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A vessel at rest explodes, breaking into three pieces

  1. Apr 1, 2008 #1
    1. Ok, i've been working on this problem for ever and it seems easy but I know that there is something wrong.
    A vessel at rest explodes, breaking into three pieces. Two pieces, having equal mass, fly off perpendicular to one another with the same speed of 90 m/s. One goes along the negative x-axis and the other along the negative y-axis. The third piece has three times the mass of each other piece. What are the direction and magnitude of its velocity immediately after the explosion? a) What is the angle with respect to the x-axis?
    b)What is the magnitude of its velocity?
    for A) I understand that the value is 45 degrees however I do not exactly understand why it is 45 degrees. and for B) I have no idea where to start...I know that momentum is involved but thats about it.


    2. P=mv ? not sure where to go from there.
     
  2. jcsd
  3. Apr 1, 2008 #2

    DaveC426913

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    Gold Member

    Leave the formulae aside for a moment and try to intuit the problem. Then the formulae should make more sense.

    The explosion is gonig to be symmetric. Since it started with 0 momentum, the sum of the pieces will also have to be 0 momentum. i.e. everything will have to cancel out. The vector sum of the known pieces m(a) and m(b) has to be equal and opposite to the unknown third piece's m(c) vector to maintain 0 momentum. This means it has to be opposite in both angle AND magnitude.


    Code (Text):

    [FONT="Courier New"]       [COLOR="SeaGreen"][B]O[/B][/COLOR] m(c)
    m(a)[COLOR="Red"]  /[/COLOR]
      90 [COLOR="Red"]/[/COLOR]
    [COLOR="seagreen"][B].[/B][/COLOR]---*
    [COLOR="silver"]|[/COLOR]  [COLOR="Blue"]/[/COLOR]|
    [COLOR="silver"]|[/COLOR] [COLOR="Blue"]/[/COLOR] | 90  
    [COLOR="silver"]|[/COLOR][COLOR="Blue"]/[/COLOR]  |
     [COLOR="silver"]---[/COLOR][COLOR="seagreen"][B].[/B][/COLOR]  m(b)


    [COLOR="Blue"]/[/COLOR] = sum of vectors of known m(a) and m(b)
    [COLOR="Red"]/[/COLOR] = vector of unknown m(c)

    angle of [COLOR="Red"]/[/COLOR] must be equal & opposite angle of [COLOR="Blue"]/[/COLOR]
    magnitude of [COLOR="Red"]/[/COLOR] must be equal to magnitude of [COLOR="Blue"]/[/COLOR] [B]times the ratio of masses[/B]
    [/FONT]



     
    Momentum comes in when we calculate the magnitude of the unknown vector. m(c)'s momentum will have to exactly balance the sum of the momenta of the m(a) and m(b). So m(c)v(c) = m(a)v(a)+m(b)v(b). Tripling the object's mass while maintaining the same momentum will require the object to go one third as far.
     
    Last edited: Apr 1, 2008
  4. Apr 1, 2008 #3

    DaveC426913

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    At least, that seems right. I only studied this in High School - 30 years ago...

    How'm I doin? :rolleyes:
     
  5. Apr 1, 2008 #4
    Thats what I believed to be true also but that would yield an answer of 60 m/s for the magnitude of the final speed which is incorrect...
    I attempted to solve using that method....3m(c)v(c)=m(a)*90 m/s + m(b)*90 m/s.
     
  6. Apr 2, 2008 #5

    Mapes

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    Gold Member

    You need to incorporate the angles. Again intuitively: if A and B flew directly apart from each other, C's velocity would have to be zero. (A and B already satisfy the conservation law, any C velocity would violate it.) Conversely, if A and B flew off in exactly the same direction, then we would have DaveC426913's equation above, and C's speed would be 60 m/s.

    Since neither of the above is exactly the case here, you're probably going to have a sine or cosine somewhere in the mix. Make sense?
     
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