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AP Physics 2 : AP Physics 2

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Example Questions

Example Question #11 : Fluids

A semi-hollow, spherical ball with an empty volume of 0.8L $\displaystyle 0.8L$ is submerged in water and has an initial mass of 0.4kg $\displaystyle 0.4kg$. The ball develops a leak and water begins entering the ball at a rate of $0.6\frac{L}{min}$ $\displaystyle 0.6\frac{L}{min}$. How long does it take before the buoyant force on the ball is equal to the gravitational force?

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

$\rho_w = 1000\frac{kg}{m^3}$ $\displaystyle \rho_w = 1000\frac{kg}{m^3}$

Possible Answers:

30s $\displaystyle 30s$

50s $\displaystyle 50s$

20s $\displaystyle 20s$

10s $\displaystyle 10s$

40s $\displaystyle 40s$

Correct answer:

40s $\displaystyle 40s$

Explanation:

We are asked when:

$\displaystyle F_g = F_b$

$m_bg = m_{w,displaced}g$ $\displaystyle m_bg = m_{w,displaced}g$

$m_b = m_{w,displaced}$ $\displaystyle m_b = m_{w,displaced}$

$m_b = \rho_wV_b$ $\displaystyle m_b = \rho_wV_b$

Now we need to develop an expression for the mass in the ball using the rate at which water enters the ball:

$m_b = m_i + \dot{m}t$ $\displaystyle m_b = m_i + \dot{m}t$

Where:

$\dot{m} = \dot{V}\rho_wt$ $\displaystyle \dot{m} = \dot{V}\rho_wt$

$m_b = m_i + \dot{V}\rho_wt$ $\displaystyle m_b = m_i + \dot{V}\rho_wt$

Plugging this into expression (1):

$m_i + \dot{V}\rho_wt= \rho_wV_b$ $\displaystyle m_i + \dot{V}\rho_wt= \rho_wV_b$

Rearranging for time, we get:

$t= \frac{\left ( \rho_wV_b- m_i \right )}{\dot{V}\rho_w}$ $\displaystyle t= \frac{\left ( \rho_wV_b- m_i \right )}{\dot{V}\rho_w}$

Plugging in our values, we get:

$t= \frac{\left ( 1\frac{kg}{L}\right )(0.8L)- (0.4kg)}{\left ( 0.6\frac{L}{min}\right )\left ( 1\frac{kg}{L}\right )}$ $\displaystyle t= \frac{\left ( 1\frac{kg}{L}\right )(0.8L)- (0.4kg)}{\left ( 0.6\frac{L}{min}\right )\left ( 1\frac{kg}{L}\right )}$

$t = \frac{2}{3}min = 40s$ $\displaystyle t = \frac{2}{3}min = 40s$

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Example Question #21 : Fluid Statics

Which of the following statements best describes the buoyant force on an object submerged in water?

Possible Answers:

The buoyant force is equal to the weight of the object submerged in water.

The buoyant force is equal to the weight of all of the water.

None of these accurately describe buoyant force.

The buoyant force is equal to the weight of the water displaced by the object.

The buoyant force is equal to the volume of water displaced by the object.

Correct answer:

The buoyant force is equal to the weight of the water displaced by the object.

Explanation:

The correct answer is that the buoyant force is equal to the weight of the water displaced by the object.

It is an upward force that is exerted on the object because of the volume of fluid displaced by that object.

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Example Question #22 : Buoyant Force

$\rho_{air}=1.225\ \frac{\textup{kg}}{\textup{m}^3}$ $\displaystyle \rho_{air}=1.225\ \frac{\textup{kg}}{\textup{m}^3}$

$\rho_{Helium}=0.164\ \frac{\textup{kg}}{\textup{m}^3}$ $\displaystyle \rho_{Helium}=0.164\ \frac{\textup{kg}}{\textup{m}^3}$

What mass of helium would be needed to lift a $1\textup{ kg}$ $\displaystyle 1\textup{ kg}$ balloon and a $5\textup{ kg}$ $\displaystyle 5\textup{ kg}$ camera off of the ground?

