AP Physics 1 Quiz: Fluids And Conservation Laws

20 questions · exam conditions

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Fluids And Conservation LawsQuestion 1 of 20

A liquid flows steadily and is incompressible through a pipe that expands from area $A$ to area $4A$ . The speed in the smaller section is $12\text{ m/s}$ . For steady, incompressible flow, what is the speed in the larger section?

48\text{ m/s}

12\text{ m/s}

6\text{ m/s}

3\text{ m/s}

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AP Physics 1 Quiz

AP Physics 1 Quiz: Fluids And Conservation Laws

Practice Fluids And Conservation Laws in AP Physics 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Fluids And Conservation Laws, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$48\text{ m/s}$
$12\text{ m/s}$
$6\text{ m/s}$
$3\text{ m/s}$ (correct answer)

Explanation: This problem applies the continuity equation to find speed in an expanding pipe section. For incompressible flow, A₁v₁ = A₂v₂ ensures constant volume flow rate. The pipe expands from area A to area 4A, with initial speed 12 m/s. Applying continuity: (A)(12 m/s) = (4A)(v₂), which gives 12A m/s = 4Av₂, so v₂ = 3 m/s. Choice A (48 m/s) incorrectly multiplies instead of dividing, misunderstanding that larger areas require slower speeds. Remember that area and speed are inversely proportional in incompressible flow: when area quadruples, speed becomes one-fourth.

Question 2

A fluid flows steadily and is incompressible through a pipe. At section 1, $A_1=A$ and $v_1=5\text{ m/s}$ . At section 2, the speed is $v_2=1\text{ m/s}$ . For steady, incompressible flow, what is $A_2$ ?

$\tfrac{1}{5}A$
$A$
$5A$ (correct answer)
$25A$

Explanation: This problem requires finding the cross-sectional area when speeds are known at two pipe sections. The continuity equation A₁v₁ = A₂v₂ applies for incompressible flow. Given A₁ = A, v₁ = 5 m/s, and v₂ = 1 m/s, we solve for A₂. Substituting: (A)(5 m/s) = (A₂)(1 m/s), which gives 5A m/s = A₂ m/s, so A₂ = 5A. Choice A (A/5) incorrectly inverts the relationship, assuming area decreases when speed decreases. For incompressible flow, remember that area and speed are inversely related: when speed decreases by a factor of 5, area must increase by the same factor.

Question 3

Water flows steadily and is incompressible through a pipe. At section 1, the radius is $r$ . At section 2, the radius is $2r$ . The speed at section 1 is $v$ . For steady, incompressible flow, what is the speed at section 2?

$\tfrac{1}{2}v$
$\tfrac{1}{4}v$ (correct answer)
$2v$
$4v$

Explanation: This problem involves the continuity equation with circular pipe cross-sections of different radii. For incompressible flow, A₁v₁ = A₂v₂, where area A = πr² for circular pipes. At section 1, radius is r and speed is v; at section 2, radius is 2r. The areas are A₁ = πr² and A₂ = π(2r)² = 4πr². Applying continuity: (πr²)(v) = (4πr²)(v₂), which simplifies to v = 4v₂, giving v₂ = v/4. Choice C (2v) incorrectly assumes speed doubles when radius doubles, ignoring that area depends on radius squared. When radius doubles, area quadruples, so speed becomes one-fourth to maintain constant flow rate.

Question 4

Oil flows steadily and is incompressible through a pipe that narrows from area $3A$ to area $A$ . The speed in the wider section is $2\text{ m/s}$ . For steady, incompressible flow, what is the speed in the narrow section?

$\tfrac{2}{3}\text{ m/s}$
$2\text{ m/s}$
$6\text{ m/s}$ (correct answer)
$3\text{ m/s}$

Explanation: This problem applies the continuity equation for incompressible fluid flow through a narrowing pipe. For steady, incompressible flow, the product of cross-sectional area and speed remains constant: A₁v₁ = A₂v₂. The pipe narrows from area 3A to area A, and the initial speed is 2 m/s. Applying continuity: (3A)(2 m/s) = (A)(v₂), which gives 6A m/s = Av₂, so v₂ = 6 m/s. Choice A (⅔ m/s) incorrectly divides instead of multiplying, misunderstanding the inverse relationship. To solve continuity problems, set up the equation A₁v₁ = A₂v₂ and solve for the unknown quantity.

Question 5

Water flows steadily through a horizontal pipe and can be treated as incompressible. At section 1 the pipe radius is $2r$ and the average speed is $v$ . Farther downstream at section 2 the pipe radius is $r$ . The flow is steady, so the volume flow rate is conserved. What is the average speed of the water at section 2?

