Question

A spaceship of mass $m_0$ moves in the absence of external forces with a constant velocity $\overrightarrow{v_0}$. To To change the motion direction, a jet engine is switched on. It starts ejecting a gas jet with velocity $u$ which is constant relative to the spaceship and directed at right angles to the spaceship motion. The engine is shut down when the mass of the spaceship decreases to $m$. Through an angle $\alpha =\frac{u}{v_0}\ln \left(\frac{x{m}_0}{m}\right)$ the motion direction of the spaceship deviates due to the jet engine operation. Find $x$.

Answer 1

As $\overrightarrow{F}=0$, from the equation of dynamics of a body with variable mass,

$m\frac{d\overrightarrow{v}}{dt}=\overrightarrow{u}\frac{dm}{dt}$ or, $d\overrightarrow{v}=\overrightarrow{u}\frac{dm}{m}$ (1)

Now $d\overrightarrow{v}\uparrow \downarrow \overrightarrow{u}$ and since \vec{u}\bot\vec{v}, we must have $|d\overrightarrow{v}|=v_{0}d\alpha$ (because $v_0$ is constant) where $d\alpha$ is the angle by which the spaceship, turns in time $dt$.

So, $-u\frac{dm}{m}=v_{0}d\alpha$ or, $d\alpha =-\frac{u}{v_0}\frac{dm}{m}$

or, $\alpha =-\frac{u}{v_0}{\int }_{m_0}^m\frac{dm}{m}=\frac{u}{v_0}\ln \left(\frac{m_0}{m}\right)$