Answer:
![m=(m_(0))/(e)](https://img.qammunity.org/2020/formulas/physics/college/sjapjeu61oy5ui8an4r1uahm8eugpbyf1j.png)
Step-by-step explanation:
Equation of the rocket is,
![m(dv)/(dt) =F-v'(dm)/(dt)](https://img.qammunity.org/2020/formulas/physics/college/ibzlhpwof7bh9pn9ts1d213439eps78trj.png)
Here, v' is the relative velocity of rocket.
In space F is zero.
So,
![m(dv)/(dt) =-v'(dm)/(dt)\\dv=-v'(dm)/(m) \\v=-v'ln(m)/(m_(0) )](https://img.qammunity.org/2020/formulas/physics/college/1udr2oruq0r7b0x8dgizvluuh6hpzr48qn.png)
Now the momentum can be obtained by multiply by m on both sides.
![P=-v'mln(m)/(m_(0) )](https://img.qammunity.org/2020/formulas/physics/college/y7fao89x4qiqwdi2fkmcceek7xxnh0s3eo.png)
Now for maxima,
![(dP)/(dm)=0](https://img.qammunity.org/2020/formulas/physics/college/kd0x6kmr120mr7wpt21nvr37kfmtom0wfz.png)
![-v'ln(m)/(m_(0) )-v'm(m_(0))/(m )m_{0=0](https://img.qammunity.org/2020/formulas/physics/college/vc3buzk0hc7gm6bon1smf8ct57hetadi0g.png)
Now,
![ln((m)/(m_(0) ) )=-1\\(m)/(m_(0) )=(1)/(e) \\m=(m_(0))/(e)](https://img.qammunity.org/2020/formulas/physics/college/oablu0r7hxunkjfk03k4qddo3ztb0lguzo.png)
Therefore, the mass of the rocket while having maximum momentum is
![(m_(0))/(e)](https://img.qammunity.org/2020/formulas/physics/college/n7ko8r67frb5sxlzyc3alvzhp7e5ut5mlm.png)