Question

If the relation between the mass of the rocket and time is given as $m=m_0{e}^{-\left(\frac{xwt}{u}\right)}$, when the rocket moves with a constant acceleration $w$, the external forces are absent, the gas escapes with a constant velocity $u$ relative to the rocket, and its mass at the initial moment equals $m_0$.

Answer 1

According to the question, $\overrightarrow{F}$ (external force) $=0$
So, $m\frac{d\overrightarrow{v}}{dt}=\frac{dm}{dt}\overrightarrow{u}$
As $d\overrightarrow{v}\uparrow \downarrow \overrightarrow{u}$,
so, in scalar form, $mdv=-udm$
or, $\frac{wdt}{u}=-\frac{dm}{m}$
Integrating within the limits for $m\left(t\right)$
$\frac{wt}{u}=-{\int }_{m_0}^m\frac{dm}{m}$ or, $\frac{wt}{u}=-ln\frac{m}{m_0}$
Hence, $m=m_0{e}^{-\left(\frac{wt}{u}\right)}$