Debating with physics teacher.

On a recent quiz I got the following question incorrect and after debating it with my teacher he would not budge.
The question is as follows: A heavy rock and a light rock of the same size are falling through the air from a tall building. The one that encounters the greatest air resistance is the
A) light rock
B) heavy rock
C) same for both

After doing some research I still stand behind my answer of C) same for both. The teachers answer was A) light rock.
I have found explanations for both equal and heavy but am baffled as to how the answer could be the light rock.
What are your opinions or thoughts?

The answer will vary depending on how the teacher defines “the same size” and “air resistance”. The most likely meaning of “the same size” is “the same shape and volume”; I’ll assume that as the intent. Then, the definition of “air resistance” may be defined either in terms of force, in which case the heavier rock will experience slightly more at any given time, as it will move faster; or in terms of acceleration, time to fall a given distance, or other motion-based measure, in which case the lighter rock will be more impacted by the effects of the air.

He might be trying to think of it from a terminal velocity perspective. The heavy rock will have a faster terminal velocity, the lighter rock will have a slower terminal velocity. So it would seem intuitive that the lighter rock is encountering more resistance because it is going slower (even though it is the heavier rock that actually has a greater force being applied to it by the air).

Overall it seems to be a poorly worded question, a more clear question would be to ask which experiences greater drag force due to air resistance.

And drag force is affected by both surface area and velocity.

If you put together a free body diagram, you may be able to solve this one out. Gravity is acting downwards (9.8m/s^2 * Mh for the heavy rock, 9.8m/s^2 * Ml for the light rock), drag force is acting upwards (I don’t have a formula handy, but it’s a function of velocity and surface area, and varies depending on shape–thankfully, rocks tend to be approximated “nicely” with a sphere).

Now all you have to do is determine which one needs a greater drag force to have Fgravity = Fdrag, and then you can determine things like which one is falling faster…

But, in this case, it’s the heavy one (Mh is by definition > Ml). The EFFECT of the air resistance is seen most on the light one (Fdrag needing to be smaller on a body that is to all intents and purposes the same surface area but less massive in order to counter Fgravity).

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

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.

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.

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:

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.

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

It’s a thing; at least in my world :wink: