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Unit 8 / Fluids

Unit 8: worked solutions

Compare the model and reasoning at each step. A matching final number is useful; a reason for every step is what lets you solve the next problem.

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Question 1 · AP topic 8.1

1. Topic 8.1

  1. Read the slope 0.90 g/mL as liquid density, then convert to 900 kg/m³.

    Why this step: One g/mL is 1000 kg/m³; slope measures added liquid mass per added volume.

  2. Identify the 45 g intercept as empty container mass.

    Why this step: At V = 0 the fitted total mass is the container alone.

  3. At 60 mL compute m_total = 0.90(60) + 45 = 99 g.

    Why this step: The balance records both liquid and container. Dividing 99 g by liquid volume would incorrectly include the container in density.

ρ = 900 kg/m³; intercept = 45 g container mass; total mass at 60 mL = 99 g.

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Question 2 · AP topic 8.2

2. Topic 8.2

  1. Subtract inside atmospheric pressure from outside pressure to get ΔP = ρgh = (1000)(9.8)(1.5) = 14,700 Pa.

    Why this step: The same atmospheric reference appears on both sides and cancels.

  2. Use net force magnitude ΔPA = (14,700)(0.020) = 294 N.

    Why this step: The pressure difference acts over the stated area; forces on opposite sides point oppositely.

  3. The force points from water toward the air side, normal to the window.

    Why this step: Pressure is scalar, but the window orientation and which side has greater pressure determine the force direction. Absolute water pressure alone would omit the air force.

Net force = 294 N from the water toward the air side, normal to the window.

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Question 3 · AP topic 8.3

3. Topic 8.3

  1. Both objects displace V, so both have F_B = ρVg.

    Why this step: They share external volume and are fully submerged in the same fluid.

  2. Apply Newton’s law: a_A = ρVg/m − g, and a_B = ρVg/(2m) − g.

    Why this step: Mass changes both weight and inertia, so buoyancy alone cannot be treated as the net force.

  3. Substitute ρVg/m = a_A + g to obtain a_B = (a_A − g)/2.

    Why this step: The extra gravity term shows why B’s acceleration is not generally half of A’s. The accelerations can even have opposite signs.

a_A = ρVg/m − g; a_B = ρVg/(2m) − g = (a_A − g)/2.

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Question 4 · AP topic 8.4

4. Topic 8.4

  1. The ideal projectile-plus-flow model gives R² = 4Hh, so expected slope is 4(0.20) = 0.80 m.

    Why this step: The jet speed comes from depth and its fall time comes from fixed outlet height.

  2. At h = 0.30 m, calculate R_ideal = √(0.80×0.30) = 0.490 m and R_fit = √(0.72×0.30) = 0.465 m.

    Why this step: The graph plots range squared, so take square roots to recover actual ranges.

  3. The slope is 10% below ideal; the range is 1 − √(0.72/0.80) = 5.1% below ideal.

    Why this step: A squared plotted quantity has a different fractional change from the quantity itself. Measurement uncertainty and possible losses are needed before naming a cause.

Expected slope = 0.80 m; ideal R = 0.490 m; fitted R = 0.465 m, about 5.1% lower.

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Question 5 · AP topic 8.1

5. Topic 8.1

  1. Use density ρ = m/V.

    Why this step: Density compares mass with the volume containing it.

  2. For the same uniform liquid at fixed conditions, tripling V triples m.

    Why this step: The composition and density remain the same in the stated comparison.

  3. Their ratios m/V are equal.

    Why this step: Increasing sample size changes total mass, not the material density.

are equal.

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Question 6 · AP topic 8.2

6. Topic 8.2

  1. Pressure is force magnitude per area and has no direction.

    Why this step: The ratio F_perpendicular/A uses the normal force magnitude and surface area; it gives a pressure value without assigning a direction.

  2. For a resting fluid, the pressure force acts normal to each surface.

    Why this step: The surface orientation supplies the direction of that force.

  3. Choose the statement that separates scalar pressure from directed force.

    Why this step: A scalar can set the magnitude of a vector force without becoming a vector itself.

Each pressure force has a direction normal to its surface, but pressure itself has no direction.

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Question 7 · AP topic 8.3

7. Topic 8.3

  1. Draw upward buoyancy and downward weight.

    Why this step: Both forces act on the released submerged block.

  2. Use F_B − mg = ma with upward positive.

    Why this step: Negligible other forces makes this the full vertical balance.

  3. Upward acceleration requires F_B > mg.

    Why this step: A positive net force changes velocity upward; equality would give zero acceleration.

Buoyancy exceeds weight if other forces are negligible.

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Question 8 · AP topic 8.4

8. Topic 8.4

  1. Use circular area A = πd²/4.

    Why this step: The comparison specifies a diameter, while continuity uses area.

  2. Halving d reduces area to A/4.

    Why this step: Area depends on the square of diameter.

  3. With Av constant, the speed quadruples.

    Why this step: One-quarter the area requires four times the speed to keep the same volume flow rate.

quadruples.

