Balloon

Classical Mechanics Level 3

A balloon with mass $m$ is descending down with an acceleration $a$ (where, $a<g$ ). How much mass should be removed from it so that it starts moving up with an acceleration $a$ ?

Note: Ignore air resistance.

\dfrac{ma}{g-a}

\dfrac{ma}{g+a}

\dfrac{2ma}{g-a}

\dfrac{2ma}{g+a}

2 solutions

A Former Brilliant Member
Sep 6, 2016

Let the retarding force acting on the balloon in vertically upward direction be $\color{#20A900}{F}$ .

When balloon is desending down with accleration $a$ ,its equation of motion will be:

$\implies ma=mg-\color{#20A900}{F} \ \ \ ...(1)$

Let the mass $M$ be removed from balloon.As it moves up with accleration $a$ , its equation if motion will be:

$\implies (m-M)a=\color{#20A900}{F}-(m-M)g$

Using $(1)$ ,

$(m-M)a=mg-ma-mg+Mg$

$2ma=M(g+a)$

$M=\boxed{\dfrac{2ma}{g+a}}$

In your future problems please be more specific. You should have clarified that we are ignoring air resistance and assuming volume (and therefore buoyancy force, not "retarding force"...) is constant

Derek Modzelewski - 4 years, 9 months ago

Erm.....the mass changes but the volume remains??

Bad SQ - 4 years, 6 months ago

Subharthi Chowdhuri
Nov 9, 2016

What is that retarding force you are referring to? You have already ignored the air resistance. It then should be buoyancy and as pointed out in one of the comments, it can only remain constant in both of the cases if the volume remains the same. That also is an absurd assumption, because then the density of the balloon itself will change. So you are also implying not the same balloon has been flown then in both the cases.

The question is extremely ill formed and you should have paid attention before posting such a problem.

Balloon

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