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Math Help - balls approaching with velocities given only.

  1. #1
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    balls approaching with velocities given only.

    I'm not sure how this can be solved if their is only two velocities?

    Two balls are approaching each other head-on. Their velocities are +9.95 and -11.4 m/s.
    (a) Determine the velocity of the center of mass of the two balls if they have the same mass.
    m/s

    (b) Determine the velocity of the center of mass of the two balls if the mass of one ball (v = 9.95 m/s) is twice the mass of the other ball (v = -11.4 m/s).
    m/s
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  2. #2
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    Quote Originally Posted by rcmango View Post
    I'm not sure how this can be solved if their is only two velocities?

    Two balls are approaching each other head-on. Their velocities are +9.95 and -11.4 m/s.
    (a) Determine the velocity of the center of mass of the two balls if they have the same mass.
    m/s

    (b) Determine the velocity of the center of mass of the two balls if the mass of one ball (v = 9.95 m/s) is twice the mass of the other ball (v = -11.4 m/s).
    m/s
    This is simply a definition:
    v_{CM} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}

    You are given the velocities for the first one (the masses cancel in this case) and the velocities and relative masses in the second case (the m's end up dropping out here, too.)

    -Dan
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  3. #3
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    I still don't see it without the masses?

    isn't this the same: m1*v1/(m1 + m2) + m2*v2/(m1 + m2)

    so v1/m2 + v2/m1

    so i still need the masses right?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rcmango View Post
    I still don't see it without the masses?

    isn't this the same: m1*v1/(m1 + m2) + m2*v2/(m1 + m2)

    so v1/m2 + v2/m1

    so i still need the masses right?
    For the first problem your masses are equal. So m1 = m2 = m. So the denominator simply becomes 2m and you can factor an m out of the top, then cancel.

    -Dan
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    it took me hours to find the equation to be 1/2(x1 + x2) in a textbook. where x1 = v1 and x2 = v2

    i didn't use the masses at all. just the velocities only.

    i wish you could've pointed me in that direction!

    so for the first one, -.725 m/s and the second one. the mass should not make a difference. but i could not reach an answer.
    Last edited by rcmango; November 17th 2007 at 01:29 AM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rcmango View Post
    it took me hours to find the equation to be 1/2(x1 + x2) in a textbook. where x1 = v1 and x2 = v2

    i didn't use the masses at all. just the velocities only.

    i wish you could've pointed me in that direction!

    so for the first one, -.725 m/s and the second one. the mass should not make a difference. but i could not reach an answer.
    So don't look for an equation from your book, derive it using the principles and definitions!! This is why I didn't just give you the final equation: I wanted you to be able to solve problems that don't have nice neat equations listed in your book. I fear this is the greatest flaw in how you are trying to solve these problems.

    Quote Originally Posted by rcmango View Post
    I'm not sure how this can be solved if their is only two velocities?

    Two balls are approaching each other head-on. Their velocities are +9.95 and -11.4 m/s.
    (a) Determine the velocity of the center of mass of the two balls if they have the same mass.
    m/s

    (b) Determine the velocity of the center of mass of the two balls if the mass of one ball (v = 9.95 m/s) is twice the mass of the other ball (v = -11.4 m/s).
    m/s
    Quote Originally Posted by topsquark View Post
    This is simply a definition:
    v_{CM} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}

    You are given the velocities for the first one (the masses cancel in this case) and the velocities and relative masses in the second case (the m's end up dropping out here, too.)

    -Dan
    For the first problem we have m_1 = m_2. So
    v_{CM} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} = \frac{m_1v_1 + m_1v_2}{m_1 + m_1}

    = \frac{m_1(v_1 + v_2)}{2m_1}

    = \frac{v_1 + v_2}{2}

    Now plug the numbers in.

    For the second problem we have m_1 = 2m_2, so
    v_{CM} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} = \frac{2m_2v_1 + m_2v_2}{2m_2 + m_2}

    You finish it.

    -Dan
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  7. #7
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    okay so thats m1 x 2 because its twice the weight of the second weight.

    so then 2*m2 (v1 + v2) / m2(2 + 1)

    which cancels and gives (2*v1 + v2) / 3

    and gives: 2.83 m/s
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