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Question 9 · AP topic 8.1

9. Topic 8.1

  1. Read the graph axes: mass vertically and volume horizontally.

    Why this step: Slope means change in the vertical quantity divided by change in the horizontal quantity.

  2. Form slope = Δm/ΔV.

    Why this step: This has mass-per-volume units.

  3. Identify the slope as density ρ.

    Why this step: The relationship m = ρV predicts this slope for a uniform liquid.

the density.

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Question 10 · AP topic 8.2

10. Topic 8.2

  1. Use P_absolute = P_atmosphere + P_gauge.

    Why this step: Gauge pressure is measured relative to the atmosphere in this open-pool case.

  2. At the surface, insert P_gauge = 0.

    Why this step: No fluid depth has been added below the chosen reference level.

  3. Absolute pressure equals atmospheric pressure.

    Why this step: Subtracting the reference to make a gauge reading does not remove the actual atmosphere.

the atmospheric pressure.

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Question 11 · AP topic 8.3

11. Topic 8.3

  1. Write F_B = ρ_fluid V_displaced g.

    Why this step: Buoyancy equals the weight of displaced fluid in this hydrostatic model.

  2. Keep density and displaced volume fixed while lowering the rigid block.

    Why this step: The incompressible fluid and fully submerged rigid object do not change these quantities.

  3. Buoyancy stays unchanged.

    Why this step: The top and bottom pressures both increase by the same amount, leaving their difference fixed.

stays unchanged.

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Question 12 · AP topic 8.4

12. Topic 8.4

  1. Write Bernoulli with pressure, kinetic, and gravitational terms.

    Why this step: Energy can shift among these contributions along the ideal streamline.

  2. Set y₁ = y₂ for the two compared locations.

    Why this step: Their gravitational terms ρgy are then equal.

  3. Cancel the difference in gravitational terms only.

    Why this step: Pressure and kinetic terms can still differ and must remain unless further conditions are given.

The difference in gravitational terms.

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Question 13 · AP topic 8.1

13. Topic 8.1

  1. Incompressible means a fixed amount keeps its volume and density.

    Why this step: This allows the same density to be used throughout the ideal-flow calculation.

  2. Nonviscous means internal frictional dissipation is neglected.

    Why this step: This is the second assumption needed for the ideal fluid model here.

  3. Choose incompressible and nonviscous.

    Why this step: Fixed shape is a solid-like feature and is not part of the ideal-fluid definition.

Incompressible and nonviscous.

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Question 14 · AP topic 8.2

14. Topic 8.2

  1. Start with P_gauge = ρgh.

    Why this step: Hydrostatic balance gives the pressure difference below the surface.

  2. Hold ρ and g constant.

    Why this step: The question specifies one uniform resting fluid.

  3. The slope versus h is ρg, in Pa/m.

    Why this step: Container bottom area cancels in the force-balance derivation.

ρg

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Question 15 · AP topic 8.3

15. Topic 8.3

  1. At free floating equilibrium, set ρ_fluid V_sub g = ρ_object V_total g.

    Why this step: Buoyancy and weight balance when no other vertical forces act.

  2. Divide to get V_sub/V_total = ρ_object/ρ_fluid.

    Why this step: The volume fraction is determined by the density ratio.

  3. Calculate 600/1000 = 0.60.

    Why this step: Sixty percent of the volume is submerged; height fraction need not be the same for an irregular shape.

0.60

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Question 16 · AP topic 8.4

16. Topic 8.4

  1. Use v_exit = √(2gh).

    Why this step: Both the open surface and small outlet have atmospheric pressure and surface speed is negligible.

  2. Replace h by 4h and take the ratio √(4h/h).

    Why this step: The square-root dependence sets the factor of change.

  3. The exit speed doubles.

    Why this step: The depth is four times larger, but speed depends on its square root.

a factor of 2.

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Question 17 · AP topic 8.2

17. Pressure below two liquid layers

  1. Each static liquid layer contributes ρgh. Atmospheric pressure cancels in the gauge value.

    Why this step: Pressure gains accumulate through successive static liquid layers; gauge pressure subtracts the shared atmospheric surface pressure.

  2. P_gauge = (850)(9.81)(0.350) + (1000)(9.81)(1.20) = 14690.5 Pa.

    Why this step: Add the oil and water density-height contributions, using each layer’s own density.

14.7 kPa gauge.

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Question 18 · AP topic 8.3

18. Buoyancy · question 1

  1. For floating equilibrium, ρ_waterV_sub g = mg, so V_sub = 50/1000 = 0.050 m³.

    Why this step: Buoyancy balances weight, not the object's total volume times water density.

  2. Divide by the horizontal face area 2² = 4 m² to get depth 0.0125 m = 12.5 mm.

    Why this step: Only a small fraction of this large, low-density cube is submerged.

12.5 mm.

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Question 19 · AP topic 8.4

19. Bernoulli’s Equation · question 1

  1. Use A₁v₁ = A₂v₂ for steady incompressible filled-pipe flow.

    Why this step: The same volume per second crosses both sections.

