Question

A rocket having initial mass 5 x 10^6 kg which include mass of fuel of mass 4 x 10^6 kg is ejecting gas with velocity 4000 m/s relative to the rocket. What will be the velocity of the rocket when entire fuel finishes?
Answer is 6437.75 m/s. I assume the question has to be solved by log.



​

Answer 1

of course you are right, this question can be solved with help of logarithm. but we derive expression to find velocity of the rocket when entire fuel finishes. OK !

obviously, you know about Newton's 2nd law of motion, $F_{ext}=\frac{d(mv)}{dt}=-mg$

here both m and v are variables so,

$\frac{d(mv)}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}=-mg$

or, $m\frac{dv}{dt}=-v_{rel}\frac{dm}{dt}-mg$

or, $\int\limits^{v_f}_{v_i}{dv}=-v_{rel}\int\limits^{m_f}_{m_i}{\frac{1}{m}}\,dm-g.\int\limits^t_0{dt}$

or, $v_f-v_i=v_{rel}ln\left[\frac{m_i}{m_f}\right]-gt$

as initial velocity of rocket = 0

and it is not given rocket moves under gravity so, we can neglect term ”-gt”

then, formula will be $v=v_{rel}ln\left[\frac{m_i}{m_f}\right]$

now, final mass of rocket is 10^6 kg

initial mass of rocket is 5 × 10^6 kg

and relative velocity is 4000 m/s

so, velocity of rocket , v = 4000 ln(5 × 10^6/10^6) = 4000 ln5

= 64377.75 m/s