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Thread: classical mechanics problem

  1. #1
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    classical mechanics problem

    A ball is thrown with initial speed $\displaystyle v_0$ up an inclined plane. the plane is at an angle $\displaystyle \phi$ and the ball's initial velocity is at angle $\displaystyle \theta$. Choose an axis with x measured up the slope, y normal, and z across it. Write down Newton's second law using these axes and find the ball's position as a function of time. Show that the ball lands a distance

    $\displaystyle R=\dfrac{2v_0^2\sin{\theta}\cos(\theta+\phi)}{\cos ^2(\phi)}$

    from it's launch point.

    Show that for a given $\displaystyle v_0$ and $\displaystyle \phi$ that the max range is $\displaystyle R_{max}=\dfrac{v_0^2}{g(1+\sin\theta)}$
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Um... have you tried anything?
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  3. #3
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    I've tried many things but none of them get me even close to getting that result. I know that the force in the x direction is $\displaystyle mg\sin\theta$ and so the displacement formula has this subed for g without the for the displacement in the x direction. The displacement in the x direction is $\displaystyle v_0\cos\theta t$ and the y is normal so there's no displacement in the y.
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  4. #4
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    Ok I'm coming back to this now.

    I took a look at the derivation for this problem in two dimensions for a projectile and tried to model my approach after it. This is how far I got.

    $\displaystyle 0=x(t)= v_0 \sin(\phi)t-\frac{1}{2}g \cos(\theta)t^2$

    then from this we can get

    $\displaystyle t=\dfrac{2v_0\sin(\phi)}{g\cos(\theta)}$

    Then as in the 2d simple case I multiplied $\displaystyle t v_0$ to obtain $\displaystyle R$

    but this doesn't give the result I was looking for.

    Can anyone help me with this?
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  5. #5
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    Quote Originally Posted by magus View Post
    A ball is thrown with initial speed $\displaystyle v_0$ up an inclined plane. the plane is at an angle $\displaystyle \phi$ and the ball's initial velocity is at angle $\displaystyle \theta$. Choose an axis with x measured up the slope, y normal, and z across it. Write down Newton's second law using these axes and find the ball's position as a function of time. Show that the ball lands a distance

    $\displaystyle R=\dfrac{2v_0^2\sin{\theta}\cos(\theta+\phi)}{g\co s^2(\phi)}$

    note the correction in your formula for R from your original post.
    $\displaystyle \displaystyle \Delta x = v_0\cos{\theta} \cdot t - \frac{1}{2}g\sin{\phi} \cdot t^2$

    $\displaystyle \displaystyle \Delta y = v_0\sin{\theta} \cdot t - \frac{1}{2}g\cos{\phi} \cdot t^2$

    since $\displaystyle \Delta y = 0$ ...

    $\displaystyle \displaystyle t = \frac{2v_0\sin{\theta}}{g\cos{\phi}}$

    substituting for $\displaystyle t$ in the $\displaystyle \Delta x$ equation ...

    $\displaystyle \displaystyle \Delta x = \frac{2v_0^2 \sin{\theta} \cos{\theta}}{g\cos{\phi}} - \frac{2v_0^2 \sin^2{\theta} \sin{\phi}}{g\cos^2{\phi}}$

    common denominator ...

    $\displaystyle \displaystyle \Delta x = \frac{2v_0^2 \sin{\theta} \cos{\theta}\cos{\phi}}{g\cos^2{\phi}} - \frac{2v_0^2 \sin^2{\theta} \sin{\phi}}{g\cos^2{\phi}}$

    combine and factor ...

    $\displaystyle \displaystyle \Delta x = \frac{2v_0^2 \sin{\theta}(\cos{\theta}\cos{\phi} - \sin{\theta}\sin{\phi})}{g\cos^2{\phi}}$

    finally, note that ...

    $\displaystyle \cos{\theta}\cos{\phi} - \sin{\theta}\sin{\phi} = \cos(\theta+\phi)$

    ... and you're there.
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  6. #6
    MHF Contributor Unknown008's Avatar
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    I think there is a 'g' missing in your first part.

    Make a sketch.

    (I'll be using v instead of vo to make it easier to type)

    The velocity of the ball up the plane is $\displaystyle v \cos\theta$
    That perpendicular to the plane is $\displaystyle v \sin\theta$

    The acceleration is down the plane and is given by $\displaystyle g\ sin\phi$
    And that perpendicular to the plane is given by $\displaystyle g \cos\phi$

    From this, the distance perpendicular to the plane where the particle lands is 0 and is given by:

    $\displaystyle s = ut + \dfrac12 at^2$

    This:

    $\displaystyle 0 = v\sin\theta t - \dfrac12 g\cos\phi t^2$

    For the distance along the plane, we have:

    $\displaystyle s = v\cos\theta t - \dfrac12 g\sin\phi t^2$

    Can you complete the first part now?

    EDIT: Didn't see you replied Skeeter
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  7. #7
    MHF Contributor Unknown008's Avatar
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    For the second part now. (there is also a mistake, see below)

    $\displaystyle R = \dfrac{2v^2\sin\theta\cos(\theta+\phi)}{g\cos^2\ph i}$

    Use the identity: $\displaystyle 2\sin A\cos B = \sin(A+B) + \sin(A-B)$

    This gives:

    $\displaystyle R = \dfrac{v^2(\sin(2\theta + \phi) + \sin(-\phi))}{g\cos^2(\phi)}$

    Simplify:

    $\displaystyle R = \dfrac{v^2(\sin(2\theta + \phi) - \sin(\phi))}{g(1+\sin\phi)(1 - \sin\phi)}$

    At the maximum range, $\displaystyle \sin(2\theta + \phi) = 1$ since only theta can vary.

    Hence we get:

    $\displaystyle R = \dfrac{v^2(1 - \sin(\phi))}{g(1+\sin\phi)(1 - \sin\phi)}$

    Something cancels out, giving:

    $\displaystyle R_{max} = \dfrac{v^2}{g(1+\sin\phi)}$

    From this, you can even find the relation between theta and phi for this value of range
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  8. #8
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    Thanks. The $\displaystyle \Delta y$ is the other equation I really needed I guess.
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