question_answer

A 5000 kg rocket is set for vertical firing. The exhaust speed is \[800\text{ }m/s.\] 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:

A)  137.5 kg/s     

B)  185.5 kg/s

C)  127.5 kg/s     

D)  187.5 kg/s

Correct Answer: D

Solution :

 Key Idea: The product of the force and the time interval is called the impulse of the force. If a constant force \[\text{\vec{F}}\]is applied on a body for a short interval of time \[\Delta t\]then the impulse of this force is \[\vec{F}\times \Delta t.\]Let m be the mass of the body.            On applying a constant force\[\text{\vec{F},}\]for a time interval \[\Delta t,\]body suffers a velocity charge \[\Delta \vec{v}.\] Then, according to Newton's Second law, we have \[\vec{F}=m\,\vec{a}=m\frac{\Delta \vec{v}}{\Delta t}\] \[\vec{F}\Delta t=m\Delta \,\vec{v}=-\vec{v}(\Delta m)\] \[\vec{F}=\vec{v}\left( -\frac{\Delta m}{\Delta t} \right)\] But \[m\Delta \,\vec{v}=\Delta \vec{p}\](change in momentum) Mass of rocket \[m=5000\,kg\] Exhaust speed\[v=800\,m/s\] Acceleration  \[a=20\,m/{{s}^{2}}\] Putting these values in above equation, we have \[\frac{\Delta m}{\Delta t}v=m(a+g)\] \[\frac{\Delta m}{\Delta t}\times 800=5000(10+20)\] \[\frac{\Delta m}{\Delta t}=\frac{5000\times 30}{800}=187.5\,kg/s\]