A balloon of mass M is floating motionless in the air. A person of mass less than M is on a rope ladder hanging from the balloon. The person begins to climbthe ladder at a uniform speed v relative to the ground. How doesthe balloon move relative to the
ground?

A. Up with speed v
B. Up with a speed less than v
C. Down with a speed less than v
D. Down with speed v

Let the mass of the person be m. Total momentum is conserved (because the exterior forces on the system are balanced), especially the component in the vertical direction.

Given that,

Mass of gallon is M

Let man mass be m

Velocity of man is v

Let velocity if ballot be Vb

When the person begin to move we have

Conservation of momentum

mv + MVb=0

MVb=-mv

Vb= -(m/M) v

Given that the mass of man is less than mass of balloon. i.e. m<M

So, if m<M, then, m/M <1

Therefore, .

Vb= -(m/M) v

Vb< -v

This implies that the velocity of balloon is less than the velocity of man and if is also moving in opposite direction

So the man is moving upward, then the balloon is moving downward and it's velocity is less than the velocity of man,

The answer is C

Down with a speed less than v

Given Information:

Mass of balloon = M

Mass of person = m    (m < M)

Speed of balloon = V

Speed of person = v

Required Information:

How does the balloon move relative to the  ground ?

The balloon is moving down with speed V that is less than speed of person v.

Explanation:

Initially, the person and balloon both are at rest and no external net force on the system is acting.

The momentum is given by

Δp = FΔt

Where F is external force and if F is 0 then Δp must be 0 so that momentum is conserved.

pi = pf

The initial momentum is 0 since they were at rest initially, so the final momentum is

MV + mv = 0

MV = -mv

V = -mv/M

since m < M then m/M < 1 so

V < -v

The balloon is moving down with speed V that is less than speed of person v.