Hot Air Balloon & Momentum

[pgoyal]

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From TBR Physics II Pg 95:

As a hot air balloon ascends upward at constant speed a package is dropped out of the balloon. At the instant the package is released, the momentum of the balloon:

a. increases
b. decreases
c. is unaffected b/c forces does not change
d. is unaffected b/c the decrease in mass is compensated by an increase in velocity.

and short explanation please?

thanks!

 

[chopes31]

I would say "B".

p = mv

Mass is constant. While ascending, velocity is positive, so momentum is some positive number. At the instant the object is released, velocity becomes zero, so momentum is 0.

 

[milski]

It's A or B, without defining whether up or down is positive direction you cannot say which. Unless there's anything about +/- in the passage, up being positive is the more likely case but it's rather lame question.

 

[milski]

I would say "B".

p = mv

Mass is constant. While ascending, velocity is positive, so momentum is some positive number. At the instant the object is released, velocity becomes zero, so momentum is 0.

When the object is dropped, the balloon will start ascending faster, it won't stop. The momentum of the object is pointed down, so the change of momentum of the balloon will be pointed up.

 

[pgoyal]

we can eliminate C ofcourse.

but I picked "d" because even though v of the balloon increases, once you drop the package doesnt the m decrease as well?

the correct answer is A btw - and its a discrete question so all the information we need is there.

 

[milski]

The total momentum of the balloon + the object is conserved. The momentum of the object decreases (since its moving downward now), which means that the momentum of the balloon has to increase. The mass does decrease but that only means that v will increase even more.

All that assumes that y increases when you go up, which somewhat reasonable but not ideal.

 

[adrakdavra]

I am not sure , but I think momentum is constant mv=c if m decrease then V increase.
another way to think about it Momentum velocity is moving upward the more mass it losses the less gravitational force on it so the more velocity upward .

 

[milski]

I am not sure , but I think momentum is constant mv=c if m decrease then V increase.
another way to think about it Momentum velocity is moving upward the more mass it losses the less gravitational force on it so the more velocity upward .

Momentum is constant only for a closed system. If you consider the balloon and the object, their total momentum should stay the same.

 

[adrakdavra]

Momentum velocity is moving upward the more mass it losses the less force of gravity on it so the more velocity upward . if we don't loss mass then momentum will not change

 

[LaughingMan]

Well OP--what is the answer?

 

[adrakdavra]

🙂I think A

 

[ricklee]

I would like to say it's C.
Think the balloon is a system, when the forces are balanced, the momentum is conserved (isolated system).

Think about what is the momentum change of the bag that is dropped? At the moment the bag is dropped, there is no change in it's velocity regardless of the gradational force.
If the momentum of the bag does not change, the balloon's moment does not change either.

 

[milski]

The answer in TBR is A.

The more I look at the problem, the less I like it. To apply conservation of momentum, all the bodies exerting forces need to be part of the system and in that case, that will include the Earth and its atmosphere. If the object is just released from the balloon (as opposed to being thrown forcefully), there is no impulse transfer between the balloon and the object and the momentum of each should not change. Ugh. I rather not get this on the exam.

Actually... the mass of the balloon decreasing will decrease its weight and the net force of it pointed up will increase, leading to an increase in momentum. Now I'm at peace with A as an answer. The extra momentum comes from atmosphere though, not from the dropped object.

 

[adrakdavra]

The extra momentum comes from atmosphere though, not from the dropped object ???

 

[milski]

The extra momentum comes from atmosphere though, not from the dropped object ???

Yes. Before the object is dropped, the net force on the balloon is zero. After the object is dropped, the buoyant force acting on the balloon stays the same but its weight decreases and as a result the net force on it is non-zero pointed up. Remember that Δp=F.Δt. In this case the force is being exerted on the balloon by the atmosphere, so the momentum is being transferred from the atmosphere to the balloon.

 

[adrakdavra]

It is the difference in air pressure causes an upward buoyant force in the air all around us. Essentially, the air pressure is greater below things than it is above things, so air pushes up more than it pushes down. But this buoyant force is weak compared to the force of gravity -- it is only as strong as the weight of the air displaced by an object. Obviously, most any solid object is going to be heavier than the air it displaces, so buoyant force doesn't move it at all. The buoyant force can only move things that are lighter than the air around them.

