Question

A spaceship of mass $m_0=500\text{kg}$ moves in the absence of an external force with a constant velocity $v_0=50\text{m/s}$. To change the direction of motion, a jet engine is switched on. It starts ejecting a gas jet with velocity $u=10\text{m/s}$ 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 $400\text{kg}$. Through what angle $\alpha$ does the direction of motion of the spaceship deviate due to the jet engine operation? (Take $\ln \left(1.25\right)=0.223$)

• A. $0.045$
• B. $0.45$
• C. $0.06$
• D. $0.08$

Answer 1

The correct option is A $0.045$

For a variable mass system with constant rate of ejection $\frac{dm}{dt}$, we have
$m\frac{d\overrightarrow{v}}{dt}=\overrightarrow{F}-\overrightarrow{u}\frac{dm}{dt}$
Given, external force $\overrightarrow{F}=0$
$\Rightarrow m\frac{d\overrightarrow{v}}{dt}=-\overrightarrow{u}\frac{dm}{dt}\text{or}d\overrightarrow{v}=-\overrightarrow{u}\frac{dm}{m}$

Velocity of ejection of gas jet $\overrightarrow{u}$ is opposite in direction to change in velocity of spaceship $d\overrightarrow{v}$.
We have $|d\overrightarrow{v}|=v_0\left(d\alpha \right)$ (because $v_0$ is constant) where $d\alpha$ is the angle by which the spaceship turns in time $dt$.
So, $v_{0}d\alpha -u\frac{dm}{m}\text{or}d\alpha =\frac{-u}{v_0}\frac{dm}{m}$
($-ve$ sign signifies decrease in mass)

Integrating both sides, we get
$\alpha =\frac{-u}{v_0}{\int }_{m_0}^m\frac{dm}{m}=\frac{u}{v_0}\ln \left(\frac{m_0}{m}\right)$
Here, initial mass $m_0=500\text{kg}$ and final mass $m=400\text{kg}$, velocity of rocket $v_0=50\text{m/s}$ and relative velocity of ejection of gas jet $u=10\text{m/s}$
Substituting the above values,
$\alpha =\frac{10}{50}\ln \left(\frac{500}{400}\right)$
$\Rightarrow \alpha =0.045$