• # 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)                 $137.5\,kg\,{{s}^{-1}}$

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)

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