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Help! Conservation of momentum: contestant running on raft. Tough

  1. Feb 26, 2008 #1
    1. The problem statement, all variables and given/known data

    A 62.7-kg woman contestant on a reality television show is at rest at the south end of a horizontal 141-kg raft that is floating in crocodile-infested waters. She and the raft are initially at rest. She needs to jump from the raft to a platform that is several meters off the north end of the raft. She takes a running start. When she reaches the north end of the raft she is running at 4.78 m/s relative to the raft. At that instant, what is her velocity relative to the water?

    [tex]v[/tex] = velocity of contestant with respect to the water
    [tex]v_{r}[/tex] = velocity of contestant with respect to the raft
    [tex]V[/tex] = velocity of raft with respect to the water

    [tex]m[/tex] = mass of contestant
    [tex]M[/tex] = mass of raft

    2. Relevant equations

    [tex]P_{I} = P_{f}[/tex]


    3. The attempt at a solution

    Here is my derivation and equation but I'm not getting the right answer! Where did I go wrong?

    1. [tex]P_{I} = P_{f}[/tex]

    2. [tex]0=MV+mv[/tex]

    We don't know V so I substitute with:


    3. [tex]0=M(v-v_{r}) + mv[/tex]

    Now solve for v (speed of contestant relative to water)

    Last edited: Feb 26, 2008
  2. jcsd
  3. Feb 27, 2008 #2
    Can anybody help? It would be awesome because this is due soon.
  4. Feb 27, 2008 #3
    I came up with:

    [tex] v=\frac{(M+m)v_{r}}{M+2m}[/tex]

    but it gives me the wrong answer.

    Not sure what else to do, I swear that equation is correct. I even derived it and compared it to a similar problem in our book and it was very similar! The only difference is that it is asking for the velocity of the raft w/ respect to the water instead of the contestant w/ respect to the water.
  5. Feb 10, 2009 #4
    This is a lot more simple than it looks. All you have to do is find the velocity of the raft. To do this take the woman's mass multiplied by her velocity (mv) and set it equal to the her mass plus that of the raft times v (m + M)v. So you get mv = (M + m)v. When you put the numbers in you get 62.7kg(4.78m/s) = (62.7 + 141)v

    Once you've solved this you get v = 1.47. Since the raft is moving in the opposite direction from the woman v is actually -1.47. To relate this to the water, all you have to do is take the woman's velocity 4.78 m/s and subtract the water's velocity 1.47 m/s. Your answer comes out to 3.31 m/s.
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