A particle of mass m moving with speed u collides perfectly inelastically with a sphere of radius R and same mass, at rest, at an impact parameter d. Find (a) Angle between their final velocities (b) Magnitude of their final velocities


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Question Number 25091 by Tinkutara last updated on 03/Dec/17

$A p a r t i c l e o f m a s s m m o v i n g w i t h$ $s p e e d u c o l l i d e s p e r f e c t l y i n e l a s t i c a l l y$ $w i t h a s p h e r e o f r a d i u s R a n d s a m e$ $m a s s, a t r e s t, a t a n i m p a c t p a r a m e t e r$ $d . F i n d$ $(a) A n g l e b e t w e e n t h e i r f i n a l v e l o c i t i e s$ $(b) M a g n i t u d e o f t h e i r f i n a l$ $v e l o c i t i e s$
	Answered by ajfour last updated on 03/Dec/17

	Commented by ajfour last updated on 04/Dec/17

	$c o n s e r v i n g l i n e a r m o m e n t u m$ $m u = 2 m v + m ω R \sin θ$ $a n d \sin θ = d / R, s o$ $2 v + ω d = u . . . (i)$ $c o n s e r v i n g a n g u l a r m o m e n t u m$ $a b o u t g r o u n d p o i n t G :$ $m u (R + d) = m v R + \frac{2}{5} {m R}^{2} ω +$ $m (v + ω R \sin θ) (R + d) + m ω {(R \cos θ)}^{2}$ $. . . . (i i)$ $a n d a s {(R \cos θ)}^{2} = (R + d) (R - d),$ $a l s o R \sin θ = d$ $\Rightarrow u (R + d) = v R + \frac{2}{5} ω R^{2} +$ $(v + ω d) (R + d) + ω (R^{2} - d^{2})$ $f r o m (i) : v + ω d = u - v, s o$ $v d = ω (\frac{7}{5} R^{2} - d^{2})$ $a n d f r o m (i) : 2 v + ω d = u$ $\Rightarrow 2 ω (\frac{7}{5} R^{2} - d^{2}) + ω d^{2} = u d$ $ω = \frac{u d}{(\frac{14}{5} R^{2} + d^{2})}; v = \frac{u (\frac{7}{5} R^{2} - d^{2})}{(\frac{14}{5} R^{2} + d^{2})}$ $\tan ϕ = \frac{ω R \cos θ}{v + ω d} = \frac{d \sqrt{R^{2} - d^{2}}}{(v d / ω) + d^{2}}$ $\tan ϕ = \frac{5 d \sqrt{R^{2} - d^{2}}}{7 R^{2}} .$ $V = \frac{ω R \cos θ}{\sin ϕ} = ω [\frac{\sqrt{R^{2} - d^{2}}}{\sin ϕ}] .$
	Commented by ajfour last updated on 03/Dec/17

	$∠ b e t w e e n t h e i r f i n a l v e l o c i t i e s$ $i s ϕ .$ $f i n a l v e l o c i t y o f s p h e r e v .$ $F i n a l v e l o c i t y o f p a r t i c l e i s V .$
	Commented by Tinkutara last updated on 05/Dec/17

	$A n s w e r i s :$ $v_{1} = \frac{u d}{R}, v_{2} = - v_{3} = \frac{u \sqrt{R^{2} - d^{2}}}{2 R}$
	Commented by ajfour last updated on 05/Dec/17

	$w h i c h a n s w e r i s c o r r e c t f o r$ $d = 0, a n d e v e n d = R ? P l e a s e$ $c h e c k o r i t h i n k i h a v e$ $m i s u n d e r s t o o d t h e q u e s t i o n .$ $m r W s i r c a n h e l p i f h e f i n d s$ $t i m e a n d n o t i c e s m y r e q u e s t .$
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