- #1

Habeebe

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## Homework Statement

I kick a puck of mass m up an incline (angle of slope = θ) with intial speed v

_{0}. There is no friction between the puck and the incline, but there is air resistance with magnitude f(v) = cv

^{2}. Write down and solve Newton's second law for the puck's velocity as a function of t on the upward journey. How long does the upward journey last?

## Homework Equations

F=ma=m*[itex]\frac{dv}{dt}[/itex]

According to Wolfram Alpha (I use this later):

[itex]\int \frac{dx}{a+bx^2} = \frac{arctan(\frac{\sqrt{b}x}{\sqrt{a}})}{\sqrt{ab}}[/itex]

## The Attempt at a Solution

I set the axes so x is along the ramp in the direction v

_{0}and y is normal to the ramp upwards. This gives force and acceleration in the x direction only.

F=weight+resistance=-mg*sin(θ)-cv

^{2}

a=[itex]\frac{dv}{dt}[/itex]=-gsin(θ)-cv

^{2}/m

Separation of variables gets me to:

[itex]\frac{-dv}{gsin(\theta)+\frac{cv^2}{m}} = dt[/itex]

I didn't know offhand how to do the integral, and it looked fishy, so I Wolfram Alpha'd it to see if I get something that makes sense before I figure out the method. Using that solution I with limits of v from v

_{0}to v and t from 0 to t I get:

[itex]\frac{ arctan(v*\sqrt{\frac{c}{ mgsin\theta }}) } {sqrt{\frac{cgsin(\theta)}{m}}} |^{v}_{v_0} = -t [/itex]

And this is a jumbly mess. I can't really tell if I'm right or not because I can't identify intuitively what parts of the expression on the left stand for what. My gut feeling is that this can't be right because the answer is so absurdly ugly.

I also tried using [itex]\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v[/itex] on my original equation, ultimately getting:

[itex]\frac{cv^2/m+gsin\theta}{cv_{0}^{2}/m+gsin\theta}=e^{-2cx/m}[/itex]

But this is a function v(x(t)) and I'm not really sure how to go about solving that for v(t).

Thanks for the help.

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