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Momentum of an object exploding into two separate objects

  1. Jul 25, 2013 #1
    1. The problem statement, all variables and given/known data

    An object with kinetic energy K explodes into two pieces, each of which moves with twice the speed of the original object.

    Find the angles of the two velocities relative to the direction of motion before the explosion.

    Modified question from "Essential University Physics 2nd Ed" by Richard Wolfson, Ch.9 Qu. 22 Page 151.

    2. Relevant equations

    Momentum: [itex]\vec{p} = m \times \vec{v} [/itex]

    Momentum is conserved within an isolated system where no external forces are acting.


    3. The attempt at a solution

    attachment.php?attachmentid=60501&stc=1&d=1374775884.png
    Above diagram shows the object before explosion as "m1",
    and the two objects created afterwards as "m2" and "m3".

    [itex] m_1 = m_2 + m_3 [/itex]

    Conservation of momentum gives us:

    Horizontal: [itex] m_1 v = m_2\ 2v\ cos(θ) + m_3\ 2v\ cos(α) [/itex]

    Vertical: [itex] m_2\ 2v\ sin(θ) = m_3\ 2v\ sin(α) [/itex]


    Intuitively, if both resulting objects have the same velocity, and with no other data given, I would tend to think that the angles are the same as each other, and both masses are equal. Apparently this is the correct answer.

    I don't understand why. I've tried a multitude of textbooks and worked examples. As far as I can gather, it is somehow related to the centre of mass of the system.

    But in my mind, if I change the two resulting masses to be uneven, you can set the angles in such a way that the resulting centre of mass is still correct.

    (As this is an isolated system with no external forces, the motion of the centre of mass remains unchanged after the explosion).

    Manipulating the equations doesn't result in anything that I thought was useful in helping me along. I was hoping someone here could nudge me in the right direction, and help to explain the concepts involved? Thanks in advance.

    Bharat.
     

    Attached Files:

  2. jcsd
  3. Jul 25, 2013 #2

    CWatters

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    Consider the momentum in the vertical direction (y axis) before and after. There is no momentum in that plane before the explosion so the net momentum afterwards must also be zero. If the masses are the same then the y component of their velocity must be the same magnitude but opposite sign.

    I'll let you do the horizontal direction (x axis).
     
  4. Jul 25, 2013 #3
    Yup, I get that, but why must their masses be the same? That's the bit I don't understand.
     
  5. Jul 25, 2013 #4

    haruspex

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    I agree - there's not enough information.
    I note that it tells you the KE beforehand, and you can determine the added energy from the explosion, but I don't see that that gets you anywhere.
    To prove there's not enough information, specify some ratio of the masses and demonstrate a solution satisfying all conditions.
     
  6. Jul 26, 2013 #5
    Not necessary the two masses have to be equal. But I think that's a given. Using that, the angles have to be calculated.
     
  7. Jul 26, 2013 #6

    mfb

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    With equal masses, it is easy, as the system is symmetric and the horizontal component is trivial to evaluate.

    The solutions with unequal masses are much more interesting.
    Bonus question: what is the maximal ratio between the two fragment masses?
     
  8. Jul 26, 2013 #7

    haruspex

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    Yes, that's an interesting case. Another is where one of the masses goes at right angles to the original velocity.
     
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