Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
7
Linear Momentum
Change chapter

7-1 and 7-2: Momentum and its Conservation
7-3: Collisions and Impulse
7-4 and 7-5: Elastic Collisions
7-6: Inelastic Collisions
7-7: Collisions in Two Dimensions
7-8: Center of Mass (CM)
7-9: CM for the Human Body
7-10: CM and Translational Motion

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 66
Q

A huge balloon and its gondola, of mass MM, are in the air and stationary with respect to the ground. A passenger, of mass mm, then climbs out and slides down a rope with speed vv, 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?

A
  1. mvm+M\dfrac{mv}{m+M}
  2. the balloon stops
Giancoli 7th Edition, Chapter 7, Problem 66 solution video poster
Padlock

In order to watch this solution you need to have a subscription.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. As this man descends along this rope attached to the gondola of the hot air balloon, he's gonna have some velocity of the man with respect to the balloon downwards and we are told what that velocity is; it's gonna be little v, let's call it negative v— negative cause it's downwards— and as he goes down, the balloon has gonna have to go up so that the center of mass of this man-balloon system stays where it is. The center of mass isn't going to move because this system starts off with zero momentum so even with a guy moving down— that's still particles moving within this closed system— is not gonna make the center of mass move at all or you can think of it in terms of forces: there's no external force being applied to any of these parts of the system; the only forces involved here are internal to the system. So there's not gonna be any acceleration, I mean there is gravity but then, you know, the gravity downwards is being compensated for equally by the buoyant force upwards in this hot air balloon and that's why it was stationary to begin with and so otherwise there's no net external force and the word 'net' is important there, okay! So the center of mass stays where it is; velocity of the center of mass is 0. Now we know that the total... I guess to be more precise the total mass of the system is m plus M although it doesn't really matter because the velocity of this center mass is 0 so this whole term is gonna be 0 which is why I have written equals 0 there. This total mass times velocity of the center of mass equals the mass of one part times its velocity plus the mass of the other part times its velocity. And I have written a couple of subscripts in the velocities here just to be clear because there's an issue with relative velocities in this question where we are given the velocity of the man with respect to the balloon and yet we have to state our answer of the velocity of the balloon with respect to the ground so yeah I'm putting some subscripts on here because the velocity of the man with respect to the ground is gonna be the velocity of the man with respect to the balloon that we are given plus the velocity of the balloon with respect to the ground. And you can think of this as: here's velocity of the man with respect to the balloon and then that's downwards and then doing some vector addition by putting the tail of this velocity of the balloon with respect to the ground on the head of the velocity of the man with respect to the balloon, you end up with velocity of the man with respect to ground. When you have the inner subscripts being the same, when you have written velocity this way— velocity of one thing with respect to the other thing— when the inner subscripts are the same, you end up with velocity of the outer subscripts— man with respect to ground. Okay that was chapter 3 for that relative velocity business there. So we can rewrite this formula by substituting this instead and that's useful because now instead of this man with respect to ground velocity, which we don't know, we can write it in terms of velocity of man with respect to balloon, which we do know, or velocity of the balloon with respect to ground which is what we are trying to find. So now we have an equation with only one variable, v BG. So distribute this m into the brackets and we have m times v MB plus mv BG plus mass of the balloon times v BG equals 0 and factor out this v BG from both of these terms and move this term to the right hand side or subtract it from both sides and you have velocity of the balloon with respect to ground is the total mass equals negative mass of the man times velocity of the man with respect to the balloon. And then divide both sides by this m plus M and you have velocity of the balloon with respect to ground equals negative of the mass of the man times his velocity with respect to the balloon divided by total mass. Now the velocity of the man with respect to the balloon is negative v—negative because it's down and v because it said so in the question. So we plug in negative v in there and we'll have velocity of the balloon with respect to ground is positive mv over m plus M and because it's positive, it must be upward. Now in part (b), if the man stops then the balloon also has to stop in order for the center of mass to not move.

COMMENTS
By rashik.raian on Sat, 5/2/2020 - 6:45 PM

Where are questions 67-85?

By sausha.c.20 on Thu, 9/29/2022 - 3:34 PM

Where are the rest of the questions??? (67-85)

By Mr. Dychko on Thu, 9/29/2022 - 4:27 PM

Hi sausha, thank you for your question. I have solved only the regular "Problems" section, not the additional "General Problems". Copied from the faq: It would be enormously time consuming to also answer all the "General Problems" so I'm limiting coverage only to the regular "Problems", of which there are still more than 1700. If you need help with a "General Problem", my suggestion would be to try and find a regular "Problem" that is similar to the "General Problem", and see if you can apply the same problem solving technique.
All the best with your studies,
Shaun

Find us on:

Facebook iconTrustpilot icon
Giancoli Answers, including solutions and videos, is copyright © 2009-2024 Shaun Dychko, Vancouver, BC, Canada. Giancoli Answers is not affiliated with the textbook publisher. Book covers, titles, and author names appear for reference purposes only and are the property of their respective owners. Giancoli Answers is your best source for the 7th and 6th edition Giancoli physics solutions.