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AP Physics 2 : Fluid Dynamics

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Example Questions

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Example Question #11 : Flow Rate

How can the velocity of fluid through a pipe be increased?

Possible Answers:

Increase the length of the pipe

Decrease the length of the pipe

Increase the diameter of the pipe

Increase the density of the fluid

Decrease the diameter of the pipe

Correct answer:

Decrease the diameter of the pipe

Explanation:

By decreasing the diameter of the pipe we increase the volume flow rate, or the velocity of the fluid which passes through the pipe according to the continuity equation.

Increasing or decreasing the length of the pipe has no effect on fluid velocity. Therefore the correct answer is to decrease the diameter of the pipe.

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Example Question #11 : Flow Rate

A pipe has fluid flowing through it. Which of the following situations will occur if a section of the pipe is compressed resulting in a small area?

Possible Answers:

More than one of these is true

The velocity of the fluid in the compressed section will decrease

The pressure of the fluid in the compressed section will increase

The velocity of the fluid in the compressed section will increase

The velocity of the fluid in the compressed section will decrease

Correct answer:

The velocity of the fluid in the compressed section will increase

Explanation:

The velocity of the fluid in the compressed section will increase and the The pressure of the fluid in the compressed section will decrease. Therefore the correct answer is: More than one of these is true.

When the area of a pipe decreases the fluid velocity increases, and an increase in fluid velocity results in the decrease of pressure.

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Example Question #11 : Flow Rate

Water is flowing through a pipe of radius $0.25\textup{ m}$ at a velocity of $4\ \frac{\textup{m}}{\textup{s}}$ . The pipe then narrows to a radius of $0.18\textup{ m}$ . Determine the new velocity.

Possible Answers:

$v_f=7.72\ \frac{\textup{m}}{\textup{s}}$

$v_f=9.81\ \frac{\textup{m}}{\textup{s}}$

$v_f=6.68\ \frac{\textup{m}}{\textup{s}}$

None of these

$v_f=4.58\ \frac{\textup{m}}{\textup{s}}$

Correct answer:

$v_f=7.72\ \frac{\textup{m}}{\textup{s}}$

Explanation:

Initial volume rate must equal final volume rate

$.5*\pi r_i^2*v_i=.5*\pi r_f^2*v_f$

Solving for v_f :

$v_f=\frac{r_i^2v_i}{r^2_f}$

Plugging in values:

$v_f=\frac{.25^2*4}{.18^2}$

$v_f=7.72\ \frac{\textup{m}}{\textup{s}}$

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Example Question #31 : Fluid Dynamics

A tank is completely full of water to the height of . On the side of the tank, at the very bottom a small hole is punctured. With what velocity does water flow though the hole at the bottom of the water tank?

Possible Answers:

$200 \frac{m}{s}$

$100 \frac{m}{s}$

$150 \frac{m}{s}$

$10 \frac{m}{s}$

$20 \frac{m}{s}$

Correct answer:

$10 \frac{m}{s}$

Explanation:

The equation for determining the velocity of fluid through a hole is as follows:

$v=\sqrt{2gh}$

This equation is actually derived from Bernoulli's principle. The is for velocity, the is the acceleration due to gravity and is the height. We solve for velocity by substituting for the values:

$v=\sqrt{2(10\frac{m}{s^{2}})(5m)}=\sqrt{100}= 10\frac{m}{s}$

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Example Question #32 : Fluid Dynamics

A syringe has a cross-sectional area of .5m^2 and the needle attached to the syringe has a cross-sectional area of .005m^2 . The fluid in the syringe is pushed with a speed of $.1\frac{m}{s}$ , with what velocity does the fluid exit the needle opening?

Possible Answers:

$5 \frac{m}{s}$

$100 \frac{m}{s}$

$1 \frac{m}{s}$

$10 \frac{m}{s}$

$50 \frac{m}{s}$

Correct answer:

$10 \frac{m}{s}$

Explanation:

The correct answer is $10 \frac{m}{s}$ because the cross-sectional area of the syringe is 100 times larger than the needle opening. Therefore, the velocity will be 100 larger as well.

