5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
Water flows steadily from a tap through a garden hose that is completely filled and out through a nozzle that is narrower than the hose. Treat the water as an ideal fluid. How does the mass of water leaving the nozzle each second compare with the mass of water entering the hose from the tap each second?
Answer and reasoning
ASmaller: the narrow nozzle lets less water through per second. A student who thinks the size of an opening alone sets the flow rate picks this. The nozzle is narrower, but the water speeds up as it passes through, so the same mass leaves each second; otherwise water would pile up in a hose that is already full.
BGreater: the water leaves faster than it enters the hose. A student who takes a faster flow to mean more water per second picks this. The water does leave faster, but through a smaller area; the mass per second, ρAv, is the same as at the tap.
CSmaller: some of the water's flow is used up along the hose. A student who thinks a flow is used up as it travels picks this. Nothing destroys water along the hose: the mass that enters each second must leave each second.
DEqual: water cannot pile up in a hose that is already full.Correct The hose is completely filled and the water is incompressible, so no extra water can collect in it, and none leaks out. Each kilogram that enters at the tap pushes a kilogram out at the nozzle, so the mass flow rates are equal.
A completely filled pipe of cross-sectional area A carries water at speed v. The pipe divides into two identical branches, each of cross-sectional area A/4, which are also completely filled. Treat the water as an ideal fluid. Which expression gives the speed of the water in each branch?
Answer and reasoning
A2.0vCorrect The volume per second entering the junction leaves through the two branches together, whose total area is 2(A/4) = A/2. Continuity gives Av = (A/2)vb, so vb = 2v.
B4.0v A student who thinks each branch carries the whole flow sets Av = (A/4)vb and gets 4v. The flow divides between the two branches, so each carries only half of it.
C0.5v A student who thinks water slows down where the pipes are narrower inverts the area ratio and gets v × (A/2)/A = 0.5v. To carry the same volume per second through a smaller total area, the water must move faster.
D1.0v A student who thinks an incompressible fluid keeps the same speed everywhere picks v. Only the volume per second is the same; the branches' total area is half the main pipe's, so the speed doubles.
Working The volume per second entering the junction equals the total leaving through both branches: Av = 2(A/4)vb = (A/2)vb, so vb = 2v = 2.0v.
Water flows steadily upward through a completely filled pipe of uniform cross-sectional area, from the ground floor of a building to the second floor. Treat the water as an ideal fluid. Which statement comparing the water at the second floor with the water at the ground floor is correct?
Answer and reasoning
AIts speed is lower at the second floor than at the ground floor. A student who thinks water rising in a pipe slows down, like a ball thrown upward, picks this. In a full pipe of uniform area the same volume passes every point each second, so the speed is the same at both floors; the energy for the rise comes from the pressure instead.
BIts pressure is the same at the second floor as at the ground floor. A student who thinks the pressure in a moving fluid depends only on its speed picks this. The speeds are equal, but the water at the second floor is higher, so ρgy is larger there and P must be smaller by ρgΔy.
CIts pressure is lower at the second floor than it is at the ground floor.Correct The pipe's area is the same throughout, so by continuity the speed is the same at both floors. Conservation of energy (Bernoulli's equation) then gives P₁ + ρgy₁ = P₂ + ρgy₂: the gain in gravitational potential energy per unit volume, ρg(y₂ − y₁), comes from a drop in pressure of the same size.
DIts pressure is greater at the second floor than at the ground floor. A student who reads the h in P = P₀ + ρgh as a height rather than a depth picks this. Pressure increases with depth, so going up the pipe the pressure decreases, by ρgΔy.
Bernoulli's equation, P₁ + ρgy₁ + (1/2)ρv₁² = P₂ + ρgy₂ + (1/2)ρv₂², applies to the steady flow of an ideal fluid. Which statement correctly describes what the equation expresses?
Answer and reasoning
AThe same mass of fluid passes every point along the flow in each second, whatever the area. A student who confuses Bernoulli's equation with the continuity equation picks this. Conservation of mass in the flow is A₁v₁ = A₂v₂; every term in Bernoulli's equation is an energy per unit volume.
BWork done by pressure differences equals the change in the fluid's kinetic and potential energy.Correct Bernoulli's equation is conservation of energy for a flowing ideal fluid, written per unit volume. The pressure terms account for the work done on the fluid by the fluid around it, and ρgy and (1/2)ρv² are its gravitational potential and kinetic energies per unit volume.
