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Momentum of a boat problem

  1. Apr 21, 2007 #1
    1. The problem statement, all variables and given/known data

    A child in a boat throws a 6.40kg package out horizontally with a speed of 10.0 m/s. Calculate the veolcity of the boat immediately after, assuming it was initially at rest. The mass of the child is 26.0kg and that of the boat is 45.0kg. Ignore water resistance.

    2. Relevant equations

    (m1+m2+m3)v(boat final) = m1v1(initial) + m2v2(initial) + m3v3(initial)

    3. The attempt at a solution

    (6.40+26+45)v(boat final) = 6.4(10)+26(0)+45(0)
    (77.4)v(boat final)=64

    Dont seem right to me. Thanks in advance!
    v(boat final)= 0.83 m/s
    i know that we're suppose to treat it like an explosion problem. but im stuck there cuz the answer dont seem to make sense to me.
    Last edited: Apr 21, 2007
  2. jcsd
  3. Apr 21, 2007 #2
    m1 = mass of package
    m2 = mass of boat+child

    Initial momentum = Final momentum
    Initial momentum = 0 = m1v1 = m2v2

    Don't take my word for it though, i'm still very new to conservation laws myself...
  4. Apr 21, 2007 #3
    the answer is the same...but what's the initial velocity of the package? is it 10m/s then? or is that the final velocity?
  5. Apr 21, 2007 #4
    Velocity of the package is, as given, 10m/s. It will be part of final momentum.

    Initial momentum = momentum when the system (boat+boy+package) is at rest
    Final momentum = momentum at the moment of throwing of package

    Do you know what the correct answer is?
  6. Apr 21, 2007 #5
    no i dont.. that's the thing.
  7. Apr 21, 2007 #6
    Reading the question it sounds like everything is at rest initially. What does this imply about the system initially? (See Siddhartmishra's post)

    Post action we have a package travelling at 10m/s, and the boat travelling at some other speed to be determined. Recall that speed cannot determine direction per se, but it can determine backwords and forwards by a negative sign, thus we don't have to worry about vectors here. If the speed of the gift is 10m/s, what does this imply about the speed of the boat?

    By answering that you'll be able to setup your equation properly and the answer should be acceptable.
  8. Apr 21, 2007 #7
    Read my post #4 and solve accordingly by equating initial and final momentum... there is nothing wrong with the answer, it seems pretty plausible.
  9. Apr 21, 2007 #8
    ok...so this is what i got.. i hope im right!

    (6.40*0)+(26*0)+(45*0)=(6.4+26+45)v(boat final)
    the answer is 0.
  10. Apr 21, 2007 #9
    you have got that toooooo wrong:rofl: :rofl: :rofl: :rofl: :rofl: :rofl:
    the boat and chils are a system having same velocity while the package will have some other velocity


    initial momentum=0=finalmomentum = 6.4 X 10 + (boat + chiild) X V

    '-' indicates boat travels in opposite direction to the package
  11. Apr 21, 2007 #10


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    Homework Helper

    But the package isn't on the boat anymore, so you must treat it separately from the boat+child! It goes off at 10 m/s, right?
    You're initial momentum is correct, you need to work on the final now.
  12. Apr 21, 2007 #11

    Your first equation is wrong.
    You should (always) start with the full equation, here:

    m1v1(final) + m2v2(final) + m3v3(final) = m1v1(initial) + m2v2(initial) + m3v3(initial)

    (assuming for example: 1=boat, 2=children, 3=package)

    and replace carefully each term with the available information.
    In particular you have:

    v1(initial) = v2(initial) = v3(initial) = 0
    v3(final) = 10 m/s
    v1(final) = v2(final)

    Hope it helps,

  13. Apr 21, 2007 #12
    thank you guys so much! i really appreciate your help!
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