$v= 100 \cdot 0.1 \frac{m}{s} = 10\frac{m}{s}$

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Example Question #33 : Fluid Dynamics

Pipe has radius $R_{1}$ , and pipe has radius $R_{2}$ . The two pipes are connected. In order for the speed of water in pipe to be times as great as the speed in pipe , what must be $\frac{R_{2}}{R_{1}}$ ?

Possible Answers:

$2\sqrt{2}$

$\frac{\sqrt{2}}{4}$

$\frac{1}{8}$

$\frac{1}{4}$

Correct answer:

$\frac{\sqrt{2}}{4}$

Explanation:

The continuity equation says that the cross sectional area of the pipe multiplied by velocity must be constant. Let be the water speed in pipe .

$\pi R_{1}^2v=\pi R_{2}^2(8v)$

$\frac{R_{2}^2}{R_{1}^2}=\frac{1}{8}$

$\frac{R_{2}}{R_{1}}=\frac{\sqrt{2}}{4}$

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Example Question #101 : Fluids

Physics pic

A fluid is forced through a pipe of changing cross sections as shown. In which section would the velocity of the fluid be a maximum?

Possible Answers:

All sections have the same velocity

III

Correct answer:

Explanation:

Flow rate is equal to the product of cross-sectional area and velocity and must remain constant. Therefore, as the cross-sectional area decreases, the velocity increases.

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Example Question #1 : Bernoulli's Equation

Water is flowing at a rate of $2 \frac{m^3}{s}$ through a tube with a diameter of 1m. If the pressure at this point is 80kPa, what is the pressure of the water after the tube narrows to a diameter of 0.5m?

$\rho_{water} = 1.0\frac{kg}{l}$

Possible Answers:

61.2 kPa

76.2kPa

93.5 kPa

100.7 kPa

31.6kPa

Correct answer:

76.2kPa

Explanation:

We need Bernoulli's equation to solve this problem:

$P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2$

The problem statement doesn't tell us that the height changes, so we can remove the last term on each side of the expression, then arrange to solve for the final pressure:

$P_2 = P_1 + \frac{1}{2}\rho (v_1^2-v_2^2)$

We know the initial pressure, so we still need to calculate the initial and final velocities. We'll use the continuity equation:

$\dot{V} = vA$

Rearrange for velocity:

$v = \frac{\dot{V}}{A}$

Where is the cross-sectional area. We can calculate this for each diameter of the tube:

$A_1 = \frac{\pi d^2}{4} = \frac{\pi(1m)^2}{4} = \frac{\pi}{4}m$

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.5m)^2}{4} = \frac{\pi}{16}m$

Now we can calculate the velocity for each diameter:

$v_1 = \frac{2\frac{m^3}{s}}{\frac{\pi}{4}m} = \frac{8}{\pi}\frac{m}{s}$

$v_2 = \frac{2\frac{m^3}{s}}{\frac{\pi}{16}m} = \frac{32}{\pi}\frac{m}{s}$

Now we have all of the values needed for Bernoulli's equation, allowing us to solve:

$P_2 = (80,000Pa)+ \frac{1}{2}(1000 \frac{kg}{m^3})(\frac{8}{\pi}-\frac{32}{\pi})$

$P_2 = (80,000Pa)+ (500)(\frac{-24}{\pi}) = 76.2kPa$

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Example Question #1 : Bernoulli's Equation

Suppose that a huge tank 50m high and filled with water is open to the atmosphere and is hit with a bullet that pierces one side of the tank, allowing water to flow out. The hole is 2m above the ground. If the hole is very small in comparison with the size of the tank, how quickly will the water flow out of the tank?