  2. Since area scales as diameter squared, v₂ = 4(1/1.5)² = 1.7778 m/s.

    Why this step: Using the unsquared diameter ratio would overestimate the outlet speed.

1.78 m/s.

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Question 20 · AP topic 8.2

20. A water-and-alcohol U-tube

  1. At the interface level, the pressure from the alcohol arm equals the pressure from the water arm.

    Why this step: The connected water is static, so pressure must agree at the same horizontal interface level in both arms.

  2. P_atm + ρ_alcohol g h = P_atm + ρ_water g(0.0800 m).

    Why this step: Both free surfaces share atmospheric pressure, leaving each column’s density times depth to determine the interface pressure.

  3. h = (1000/790)(0.0800) = 0.101266 m.

    Why this step: Cancel atmospheric pressure and g, then solve for the taller alcohol column needed with its lower density.

Alcohol-column height = 10.1 cm.

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Question 21 · AP topic 8.3

21. Buoyancy · question 2

  1. On the boat, the anchor's added weight requires displaced volume m_anchor/ρ_water.

    Why this step: The boat supports the anchor through buoyancy.

  2. On the bottom, the dense anchor displaces only its physical volume m_anchor/ρ_anchor, which is smaller.

    Why this step: Less water displacement means a lower pond level; this assumes the sunk anchor is no longer pulling down on the boat.

The water level falls.

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Question 22 · AP topic 8.4

22. Bernoulli’s Equation · question 2

  1. Write A₂/A₁ = v₁/v₂ = 3/12 = 1/4.

    Why this step: Continuity links areas and mean speeds inversely.

  2. Check that the final section has one quarter of the initial area.

    Why this step: Its speed is four times the initial speed while volume flow rate stays constant.

1:4.

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Question 23 · AP topic 8.3

23. Buoyancy · question 3

  1. Find full volume π(0.250)²(0.100) ≈ 0.019635 m³.

    Why this step: All given dimensions must use metres.

  2. Even completely submerged it displaces only about 19.635 kg of water, much less than its 100 kg mass.

    Why this step: The maximum buoyant force is smaller than its weight; this is sinking rather than a half-submerged floating equilibrium.

Yes; the cylinder sinks, so the hole becomes submerged.

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Question 24 · AP topic 8.4

24. Bernoulli’s Equation · question 3

  1. Use P + ½ρv² = constant along the horizontal ideal streamline.

    Why this step: The elevation term is unchanged and no energy is lost.

  2. If v increases, the kinetic-energy-density term increases, so P must decrease.

    Why this step: The conclusion depends on the stated ideal assumptions, not on all possible real pipe systems.

Static pressure decreases.

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Question 25 · AP topic 8.3

25. Buoyancy · question 4

  1. For floating ice, V_sub/V_total = ρ_ice/ρ_water = 0.92.

    Why this step: Equating buoyancy and weight cancels g.

  2. For the upright cube, depth = 0.92(5 mm) = 4.6 mm.

    Why this step: The remaining 0.4 mm sits above the water surface.

4.6 mm.

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Question 26 · AP topic 8.4

26. Bernoulli’s Equation · question 4

  1. Read the widths as decreasing from A to D.

    Why this step: The same flow rate passes every filled section.

  2. Apply v = Q/A: smallest area D has largest speed, followed by C, B and A.

    Why this step: The rank labels refer to greatest through least, not the alphabetical pipe labels.

Speeds: D > C > B > A; targets I–IV match that order.

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Question 27 · AP topic 8.3

27. Buoyancy · question 5

  1. Use submerged volume = (1 m)(0.2 m)(0.02 m) = 0.004 m³.

    Why this step: The long axis is horizontal and the submerged depth is smaller than the 0.2 m side.

  2. Floating equilibrium gives m = ρ_waterV_sub = (1000)(0.004) = 4 kg.

    Why this step: The full box volume is not the displaced volume.

4 kg.

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Question 28 · AP topic 8.4

28. Bernoulli’s Equation · question 5

  1. From continuity, speed rises as area decreases from A to D.

    Why this step: The horizontal lossless-flow conditions permit Bernoulli comparison.

  2. Use P = constant − ½ρv², so the widest, slowest part has the highest pressure.

    Why this step: Pressure and speed rank oppositely only under the assumptions supplied here.

Pressures: A > B > C > D; targets I–IV match that order.

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Question 29 · AP topic 8.3

29. Work it out

  1. The diagram shows a downward tether. V = (4/3)πr³ = (4/3)π(0.15 m)³ = 0.0141372 m³.

    Why this step: The whole sphere is submerged, so its entire geometric volume displaces water; the lower tether can pull only downward.

  2. Buoyancy is upward: F_B = ρgV = 138.686 N. Weight is downward: mg = 98.1 N.

    Why this step: Archimedes’ force equals displaced-water weight and points up, opposite the sphere’s own weight.

