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This is the problem I am currently attempting. So far, I've resolved the velocity parallel and perpendicular to the plane to get, perpendicular: $u \sin \theta$ upon launch and $-u\sin \theta$ on landing.

Parallel: $u \cos \theta - 2u \sin \theta \tan \alpha$.

Where do I go from here?

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  • $\begingroup$ General advice: try drawing a diagram $\endgroup$
    – user854
    Commented Jan 28, 2016 at 13:53
  • $\begingroup$ @barrycarter I've updated my question with my latest try. $\endgroup$
    – Zain Patel
    Commented Jan 28, 2016 at 14:06
  • $\begingroup$ I still don't see a diagram? Even a hand drawn one would be fine. $\endgroup$
    – user854
    Commented Jan 28, 2016 at 15:06
  • $\begingroup$ @ZainPatel Try and present a diagram for the problem. $\endgroup$ Commented Jan 28, 2016 at 15:29
  • $\begingroup$ Hi Zain. Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$ Commented Jan 29, 2016 at 6:57

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@EDIT: SOLVED IT! :) The key part I was missing was that $e$ only acts on the component perpendicular to the slope (the y-component), i.e $u_{r,x} = v_x$ and $u_{r,x} = -e v_y$. Huge thanks to @Floris for spotting this, and all the help!!

Always start with a diagram :)

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I tried solving this two ways: relative to vertical/horizontal and relative to the slope. The reflection for vertical/horizontal is horrendous, and it works out much neater to just resolve relative to the slope.

Relative to the slope of the plane


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Projection

\begin{align} {v_x \choose v_y} &= {u_x + a_x t_1\choose u_y + a_y t_1} \\ {s_x \choose 0} &= {u_x t_1 + \frac{1}{2}a_x t_1^2\choose u_y t_1 + \frac{1}{2}a_y t_1^2} \end{align}

where $u_x = u\cos(\theta)$, $u_y = u\sin(\theta)$, $a_x = -mg \sin(\alpha)$, $a_y = -mg \cos(\alpha)$ and $s_y = 0$.

Reflection

$$ \vec{u_{r}} = \vec{v} - 2(\vec{v} \cdot \hat{n})\hat{n} $$ where $\hat{n} = {0 \choose 1}$ is the unit normal vector to the slope. $$ {u_{r,x} \choose u_{r,y}} = {v_x \choose v_y} - 2 ({v_x \choose v_y} \cdot {0 \choose 1}) {0 \choose 1} $$ Rearrange the equation and remember that the velocity perpendicular to the plane is reduced by a factor of $e$. $$ {u_{r,x} \choose u_{r,y}} = {v_x \choose -ev_y} $$

Rebound \begin{align} {-s_x \choose -s_y} &= {u_{r,x} t_2 + \frac{1}{2}a_x t_2^2\choose u_{r,y} t_2 + \frac{1}{2}a_y t_2^2} \\ {s_x \choose 0} &= {-v_x t_2 - \frac{1}{2}a_x t_2^2 \choose -ev_y t_2 + \frac{1}{2}a_y t_2^2} \end{align}

By using $s_y = 0$, we can immediately solve for $t_1$ and $t_2$ $$ t_1 = -\frac{2u_y}{a_y} \quad t_2 = \frac{2ev_y}{a_y} $$ By plugging our solution for $t_1$ into $v_y = u_y + a_y t_1$, we get $$ v_y = -u_y $$ which in hindsight is obvious, because acceleration perpendicular to the plane is constant.

