A rocket with an initial mass m_0 is going up | vedclass

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(A) The equation of motion for a rocket with variable mass is given by $F_{	ext{ext}} + v_{	ext{rel}} \frac{dm}{dt} = m \frac{dv}{dt}$.Given that th

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$m=m_0 e^{-\frac{a t}{v}}$

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(A) The equation of motion for a rocket with variable mass is given by $F_{	ext{ext}} + v_{	ext{rel}} \frac{dm}{dt} = m \frac{dv}{dt}$.Given that the rocket moves with constant acceleration $a$,we have $\frac{dv}{dt} = a$.Assuming no external forces $(F_{	ext{ext}} = 0)$,the equation becomes $v_{	ext{rel}} \frac{dm}{dt} = m a$.Here,$v_{	ext{rel}} = v$ (the exhaust velocity relative to the rocket).Rearranging the terms,we get $\frac{dm}{m} = \frac{a}{v} dt$.Integrating both sides from initial mass $m_0$ at $t=0$ to mass $m$ at time $t$:$\int_{m_0}^{m} \frac{dm}{m} = \int_{0}^{t} \frac{a}{v} dt$.$\ln \left( \frac{m}{m_0} ight) = \frac{a t}{v}$.Taking the exponential of both sides,we get $m = m_0 e^{-\frac{at}{v}}$.

$A$ rocket with an initial mass $m_0$ is going up with a constant acceleration $a$ by exhausting gases with a velocity $v$ relative to the rocket motion. The mass of the rocket at any instant of time $t$ is (assume that no other forces act on it).

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