Chapter 7, Problem 61P

To determine

The speed and direction (relative to Earth) of balloon motion.

To explain: The effect if the passenger stops.

Answer to Problem 61P

The speed of the balloon (relative to Earth) is $\frac{mv}{m+M}$ moving upwards.

The whole system will stop if the passenger stops.

Explanation of Solution

Given:

Mass of balloon and its gandola is M while mass of passenger is m .

Formula used:

According to the conservation of momentum principle, the total momentum before and after momentum remains conserved that is $p_i=p_f$

Calculation:

Considering the upward direction as positive reference frame.

Velocity of the man with respect to the ground is

  $v_{MG}=-v+v_{BG}$

Where,

  $v_{BG}$ is the velocity of balloon with respect to ground.

  $-v$ is the velocity of passenger with respect to balloon.

Using momentum conservation principle,

  $\begin{array}{l}0+0=m{v}_{MG}+M{v}_{BG}\\ 0=m(-v+v_{BG})+M{v}_{BG}\\ v_{BG}=\frac{mv}{m+M}\end{array}$

Here, velocity is positive indicating the velocity of the balloon with respect to the ground going upwards.

If the passenger stops then balloon will also remain stationary and center of mass will remains at rest. Thus, the whole system will stop.

Conclusion:

Speed of the balloon (relative to Earth) is $\frac{mv}{m+M}$ moving upwards. If passenger stops then balloon with respect to ground will also remain stationary.