Question-29633

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

$${what}\:{are}\:{the}\:{masses}\:{of}\:{n}\:{blocs}? \\ $$

Commented by mrW2 last updated on 11/Feb/18

$${Just}\:{a}\:{thought},\:{not}\:{a}\:{proof}: \\ $$$${max}.\:{possible}\:{speed}\:{of}\:{last}\:{mass}\:{m}_{\mathrm{2}} : \\ $$$${u}={v}\:\left[\frac{\mathrm{2}}{\mathrm{1}+\left(\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} }\right)^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} }\right]^{{n}+\mathrm{1}} \\ $$$$ \\ $$$${e}.{g}.\:{m}_{\mathrm{2}} =\mathrm{2}{m}_{\mathrm{1}} ,{n}=\mathrm{10}: \\ $$$${u}\approx\mathrm{0}.\mathrm{703}\:{v} \\ $$$$ \\ $$$${e}.{g}.\:{m}_{\mathrm{2}} =\mathrm{0}.\mathrm{5}{m}_{\mathrm{1}} ,{n}=\mathrm{10}: \\ $$$${u}\approx\mathrm{1}.\mathrm{406}\:{v} \\ $$$$ \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{u}=\sqrt{\frac{{m}_{\mathrm{1}} }{{m}_{\mathrm{2}} }}\:{v} \\ $$$${this}\:{is}\:{the}\:{max}.\:{possible}\:{speed}\:{which} \\ $$$${the}\:{mass}\:{m}_{\mathrm{2}} \:{can}\:{achieve}\:{by}\:{placeing} \\ $$$${infinitely}\:{many}\:{blocks}\:{between}\:{m}_{\mathrm{1}} \\ $$$${and}\:{m}_{\mathrm{2}} . \\ $$

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

$${what}\:{are}\:{the}\:{masses}\:{of}\:{n}\:{blocs} \\ $$$${any}\:{info}\:{given}.\:{plz}\:{tell}? \\ $$

Commented by Tinkutara last updated on 11/Feb/18

Masses of all blocks are different.

Commented by Tinkutara last updated on 11/Feb/18

$${But}\:{answer}\:{should}\:{be}\:\frac{\mathrm{2}^{{n}+\mathrm{1}} {v}}{\left(\mathrm{1}+{a}\right)^{{n}+\mathrm{1}} }\:{or} \\ $$$${you}\:{are}\:{getting}\:{another}\:{answer}? \\ $$

Commented by mrW2 last updated on 11/Feb/18

$${Yes},\:{answer}\:{should}\:{be}\:\frac{\mathrm{2}^{{n}+\mathrm{1}} {v}}{\left(\mathrm{1}+{a}\right)^{{n}+\mathrm{1}} },\:{this} \\ $$$${is}\:{also}\:{my}\:{answer}.\:{But}\:{the}\:{answer}\:{you} \\ $$$${gave}\:{from}\:{the}\:{book}\:{is} \\ $$$$\frac{\mathrm{2}^{{n}+\mathrm{1}} \:{a}^{\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}} {v}}{\left(\mathrm{1}+{a}\right)^{{n}+\mathrm{1}} }.\:{This}\:{is}\:{not}\:{correct},\:{I}\:{think}. \\ $$

