Answer:
[tex]m=\frac{m_{0}}{e}[/tex]
Explanation:
Equation of the rocket is,
[tex]m\frac{dv}{dt} =F-v'\frac{dm}{dt}[/tex]
Here, v' is the relative velocity of rocket.
In space F is zero.
So,
[tex]m\frac{dv}{dt} =-v'\frac{dm}{dt}\\dv=-v'\frac{dm}{m} \\v=-v'ln\frac{m}{m_{0} }[/tex]
Now the momentum can be obtained by multiply by m on both sides.
[tex]P=-v'mln\frac{m}{m_{0} }[/tex]
Now for maxima, [tex]\frac{dP}{dm}=0[/tex]
[tex]-v'ln\frac{m}{m_{0} }-v'm\frac{m_{0}}{m }m_{0=0[/tex]
Now,
[tex]ln(\frac{m}{m_{0} } )=-1\\\frac{m}{m_{0} }=\frac{1}{e} \\m=\frac{m_{0}}{e}[/tex]
Therefore, the mass of the rocket while having maximum momentum is [tex]\frac{m_{0}}{e}[/tex]