 

[milski]

It is the difference in air pressure causes an upward buoyant force in the air all around us. Essentially, the air pressure is greater below things than it is above things, so air pushes up more than it pushes down. But this buoyant force is weak compared to the force of gravity -- it is only as strong as the weight of the air displaced by an object. Obviously, most any solid object is going to be heavier than the air it displaces, so buoyant force doesn't move it at all. The buoyant force can only move things that are lighter than the air around them.

Yes, and that's exactly how a hot air balloon accelerates up - by being lighter that the air it displaces.

 

[adrakdavra]

So, to lower air density in a balloon without losing air pressure, you simply need to increase the speed of the air particles. You can do this very easily by heating the air. The air particles absorb the heat energy and become more excited. This makes them move faster, which means they collide with a surface more often, and with greater force. So a hot air balloon rises because it is filled with hot, less dense air and is surrounded by colder, more dense air.

 

[pgoyal]

yes, the TBR answer is A.

i realized, at least for me, the trick was the "wording" of the question.

that is the question asks for the momentum of the balloon. i kept thinking MV initial = MV final which is true but for the system.

so for the Mathies out there like myself the equation still works just know what each of the quantities are:

p(ballon initial) + p(package initial) = p(balloon final) + p(package final)

it just so happens that the velocity of the package & balloon are the same initially so we group them together. that is: (Mballoon + Mpackage)Vinitial = Mballon*vfinalballon + Mpackage*vfinalpackage

therefore the values you need to compare for the balloon and NOT the system are:
MballoonVinitial versus MbaloonVfinalballoon. clearly Vfinalballoon > Vinitial therefore the momentum of the balloon increases. while momentum of the system is conserved.

 

[milski]

Here is something to think about:
In your equation you have the Vfinal for the balloon dependent on Vfinal for the package. What happens when the package hits the ground? Its velocity becomes zero, should the velocity of the balloon change at that time? And if it does not, how is then momentum preserved? 😉

 

[pgoyal]

again this is because momentum of the system is preserved.

milski you must remember that when the book hits the group an inelastic collision is taking place, therefore the momentum of the book is transferred to the ground (in form of heat, vibrational energy, etc) so momentum of the system is infact conserved. therefore velocity of the balloon does NOT need to become 0 in order for momentum to be conserved.

as a side if someone is wondering about energy and kinetic energy, energy of the system is conserved however kinetic energy by itself is not conserved in inelastic collisions.

 

[milski]

You are right about that part, never mind.

 

[mehc012]

This was the most civil, straightforward, interesting MCAT discussion I've yet seen on here! And now I understand momentum better. Win/win/win!! 👍

 

[Vicodeen]

i don't understand the logic for this question. It asks at the MOMENT it is released, and at that moment the mass of the balloon will have decreased while the velocity will remained relatively unchanged. And as you all know momentum = m * v. m decreases and v stays about the same for the balloon at the moment the package is released. Furthermore change in momentum is F * dTime, and yes net force increases on the balloon but change in time is infinitesimal, limit-->0

My argument would be for D. Any thoughts?

 

[FeinMS]

i don't understand the logic for this question. It asks at the MOMENT it is released, and at that moment the mass of the balloon will have decreased while the velocity will remained relatively unchanged. And as you all know momentum = m * v. m decreases and v stays about the same for the balloon at the moment the package is released. Furthermore change in momentum is F * dTime, and yes net force increases on the balloon but change in time is infinitesimal, limit-->0

My argument would be for D. Any thoughts?

When you fire a gun, you are pushed back to conserve momentum. It's a similar situation.

 

[Vicodeen]

When you fire a gun, you are pushed back to conserve momentum. It's a similar situation.

Its not exactly the same situation, if you threw a box down, yes there would be an equal and opposite reaction but in this case, the system is not isolated and earth is what is pulling the box down.

