Question

A balloon of mass M is floating motionless in the air. A person ofmass 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

Answer 1

Correct answer letter C.

Step-by-step explanation:

As we see in this problem, the total linear momentum is conserved, since there are no external forces acting in that system.

So, we can say that
`\Delta p = 0` or
`p_(initial) = p_(final)`

Now, the initial momentum is zero, because before the person starts climbing, the system (balloon plus person) are floating motionless, therefore we have:

`0=m_(person)v_(person)+M_(balloon)V_(balloon)`

`V_(balloon)=-(m_(person))/(M_(balloon))v_(person)` (1)

We know that M > m, therefore, m/M < 1 and using (1) we can deduce that V(balloon) < v(person).

Finally, we can say that the balloon moves down (because of the minus sign of V(balloon)) and its speed is less than v(person). Correct answer letter C.

I hope it helps you!