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Fluids practice problems AP Physics 1, Unit 8

Written by STEM-Tutor's tutors · Updated October 2026 · 10 questions and 1 free-response question

Density, pressure, buoyancy, the continuity equation and Bernoulli's equation. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.

On this page
  1. Key equations
  2. Warm-up questions
  3. Exam-style questions
  4. Free-response practice
  5. Common mistakes
  6. Keep going

Key equations for fluids

ρ = m / Vdensity
P = P0 + ρghpressure at depth h
Fb = ρVgbuoyant force: the weight of fluid displaced
A1v1 = A2v2continuity, for an incompressible fluid
P1 + ρgy1 + ½ρv12 = P2 + ρgy2 + ½ρv22Bernoulli's equation

All questions use g = 9.8 m/s², as on the AP Physics 1 equation sheet. A calculator is allowed, as in the exam. The full sheet is in our AP Physics 1 equation sheet.

Warm-up questions

Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.

Q1Warm-up
A 2.0 kg block has a volume of 0.0025 m³. What is its density?
kg/m³

Hint

Density is mass per unit volume.

Worked solution
  1. ρ = m / V = 2.0 / 0.0025.
  2. ρ = 800 kg/m³, so it floats in water.
Q2Warm-up
What is the gauge pressure 5.0 m below the surface of water (density 1000 kg/m³)?
Pa

Hint

Gauge pressure leaves out atmospheric pressure: it is just ρgh.

Worked solution
  1. Pgauge = ρgh = 1000 × 9.8 × 5.0.
  2. Pgauge = 49,000 Pa, about half an atmosphere.
Q3Warm-up
A block floats at rest on water. How does the buoyant force on it compare with its weight?

Hint

The block is in equilibrium.

Worked solution
  1. The block is at rest, so the net force is zero.
  2. The only forces are its weight (down) and the buoyant force (up), so they are equal.
  3. The answer is B. How deep it floats adjusts itself until this is true.
Q4Warm-up
Water flows at 2.0 m/s through a pipe of cross-sectional area 0.040 m². The pipe narrows to 0.010 m². How fast does the water flow in the narrow part?
m/s

Hint

The volume flow rate Av is the same all along the pipe.

Worked solution
  1. A1v1 = A2v2.
  2. 0.040 × 2.0 = 0.010 × v2.
  3. v2 = 8.0 m/s: a quarter of the area means four times the speed.

Exam-style questions

Questions at the level of the AP Physics 1 exam. The ones marked Challenge combine more than one idea.

Q5Exam-style
An object of volume 0.0020 m³ is held completely under water (density 1000 kg/m³). What is the buoyant force on it?
N

Hint

The buoyant force equals the weight of the water displaced.

Worked solution
  1. Fb = ρwaterVg = 1000 × 0.0020 × 9.8.
  2. Fb = 19.6 N, whatever the object is made of.
Q6Exam-style
A block of wood with density 600 kg/m³ floats in water (1000 kg/m³). What fraction of its volume is under the water? Give your answer as a decimal.

Hint

Set the buoyant force equal to the weight, and write both in terms of densities and volumes.

Worked solution
  1. Floating: ρwaterVunderg = ρwoodVtotalg.
  2. Vunder / Vtotal = ρwood / ρwater = 600 / 1000.
  3. Fraction under water = 0.60.
Q7Exam-style
Water flows through a horizontal pipe. Its speed is 2.0 m/s at point 1 and 6.0 m/s at point 2. By how much is the pressure at point 1 greater than at point 2?
Pa

Hint

In a horizontal pipe the ρgy terms cancel in Bernoulli's equation.

Worked solution
  1. P1 + ½ρv12 = P2 + ½ρv22.
  2. P1 − P2 = ½ × 1000 × (6.02 − 2.02) = 500 × 32.
  3. P1 − P2 = 16,000 Pa. Faster flow, lower pressure.
Q8Exam-style
Water leaks from a small hole in the side of a large open tank, 1.8 m below the water surface. How fast does the water leave the hole?
m/s

Hint

Apply Bernoulli's equation between the surface and the hole. Both are at atmospheric pressure, and the surface barely moves.

Worked solution
  1. The pressure terms cancel and the surface speed is about zero, so ρgh = ½ρv2.
  2. v = √(2gh) = √(2 × 9.8 × 1.8) = √35.28.
  3. v = 5.94 m/s, the same speed as an object falling 1.8 m.
Q9Exam-style
A ship sails from a river (fresh water) into the sea (denser salt water). What happens?

Hint

The buoyant force must still equal the ship's weight. How much denser water must it displace for that?

Worked solution
  1. Floating, the buoyant force equals the weight in both cases, so the buoyant force does not change.
  2. Fb = ρVg: with larger ρ, a smaller volume V is needed.
  3. The ship displaces less water, so it rises: B.
Q10Challenge
A rock weighs 30 N in air and 20 N when hung fully under water (1000 kg/m³). What is the density of the rock?
kg/m³

Hint

The loss in weight is the buoyant force, which tells you the rock's volume.

Worked solution
  1. Buoyant force = 30 − 20 = 10 N = ρwaterVg.
  2. The rock's weight is ρrockVg = 30 N.
  3. Divide: ρrock / ρwater = 30 / 10 = 3.0.
  4. ρrock = 3000 kg/m³.

Free-response practice

One original question in the style of the exam's free-response section. Answer it in full, with working, before you open the scoring guide.

Translation between representations style · 5 points · about 20 minutes

A block hangs from a spring scale. The scale reads W in air. The block is then lowered slowly into a beaker of water until it is completely under water, and then lowered further without touching the bottom.

depth of bottom facescale readingWW′h (fully under)
Scale reading against depth: it falls steadily while the block goes under, then stays constant.
  1. (a) Sketch, or describe, a graph of the scale reading against the depth of the block's bottom face below the water surface.
  2. (b) Derive an expression for the density of the block, in terms of W, the fully submerged reading W′, and the density of water ρw.
  3. (c) W = 12 N and W′ = 7.5 N. Calculate the density of the block.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

The reading starts at W. As the block goes under, the buoyant force grows in proportion to the volume under water, so the reading falls in a straight line. Once the block is fully under water, the displaced volume stops changing, so the reading stays constant at a lower value W′.

Part (b)

Buoyant force when fully submerged: W − W′ = ρwVg. Weight: W = ρVg. Dividing gives ρ = ρwW / (W − W′).

Part (c)

ρ = 1000 × 12 / 4.5 = 2670 kg/m³.

Your mark: 0 of 5

Common mistakes in fluids

Keep going

Revise the ideas behind these questions:

Then try real exam questions: AP Physics 1 past papers, with official scoring guidelines. · All AP Physics 1 units