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Question Number 54396 by mr W last updated on 03/Feb/19

Commented by mr W last updated on 03/Feb/19

$b l o c k 1 w i t h m a s s m_{1} m o v e s w i t h$ $v e l o c i t y u t o w a r d s b l o c k 2 w i t h m a s s$ $m_{2} . a t h i r d b l o c k i s p l a c e d b e t w e e n t h e m .$ $f i n d t h e m a s s o f t h e t h i r d b l o c k s u c h$ $t h a t t h e v e l o c i t y o f b l o c k 2 i s m a x i m u m .$ $f i n d a l s o v_{m a x} .$ $c a s e 1 : a l l c o l l i s i o n s a r e e l a s t i c .$ $c a s e 2 : a l l c o l l i s i o n s h a v e a c o e f f i c i e n t$ $o f r e s t i t u t i o n o f e .$ $w h a t i f w e p l a c e m o r e t h a n o n e b l o c k$ $b e t w e e n t h e m ?$
	Answered by mr W last updated on 06/Feb/19

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