Travelling object that explodes into 3 separate pieces

  1. Oct 31, 2015 #1
    1. The problem statement, all variables and given/known data
    An object, initially travelling southeast with a speed of v_0, explodes into three separate pieces. A fragment of mass m_1 travels due south with a speed of v_1. Another fragment, with a mass of m_2 travels travels northeast at v_2. The last fragment has a mass of m_3, but travels in at an unknown rate in an unknown direction.
    What is the magnitude and direction of the third fragments velocity? (Only variables allowed are v_0, v_1, v_2, m_1, m_2, m_3)

    2. Relevant equations
    Well, I think that due to conservation of momentum, the initial MV should equal the final MV.
    And I think p=mv will also be useful.

    3. The attempt at a solution
    So I started off by drawing a picture to try and visualize the problem. I have the vector going southeast (-45degrees), then from there I drew the first fragment going south (so -another 45degrees to end up in the -90degree direction), and then I drew the second fragment going northeast (+45degrees) which I noticed also happens to be perpendicular to the initial velocity direction.

    So, now I have to figure out what direction the third piece is going and how fast.

    I decided to go with P_initial=P_final, and p=mv. So...
    (M_initial)(V_0)=(m_1)(v_1)+(m_2)(v_2)+(m_3)(v_3)
    M initial is just m_1+m_2+m_3, and I want to find v_3 so I changed it up to.

    ((M_i)(v_0)-(m_1)(v_1)-(m_2)(v_2))/(m_3)=v_3

    Since those are the variables I can use for the solution, that's where I stopped for that.

    For the direction, I wasn't entirely sure but I think the last piece continued going off in the southeast direction at the new speed.

    I have a test this week, and this is practice for that test. There is however no way of knowing if I'm correct or wrong on ANY of these problems because my professor is a very busy man and takes up to a week to answer emails, and waiting until next class seems wasteful. So, if anyone here can help me out and tell me what I'm doing right, or wrong, or just any advice at all it will be very appreciated.
     
  2. jcsd
  3. Oct 31, 2015 #2

    mfb

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    The velocities are vectors, v_3 will have some direction. You can express the angle as function of this v_3 only, but it is probably easier to consider the velocity components (south/north and east/west) separately.
    The direction will depend on the given quantities (m1, m2, m3, v0, v1, v2) , the answer will include them in some way.
     
  4. Oct 31, 2015 #3

    Simon Bridge

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    You can try breaking the vectors into components or sketch the vectors head to tail.
     
  5. Oct 31, 2015 #4
    Ooohhh. I had completely forgotten about breaking vectors into components.

    Just to make sure, that's when instead of ((M_i)(v_0)-(m_1)(v_1)-(m_2)(v_2))/(m_3)=v_3
    I would have ((M_i)(v_0*cos)-(m_1)(v_1*cos)-(m_2)(v_2*cos))/(m_3)=v_3*cos and then for the vertical components it would be sin instead of cos, correct?

    And then to find the actual magnitude of v_3 would be v_3 = Sqrt((v_3x)^2 + (v_3y)^2)

    But wait, wouldn't this give me 2 unknowns on the above ((M_i)(v_0*cos)-(m_1)(v_1*cos)-(m_2)(v_2*cos))/(m_3)=v_3*cos ?? v_3 and its angle.
     
  6. Oct 31, 2015 #5
    Ooohhh. I had completely forgotten about breaking vectors into components.
    Just to make sure, that's when instead of ((M_i)(v_0)-(m_1)(v_1)-(m_2)(v_2))/(m_3)=v_3
    I would have ((M_i)(v_0*cos)-(m_1)(v_1*cos)-(m_2)(v_2*cos))/(m_3)=v_3*cos and then for the vertical components it would be sin instead of cos, correct?

    And then to find the actual magnitude of v_3 would be v_3 = Sqrt((v_3x)^2 + (v_3y)^2)

    But wait, wouldn't this give me 2 unknowns on the above ((M_i)(v_0*cos)-(m_1)(v_1*cos)-(m_2)(v_2*cos))/(m_3)=v_3*cos ?? v_3 and its angle.
     
  7. Nov 1, 2015 #6

    mfb

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    cosine of what?
    You know most of those angles (they are different) and their sines and cosines.

    You get two unknowns, but you also have two equations.
     
  8. Nov 1, 2015 #7
    Alright so solved for the x and y components to get two equations and I isolated v3 on one, and theta on the other. However, that would end up making everything WAY to complicated cause I'd have something like
    v_3 = ((M_i)(v_0*cos(-45))-(m_1)(v_1*cos(-90))-(m_2)(v_2*cos(45)))/(m_3)(sin(arccos(((M_i)(v_0)(cos(-45))-(m_2)(v_2)(cos45))/m_3*v_3)

    Which looks like a nightmare, so then I thought of two different ways that I might be able to do it, and my first idea was to isolate v_3 in both the x and y component equations, because I THINK v_3 would be the same in those equations, the thing that changes that v value is the cos or sin of the angle.

    So after that, since they're both equal to v_3, then I set them equal to each other, which I'm not sure I could or should have done, and isolated theta to get another long equation but that was much easier to work with.

    So then AFTER that I was also thinking that, what if I used P_i = P_f, without the x and y components first, to get v_3 = ((M_i)(v_0) - m1v1 - m2v2) / m3
    After getting that THEN I could have just used that v3 definition to plug into the x and y component equations to THEN isolate theta in one of them to get theta, and THEN also use it to find the magnitude of v3.

    So, I did those 3 potential ways, but I'm not sure which one or ones are the correct way, (if any) and I was wondering if you could give me some more insight on that.
     
  9. Nov 2, 2015 #8

    mfb

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    What is cos(-90), for example?
    If you define it as magnitude of the velocity, it is the same, sure.
    That equation works with vectors for the velocities only, you cannot plug in the magnitude here.
     
  10. Nov 2, 2015 #9
    Thank you so much for all your help!
     
  11. Nov 4, 2015 #10

    Simon Bridge

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    I'd have made SE the ##+\vec\imath## (x) direction and NE the ##+\vec\jmath## (y) direction, makes the math easier but you have to remember to translate back to compass directions when you write the answer. If you've done any work with objects sliding down a slope, the approach should be familiar.
     
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