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Newton's laws&Conservation of Momentum

  1. Sep 22, 2007 #1
    1. The problem statement, all variables and given/known data
    A boat of mass 250KG is coasting, with its engine in neutral, through the water at 3.00 m/s when it starts to rain. The rain falls vertically and accumulates in the boat at 10Kg/hr. The drag on the boat is proportional to the speec of the boast in the form Fd=0.5v^2. What the acceleration of the boat just after the rain starts. Take the posititive x-axis to be along the direction of motion. Give your answer in meters per second.

    Please help me! Thanks a lot for your help and continued effort. I really appreciate it.

    2. Relevant equations

    When the rain "just starts," Fd = 0.5 x 3 x 3 = 4.5N.
    Thus, a = 4.5/250 = 0.018m/s^s. However, we need to write -0.018 since the acceleration will be opposite to direction of motion. Am I right? The answer to this question is not given so I need help determining whether I am right or not.

    Thank-you very much for your time and effort!
    3. The attempt at a solution
  2. jcsd
  3. Sep 22, 2007 #2


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    For a system of variable mass (a so-called geometric system, rather than material system), Newton's second law reads:
    where [itex]\vec{F}[/itex] is the net force acting on the particles contained within the system, [itex]\vec{p}[/itex] is the amount of momentum contained within the system, and [itex]\vec{m}[/itex] is the momentum flux out of the system.

    In this case, the momentum flux is strictly in the perpendicular direction of the direction of motion, so limiting ourself to the component in the horizontal direction, we get:
    [tex]F=\frac{dp}{dt} (*)[/tex]
    where F is the net sum of horizontally acting forces, and p the horizontal momentum of the system, p=m(t)v(t), where m(t) is the mass of the system as a function of time, whereas v(t) is the horizontal velocity as a function of time.

    Use (*) as the basis for your calculations.
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