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Conservation of Momentum and Thrown Objects

  1. Feb 6, 2008 #1
    [SOLVED] Conservation of Momentum and Thrown Objects

    As part of our physics class, we've been given lots of problems typically solved by taking into account the law of conservation of momentum. I've had little-to-no trouble solving them, but one problem in particular is driving me nuts.

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

    There is a 50 kg girl in a 50 kg stationary canoe holding 2 10 kg cannon balls. She throws a cannon ball at 5 m/s, and then throws the second at 5 m/s w.r.t the boat. What is the canoe's final velocity if the (incorrect) assumption is made that their is no friction between the canoe and the water? Answer: 0.87 m/s

    2. Relevant equations

    [tex]
    \vec{p} = \vec{p \prime}
    [/tex]

    3. The attempt at a solution

    Ok, so I first calculate the speed of the canoe w.r.t the ground using the law.
    [tex]
    0 = \vec{p_c} + \vec{p_b}
    [/tex]

    [tex]
    0 = 110 v_c + 10(5)
    [/tex]

    [tex]
    v_c = -5/11 m/s \approx -0.454545454... {m/s}
    [/tex]

    Next, my idea was that the fact the second ball is being thrown at 5 m/s w.r.t. the canoe while the canoe was moving was important. So I used the L. of C. of M. again but in the frame of reference of the canoe.

    [tex]
    0 = 100v_c + 5(10)
    [/tex]

    [tex]
    v_c = -1/2
    [/tex]

    So I figured I would just add the speed in the frame of reference of the boat to the speed it was traveling before. That gives me
    [tex]
    v_c \approx -0.95454545... m/s
    [/tex]

    Which is clearly not the right answer. Am I on the wrong track completely, or is there something simple I've missed? Any help would be appreciated. :)
     
  2. jcsd
  3. Feb 6, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    When she throws the balls at 5 m/s with respect to the canoe, they mean with respect to the canoe after the ball is thrown. If the canoe's velocity (after the throw) is Vc (negative, of course), the ball's velocity is Vc + 5.
     
  4. Feb 6, 2008 #3
    Thanks Doc Al. If I'm not mistaken, that leads me to this answer:

    [tex]
    \vec{p} = (-5/11)(110) = \vec{p \prime} = 100v_c + 10(-5/11 + 5)
    [/tex]

    [tex]
    -50 = 100v_c + 500/11
    [/tex]

    [tex]
    v_c = -0.9545
    [/tex]

    Which is the same as my previous attempt. I think that makes sense, as I've just switched frames of reference. What am I doing wrong?
     
  5. Feb 6, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Looks to me like you're adding 5 m/s to the speed of the canoe before the throw, not after like I suggested.

    I'd do it in two steps, having both throws be 5 m/s with respect to the canoe after the the throw. (The idea of changing frames is a good one; use that for the second throw.)
     
  6. Feb 6, 2008 #5
    When you say after the throw, do you mean after the first ball is thrown (before which, the canoe is stationary), or after the second ball is thrown by the girl on the moving canoe?
     
  7. Feb 6, 2008 #6

    Doc Al

    User Avatar

    Staff: Mentor

    After each throw. (That's how I would do it.)

    Find the speed after the first throw (with the ball being 5 m/s w.r.t. the canoe after the throw). Then repeat (after switching frames) for the second throw.
     
  8. Feb 6, 2008 #7
    Doc Al! Thank you so much for that help. It was exactly what I needed: it gave strong hints, and forced me to think.

    In case anyone is interested:
    [tex]
    0 = 110 v_c + 10(v_c + 5)
    [/tex]

    [tex]
    v_c \approx -0.417 m/s
    [/tex]

    In the frame of reference of the boat...
    [tex]
    0 = 100v_c + 10(v_c + 5)
    [/tex]

    [tex]
    v_c = -5/12 m/s
    [/tex]

    Get back to the ground frame...
    [tex]
    v_c = -5/12 - 5/11 \approx -0.87 m/s
    [/tex]

    Yay!

    EDIT: it could arguably be solved more simply in the ground frame, but I find it easier to visualize in the canoe frame.
     
    Last edited: Feb 6, 2008
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