CThe fluid's pressure is greater wherever it moves faster, because faster fluid pushes harder. A student who thinks faster fluid has a greater pressure picks this. The equation says the opposite: at one height, a larger (1/2)ρv² must be balanced by a smaller P.
DThe pressure has the same value at every point in the fluid, as the fluid is ideal. A student who thinks the pressure is the same everywhere in a fluid picks this. The equation keeps the sum P + ρgy + (1/2)ρv² constant; P itself changes wherever the height or the speed changes.
An open tank of water stands on the floor. The water surface is 1.25 m above the floor, and water flows out through a small hole in the side of the tank 0.45 m above the floor. Treat the water as an ideal fluid, and assume the water level falls negligibly slowly. Use g = 10 m/s². With what speed does the water leave the hole?
Answer and reasoning
A3.0 m/s A student who thinks the exit speed depends on how high the hole is above the floor uses 0.45 m and gets √(2 × 10 × 0.45) = 3.0 m/s. The speed comes from the drop from the water surface down to the hole, 0.80 m.
B5.0 m/s A student who uses the whole height of the water above the floor, 1.25 m, gets √(2 × 10 × 1.25) = 5.0 m/s. The water leaving the hole has come down only from the surface to the hole, 0.80 m.
C4.0 m/sCorrect Torricelli's theorem uses the height of the water surface above the hole: Δy = 1.25 − 0.45 = 0.80 m. Then v = √(2gΔy) = √(2 × 10 × 0.80) = 4.0 m/s.
D2.8 m/s A student who drops the 1/2 from the kinetic energy term sets ρgΔy = ρv² and gets v = √(gΔy) = √(10 × 0.80) = 2.8 m/s. With ρgΔy = (1/2)ρv², v = √(2gΔy) = 4.0 m/s.
Working Torricelli's theorem: Δy is the height of the water surface above the hole, 1.25 m − 0.45 m = 0.80 m. v = √(2gΔy) = √(2 × 10 m/s² × 0.80 m) = √(16 m²/s²) = 4.0 m/s.
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
8.4.A.1 Pressure difference as the cause of flow Fix
Pressure difference as the cause of flow
A fluid flows from a region of higher pressure toward a region of lower pressure, because the fluid on the high-pressure side pushes a portion of fluid harder than the fluid on the low-pressure side pushes back. The pressure that counts includes any gas pressing on the fluid's surface as well as the depth of the fluid.
Mass flow rate
The mass of fluid passing a cross-section of a pipe per unit time. For a completely filled tube open at both ends, the mass flow rate into the tube equals the mass flow rate out, because no fluid can collect in the tube or be destroyed in it. Unit: kg/s.
Volume flow rate (V/t = Av)
The volume of fluid passing a cross-section per unit time, equal to the cross-sectional area of the flow multiplied by the speed of the fluid there: V/t = Av. It depends on both the area and the speed, not on either alone. Unit: m³/s.
Cross-sectional area of a pipe
The area of a section cut through a pipe perpendicular to the flow. For a circular pipe of diameter d, A = πd²/4, so halving the diameter makes the area one-fourth as large. Unit: m².
Students often think Water always flows from the container with the higher water surface toward the lower one, whatever presses on the surfaces. In fact No. Water flows from higher to lower pressure. For two containers open to the same air, the higher surface does give the greater pressure at the bottom, but a gas trapped above one surface adds its own pressure and can reverse the flow.
Students often think A fluid flows only downhill; along a level pipe nothing makes it move, because gravity is what makes fluids flow. In fact No. A fluid flows wherever there is a pressure difference, which can push it along a level pipe or even uphill. Gravity causes flow only by producing pressure differences.
8.4.A.2 Incompressible fluid Fix
Incompressible fluid
A fluid whose density does not change, so a given mass always occupies the same volume. For such a fluid, conservation of mass flow rate is the same as conservation of volume flow rate.
Continuity equation (A₁v₁ = A₂v₂)
For an incompressible fluid completely filling a pipe, the volume flow rate is the same at every cross-section, so A₁v₁ = A₂v₂: the fluid moves faster where the pipe is narrower. Where a pipe branches, the flow rate into the junction equals the total flow rate out of all branches.
