# Hockey puck

1. Oct 13, 2004

### Speags

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 $$cv^2$$ , 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 $$v_{o}$$ 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 $$F=-(f+cv^2)$$
$$m\frac{dv}{dx}\frac{dx}{dt}=-(f+cv^2)$$
$$mvdv=-(f+cv^2)$$
$$\frac{mvdv}{(f+cv^2)}=-dx$$
takeing the intergral of both sides you get (i think)
$$\frac{m}{2c} ln(f+cv^2)=-x$$

$$x=-\frac{m}{2c}ln(f+cv^2)$$

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 $$v$$ into $$\frac{dx}{dt}$$
but i'm not sure where to start or how

2. Oct 13, 2004

### HallsofIvy

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

Now, what is x when v= 0?

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

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