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Question Number 29633 by Tinkutara last updated on 10/Feb/18

Commented by Tinkutara last updated on 10/Feb/18
http://ibb.co/cyaemS
Commented by 33 last updated on 11/Feb/18

$w h a t a r e t h e m a s s e s o f n b l o c s ?$
Commented by mrW2 last updated on 11/Feb/18

$J u s t a t h o u g h t, n o t a p r o o f :$ $m a x . p o s s i b l e s p e e d o f l a s t m a s s m_{2} :$ $u = v {[\frac{2}{1 + {(\frac{m_{2}}{m_{1}})}^{\frac{1}{n + 1}}}]}^{n + 1}$ $e . g . m_{2} = 2 m_{1}, n = 10 :$ $u \approx 0.703 v$ $e . g . m_{2} = 0.5 m_{1}, n = 10 :$ $u \approx 1.406 v$ $\lim_{n \to \infty} u = \sqrt{\frac{m_{1}}{m_{2}}} v$ $t h i s i s t h e m a x . p o s s i b l e s p e e d w h i c h$ $t h e m a s s m_{2} c a n a c h i e v e b y p l a c e i n g$ $i n f i n i t e l y m a n y b l o c k s b e t w e e n m_{1}$ $a n d m_{2} .$
Commented by Tinkutara last updated on 11/Feb/18

Commented by Tinkutara last updated on 11/Feb/18
This is the answer. I am getting a slightly different one. I am not getting any exponent of a in numerator rest is the same.
Commented by 33 last updated on 11/Feb/18

$w h a t a r e t h e m a s s e s o f n b l o c s$ $a n y i n f o g i v e n . p l z t e l l ?$
Commented by Tinkutara last updated on 11/Feb/18
Masses of all blocks are different.
Commented by Tinkutara last updated on 11/Feb/18

$B u t a n s w e r s h o u l d b e \frac{2^{n + 1} v}{{(1 + a)}^{n + 1}} o r$ $y o u a r e g e t t i n g a n o t h e r a n s w e r ?$
Commented by mrW2 last updated on 11/Feb/18

