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AP Physics 2 : AP Physics 2

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Example Questions

Example Question #11 : Flow Rate

Water is flowing through a fully-filled vertical, circular pipe. If the water has an initial velocity of $4\frac{m}{s}$ and a final velocity of $10\frac{m}{s}$ after a height change of , what is the ratio of the final cross-sectional area of the pipe to the initial cross-sectional area?

Assume $g = 10\frac{m}{s^2}$

Possible Answers:

0.4

0.8

1.0

1.2

2.4

Correct answer:

1.2

Explanation:

We will begin this problem with the expression for conservation of energy:

E = U_i + K_i = U_f + K_f

If we assume a final height of 0, we get:

U_i + K_i = K_f

Plugging in expressions, we get:

$mgh_i = \frac{1}{2}mv_i^2=\frac{1}{2}mv_f^2$

Rearranging for final velocity, we get:

$v_f = \sqrt{2gh_i+v_i^2}$

$v_f = 11.66\frac{m}{s}$

This would be the velocity if the cross-sectional area of the tube stayed constant. However, that is not the case, so we can use the following expression to adjust for the difference of the cross-sectional area:

$\dot{V_1}=\dot{V_2}$

v_1A_1=v_2A_2

Rearranging for the ratio we are looking for:

$\frac{A_2}{A_1}=\frac{v_1}{v_2}$

Where:

$v_1 = 11.66\frac{m}{s}$

$v_2 = 10\frac{m}{s}$

Plugging these in, we get:

$\frac{A_2}{A_1} = 1.2$

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Example Question #11 : Flow Rate

Water is flowing through a pipe of radius $0.5\textup{ m}$ at a velocity of $2\ \frac{\textup{m}}{\textup{s}}$ . The piper then narrows to a radius of $0.23\textup{ m}$ . Determine the new velocity

Possible Answers:

$2.3\ \frac{\textup{m}}{\textup{s}}$

$8.11\ \frac{\textup{m}}{\textup{s}}$

$15\ \frac{\textup{m}}{\textup{s}}$

$5.05\ \frac{\textup{m}}{\textup{s}}$

$9.45\ \frac{\textup{m}}{\textup{s}}$

Correct answer:

$9.45\ \frac{\textup{m}}{\textup{s}}$

Explanation:

Initial volume rate must equal final volume rate:

Q_1=Q_2

v_iA_i=v_fA_f

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

Solving for v_f

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

Plugging in values:

$v_f=\frac{.5^2*2}{.23^2}$

$v_f=9.45\frac{m}{s}$

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Example Question #11 : Flow Rate

Water is flowing through a horizontal cylindrical tube. By what factor does the velocity change by if the circumference of the tube doubles?

Possible Answers:

$\frac{1}{2}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{4}$

Explanation:

We can model the volumetric flow through the tube as the following expression:

$\dot{V}=vA$

Where:

$A= \pi r^2$

so:

$\dot{V}=v\pi r^2$

Applying this to the first and second scenario, we get:

$\dot{V_1}=v_1\pi r_1^2$

$\dot{V_2}=v_2\pi r_2^2$

According to the law of continuity, we know that the volumetric flow through the tube (when neglecting friction and assuming that it is horizontal) is constant. Therefore, we can say:

$v_1\pi r_1^2=v_2\pi r_2^2$

Rearranging for the ratio $\frac{v_2}{v_1}$ :

$\frac{v_2}{v_1}=\frac{r_1^2}{r_2^2}$

From the problem statement, we are told that the circumference is doubled. Thus, we know that the radius of the tube doubles as well:

r_2 = 2r_1

Plugging this into the expression, we get:

$\frac{v_2}{v_1}=\frac{r_1^2}{(2r_1)^2}$

$\frac{v_2}{v_1}=\frac{1}{4}$

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Example Question #11 : Flow Rate

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

Possible Answers:

Increase the density of the fluid

Increase the diameter of the pipe

Decrease the length of the pipe

Increase the length of the pipe

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 #12 : 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:

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

None of these

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

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

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

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:

$100 \frac{m}{s}$

$150 \frac{m}{s}$

$20 \frac{m}{s}$

$200 \frac{m}{s}$

$10 \frac{m}{s}$

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:

$100 \frac{m}{s}$

$50 \frac{m}{s}$

$5 \frac{m}{s}$

$1 \frac{m}{s}$

$10 \frac{m}{s}$

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 #621 : Ap Physics 2

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:

III

All sections have the same velocity

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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AP Physics 2 : AP Physics 2

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 #11 : Flow Rate

Example Question #11 : Flow Rate

Example Question #12 : Flow Rate

Example Question #31 : Fluid Dynamics

Example Question #32 : Fluid Dynamics

Example Question #33 : Fluid Dynamics

Example Question #621 : Ap Physics 2

All AP Physics 2 Resources

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