Possible Answers:

This will be impossible

$3.082\textup{ kg}$ $\displaystyle 3.082\textup{ kg}$

$0.245\textup{ kg}$ $\displaystyle 0.245\textup{ kg}$

$10.35\textup{ kg}$ $\displaystyle 10.35\textup{ kg}$

$0.889\textup{ kg}$ $\displaystyle 0.889\textup{ kg}$

Correct answer:

$0.889\textup{ kg}$ $\displaystyle 0.889\textup{ kg}$

Explanation:

$F_{net}=F_{gravity}+F_{buoyant}$ $\displaystyle F_{net}=F_{gravity}+F_{buoyant}$

$F_{gravity}=g*(m_{camera}+m_{balloon}+m_{Helium})$ $\displaystyle F_{gravity}=g*(m_{camera}+m_{balloon}+m_{Helium})$

$F_{buoyant}=V_{balloon}*\rho_{air}*-g$ $\displaystyle F_{buoyant}=V_{balloon}*\rho_{air}*-g$

$m_{Helium}=\rho_{Helium}*V_{Helium}$ $\displaystyle m_{Helium}=\rho_{Helium}*V_{Helium}$

$V_{Helium}=V_{balloon}$ $\displaystyle V_{Helium}=V_{balloon}$

Combining equations:

$F_{net}=g*(m_{camera}+m_{balloon}+m_{Helium})+\frac{m_{Helium}}{\rho_{Helium}}*\rho_{air}*-g$ $\displaystyle F_{net}=g*(m_{camera}+m_{balloon}+m_{Helium})+\frac{m_{Helium}}{\rho_{Helium}}*\rho_{air}*-g$

Plugging in values:

$0=-9.8*(5+1+m_{Helium})+\frac{m_{Helium}}{.164}*1.225*9.8$ $\displaystyle 0=-9.8*(5+1+m_{Helium})+\frac{m_{Helium}}{.164}*1.225*9.8$

Solving for $m_{Helium}$ $\displaystyle m_{Helium}$

$m_{Helium}=0.889\textup{ kg}$ $\displaystyle m_{Helium}=0.889\textup{ kg}$

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Example Question #23 : Fluid Statics

A block weighing 150 N $\displaystyle 150 N$ is suspended by a string and submerged in water. The block is $\displaystyle 5$ times denser than water. What is the apparent weight of the block in water?

Possible Answers:

30 N $\displaystyle 30 N$

120 N $\displaystyle 120 N$

150 N $\displaystyle 150 N$

180 N $\displaystyle 180 N$

80 N $\displaystyle 80 N$

Correct answer:

120 N $\displaystyle 120 N$

Explanation:

The correct answer is 120 N $\displaystyle 120 N$

We know that the block is five times denser than the water. This means that a weight $\frac{1}{5}$ $\displaystyle \frac{1}{5}$ of the block would be the buoyant force which can be calculated as the following:

$\frac{150N}{5}=30N$ $\displaystyle \frac{150N}{5}=30N$

We can subtract the 30 N $\displaystyle 30 N$of weight from the original weight of the block which was 150 N $\displaystyle 150 N$

$\displaystyle 150 N - 30 N = 120 N$

We obtain 120 N $\displaystyle 120 N$

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Example Question #21 : Buoyant Force

A block with a volume of 20 L $\displaystyle 20 L$ sinks to the bottom of a water tank. What is the buoyant force on the block?

Possible Answers:

100 N $\displaystyle 100 N$

There is no buoyant force on the block

200 N $\displaystyle 200 N$

10 N $\displaystyle 10 N$

20 N $\displaystyle 20 N$

Correct answer:

200 N $\displaystyle 200 N$

Explanation:

The correct answer is 200 N $\displaystyle 200 N$ because although the block has sunk, there is still a buoyant force. This buoyant force is the result of the block displacing a volume of water, equal to the block's volume. The weight of the water volume displaced is 20 kg $\displaystyle 20 kg$ because 20L $\displaystyle 20L$ was displaced. The weight of 20 kg $\displaystyle 20 kg$ is 200 N $\displaystyle 200 N$ which is equal to the buoyant force on the block.

