Question

A rocket of mass $m$ is fired vertically upward and after the fuel burning it weighs $m^{\prime }$. Ejection of fuel gas is at a constant rate of $m_0$ per second with a constant velocity of $u_{rel}$ relative to the rocket. Final speed of rocket after the complete burn out of the fuel is given by $v=$

• A. $u_{rel}{\log }_e\frac{m}{m^{\prime }}$
• B. $u_{rel}{\log }_e\frac{m_0}{m}$
• C. $-u_{rel}{\log }_e\frac{m_0}{m^{\prime }}$
• D. $-u_{rel}\frac{dm}{m}$

Answer 1

Answer: (A)

$u_{rel}{\log }_e\frac{m}{m^{\prime }}$

Let, mass of the rocket at $t+dt=m-d{m}^{\prime }$
Velocity of rocket at $t+dt=u+du$
mass of exhaust $=d{m}^{\prime }$
velocity of exhaust $=(u+du)-u_{rel}$
According to momentum conservation,
$mu=(m-d{m}^{\prime })(u+du)+d{m}^{\prime }(u+du-u_{rel})$
or, $mu=mu-ud{m}^{\prime }+mdu-d{m}^{\prime }du+ud{m}^{\prime }+d{m}^{\prime }du-u_{rel}d{m}^{\prime }$
or,$0=mdu-u_{rel}d{m}^{\prime }$
A positive exhaust mass is a mass lost from the rocket so we say that $d{m}^{\prime }=-dm$
thus, $mdu+u_{rel}dm=0$

${\int }_0^{v}du=-u_{rel}{\int }_m^{m^{\prime }}\frac{dm}{m}$

$\therefore v=-u_{rel}[lo{g}_e{m}^{\prime }-lo{g}_{e}m]=u_{rel}lo{g}_e\left(\frac{m}{m^{\prime }}\right)$