Air Resistance on MCAT?

[phillips101]

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Is air resistance even at all tested on the MCAT anymore? I didn't see it in the AAMC's physical sciences list, however, it seems to be one of EK's favorite topics to bring up.

 

[Wheee]

My test had a discrete about a falling drop, and I think it told me to ignore air resistance. That was all that ever had to do with it. You generally work out fine if you assume none, or start out assuming none (ie g=10 m/s^2) and then tack on a little extra time. The answers probably will be far apart enough for it to not matter. BUT mine was just one test, maybe they do test it sometimes? Who knows.

 

[kangar00]

Does anyone have an answer to this question? It was also stressed a good amount in Berkeley Review Physics books.

 

[eagles22]

I didn't see it on any of my practice tests. My Kaplan instructor said it would probably not come up in a quantitative manner. My advice, spend your time worrying about higher yield concepts.

 

[BerkReviewTeach]

Does anyone have an answer to this question? It was also stressed a good amount in Berkeley Review Physics books.

In the brand new physics book it is emphasized less than before, but is still there as a conceptual twist on free fall. If you have a basic idea what terminal velocity is and a working definition of viscosity, you should be fine. Don't worry about the math, but at least know that objects with a high surface area to mass ratio move slower in the air and that total energy of a projectile decreases as it moves through the air. That's really not that much and is mostly common sense.

 

[Shaptown]

In the brand new physics book it is emphasized less than before, but is still there as a conceptual twist on free fall. If you have a basic idea what terminal velocity is and a working definition of viscosity, you should be fine. Don't worry about the math, but at least know that objects with a high surface area to mass ratio move slower in the air and that total energy of a projectile decreases as it moves through the air. That's really not that much and is mostly common sense.

sorry but I would argue that your last statement is incorrect:

-the total energy of a projectile remains CONSTANT as it travels along the vector that is its flight path.

This would be primarily due to conservation of momentum. Since the mass of our projectile remains constant, the velocity of the object must remain constant as well to satisfy the conservation of momentum, assuming the projectile is not acted upon by an external force. Because our velocity at the two subsequent time intervals remains constant, the delta(KE) must be equal to zero. If this is the case, the differential gravitational potential component of total energy of the projectile must again be zero, implying that no change of vertical height would be experienced if the projectile had properties v(1), y(1), and m(1) and time 1, and v(2), y(2), and m(2) at time 2.


EDIT: However, perhaps you are right......I don't see how my logic is flawed though....

 

[vickpick]

u dont need to know the math. Just know that a bigger object of the same shape has higher air resistance regardless of the weight. If both objects have same shape and DENSITY, their terminal velocity is same fore both. - Density takes care of the weight part. Higher weight same size/shape object has higher terminal velocity than lower weight object. Terminal velocity is reached when Air resitance = gravitional force. Air resitance increases w/ velocity.

 

[BerkReviewTeach]

sorry but I would argue that your last statement is incorrect:

-the total energy of a projectile remains CONSTANT as it travels along the vector that is its flight path.

This would be primarily due to conservation of momentum. Since the mass of our projectile remains constant, the velocity of the object must remain constant as well to satisfy the conservation of momentum, assuming the projectile is not acted upon by an external force. Because our velocity at the two subsequent time intervals remains constant, the delta(KE) must be equal to zero. If this is the case, the differential gravitational potential component of total energy of the projectile must again be zero, implying that no change of vertical height would be experienced if the projectile had properties v(1), y(1), and m(1) and time 1, and v(2), y(2), and m(2) at time 2.


EDIT: However, perhaps you are right......I don't see how my logic is flawed though....

Air resistance IS the external force you mentioned. It is doing work on the object in motion, draining kientic energy and reducing momentum. When we consider air resistance, the projectile will not reach the same maximum height as it would without air resistance and the range will be reduced as well.

Also, for a projectile in motion, the velocity and momentum are constantly changing because there is the external force of gravity acting on them. This is why we generally use the work-energy theorem to analyze projectile motion.

Hope this clears it up a little.

 

[BerkReviewTeach]

u dont need to know the math. Just know that a bigger object of the same shape has higher air resistance regardless of the weight. If both objects have same shape and DENSITY, their terminal velocity is same fore both. - Density takes care of the weight part. Higher weight same size/shape object has higher terminal velocity than lower weight object. Terminal velocity is reached when Air resitance = gravitional force. Air resitance increases w/ velocity.

Be careful in using the term density here. It's actually mass to contact area ratio. A perfect example is a single piece of paper versus a stack of paper. They each have the same density and same rectangular shape, but the stack falls much faster. The stack and single sheet have the same contact area, but different masses. You have the right concept, so I apologize for being nit-picky on the wording, but it's mass-to-contact area as opposed to mass-to-volume.

Another example is turning a piece of paper sideways (thin edge down) and dropping it. Same object with the same density and exact same shape, but aligned with different contact area will fall at drastically different speeds. Horizontal (parallel to the ground) falls slowly while vertical (perpendicular to the ground) falls quickly.

 

[Hoodat]

In the brand new physics book it is emphasized less than before, but is still there as a conceptual twist on free fall. If you have a basic idea what terminal velocity is and a working definition of viscosity, you should be fine. Don't worry about the math, but at least know that objects with a high surface area to mass ratio move slower in the air and that total energy of a projectile decreases as it moves through the air. That's really not that much and is mostly common sense.

In the new BR physics book, there was a tricky example about a ball being launched at an angle from a slingshot. It asked you to consider whether a ball placed higher up in the sling (so that the sling imparted some backspin) would end up with a higher maximum height and total flight time. At first, I thought that the maximum height would be somewhat less, since some of the kinetic energy transferred to the ball would be rotational, so that the translational energy would be slightly less. However, the correct answer was that the maximum height would be higher, due to the lift created from backspin. I found this explanation a bit unsatisfying, as this was in the kinematics lecture (Chapter 1), and the book thus far had hardly addressed the issue of air at all - and what it had addressed had to due with a conceptual idea of drag force only.