Answer:
Let M = mass of the
system at \[t=0\].
\[\upsilon
=\] speed
of the system at time \[t\]
At
time \[t+\Delta t,\] mass
of the system \[=M-\Delta m\]. When rocket ejects gases, its mass decreases and
speed increases. Let \[(\upsilon +\Delta u)\] be the speed of the system at
time \[(t+\Delta t)\]. Speed of the ejected gas with respect to the system or
rocket \[=\,(\upsilon -u)\]
K.E.
of the system at time \[t,\,{{k}_{1}}=\frac{1}{2}M{{\upsilon }^{2}}\]
K.E.
of die system at time \[(t+\Delta t)\],
\[{{k}_{2}}=\frac{1}{2}(M-\Delta
m){{(\upsilon +\Delta \upsilon )}^{2}}\]\[+\frac{1}{2}\Delta m\,{{(\upsilon
-u)}^{2}}\]
\[=\,\frac{1}{2}\,[M-\,\Delta
m]\,\,[{{\upsilon }^{2}}+\Delta {{\upsilon }^{2}}+\,2\upsilon \Delta \upsilon
]\,+\frac{1}{2}\]
\[\Delta
m\,[{{\upsilon }^{2}}+{{u}^{2}}-2u\,\upsilon ]\]
Since
\[\Delta \upsilon \]is very small, so neglect \[\Delta {{\upsilon }^{2}}\]
\[\therefore
\]\[{{K}_{2}}=\,\frac{1}{2}\,(M-\,\Delta m)\,({{\upsilon }^{2}}+2\upsilon
\Delta \upsilon )\,+\frac{1}{2}\]
\[\Delta
m\,({{\upsilon }^{2}}+{{u}^{2}}-2u\,\upsilon )\]
\[=\frac{1}{2}\,M{{\upsilon
}^{2}}+M\upsilon \Delta \upsilon +\frac{1}{2}\,\Delta m\,{{u}^{2}}\,-\Delta
m\,u\upsilon \]
(neglecting
a term having\[\Delta m\,\,\Delta \,\upsilon \])
Work
done = change in K.E.
\[={{K}_{2}}-{{K}_{1}}=\frac{1}{2}\Delta
m{{u}^{2}}+\upsilon (M\Delta \upsilon -\Delta m\upsilon )\]
Since
\[\Delta \upsilon \] and
\[\Delta m\] are very small,
So \[(M\Delta
\upsilon -\Delta mu)\] is negligible.
Hence,
Work done\[=\frac{1}{2}\,\Delta m\,\,{{\upsilon }^{2}}\]
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