91Ó°ÊÓ

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Thus, using the free body diagram the mass that must be thrown out of the hot-air balloon to give it an upward acceleration of magnitude a can be found.

Formula:

It is given that,the mass of hot air balloon is M and the acceleration of the hot air balloon is –a, thus the Newton’s second law is,

$F_{net}=\underset{}{\overset{}{∑Ma}}$ (i)

Let mass m be thrown out of the hot air balloon.

FBD for the system before the mass is thrown from the balloon is,

Let, mass m be thrown out of the hot air balloon.

$$\begin{gathered} F_a-\left(M-m\right)g=\left(M-m\right)a \\ {F}_a=\left(M-m\right)a+\left(M-m\right)g \end{gathered}$$FBD for the system after the mass is thrown from the balloon is,

From the first FBD and using notion in equation (i),

$Mg-F_a=Ma$ (ii)

From the second FBD,

$$\begin{gathered} F_a-\left(M-m\right)g=\left(M-m\right)a \\ {F}_a=\left(M-m\right)a+\left(M-m\right)g \end{gathered}$$ (iii)

Substituting equation (iii) in equation (ii),

$$\begin{gathered} Mg-(M-m)a-(M-m)g=Ma \\ Mg-Ma+ma-Mg+mg=Ma \\ ma+mg=Ma+Ma \\ m(a+g)=2Ma \\ m=\frac{2Ma}{a+g} \end{gathered}$$

Therefore, the mass is $m=\frac{2Ma}{a+g}$ , that must be thrown out of the hot-air balloon to give it an upward acceleration of magnitude a.