Question

A baloon  with mass 'm' is descending down with an acceleration  'a' . How much mass should be removed from it  so that it so that it starts moving up with an acceleration'a'?

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

${\green{\therefore{\text{ x=}}\frac{2ma}{g+a}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a balloon is descending with mass m.

• We have to find the minimum weight that can be remove from ballon so that the balloon goes up with acceration a.

$\green{\underline \bold{Given : }} \\ :\implies \text{Mass \: of \: balloon = m} \\\\ :\implies \text{Force \: acting \: upward = f} \\\\ :\implies \text{Downward \: force = mg} \\\\ :\implies a \lt g \\\\ :\implies \text{Let \: remove \: mass = x}\\ \\ \bold{ Removing \: x \: mass } \\\\ \bold{So \: downward \: force = xg }\\ \\ \red{\underline \bold{To \: Find : }} \\\\ :\implies \text{Extract \: mass = x}$

• According to given question :

$\bold {Case \:1} \\\\ \bold{When \: balloon \: weight \: is \: m}\\\\ :\implies mg - f = ma - - - - - (1) \\ \\ \bold {Case \:2} \\\\ \bold{when \: x \: mass \: remove \: from \: balloon} \\\\ :\implies f - mg - xg = ma -xa \\\\ :\implies f - (m - x)g = (m - x)a - - - - - (2) \\ \\ \bold {Adding \: (1) \: and \: (2)} \\\\ :\implies mg - f + f - (m - x)g = ma + (m - x)a \\\\ :\implies mg - mg + xg = ma + ma - xa\\\\ :\implies xg + ax = 2ma \\\\ :\implies x(g + a) = 2ma \\\\ \green{:\implies x = \frac{2ma}{g + a} }$

Answer 2

Answer:

$\bullet \; \large{\tt \Delta \; m = \dfrac{2 \; ma}{g + a}}$

Given:

1. Mass of balloon = m
2. Acceleration = a
3. Acceleration due to gravity = g  

Explanation:

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Case - 1

Let the Resistive  Force exerted by Air be denoted as " B ".

Now, if the Body is moving Down, the Resistive Force (B) will be Less than weight of the Balloon [ B < Mg ]

Now,

When the Balloon is descending down with Acceleration " a "

$\longmapsto\large{\tt F_{(net)} = W - B}$

$\longmapsto\large{\tt m \times a = mg - B \; \; \; \because\left[F = ma \; , W = mg \right]}$

Now,

$\longmapsto\large{\tt mg - B = ma \; \quad\dfrac{\quad}{}[1]}$

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Case - 2

Assumption:

Here, we should assume that while removing some mass the volume of balloon and hence Resistive Force will not change.

Let the new mass of the balloon is " m' "

Now, if the Body is moving up Now, the Resistive Force (B) will be More than weight of the Balloon [ B > m'g ]

Now,

When the Balloon is ascending down with Acceleration " a "

$\longmapsto\large{\tt F_{(net)} = B - W}$

$\longmapsto\large{\tt m' \times a = m'g - B \; \; \; \because\left[F = m'a \; , W = m'g \right]}$

Now,

$\longmapsto\large{\tt B - m'g = m'a \; \quad\dfrac{\quad}{}[2]}$

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Adding Equations [1] & [2]

Therefore,

$+\large{\begin{array}{c c c c c} \tt{mg} & - & \tt{B} & = & \tt{ma} \\ \\ \tt{B} & - & \tt{m'g} & = & \tt{m'a}\end{array}} \\\\ \rule{120}{0.5} \\\\ \tt mg - B + B - m'g = ma + m'a \\\\ \rule{120}{0.5}$

We get,

$\longmapsto\large{\tt mg - B + B - m'g = ma + m'a}$

$\longmapsto\large{\tt mg \; \cancel{- \; B\; } \cancel{\; + B} - m'g = ma + m'a}$

$\longmapsto\large{\tt mg - m'g = ma + m'a}$

Rearranging,

$\longmapsto\large{\tt mg - ma = m'g + m'a}$

$\longmapsto\large{\tt m(g - a) = m'(g + a)}$

Now,

$\longmapsto\large{\underline{\boxed{\tt m' = \dfrac{m(g - a)}{g + a}}}}$

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Mass removed = Δ m

$\longmapsto\large{\tt \Delta \; m = m - m'}$

Substituting the values,

$\longmapsto\large{\tt \Delta \; m = m - \dfrac{m(g - a)}{g + a}}$

$\longmapsto\large{\tt \Delta \; m = m\left[1 - \dfrac{(g - a)}{g + a}\right]}$

$\longmapsto\large{\tt \Delta \; m = m\left[\dfrac{g + a - (g - a)}{g + a}\right]}$

$\longmapsto\large{\tt \Delta \; m = m\left[\dfrac{g + a - g + a}{g + a}\right]}$

$\longmapsto\large{\tt \Delta \; m = m\left[\dfrac{a + a}{g + a}\right]}$

$\longmapsto\large{\tt \Delta \; m = m \times \left[\dfrac{2a}{g + a}\right]}$

$\longmapsto\large{\underline{\boxed{\red{\tt \Delta \; m = \dfrac{2 \; ma}{g + a}}}}}$

Hence Derived !!

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