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Momentum and impulse

  1. Oct 1, 2007 #1
    1. The problem statement, all variables and given/known data

    A 12 kg shell is launched at an angle of 55 degrees above horizontal with v_O = 150 m/s. When at highest point, it explodes into two fragments, M_1 with mass 9 kg and M_2 with mass 3 kg. The fragments reach the ground at the same time. The heavier fragment lands back at the same point - where does the lighter fragment land?

    3. The attempt at a solution

    Ok, I know that the speed at the top is V_x = 86 m/s. For the heavier fragment to land at the same point, it must have the same v_initial (150 m/s) but opposite direction, so v_x must be changed -2 times (so it's -86 m/s) and v_y = -122 m/s.
    THen I say that momentum of single fragment before explosion equals total momentum of the two fragments - but I get that the speed of the lighter fragment is very high!

    Where am I wrong?
  2. jcsd
  3. Oct 1, 2007 #2
    I got it.. m_1 = 3*m_2 and v_2 = 3*v_1
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