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Explosive Particle and momentum

  1. Mar 20, 2009 #1

    TFM

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    1. The problem statement, all variables and given/known data

    A stationary mass of 10 kg disintegrates explosively, with the evolution of [tex]4.2.10^5[/tex] J, into three parts, two having equal masses of 3 kg. These two masses move at right angles to each other. Calculate the velocity of the third part.

    2. Relevant equations

    [tex] E_k = \frac{1}{2}mv^2 [/tex]

    Momentum = mv

    3. The attempt at a solution

    Okay so I know that this is a momentum question. And it needs to be constant, therefore

    [tex] mv_{before} = mv_{after} [/tex]

    the initial momentum before is 0, so the total final miomentum also equals 0. this means that since the two known partiles are travelling perpendicular, the third, which mus have mass 4kg, is travelling at an angle opposite, probably 45. However, I am slightly uncertain, since they don't give the speed, although they give with the evolution of [tex]4.2.10^5[/tex] J. I am slightly unsure what this is, do I need to calculate the velocities of the two 3kg particles using the kinetic energy formula and the energy given above?
     
  2. jcsd
  3. Mar 20, 2009 #2

    LowlyPion

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    Homework Helper

    Initially at rest looks like it means the center of mass will be preserved at no velocity.

    So ∑ mv = 0

    Since you have 3V1 i + 3V2 j for the 2 pieces, then you have 4V3(i,j) equals the sum of those 2.

    3V1 i + 3V2 j = 4V3x i + 4V3y j

    Where V3x2 + V3y2 = V32

    The energy also tells you that 1/2(3)V12 + 1/2(3)V22 + 1/2(4)V32 = 4.2*105
     
  4. Mar 20, 2009 #3

    TFM

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    Okay, so would I now I find V1 in terms of the energy and V 2 like so:

    [tex] 4v_3 = 3v_1 + 3v_2 [/tex]

    [tex] (4v_3)^2 = (3v_1)^2 + (3v_2)^2 [/tex]

    [tex] (v_3)^2 = \frac{(3v_1)^2}{16} + \frac{(3v_2)^2}{16} [/tex]


    [tex] \frac{1}{2}3v_1^2 + \frac{1}{2}3v_2^2 + \frac{1}{2}4v_3^2 [/tex]

    [tex] \frac{1}{2}3v_1^2 + \frac{1}{2}3v_2^2 + \frac{1}{2}4\frac{(3v_1)^2}{16} + \frac{(3v_2)^2}{16}
    [/tex]

    Rearrangeing to give:

    [tex] \frac{21}{8}v_1^2 + \frac{21}{8}v_2^2 = E [/tex]

    [tex] v_1^2 + v_2^2 = \frac{8}{21}E [/tex]

    [tex] v_1^2 = \frac{8}{21}E - v_2^2[/tex]
     
  5. Mar 20, 2009 #4

    LowlyPion

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    That's sort of where I was heading, but I was intending that

    1.5V12 + 1.5V12 + 2V32 = 4.2*105

    Recognizing that V1 = -4/3V3x and V2 = -4/3V3y

    1.5(-4/3V3x)2 + 1.5*(-4/3V3y)2 + 2V32 = 4.2*105

    2*V3x2 + 2*V3y2 +2V32 = 4.2*105

    But V3x2 + V3y2 = V32

    Substituting I get 2*V32 + 2V32 = 4V32 = 4.2*105
     
  6. Mar 21, 2009 #5

    TFM

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    Okay, I am not quite getting the same answer.

    [tex] v_1 = \frac{4}{3}v_{3(x)} [/tex]

    [tex] v_2 = \frac{4}{3}v_{3(y)} [/tex]

    inserting:


    [tex] \frac{3}{2}v_1^2 + \frac{3}{2}v_2^2 + 2v_3^2 = E [/tex]

    [tex] \frac{3}{2}(\frac{4}{3}v_{3(x)})^2 + \frac{3}{2}(\frac{4}{3}v_{3(x)})^2 + 2v_3^2 = E [/tex]


    [tex] \frac{3}{2}\frac{16}{9}v_{3(x)}^2 + \frac{3}{2}\frac{16}{9}v_{3(x)}^2 + 2v_3^2 = E [/tex]

    giving:

    [tex] \frac{48}{18}v_{3(x)}^2 + \frac{48}{18}v_{3(x)}^2 + 2v_3^2 = E [/tex]

    ?
     
  7. Mar 21, 2009 #6

    LowlyPion

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    Sorry, I jotted down the 4/3 without squaring. You have the right equation. I was apparently more focused on the method.
     
  8. Mar 21, 2009 #7

    TFM

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    Well, it gives me a nice answer of 300m/s. This seems rather big, though...
     
  9. Mar 21, 2009 #8

    LowlyPion

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    Roughing it out V1 = V2 ~ V3

    So roughly speaking the masses are moving with similar velocities, which suggests that

    420,000 J ~ 1/2*Σ m * V2

    V2 ~ 84,000 or

    V ~ 290 m/s

    This isn't so far from the calculated 239, 319, 319 now is it?
     
  10. Mar 21, 2009 #9

    TFM

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    Nope, not too bad at all.

    Many Thanks, LowlyPion :smile:
     
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