A) \[127.5\text{ }kg\text{ }{{s}^{-1}}\]
B) \[187.5\text{ }kg\text{ }{{s}^{-1}}\]
C) \[185.5\text{ }kg\text{ }{{s}^{-1}}\]
D) \[137.5\text{ }kg\text{ }{{s}^{-1}}\]
Correct Answer: B
Solution :
Key idea: Thrust force on the rocket balances the weight of the rocket. |
Thrust force on the rocket. |
\[{{F}_{t}}={{v}_{r}}\left( -\frac{dm}{dt} \right)(upward)\] |
Weight of the rocket |
\[w=mg\] (downward) |
Net force on the rocket |
\[{{F}_{net}}={{F}_{t}}-w\] |
\[\Rightarrow \] \[ma={{v}_{r}}\left( \frac{-dm}{dt} \right)-mg\] |
\[\Rightarrow \] \[\left( \frac{-dm}{dt} \right)=\frac{m\,(g+a)}{{{v}_{r}}}\] |
\[\therefore \] Rate of gas ejected per second |
\[=\frac{500\,(10+20)}{800}=\frac{5000\times 30}{800}\] |
\[=187.5\,kg\,{{s}^{-1}}\] |
Note: Problems related to variable mass can be solved in following three steps: |
Make a list of all the forces acting on the main mass and apply them on it. |
Apply an additional thrust force \[{{\vec{F}}_{t}}\] on the mass, the magnitude of which is \[\left| {{{\vec{v}}}_{r}}\left( \pm \frac{dm}{dt} \right) \right|\] and direction is given by the direction of \[{{\vec{v}}_{r}}\] in case the mass is increasing and otherwise the direction of \[-{{\vec{v}}_{r}}\] if it is decreasing. |
Find net force on the mass and apply \[{{\vec{F}}_{net}}=m\frac{d\,\vec{v}}{dt}\] (\[m=\] mass at that particular instat) |
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