$v$
$2v$
$4v$ (correct answer)
$\dfrac{v}{4}$

Explanation: This question assesses the application of the continuity equation in fluid dynamics for incompressible fluids. For incompressible fluids in steady flow, the mass flow rate is conserved, which means the volume flow rate is the same at every cross-section since density is constant. The volume flow rate is given by Q = A v, where A is the cross-sectional area and v is the average speed. Therefore, A1 v1 = A2 v2; here, A1 = π (2r)^2 = 4 π r², A2 = π r², so v2 = (A1 / A2) v1 = 4 v. A common distractor is B, 2v, which might result from incorrectly using the radius ratio instead of the area ratio, since radius halves, but area quarters. To solve similar problems, always remember to use the continuity equation Q1 = Q2 and calculate areas properly from given dimensions.

Question 6

A steady, incompressible stream of water flows through a pipe of area $A$ . The pipe then splits into two identical branches, each of area $\frac{A}{2}$ . What is the speed in each branch compared to the original speed $v$ ?

$\frac{v}{2}$
$v$ (correct answer)
$2v$
$4v$

Explanation: This question explores the continuity equation in branching pipes for incompressible flow, based on mass conservation. In steady flow, the mass flow rate into the split equals the sum out, and with constant density, volumetric rates conserve similarly. When splitting into identical branches, each carries half the flow, but speeds depend on areas. Qualitatively, same area halves would maintain speed if flow halves per branch. Choice C, 2v, could be chosen if someone assumes speed doubles when area halves without considering the split. A key strategy is to calculate total flow rate and divide appropriately among branches, then apply A v per branch.

Question 7

Water flows steadily through a horizontal pipe and is incompressible. At section 1 the cross-sectional area is $2A$ and the speed is $v$ . At section 2 the area is $A$ . For this steady, incompressible flow, what is the speed at section 2?

$v$
$2v$ (correct answer)
$\tfrac{1}{2}v$
$4v$

Explanation: This problem tests the continuity equation for incompressible fluids flowing through pipes with varying cross-sectional areas. For steady, incompressible flow, the volume flow rate must remain constant throughout the pipe, meaning A₁v₁ = A₂v₂. At section 1, we have area 2A and speed v, while at section 2, the area is A. Substituting into the continuity equation: (2A)(v) = (A)(v₂), which simplifies to 2Av = Av₂, giving v₂ = 2v. Choice C (½v) incorrectly assumes speed decreases when area decreases, reversing the relationship. When solving continuity problems, remember that speed and area are inversely proportional for incompressible flow.

Question 8

Water flows steadily and is incompressible through a pipe with two sections. At section 1, $A_1=8\text{ cm}^2$ and $v_1=10\text{ cm/s}$ . At section 2, $v_2=40\text{ cm/s}$ . For steady, incompressible flow, what is $A_2$ ?

$2\text{ cm}^2$ (correct answer)
$32\text{ cm}^2$
$8\text{ cm}^2$
$0.5\text{ cm}^2$

Explanation: This problem tests the continuity equation with numerical values for areas and speeds. For incompressible flow, the volume flow rate A₁v₁ equals A₂v₂ at all pipe sections. Given A₁ = 8 cm², v₁ = 10 cm/s, and v₂ = 40 cm/s, we find A₂. Applying continuity: (8 cm²)(10 cm/s) = (A₂)(40 cm/s), which gives 80 cm³/s = 40A₂ cm/s, so A₂ = 2 cm². Choice B (32 cm²) incorrectly multiplies areas instead of recognizing the inverse relationship with speed. When speed quadruples in incompressible flow, area must become one-fourth to conserve mass flow rate.

Question 9

Incompressible water flows steadily through a pipe. At section 1 the diameter is $D$ and the average speed is $8\text{ m/s}$ . At section 2 the diameter is $2D$ . Assuming steady flow, what is the average speed at section 2?

$16\text{ m/s}$
$8\text{ m/s}$
$4\text{ m/s}$
$2\text{ m/s}$ (correct answer)

Explanation: This question assesses conservation of volume flow rate using pipe diameters. For incompressible steady flow, mass conservation implies A1 v1 = A2 v2. Areas are proportional to diameter squared, so A2 / A1 = (2D/D)^2 = 4. Thus v2 = v1 (A1 / A2) = 8 / 4 = 2 m/s. Distractor A 16 m/s might come from incorrectly doubling the speed instead of quartering it. Always compute the area ratio using (d2/d1)^2 when diameters are given.

Question 10

In steady incompressible flow, a pipe splits into branches X and Y. Branch X has flow rate $2\ \text{L/s}$ and branch Y has $5\ \text{L/s}$ . What is the flow rate in the main pipe before the split?

