AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

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Example Questions

Example Question #14 : Submersion

Two identical balloons, balloon A and balloon B, are filled with gas A and gas B, respectively. In which described scenario will the balloon rise the fastest?

Possible Answers:

Balloon B at 

All scenarios describe equal rising rates

Balloon B at

Balloon A at

Balloon A at 

Correct answer:

Balloon B at 

Explanation:

As temperature increases, density decreases.

In a free body diagram of each balloon, the only forces acting in the vertical direction are gravity and buoyant force.

These forces act in opposite directions, with the buoyant force pulling up and gravity pulling down.

As gas B is heated to the higher temperature, the density decreases. Gas B already has a lower density than gas A; heating it simply intensifies this discrepancy. As the density decreases, the force of gravity on the balloon decreases as well, allowing a greater net force in the upward direction due to the buoyant force.

The net upward force experienced by balloon B at the highest temperature will be greater than any of the other scenarios listed, resulting in the greatest upward acceleration.

Example Question #1 : Buoyancy And Displacement

An irregular solid has a mass of  on a laboratory balance. It is suspended by a thread from a spring scale and fully immersed in water. The spring scale reads . What is the buoyant force on the solid?

Possible Answers:

Correct answer:

Explanation:

The difference between the mass as measured in air and the mass as measured when it is suspended in water is the mass of the displaced water, easily reduced to the volume of the solid.

First calculate the weight of the solid in air.

In water, it weighs  The difference between these values is the buoyant force acting on the solid.

Recall that buoyant force is equal to the weight of water displaced by the object.

Example Question #21 : Submersion

Two students in a physics class are conducting an experiment to see how different objects displace water in a container. Student A places a lead spherical marble with a radius of  in the container. Student B places a considerably lighter solid glass sphere in the container, with twice the volume of the lead marble. Which of the following statements is true?

I. The amount of displaced water depends on the mass of a fully immersed object

II. The amount of water displaced depends on the volume of a fully immersed object

III. Student A's object will make the water level rise more than Student B's

IV. Student B's object will make the water level rise more than Student A's

V. From the information given, one cannot determine which object would displace more water

Possible Answers:

II and IV

V only

I and III

IV only

II only

Correct answer:

II and IV

Explanation:

This question relates directly to Archimedes' principle considering buoyancy. The take home points of this question are given in the answers:

1) The more volume an object occupies, the more water it can displace.
2) The mass of an object has no relevance to the amount of water displaced as long the object is fully immersed.

From the second of these two points, we can eliminate statement I since the amount of water displaced is independent of mass. We can also eliminate statements III and V, since the marble with greater volume will displace a greater volume of water. This leaves the correct statements: II and IV.

Example Question #1 : Pressure

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise?

Atmospheric pressure = 1.01 * 10Pa

g = 9.81 m/s2

Density of water = 1000 kg/m3)

Possible Answers:

4.2m

12.6cm

7.6cm

13.2m

8.3m

Correct answer:

13.2m

Explanation:

The pressure exerted by the air on the top of the ethanol must equal the fluid pressure in the column of liquid at equilibrium. Fluid pressure is the product of density, gravity, and height of the fluid.

The density of ethanol, , may be calculated by multiplying specific gravity times the reference density of water, as specific gravity is simply a ratio of a compound's density to that of water.

Using this density in the pressure equation allows us to solve for the height of the ethanol.

1.01 * 10^{5} Pa = \left ( 780\frac{kg}{m^{3}} \right )\left ( 9.81\frac{m}{s^{2}} \right )h

After calculating, h = 13.2 m

Example Question #1 : Pressure

A diver in a lake rises from  deep, where the gauge pressure reads , to the surface, where the gauge pressure reads . Upon reaching the surface, by what percent has the pressure experienced by the diver changed?

Possible Answers:

Correct answer:

Explanation:

When the diver is submerged, the pressure experienced is the sum of the gauge pressure and the atmospheric pressure above the surface.

 

At the surface, the gauge pressure will be zero.

 

The percent change in pressure will be equal to the ratio of the change in pressure to the pressure while submerged.

Example Question #21 : Fluids And Gases

A tank is filled to a depth of with water. Oil with a density of  is added to the tank until the final depth is . What is the reading of the pressure gauge at the bottom of the tank?

Ignore the contribution of atmospheric pressure.

Possible Answers:

Correct answer:

Explanation:

The pressure contribution of the liquid can be found using the equation:

The product of the density, gravity, and height of the column of liquid will give the pressure. Start by calculating the pressure of the water.

Then calculate the pressure of the oil.

Sum the pressure of each liquid to find the total pressure at the bottom of the tank.

Example Question #24 : Fluids And Gases

Two coins lie at the bottom of a pool with an uneven bottom. One is under  of the water, and the second is under  of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

Possible Answers:

Correct answer:

Explanation:

Atmospheric pressure always equals  on earth, assuming normal elevation.

Gauge pressure always equals , where is the density of the medium, is gravity, and is the distance to the surface.

Example Question #2 : Flow

Diffusion can be defined as the net transfer of molecules down a gradient of differing concentrations. This is a passive and spontaneous process and relies on the random movement of molecules and Brownian motion. Diffusion is an important biological process, especially in the respiratory system where oxygen diffuses from alveoli, the basic unit of lung mechanics, to red blood cells in the capillaries.

