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Center of mass project exploding at max height

  1. Jul 7, 2015 #1
    1. The problem statement, all variables and given/known data
    A shell is shot with an initial velocity http://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2529559entrance1_N10030.mml?size=14&ver=1436312588112 [Broken] of 17 m/s, at an angle of θ0 = 63° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (see the figure). One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?

    http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c09/fig09_46new.gif

    2. Relevant equations
    m2/M d = xcom

    3. The attempt at a solution
    I have found the correct answer for this one its 35.77m which i solve by inspection and symmetry. My problem is how do i find the distance of the other particle if it was not symmetric? I tried using,
    xcom = m1x1 + m2x2 / M​
    where M = sum of all masses. And solving this eq for x2 But i keep getting 23.86m which is the distance of xcom.
    20150707_185705_zpstm5laruz.jpg
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jul 7, 2015 #2
    The problem states that it falls straight down from the top of its trajectory. Can you find the x coordinate of the maximum height?
     
  4. Jul 7, 2015 #3

    haruspex

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    In the middle of your working you have a line that ends with xcom=17....., but I have no idea how you came to that. What relative masses are you assuming?
     
  5. Jul 7, 2015 #4

    haruspex

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    J-dizzal has solved the original problem. The question now is how to solve it if the masses are not equal?
     
  6. Jul 7, 2015 #5
    Ah. Thanks. I misread his question.

    To be clear: you mean that the shell still explodes at apex in the horizontal direction and that that one of the pieces comes to rest and falls straight down?
     
  7. Jul 7, 2015 #6

    collinsmark

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    Yes, that's my interpretation. The difference is that the mass fragments' masses might not necessarily be the same this time.

    The thing is, whether or not the mass fragments are of equal mass or not, it doesn't change where the "center of mass" lands. In other words, it won't have a effect on [itex] x_{COM} [/itex].

    Given the diagram in the attachment, the OP has already figured out that [itex] x_{COM} [/itex] = 23.84 m. But as @haruspex points out, the OP later treats the [itex] x_{COM} [/itex] as 17.89 meters later in the calculations. 'Not sure where this 17.89 m value is coming from.
     
    Last edited: Jul 7, 2015
  8. Jul 7, 2015 #7
    In that case my suggestion would be to work symbolically and perhaps write the total mass M as the sum of two masses m and nm where n is the factor that relates the two masses.
     
  9. Jul 7, 2015 #8
    ok i get it now, my mistake was setting m2= the mass as if it were equal to m1 and using this eq;
    x2= (M(xcom) - m1x2) / (m2).​
    I believe this eq will work if i had the mass of m2.
     
  10. Jul 7, 2015 #9
    sorry i should of erased that equation, but 17.895m was the xcom when setting the coordinate system to x=0 at the center of the parabola along the x-axis. it was my initial attempt to find xcom using the eq xcom= (m2/M)d
    so then 17.895 + 5.965 = 23.84
     
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