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Conservation of linear momentum/Galilean relativity problem

  1. Apr 23, 2008 #1
    [SOLVED] Conservation of linear momentum/Galilean relativity problem

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


    A 45-kg girl is standing on a plank that has a mass of 150 kg. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless supporting surface. The girl begins to walk along the plank at a constant velocity of 1.5 m/s relative to the plank.

    (a) What is her velocity relative to the ice surface?

    (b) What is the velocity of the plank relative to the ice surface?


    2. Relevant equations


    [tex]m_1v_1 = m_2v_2[/tex]
    [tex]\overrightarrow{p_{1i}} + \overrightarrow{p_{2i}} = \overrightarrow{p_{1f}} + \overrightarrow{p_{2f}}[/tex]

    [tex]\overrightarrow{r}' = \overrightarrow{r} - \overrightarrow{u}t[/tex]
    [tex]\overrightarrow{v}' = \overrightarrow{v} - \overrightarrow{u}[/tex]


    3. The attempt at a solution


    mass of girl:
    [tex]m_g = 45 kg[/tex]

    mass of plank:
    [tex]m_p = 150 kg[/tex]

    velocity of girl relative to plank:
    [tex]v_{gp} = 1.5 m/s[/tex]

    velocity of plank relative to girl:
    [tex]v_{pg} = ?[/tex]

    conservation of momentum:
    [tex]m_gv_{gp} = m_pv_{pg} \Rightarrow[/tex]
    [tex]v_{pg} = \frac{m_gv_{gp}}{m_p} = \frac{(45 kg)(1.5 m/s)}{150kg} = 0.450 m/s[/tex]


    So I have the velocity of the plank relative to the girl; but I have no idea how to find the velocity of the girl or the plank relative to the ice. Can you help? Thank you.
     
  2. jcsd
  3. Apr 23, 2008 #2
    I should also note that in the back of the book, the answer for (a) is 1.15 m/s and the answer for (b) is -0.346 m/s.
     
  4. Apr 23, 2008 #3

    Dick

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    Science Advisor
    Homework Helper

    v_pg and v_gp are equal and opposite in sign. They HAVE to be, just think about it. They are the same relative velocity measured in two different frames. You can't combine them in a single conservation of momentum equation. If vg and vp are the velocities relative to the ice, conservation of momentum tells you mg*vg+mp*vp=0. v_gp is just vg-vp.
     
  5. Apr 24, 2008 #4
    That makes sense. But now I'm really confused about how to proceed ...
     
  6. Apr 24, 2008 #5

    Doc Al

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    Staff: Mentor

    Just write an expression for conservation of momentum with all velocities with respect to the ground. The only unknown will be the velocity of the plank. Hint: Assume the plank goes left (negative) and the girl goes right (positive).
     
  7. Apr 24, 2008 #6
    How can the only unknown be the velocity of the plank? What is the velocity of the girl relative to the ground?
     
  8. Apr 24, 2008 #7

    Doc Al

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    Staff: Mentor

    You can express the girl's velocity with respect to the ground in terms of the plank's velocity.
     
  9. Apr 24, 2008 #8
    Now I get it! ...

    [tex]v_{gi} = v_{gp} - v_{pi}[/tex]
    [tex]m_gv_{gi} = m_pv_{pi}[/tex]
    [tex]\Rightarrow m_g(v_{gp} - v_{pi}) = m_pv_{pi}[/tex]
    [tex]\Rightarrow m_gv_{gp} - m_gv_{pi} = m_pv_{pi}[/tex]
    [tex]\Rightarrow m_gv_{gp} = m_gv_{pi} + m_pv_{pi}[/tex]
    [tex]\Rightarrow v_{pi} = \frac{m_gv_{gp}}{m_g + m_p}[/tex]
    [tex]\Rightarrow v_{pi} = \frac{(45 kg)(1.50 m/s)}{45 kg + 150 kg} = 0.346 m/s[/tex]

    So the velocity of the plank with respect to the ice is 0.346 m/s to the left. This is the answer for part (b).

    To solve part (a) ...
    [tex]v_{gi} = v_{gp} - v_{pi}[/tex]
    [tex]\Rightarrow v_{gi} = 1.50 m/s - 0.346 m/s = 1.15 m/s (to the right)[/tex]

    Thank you all for your time and patience! Especially Doc Al!
     
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