Students often think A fluid moves more slowly where a pipe is narrower, because the narrow section holds it back, like a traffic jam at a bottleneck. In fact No. By continuity, A₁v₁ = A₂v₂, so for the same volume per second to pass through a smaller area, the fluid must move faster there.
Students often think An incompressible fluid moves at the same speed at every point in a pipe, whatever the pipe's width. In fact No. Incompressibility keeps the volume flow rate Av the same everywhere in a full pipe. If the area changes, the speed must change in inverse proportion.
8.4.B.1 Energy per unit volume in a fluid Fix
Energy per unit volume in a fluid
For a fluid, the gravitational potential energy per unit volume is ρgy and the kinetic energy per unit volume is (1/2)ρv². Both have the unit J/m³, which is the same as the pascal (N/m²), the unit of pressure.
Pressure change with height in flowing fluid
Where an ideal fluid flows at the same speed at two heights, conservation of energy requires P₁ + ρgy₁ = P₂ + ρgy₂: the pressure is lower at the higher point by ρg(y₂ − y₁).
Students often think Water rising through a pipe slows down as it goes up, like a ball thrown upward. In fact No. In a full pipe of uniform area, continuity requires the same volume per second, and so the same speed, at every height. The energy needed to raise the water comes from a decrease in pressure, not from a decrease in speed.
Students often think The pressure in a moving fluid depends only on its speed, so where the speeds are equal the pressures are equal, whatever the heights. In fact No. By Bernoulli's equation, P + ρgy + (1/2)ρv² is the same at both points. With equal speeds, the pressure is lower at the higher point by ρgΔy.
8.4.B.2 Bernoulli's equation Fix
Bernoulli's equation
P₁ + ρgy₁ + (1/2)ρv₁² = P₂ + ρgy₂ + (1/2)ρv₂² for two points along the steady flow of an ideal fluid. It is conservation of energy per unit volume: work done on the fluid by pressure differences equals the change in its kinetic plus gravitational potential energy.
Ideal fluid
A fluid that is incompressible and has no viscosity (no internal friction), so no mechanical energy is transformed into internal energy as it flows. In AP Physics 1, fluids are treated as ideal and pipes as completely filled unless stated otherwise.
Students often think A faster-moving fluid exerts a greater pressure, because it pushes harder on whatever it touches. In fact No. In the steady flow of an ideal fluid at one height, the pressure is lower where the fluid moves faster: P + (1/2)ρv² is constant, so a larger v means a smaller P.
Students often think The change in kinetic energy per unit volume is (1/2)ρ(v₂ − v₁)², the square of the change in speed. In fact No. The change is (1/2)ρv₂² − (1/2)ρv₁² = (1/2)ρ(v₂² − v₁²). Squaring the difference of the speeds gives a different, smaller value.
8.4.B.3 Torricelli's theorem (v = √(2gΔy)) Fix
Torricelli's theorem (v = √(2gΔy))
The speed of fluid leaving a small opening in an open container, where Δy is the height of the fluid's top surface above the opening. It follows from Bernoulli's equation because the pressure is atmospheric at both the surface and the opening and the surface moves negligibly slowly, so ρgΔy = (1/2)ρv². Unit of v: m/s.
Students often think Fluid leaves an opening faster the higher the opening is above the ground, because the fluid there has more gravitational potential energy. In fact No. By Torricelli's theorem, the exit speed depends on the height Δy of the fluid's top surface above the opening. A higher opening has less fluid above it, so the fluid leaves it more slowly.
Students often think In v = √(2gΔy), Δy is the total height of the fluid in the container (its surface above the bottom or the ground), wherever the opening is. In fact No. Δy is the height of the top surface above the opening. Fluid leaving an opening partway up a container has fallen only from the surface to the opening.
8 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 8
The diagram shows two tanks of water joined at their bottoms by a horizontal pipe that is completely filled with water and closed by a valve. Tank X is wide and open to the air. Tank Y is narrow and sealed, and the air trapped above its water has the gauge pressure shown. The depths of the water surfaces above the pipe are labeled. Treat the water as an ideal fluid (density 1000 kg/m³; use g = 10 m/s²). Which statement correctly describes what happens when the valve is opened?
