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💧 Unit 8 · Fluids 🗂 Flashcards 🗺 Cheat Sheet Essentials 🎙 Podcast 🎨 Visual Review 📝 MC Practice FRQ Practice

AP Physics 1 Unit 8 FRQ Practice

A buoyancy free-response question with a model answer and AP-style scoring guide. Use g = 10 m/s² and ρ_water = 1000 kg/m³.

← Back to Unit 8 hub

Question: Floating Wooden Block

A wooden block has mass 0.50 kg and volume 1.0 × 10⁻³ m³ (which is 1 liter, since 1 L = 10⁻³ m³). The block is placed gently on the surface of a tub of water and allowed to come to rest, where it floats.

Use g = 10 m/s² and ρ_water = 1000 kg/m³ throughout this problem.

A
What fraction of the block's volume is submerged when it floats? Show your reasoning.

Method 1 — Use density ratio:

The block's density is ρ_block = m/V = 0.50 / (1.0 × 10⁻³) = 500 kg/m³.+1 point

For a floating object, the fraction submerged equals the density ratio:

fraction = ρ_block / ρ_water = 500 / 1000 = 0.50 (50%)+1 point

Method 2 — Use Newton's 2nd law (full justification):

For a floating object at rest: F_b = mg. So ρ_water · V_submerged · g = mg.

V_submerged = m / ρ_water = 0.50 / 1000 = 5.0 × 10⁻⁴ m³

Fraction submerged = V_submerged / V_total = 5.0 × 10⁻⁴ / 1.0 × 10⁻³ = 0.50 (50%)

Either method earns the same points: density ratio set up correctly (+1), correct fraction with reasoning (+1).

B
Calculate the buoyant force on the block while it is floating. Show your work and confirm the result is consistent with the block's weight.

Buoyant force formula: F_b = ρ_water · V_displaced · g+1 point

V_displaced = V_submerged = 5.0 × 10⁻⁴ m³ (from part A)

F_b = (1000)(5.0 × 10⁻⁴)(10) = 5.0 N+1 point

Consistency check: The block's weight is W = mg = (0.50)(10) = 5.0 N. Since the block is floating (in equilibrium), the buoyant force must equal the weight. F_b = mg = 5.0 N ✓+1 point

This confirms our answer to part A: the block displaces just enough water to balance its weight.

C
A student claims: "If we pushed the block fully under water (so it was completely submerged), the buoyant force on it would stay the same." Is the student correct? Explain. If the student is incorrect, calculate the new buoyant force AND state the magnitude and direction of the net force on the block when fully submerged.

The student is INCORRECT.+1 point

Explanation: The buoyant force depends on the volume of fluid DISPLACED (F_b = ρ_water · V_displaced · g). When the block was floating, only half its volume was submerged, so it displaced 5.0 × 10⁻⁴ m³. When fully submerged, it displaces its ENTIRE volume of 1.0 × 10⁻³ m³ — twice as much fluid. Therefore the buoyant force DOUBLES.+1 point

New buoyant force (fully submerged):

F_b = ρ_water · V_total · g = (1000)(1.0 × 10⁻³)(10) = 10 N+1 point

Net force on the fully submerged block:

F_net = F_b − mg = 10 − 5.0 = 5.0 N, directed UPWARD+1 point

The block would accelerate upward (and pop back to the surface) if released, because the buoyant force exceeds gravity when fully submerged. The block can only stay submerged if something holds it down.

FRQ scoring tips

Always show your work. Even if you get the wrong final number, you can earn partial credit if you set up the right equation. Write F_b = ρVg before you plug in numbers.

Use the FLUID's density in F_b = ρVg. The most common Unit 8 error is using the OBJECT's density. The fluid is what's doing the pushing.

Distinguish submerged volume from total volume. For a floating object, V_displaced is the SUBMERGED part. For a fully submerged object, V_displaced is the entire object.

For "is the student correct?" questions, clearly state YES or NO first, then justify. The justification is worth more points than the yes/no.

Include units throughout. Reading the final value as 5 instead of 5.0 N can lose a point.