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Question Number 104987 by mr W last updated on 25/Jul/20

Commented by mr W last updated on 25/Jul/20

$a b l o c k w i t h m a s s m_{1} a n d v e l o c i t y v_{1}$ $c o l l i d e s w i t h a n o t h e r b l o c k w i t h$ $m a s s m_{2} i n r e s t . a f t e r t h e c o l l i s i o n$ $t h e b l o c k m_{2} h a s t h e v e l o c i t y v_{2} .$ $i f t h e c o l l i s i o n i s e l a s t i c, f i n d v_{2} .$ $i f w e p l a c e a n a d d i t i o n a l b l o c k b e t w e e n$ $m_{1} a n d m_{2}, f i n d t h e m a x i m a l$ $p o s s i b l e v e l o c i t y w h i c h t h e b l o c k m_{2}$ $m a y g e t .$ $w h a t i s t h e a n s w e r i f w e p l a c e n (n ⩾ 1)$ $a d d i t i o n a l b l o c k s b e t w e e n m_{1} a n d m_{2} ?$
Commented by ajfour last updated on 25/Jul/20

$G r e a t q u e s t i o n S i r, i c a n o n l y h o p e$ $t o c o m p r e h e n d t h e s o l u t i o n w h e n$ $y o u s h a l l p o s t i t, t o u g h o n e f o r m e$ $t o e v e n t r y a n d a t t e m p t . .$
	Answered by mr W last updated on 28/Jul/20

	$c o l l i s i o n m_{1} w i t h m_{2} d i r e c t l y :$ $l e t u_{1} = s p e e d o f m_{1} a f t e r c o l l i s i o n$ $v_{1} = v_{2} - u_{1}$ $m_{1} v_{1} = m_{1} u_{1} + m_{2} v_{2}$ $m_{1} v_{1} = m_{1} (v_{2} - v_{1}) + m_{2} v_{2}$ $2 v_{1} = (1 + \frac{m_{2}}{m_{1}}) v_{2}$ $\Rightarrow v_{2} = \frac{2 v_{1}}{1 + \frac{m_{2}}{m_{1}}}$ $w i t h a n a d d i t i o n a l b l o c k M_{1} :$ $V_{1} = v e l o c i t y o f M_{1} a f t e r c o l l i s i o n$ $V_{1} = \frac{2 v_{1}}{1 + \frac{M_{1}}{m_{1}}}$ $v_{2} = \frac{2 V_{1}}{1 + \frac{m_{2}}{M_{1}}} = \frac{2^{2} v_{1}}{(1 + \frac{M_{1}}{m_{1}}) (1 + \frac{m_{2}}{M_{1}})}$ $i n g e n e r a l w i t h n a d d i t i o n a l b l o c k s :$ $M_{1}, M_{2}, . . ., M_{n}$ $v_{2} = \frac{2^{n + 1} v_{1}}{(1 + \frac{M_{1}}{m_{1}}) (1 + \frac{M_{2}}{M_{1}}) (1 + \frac{M_{3}}{M_{2}}) . . . (1 + \frac{m_{2}}{M_{n}})}$ $v_{2} i s m a x i m u m, i f$ $\frac{M_{1}}{m_{1}} = \frac{M_{2}}{M_{1}} = \frac{M_{3}}{M_{2}} = . . . = \frac{m_{2}}{M_{n}} = k$ $\Rightarrow \frac{m_{2}}{m_{1}} = k^{n + 1}$ $\Rightarrow k = \sqrt[n + 1]{\frac{m_{2}}{m_{1}}}$ $\Rightarrow M_{1} = {k m}_{1}$ $\Rightarrow M_{2} = k^{2} m_{1}$ $\Rightarrow M_{n} = k^{n} m_{1}$ $\Rightarrow v_{2, m a x} = \frac{2^{n + 1} v_{1}}{{(1 + k)}^{n + 1}} = {(\frac{2}{1 + \sqrt[n + 1]{\frac{m_{2}}{m_{1}}}})}^{n + 1} v_{1}$
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