Answer and reasoning
AWater flows from X to Y, since the water surface in tank X is the higher of the two. A student who thinks water always flows toward the lower surface picks this. That is true for two tanks open to the same air, but the trapped air in Y adds 6.0 × 10³ Pa, so the pressure at the valve is greater on Y's side: 1.2 × 10⁴ Pa against 1.0 × 10⁴ Pa.
BNo water flows, since the pipe is level and gravity cannot push the water along it. A student who thinks water flows only downhill picks this. A fluid flows wherever there is a pressure difference; here the pressure at the valve is 2.0 × 10³ Pa greater on Y's side, so water is pushed along the level pipe from Y to X.
CWater flows from Y to X, since the pressure at the valve is higher on Y's side.Correct A fluid flows from higher toward lower pressure. At the valve the gauge pressure on X's side is ρgh = (1000)(10)(1.0) = 1.0 × 10⁴ Pa; on Y's side it is 6.0 × 10³ + (1000)(10)(0.60) = 1.2 × 10⁴ Pa. The net force on the water at the valve points from Y toward X.
DWater flows from X to Y, since tank X holds a larger amount of water than tank Y. A student who thinks the pressure depends on the amount of water in a tank picks this. The pressure at the valve depends on the depth of the water and the pressure on its surface, not on the tank's width: 1.0 × 10⁴ Pa from X against 1.2 × 10⁴ Pa from Y.
Working Gauge pressure at the valve on X's side: ρgh = (1000)(10)(1.0) = 1.0 × 10⁴ Pa. On Y's side: 6.0 × 10³ Pa + (1000)(10)(0.60) = 6.0 × 10³ + 6.0 × 10³ = 1.2 × 10⁴ Pa. The pressure is greater on Y's side, so the net force on the water at the valve points from Y toward X, and water flows from Y to X.
A student fills the same 2.0 L bottle from three different nozzles, P, Q and R, each fed by a completely filled hose. For each nozzle she records the area of its opening, the speed of the water leaving it, and the time to fill the bottle, as shown in the table. Treat the water as an ideal fluid. Which claim is supported by the data?
Answer and reasoning
AThe greater the speed at which the water leaves, the shorter the time needed to fill the bottle. A student who judges the flow rate by the speed alone picks this. P's water leaves twice as fast as Q's, yet both fill the bottle in 5.0 s, because Q's opening has twice the area.
BThe greater the product of area and exit speed, the shorter the time needed to fill the bottle.Correct The volume per second leaving a nozzle is V/t = Av. P and Q have the same product Av (4.0 × 10⁻⁴ m³/s) and fill the bottle in the same time, 5.0 s; R has half that product and takes twice as long, 10 s. The data fit a fill time that decreases as Av increases.
CThe greater the area of the nozzle's opening, the shorter the time needed to fill the bottle. A student who judges the flow rate by the size of the opening alone picks this. Q's opening has twice the area of P's, yet both take 5.0 s, because the water leaves Q at half the speed.
DThe smaller the nozzle's opening, the shorter the time needed to fill the bottle. A student who thinks narrowing an opening delivers more water picks this. R has a smaller opening than P and the same exit speed, and it takes twice as long: a smaller area at the same speed carries less water each second.
Working Volume flow rate V/t = Av. P: (1.0 × 10⁻⁴ m²)(4.0 m/s) = 4.0 × 10⁻⁴ m³/s, so 2.0 × 10⁻³ m³ takes 5.0 s. Q: (2.0 × 10⁻⁴)(2.0) = 4.0 × 10⁻⁴ m³/s, 5.0 s. R: (0.50 × 10⁻⁴)(4.0) = 2.0 × 10⁻⁴ m³/s, 10 s. Equal products Av give equal times (P, Q), and the smaller product gives the longer time (R). P and Q have different speeds but equal times, and different areas but equal times, so neither speed nor area alone decides the time; R has a smaller area than P but a longer time.
Water leaves the end of hose 1 with speed v. Hose 2 has half the inner diameter of hose 1, and water leaves its end with speed 2v. Both hoses are completely filled, and the water can be treated as an ideal fluid. The volume of water leaving hose 2 each second is what multiple of the volume leaving hose 1 each second?
Answer and reasoning
A×½Correct The volume per second is V/t = Av. Half the diameter gives one-fourth of the area, because A = πd²/4, and the speed is doubled, so the flow rate changes by (1/4)(2) = ½.
