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Conservation of Momentum question (find final velocities)

  1. Feb 19, 2013 #1
    1. The problem statement, all variables and given/known data

    A boy and girl are standing on ice skates and pulling on opposite ends of a rope.

    Find their velocities the instant they meet.
    How long did it take?

    Relative speed of approach is 2 m/s
    mass of boy = 60kg
    mass of girl = 50kg
    distance = 5m


    2. Relevant equations
    m1v1 + m2v2 = (m1+m2)v


    v1 & m1 = boy
    v2 & m2 = girl

    3. The attempt at a solution

    I know that it takes them 2.5 seconds to meet (t = d/v).

    I cannot figure out their individual final velocities.

    v1+v2 = v_rel
    v2 = v_rel - v1

    m1(v_rel - v1) + m2v2 = (m1+m2)v

    But this leaves me with 2 equations, three unknowns.
     
  2. jcsd
  3. Feb 19, 2013 #2

    vela

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    What does v represent here?

    You substituted v2 in for v1 on the lefthand side.
     
  4. Feb 19, 2013 #3
    Well, I'm assuming since they embrace that they must be as one body, so shouldn't they have one velocity?
    I know this can't be true because it's asking for their individual final velocities.



    My mistake. That was simply a typing error I must have missed. I have it written correctly on my paper.

    m1v1 + m2(v_rel - v1) = ?
     
    Last edited: Feb 19, 2013
  5. Feb 19, 2013 #4

    vela

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    You also need to be a bit more careful with what ##v_1## and ##v_2## represent. Are those variables speeds or velocities? When you wrote ##m_1 v_1 + m_2 v_2##, it seems you're summing their momenta and are therefore assuming the ##v##'s are velocities. On the other hand, when you wrote ##v_1 + v_2 = v_\text{rel}##, it appears you're assuming they are speeds.
     
  6. Feb 19, 2013 #5
    Okay, thanks.
    I'm assuming they are velocities.
    I'm not sure if the equation relating the velocities to the relative velocity is actually correct. I can't find anything like that in my textbook.
    I'm assuming that the difference in the velocities is equal to the relative velocity. The reason they are being added is because they are headed in opposite directions (towards each other).

    So v1 - (-v2) = v_rel
    which gives me the the v1 + v2 = v_rel
     
  7. Feb 19, 2013 #6

    vela

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    In terms of vectors, the sum of the momenta is given by ##m_1 \vec{v}_1 + m_2\vec{v}_2## and the relative velocity is ##\vec{v}_1 - \vec{v}_2##. When working in one dimension, sometimes we're sloppy and just drop the arrows. The sign of the answers we get for ##v_1## and ##v_2## tell us the directions. Often, though, it's easier to think about things in terms of speeds. We assume ##v_1## and ##v_2## to be positive and put the directions into the equations explicitly. In this case, you'd have ##m_1 v_1 + m_2 (-v_2) = m_1 v_1 - m_2 v_2## and ##v_1 - (-v_2) = v_1+v_2##. Either way is fine. You just have to be consistent.

    Sticking with the vector notation for now, the velocity of the center of mass is given by
    $$\vec{v}_\text{cm} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1+m_2}.$$ What's the center of mass doing in this problem?
     
  8. Feb 19, 2013 #7
    Hmmm, I'm not positive.
    Would the center of mass be staying in the same spot?
     
  9. Feb 19, 2013 #8

    vela

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    Yup. If it wasn't moving before, it would only begin moving in response to an external force acting on the boy-rope-girl system, but there aren't any in this situation.
     
  10. Feb 19, 2013 #9
    Maybe I'm just not seeing it, but how does this relate to their final velocities?
    Do I need to find where the center of mass is and know they'll both meet there. Then try to use the time, the distance they each traveled, and their respective masses to figure out their final velocities?
     
  11. Feb 19, 2013 #10

    vela

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    You seem to be thinking this is a collision problem. It's not, so there's no final velocity to speak of. The problem does involve conservation of momentum, however. No external forces means momentum of the system is conserved which means the center of mass doesn't move.

    If the center of mass isn't moving, you have ##v_\text{cm} = 0##. So now you have two equations and two unknowns.
     
  12. Feb 19, 2013 #11
    Okay, so to find the final velocity of the boy and the girl:

    $$\vec{v}_\text{cm} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1+m_2} = 0$$

    m1v1 +m2v2 = m1v`1 + m2v`2

    (not a collision problem so not right hand side = (m1+m2)v)

    and v1 = v_rel - v2
    and then using from the COM equation that m1v1 = - m2v2 or v2 = - m1v1 / m2

    so

    v1 = v_rel - (- m1v1/m2)

    v1 = m2*v_rel / m1

    same process gives
    v1 = -m2v2 / m1
    v2 = v_rel - (-m2v2 / m1)

    v2 = m1*v_rel/m2
    so substituting that in gives:

    m1* [m2*v_rel / m1] + m2* [m1*v_rel/m2] = m1v`1 + m2v`2

    That still leaves me with two unknowns of v`1 and v`2.
    Unless I am supposed to use the right hand side as some (m1+m2)*v and then use that total final v to find their respective v's.

    Sorry I'm having such a mental block with this problem. I just can't get my answers to match the book.
     
  13. Feb 19, 2013 #12
    Solved it :)
     
  14. Feb 19, 2013 #13

    vela

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    Congrats! I figured you just needed some time to futz around with it.
     
  15. Feb 19, 2013 #14
    Thanks so much for your help!
     
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