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Example Question #21 : Fluid Statics

An 1.2kg $\displaystyle 1.2kg$ paint can of volume 1550ml $\displaystyle 1550ml$ is floating on top of a lake, but water is slowly leaking in. Determine the volume of water present in the can when it finally sinks.

Possible Answers:

None of these

.775L $\displaystyle .775L$

.560L $\displaystyle .560L$

1.550L $\displaystyle 1.550L$

.350L $\displaystyle .350L$

Correct answer:

.350L $\displaystyle .350L$

Explanation:

When the can sinks, the net force of the water in the can plus gravity will be greater than the buoyant force. So the point of sinking is when they are equal.

$m_{can}*g+m_{water}*g=\rho_{water}*g*V_{can}$ $\displaystyle m_{can}*g+m_{water}*g=\rho_{water}*g*V_{can}$

Solving for $m_{water}$ $\displaystyle m_{water}$

$-m_{can}+\rho_{water}*V_{can}=m_{water}$ $\displaystyle -m_{can}+\rho_{water}*V_{can}=m_{water}$

Plugging in values:

$-1.2kg+1\frac{kg}{L}*1.550L=m_{water}$ $\displaystyle -1.2kg+1\frac{kg}{L}*1.550L=m_{water}$

$m_{water}=.350kg$ $\displaystyle m_{water}=.350kg$

Using the density of water,

$V_{water}=.350L$ $\displaystyle V_{water}=.350L$

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Example Question #26 : Fluid Statics

Seawater density: $1.03\frac{g}{cm^3}$ $\displaystyle 1.03\frac{g}{cm^3}$

A baseball has a mass of 145g $\displaystyle 145g$ and a circumference of 230mm $\displaystyle 230mm$. Determine what percentage of a baseball will be submerged when it is floating in seawater after a home run?

Possible Answers:

$69\%$ $\displaystyle 69\%$

$50\%$ $\displaystyle 50\%$

$85\%$ $\displaystyle 85\%$

$23\%$ $\displaystyle 23\%$

None of these

Correct answer:

$69\%$ $\displaystyle 69\%$

Explanation:

$F_{net}=F_{gravity}+F_{buoyant}$ $\displaystyle F_{net}=F_{gravity}+F_{buoyant}$

$0=\rho_{seawater}*V_{submerged}*g-m*g$ $\displaystyle 0=\rho_{seawater}*V_{submerged}*g-m*g$

Solving for $V_{submerged}$ $\displaystyle V_{submerged}$

$\frac{m_{ball}}{\rho_{seawater}}=V_{submerged}$ $\displaystyle \frac{m_{ball}}{\rho_{seawater}}=V_{submerged}$

Plugging in values

$\frac{145}{1.03}=V_{submerged}$ $\displaystyle \frac{145}{1.03}=V_{submerged}$

$V_{submerged}=141cm^3$ $\displaystyle V_{submerged}=141cm^3$

If the circumference is 230mm $\displaystyle 230mm$, it is 23cm $\displaystyle 23cm$

$C=2\pi r=23cm$ $\displaystyle C=2\pi r=23cm$

$\displaystyle r=3.66cm$

$V_{ball}=\frac{4}{3}\pi r^3$ $\displaystyle V_{ball}=\frac{4}{3}\pi r^3$

$V_{ball}=\frac{4}{3}\pi *3.66^3$ $\displaystyle V_{ball}=\frac{4}{3}\pi *3.66^3$

$V_{ball}=205cm^3$ $\displaystyle V_{ball}=205cm^3$

$\frac{V_{submerged}}{V_{ball}}=\frac{141cm^3}{205cm^3}=69\%$ $\displaystyle \frac{V_{submerged}}{V_{ball}}=\frac{141cm^3}{205cm^3}=69\%$

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Example Question #541 : Ap Physics 2

A ship has mass $\displaystyle 7.4*10^7kg$. What volume of water would it need to be able to displace in order to be able to float?

