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A shell explodes into three fragments of equal masses.

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data
    A shell explodes into three fragments of equal masses. If two fragments travel at right angles to one another with equal speeds, 'v', what is the direction and speed of the third fragment?


    2. Relevant equations
    doing ratios?


    3. The attempt at a solution
    i dont know what it mean by two fragments travel at right angles.
     
  2. jcsd
  3. Dec 4, 2011 #2

    BruceW

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    Re: Momentum

    Travel at right angles means there is 90 degrees between their paths. For example, one might travel vertically, while the other travels horizontally.

    For the question, think about what quantity will be conserved.
     
  4. Dec 4, 2011 #3
    Re: Momentum

    what does conserved really mean cause when i hear the word conserved i think of something being stored
     
  5. Dec 4, 2011 #4

    BruceW

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    Re: Momentum

    In physics, it roughly means 'stays the same'. Is this homework set by your teacher? Name me some quantities that you think might be conserved in this situation.
     
  6. Dec 4, 2011 #5
    Re: Momentum

    p=mv?
     
  7. Dec 4, 2011 #6

    BruceW

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    Re: Momentum

    Yes, that's the one! You might also think of energy, but when the shell explodes it turns an unknown amount of chemical energy into kinetic energy, so energy conservation doesn't help us here.

    So you're right to think of momentum conservation. After the shell explodes, there are 3 pieces, so you need to think of the equation for the total momentum due to several objects.
     
  8. Dec 5, 2011 #7
    Re: Momentum

    i got the velocity to be v2= root square of v^2(m1^2+m2^2) / m3

    and the angle to be tan^-1 ( m2/m1)

    am i right?
     
  9. Dec 6, 2011 #8

    BruceW

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    Re: Momentum

    The angle is right. I'm not totally sure I understand what your answer for the velocity is supposed to look like... Does it mean:
    [tex]v_{third} = \frac{ \sqrt{v^2(m_1^2+m_2^2)} }{m_3} [/tex]
    If this is what you meant, then yes, you've got this right as well.

    The next step is to use the knowledge that all the masses are the same.
     
  10. Dec 6, 2011 #9
    Re: Momentum

    ok thank you for the help!
     
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