Possible Answers:

There is not enough information

$21.69\frac{m}{s}$

$112.8\frac{m}{s}$

$940.8\frac{m}{s}$

$30.67\frac{m}{s}$

Correct answer:

$30.67\frac{m}{s}$

Explanation:

To begin with, it will be necessary to make use of Bernoulli's equation:

$P_{1}+\frac{1}{2} \rho v^{2}_{1}+\rho gh_{1}=P_{2}+\frac{1}{2} \rho v^{2}_{2}+\rho gh_{2}$

For the situation described in the question stem, we'll designate the top of the container as point 1, and the hole where water is flowing out as point 2.

To begin simplifying things, it's important to realize a few things. First, both points are open to the atmosphere. Therefore, the term on each side of the above equation is equal to 1atm and thus can cancel out. Secondly, since the size of the hole on the side of the tank is so small compared to the rest of the tank, the velocity of the water at point 1 is nearly equal to 0. Hence, we can cancel out the $\frac{1}{2} \rho v^{2}_{1}$ term on the left side of the equation. Thus far, we have:

$\rho gh_{1}=\frac{1}{2} \rho v^{2}_{2}+\rho gh_{2}$

Dividing everything by $\rho$ , we obtain:

$gh_{1}=\frac{1}{2} v^{2}_{2}+ gh_{2}$

And rearranging:

$\frac{1}{2} v^{2}_{2}= gh_{1}-gh_{2}$

$\frac{1}{2} v^{2}_{2}= g(h_{1}-h_{2})=g\Delta h$

$v_{2}=\sqrt{2g\Delta h}=\sqrt{2(9.8\frac{m}{s^{2}})(50m-2m)}=30.67\frac{m}{s}$

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Example Question #1 : Bernoulli's Equation

Below is a picture of a pipe with a fluid running through it.

Bernoullipipe

If the initial velocity of the fluid is $5\frac{m}{s}$ , the final velocity is $10\frac{m}{s}$ , and the initial pressure is 110 kPa (with an unchanging potential energy), what is the value of the final pressure?

$\rho=1\frac{kg}{m^3}$

Possible Answers:

72.5kPa

147.5kPa

None of the other answers is correct

185kPa

35kPa

Correct answer:

72.5kPa

Explanation:

The equation relating fluid pressure, kinetic energy, and potential energy from state to state is known as the Bernoulli equation, and is as follows:

$P_1 + \frac{1}{2}\rho v_1^2 +\rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 +\rho gh_2$

Our potential energy is the same, so we can remove that part from the equation.

$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$

$\rho$ is the density, which in our case is equal to 1, so it doesn't change anything here.

$P_1 + \frac{1}{2} v_1^2 = P_2 + \frac{1}{2}v_2^2$

We have the values for the initial pressure, initial velocity, and final velocity, so we can rearrange our equation to equal final pressure.

$P_1 + \frac{1}{2} v_1^2 &= P_2 + \frac{1}{2}v_2^2$

$P_1 + \frac{1}{2}v_1^2-\frac{1}{2}v_2^2&= P_2$

$P_1 + \frac{1}{2}(v^2_1-v_2^2)&= P_2$

Now, we can plug in our values.

$P_2&=P_1 + \frac{1}{2}(v^2_1-v_2^2)$

$P_2=110 +\frac{5^2+10^2}{2}$

$P_2= 110 + ({-}37.5)$

P_2=72.5

Therefore, the final pressure is equal to 72.5kPa .

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AP Physics 2 : Fluid Dynamics

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #11 : Flow Rate

Example Question #11 : Flow Rate

Example Question #11 : Flow Rate

Example Question #31 : Fluid Dynamics

Example Question #32 : Fluid Dynamics

Example Question #33 : Fluid Dynamics

Example Question #101 : Fluids

Example Question #1 : Bernoulli's Equation

Example Question #1 : Bernoulli's Equation

Example Question #1 : Bernoulli's Equation

All AP Physics 2 Resources

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