All AP Physics 2 Resources
Example Questions
Example Question #1 : Pressure
Suppose that a force of is acting upon a surface of area . What is the resulting pressure that is acting on this surface?
This is a fairly straight-forward question, providing us with a force vector that is acting upon a surface and asking us for the pressure.
First and foremost, we can examine the equation for pressure:
Next, we'll need to figure out the sign convention for pressure. Based on the above equation, it may seem like if we plug in a negative value for pressure and a positive value for area, we'll get a negative value for pressure. However, this is not the case.
As it turns out, pressure is a scalar quantity. The reason for this is that when any force acts on a surface, it is acting perpendicularly to that surface. Therefore, the direction in which the force is acting is dependent on how the surface is oriented. No matter which way you orient the surface, the force that is acting on that surface will always be perpendicular. Consequently, pressure is essentially just a proportionality between force and area that has no bearing on direction.
Example Question #9 : Pressure
Suppose that you accidentally let go of a helium balloon, and it begins to rise higher and higher into the sky. The balloon will continue to rise until which of the following takes place?
The temperature of the air inside the balloon is equal to the temperature of the surrounding air.
The balloon will continue to rise indefinitely.
The mass of the air inside the balloon is equal to the mass of the surrounding air.
The weight of the air inside the balloon is equal to the weight of the surrounding air.
The density of the air inside the balloon is equal to the density of the surrounding air.
The density of the air inside the balloon is equal to the density of the surrounding air.
To answer this question, it's useful to think of this conceptually as the balloon being "submerged" in the atmosphere. Remember, buoyant forces refer to the force caused by the displacement of a fluid, and both liquids and gasses count as fluids. Thus, in this case, the balloon can be thought of as submerged in a fluid in the same way that any object can be submerged in a liquid such as water.
We're trying to look for the point at which the balloon stops rising. This is analogous to a situation in which the balloon is "floating" in the fluid. In such a case, the upward and downward forces are equal and balance each other out.
So now, we need to identify the upward and downward forces acting on the balloon and set them equal to each other. The upward force will be the buoyant force due to the displacement of air from the atmosphere. The downward force will be the weight of the balloon.
We can further rewrite the mass of the balloon in terms of its weight and density.
Canceling out the gravity term, we obtain:
Next, it's important to realize that the volume of atmosphere being displaced is exactly the same as the volume of the balloon, since it is the balloon that is causing the displacement of air in the atmosphere. Thus:
Furthermore, because these two values are equal, we can cancel them out in the above expression to obtain the following.
Here, we have shown that once the density of the balloon becomes equal to that of the surrounding atmosphere, it will cease rising.
Example Question #1 : Pressure
Consider the three differently shaped containers, as shown. Each container is filled with water to a depth of . In which container is the pressure the greatest at the bottom?
It cannot be determined without knowing the volume of water in each container.
Each container has the same pressure at the bottom
Container B
Container C
Container A
Each container has the same pressure at the bottom
In this question, we're told that three containers of different shapes are filled with water. We're also told that each container has the same depth of water.
To find the pressure at the bottom of any of the containers, we'll need to remember the equation for pressure.
Since each of the containers is open to the atmosphere, we can disregard the term for atmospheric pressure in the above expression. Therefore, the absolute pressure becomes the gauge pressure.
Also, since each container is filled with water, the density of the fluid in each container is identical. Moreover, the depth we are considering for each container is also the same. Therefore, the pressure at the bottom of each container is exactly the same.
Example Question #11 : Pressure
A ball with radius is submerged in syrup at a depth of . What is the total force from pressure acting on the ball?
The total pressure on the ball includes both hydrostatic and atmospheric pressure:
We are given the atmospheric pressure, so we just need to determine the hydrostatic pressure using the following expression:
Plugging in values:
Therefore,
Now to determine the force on the ball, we need it's surface area. For a sphere:
Then,
Example Question #11 : Pressure
Pressure exerts a force of spherical ball with radius . The ball is submerged in the ocean which has a density of . How deep is the ball?
We can calculate the total pressure on the ball from the given force and radius:
Where for a sphere:
There are two pressures that combine to the total pressure on the ball: hydrostatic and atmospheric.
Since the ball is submerged in the ocean, we know that the surface of the water is at sea level and thus has a pressure:
Also, we can calculate the hydrostatic pressure with the following expression:
Where density is of the ocean and the height is the depth of the ball. Plugging this into our last expression, we get:
Rearranging for height:
Example Question #11 : Pressure
A ship has crashed and is currently sinking to the bottom of the ocean. At time , the ship is at a depth of and has reached a terminal velocity of downward. What is the hydrostatic pressure on the ship at time ?
To determine the hydrostatic pressure, we need to know the depth of the ship at time t = 12s. We can determine this with the following expression:
Then using the expression for hydrostatic pressure:
Example Question #44 : Fluid Statics
Someone who is down on their luck throws a dime down a deep well. At time , the dime's velocity is immediately reduced to as it hits the water and begins accelerating down at a rate . How much time has passed when the hydrostatic pressure on the coin is ?
Given the hydrostatic pressure, we can calculate the depth that this occurs at:
Rearranging for height:
Plugging in our values, we get:
We can then use the following kinematics equation to determine how much time has passed:
If we designate the downward direction as positive and plugging in values to the kinematics equation, we get:
Rearranging, we get:
Since we can't have a negative time, the first one is the answer.
Example Question #11 : Pressure
A vertical, cylindrical tube is filled to a height of with mercury. Then, the tube is filled to a total height of with water. What is the hydrostatic pressure in the tube at a height of ?
We will use the expression for hydrostatic pressure for this problem:
In this scenario, the pressures from each material are additive. Therefore:
Plugging in expressions:
Plugging in values:
Note how we used 0.9m for the height of mercury since we are asked for the pressure at a height of 0.5m.
Example Question #11 : Pressure
A U-shaped tube is filled with water, however the openings on either ends have different cross-sectional areas of and . If a force of is applied to the opening that is in area, how much force will be exerted on the other end of the tube?
The following formula on pressure and area is used:
We substitute our known values and solve for F2 to obtain the output force:
Therefore the correct answer is of force.
Example Question #47 : Fluid Statics
A cube with a mass of and sides of length rests on a table. What pressure does this cube exert on the table?
The formula for pressure is given as:
Where is pressure in pascals, is force in newtons, and is area in meters squared. By substituting our known values we can solve for pressure:
Therefore the correct answer is