
Suppose a rocket with an initial mass \[{M_0}\] eject a mass \[\Delta m\] in the form of gases in time \[\Delta t\], then the mass of the rocket after time t is :-
A. \[{M_0} - \dfrac{{\Delta m}}{{\Delta t}} \cdot t\]
B. \[{M_0} - \dfrac{{\Delta m}}{{\Delta t}}\]
C. \[{M_0} + \dfrac{{\Delta m}}{{\Delta t}}\]
D. \[{M_0}\]
Answer
586.5k+ views
Hint:The above problem can be resolved using the concepts of rocket propulsion. The rocket has some initial mass when it tends to take off, some part of that initial mass is being ejected in the form of gases, as soon the rocket begins its flight. This mass is known as the ejected mass, and the mathematical relationship is given by dividing the value of ejected mass and the time interval for this ejection. The rest mass can then be calculated by taking the difference of the rocket's initial mass and the calculated value of the rest mass.
Complete step by step answer:
Given:
The initial mass of the rocket is \[{M_0}\].
The mass ejected in the form of gases is \[\Delta m\].
The time interval is \[\Delta t\].
The expression for the rate of mass ejected is,
\[{m_1} = \dfrac{{\Delta m}}{{\Delta t}}\]
Here, \[{m_1}\] is the rate of ejected mass.
The value of the mass ejected is given as,
\[{m_2} = {m_1} \times t\]
Here, t denotes the time and \[{m_2}\] is the ejected mass.
On substituting the value in above equation, we get,
\[\begin{array}{l}
{m_2} = {m_1} \times t\\
{m_2} = \left( {\dfrac{{\Delta m}}{{\Delta t}}} \right) \times t
\end{array}\]
Now, the value of the rest mass is,
\[{m_3} = {M_0} - {m_2}\]
On substituting the value as,
\[\begin{array}{l}
{m_3} = {M_0} - \left[ {\left( {\dfrac{{\Delta m}}{{\Delta t}}} \right) \times t} \right]\\
{m_3} = {M_0} - \left( {\dfrac{{\Delta m}}{{\Delta t}}} \right)t
\end{array}\]
This is the value of mass after the time t.
Therefore, the mass of the rocket after time t is \[{M_0} - \dfrac{{\Delta m}}{{\Delta t}} \cdot t\] and option (A) is correct.
Note: Try to understand the concept of rocket propulsion, along with the various fundamentals like rest mass and ejected mass, must be considered. Moreover, the practical applications of rocket propulsion are based on Newton's third law of motion.
Complete step by step answer:
Given:
The initial mass of the rocket is \[{M_0}\].
The mass ejected in the form of gases is \[\Delta m\].
The time interval is \[\Delta t\].
The expression for the rate of mass ejected is,
\[{m_1} = \dfrac{{\Delta m}}{{\Delta t}}\]
Here, \[{m_1}\] is the rate of ejected mass.
The value of the mass ejected is given as,
\[{m_2} = {m_1} \times t\]
Here, t denotes the time and \[{m_2}\] is the ejected mass.
On substituting the value in above equation, we get,
\[\begin{array}{l}
{m_2} = {m_1} \times t\\
{m_2} = \left( {\dfrac{{\Delta m}}{{\Delta t}}} \right) \times t
\end{array}\]
Now, the value of the rest mass is,
\[{m_3} = {M_0} - {m_2}\]
On substituting the value as,
\[\begin{array}{l}
{m_3} = {M_0} - \left[ {\left( {\dfrac{{\Delta m}}{{\Delta t}}} \right) \times t} \right]\\
{m_3} = {M_0} - \left( {\dfrac{{\Delta m}}{{\Delta t}}} \right)t
\end{array}\]
This is the value of mass after the time t.
Therefore, the mass of the rocket after time t is \[{M_0} - \dfrac{{\Delta m}}{{\Delta t}} \cdot t\] and option (A) is correct.
Note: Try to understand the concept of rocket propulsion, along with the various fundamentals like rest mass and ejected mass, must be considered. Moreover, the practical applications of rocket propulsion are based on Newton's third law of motion.
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