Question

A 6000 kg rocket is set for vertical firing. If the exhaust speed is $1000\,{m}/{\sec }\;$. How much gas must be ejected each second to supply the thrust needed to overcome the weight of the rocket? (consider $g=10\,{m}/{{{\operatorname{s}}^{2}}}\;$acceleration due to gravity).

& A.\,6\,{kg}/{\sec }\; \\\ & B.\,60\,{kg}/{\sec }\; \\\ & C.\,600\,{kg}/{\sec }\; \\\ & D.\,6000\,{kg}/{\sec }\; \\\ \end{aligned}$$

Answer 1

The exhaust speed of the rocket represents the velocity of the gas that gets ejected. The rate of change of mass equals the product of the mass of the rocket and acceleration due to gravity by the velocity of the gas.
Formula used:
$F=ma=(u-{{V}_{0}})\dfrac{dm}{dt}$

Complete step by step answer:
The thrust force of the rocket is given as follows.
$F=(u-{{V}_{0}})\dfrac{dm}{dt}$
Where u is the velocity at which the gases leave the exhaust, ${{V}_{0}}$is the final velocity of the rocket and $\dfrac{dm}{dt}$is the rate of change of mass of the rocket.
From the given information, we have the data as follows.
A 6000 kg rocket is set for vertical firing. The exhaust speed is $1000\,{m}/{\sec }\;$.
The momentum of the rocket system equals zero as there will be no external force acting on it. ${{P}_{i}}=0$.
The final momentum equals the initial momentum. ${{P}_{f}}={{P}_{i}}$.
The sum of the momentum of the gas and the momentum of the rocket equals zero.

& {{m}_{g}}{{v}_{g}}+{{m}_{r}}{{v}_{r}}=0 \\\ & \Rightarrow {{m}_{g}}{{v}_{g}}=-{{m}_{r}}{{v}_{r}} \\\ & \therefore \left| {{m}_{g}}{{v}_{g}} \right|=\left| {{m}_{r}}{{v}_{r}} \right| \\\ \end{aligned}$$ The thrust force is given as follows. $$TF=\dfrac{d}{dt}\left( {{m}_{r}}{{v}_{r}} \right)=\dfrac{d}{dt}\left( {{m}_{g}}{{v}_{g}} \right)$$ Thus, the thrust force equals the momentum of the gas. $$TF={{v}_{g}}\dfrac{dm}{dt}+{{m}_{g}}\dfrac{dv}{dt}$$ The change in velocity equals zero. $$\begin{aligned} & TF={{v}_{g}}\dfrac{dm}{dt}+{{m}_{g}}\left( 0 \right) \\\ & \therefore TF={{v}_{g}}\dfrac{dm}{dt} \\\ \end{aligned}$$ The weight of the gas equals the thrust force. $$\begin{aligned} & {{m}_{r}}g={{v}_{g}}\dfrac{dm}{dt} \\\ & \therefore \dfrac{dm}{dt}=\dfrac{{{m}_{r}}g}{{{v}_{g}}} \\\ \end{aligned}$$ Substitute the values in the above equation. $$\begin{aligned} & \dfrac{dm}{dt}=\dfrac{6000\times 10}{1000} \\\ & \therefore \dfrac{dm}{dt}=60{kg}/{s}\; \\\ \end{aligned}$$ **So, the correct answer is “Option B”.** **Note:** The weight of the gas equals the velocity of gas times the rate of change of mass. The mass is a constant quantity, when it changes with respect to the time, then, it’s called the weight. The acceleration due to gravity equals the velocity of the gas.