Terminal velocity

[TheMightyTexan]

TBR ch2, passage 1 (of the 52 questions), question #3

Which has a greater terminal velocity when falling through the air, a 1000-kg boulder (with a cross sectional area of 5 m^2) or a 1.0-gm pebble (with a cross-sectional area of 5 cm^2)?

A. The boulder
B. The pebble
C. Both the pebble and the boulder have the same terminal velocity
D. There is insufficient information given to determine the relative terminal velocities of the boulder and pebble


I picked A. Mainly due to intuition, however the passage does discuss terminal velocity a little bit and give the equation as Vt = sqrt (2mg / (C x density x cross sectional area) , where C is dimensionless and is due to the objects shape.

The answer in the back states D is the best answer, and states that since we don't know C (the drag coefficient) we don't have enough information. I'm arguing that we do, considering C and mass are proportional and the mass of the boulder is way higher than C could ever be making C negligible. Is this wrong, am i missing something here?

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[Cawolf]

I would agree with the listed answer.

If you work it out the Vt of the boulder is sqrt(200/c) and the pebble is sqrt(2/c).

Technically we don't have enough information - making D the best answer. The drag coefficient is determined by the shape of the object - not it's mass. So even though intuitively, you are correct - if the boulder was a not aerodynamic and the pebble was some streamlined object - they could have the same terminal velocity.

 

[TheMightyTexan]

I would agree with the listed answer.

If you work it out the Vt of the boulder is sqrt(200/c) and the pebble is sqrt(2/c).

Technically we don't have enough information - making D the best answer. The drag coefficient is determined by the shape of the object - not it's mass. So even though intuitively, you are correct - if the boulder was a not aerodynamic and the pebble was some streamlined object - they could have the same terminal velocity.

Oh ok, now that i thought about it more, the boulder could be spread thin like a huge pancake and the pebble could be super thin and arrow shaped. So the drag coefficient really could be of a large magnitude. Thanks Cawolf!

 

[labqi]

Vt = sqrt (2mg/pAC) a = area, p = density of the fluid the obj passes thru, c=drag coefficient

 

[shouldvestudied]

This is just a terrible question from TBR. From a practical standpoint, the shapes of the Boulder and pebble will not be dissimilar enough to overcome the two orders of magnitude difference in the boulder's favor. For example, the difference in drag coefficients between an edge-leading airplane wing and a face-leading flat plate (0.05 and 2, respectively) isn't even enough to overcome the density*cross-section discrepancy . This is one of those cases where having more knowledge than the absolute minimum for the exam totally screws you over.

 

[TheMightyTexan]

So true..this question really bugged me but then i saw that *technically* it was actually possible...though i would hardly deem the possibilities fitting to be called such common names

 

[shouldvestudied]

Welcome to the MCAT, where knowing more than the absolute bare minimum can only serve to screw you over.