The problems are on the fluid mechanics worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Fluids obey a few linked rules: density is mass per volume, pressure rises with depth as P = P₀ + ρgh, buoyant force equals the weight of fluid displaced, continuity says Av is constant in a pipe, and Bernoulli shows that faster flow means lower pressure. These worked answers correct the 20 problems of the AP Physics 1 worksheet on fluid mechanics, from density of a metal block to loaded barge: each solution shows the formula, the substitution and the final answer with its unit.
Check each answer against your own working. Marks in AP-style grading go to the correct law, a clean substitution and a final value with the right unit, so compare your method as well as your number.
Level: AP Physics 1 (algebra-based) · Curriculum: AP P1 Unit 8
Problem 1 Density of a metal block
1. kg/m³.2. ρ = m/V = 0.54 / 2.0 × 10⁻⁴ = 2.7 × 10³ kg/m³.
3. 1 g/cm³ = 1000 kg/m³, so the density is 2.7 g/cm³.
Problem 2 Drawing pin pressure
1. P = F/A = 20 / 5.0 × 10⁻⁷ = 4.0 × 10⁷ Pa.2. P = 20 / 1.0 × 10⁻⁴ = 2.0 × 10⁵ Pa.
3. The same force acts over a tiny area at the tip, so the pressure is 200 times larger and exceeds what the board can withstand.
Problem 3 Tyre pressure
1. 250 kPa = 2.5 × 10⁵ Pa.2. P = P₀ + Pgauge = 1.01 × 10⁵ + 2.20 × 10⁵ = 3.21 × 10⁵ Pa.
3. It reads gauge pressure: the pressure above atmospheric pressure.
Problem 4 Pressure in a water tank
1. Pgauge = ρgh = 1000 × 9.81 × 5.0 = 4.9 × 10⁴ Pa.2. P = 1.01 × 10⁵ + 0.49 × 10⁵ = 1.50 × 10⁵ Pa.
Problem 5 Aquarium base
1. P = ρgh = 1000 × 9.81 × 0.40 = 3.9 × 10³ Pa.2. A = 0.60 × 0.30 = 0.18 m², F = PA = 3924 × 0.18 = 7.1 × 10² N.
3. W = ρVg = 1000 × (0.18 × 0.40) × 9.81 = 706 N, equal to the force on the base because the sides are vertical.
Problem 6 Fully submerged stone
1. A fully submerged object displaces its own volume: 3.0 × 10⁻⁴ m³.2. FB = ρVg = 1000 × 3.0 × 10⁻⁴ × 9.81 = 2.9 N.
Problem 7 Flow rate in a pipe
1. Q = Av = 4.0 × 10⁻⁴ × 1.5 = 6.0 × 10⁻⁴ m³/s.2. V = Qt = 6.0 × 10⁻⁴ × 60 = 0.036 m³ = 36 L.
Problem 8 Pressure against depth graph
1. Gradient = (160 − 101) / 6 = 9.8 kPa/m.2. The gradient is ρg, so ρ = 9.8 × 10³ / 9.81 = 1.0 × 10³ kg/m³, which is water.
3. P = 101 + 9.8 × 8 = 1.8 × 10² kPa (179 kPa).
Problem 9 Float or sink
1. 500 kg/m³ floats, 1000 kg/m³ is neutrally buoyant, 1200 kg/m³ sinks.2. Fraction = ρobject / ρfluid = 500/1000 = 0.50.
3. Its weight equals the buoyant force when fully submerged, wherever it is, so the net force is zero at any depth.
Problem 10 Iceberg
1. Fraction = 917 / 1025 = 0.89.2. Above the surface: 1 − 0.89 = 0.11, about 11 %.
Problem 11 Weight in air and in water
1. FB = 4.90 − 4.20 = 0.70 N.2. V = FB/(ρg) = 0.70 / (1000 × 9.81) = 7.1 × 10⁻⁵ m³.
3. m = 4.90 / 9.81 = 0.500 kg, so ρ = 0.500 / 7.14 × 10⁻⁵ = 7.0 × 10³ kg/m³.
Problem 12 Garden hose nozzle
1. A1/A2 = (1.8/0.60)² = 9, so v₂ = 9 × 1.2 = 11 m/s (10.8 m/s).2. Q = π(0.0090)² × 1.2 = 3.05 × 10⁻⁴ m³/s = 18 L/min.
Problem 13 Hydraulic car lift
1. P = 150 / 4.0 × 10⁻³ = 3.75 × 10⁴ Pa.2. F₂ = PA₂ = 3.75 × 10⁴ × 0.16 = 6.0 × 10³ N.
3. Equal volumes: 4.0 × 10⁻³ × 0.80 = 0.16 × d, so d = 0.020 m. The work in (120 J) equals the work out.
Problem 14 Pressure drop in a fast section
1. P₂ = P₁ + ½ρ(v₁² − v₂²) = 2.40 × 10⁵ + 500 × (1.0 − 25) = 2.28 × 10⁵ Pa.2. The drop is 1.2 × 10⁴ Pa.
Problem 15 Roof in a strong wind
1. ΔP = ½ρv² = 0.5 × 1.2 × 20² = 240 Pa.2. F = ΔP × A = 240 × 8.0 = 1.9 × 10³ N upward.
3. W = 150 × 9.81 = 1.47 × 10³ N. Since 1920 N is more than 1472 N, the net force is about 4.5 × 10² N upward and the roof lifts.
Problem 16 Jet from a tank
1. P and the tank-top speed are negligible compared with the exit: v = √(2gh) = √(2 × 9.81 × 3.2) = 7.9 m/s.2. t = √(2 × 1.0 / 9.81) = 0.45 s.
3. x = vt = 7.92 × 0.4515 = 3.6 m.
Problem 17 Venturi flow meter
1. A₁v₁ = A₂v₂ gives v₂ = 4v₁.2. ΔP = ½ρ(v₂² − v₁²) = 500 × 15 v₁² = 7500 v₁², so v₁² = 2.0 and v₁ = 1.4 m/s.
3. Q = A₁v₁ = 2.0 × 10⁻³ × 1.414 = 2.8 × 10⁻³ m³/s.
Problem 18 Ball held under water
1. FB = 1000 × 2.0 × 10⁻³ × 9.81 = 19.6 N.2. Forces: B up, weight and tension down. W = 0.50 × 9.81 = 4.9 N, so T = 19.62 − 4.91 = 14.7 N.
3. a = (B − W)/m = 14.7 / 0.50 = 29 m/s² upward.
Problem 19 Oil in a U-tube
1. Points at the same level in the same connected water: the oil-water interface in the right arm and the point at that level in the left arm.2. ρw g h = ρo g (0.120) so h = 800 × 0.120 / 1000 = 0.096 m.
3. 12.0 − 9.6 = 2.4 cm.
Problem 20 Loaded barge
1. V = m/ρ = 3.0 m³; area = 12 m², so depth = 0.25 m.2. Total 5400 kg: V = 5.4 m³, depth = 5.4 / 12 = 0.45 m.
3. Maximum displaced mass = 1000 × 12 × 0.80 = 9600 kg, so the load is 9600 − 3000 = 6.6 × 10³ kg.
4. Depth = 5400 / (1025 × 12) = 0.44 m, slightly higher in the water.
