A balloon with total mass m (including its load) falls with…

A balloon with total mass $m$ (including its load) falls with acceleration $a$ under the action of the lift (buoyant) force $F_p$. What mass $\Delta m$ of ballast must be dropped for the balloon to rise with an acceleration of the same magnitude? (Neglect air resistance; the lift stays the same.)

Explanation of Method

Ballast to turn a falling balloon into a rising one.

Step 1: Falling

\[ mg - F_p = ma \;\Longrightarrow\; F_p = m\,(g - a). \]

Step 2: Rising

\[ F_p - (m - \Delta m)g = (m - \Delta m)a, \]

\[ m\,(g - a) = (m - \Delta m)(g + a). \]

Step 3: Ballast

\[\Delta m = \frac{2ma}{g + a}\]

Check

$a = 0$: $\Delta m = 0$, nothing to drop $\checkmark$

Example $a = 1\ \mathrm{m/s^2}$: $\Delta m = 2m/10.81 \approx 0.185\,m$, about 18.5% of the mass

$a$ $F_p$ $(m)g$ $\text{falls}$ $a$ $F_p$ $(m - \Delta m)g$ $\text{rises}$

The lift $F_p$ is the same in both cases. With the full load, gravity wins by $ma$; after dropping $\Delta m$, the lift must win by $(m - \Delta m)a$.

Result

\[\boxed{\,\Delta m = \frac{2ma}{g + a}\,}\]

Remark

Check. The original result is correct.