question_answer

A \[5000\,\,kg\] rocket is set for vertical, firing. The exhaust speed, is \[800\,\,m/s\]. 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 \[\overset{\to }{\mathop{\mathbf{F}}}\,\] is applied on a body for a short interval of time \[\Delta t\] then the impulse of this force is\[\overset{\to }{\mathop{\mathbf{F}}}\,\times \Delta t\]. Let \[m\] be the mass of the body. On applying a constant force\[\overset{\to }{\mathop{\mathbf{F}}}\,\], for a time interval\[\Delta t\], body suffers a velocity charge\[\Delta \overset{\to }{\mathop{\mathbf{v}}}\,\]. Then, according to Newton's second law, we have                 \[\overset{\to }{\mathop{\mathbf{F}}}\,=m\overset{\to }{\mathop{\mathbf{a}}}\,=m\frac{\Delta \overset{\to }{\mathop{\mathbf{v}}}\,}{\Delta t}\]                 \[\overset{\to }{\mathop{\mathbf{F}}}\,\Delta t=m\Delta \overset{\to }{\mathop{\mathbf{v}}}\,=-\overset{\to }{\mathop{\mathbf{v}}}\,(\Delta m)\] But\[m\Delta \overset{\to }{\mathop{\mathbf{v}}}\,=\Delta \overset{\to }{\mathop{\mathbf{p}}}\,\](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                 \[\frac{\Delta m}{\Delta t}=m(a+g)\]      \[\frac{\Delta m}{\Delta t}\times 800=5000(10+20)\]               \[\frac{\Delta m}{\Delta t}=\frac{5000\times 30}{800}=187.5\,\,kg/s\]