Hockey Puck Motion: Determining Position & Time to Rest

[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

 

[HallsofIvy]

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?

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}[/tex] <br /> but I&#039;m not sure where to start or how

<br /> <br /> The simplest thing to do is go back to your original equation:<br /> m\frac{dv}{dt}= -(f+cv^2)<br /> and <b>don&#039;t</b> convert to x. You get<br /> m\frac{dv}{f+cv^2}= -dt<br /> Can you integrate that? (Think: arctangent.) <br /> Remember that v= v₀ when t= 0 and solve for t when v= 0.

 

[wais]

to solve for t

To determine the time at which the puck comes to rest, we can use the fact that when the puck comes to rest, its velocity will be equal to 0. So, we can set v=0 in the equation we derived in part a):

\frac{mvdv}{(f+cv^2)}=-dx

0=-dx

We can then integrate both sides to solve for t:

\int_{0}^{t}\frac{mvdv}{(f+cv^2)}=\int_{0}^{t}-dx

\frac{m}{2c}ln(f+cv^2)|_{0}^{t}=-x|_{0}^{t}

\frac{m}{2c}(ln(f+cv^2)|_{t}-ln(f+cv^2)|_{0})=-x|_{t}-x|_{0}

Since x|_{0}=0, we can simplify to:

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

Now, we can substitute in the values given in the problem to solve for t:

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

\frac{m}{2c}ln(f+ct^2)|_{t}=-x|_{t}

\frac{m}{2c}ln(f+ct^2)|_{t}=-\frac{m}{2c}ln(f+cv^2)|_{t}

\frac{m}{2c}ln(f+ct^2)=\frac{m}{2c}ln(f+cv^2)

ln(f+ct^2)=ln(f+cv^2)

f+ct^2=f+cv^2

ct^2=cv^2

t^2=\frac{cv^2}{c}

t=\sqrt{\frac{v^2}{c}}

Therefore, the time at which the puck comes to rest is:

t=\sqrt{\frac{v^2}{c}}

We can simplify this further by substituting in the value for v from the given information:

t=\sqrt{\frac{v_o^2}{c}}

So, the puck will come to rest after a time of t=\sqrt{\frac{