A rocket accelerates straight up by ejecting gas downwards. In a small time interval $\Delta t$, it ejects a gas of mass $\Delta m$ at a relative speed $u$. Calculate $KE$ of the entire system at $t + \Delta t$ and $t$ and show that the device that ejects gas does work $=(\frac {1}{2})\Delta mu^2$ in this time interval (negative gravity).
A rocket accelerates straight up by ejecting gas downwards. In a small time interval $\Delta t$, it ejects a gas of mass $\Delta m$ at a relative speed $u$. Calculate $KE$ of the entire system at $t + \Delta t$ and $t$ and show that the device that ejects gas does work $=(\frac {1}{2})\Delta mu^2$ in this time interval (negative gravity).
$\mathrm{M}=$ mass of rocket at $t$
$v=$ velocity of rocket at $t$
$\Delta m=$ mass of ejected gas in $\Delta t$
$u=$ relative speed of ejected gas
Consider at time $t+\Delta t$
$(\mathrm{KE})_{t}+(\mathrm{KE})_{\Delta t}=\mathrm{KE}$ of rocket $+\mathrm{KE}$ of gas
$=\frac{1}{2}(\mathrm{M}-\Delta m)(v+\Delta v)^{2}+\frac{1}{2} \Delta m(v-u)^{2}$
$=\frac{1}{2} \mathrm{M} v^{2}+\mathrm{M} v \Delta v-\Delta m v u+\frac{1}{2} \Delta m u^{2}$
$(\mathrm{KE})_{t}=\mathrm{KE}$ of the rocket at time $t=\frac{1}{2} \mathrm{M} v^{2}$
$\Delta \mathrm{K} =(\mathrm{KE})_{t}+(\mathrm{KE})_{\Delta t}-(\mathrm{KE})_{t}$
$=(\mathrm{M} \Delta v-\Delta m u) v+\frac{1}{2} \Delta m u^{2}$
Hence, $\mathrm{M} \frac{d v}{d t}=\frac{d c}{d t}|u|$
$\therefore \mathrm{M} \Delta v=\Delta m u$
$\therefore \Delta \mathrm{K}=\frac{1}{2} \Delta m u^{2}$
Now, by work-energy theorem,
$\Delta \mathrm{K} =\Delta \mathrm{W}$
$\therefore \Delta \mathrm{W} =\frac{1}{2} \Delta m u^{2}$
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