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Homework Help: 2D momentum question. I have my answers I just want to cheak with you guys

  1. May 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Two objects are involved in an elastic collision object 1 has a mass of 20Kg while object 2 has mass of 25Kg the initial velocities of the objects are unknown but neither of the objects are stationary Object 1 is moving 35m/sN30E after collision and object 2 is moving 15 m/s S30E after collision find initial velocities of both objects

    2. Relevant equations



    3. The attempt at a solution

    V2=V2=30m/s E33N

    V1=20m/s E28S

    Dont bother writing the whole work if its wrong it too long it took me a whole page to do this. But just tell me it wrong and where I might have went wrong.

    Thanks
     
  2. jcsd
  3. May 25, 2010 #2

    vela

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    The problem isn't solvable with the information you've provided. There are four unknowns but only three equations.
     
  4. May 25, 2010 #3
    There is only 2 unknowns v1 and v2.

    v1=v2+v2'-v1'
    v2=v1+v1'-v2'

    Everything is here but I just want to double cheak if what I got was right
     
  5. May 25, 2010 #4

    vela

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    There are two components for each v1 and v2. That's four altogether.
     
  6. May 25, 2010 #5
    so

    v1x=v2x+v2'x-v1'x
    v1y=v2y+v2'y-v2'y

    v1x^2 + v1y^2 = v1^2

    and same for v2
     
  7. May 25, 2010 #6

    vela

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    Those should be momenta, not velocities.
    This isn't really an equation you can use to solve for the velocity of the first object. It just allows you to convert between representations -- component form or magnitude and direction.

    The only equations you have are two from conservation of momentum, one for each axis, and the conservation of kinetic energy equation. That's three total. You can't solve for four unknowns without another equation.
     
  8. May 25, 2010 #7
    Yes your right so when I solve for v1x/y and v2x/y

    I will put them into

    m1v1+m2v2=m1v1' + m2v2'

    I will have 4 different answers
    x and y for v1 and v2
     
  9. May 26, 2010 #8

    collinsmark

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    I have to agree with Vela on this. You start with 5 equations,

    [tex] \mbox{\{ something known \}} = m_1 v_{1x} + m_2 v_{2x} [/tex]

    [tex] \mbox{\{ something else known \}} = m_1 v_{1y} + m_2 v_{2y} [/tex]

    [tex] \mbox{\{ something also known \}} = m_1 v_{1}^2 + m_2 v_{2}^2 [/tex]

    [tex] v_1 = \sqrt{v_{1x}^2 + v_{1y}^2} [/tex]

    [tex] v_2 = \sqrt{v_{2x}^2 + v_{2y}^2} [/tex]

    leaving you with 6 unkowns: [itex] v_{1x} [/itex], [itex] v_{1y} [/itex], [itex] v_{2x} [/itex], [itex] v_{2y} [/itex], [itex] v_1 [/itex], and [itex] v_2 [/itex].

    That's 5 equations and 6 unknowns. Combining any of the above equations to get the 6th equation will merely lead to degeneracy, and you'll eventually end up with something rather meaningless like [itex] 2 v_1 = 2 v_1 [/tex], which doesn't help (this is because the 6th equation is not independent). And you can't use the equations of the form [itex] v_{1x} = v_1 cos \theta_1 [/itex]. That could get you a couple more equations, but a couple more unknowns to go along with them.

    There needs to be just one additional piece of information somewhere. Something like the initial angle of the first object. Or maybe some little tidbit of more information such as one of the objects was initially at rest. 'Something like that anyway. It wouldn't take much; just some little detail not expressed already in the above equations.

    And by the way, on a different note, you can check that your original answer (see your original post) is incorrect. The total kinetic energy before the collision does not match the total kinetic energy after, according to your result (V2=30m/s E33N, V1=20m/s E28S).

    If we're missing something, and you feel that the problem can be solved as it is stated, maybe now would be a good time to show us your work.
     
  10. May 26, 2010 #9

    OK well I redid it since it is wrong.

    Kinetic energy is conserved

    so v1=v2+v2'-v1'

    conservation of momentum

    20v1+25(v1+35cos60-15cos60)=350+187.5

    v1x=+6.4m/s or 6.4m/s E

    20v1+25(v1+35sin60+15sin60)=606.2-324.8
    v1y=-17.8 m/s or 17.8m/s S

    so V1=18.9m/s 19.8 E of S

    I did exact same for V2 and got

    21.6m/s 49.5 N of E
     
  11. May 26, 2010 #10

    collinsmark

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    Sorry, but there's still something not right.

    The objects' final kinetic energy, Tf, (using the information in the problem statement) is:

    [tex] T_f = \frac{1}{2} 20(35)^2 \ [J] + \frac{1}{2} 25(15)^2 \ [J] = 15.0625 \ kJ [/tex]

    But the initial kinetic energy, Ti, based on your results is:

    [tex] T_i = \frac{1}{2} 20(18.9)^2 \ [J] + \frac{1}{2} 25(21.6)^2 \ [J] = 9.4041 \ kJ [/tex]

    Maybe the above is a typo? :uhh: Since you're not using any squares, the above equation is actually more representative of a momentum equation. But if that were the case, don't you mean,

    [tex] m_1 \vec v_1 = {\color{red}{-}} m_2 \vec v_2 + m_2 \vec{\acute{v_2}} \ {\color{red}{+}} \ m_1 \vec{\acute{v_1}} [/tex]

    Notice the signs. Also notice that the masses do not cancel. The above simply comes from vector based conservation of momentum,

    [tex] m_1 \vec v_1 + m_2 \vec v_2 = m_1 \vec{\acute{v_1}} + m_2 \vec{\acute{v_2}} [/tex]

    where the variables above are vectors and need to be added that way. The above actually represents two equations,

    [tex] m_1 v_{1x} + m_2 v_{2x} = m_1 \acute{v_{1x}} + m_2 \acute{v_{2x}} [/tex]

    [tex] m_1 v_{1y} + m_2 v_{2y} = m_1 \acute{v_{1y}} + m_2 \acute{v_{2y}} [/tex]

    Notice I didn't use the vector symbol since the above equations deal with the vector components directly (which are scalars, by themselves).

    In terms of conservation of energy, the equation you want to use is:

    [tex] m_1 v_{1}^2 + m_2 v_{2}^2 = m_1 \acute{v_{1}}^2 + m_2 \acute{v_{2}}^2 [/tex]

    All the "1/2" terms in the kinetic energy equation cancel out which simplifies things. But the masses do not cancel.

    But conservation of kinetic energy does not imply that [itex]
    v_1 = v_2 + \acute{v_2} - \acute{v_1} [/itex].


    I didn't check any more of the above (except to check that it's not quite right).

    [Edit: Last blurb I wrote deleted. I'll follow up with something more correct in my next post.]

    [Second edit: It's getting late, and I don't think I'll be able to give a detailed response in time. But I still don't think this problem can be solved given the way it is stated. The part I deleted in my last post is I think there may be an infinite number of solutions to the problem (given the way it is stated). There needs to be some other tidbit of information somewhere to narrow down an exact solution.]
     
    Last edited: May 27, 2010
  12. May 26, 2010 #11
    So what do I tell my teacher?
     
  13. May 26, 2010 #12

    vela

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    This relationship is actually true for one-dimensional elastic collisions, but as this is a two-dimensional problem, it doesn't apply.
     
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