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Step-by-Step Solution
Step 1: Identify the Forces on the Balloon During Descent
When the balloon of mass $m$ is descending with acceleration $a$, the forces acting on it are:
Weight (downwards) = $mg$
Upthrust from air (upwards) = $F_a$
Since it is descending (moving downward) with acceleration $a$ (where $a < g$), we can write the net force.
Step 2: Net Force Equation for Downward Motion
Using Newton's second law ($F_{\text{net}} = ma$) in the downward direction:
$mg - F_a = ma$
Hence:
$F_a = mg - ma$
Step 3: Condition for Upward Motion with the Same Acceleration $a$
Now we want the balloon to accelerate upwards with the same magnitude of acceleration $a$. Suppose we remove a mass $\Delta m$ from the balloon. Then its new mass becomes $m - \Delta m$.
Step 4: Net Force Equation for Upward Motion
When the balloon moves upward, the net upward force is:
$F_a - (m - \Delta m)g = (m - \Delta m) \, a$
We already found that $F_a = mg - ma$. Substituting this in:
$(mg - ma) - (m - \Delta m)g = (m - \Delta m)a$
Step 5: Solve for $\Delta m$
Expand and simplify the expression:
$mg - ma - mg + \Delta m g = ma - \Delta m\,a$
Notice that $mg - mg$ cancels out, leaving:
$-ma + \Delta m\,g = ma - \Delta m\,a$
Rearrange to group terms involving $\Delta m$ on one side:
$\Delta m \, g + \Delta m \, a = ma + ma$
Factor out $\Delta m$ on the left side:
$\Delta m (g + a) = 2ma$
Therefore:
$\Delta m = \dfrac{2ma}{g + a}$
Step 6: Final Answer
Thus, the mass that must be removed from the balloon so that it starts moving up with acceleration $a$ is:
$\boxed{\dfrac{2ma}{g + a}}$
