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

Answer:

Correct answer letter C.

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:

$0=m_{person}v_{person}+M_{balloon}V_{balloon}$

$V_{balloon}=-\frac{m_{person}}{M_{balloon}}v_{person}$  (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!