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Question Number 23066 by Tinkutara last updated on 25/Oct/17

A rocket accelerates straight up by ejecting gas downwards. In a small time interval Δt, it ejects a gas of mass Δm at a relative speed u. Calculate KE of the entire system at t + Δt and t and show that the device that ejects gas does work = ((1/2))Δmu^2 in this time interval (neglect gravity).

$$\mathrm{A}\:\mathrm{rocket}\:\mathrm{accelerates}\:\mathrm{straight}\:\mathrm{up}\:\mathrm{by} \\ $$$$\mathrm{ejecting}\:\mathrm{gas}\:\mathrm{downwards}.\:\mathrm{In}\:\mathrm{a}\:\mathrm{small} \\ $$$$\mathrm{time}\:\mathrm{interval}\:\Delta{t},\:\mathrm{it}\:\mathrm{ejects}\:\mathrm{a}\:\mathrm{gas}\:\mathrm{of}\:\mathrm{mass} \\ $$$$\Delta{m}\:\mathrm{at}\:\mathrm{a}\:\mathrm{relative}\:\mathrm{speed}\:{u}.\:\mathrm{Calculate}\:\mathrm{KE} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{entire}\:\mathrm{system}\:\mathrm{at}\:{t}\:+\:\Delta{t}\:\mathrm{and}\:{t}\:\mathrm{and} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{device}\:\mathrm{that}\:\mathrm{ejects}\:\mathrm{gas} \\ $$$$\mathrm{does}\:\mathrm{work}\:=\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Delta{mu}^{\mathrm{2}} \:\mathrm{in}\:\mathrm{this}\:\mathrm{time} \\ $$$$\mathrm{interval}\:\left(\mathrm{neglect}\:\mathrm{gravity}\right). \\ $$

Commented by Physics lover last updated on 26/Oct/17

“Education is not learning facts but training of mind to think”, so truely said by Mr. Albert Einstien.

$$``{Education}\:{is}\:{not}\:{learning}\:{facts} \\ $$$${but}\:{training}\:{of}\:{mind}\:{to}\:{think}'', \\ $$$${so}\:{truely}\:{said}\:{by}\:{Mr}.\:{Albert}\:{Einstien}. \\ $$$$ \\ $$

Commented by math solver last updated on 26/Oct/17

hey, rocket repulsion is not in JEE syllabus :)

$$\mathrm{hey},\:\mathrm{rocket}\:\mathrm{repulsion}\:\mathrm{is}\:\mathrm{not}\:\mathrm{in}\:\mathrm{JEE} \\ $$$$\left.\mathrm{syllabus}\::\right) \\ $$

Commented by Physics lover last updated on 26/Oct/17

seriously???????

$${seriously}??????? \\ $$

Commented by math solver last updated on 26/Oct/17

yup , i am serious you can check on net also. or else ask your teachers !!

$$\mathrm{yup}\:,\:\mathrm{i}\:\mathrm{am}\:\mathrm{serious}\: \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{check}\:\mathrm{on}\:\mathrm{net}\:\mathrm{also}. \\ $$$$\mathrm{or}\:\mathrm{else}\:\mathrm{ask}\:\mathrm{your}\:\mathrm{teachers}\:!! \\ $$

Commented by Physics lover last updated on 26/Oct/17

And i really believe its not only about getting good marks or rank. Its about enjoying in learning and satisfy your curiosity.

$${And}\:{i}\:{really}\:{believe}\:{its}\:{not}\:{only} \\ $$$${about}\:{getting}\:{good}\:{marks}\:{or}\:{rank}. \\ $$$${Its}\:{about}\:{enjoying}\:{in}\:{learning} \\ $$$${and}\:{satisfy}\:{your}\:{curiosity}. \\ $$

Commented by Physics lover last updated on 26/Oct/17

Alright,Thanks.

$${Alright},{Thanks}. \\ $$

Commented by Physics lover last updated on 26/Oct/17

Anyways, we should clear our concept,no matter whether its in syllabus or not.

$${Anyways},\:{we}\:{should}\:{clear}\:{our}\: \\ $$$${concept},{no}\:{matter}\:{whether}\:{its} \\ $$$${in}\:{syllabus}\:{or}\:{not}. \\ $$

Answered by ajfour last updated on 26/Oct/17

Rocket equation ((Mdv)/dt) = −u((dM/dt))+F_(ext) ⇒ dv=((u𝚫m)/M) .....(i) and dM=−Δm and F_(ext) =0 here Change in Kinetic energy of rocket and of gas between t+Δt and t is ΔK=Mvdv+(1/2)(dM)v^2 +(1/2)(Δm)(v−u)^2 =Mvdv−((v^2 Δm)/2)+((v^2 Δm)/2) +((u^2 𝚫m)/2)−uvΔm W=𝚫K=((u^2 𝚫m)/2)+Mv(((u𝚫m)/M)) −uv𝚫m =((u^2 𝚫m)/2) .

$$\:{Rocket}\:{equation} \\ $$$$\frac{{Mdv}}{{dt}}\:=\:−{u}\left(\frac{{dM}}{{dt}}\right)+{F}_{{ext}} \: \\ $$$$\Rightarrow\:\boldsymbol{{dv}}=\frac{\boldsymbol{{u}\Delta{m}}}{\boldsymbol{{M}}}\:\:\:.....\left({i}\right) \\ $$$${and}\:\:{dM}=−\Delta{m}\:\:\:\:{and}\:{F}_{{ext}} =\mathrm{0}\:{here} \\ $$$${Change}\:{in}\:{Kinetic}\:{energy}\:{of} \\ $$$${rocket}\:{and}\:{of}\:{gas}\:{between}\:{t}+\Delta{t} \\ $$$${and}\:{t}\:{is} \\ $$$$\Delta{K}={Mvdv}+\frac{\mathrm{1}}{\mathrm{2}}\left({dM}\right){v}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\left(\Delta{m}\right)\left({v}−{u}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:={Mvdv}−\frac{{v}^{\mathrm{2}} \Delta{m}}{\mathrm{2}}+\frac{{v}^{\mathrm{2}} \Delta{m}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\boldsymbol{{u}}^{\mathrm{2}} \boldsymbol{\Delta{m}}}{\mathrm{2}}−{uv}\Delta{m} \\ $$$$\:\:\:\boldsymbol{{W}}=\boldsymbol{\Delta{K}}=\frac{\boldsymbol{{u}}^{\mathrm{2}} \boldsymbol{\Delta{m}}}{\mathrm{2}}+\boldsymbol{{M}}{v}\left(\frac{\boldsymbol{{u}\Delta{m}}}{\boldsymbol{{M}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\boldsymbol{{uv}\Delta{m}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\boldsymbol{{u}}^{\mathrm{2}} \boldsymbol{\Delta{m}}}{\mathrm{2}}\:\:. \\ $$

Commented by Tinkutara last updated on 26/Oct/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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