Commented by Tinkutara last updated on 11/Feb/18

Thanks! I got it now. ����☺

Answered by mrW2 last updated on 12/Feb/18

$${We}\:{have}\:{totally}\:{n}+\mathrm{1}\:{masses}: \\ $$$${M}_{\mathrm{0}} ,\:{M}_{\mathrm{1}} ,{M}_{\mathrm{2}} ,…,{M}_{{n}} ,{M}_{{n}+\mathrm{1}} \\ $$$${with}\:{M}_{\mathrm{0}} ={m}_{\mathrm{1}} \:{and}\:{M}_{{n}+\mathrm{1}} ={m}_{\mathrm{2}} . \\ $$$${M}_{\mathrm{1}} ,{M}_{\mathrm{2}} ,…,{M}_{{n}} \:{are}\:{unknown}\:{and}\:{may} \\ $$$${be}\:{freely}\:{determined}. \\ $$$$ \\ $$$${For}\:{mass}\:{M}_{{i}} \:\left({i}=\mathrm{0},\mathrm{1},\mathrm{2},…,{n}+\mathrm{1}\right)\:{let}'{s} \\ $$$${assume}: \\ $$$${v}_{{i},\mathrm{1}} ={speed}\:{after}\:{collision}\:{with}\:{M}_{{i}−\mathrm{1}} \\ $$$${v}_{{i},\mathrm{2}} ={speed}\:{after}\:{collision}\:{with}\:{M}_{{i}+\mathrm{1}} \\ $$$${v}_{\mathrm{0},\mathrm{1}} ={V}_{\mathrm{1}} ={initial}\:{speed}\:{of}\:{first}\:{mass}\:{m}_{\mathrm{1}} \\ $$$${v}_{{n}+\mathrm{1},\mathrm{1}} ={V}_{\mathrm{2}} ={final}\:{speed}\:{of}\:{last}\:{mass}\:{m}_{\mathrm{2}} \\ $$$$ \\ $$$${Collision}\:{M}_{{i}} \:{with}\:{M}_{{i}+\mathrm{1}} \:\left({perfectly}\:{elastic}\right): \\ $$$${M}_{{i}} {v}_{{i},\mathrm{1}} ={M}_{{i}} {v}_{{i},\mathrm{2}} +{M}_{{i}+\mathrm{1}} {v}_{{i}+\mathrm{1},\mathrm{1}} \\ $$$$\Rightarrow{M}_{{i}} \left({v}_{{i},\mathrm{1}} −{v}_{{i},\mathrm{2}} \right)={M}_{{i}+\mathrm{1}} {v}_{{i}+\mathrm{1},\mathrm{1}} \:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{M}_{{i}} {v}_{{i},\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{M}_{{i}} {v}_{{i},\mathrm{2}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{M}_{{i}+\mathrm{1}} {v}_{{i}+\mathrm{1},\mathrm{1}} ^{\mathrm{2}} \:\: \\ $$$$\Rightarrow{M}_{{i}} \left({v}_{{i},\mathrm{1}} ^{\mathrm{2}} −{v}_{{i},\mathrm{2}} ^{\mathrm{2}} \right)={M}_{{i}+\mathrm{1}} {v}_{{i}+\mathrm{1},\mathrm{1}} ^{\mathrm{2}} \:\:\:\:…\left({ii}\right) \\ $$$$\left({ii}\right)/\left({i}\right): \\ $$$$\Rightarrow{v}_{{i},\mathrm{1}} +{v}_{{i},\mathrm{2}} ={v}_{{i}+\mathrm{1},\mathrm{1}} \\ $$$${take}\:{this}\:{into}\:\left({i}\right): \\ $$$$\Rightarrow{M}_{{i}} \left(\mathrm{2}{v}_{{i},\mathrm{1}} −{v}_{{i}+\mathrm{1},\mathrm{1}} \right)={M}_{{i}+\mathrm{1}} {v}_{{i}+\mathrm{1},\mathrm{1}} \: \\ $$$$\Rightarrow\mathrm{2}{M}_{{i}} {v}_{{i},\mathrm{1}} =\left({M}_{{i}} +{M}_{{i}+\mathrm{1}} \right){v}_{{i}+\mathrm{1},\mathrm{1}} \: \\ $$$$\Rightarrow\frac{{v}_{{i}+\mathrm{1},\mathrm{1}} }{{v}_{{i},\mathrm{1}} }=\frac{\mathrm{2}}{\mathrm{1}+\frac{{M}_{{i}+\mathrm{1}} }{{M}_{{i}} }} \\ $$$$\frac{{v}_{\mathrm{1},\mathrm{1}} }{{v}_{\mathrm{0},\mathrm{1}} }×\frac{{v}_{\mathrm{2},\mathrm{1}} }{{v}_{\mathrm{1},\mathrm{1}} }×…×\frac{{v}_{{n}+\mathrm{1},\mathrm{1}} }{{v}_{{n},\mathrm{1}} }=\frac{\mathrm{2}}{\mathrm{1}+\frac{{M}_{\mathrm{1}} }{{M}_{\mathrm{0}} }}×\frac{\mathrm{2}}{\mathrm{1}+\frac{{M}_{\mathrm{2}} }{{M}_{\mathrm{1}} }}×…×\frac{\mathrm{2}}{\mathrm{1}+\frac{{M}_{{n}+\mathrm{1}} }{{M}_{{n}} }} \\ $$$$\Rightarrow\frac{{v}_{{n}+\mathrm{1},\mathrm{1}} }{{v}_{\mathrm{0},\mathrm{1}} }=\frac{\mathrm{2}^{{n}+\mathrm{1}} }{\left(\mathrm{1}+\frac{{M}_{\mathrm{1}} }{{M}_{\mathrm{0}} }\right)\left(\mathrm{1}+\frac{{M}_{\mathrm{2}} }{{M}_{\mathrm{1}} }\right)…\left(\mathrm{1}+\frac{{M}_{{n}+\mathrm{1}} }{{M}_{{n}} }\right)}\: \\ $$$$\Rightarrow\frac{{V}_{\mathrm{2}} }{{V}_{\mathrm{1}} }=\frac{\mathrm{2}^{{n}+\mathrm{1}} }{\left(\mathrm{1}+\frac{{M}_{\mathrm{1}} }{{M}_{\mathrm{0}} }\right)\left(\mathrm{1}+\frac{{M}_{\mathrm{2}} }{{M}_{\mathrm{1}} }\right)…\left(\mathrm{1}+\frac{{M}_{{n}+\mathrm{1}} }{{M}_{{n}} }\right)}=\mathrm{2}^{{n}+\mathrm{1}} {F}\left({M}_{\mathrm{1}} ,{M}_{\mathrm{2}} ,…,{M}_{{n}} \right)\: \\ $$$${with} \\ $$$${F}\left({M}_{\mathrm{1}} ,{M}_{\mathrm{2}} ,…,{M}_{{n}} \right)=\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{{M}_{\mathrm{1}} }{{M}_{\mathrm{0}} }\right)\left(\mathrm{1}+\frac{{M}_{\mathrm{2}} }{{M}_{\mathrm{1}} }\right)…\left(\mathrm{1}+\frac{{M}_{{n}+\mathrm{1}} }{{M}_{{n}} }\right)} \\ $$$${or} \\ $$$${F}\left({M}_{\mathrm{1}} ,{M}_{\mathrm{2}} ,…,{M}_{{n}} \right)=\frac{{M}_{\mathrm{0}} {M}_{\mathrm{1}} {M}_{\mathrm{2}} …{M}_{{n}} }{\left({M}_{\mathrm{0}} +{M}_{\mathrm{1}} \right)\left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right)…\left({M}_{{n}} +{M}_{{n}+\mathrm{1}} \right)} \\ $$$$\frac{\partial{F}}{\partial{M}_{{i}} }=\frac{{M}_{\mathrm{0}} {M}_{\mathrm{1}} {M}_{\mathrm{2}} …{M}_{{n}} }{\left({M}_{\mathrm{0}} +{M}_{\mathrm{1}} \right)\left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right)…\left({M}_{{n}} +{M}_{{n}+\mathrm{1}} \right){M}_{{i}} }−\frac{{M}_{\mathrm{0}} {M}_{\mathrm{1}} {M}_{\mathrm{2}} …{M}_{{n}} \left({M}_{{i}−\mathrm{1}} +{M}_{{i}} +{M}_{{i}} +{M}_{{i}+\mathrm{1}} \right)}{\left({M}_{\mathrm{0}} +{M}_{\mathrm{1}} \right)\left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right)…\left({M}_{{n}} +{M}_{{n}+\mathrm{1}} \right)\left({M}_{{i}−\mathrm{1}} +{M}_{{i}} \right)\left({M}_{{i}} +{M}_{{i}+\mathrm{1}} \right)} \\ $$$$\frac{\partial{F}}{\partial{M}_{{i}} }={F}\left({M}_{\mathrm{1}} ,{M}_{\mathrm{2}} ,…,{M}_{{n}} \right)\left\{\frac{\mathrm{1}}{{M}_{{i}} }−\frac{{M}_{{i}−\mathrm{1}} +\mathrm{2}{M}_{{i}} +{M}_{{i}+\mathrm{1}} }{\left({M}_{{i}−\mathrm{1}} +{M}_{{i}} \right)\left({M}_{{i}} +{M}_{{i}+\mathrm{1}} \right)}\right\} \\ $$$$\frac{\partial{F}}{\partial{M}_{{i}} }={F}\left({M}_{\mathrm{1}} ,{M}_{\mathrm{2}} ,…,{M}_{{n}} \right)\left\{\frac{{M}_{{i}−\mathrm{1}} {M}_{{i}} +{M}_{{i}} ^{\mathrm{2}} +{M}_{{i}−\mathrm{1}} {M}_{{i}+\mathrm{1}} +{M}_{{i}} {M}_{{i}+\mathrm{1}} −{M}_{{i}−\mathrm{1}} {M}_{{i}} −\mathrm{2}{M}_{{i}} ^{\mathrm{2}} −{M}_{{i}} {M}_{{i}+\mathrm{1}} }{{M}_{{i}} \left({M}_{{i}−\mathrm{1}} +{M}_{{i}} \right)\left({M}_{{i}} +{M}_{{i}+\mathrm{1}} \right)}\right\} \\ $$$$\frac{\partial{F}}{\partial{M}_{{i}} }={F}\left({M}_{\mathrm{1}} ,{M}_{\mathrm{2}} ,…,{M}_{{n}} \right)\left\{\frac{{M}_{{i}−\mathrm{1}} {M}_{{i}+\mathrm{1}} −{M}_{{i}} ^{\mathrm{2}} }{{M}_{{i}} \left({M}_{{i}−\mathrm{1}} +{M}_{{i}} \right)\left({M}_{{i}} +{M}_{{i}+\mathrm{1}} \right)}\right\} \\ $$$${For}\:{F}\left({M}_{\mathrm{1}} ,{M}_{\mathrm{2}} ,…,{M}_{{n}} \right)\:{to}\:{be}\:{maximum},\:\frac{\partial{F}}{\partial{M}_{{i}} }=\mathrm{0},\:{i}=\mathrm{1},\mathrm{2},..,{n} \\ $$$$\Rightarrow{M}_{{i}−\mathrm{1}} {M}_{{i}+\mathrm{1}} −{M}_{{i}} ^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\frac{{M}_{{i}} }{{M}_{{i}−\mathrm{1}} }=\frac{{M}_{{i}+\mathrm{1}} }{{M}_{{i}} },\:{i}=\mathrm{1},\mathrm{2},..,{n} \\ $$$${i}.{e}.\:{M}_{\mathrm{0}} ,\:{M}_{\mathrm{1}} ,{M}_{\mathrm{2}} ,…,{M}_{{n}} ,{M}_{{n}+\mathrm{1}} \:{must}\:{buid} \\ $$$${a}\:{G}.{P}.\:{to}\:{get}\:{maximum}\:{for}\:\frac{{V}_{\mathrm{2}} }{{V}_{\mathrm{1}} }, \\ $$$${i}.{e}.\:\frac{{M}_{\mathrm{1}} }{{M}_{\mathrm{0}} }=\frac{{M}_{\mathrm{2}} }{{M}_{\mathrm{1}} }=…=\frac{{M}_{{n}+\mathrm{1}} }{{M}_{{n}} }={a}\:\left({say}\right) \\ $$$$\Rightarrow\frac{{M}_{\mathrm{1}} }{{M}_{\mathrm{0}} }×\frac{{M}_{\mathrm{2}} }{{M}_{\mathrm{1}} }×\frac{{M}_{\mathrm{3}} }{{M}_{\mathrm{2}} }×…×\frac{{M}_{{n}+\mathrm{1}} }{{M}_{{n}} }={a}^{{n}+\mathrm{1}} \\ $$$$\Rightarrow\frac{{M}_{{n}+\mathrm{1}} }{{M}_{\mathrm{0}} }={a}^{{n}+\mathrm{1}} =\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} } \\ $$$$\Rightarrow{a}=\left(\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} }\right)^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} \\ $$$$\Rightarrow\left(\frac{{V}_{\mathrm{2}} }{{V}_{\mathrm{1}} }\right)_{{max}} =\frac{\mathrm{2}^{{n}+\mathrm{1}} }{\left(\mathrm{1}+{a}\right)^{{n}+\mathrm{1}} }\:=\left(\frac{\mathrm{2}}{\mathrm{1}+{a}}\right)^{{n}+\mathrm{1}} \\ $$$$\Rightarrow{max}.\:{V}_{\mathrm{2}} =\left[\frac{\mathrm{2}}{\mathrm{1}+\left(\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} }\right)^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} }\right]^{{n}+\mathrm{1}} {V}_{\mathrm{1}} \\ $$

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

$${That}'{s}\:{possible}\:{and}\:{easier}.\:{I}\:{wanted}\:{a} \\ $$$${general}\:{solution}\:{for}\:{any}\:{number}\:{of}\: \\ $$$${blocks}. \\ $$

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