AP Physics 2 Exam Question:

A container is filled with an ideal fluid. The container is in the shape of a cylinder with a radius of 0.5 m and a height of 2 m. A small hole is made at the bottom of the container, and the fluid starts to flow out of the hole. At a certain instant, the fluid is observed to be flowing out of the hole with a speed of 10 m/s. Calculate the pressure difference between the top and bottom of the container.

Solution:

To calculate the pressure difference between the top and bottom of the container, we can use Bernoulli's equation, which states that the sum of pressure, kinetic energy per unit volume, and potential energy per unit volume remains constant for an ideal fluid in steady flow.

The equation for Bernoulli's principle is given as:

P1 + 1/2 ρv1^2 + ρgh1 = P2 + 1/2 ρv2^2 + ρgh2

where: P1 and P2 are the pressures at points 1 and 2, respectively, ρ is the density of the fluid, v1 and v2 are the velocities at points 1 and 2, respectively, g is the acceleration due to gravity, and h1 and h2 are the heights at points 1 and 2, respectively.

In this scenario, we are given the following known values: v1 (velocity at the top of the container) = 0 m/s (because the fluid is at rest) v2 (velocity at the hole) = 10 m/s h1 (height at the top of the container) = 2 m h2 (height at the hole) = 0 m (since the hole is at the bottom)

We are required to find the pressure difference, which is P1 - P2.

Substituting the known values into Bernoulli's equation, we get:

P1 + 1/2 ρ(0 m/s)^2 + ρ(9.8 m/s^2)(2 m) = P2 + 1/2 ρ(10 m/s)^2 + ρ(9.8 m/s^2)(0 m)

Since the fluid is at rest at the top, the term 1/2 ρ(0 m/s)^2 simplifies to 0.

Thus, the equation becomes:

P1 + ρgh1 = P2 + 1/2 ρv2^2

P1 - P2 = 1/2 ρv2^2 - ρgh1

P1 - P2 = 1/2 ρ(10 m/s)^2 - ρ(9.8 m/s^2)(2 m)

Next, we substitute the known values for density (ρ = 1000 kg/m^3) and solve for the pressure difference:

P1 - P2 = 1/2 (1000 kg/m^3)(10 m/s)^2 - (1000 kg/m^3)(9.8 m/s^2)(2 m)

P1 - P2 = 1/2 (1000 kg/m^3)(100 m^2/s^2) - (1000 kg/m^3)(19.6 m^2/s^2)

P1 - P2 = 50000 N/m^2 - 19600 N/m^2

P1 - P2 = 30400 N/m^2

Therefore, the pressure difference between the top and bottom of the container is 30400 N/m^2.

Note: The unit of pressure is Pascal (Pa), which is equivalent to N/m^2.

Answer: 30400 N/m^2