$3\ \text{L/s}$
$7\ \text{L/s}$ (correct answer)
$10\ \text{L/s}$
$2.5\ \text{L/s}$

Explanation: This question assesses mass conservation in splitting pipes for steady incompressible flow, requiring total flow rate equality. Since the fluid is incompressible, the volume flow rate before the split must equal the sum after, preventing mass buildup. This conservation law applies at junctions, adding branch rates to get the main. Qualitatively, higher branch flows imply a proportionally larger main flow. Choice C, 10 L/s, might be a distractor if someone multiplies instead of adding the branches. For similar problems, always add or subtract flow rates at junctions while remembering total conservation.

Question 11

A steady, incompressible fluid flows through a pipe that splits into two branches. The main pipe carries volumetric flow rate $Q$ . One branch carries $\frac{1}{4}Q$ . What flow rate must the other branch carry?

$\frac{1}{4}Q$
$\frac{3}{4}Q$ (correct answer)
$Q$
$\frac{5}{4}Q$

Explanation: This question tests conservation of mass in branching pipes for incompressible flow, where total flow rate is preserved. In incompressible fluids, the volumetric flow rate into a junction equals the sum of flow rates out, as mass can't accumulate in steady flow. This means the main pipe's flow splits among branches without loss. Qualitatively, if one branch takes a fraction, the other takes the remainder to conserve the total. Choice A, 1/4 Q, is a distractor possibly chosen if someone subtracts incorrectly or confuses addition. A general strategy is to sum the branch flow rates to find the main or vice versa, ensuring conservation.

Question 12

Incompressible water flows steadily through a pipe that splits into two branches. The incoming volume flow rate is $9\text{ L/s}$ . One branch carries $4\text{ L/s}$ . Assuming steady flow, what volume flow rate must the other branch carry?

$5\text{ L/s}$ (correct answer)
$9\text{ L/s}$
$13\text{ L/s}$
$\dfrac{9}{4}\text{ L/s}$

Explanation: This question focuses on conservation of volume flow rate in branching pipes for incompressible fluids. When a pipe splits, the total volume flow rate into the junction equals the sum of flow rates out of the branches. Since the fluid is incompressible and flow is steady, no accumulation occurs, so Qin = Qbranch1 + Qbranch2. Here, 9 L/s = 4 L/s + Q2, so Q2 = 5 L/s. A distractor like C, 13 L/s, could be from adding instead of subtracting. Always apply the continuity principle at junctions by summing the outgoing flows to match incoming.

Question 13

Incompressible water flows steadily through a pipe that narrows from cross-sectional area $5A$ to $A$ . The average speed in the wide section is $0.6\text{ m/s}$ . Using conservation of volume flow rate, what is the speed in the narrow section?

$0.12\text{ m/s}$
$0.6\text{ m/s}$
$3.0\text{ m/s}$ (correct answer)
$5.0\text{ m/s}$

Explanation: This question applies conservation of volume flow rate in a narrowing pipe. Incompressible fluid in steady flow has constant Q = A v. As area decreases, speed increases proportionally. Here, area ratio 5A/A =5, v2=5*0.6=3.0 m/s. Distractor B 0.6 m/s might be chosen if forgetting the area change affects speed. Consistently use the continuity equation to relate speeds and areas in different sections.

Question 14

A steady, incompressible fluid flows through a pipe that splits into two equal-area branches. The main pipe carries volume flow rate $Q$ . Each branch has the same cross-sectional area. Assuming steady flow, what is the volume flow rate in each branch?

$Q$
$\dfrac{Q}{2}$ (correct answer)
$2Q$
$\dfrac{Q}{4}$

Explanation: This question examines flow rate conservation in a splitting pipe with equal branches. For steady incompressible flow, the incoming flow rate equals the sum of branch flow rates. If branches have equal area and are symmetric, the flow splits equally. Thus, each branch carries $\dfrac{Q}{2}$ . Distractor C $2Q$ might be from misunderstanding and doubling instead. Always verify assumptions like equal splitting when branches are identical.

Question 15

A steady, incompressible fluid flows through a pipe. At section 1, $A_1=2A$ and $v_1=v$ . At section 2, $A_2=\dfrac{A}{2}$ . By conservation of volume flow rate, what is $v_2$ ?

$\dfrac{v}{4}$
$v$
$2v$
$4v$ (correct answer)

Explanation: This question tests understanding of continuity for incompressible flow. Conservation of mass implies constant volume flow rate for constant density. Q = A1 v1 = A2 v2. Given A1=2A, A2=A/2, v1=v, v2 = (2A v)/(A/2) = 4v. Distractor A v/4 might come from inverting the area ratio incorrectly. A transferable strategy is to express all areas in terms of a common variable and solve for the unknown speed.