  3. Before cutting, equilibrium gives F_B − mg − T = 0, so T = 40.5856 N.

    Why this step: At rest, the downward tether supplies the extra downward force needed to balance buoyancy greater than weight.

  4. Immediately after cutting, with upward positive, ma = F_B − mg, so a = 40.5856/10 = 4.05856 m/s² upward.

    Why this step: Cutting the tether removes T but leaves buoyancy and weight; their upward difference accelerates the sphere in the stated no-drag model.

T = 40.6 N downward on the ball; initial acceleration = 4.06 m/s² upward.

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Question 30 · AP topic 8.4

30. Work it out

  1. The source image states 40 cm, so Δz = 0.40 m. Continuity gives v₂ = v₁(d₁/d₂)² = 0.09(8.0/3.4)² = 0.498270 m/s.

    Why this step: The rise is 40 cm as printed; incompressible continuity makes speed grow by the inverse cross-sectional area ratio.

  2. Bernoulli gives P₂ = P₁ + (ρ/2)(v₁² − v₂²) − ρgΔz.

    Why this step: Along the lossless streamline, increased speed and height both take energy from the pressure term.

  3. Convert P₁ = 4.6(101325) = 466095 Pa. The speed term is −120.086 Pa and the elevation term is −3924 Pa.

    Why this step: Convert atmospheres to pascals before combining pressure with the kinetic and elevation terms.

  4. Thus P₂ = 462050.914 Pa = 4.560 atm. If the intended rise were 40 m, the alternative pressure would be 73574.914 Pa = 0.7261 atm.

    Why this step: Add the two pressure changes for the stated 0.40 m rise; the 40 m alternative illustrates why that source dimension matters.

For the written 40 cm rise: v₂ = 0.498 m/s and P₂ = 4.62 × 10⁵ Pa = 4.56 atm.

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Explain and apply: worked solutions

Give credit for equivalent correct reasoning. Use the rubric to find the next step to improve, even when your final number matches.

Explain shape using particle interactions

A solid sample, a liquid sample, and a dilute gas sample of the same substance are considered under suitable conditions for each phase. Each is moved to a differently shaped container. Separately, a 0.180 kg sample of ideal incompressible liquid has volume 200 mL before it is poured from a narrow cylinder into a wide bowl. Ignore evaporation and spilling.

Part a

Solid particles interact so that neighbors remain near particular relative positions while vibrating; the sample ordinarily keeps its own shape and nearly fixed volume. Liquid particles remain close and interact but can move past one another and change neighbors; the sample takes a container’s shape while retaining nearly fixed volume. In a dilute gas, particles are widely separated and interactions between encounters are less influential, so motion spreads the sample through the available volume and it has no fixed shape or fixed sample volume. Liquids and gases are fluids because neither maintains a fixed shape. Individual particle size need not change. The explanation does not require all particles in one phase always to move faster than those in another.

  1. Link the solid response to neighbors remaining near fixed relative positions.
  2. Link liquid flow to close particles changing neighbors.
  3. Link gas expansion to widely separated particles moving through available space.
  4. Use the no-fixed-shape definition to classify the two fluid phases.
Scoring criteria
  • 1 point: explains solid behavior using interactions and relative positions, allowing vibration.
  • 1 point: explains liquid shape change with close particles that can rearrange.
  • 1 point: explains gas shape/volume response with dispersed moving particles.
  • 1 point: classifies liquid and gas as fluids and states lack of fixed shape.

Part b

200 mL = 2.00 × 10⁻⁴ m³. Before pouring, ρ = 0.180/(2.00 × 10⁻⁴) = 900 kg/m³. After pouring the mass remains 0.180 kg and the volume remains 200 mL, so ρ remains 900 kg/m³. Particles rearrange to a new container shape without changing the amount or modeled volume.

  1. Convert milliliters to cubic meters.
  2. Use ρ = m/V.
  3. Apply no spilling and incompressibility separately to conserve mass and volume.
Scoring criteria
  • 1 point: correct conversion and 900 kg/m³.
  • 1 point: after-pouring mass 0.180 kg, volume 200 mL and unchanged density with an explanation.

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Pressure changes without compressing the ideal sample

A sealed 300 mL sample of ideal incompressible liquid has mass 0.240 kg. Its pressure is changed from 100 kPa to 250 kPa at fixed temperature. A separate small vertical plate has area 0.0060 m², approximately uniform liquid pressure 130 kPa on its left face and air pressure 100 kPa on its right face. The two faces are at the same height; treat all stated pressures as absolute.

Part a

At both pressures V = 300 mL = 3.00 × 10⁻⁴ m³ and ρ = 0.240/(3.00 × 10⁻⁴) = 800 kg/m³. The volume graph is horizontal at 300 mL, and the density graph is horizontal at 800 kg/m³, with pressure in kPa on the horizontal axis from 100 to 250. Incompressibility holds volume and density fixed for the given amount; pressure itself changes. Real fluid compression can require a different model.

