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).


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

$A rocket accelerates straight up by$ $ejecting gas downwards . In a small$ $time interval Δ t, it ejects a gas of mass$ $Δ m at a relative speed u . Calculate KE$ $of the entire system at t + Δ t and t and$ $show that the device that ejects gas$ $does work = (\frac{1}{2}) Δ {m u}^{2} in this time$ $interval (neglect gravity) .$
Commented by Physics lover last updated on 26/Oct/17

$‘ ‘ E d u c a t i o n i s n o t l e a r n i n g f a c t s$ $b u t t r a i n i n g o f m i n d t o {t h i n k}^{″},$ $s o t r u e l y s a i d b y M r . A l b e r t E i n s t i e n .$
Commented by math solver last updated on 26/Oct/17

$hey, rocket repulsion is not in JEE$ $syllabus :)$
Commented by Physics lover last updated on 26/Oct/17

$s e r i o u s l y ? ? ? ? ? ? ?$
Commented by math solver last updated on 26/Oct/17

$yup, i am serious$ $you can check on net also .$ $or else ask your teachers!!$
Commented by Physics lover last updated on 26/Oct/17

$A n d i r e a l l y b e l i e v e i t s n o t o n l y$ $a b o u t g e t t i n g g o o d m a r k s o r r a n k .$ $I t s a b o u t e n j o y i n g i n l e a r n i n g$ $a n d s a t i s f y y o u r c u r i o s i t y .$
Commented by Physics lover last updated on 26/Oct/17

$A l r i g h t, T h a n k s .$
Commented by Physics lover last updated on 26/Oct/17

$A n y w a y s, w e s h o u l d c l e a r o u r$ $c o n c e p t, n o m a t t e r w h e t h e r i t s$ $i n s y l l a b u s o r n o t .$
	Answered by ajfour last updated on 26/Oct/17

	$R o c k e t e q u a t i o n$ $\frac{M d v}{d t} = - u (\frac{d M}{d t}) + F_{e x t}$ $\Rightarrow d v = \frac{u Δ m}{M} . . . . . (i)$ $a n d d M = - Δ m a n d F_{e x t} = 0 h e r e$ $C h a n g e i n K i n e t i c e n e r g y o f$ $r o c k e t a n d o f g a s b e t w e e n t + Δ t$ $a n d t i s$ $Δ K = M v d v + \frac{1}{2} (d M) v^{2}$ $+ \frac{1}{2} (Δ m) {(v - u)}^{2}$ $= M v d v - \frac{v^{2} Δ m}{2} + \frac{v^{2} Δ m}{2}$ $+ \frac{u^{2} Δ m}{2} - u v Δ m$ $W = Δ K = \frac{u^{2} Δ m}{2} + M v (\frac{u Δ m}{M})$ $- u v Δ m$ $= \frac{u^{2} Δ m}{2} .$
	Commented by Tinkutara last updated on 26/Oct/17

	$Thank you very much Sir!$
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