 

[mehc012]

i don't understand the logic for this question. It asks at the MOMENT it is released, and at that moment the mass of the balloon will have decreased while the velocity will remained relatively unchanged. And as you all know momentum = m * v. m decreases and v stays about the same for the balloon at the moment the package is released. Furthermore change in momentum is F * dTime, and yes net force increases on the balloon but change in time is infinitesimal, limit-->0

My argument would be for D. Any thoughts?

If you're going to get nitpicky about exact timings...at the instant you have released the package, its velocity has not changed, the balloon's velocity has not changed, etc...and so it would be a meaningless question. It is implied that they are, in fact, asking about the first instant in which anything has changed (the instant after you have released it.) So yeah, they should probably phrase it as "the instant after the mass is released, the momentum of the balloon would be..." but being that picky makes you lose sight of what the question is trying to test.

Either way, just keep in mind the split equation that has been posted above: v_iBalloon·m_balloon + v_iObject·m_object = v_fBalloon·m_balloon + v_fObject·m_object

The point is that any change in v_object requires a change in the opposite direction of v_balloon. THAT is what the question is trying to test you on, and you should pick the answer that best reflects an understanding of that concept, even if you can find vagueness in their prompt if you dissect the details out enough.

 

[mehc012]

Its not exactly the same situation, if you threw a box down, yes there would be an equal and opposite reaction but in this case, the system is not isolated and earth is what is pulling the box down.

Yes, the earth is what pulls the box down, but it was ALWAYS pulling the box down...the balloon was just counteracting it.

The earth's gravitational force on the box does not change.
The buoyant force on the box does not change.
The earth's gravitational force on the balloon does not change.
The buoyant force on the balloon does not change.

What DOES change is that initially, the box was pushing down on the balloon (weight), and the balloon was pushing up on the box (normal force, equal to the weight).

Now you've removed those forces...so while it's not exactly a case of equal and opposite reaction, it IS an instance where the equal and opposite forces are actually removed, and the principles are the same.

 

[KoalaT]

Reviving this because I had the same problem and all the answers in this thread were not satisfying. I figured it out, and the solution is the question is garbage (except it is useful as food for thought). Here's the solution:

The hot air balloon moves at constant velocity and has no acceleration acting on it because the Force gravity (mg) and Force buoyancy (pVg) are equal. When the package is dropped, force gravity decreases and Force buoyancy will now accelerate the balloon upward with a decreased mass and increasing velocity. Notice the acceleration will continue to accelerate the balloon. The velocity will increase, increase, increase, and increase.

With this in mind, choice A, B or D can actually be true and choice B is TECHNICALLY the best answer.

The INSTANT the package is dropped the acceleration has not increased the velocity at all. Only the mass has decreased and thus momentum has decreased (Answer B)

At a short period of time (depending on the acceleration), the velocity will increase to an amount that is exactly proportional to the decrease in mass (answer D).

Since the velocity is just increasing and increasing because we now have an acceleration, the momentum of the balloon will continue to increase (Answer A). You can equate the increasing momentum of the balloon to the momentum of the package over time. If someone dropped a package on your head from 1cm above you, it wouldn't hurt much because it has little momentum. If they dropped it a few feet above, it might give you a good bump because it gained momentum. If a package dropped from a hot air balloon and hit you in the head, you just got knocked out by a massive amount of momentum. The acceleration is constant and velocity continues to increase. The balloon is doing the same.

Answer choice C is completely wrong because the force of buoyancy changed. In conclusion, this question is garbage. I put D originally but technically (by the wording of INSTANT) it should be B. And yet, TBR says its A which makes some sense considering it will continue to gain momentum forever (hypothetically)

 

[aldol16]

Actually, the question can be simplified just from a simple conservation of momentum argument - the result is that the momentum of the balloon must change. At the instant you release the object, it has the same velocity as the balloon because gravity hasn't had time to decelerate it yet. So instead of a momentum expression like this: p = Mv*v where M = mass object plus mass balloon, it looks like p = mass object*v + mass balloon*v. This means the momentum of the balloon itself has decreased - it is now only a fraction of the total momentum of the system. Think about it just like one object exploding into a bunch of pieces and flying in multiple directions.