$Y e s, a n s w e r s h o u l d b e \frac{2^{n + 1} v}{{(1 + a)}^{n + 1}}, t h i s$ $i s a l s o m y a n s w e r . B u t t h e a n s w e r y o u$ $g a v e f r o m t h e b o o k i s$ $\frac{2^{n + 1} a^{\frac{n (n - 1)}{2}} v}{{(1 + a)}^{n + 1}} . T h i s i s n o t c o r r e c t, I t h i n k .$
Commented by Tinkutara last updated on 11/Feb/18
Thanks! I got it now. ��☺
	Answered by mrW2 last updated on 12/Feb/18
	$We have totally n+1 masses: M_0 , M_1 ,M_2 ,...,M_n ,M_(n+1) with M_0 =m_1 and M_(n+1) =m_2 . M_1 ,M_2 ,...,M_n are unknown and may be freely determined. For mass M_i (i=0,1,2,...,n+1) let′s assume: v_(i,1) =speed after collision with M_(i−1) v_(i,2) =speed after collision with M_(i+1) v_(0,1) =V_1 =initial speed of first mass m_1 v_(n+1,1) =V_2 =final speed of last mass m_2 Collision M_i with M_(i+1) (perfectly elastic): M_i v_(i,1) =M_i v_(i,2) +M_(i+1) v_(i+1,1) ⇒M_i (v_(i,1) −v_(i,2) )=M_(i+1) v_(i+1,1) ...(i) (1/2)M_i v_(i,1) ^2 =(1/2)M_i v_(i,2) ^2 +(1/2)M_(i+1) v_(i+1,1) ^2 ⇒M_i (v_(i,1) ^2 −v_(i,2) ^2 )=M_(i+1) v_(i+1,1) ^2 ...(ii) (ii)/(i): ⇒v_(i,1) +v_(i,2) =v_(i+1,1) take this into (i): ⇒M_i (2v_(i,1) −v_(i+1,1) )=M_(i+1) v_(i+1,1) ⇒2M_i v_(i,1) =(M_i +M_(i+1) )v_(i+1,1) ⇒(v_(i+1,1) /v_(i,1) )=(2/(1+(M_(i+1) /M_i ))) (v_(1,1) /v_(0,1) )×(v_(2,1) /v_(1,1) )×...×(v_(n+1,1) /v_(n,1) )=(2/(1+(M_1 /M_0 )))×(2/(1+(M_2 /M_1 )))×...×(2/(1+(M_(n+1) /M_n ))) ⇒(v_(n+1,1) /v_(0,1) )=(2^(n+1) /((1+(M_1 /M_0 ))(1+(M_2 /M_1 ))...(1+(M_(n+1) /M_n )))) ⇒(V_2 /V_1 )=(2^(n+1) /((1+(M_1 /M_0 ))(1+(M_2 /M_1 ))...(1+(M_(n+1) /M_n ))))=2^(n+1) F(M_1 ,M_2 ,...,M_n ) with F(M_1 ,M_2 ,...,M_n )=(1/((1+(M_1 /M_0 ))(1+(M_2 /M_1 ))...(1+(M_(n+1) /M_n )))) or F(M_1 ,M_2 ,...,M_n )=((M_0 M_1 M_2 ...M_n )/((M_0 +M_1 )(M_1 +M_2 )...(M_n +M_(n+1) ))) (∂F/∂M_i )=((M_0 M_1 M_2 ...M_n )/((M_0 +M_1 )(M_1 +M_2 )...(M_n +M_(n+1) )M_i ))−((M_0 M_1 M_2 ...M_n (M_(i−1) +M_i +M_i +M_(i+1) ))/((M_0 +M_1 )(M_1 +M_2 )...(M_n +M_(n+1) )(M_(i−1) +M_i )(M_i +M_(i+1) ))) (∂F/∂M_i )=F(M_1 ,M_2 ,...,M_n ){(1/M_i )−((M_(i−1) +2M_i +M_(i+1) )/((M_(i−1) +M_i )(M_i +M_(i+1) )))} (∂F/∂M_i )=F(M_1 ,M_2 ,...,M_n ){((M_(i−1) M_i +M_i ^2 +M_(i−1) M_(i+1) +M_i M_(i+1) −M_(i−1) M_i −2M_i ^2 −M_i M_(i+1) )/(M_i (M_(i−1) +M_i )(M_i +M_(i+1) )))} (∂F/∂M_i )=F(M_1 ,M_2 ,...,M_n ){((M_(i−1) M_(i+1) −M_i ^2 )/(M_i (M_(i−1) +M_i )(M_i +M_(i+1) )))} For F(M_1 ,M_2 ,...,M_n ) to be maximum, (∂F/∂M_i )=0, i=1,2,..,n ⇒M_(i−1) M_(i+1) −M_i ^2 =0 ⇒(M_i /M_(i−1) )=(M_(i+1) /M_i ), i=1,2,..,n i.e. M_0 , M_1 ,M_2 ,...,M_n ,M_(n+1) must buid a G.P. to get maximum for (V_2 /V_1 ), i.e. (M_1 /M_0 )=(M_2 /M_1 )=...=(M_(n+1) /M_n )=a (say) ⇒(M_1 /M_0 )×(M_2 /M_1 )×(M_3 /M_2 )×...×(M_(n+1) /M_n )=a^(n+1) ⇒(M_(n+1) /M_0 )=a^(n+1) =(m_2 /m_1 ) ⇒a=((m_2 /m_1 ))^(1/(n+1)) ⇒((V_2 /V_1 ))_(max) =(2^(n+1) /((1+a)^(n+1) )) =((2/(1+a)))^(n+1) ⇒max. V_2 =[(2/(1+((m_2 /m_1 ))^(1/(n+1)) ))]^(n+1) V_1$