Question 16

Oil flows steadily through a pipe and is incompressible. At a wide section the cross-sectional area is $6A$ and the average speed is $2\text{ m/s}$ . At a narrow section the area is $3A$ . The flow is steady, so volume flow rate is conserved. What is the average speed in the narrow section?

$1\text{ m/s}$
$2\text{ m/s}$
$4\text{ m/s}$ (correct answer)
$12\text{ m/s}$

Explanation: This question evaluates understanding of volume flow rate conservation for incompressible fluids. In steady flow of incompressible fluids, mass conservation leads to constant volume flow rate along the pipe. Q = A v remains constant, so in a narrowing pipe, speed increases inversely with area. Here, Q = 6A * 2 = 12A, v_narrow = 12A / 3A = 4 m/s. Distractor D, 12 m/s, might come from multiplying areas instead of using the ratio correctly. A useful strategy is to calculate the flow rate in one section and then solve for the unknown in the other.

Question 17

A liquid flows steadily and is incompressible through a pipe that narrows from radius $3r$ to radius $r$ . The speed in the wider section is $v$ . For steady, incompressible flow, what is the speed in the narrow section?

$3v$
$9v$ (correct answer)
$\tfrac{1}{3}v$
$\tfrac{1}{9}v$

Explanation: This problem involves the continuity equation for a pipe that narrows from radius 3r to radius r. For incompressible flow, A₁v₁ = A₂v₂, where circular area A = πr². The initial area is A₁ = π(3r)² = 9πr² and final area is A₂ = πr², with initial speed v. Applying continuity: (9πr²)(v) = (πr²)(v₂), which simplifies to 9v = v₂. Choice A (3v) incorrectly uses the radius ratio instead of the area ratio, forgetting that area varies as radius squared. When radius decreases by a factor of 3, area decreases by a factor of 9, so speed must increase ninefold.

Question 18

A fluid flows steadily and is incompressible through two connected sections. At section 1, $A_1=4\text{ cm}^2$ and $v_1=30\text{ cm/s}$ . At section 2, $A_2=6\text{ cm}^2$ . For steady, incompressible flow, what is $v_2$ ?

$20\text{ cm/s}$ (correct answer)
$30\text{ cm/s}$
$45\text{ cm/s}$
$50\text{ cm/s}$

Explanation: This problem involves applying the continuity equation for incompressible flow between two pipe sections. The continuity equation states that A₁v₁ = A₂v₂ for steady, incompressible flow. Given A₁ = 4 cm², v₁ = 30 cm/s, and A₂ = 6 cm², we need to find v₂. Substituting: (4 cm²)(30 cm/s) = (6 cm²)(v₂), which gives 120 cm³/s = 6v₂ cm², so v₂ = 20 cm/s. Choice C (45 cm/s) incorrectly multiplies 30 by 1.5 instead of recognizing the inverse relationship between area and speed. Remember that when area increases, speed must decrease proportionally to maintain constant flow rate.

Question 19

Oil flows steadily and incompressibly through a pipe that widens from area $A$ to $3A$ . If the speed in the narrow section is $6\ \text{m/s}$ , what is the speed in the wide section?

$18\ \text{m/s}$
$2\ \text{m/s}$ (correct answer)
$6\ \text{m/s}$
$9\ \text{m/s}$

Explanation: This question assesses understanding of the continuity equation in incompressible flow, based on mass conservation. For incompressible fluids, the volume flow rate remains constant because density doesn't change, so A v is the same everywhere. When the pipe widens, the fluid slows down to keep the flow rate steady. This ensures the same amount of fluid volume passes through wider and narrower sections in the same time. Choice A, 18 m/s, could be a distractor if someone multiplies instead of dividing the area ratio. A useful strategy for these problems is to identify the ratio of areas and inversely apply it to the velocities while conserving A v.

Question 20

Water flows steadily and incompressibly through a pipe with $A_1=3\ \text{cm}^2$ and $v_1=4\ \text{m/s}$ . If $A_2=6\ \text{cm}^2$ , what is $v_2$ ?

$8\ \text{m/s}$
$4\ \text{m/s}$
$2\ \text{m/s}$ (correct answer)
$\frac{1}{2}\ \text{m/s}$

Explanation: This question examines the continuity equation for incompressible fluids, stemming from the conservation of mass. For incompressible flow, density is uniform, so mass conservation requires the volumetric flow rate to be constant throughout the pipe. When the cross-sectional area increases, the velocity decreases proportionally to maintain this flow rate. This qualitative understanding helps predict that doubling the area halves the speed. Choice A, 8 m/s, might be selected if someone doubles the speed instead of halving it due to confusion with area ratios. To tackle similar questions, consistently use A1 v1 = A2 v2 and verify units for consistency.