  1. Convert the initial volume and find m/V.
  2. Use the explicitly stated incompressible model for the fixed amount.
  3. Draw horizontal lines rather than pressure-dependent compression curves.
Scoring criteria
  • 1 point: volume 300 mL and density 800 kg/m³ at both pressures.
  • 1 point: both horizontal graphs with numerical axes and units.
  • 1 point: distinguishes fixed volume/density from changing pressure and states model scope.

Part b

Many particles interact with the plate, transferring momentum and force. Their collective normal force per surface area gives the macroscopic pressure; liquid particle interactions also transmit forces through the fluid. The 130,000 Pa liquid exerts (130,000)(0.0060) = 780 N right, and the 100,000 Pa air exerts 600 N left. The resultant is 180 N right, toward the air. Pressure is scalar, but each surface normal supplies its force direction. The air force must be included; 780 N is only one face’s force, not the plate’s resultant.

  1. Relate many particle-surface interactions to normal force per area.
  2. Multiply each pressure by the same face area.
  3. Place the force arrows in opposite normal directions and subtract their magnitudes.
Scoring criteria
  • 1 point: explicitly explains pressure as collective particle-surface interactions.
  • 1 point: correct labeled 780 N right and 600 N left arrows.
  • 1 point: resultant 180 N right and explanation of scalar pressure versus directed force.

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Describe every way fluid velocity can change

Follow a material parcel of ideal fluid of constant mass 0.10 kg. For a short 0.20 s interval its opposing left and right faces each have area 0.0020 m², with uniform pressures 110,000 Pa on the left and 109,500 Pa on the right. These face forces are approximately constant during the interval; other horizontal forces are negligible. Right is positive. Also consider a separate parcel traveling around a bend at constant speed and a closed material sample containing two neighboring interacting fluid particles.

Part a

A nonzero net force causes a fluid parcel’s bulk velocity to change according to F_net = ma. Zero net force leaves velocity constant; that constant may be zero. A component along the current velocity increases speed, an opposite component initially decreases speed, and a perpendicular component turns velocity without instantaneously changing its magnitude. An opposite force sustained long enough may stop and reverse a parcel. At the bend the speed can be constant while the velocity changes direction, requiring a net force toward the curve’s inside from pressure forces and wall interactions. Gravity must also be included where relevant; a static parcel has all its forces balanced. Microscopic particles obey Newton’s laws too, though their random velocities differ from the parcel’s bulk velocity.

  1. Treat velocity as a vector rather than just speed.
  2. Apply Newton’s first law to balanced forces and second law to unbalanced forces.
  3. Resolve the net force relative to the current velocity direction.
  4. Use the changing velocity arrows around the bend to identify inward acceleration.
Scoring criteria
  • 1 point: nonzero net force changes bulk velocity; zero net force preserves it.
  • 1 point: correctly explains along/opposite force effects on speed, including the initial nature of slowing.
  • 1 point: explains constant-speed turning as velocity change requiring inward net force.
  • 1 point: identifies physical pressure/wall/gravity forces as applicable rather than treating a bend as a new force.

Part b

Neighboring particles interact and exert forces on one another; these interactions transmit pressure and redistribute momentum among particles or parcels, producing the observed macroscopic response. If particle 1 exerts F on particle 2, particle 2 exerts −F on particle 1. Both are internal to the chosen closed sample, so their contributions cancel when summing the sample’s momentum change. Gravity from Earth, contact from a wall, and pressure forces from fluid outside the chosen sample are external sources; any two suffice. The sum of external forces determines the whole sample’s momentum change. Internal forces still alter individual particles’ momenta and convey stresses even though their paired total is zero. Random microscopic motion does not automatically mean the sample’s bulk velocity is changing.

  1. Choose the system boundary before labeling an interaction internal or external.
  2. Name both recipients in the equal-opposite internal pair.
  3. Sum the two momenta to cancel the internal forces.
  4. Separate total-momentum change from redistribution within the sample.
Scoring criteria
  • 1 point: describes neighboring particle/parcel interactions and pressure transmission or momentum redistribution.
  • 1 point: identifies the two opposite internal forces and their cancellation for the whole sample.
  • 1 point: gives two appropriate external sources and states their role in total momentum change.
  • 1 point: explains why internal cancellation does not erase local particle effects.

Part c

The left-face pressure force is 220 N right; the right-face force is 219 N left. Their resultant is +1.0 N. Thus a = +1.0/0.10 = +10 m/s² and Δv = aΔt = +2.0 m/s. Starting at +1.0 m/s gives v_final = +3.0 m/s, speeding up rightward. Starting at −1.0 m/s gives v_final = +1.0 m/s: it slows leftward, stops after 0.10 s, then reverses and speeds up rightward. The idealized forces are assumed constant only for the stated short interval.