	$W e h a v e t o t a l l y n + 1 m a s s e s :$ $M_{0}, M_{1}, M_{2}, . . ., M_{n}, M_{n + 1}$ $w i t h M_{0} = m_{1} a n d M_{n + 1} = m_{2} .$ $M_{1}, M_{2}, . . ., M_{n} a r e u n k n o w n a n d m a y$ $b e f r e e l y d e t e r m i n e d .$ $F o r m a s s M_{i} (i = 0, 1, 2, . . ., n + 1) {l e t}^{'} s$ $a s s u m e :$ $v_{i, 1} = s p e e d a f t e r c o l l i s i o n w i t h M_{i - 1}$ $v_{i, 2} = s p e e d a f t e r c o l l i s i o n w i t h M_{i + 1}$ $v_{0, 1} = V_{1} = i n i t i a l s p e e d o f f i r s t m a s s m_{1}$ $v_{n + 1, 1} = V_{2} = f i n a l s p e e d o f l a s t m a s s m_{2}$ $C o l l i s i o n M_{i} w i t h M_{i + 1} (p e r f e c t l y e l a s t i c) :$ $M_{i} v_{i, 1} = M_{i} v_{i, 2} + M_{i + 1} v_{i + 1, 1}$ $\Rightarrow M_{i} (v_{i, 1} - v_{i, 2}) = M_{i + 1} v_{i + 1, 1} . . . (i)$ $\frac{1}{2} M_{i} v_{i, 1}^{2} = \frac{1}{2} M_{i} v_{i, 2}^{2} + \frac{1}{2} M_{i + 1} v_{i + 1, 1}^{2}$ $\Rightarrow M_{i} (v_{i, 1}^{2} - v_{i, 2}^{2}) = M_{i + 1} v_{i + 1, 1}^{2} . . . (i i)$ $(i i) / (i) :$ $\Rightarrow v_{i, 1} + v_{i, 2} = v_{i + 1, 1}$ $t a k e t h i s i n t o (i) :$ $\Rightarrow M_{i} (2 v_{i, 1} - v_{i + 1, 1}) = M_{i + 1} v_{i + 1, 1}$ $\Rightarrow 2 M_{i} v_{i, 1} = (M_{i} + M_{i + 1}) v_{i + 1, 1}$ $\Rightarrow \frac{v_{i + 1, 1}}{v_{i, 1}} = \frac{2}{1 + \frac{M_{i + 1}}{M_{i}}}$ $\frac{v_{1, 1}}{v_{0, 1}} \times \frac{v_{2, 1}}{v_{1, 1}} \times . . . \times \frac{v_{n + 1, 1}}{v_{n, 1}} = \frac{2}{1 + \frac{M_{1}}{M_{0}}} \times \frac{2}{1 + \frac{M_{2}}{M_{1}}} \times . . . \times \frac{2}{1 + \frac{M_{n + 1}}{M_{n}}}$ $\Rightarrow \frac{v_{n + 1, 1}}{v_{0, 1}} = \frac{2^{n + 1}}{(1 + \frac{M_{1}}{M_{0}}) (1 + \frac{M_{2}}{M_{1}}) . . . (1 + \frac{M_{n + 1}}{M_{n}})}$ $\Rightarrow \frac{V_{2}}{V_{1}} = \frac{2^{n + 1}}{(1 + \frac{M_{1}}{M_{0}}) (1 + \frac{M_{2}}{M_{1}}) . . . (1 + \frac{M_{n + 1}}{M_{n}})} = 2^{n + 1} F (M_{1}, M_{2}, . . ., M_{n})$ $w i t h$ $F (M_{1}, M_{2}, . . ., M_{n}) = \frac{1}{(1 + \frac{M_{1}}{M_{0}}) (1 + \frac{M_{2}}{M_{1}}) . . . (1 + \frac{M_{n + 1}}{M_{n}})}$ $o r$ $F (M_{1}, M_{2}, . . ., M_{n}) = \frac{M_{0} M_{1} M_{2} . . . M_{n}}{(M_{0} + M_{1}) (M_{1} + M_{2}) . . . (M_{n} + M_{n + 1})}$ $\frac{\partial F}{\partial M_{i}} = \frac{M_{0} M_{1} M_{2} . . . M_{n}}{(M_{0} + M_{1}) (M_{1} + M_{2}) . . . (M_{n} + M_{n + 1}) M_{i}} - \frac{M_{0} M_{1} M_{2} . . . M_{n} (M_{i - 1} + M_{i} + M_{i} + M_{i + 1})}{(M_{0} + M_{1}) (M_{1} + M_{2}) . . . (M_{n} + M_{n + 1}) (M_{i - 1} + M_{i}) (M_{i} + M_{i + 1})}$ $\frac{\partial F}{\partial M_{i}} = F (M_{1}, M_{2}, . . ., M_{n}) {\frac{1}{M_{i}} - \frac{M_{i - 1} + 2 M_{i} + M_{i + 1}}{(M_{i - 1} + M_{i}) (M_{i} + M_{i + 1})}}$ $\frac{\partial F}{\partial M_{i}} = F (M_{1}, M_{2}, . . ., M_{n}) {\frac{M_{i - 1} M_{i} + M_{i}^{2} + M_{i - 1} M_{i + 1} + M_{i} M_{i + 1} - M_{i - 1} M_{i} - 2 M_{i}^{2} - M_{i} M_{i + 1}}{M_{i} (M_{i - 1} + M_{i}) (M_{i} + M_{i + 1})}}$ $\frac{\partial F}{\partial M_{i}} = F (M_{1}, M_{2}, . . ., M_{n}) {\frac{M_{i - 1} M_{i + 1} - M_{i}^{2}}{M_{i} (M_{i - 1} + M_{i}) (M_{i} + M_{i + 1})}}$ $F o r F (M_{1}, M_{2}, . . ., M_{n}) t o b e m a x i m u m, \frac{\partial F}{\partial M_{i}} = 0, i = 1, 2, . ., n$ $\Rightarrow M_{i - 1} M_{i + 1} - M_{i}^{2} = 0$ $\Rightarrow \frac{M_{i}}{M_{i - 1}} = \frac{M_{i + 1}}{M_{i}}, i = 1, 2, . ., n$ $i . e . M_{0}, M_{1}, M_{2}, . . ., M_{n}, M_{n + 1} m u s t b u i d$ $a G . P . t o g e t m a x i m u m f o r \frac{V_{2}}{V_{1}},$ $i . e . \frac{M_{1}}{M_{0}} = \frac{M_{2}}{M_{1}} = . . . = \frac{M_{n + 1}}{M_{n}} = a (s a y)$ $\Rightarrow \frac{M_{1}}{M_{0}} \times \frac{M_{2}}{M_{1}} \times \frac{M_{3}}{M_{2}} \times . . . \times \frac{M_{n + 1}}{M_{n}} = a^{n + 1}$ $\Rightarrow \frac{M_{n + 1}}{M_{0}} = a^{n + 1} = \frac{m_{2}}{m_{1}}$ $\Rightarrow a = {(\frac{m_{2}}{m_{1}})}^{\frac{1}{n + 1}}$ $\Rightarrow {(\frac{V_{2}}{V_{1}})}_{m a x} = \frac{2^{n + 1}}{{(1 + a)}^{n + 1}} = {(\frac{2}{1 + a})}^{n + 1}$ $\Rightarrow m a x . V_{2} = {[\frac{2}{1 + {(\frac{m_{2}}{m_{1}})}^{\frac{1}{n + 1}}}]}^{n + 1} V_{1}$
	Commented by Tinkutara last updated on 11/Feb/18
	How can you say that maximum occurs when masses are in GP? It also requires a proof.
	Commented by mrW2 last updated on 11/Feb/18
	I have added the proof. But that was obvious.
	Commented by Tinkutara last updated on 11/Feb/18
	Thanks but we can also maximize the expression of velocity acquired by last block when there are 3 such different blocks to get relation.
	Commented by mrW2 last updated on 11/Feb/18

	${T h a t}^{'} s p o s s i b l e a n d e a s i e r . I w a n t e d a$ $g e n e r a l s o l u t i o n f o r a n y n u m b e r o f$ $b l o c k s .$
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