Question

Find the mass of the rocket as a function of time, if it moves with constant acceleration $a$, in absence of external forces. The gas escaped with a constant velocity $u$ relative to the rocket and its mass initially was ${m_o}$

Answer 1

Hint:You can solve this question quite easily by using the fact that since there are no external forces, we can imply Newton's second law of motion and find a relation between the derivative of mass and acceleration. In other words, since net external force is absent which means rocket is propelled by relative change of momentum of exhaust gases with respect to rocket. Hence net force is equal to the rate of change of momentum of exhaust.

Complete step by step answer:
We will proceed with the solution exactly as told in the hint section of the solution to the question.
We’ve been given in the question that initial mass is ${m_o}$
Acceleration is constant and has a value of $a$ which is nothing but derivative of speed with respect to time, or $\dfrac{{dv}}{{dt}}$
The constant velocity with respect to the rocket with which the exhaust gases are escaping the rocket is given as $u$
Using Newton’s second law, we can write:
$F = \dfrac{{d\left( {mv} \right)}}{{dt}}$
Using the product rule of differentiation:
$F = m\dfrac{{dv}}{{dt}} + v\dfrac{{dm}}{{dt}}$.............................(i)
In the given question, it is given that:
$\dfrac{{dv}}{{dt}} = a$
$F = 0$ and,
$v = u$
Substituting the above-mentioned results in the equation (i):
$0 = ma + u\dfrac{{dm}}{{dt}}$
After transposing, we get:
$\dfrac{{dm}}{m} = - \dfrac{a}{u}dt$
Now, we can see that to reach the answer, we simply need to integrate the equation with proper limits.
At any given time, the value of time has gone from $0$ to $t$ and the value of mass has gone from ${m_o}$ to $m$. Now, integrating the found equation with the same limits, we get:
$\int\limits_{{m_o}}^m {\dfrac{{dm}}{m}} = \int\limits_0^t { - \dfrac{a}{u}dt} $
After integrating, we get:
$\ln \left( {\dfrac{m}{{{m_o}}}} \right) = - \dfrac{{at}}{u}$
After we transpose this further, we get the resulting equation as:
$m = {m_o}{e^{ - \dfrac{{at}}{u}}}$
And this is the equation of mass as a function of time as asked to us by the question.

Note: Many students take the limit as wrong as they don’t understand how mass can be reduced, we need to see that in the equations, the mass that has been talked about is not just the mass of the body of the rocket but the mass of the whole system, body of rocket and fuel combined.