Aballoon 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

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 or

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

 (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!  

no. dot product   =0 when vecs at right angles