Question

A hot-air balloon of massis descending vertically with downward acceleration of magnitude a. How much mass (ballast) must be thrown out to give the balloon an upward acceleration of magnitude a? Assume that the upward force from the air (the lift) does not change because of the decrease in mass.

Short answer

The mass must be thrown out of the hot-air balloon to give it an upward acceleration of magnitude a, that is,

$m=\frac{2Ma}{a+g}$

Step-by-step solution

Determining the concept

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,

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

Determining the Free Body Diagram (FBD)

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.

Determining the mass

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