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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, then the mass of the rocket at any instant of time is (assume that no other forces act on it)
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Verified Answer
The correct answer is:
$m=m_0 e^{-\frac{a t}{v}}$
For a rocket,
$$
m \frac{d v}{d t}=F_{\text {ext }}+v \frac{d m}{d t}
$$
Without any external force $\left(F_{\text {ext }}=0\right)$,
$$
\begin{array}{ll}
& m \frac{d v}{d t}=v \frac{d m}{d t} \Rightarrow \int \frac{d v}{v}=\int \frac{d m}{m} \\
\Rightarrow \quad & \log v=\log m+C
\end{array}
$$
When, $v=0, m=m_0$
$$
\therefore \quad C=-\log m_0
$$
So, we have
$$
\frac{\Delta v}{v}=\log \frac{m}{m_0} \cdot d t \text { or } m=m_0 e^{-a t / v}
$$
$$
m \frac{d v}{d t}=F_{\text {ext }}+v \frac{d m}{d t}
$$
Without any external force $\left(F_{\text {ext }}=0\right)$,
$$
\begin{array}{ll}
& m \frac{d v}{d t}=v \frac{d m}{d t} \Rightarrow \int \frac{d v}{v}=\int \frac{d m}{m} \\
\Rightarrow \quad & \log v=\log m+C
\end{array}
$$
When, $v=0, m=m_0$
$$
\therefore \quad C=-\log m_0
$$
So, we have
$$
\frac{\Delta v}{v}=\log \frac{m}{m_0} \cdot d t \text { or } m=m_0 e^{-a t / v}
$$
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