  1. Multiply each pressure by 0.0020 m² and draw opposite normal arrows.
  2. Subtract the force magnitudes and divide by the parcel mass.
  3. Use v_final = v_initial + aΔt with signed initial velocities.
  4. For the second case solve 0 = −1.0 + 10t to locate the reversal.
Scoring criteria
  • 1 point: correct labeled opposing 220 N and 219 N forces and +1.0 N resultant.
  • 1 point: +10 m/s² acceleration and +2.0 m/s velocity change.
  • 1 point: final velocities +3.0 and +1.0 m/s.
  • 1 point: explains slowing, stopping at 0.10 s and rightward reversal in the second run.

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Build buoyancy from the surface forces

A rigid rectangular block is fully submerged in resting uniform water, with horizontal face area A = 0.0080 m² and height H = 0.25 m. The top is 0.40 m below the free surface. Water density is 1000 kg/m³, g = 9.8 m/s², and the surface absolute pressure is 101,000 Pa. The block mass is 2.50 kg and it is held stationary by an upward support force. Ignore surface tension and contact with the tank.

Part a

Water particles collectively exert normal forces on each face. The bottom lies at 0.65 m depth, so P_top = 101,000 + (1000)(9.8)(0.40) = 104,920 Pa and P_bottom = 107,370 Pa. The top force is 839.36 N down and the bottom force is 858.96 N up. Opposite vertical sides have equal pressure at corresponding depths and their horizontal forces cancel. The remaining surface-force resultant is (P_bottom − P_top)A = 19.60 N up. In general, P_bottom − P_top = ρgH, so the resultant is ρgHA = ρgV_displaced. Here V_displaced = 0.0020 m³, and the displaced water weight is 19.60 N. Buoyancy is this resultant, not another force to add to the face forces.

  1. Find bottom depth by adding the block height to top depth.
  2. Apply hydrostatic pressure at each horizontal face.
  3. Multiply each pressure by area and assign opposing normal directions.
  4. Cancel opposite side forces and subtract top from bottom.
  5. Replace HA by displaced volume to derive Archimedes’ expression.
Scoring criteria
  • 1 point: explicitly connects collective particle forces to pressure forces on the faces.
  • 1 point: both correct absolute pressures and labeled normal arrows.
  • 1 point: force difference 19.60 N upward with side-force cancellation.
  • 1 point: derives ρgHA = ρgV_displaced and avoids double-counting buoyancy.

Part b

Weight is mg = 24.50 N, so support = 24.50 − 19.60 = 4.90 N upward. Lowering the block adds 2940 Pa to both face pressures, increasing each face force magnitude by (2940)(0.0080) = 23.52 N. The new top and bottom forces are 862.88 N down and 882.48 N up. Their difference remains 19.60 N, so support remains 4.90 N. Larger absolute pressures add equal opposing contributions; the hydrostatic difference depends on H and density, which are unchanged for a rigid block in incompressible uniform water.

  1. Use support + buoyancy − weight = 0.
  2. Find the common hydrostatic pressure increase.
  3. Compute the equal increase in opposite face-force magnitudes.
  4. Subtract again to show unchanged buoyancy.
Scoring criteria
  • 1 point: initial support 4.90 N with a correct vertical force balance.
  • 1 point: pressure increase 2940 Pa and face-force increase 23.52 N each.
  • 1 point: unchanged 19.60 N buoyancy and 4.90 N support explained by the equal opposing additions.

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Follow the mass and identify the driving force

Ideal water of density 1000 kg/m³ flows steadily through a rigid filled pipe with no branches or leaks. Inlet area is 0.0040 m² and average inlet speed is 2.0 m/s. Outlet area is 0.0016 m². Separately, consider a horizontal uniform-area fluid segment initially at rest with end area 0.0040 m², pressure 115,000 Pa on its left end and 110,000 Pa on its right. There is no other horizontal force on that segment. All pipes are completely filled; dissipative forces are neglected.

Part a

In Δt the fluid advances length vΔt, so the crossing volume is ΔV = A(vΔt), giving Q = ΔV/Δt = Av. Q = (0.0040)(2.0) = 0.0080 m³/s = 8.0 L/s. Mass flow rate is ρQ = (1000)(0.0080) = 8.0 kg/s. In 3.0 s the crossing volume is 0.024 m³, or 24 L.

  1. Construct the crossing cylinder with area A and length vΔt.
  2. Divide volume by time to obtain Av.
  3. Convert cubic meters per second to liters per second.
  4. Multiply Q by density for mass rate and by time for total volume.
Scoring criteria
  • 1 point: derives ΔV = AvΔt and Q = Av.
  • 1 point: correct 0.0080 m³/s and 8.0 L/s.
  • 1 point: correct 8.0 kg/s and 0.024 m³ in 3.0 s.

Part b

For a rigid completely filled pipe with constant incompressible density, the stored mass is fixed. Steady flow with no branch or leak therefore requires mass in per time = mass out per time. Thus ρA₁v₁ = ρA₂v₂, and constant density gives A₁v₁ = A₂v₂. Outlet speed is 0.0080/0.0016 = 5.0 m/s. The reported mismatch would accumulate 1.0 kg every second, so at least a measurement or a stated steady/filled/incompressible/no-leak assumption must be wrong; it cannot persist under all the given assumptions.

