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Homework Help: Conservation of Momentum, dropped vase

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data

    I've been trying to solve this question for hours now, NEED HELP

    You are standing in a shop, holding an expensive vase of mass m. The vase accidentally slips from your hand and falls to the floor. It breaks into three pieces, two of mass 1/4m and one of mass 1/2m. The two pieces of 1/4m slide along the floor with speed v=20 cm/s. If one of these pieces moves east and the other piece moves southeast (i.e. 45 degrees below the +x-axis), what is the velocity (magnitude and direction) of the third piece?

    2. Relevant equations

    Law of Conservation of Momentum


    3. The attempt at a solution

    I got 0.14m/s for the velocity which I am pretty sure is right. The direction is what I am unsure about. I did get Northeast 45 degrees above x-axis but I think that is wrong. Any help?
  2. jcsd
  3. Nov 13, 2011 #2


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    Add the momenta of the two small pieces. Clearly it will be in a direction half way between the 0 and -45 degrees of the pieces, so -22.5 degrees. The big piece must go in the opposite direction (add 180 degrees).

    I'm not getting .14 for the speed. Best to write your mv + mv = 0 and figure it out carefully. Take the component of each 20 cm/s in the 22.5 degree direction.
  4. Nov 13, 2011 #3
    22.5 degrees west of north?

    For the speed, is this correct?

    momentum = (m/2)V

    (m/2)V = sq rt[{(m/4)^2}(0.20^2) + {(m/4)^2}(0.20^2) + {2*(m/4)^2}(0.20^2)cos 45]
    = sq rt(2)*(m/4)*sq rt[1+cos 45] = sq rt(2)*(m/4)*sq rt[2*cos^2(45/2)] = (m/2)*cos (22.5)
    So V = cos(22.5) = 0.9238 m/s
  5. Nov 13, 2011 #4


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    I agree with the 22.5 degrees west of north.

    That momentum calc seems unnecessarily complicated. Consider one axis along the combined momentum of the two smaller pieces at 22.5 degrees south of east. The components of momentum in the direction perpendicular to this cancel out. So the momentum of these pieces is 2*.25m*20cos(22.5) and this must be equal to the momentum of the .5m in the opposite direction. So
    .5mv = 2*.25m*20cos(22.5)
    Intuitively the answer must be not much smaller than 20 cm/s because the cos(22.5) is close to 1.
  6. Nov 13, 2011 #5
    The 1/2m cancel so v=20cos(22.5). From that I get 18.48 cm/s.
  7. Nov 14, 2011 #6
    Just to make sure I have done this correctly


    0 = (.25m)(20) + (.25m)(20)(cos 45) + (.5m)(v3)(cos θ) - (.5m)(v3)(cos θ)
    cos θ = (10 + 5sqrt2)/v3


    0 = (.25m)(20)(sin 45) +(.5m)( v3)(sin θ) - (.5m)(v3)(sin θ)
    sin θ = 5sqrt2/v3

    tan θ = (5sqrt2)/(10 + 5sqrt2) = 0.414
    θ = inv tan (0.414) = 22.5 degrees
  8. Nov 14, 2011 #7


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    I agree with your 18.48 cm/s and 22.5 degrees.
    I did my solution with a different set of axes, so it is difficult for me to check your solution. It looks like you have 4 moving objects in the first line of your solution for the x component, which doesn't seem right. Probably a typo; getting the same answers two different ways strongly suggests both are correct.
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