Question

A rocket of mass $3000\text{kg}$ is launched with an exhaust speed of the gas $v=50\text{m/s}$. What will be the acceleration of the rocket after $3sec$, if the rate of consumption of fuel is $600\text{kg/s}$?
(Take $g=10{\text{m/s}}^2)$

• A. $0{\text{m/s}}^2$
• B. $35{\text{m/s}}^2$
• C. $6{\text{m/s}}^2$
• D. $15{\text{m/s}}^2$

Answer 1

The correct option is D $15{\text{m/s}}^2$

Given,
Mass of rocket; $M=3000\text{kg}$
Exhaust speed of gas, $v=50\text{m/s}$
Loss of mass of rocket in $1\text{sec}$ is $600\text{kg}$
so,
$\therefore$Loss of mass of rocket in $3\text{sec}$ is $3\times 600=1800\text{kg}$
So mass of rocket after $3\text{sec};M^{\prime }=3000-1800=1200\text{kg}$
$\therefore M^{\prime }=1200\text{kg},v=50\text{m/s}$
$\frac{dm}{dt}=600\text{kg/s},g=10{\text{m/s}}^2$
As we know,
$v.\frac{dm}{dt}-M^{\prime }g=M^{\prime }a$
$50\times \left(600\right)-\left(1200\right)\times 10=\left(1200\right)\times a$
$a=15{\text{m/s}}^2$