  1. Use unchanged pipe volume and density to fix the mass stored inside.
  2. Set entering and leaving mass rates equal.
  3. Cancel the common density and solve for outlet speed.
  4. Subtract the claimed rates to diagnose accumulation.
Scoring criteria
  • 1 point: explains equal mass rates from no accumulation, rather than asserting equal speed.
  • 1 point: derives continuity and obtains 5.0 m/s.
  • 1 point: identifies 1.0 kg/s accumulation and the contradiction with the setup.

Part c

Left pressure gives (115,000)(0.0040) = 460 N right; right pressure gives 440 N left. The resultant is 20 N right, so the initially resting fluid accelerates and begins flowing toward the lower right-side pressure. Pressure is not an extra kind of matter: unequal forces transferred by the surrounding particles change the fluid’s momentum. A stationary vertical fluid can have a pressure difference whose net upward pressure force balances its weight, leaving zero net force and no flow. A pressure difference must be interpreted with all relevant forces and the model geometry.

  1. Draw forces normal to both end faces.
  2. Compute their difference using the common area.
  3. Use Newton’s second law to connect the unbalanced force with the initial velocity change.
  4. Include gravity when interpreting the hydrostatic vertical comparison.
Scoring criteria
  • 1 point: opposite 460 N and 440 N arrows, giving 20 N right.
  • 1 point: initial flow toward lower pressure explained through an unbalanced force.
  • 1 point: explains why gravity can balance a vertical hydrostatic pressure difference.

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Keep all three terms in a rising pipe

Steady ideal incompressible water, density 1000 kg/m³, flows through a completely filled pipe without a pump or dissipative loss between two points on one streamline. At point 1, A₁ = 0.0040 m², v₁ = 2.0 m/s, y₁ = 0 and absolute P₁ = 170,000 Pa. At point 2, A₂ = 0.0020 m² and y₂ = +1.50 m. Take g = 9.8 m/s².

Part a

A₁v₁ = A₂v₂ gives v₂ = 4.0 m/s. A 1.0 L parcel has volume 0.0010 m³ and mass 1.0 kg. ΔK = ½(1.0)(4.0² − 2.0²) = +6.0 J. ΔU_g = (1.0)(9.8)(1.50) = +14.7 J. A speed-only comparison would account for only 6.0 J and miss the energy required to raise the parcel. The total increase is 20.7 J.

  1. Use equal volume flow at both cross sections.
  2. Convert one liter to cubic meters and multiply by density.
  3. Calculate the difference of kinetic energies, not just final K.
  4. Add the gravitational increase from the stated height difference.
Scoring criteria
  • 1 point: v₂ = 4.0 m/s and parcel mass 1.0 kg.
  • 1 point: ΔK = +6.0 J and ΔU_g = +14.7 J.
  • 1 point: explains the incomplete speed-only balance.

Part b

With parcel mass ρΔV, the work equation is (P₁ − P₂)ΔV = ½ρΔV(v₂² − v₁²) + ρΔVg(y₂ − y₁). Divide by ΔV and rearrange to obtain P₁ + ½ρv₁² + ρgy₁ = P₂ + ½ρv₂² + ρgy₂. Thus P₂ = 170,000 − 6000 − 14,700 = 149,300 Pa. Point 1 has P = 170,000 Pa, ½ρv² = 2000 Pa, ρgy = 0 and sum 172,000 Pa. Point 2 has P = 149,300 Pa, ½ρv² = 8000 Pa, ρgy = 14,700 Pa and the same sum 172,000 Pa. Each term is energy per volume, in Pa = J/m³. The equation applies along a streamline for steady ideal incompressible flow, with filled pipe, no intervening pump and negligible dissipative losses.

  1. Replace parcel mass by density times volume in both energy changes.
  2. Divide the work balance by the shared volume.
  3. Move initial-state terms to one side and final-state terms to the other.
  4. Retain the nonzero height term and solve for the final pressure.
  5. Check conservation by adding all three numerical terms at each point.
Scoring criteria
  • 1 point: correctly derives the complete Bernoulli equation from pressure work.
  • 1 point: P₂ = 149,300 Pa with the height and kinetic contributions retained.
  • 1 point: correct two-row ledger totaling 172,000 Pa at each point.
  • 1 point: states streamline, steady, incompressible/ideal and no-pump/loss scope.

Part c

In the rising pipe, net pressure work of 20.7 J per 1.0 L supplies a 6.0 J kinetic gain and a 14.7 J gravitational gain. The pressure contribution falls while both other terms rise. If y₂ = −1.50 m, P₂ = 170,000 − 6000 + 14,700 = 178,700 Pa. Gravitational potential now falls by 14.7 J, enough to provide the 6.0 J kinetic increase and an 8.7 J-per-liter increase in the pressure contribution. Thus the faster downstream stream can have higher pressure when it descends enough. Pressure represents work by surrounding fluid, not a separate parcel kinetic energy.

