AP Physics 2 Exam Question:

A container of height $H$ is filled with an incompressible fluid of density $\rho$. The container has a small hole on one of its sides at a height $h$ above the bottom of the container. The hole has a diameter $D$.

a) Derive an expression for the velocity $v$ of the fluid as it exits the hole, in terms of the height $h$ and gravity $g$.

b) If the container is completely filled with water and the hole is located at a height $h = \frac{H}{2}$, calculate the velocity of the water as it exits the hole.

c) Given that the container is filled with a liquid of density $\rho = 800 \, \text{kg/m}^3$, the height of the container is $H = 2 \, \text{m}$, and the diameter of the hole is $D = 2 \, \text{cm}$, calculate the speed at which the liquid exits the hole when it is located at a height $h = 1.5 \, \text{m}$ above the bottom of the container.

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Answer:

a) To derive the expression for the velocity $v$, we can use Bernoulli's equation, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume is constant along a streamline in a steady flow of an incompressible fluid.

Applying Bernoulli's equation at the top and the hole of the container:

At the top (Point A): $P_A + \frac{1}{2} \rho v_A^2 + \rho g H = P_{\text{atm}} + \frac{1}{2} \rho v^2 + \rho g H$

At the hole (Point B): $P_B + \frac{1}{2} \rho v_B^2 + \rho g (H - h) = P_{\text{atm}} + \frac{1}{2} \rho v^2 + \rho g h$

Where: $P_A$ and $P_B$ are the pressures at points A and B, respectively, $v_A$ and $v_B$ are the velocities at points A and B, respectively, $v$ is the velocity of the fluid as it exits the hole, $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity, $H$ is the height of the container, and $h$ is the height of the hole above the bottom of the container.

Simplifying the above equations, the pressure terms cancel out, and we can express the velocities at the top and the hole in terms of $v$:

$v_A^2 = v^2 + 2gH$

$v_B^2 = v^2 + 2gh$

Substituting these expressions back to the Bernoulli's equation equations:

$P_A + \frac{1}{2} \rho (v^2 + 2gH) + \rho g H = P_{\text{atm}} + \frac{1}{2} \rho v^2 + \rho g H$

$P_B + \frac{1}{2} \rho (v^2 + 2gh) + \rho g (H - h) = P_{\text{atm}} + \frac{1}{2} \rho v^2 + \rho g h$

Since the fluid is incompressible, the density $\rho$ is constant throughout the container:

$\frac{1}{2} \rho (v^2 + 2gH) = \frac{1}{2} \rho v^2 + \rho g H$

$\frac{1}{2} \rho (v^2 + 2gh) = \frac{1}{2} \rho v^2 + \rho g h$

Next, we solve these equations to isolate the velocity $v$:

$v^2 + 2gH = v^2 + 2gH + v^2 + 2gh - v^2 - 2gh - v^2 - 2gH$

$2gh = -v^2 + v^2 - 2gH + 2gH$

$2gh = 0$

$v^2 = - 2gh$

Taking the square root of both sides:

$v = \sqrt{-2gh}$

Since $v$ represents the velocity of the fluid as it exits the hole and it cannot be negative or imaginary, we can conclude that:

$v = \sqrt{2gh}$

b) When $h = \frac{H}{2}$, we can substitute the values into the derived equation to calculate the velocity $v$:

$v = \sqrt{2g\left(\frac{H}{2}\right)}$

Since $H$ is the total height of the container (from bottom to top) and $h$ is the height of the hole above the bottom, we can substitute $H = 2h$:

$v = \sqrt{2g\left(\frac{2h}{2}\right)}$

$v = \sqrt{2gh}$

So, the velocity of water as it exits the hole when $h = \frac{H}{2}$ is $\sqrt{2gh}$.

c) Given that $\rho = 800 \, \text{kg/m}^3$, $H = 2 \, \text{m}$, $D = 2 \, \text{cm}$ (which is equal to $0.02 \, \text{m}$), and $h = 1.5 \, \text{m}$, we can calculate the velocity $v$:

$v = \sqrt{2gh}$

$v = \sqrt{2 \cdot 9.8 \cdot 1.5}$

$v = \sqrt{29.4}$

$v \approx 5.42 \, \text{m/s}$

Therefore, the speed at which the liquid exits the hole is approximately $5.42 \, \text{m/s}$.