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buoyancy conceptual question - physics mcat

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MC789

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hey guys,

I'm confused on the answer to this question:
Q: as a hot air balloon ascends upward at a constant speed, a package is dropped out of the balloon. at the instant the package is released, the momentum of the balloon:

correct answer: momentum of balloon increases
my answer: the momentum of balloon decreases

in the past I noticed I intend to ignore the given equations in the passage to answer the questions and thats the safest way to go. so I used the buoyancy equation B = density of fluid times volume of fluid times gravity.

I chose momentum decreases because when mass is thrown out, the mass of the balloon (mass = density times volume) decreases and gravity is constant. so B = m times g, and m decreases and g is constant so B should decrease. when bouyant force decreases, there isn't much of a push on the balloon so its momentum decreases.

could someone explain why I am wrong?
 

Hi_I'mPaul

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hey guys,

I'm confused on the answer to this question:
Q: as a hot air balloon ascends upward at a constant speed, a package is dropped out of the balloon. at the instant the package is released, the momentum of the balloon:

correct answer: momentum of balloon increases
my answer: the momentum of balloon decreases

in the past I noticed I intend to ignore the given equations in the passage to answer the questions and thats the safest way to go. so I used the buoyancy equation B = density of fluid times volume of fluid times gravity.

I chose momentum decreases because when mass is thrown out, the mass of the balloon (mass = density times volume) decreases and gravity is constant. so B = m times g, and m decreases and g is constant so B should decrease. when bouyant force decreases, there isn't much of a push on the balloon so its momentum decreases.

could someone explain why I am wrong?
Hot Air Balloon & Momentum

Hopefully that link works. There you go.
 
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