  1. Describe the pressure-work source for both positive energy changes.
  2. Reverse the height term sign for the downhill case.
  3. Compare the 14.7 J gravitational decrease with the 6.0 J kinetic increase.
Scoring criteria
  • 1 point: correctly explains pressure work supplying both initial gains.
  • 1 point: downhill P₂ = 178,700 Pa.
  • 1 point: justifies higher pressure with faster flow by the gravitational energy release.

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Derive the reservoir speed and test its assumptions

An open tank drains through a small opening at depth h below its free surface. The free-surface area is much larger than the opening area. Follow a streamline from the free surface to the emerging jet, assume locally steady ideal incompressible flow and negligible losses, and use g = 9.8 m/s². Both exposed points are at atmospheric pressure. For a later comparison the liquid is water of density 1000 kg/m³ and a sealed tank has surface pressure 5000 Pa above the exit pressure.

Part a

P_s + ½ρv_s² + ρgy_s = P_e + ½ρv_e² + ρgy_e. The exposed surface and jet have equal atmospheric pressure, so P_s = P_e cancels. Continuity A_s v_s = A_e v_e and A_s ≫ A_e imply v_s ≪ v_e, making the surface kinetic term negligible. Since y_s − y_e = h, the remaining balance is ρgh = ½ρv_e². Cancel density, multiply by 2 and take the positive speed to obtain v_e = √(2gh). For a parcel of mass m this is the conversion mgh = ½mv_e²; pressure work has zero net contribution under the equal-pressure comparison.

  1. Write all pressure, speed and height terms before deleting any.
  2. Use equal external atmospheric pressures for the first cancellation.
  3. Use continuity and the area ratio for the surface-speed approximation.
  4. Move the height terms to form ρgh.
  5. Solve the remaining mechanical-energy equation for positive exit speed.
Scoring criteria
  • 1 point: writes the complete two-point Bernoulli balance.
  • 1 point: separately explains equal-pressure cancellation and the large-area surface-speed approximation.
  • 1 point: algebraically derives v_e = √(2gh) rather than quotes it.
  • 1 point: explains height-energy conversion and states ideal/loss-free scope.

Part b

At h = 0.50 m, v_exit² = 9.80 m²/s² and v_exit = 3.13 m/s. The graph points (h in m, v² in m²/s²) are (0.20, 3.92), (0.50, 9.80) and (0.80, 15.68). The model line passes through zero with slope 19.6 m/s² = 2g. Plot h horizontally and v² vertically with units, numerical scales and these collinear points. It tests the conversion of gravitational height energy per mass gh into kinetic energy per mass ½v² for the stated open large-tank model.

  1. Square the derived theorem to obtain v² = 2gh.
  2. Calculate the three values and the square root at 0.50 m.
  3. Read the slope units as (m²/s²)/m = m/s².
Scoring criteria
  • 1 point: 3.13 m/s at 0.50 m.
  • 1 point: correct three-point labeled graph and intercept zero.
  • 1 point: slope 19.6 m/s² with the gravitational-to-kinetic interpretation.

Part c

Pressure no longer cancels because P_s − P_e = ΔP = 5000 Pa. Rearranging the full balance gives ½ρv_e² = ΔP + ρgh, so v_e² = 2gh + 2ΔP/ρ. Here v_e² = 9.80 + 10.0 = 19.80 m²/s² and v_e = 4.45 m/s. At constant ΔP and density, the graph retains slope 2g = 19.6 m/s² but has vertical intercept 2ΔP/ρ = 10.0 m²/s². Extra pressure work adds to the released gravitational energy. These expressions still neglect surface speed and dissipative losses.

  1. Keep the unequal pressure terms in Bernoulli.
  2. Divide the pressure difference by density while isolating v².
  3. Evaluate both the height and pressure contributions.
  4. Identify the graph’s coefficient of h and its constant term.
Scoring criteria
  • 1 point: correct modified expression and failed pressure cancellation.
  • 1 point: speed 4.45 m/s.
  • 1 point: unchanged slope and intercept 10.0 m²/s² explained by added pressure work.

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Can faster flow have greater pressure?

Assume a filled pipe, ideal incompressible water, steady flow, no intervening pump and no dissipative loss. Compare the specified points along one streamline.

Part a

P₂ exceeds P₁ by 15,600 Pa. The gravitational term decreases by 19,600 Pa while the kinetic term increases by 4000 Pa; conservation requires the pressure term to increase by the remaining 15,600 Pa.

  1. Calculate kinetic-term increase: ½(1000)(3.0² − 1.0²) = 4000 Pa.
  2. Calculate height-term decrease: (1000)(9.8)(−2.0) = −19,600 Pa.
  3. Use ΔP + Δ(½ρv²) + Δ(ρgy) = 0 to get ΔP = +15,600 Pa.
Scoring criteria
  • 1 point: selects the 15,600 Pa pressure increase.

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