• question_answer
                    A 5000 kg rocket is set for vertical firing. The exhaust speed is \[800\,m{{s}^{-1}}\]. To give an initial upward acceleration of \[20\,m/{{s}^{2}},\] the amount of gas ejected per second to supply the needed thrust will be: \[(g=10\,m{{s}^{-2}})\]  

    A)                 \[127.5\,\,kg\,{{s}^{-1}}\]

    B)                 \[187.\,5\,kg\,{{s}^{-1}}\]

    C)                 \[185.5\,kg\,{{s}^{-1}}\]

    D)                 \[137.5\,kg\,{{s}^{-1}}\]

    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: 1.            Make a list of all the forces acting on the main mass and apply them on it. 2.            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. 3.            Find net force on the mass and apply \[{{\vec{F}}_{net}}=m\frac{d\,\vec{v}}{dt}\] (\[m=\] mass at that particular instat)                        

You need to login to perform this action.
You will be redirected in 3 sec spinner