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Physics help on speed of boat

  1. Apr 1, 2005 #1
    A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 3.00 m/s when it starts to rain. The rain is falling vertically, and it accumulates in the boat at the rate of 10.0 kg/hr.

    using this formula: mb*vb/(mb+k0+t1)

    i got an answer of 2.86, but it's incorrect. it tells me im close, but wrong. do i need to convert the units or something?
     
  2. jcsd
  3. Apr 1, 2005 #2
    What is the question asking you? Whats k_0? t_1?
     
  4. Apr 1, 2005 #3
    oops, so sorry. i forgot to post the question.

    What is the speed of the boat after time 2.00 hr has passed? Assume that the water resistance is negligible.

    T_1 is just time(2hr) and K_0 is 10.0 kg/hr
     
  5. Apr 1, 2005 #4
    Is this calc based or algebra based?
    And how did you derive your eq?
     
  6. Apr 1, 2005 #5

    xanthym

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    You seem to correctly be trying to use conservation of horizontal momentum:
    vb(t=2 hr) = mb(t=0)*vb(t=0)/mb(t=2 hr)
    Recalculate your value. (Should be about 2.778 m/sec).


    ~~
     
    Last edited: Apr 1, 2005
  7. Apr 1, 2005 #6

    dextercioby

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    Hold on,a second.U can't add things with different units.A flow (Kg/hr) with a mass (Kg) with a time (s)...

    What's the initial momentum of the boat?What's the boat's momentum after 2hrs??

    Daniel.
     
  8. Apr 1, 2005 #7
    thanks alot, 2.778 was correct and i know how you got it, but i dont UNDERSTAND how it works, can someone help?

    "What's the initial momentum of the boat?What's the boat's momentum after 2hrs??"
    p=mv
    initial momentum = 250*3
    after 2hr = 250*(10*2*3) (for this one, do you use the momentum formula also?)
    cause what would be v?
     
  9. Apr 1, 2005 #8
    Well this is basically an inelastic collision right? You were going a certain speed and you gained mass. What is your new speed?

    [tex] m_iv_i = m_fv_f [/tex]

    Your final momentum is [tex] m_fv_f [/tex]. Your final mass is not 250(10*2*3), that would mean you gained about 50 times your mass. In fact your final mass is [tex] mass_{boat} + mass_{rain} [/tex]. How much mass did the rain add to you?

    using [tex] p_f = m_fv_f [/tex] you can find your final momentum.

    Do you see how this works?
     
  10. Apr 1, 2005 #9
    ahh... thanks alot!
     
  11. Apr 9, 2005 #10
    I got stuck on the 3rd part of this question:

    Assume that the boat is subject to a drag force F due to water resistanceThe drag is proportional to the square of the speed of the boat, in the form F=0.5v^2. What is the acceleration of the boat just after the rain starts?

    I don't even know how to go about approaching this part
     
  12. Apr 9, 2005 #11

    dextercioby

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    Does that mean we can neglect varying mass & momentum of the boat due to the water falling down?

    Daniel.
     
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