A huge balloon and its gondola, of mass M, are in the air and stationary with respect to the ground. A passenger, of mass m, then climbs out and slides down a rope with speed v, measured with respect to the balloon. With what speed and direction (relative to Earth) does the balloon then move? What happens if the passenger stops?

The momentum of an object depends on its velocity and mass. It varies linearly with the mass of the object. Hence, for a heavier object, the momentum is also higher.

The mass of the balloon and gondola is\(M\).

The mass of the passenger is\(m\).

The velocity of the passenger with respect to the balloon is \(v\).

Suppose the balloon, gondola, and the passengers as a system. The center of mass of the system is at rest, so the total momentum of the system with respect to the ground is zero.

When the passenger slides down the rope, the momentum does not change. Thus, the center of mass of the system stays at rest.

Let the upward direction be positive. Then the velocity of the passenger with respect to the balloon will be\( - v\), and the velocity of the balloon with respect to the ground will be\({v_{{\rm{BG}}}}\).

The equation for the velocity of the passenger with respect to the ground is

\({v_{{\rm{MG}}}} = - v + {v_{{\rm{BG}}}}\). … (i)

Since the momentum of the system of particles is equal to the product of the total mass and the velocity of the center of mass of the system, we can write

\(m{v_{{\rm{MG}}}} + M{v_{{\rm{BG}}}} = 0\).

Substitute the value of equation (i) in the above equation.

\(\begin{array}{c}m\left( { - v + {v_{{\rm{BG}}}}} \right) + M{v_{{\rm{BG}}}} = 0\\ - mv + m{v_{BG}} + M{v_{BG}} = 0\\ - mv + {v_{BG}}\left( {m + M} \right) = 0\\{v_{BG}}\left( {m + M} \right) = mv\\{v_{{\rm{BG}}}} = v\left( {\frac{m}{{m + M}}} \right)\end{array}\)

Thus, the speed of the balloon with respect to the Earth is \(v\left( {\frac{m}{{m + M}}} \right)\) in the upward direction.

If the passenger stops, the balloon also stops. Also, the CM of the system remains at rest.