Debating with physics teacher.

[FrankJ]

At best a poorly worded question. It would be better as a short answer, but in that case any could be correct with the right reasoning. The best answer is B the heavier rock. For a given velocity air resistance is independent of mass. Air resistance is a function of the cube of the velocity. Since the heavier rock has greater force of gravity acting on it, it will have a greater terminal velocity thereby a greater force from air resistance.

Or once again at terminal velocity: do a force balance
F=MAg = F(air resistance). Since the more massive rock has a greater force from gravity, the counteracting force from air resistance will be greater. At less than terminal velocity, the more massive rock will have a greater acceleration, but at any given velocity the force from air resistance will be the same on both rocks. A=(MAg-F(air resistance))/M

If you measure the acceleration of the rocks and the terminal velocities, the effect of the air resistance might seem to be greater on the lighter rock since it will accelerate slower & have a lower terminal velocity, but that was not the question was it?
F=force, M=mass, Ag= acceleration of gravity, F(air resistance) Force as a function of air resistance

[wilsonmw04]

In a high school physics classroom (or intro college level for that matter), why even bring up the issue of air resistance? As illustrated by the complexity of the answers in this thread, it muddies the waters in what should be an application of a simple model.

[JamesBrown]

Originally Posted by wilsonmw04

In a high school physics classroom (or intro college level for that matter), why even bring up the issue of air resistance? As illustrated by the complexity of the answers in this thread, it muddies the waters in what should be an application of a simple model.

Because it is important to understand why the model does not match real world results. Understanding what other forces are in play is important, even if understanding how to quantify those forces is beyond the scope of the course.

[wilsonmw04]

Originally Posted by JamesBrown

Because it is important to understand why the model does not match real world results. Understanding what other forces are in play is important, even if understanding how to quantify those forces is beyond the scope of the course.

From an engineering standpoint you may be correct, but from a physics education standpoint, it is trivial and over complicates the treatment of the relationship between position, velocity and acceleration. Even when dealing with net forces I don't even deal with it. We discuss why our model breaks down when comparing real world situations, but it's not worth dealing with it even on a qualitative level. Just my opinion.

[Michael Hill]

I agree...poorly worded. It really should say something about assuming they're the same shape of the same size. I read "air resistance" from a force perspective (coming from an aeronautics background). So if they're the same size and shape, the heavy rock will experience more force. You can assume they're spheres if you want (it turns out the specific shape doesn't really matter). The aerodynamic force on a sphere will be a function of shape, surface finish, and velocity. See this: https://www.grc.nasa.gov/www/K-12/ai...ragsphere.html

Because the heavier rock has a greater gravitational force, it needs a greater aerodynamic force to counteract the gravitational force (at terminal velocity). Since the aerodynamic force is greater, the velocity is greater (this is why heavier objects of the same shape have a higher terminal velocity). Since the aerodynamic force is greater, it experiences more air resistance.

[FrankJ]

Of course you could always do it in a friction less vacuum

[wilsonmw04]

Originally Posted by FrankJ

Of course you could always do it in a friction less vacuum

It's a thing; at least in my world ;-)