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Hockey puck

  1. Oct 13, 2004 #1
    a hockey puck of mass m is sliding in the +x direction across a horizontal ice surface. while sliding, the puck is subject to two forces that oppose its motion: a constant sliding friction force of magnitude f, and a air resistance force of magnitude [tex] cv^2 [/tex] , where c is a constant and v is the puck's velocity. At time t=0, the puck's position is x=0, and it's velocity is [tex] v_{o} [/tex] In terms of the given parameters (m,f,c, and v_o), determine:
    a) how far the puck slides, that is determine it's position x when it comes to rest;
    for a) i got [tex] F=-(f+cv^2) [/tex]
    [tex]m\frac{dv}{dx}\frac{dx}{dt}=-(f+cv^2) [/tex]
    [tex] mvdv=-(f+cv^2) [/tex]
    [tex] \frac{mvdv}{(f+cv^2)}=-dx [/tex]
    takeing the intergral of both sides you get (i think)
    [tex] \frac{m}{2c} ln(f+cv^2)=-x [/tex]

    [tex] x=-\frac{m}{2c}ln(f+cv^2) [/tex]

    now the b) part asks how long does the puck slide, that is, determine the time t at which it comes to rest.
    i think i need to turn a [tex] v [/tex] into [tex] \frac{dx}{dt} [/tex]
    but i'm not sure where to start or how
  2. jcsd
  3. Oct 13, 2004 #2


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    You did fine up to this point:
    [tex] \frac{m}{2c} ln(f+cv^2)=-x [/tex]
    Should be
    [tex] \frac{m}{2c} ln(f+cv^2)=-x + C[/tex]
    where "C" is a constant of integration. If we take x=0 initially, then
    [tex]\frac{m}{2c}ln(f+cv_0^2)= C[/tex]
    so you have
    [tex] \frac{m}{2c} ln(f+cv^2)=-x+ \frac{m}{2c}ln(f+cv_0^2) [/tex]
    and you may want to write that as
    [tex]\frac{m}{2c}(ln(f+cv^2)-ln(f+cv_0^2))= -x[/tex]
    [tex]x= \frac{m}{2c}(ln(\frac{f+cv_0^2}{f+cv^2})[/tex]

    Now, what is x when v= 0?

    The simplest thing to do is go back to your original equation:
    [tex]m\frac{dv}{dt}= -(f+cv^2)[/itex]
    and don't convert to x. You get
    [tex] m\frac{dv}{f+cv^2}= -dt[/tex]
    Can you integrate that? (Think: arctangent.)
    Remember that v= v0 when t= 0 and solve for t when v= 0.
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