In Fig. 9-79, an 80 kgman is on a ladder hanging from a balloon that has a total mass of 320 kg(including the basket passenger). The balloon is initially stationary relative to the ground. If the man on the ladder begins to climb at 2.5m/srelative to the ladder, (a) in what direction and (b) at what speed does the balloon move? (c) If the man then stops climbing, what is the speed of the balloon?

i) Mass of man on the ladder, m is 80 kg .

ii) Mass of balloon with the person, M is 320 kg .

iii) Speed of man on ladder relative to the ground, $v_{m}is2.5m/s$.

You use the concept of center of mass. When the man climbs on the ladder, the center of mass of man – the balloon system will not move. You can find the speed of the balloon by using the equation of center of mass of velocity. The formula required is given below.

$V_{cm}=\frac{m{v}_m-M{v}_b}{m+M}$

As the man on theladder is moving upward, the center of mass of the man-balloon system will not move.

Thus, the direction of motion of the balloon will be downward.

We can use the equation of center of mass for velocity,

$V_{cm}=\frac{m{v}_m-M{v}_b}{m+M}$

Where,$v_m$is the speed of the man relative to the ground and$v_b$is the speed of the balloon, and v is the speed of a man relative to the ladder.

$v_m=v-v_b$

Plugging the values, we get,

$$\begin{gathered} V_{cm}=\frac{m\left(v-v_b\right)-M{v}_b}{m+M} \\ \frac{mv-\left(m+M\right)v_b}{m+M}=0 \\ {v}_b=\frac{mv}{m+M} \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} v_b=\frac{\left(80\times 2.5\right)}{\left(80+320\right)} \\ {v}_b=\frac{200}{400} \\ =0.5m/s \end{gathered}$$

Thus, the speed of the balloon, $v_{b}is0.5m/s$.

When the man stops climbing, there is no relative motion between the man on the ladder and the balloon.

Thus, the speed of the balloon will be zero.