A-rocket-accelerates-straight-up-by-ejecting-gas-downwards-In-a-small-time-interval-t-it-ejects-a-gas-of-mass-m-at-a-relative-speed-u-Calculate-KE-of-the-entire-system-at-t-t-and-t-and-show-th

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

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

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

$$“Educationisnotlearningfacts$$$$buttrainingofmindto{think}^{″},$$$$sotruelysaidbyMr.AlbertEinstien.$$$$$$

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

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

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

$$seriously???????$$

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

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

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

$$Andireallybelieveitsnotonly$$$$aboutgettinggoodmarksorrank.$$$$Itsaboutenjoyinginlearning$$$$andsatisfyyourcuriosity.$$

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

$$Alright,Thanks.$$

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

$$Anyways,weshouldclearour$$$$concept,nomatterwhetherits$$$$insyllabusornot.$$

Answered by ajfour last updated on 26/Oct/17

$$Rocketequation$$$$\frac{Mdv}{dt}=-u\left(\frac{dM}{dt}\right)+F_{ext}$$$$\Rightarrow \mathit{d}\mathit{v}=\frac{\mathit{u}\mathbf{\Delta }\mathit{m}}{\mathit{M}}\dots ..\left(i\right)$$$$anddM=-\mathrm{\Delta }mand{F}_{ext}=0here$$$$ChangeinKineticenergyof$$$$rocketandofgasbetweent+\mathrm{\Delta }t$$$$andtis$$$$\mathrm{\Delta }K=Mvdv+\frac{1}{2}\left(dM\right)v^2$$$$+\frac{1}{2}\left(\mathrm{\Delta }m\right){(v-u)}^2$$$$=Mvdv-\frac{v^2\mathrm{\Delta }m}{2}+\frac{v^2\mathrm{\Delta }m}{2}$$$$+\frac{{\mathit{u}}^2\mathbf{\Delta }\mathit{m}}{2}-uv\mathrm{\Delta }m$$$$\mathit{W}=\mathbf{\Delta }\mathit{K}=\frac{{\mathit{u}}^2\mathbf{\Delta }\mathit{m}}{2}+\mathit{M}v\left(\frac{\mathit{u}\mathbf{\Delta }\mathit{m}}{\mathit{M}}\right)$$$$-\mathit{u}\mathit{v}\mathbf{\Delta }\mathit{m}$$$$=\frac{{\mathit{u}}^2\mathbf{\Delta }\mathit{m}}{2}.$$

Commented by Tinkutara last updated on 26/Oct/17

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

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