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Question Number 111340 by ajfour last updated on 03/Sep/20

Commented by ajfour last updated on 03/Sep/20

$D i a g r a m s s h o w c o l l i s i o n o f t w o b a l l s$ $o n f l a t f r i c t i o n l e s s s u r f a c e b e f o r e$ $c o l l i s i o n a n d j u s t a f t e r c o l l i s i o n .$ $c o e f f i c i e n t o f r e s t i t u t i o n b e i n g e .$ $F i n d v, V, θ, ϕ i n t e r m s o f$ $u, b, r, R, a n d e .$
Commented by ajfour last updated on 04/Sep/20

$m r W S i r w o n t y o u t r y t h i s, p l e a s e ?$
	Answered by mr W last updated on 15/Sep/20

	$μ = \frac{M}{m}$ $\sin θ = \frac{b}{r + R}$ $V - v \cos (θ + ϕ) = e u \cos θ$ $\Rightarrow V = v \cos (θ + ϕ) + e u \cos θ$ $m u \cos θ = M V + m v \cos (θ + ϕ)$ $\Rightarrow u \cos θ = μ V + v \cos (θ + ϕ)$ $m u \sin θ = m v \sin (θ + ϕ)$ $\Rightarrow u \sin θ = v \sin (θ + ϕ)$ $\Rightarrow \frac{v}{u} = \frac{1}{\cos ϕ + \frac{\sin ϕ}{\tan θ}}$ $u \cos θ = μ v \cos (θ + ϕ) + μ e u \cos θ + v \cos (θ + ϕ)$ $\Rightarrow (1 - μ e) = \frac{v}{u} (1 + μ) (\cos ϕ - \tan θ \sin ϕ)$ $\Rightarrow \frac{1 - μ e}{1 + μ} = \frac{\cos ϕ - \tan θ \sin ϕ}{\cos ϕ + \frac{\sin ϕ}{\tan θ}}$ $\Rightarrow \frac{1 - μ e}{1 + μ} = \frac{1 - \tan θ \tan ϕ}{1 + \frac{\tan ϕ}{\tan θ}}$ $\Rightarrow [\tan θ + \frac{1 - μ e}{(1 + μ) \tan θ}] \tan ϕ = \frac{μ (1 + e)}{1 + μ}$ $\Rightarrow \tan ϕ = \frac{μ (1 + e)}{(1 + μ) \tan θ + \frac{1 - μ e}{\tan θ}}$
	Commented by ajfour last updated on 23/Sep/20

	$t h a n k s S i r!$
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