Question

A rocket motor consumes 100 kg of fuel per second, exhausting it with a speed of $5\times {{10}^{3}}m{{s}^{-1}}$.
(i) What force is exerted on the rocket?
(ii) What will be the velocity of the rocket at the instant its mass is reduced (1/20)th of its initial mass, its initial velocity being zero. Neglect gravity.

Answer 1

The force exerted on the rocket is the force exerted by the burning fuels to make the rocket start its motion. The formula for the velocity of the rocket can be obtained by integrating the formula of the thrust force of the rocket. The amount of fuel consumed will be in terms of kg/s, so the speed of the rocket should also be converted to m/s if not given in that unit form.

Formula Used:
$TF=\dfrac{dm}{dt}\times v$

$v=u{{\log }_{e}}\left( \dfrac{{{m}_{0}}}{m} \right)$

Complete step-by-step answer :
From given, we have the data,
The amount of fuel consumed by a rocket motor is, $100{kg}/{s}\;$
The speed of the rocket motor (fuel burnt) is, $v=5\times {{10}^{3}}m{{s}^{-1}}$

Firstly compute the thrust force of the rocket, that is, the force exerted on the rocket.
The thrust force of the rocket is given by the formula,
$TF=\dfrac{dm}{dt}\times v$
Where $\dfrac{dm}{dt}$is the mass ejected and v is the velocity
Substitute the given values in the above equation to find the value of the force exerted on the rocket.
Thus, we have,

& TF=\dfrac{dm}{dt}\times v \\\ & TF=100\times 5\times {{10}^{3}} \\\ & TF=5\times {{10}^{5}}N \\\ \end{aligned}$$ Therefore, the force exerted on the rocket is $$5\times {{10}^{5}}N$$. Now compute the velocity of the rocket at the instant its mass is reduced (1/20)th of its initial mass, its initial velocity being zero. The velocity of a rocket is given by the formula, $$v=u{{\log }_{e}}\left( \dfrac{{{m}_{0}}}{m} \right)$$ Where u is the speed of the gas ejection and m is the mass. Convert the base log e to base log 10, that is, $${{\log }_{e}}=2.303\times {{\log }_{10}}$$ Now the formula becomes, $$v=2.303\times u{{\log }_{10}}\left( \dfrac{{{m}_{0}}}{m} \right)$$ Substitute the given values in the above equation to find the value of the velocity. Thus, we get, $$\begin{aligned} & v=2.303\times 5\times {{10}^{3}}\times {{\log }_{10}}\left( \dfrac{{{m}_{0}}}{{}^{1}/{}_{20}{{m}_{0}}} \right) \\\ & v=2.303\times 5\times {{10}^{3}}\times {{\log }_{10}}20 \\\ & v=14981.3 \\\ & v=14.98\times {{10}^{3}}\,{m}/{s}\; \\\ \end{aligned}$$ The force exerted on the rocket is $$5\times {{10}^{5}}\,N$$ and the velocity of the rocket at the instant its mass is reduced (1/20)th of its initial mass, its initial velocity being zero is $$14.98\times {{10}^{3}}{m}/{s}\;$$. **Note:** The things to be on your finger-tips for further information on solving these types of problems are: Conversion of log to the base e to log to the base 10 makes the problem solving easier.