Density of seawater: $1.03\frac{kg}{L}$ $\displaystyle 1.03\frac{kg}{L}$

Possible Answers:

$6.66*10^{7}L$ $\displaystyle 6.66*10^{7}L$

$4.40*10^{7}L$ $\displaystyle 4.40*10^{7}L$

$3.95*10^{7}L$ $\displaystyle 3.95*10^{7}L$

None of these

$7.18*10^{7}L$ $\displaystyle 7.18*10^{7}L$

Correct answer:

$7.18*10^{7}L$ $\displaystyle 7.18*10^{7}L$

Explanation:

$F_{net}=0=F_{gravity}+F_{buoyancy}$ $\displaystyle F_{net}=0=F_{gravity}+F_{buoyancy}$

$0=mg+\rho_{water}*-g*V$ $\displaystyle 0=mg+\rho_{water}*-g*V$

Solving for $\displaystyle V$

$V=\frac{m}{\rho_{water}}$ $\displaystyle V=\frac{m}{\rho_{water}}$

$V=\frac{7.4*10^7 kg}{1.03\frac{kg}{L}}$ $\displaystyle V=\frac{7.4*10^7 kg}{1.03\frac{kg}{L}}$

$V=7.18*10^{7}L$ $\displaystyle V=7.18*10^{7}L$

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Example Question #28 : Fluid Statics

Determine the volume of water that needs to be displaced in order for a 90kg $\displaystyle 90kg$ man to be able to float.

Possible Answers:

180L $\displaystyle 180L$

90mL $\displaystyle 90mL$

90L $\displaystyle 90L$

None of these

45L $\displaystyle 45L$

Correct answer:

90L $\displaystyle 90L$

Explanation:

When floating, the net force on the man is zero. The net force is made up of buoyant force and gravity.

$F_{net}=0=F_{gravity}+F_{buoyant}$ $\displaystyle F_{net}=0=F_{gravity}+F_{buoyant}$

$0=mg-\rho*g*V$ $\displaystyle 0=mg-\rho*g*V$

Solving for volume

$V=\frac{m}{\rho}$ $\displaystyle V=\frac{m}{\rho}$

Plugging in values:

$V=\frac{90}{1}$ $\displaystyle V=\frac{90}{1}$

V=90L $\displaystyle V=90L$

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Example Question #21 : Fluids

A ship has mass $\displaystyle 4.8*10^7kg$. What volume of water would it need to be able to displace in order to be able to float?

$\rho_{seawater}=1.03\frac{kg}{L}$ $\displaystyle \rho_{seawater}=1.03\frac{kg}{L}$

Possible Answers:

$4.11*10^{7}L$ $\displaystyle 4.11*10^{7}L$

None of these

$5.55*10^{7}L$ $\displaystyle 5.55*10^{7}L$

$4.66*10^{7}L$ $\displaystyle 4.66*10^{7}L$

$4.88*10^{7}L$ $\displaystyle 4.88*10^{7}L$

Correct answer:

$4.66*10^{7}L$ $\displaystyle 4.66*10^{7}L$

Explanation:

$F_{net}=0=F_{gravity}+F_{buoyancy}$ $\displaystyle F_{net}=0=F_{gravity}+F_{buoyancy}$

$0=mg+\rho_{water}*-g*V$ $\displaystyle 0=mg+\rho_{water}*-g*V$

Solving for $\displaystyle V$

$V=\frac{m}{\rho_{water}}$ $\displaystyle V=\frac{m}{\rho_{water}}$

$V=\frac{4.8*10^7 kg}{1.03\frac{kg}{L}}$ $\displaystyle V=\frac{4.8*10^7 kg}{1.03\frac{kg}{L}}$

$V=4.66*10^{7}L$ $\displaystyle V=4.66*10^{7}L$

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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 : Fluids

Example Question #21 : Fluid Statics

Example Question #22 : Buoyant Force

Example Question #23 : Fluid Statics

Example Question #21 : Buoyant Force

Example Question #21 : Fluid Statics

Example Question #26 : Fluid Statics

Example Question #541 : Ap Physics 2

Example Question #28 : Fluid Statics

Example Question #21 : Fluids

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