(III) A huge balloon and its gondola, of mass M, are in the ai...

Question

(III) A huge balloon and its gondola, of mass M, are in the air and stationary with respect to the ground. A passenger, of mass m, then climbs out and slides down a rope with speed v, 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?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A research balloon equipped for atmospheric studies is stationary at a high altitude above the earth's surface. It has a total mass M one including both the balloon and the attached scientific instruments. For specific data collection, one instrument package with a mass M two needs to be lowered to a different altitude. This instrument package is allowed to slide down a rope at a speed V relative to the balloon. Calculate the velocity VB of the balloon relative to the earth. Immediately after the package starts sliding. What happens to the balloon's velocity relative to the earth if the instrument stops sliding? So that's our end goal. So ultimately, we're trying to solve for two separate answers. Our first answer, we're trying to figure out what the velocity VB of the balloon is relative to earth. Immediately after the package starts sliding. And our second answer we're trying to solve for it is we're trying to figure out what happens to the balloon's overall velocity relative to earth if the instrument stops sliding. Awesome. So with that in mind, let's look at our multiple choice answers to see what our final answer might be. Our answer pair, I should say might be A, is VB equals M two multiplied by V divided by M one plus M two. The balloon stops moving upward. V is VB equals negative M two multiplied by V all divided by M one M two. The balloon moves downward. C is VB equals V divided by M one M two. The balloon stops moving downward. And finally D is VB equals M one multiplied by V divided by M one M two, the balloon moves upward. Awesome. So first off, let us assume that the system which is focusing on the balloon plus the instrument package is closed and that there are no external forces acting on it. Therefore, the total momentum of the system will be conserved. Let us note that initially the total momentum is zero since the system is at rest. Therefore, we can recall and write that the total momentum which we're gonna write as PT. So I'm gonna denote total as lowercase T. So the total momentum which is denoted by P is equal to M one multiplied by. So it's M one multiplied by V one plus M two multiplied by V two. And to be specific, when I say V one and V two, these are lower case fees, which is equal to M one multiplied by zero plus M two multiplied by zero, which is all equal to zero. So now we must note that when the package starts moving downward with some velocity of negative V relative to the balloon, the balloon as a result must move in the opposite direction. In order to conserve momentum, let us call capital VB to be the velocity of the balloon relative to earth. The package's velocity relative to the earth combines the balloon's velocity VB and its own velocity relative to the balloon, which is negative V which will result in VB minus V for the package. Therefore, we could write this equation as M one multiplied by capital VB plus M two multiplied by capital VB minus lowercase V equals zero. So now we need to rearrange this equation to isolate and sulfur VB. So let me do that and to reiterate this is capital VB. So capital VB multiplied by M one plus M two is all equal to M two multiplied by lowercase V. So finally, when we rearrange and get capital VB all by itself using a little bit of algebra, we will get that VB is equal to M two multiplied by lower case V all divided by M one plus M two. Hooray. We did it, we found our answer for the first answer. So let us also note that the direction of VB is in the upwards direction. So therefore, if the instrument package stops sliding. This will mean that V equals zero. So V equals 0 m per second. And this is if the instrument package stops sliding, therefore, the balloon will stop moving relative to earth. So with that in mind, that will mean our final answer has to be the multiple choice answer letter A capital VB equals M two, multiplied by V all divided by M one plus M two and the balloon will stop moving upward. So balloon stops moving upward and that's it. That's how you solve for this problem. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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