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Resulting velocity of 2 particles

  1. Dec 29, 2008 #1
    [solved] resulting velocity of 2 particles

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

    [​IMG]


    2. Relevant equations

    conservation of momentum

    [tex]m_Av_A'=m_Bv_B'[/tex]

    [tex]v_A'=\frac{m_B}{m_A}v_B'[/tex]

    conservation of energy

    [tex]V=\frac{1}{2}m_A(v_A')^2+\frac{1}{2}m_B(v_B')^2=\frac{1}{2}m_A\left(\frac{m_B}{m_A}v_B'\right)^2+\frac{1}{2}m_B(v_B')^2[/tex]

    [tex]v_B'=\sqrt{\frac{2m_AV}{m_B(m_A+m_B)}}[/tex]

    3. The attempt at a solution

    [tex]m_A=0.9Kg[/tex]

    [tex]m_B=1.35Kg[/tex]

    [tex]V=1.35Nm[/tex]

    [tex]v_B'=\sqrt{\frac{2\cdot0.9\cdot1.35}{1.35(0.9+1.35)}}=\sqrt{0.8}\approx{0.8944[m/s]}[/tex]

    [tex]v_A'=\frac{1.35}{0.9}\cdot0.8944=1.3416[m/s][/tex]

    [​IMG]

    from the law of cosines

    [tex]v_B=\sqrt{6^2+0.8944^2-2\cdot6\cdot0.8944\cdot\cos{60^o}}\approx5.607[m/s][/tex]

    from the law of sines

    [tex]\sin{\beta}=\frac{\sin{60^o}\cdot0.8944}{5.607}\approx0.1381[/tex]

    [tex]\beta\approx7.94^o[/tex]

    As you can see that result is totally different from the solution given by my teacher. I also get different results for the other velocity. So either my calculations are incorrect or my teacher gave us wrong results. I'll be grateful for any help, thank you.
     
    Last edited: Dec 30, 2008
  2. jcsd
  3. Jan 1, 2009 #2

    nvn

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    disclaimer: Nice work. Your answer for vB, 5.6066 m/s at -7.9413 deg, is correct. The solution of your teacher is wrong.

    We can see that the solution of your teacher is impossible, because in the solution of your teacher, the final kinetic energy of particle A greatly exceeds the total initial energy of the system, which would violate conservation of energy.
     
  4. Jan 2, 2009 #3
    Thanks for your reply, nvn. Actually I figured out where my teacher made a mistake. Obviously he didn't multiply 1.35Nm by 20 while converting the unit from ft*lb.

    [tex]1{ft}\cdot{lb}\approx1.355817456Nm[/tex]

    That said, if I assume the potential energy to be 27Nm, then my results are almost the same as those given by my teacher.
     
  5. Jan 2, 2009 #4

    nvn

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    I interpreted what you wrote as meaning, "Obviously, he didn't multiply 20 ft*lbf by 1.35582 N*m/(ft*lbf) while converting from ft*lbf."

    By the way, you and/or your teacher made some mistakes writing unit symbols.
    (1) There should always be a space between the numeric value and its following unit symbol. E.g., 6 m/s, not 6m/s.
    (2) N*m needs to have an intervening space or symbol, not Nm.
    (3) The unit symbol for kilogram is spelled kg, not Kg (whereas Kg would mean something like kelvin-gram).
    (4) ft*lb should be ft*lbf, whereas lb means pound mass (lbm).
    (5) [m/s] should be m/s, whereas [m/s] implies dimensional analysis, which uses different symbols than the unit symbols.

    See international standard for writing units; i.e., ISO 31-0. See NIST for the correct spelling of any unit symbol.
     
    Last edited: Jan 2, 2009
  6. Jan 2, 2009 #5
    Frankly, I've only seen lb for both force and mass used in books. It always seemed quite confusing to me.
     
  7. Jan 3, 2009 #6

    nvn

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    Good point. And lbf versus lbm makes a difference in mathematics applications (or at least some mathematics applications), not to mention lbm is an incoherent physical unit, and can't be used as-is for this problem, as it would need to be converted to slugs.

    And here is one more mistake by your teacher.
    (6) ft/lbf is a unit of flexibility, not work; work is ft*lbf, or N*m, or J.

    I know you caught that one. Also, in all the examples I typed in this thread, the asterisk can instead be a half-high dot, if available.
     
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