Capture

Figure 1 depicts this process, showing an alveoli separated from neighboring cells by a capillary with red blood cells. The partial pressures of oxygen and carbon dioxide are given. One such equation used in determining gas exchange is Fick's law, given by:

ΔV = (Area/Thickness) · Dgas · (P1 – P2)

Where ΔV is flow rate and area and thickness refer to the permeable membrane through which the gas passes, in this case, the wall of the avlveoli. P1 and P2 refer to the partial pressures upstream and downstream, respectively. Further, Dgas­, the diffusion constant of the gas, is defined as:

Dgas = Solubility / (Molecular Weight)^(1/2)

Under certain conditions, alveoli enlarge or constrict. How would either change affect ΔV?

Possible Answers:

ΔV increases if enlarged, decreases if constricted.

ΔV decreases if enlarged, decreases if constricted.

ΔV decreases if enlarged, increases if constricted.

Has no effect

Correct answer:

ΔV increases if enlarged, decreases if constricted.

Explanation:

By Fick's law, we see that ΔV, or flow rate, is directly proportional to surface area. Thus, if alveoli enlarge, so will surface area, and therefore flow rate will also increase. Similarly, if alveoli constrict, so does the total surface area, and thus flow rate will also decrease.

This problem mainly tests to see if the test-taker can quickly see the relationship between ΔV and surface area. Also, intuitively, one might guess that the larger the surface area, the more easily and the more areas gas can diffuse.

Example Question #1 : Other Flow Principles

Diffusion can be defined as the net transfer of molecules down a gradient of differing concentrations. This is a passive and spontaneous process and relies on the random movement of molecules and Brownian motion. Diffusion is an important biological process, especially in the respiratory system where oxygen diffuses from alveoli, the basic unit of lung mechanics, to red blood cells in the capillaries.

Capture

Figure 1 depicts this process, showing an alveoli separated from neighboring cells by a capillary with red blood cells. The partial pressures of oxygen and carbon dioxide are given. One such equation used in determining gas exchange is Fick's law, given by:

ΔV = (Area/Thickness) · Dgas · (P1 – P2)

Where ΔV is flow rate and area and thickness refer to the permeable membrane through which the gas passes, in this case, the wall of the avlveoli. P1 and P2 refer to the partial pressures upstream and downstream, respectively. Further, Dgas­, the diffusion constant of the gas, is defined as:

Dgas = Solubility / (Molecular Weight)^(1/2)

In higher altitudes, a decrease in which factors of Fick's law can change in order to achieve the same flow rate at lower altitudes.

Possible Answers:

Solubility

Area

P1

Thickness

Correct answer:

Thickness

Explanation:

It should be known that in higher altitudes, the partial pressure of oxygen falls. That is, the partial pressure in the alveoli will fall. Thus, flow rate will decrease, and we will need changes to increase flow rate.

By looking at Fick's equation, we can see that a decrease in thickness can help restore flow rate. Biologically speaking, this is less likely to happen, and more correctly, hemoglobin concentration and binding affinity to oxygen increases; however, this is extraneous information not needed for the MCAT.

Example Question #1 : Other Flow Principles

Diffusion can be defined as the net transfer of molecules down a gradient of differing concentrations. This is a passive and spontaneous process and relies on the random movement of molecules and Brownian motion. Diffusion is an important biological process, especially in the respiratory system where oxygen diffuses from alveoli, the basic unit of lung mechanics, to red blood cells in the capillaries.

Capture

Figure 1 depicts this process, showing an alveoli separated from neighboring cells by a capillary with red blood cells. The partial pressures of oxygen and carbon dioxide are given. One such equation used in determining gas exchange is Fick's law, given by:

ΔV = (Area/Thickness) · Dgas · (P1 – P2)

Where ΔV is flow rate and area and thickness refer to the permeable membrane through which the gas passes, in this case, the wall of the avlveoli. P1 and P2 refer to the partial pressures upstream and downstream, respectively. Further, Dgas­, the diffusion constant of the gas, is defined as:

Dgas = Solubility / (Molecular Weight)^(1/2)

Conceptually, if alveoli are considered to be perfectly spherical and assuming that its entire surface exchanges gas, which new relationship, introducing the variable, r, the radius of an alveoli, correctly describes Fick's Equation, assuming partial pressures remain constant?

Possible Answers:

ΔV = r3/Thickness · Dgas · (P1 – P2)

ΔV = (1/2 · r1/2)/Thickness · Dgas · (P1 – P2)

ΔV = πr2/Thickness · Dgas · (P1 – P2)

ΔV = 4πr2/Thickness · Dgas · (P1 – P2)

Correct answer:

ΔV = 4πr2/Thickness · Dgas · (P1 – P2)

Explanation:

The quickest approach to this equation is to see what variable in Fick's law might be affected by a change in the radius of an alveoli. Pressure is constant and can be ruled out. The diffusion constant would not be effected by radius (hence it being a constant).

The question stem mentions nothing about a change in thickness, therefore, we are left with area.

The surface area of a sphere is measured by 4πr2 and can be substituted. If you had difficulty remembering how to measure surface area, remember that the area of a circle is πr2 andlogically it follows that the surface area of an entire sphere must be greater due to it being three dimensional.

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