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Question: a balloon with mass ‘\[m\]’ is descending down with an acceleration ‘\[a\]’ (where \[a < g\]). How m...

a balloon with mass ‘mm’ is descending down with an acceleration ‘aa’ (where a<ga < g). How much mass should be removed from it so that it starts moving up with an acceleration ‘aa’?
A. 2mag+a\dfrac{{2ma}}{{g + a}}
B. 2maga\dfrac{{2ma}}{{g - a}}
C. mag+a\dfrac{{ma}}{{g + a}}
D. maga\dfrac{{ma}}{{g - a}}

Explanation

Solution

Use Newton’s second law of motion. Apply Newton’s second law of motion to the balloon in the vertical direction by considering a force F resultant of all the forces like air resistance, thrust, etc acting on the balloon in both situations when the balloon is coming downward and going upward.

Formula used:
The expression for Newton’s second law of motion is
Fnet=ma{F_{net}} = ma …… (1)
Here, Fnet{F_{net}} is the net force on the object, mm is the mass of the object and aa is the acceleration of the object.

Complete step by step answer:
The mass of the balloon is mm.
The balloon is coming down with an acceleration of aa. A force FF also acts on the balloon in the upward direction. But since the balloon is moving in a downward direction, the force FF is less than the weight of the balloon mgmg.
Apply Newton’s second law of motion to the balloon in the vertical direction.
mgF=mamg - F = ma …… (2)

Suppose the mass mm' is removed from the balloon. Hence, the weight of the balloon becomes (mm)g\left( {m - m'} \right)g.
The mass mm' is removed from the balloon such that the force FF becomes more than the weight (mm)g\left( {m - m'} \right)g of the balloon and the balloon stars moving upward with an acceleration aa.
Apply Newton’s second law of motion to the balloon moving upward in vertical direction.
F(mm)g=(mm)aF - \left( {m - m'} \right)g = \left( {m - m'} \right)a …… (3)
Add equations (2) and (3) and solve it for mm'.
mgF+F(mm)g=ma+(mm)amg - F + F - \left( {m - m'} \right)g = ma + \left( {m - m'} \right)a
mgmg+mg=ma+mama\Rightarrow mg - mg + m'g = ma + ma - m'a
mg=2mama\Rightarrow m'g = 2ma - m'a
mg+ma=2ma\Rightarrow m'g + m'a = 2ma
m(g+a)=2ma\Rightarrow m'\left( {g + a} \right) = 2ma
m=2mag+a\therefore m' = \dfrac{{2ma}}{{g + a}}
Therefore, the mass that should be removed from the balloon is 2mag+a\dfrac{{2ma}}{{g + a}}.

So, the correct answer is “Option A”.

Note:
The force F acting on the balloon when the balloon is going upward and coming downward is considered as resultant of the all forces like air resistance, thrust, etc acting on the balloon.
If the balloon is moving in a downward direction, the force FF is less than the weight of the balloon mgmg.