Now for the fun part: setting $s_x$ during projection equal to the $s_x$ during the rebound. $$ u_x t_1 + \frac{1}{2}a_x t_1^2 = -v_x t_2 - \frac{1}{2}a_x t_2^2 $$ Plug in for time $$ [u_x + \frac{1}{2}a_x (-\frac{2u_y}{a_y}) ](-\frac{2u_y}{a_y}) = [-v_x - \frac{1}{2}a_x \frac{2ev_y}{a_y}] \frac{2ev_y}{a_y} $$ Cancel $\frac{2}{a_y}$ from both sides $$ [- u_x + u_y\frac{a_x}{a_y} ]u_y = [-v_x - ev_y \frac{a_x}{a_y}] ev_y $$ Multiply both sides by $-1$ and factor out $u_y$ from the left and $v_y$ from the right. $$ [\frac{u_x}{u_y} - \frac{a_x}{a_y} ]u_y^2 = [\frac{v_x}{v_y} + e\frac{a_x}{a_y}] ev_y^2 $$ Remember that $v_y = -u_y$, so we can cancel $v_y^2$ and $u_y^2$ from both sides. $$ \frac{u_x}{u_y} - \frac{a_x}{a_y} = e\frac{v_x}{v_y} + e^2\frac{a_x}{a_y} $$ Now we want to plug in for $v_x/v_y$ $$ v_x = u_x + a_x t_1 = u_x -2 \frac{a_x}{a_y}u_y = (\frac{u_x}{u_y} - 2 \frac{a_x}{a_y})u_y \\ \therefore \frac{v_x}{v_y} = 2 \frac{a_x}{a_y} - \frac{u_x}{u_y} $$ Plug this back in $$ \frac{u_x}{u_y} - \frac{a_x}{a_y} = e(2 \frac{a_x}{a_y} - \frac{u_x}{u_y}) + e^2\frac{a_x}{a_y} $$ This is a quadratic in $e$, so rearrange into an obviously quadratic form $$ \frac{a_x}{a_y} e^2 + (2 \frac{a_x}{a_y} - \frac{u_x}{u_y})e + \frac{a_x}{a_y} - \frac{u_x}{u_y} = 0 $$ Solve using the quadratic formula $e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ \begin{align} b^2 - 4ac &= (2 \frac{a_x}{a_y} - \frac{u_x}{u_y})^2 - 4 (\frac{a_x}{a_y})(\frac{a_x}{a_y} - \frac{u_x}{u_y}) \\ &= 4 (\frac{a_x}{a_y})^2 - 4 \frac{a_x}{a_y}\frac{u_x}{u_y} + (\frac{u_x}{u_y})^2 - 4(\frac{a_x}{a_y})^2 + 4\frac{a_x}{a_y}\frac{u_x}{u_y} \\ &= (\frac{u_x}{u_y})^2 \end{align} $$ e = \frac{(\frac{u_x}{u_y} - 2 \frac{a_x}{a_y}) \pm \frac{u_x}{u_y}}{2\frac{a_x}{a_y}} \\ \therefore \quad e_- = -1, \quad e_+ = \frac{u_x}{u_y}\frac{a_y}{a_x} - 1 $$ Remember the definitions from the start: $u_x = u\cos(\theta)$, $u_y = u\sin(\theta)$, $a_x = -mg \sin(\alpha)$, $a_y = -mg \cos(\alpha)$, so $$ \frac{u_x}{u_y} = \cot(\theta) \quad \frac{a_x}{a_y} = \tan(\alpha) $$ Plug this into our expression for $e_+$ and voila!

$$ e_+ = \cot(\theta) \cot(\alpha) - 1 $$

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    $\begingroup$ Upvoting for the valiant effort and the "always start with a diagram" comment. I think that you may have made this overly complicated by rotating the frame of reference; and I don't think that $v_y = - u_y$ when the coefficient of restitution is less than 1. It is an open question in my mind whether the velocity along the surface should be unchanged during the collision - I suppose if it is a point mass without friction, the answer is "yes". $\endgroup$
    – Floris
    Commented Mar 7, 2016 at 21:05
  • $\begingroup$ Thanks Floris! I tried it both ways: with and without the frame rotation. Without the frame rotation, the reflection expression is worse but the acceleration vector simplifies nicely. I'll have another bash at it from that viewpoint. With regards to $v_y = - u_y$, I believe that's correct although admittedly confusing - my $v$ is the velocity at impact at point A, not the rebound velocity. So I believe the coefficient of restitution doesn't get involved yet ( I let $u_r$ be the rebound velocity). $\endgroup$
    – Judge
    Commented Mar 7, 2016 at 21:12
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    $\begingroup$ From the diagram it looks like $u$ is the inbound velocity, and $v$ the rebound... $\endgroup$
    – Floris
    Commented Mar 7, 2016 at 21:45
  • $\begingroup$ Sorry! You're very right - the first diagram's notation is incorrect. Please use the "relative to the slope of the plane" diagram, until I get round to editing the first diagram. Thanks for the help! :) $\endgroup$
    – Judge
    Commented Mar 7, 2016 at 21:48

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