B×1 A student who takes the area to be proportional to the diameter halves the area instead of quartering it and gets (1/2)(2) = 1. The area of a circle is proportional to the square of the diameter.
C×2 A student who judges the flow rate by the speed alone picks ×2. The water does leave twice as fast, but through one-fourth of the area.
D×¼ A student who judges the flow rate by the size of the opening alone picks ×¼, the ratio of the areas. The flow rate is proportional to both the area and the speed, and the speed is doubled.
Working V/t = Av, with A = πd²/4. Halving the diameter gives A₂ = A₁/4. So (V/t)₂ = (A₁/4)(2v) = (1/2)A₁v = (1/2)(V/t)₁: the factor is ×½.
Water flows steadily to the right through the completely filled horizontal pipe shown, which has three sections with the cross-sectional areas labeled. Treat the water as an ideal fluid. Which of the following correctly ranks the water's speeds v₁, v₂ and v₃ at points 1, 2 and 3?
Answer and reasoning
Av₁ > v₃ > v₂ A student who thinks water slows down in a narrow section, as traffic does at a bottleneck, reverses the ranking. For the same volume per second to pass through a smaller area, the water must move faster there.
Bv₁ > v₂ > v₃ A student who thinks the water loses speed as it travels along the pipe ranks the points in order along the flow. For an ideal fluid in a full pipe, the speed depends only on the local area: point 3 has a smaller area than point 1, so the water there is faster.
Cv₁ = v₂ = v₃ A student who thinks an incompressible fluid moves at the same speed everywhere picks this. Incompressibility makes the volume per second, Av, the same at every point; since the areas differ, the speeds must differ.
Dv₂ > v₃ > v₁Correct The same volume of water passes each point every second, so (3A)v₁ = (A)v₂ = (2A)v₃. The speed is inversely proportional to the area: v₂ = 3v₁ and v₃ = 1.5v₁, so v₂ > v₃ > v₁.
Working Continuity: (3A)v₁ = (A)v₂ = (2A)v₃. So v₂ = 3v₁ and v₃ = 1.5v₁, giving v₂ > v₃ > v₁.
Water flows steadily through a completely filled pipe. At point 1 the gauge pressure is 5.0 × 10⁴ Pa and the water's speed is 2.0 m/s. Point 2 is 3.0 m higher than point 1, where the pipe is narrower and the water's speed is 4.0 m/s. Treat the water as an ideal fluid (density 1000 kg/m³; use g = 10 m/s²). What is the gauge pressure at point 2?
Answer and reasoning
A4.4 × 10⁴ Pa A student who thinks the pressure in a moving fluid depends only on its speed includes the change in speed but leaves out the 3.0 m rise, getting 5.0 × 10⁴ − 6.0 × 10³ = 4.4 × 10⁴ Pa. The ρgy term must be included: raising the water costs a further 3.0 × 10⁴ Pa.
B1.4 × 10⁴ PaCorrect Bernoulli's equation gives P₂ = P₁ − ρg(y₂ − y₁) − (1/2)ρ(v₂² − v₁²). Rising 3.0 m costs (1000)(10)(3.0) = 3.0 × 10⁴ Pa and speeding up from 2.0 to 4.0 m/s costs (1/2)(1000)(16 − 4) = 6.0 × 10³ Pa, so P₂ = 5.0 × 10⁴ − 3.0 × 10⁴ − 6.0 × 10³ = 1.4 × 10⁴ Pa.
C7.4 × 10⁴ Pa A student who thinks pressure increases with height adds ρgΔy instead of subtracting it: 5.0 × 10⁴ + 3.0 × 10⁴ − 6.0 × 10³ = 7.4 × 10⁴ Pa. Point 2 is higher, so its ρgy term is larger and its pressure must be smaller.
D2.6 × 10⁴ Pa A student who thinks faster water has a greater pressure adds the kinetic energy term instead of subtracting it: 5.0 × 10⁴ − 3.0 × 10⁴ + 6.0 × 10³ = 2.6 × 10⁴ Pa. The water speeds up between the points, so the pressure must fall to pay for its extra kinetic energy.
Working Bernoulli: P₁ + ρgy₁ + (1/2)ρv₁² = P₂ + ρgy₂ + (1/2)ρv₂², so P₂ = P₁ − ρg(y₂ − y₁) − (1/2)ρ(v₂² − v₁²) = 5.0 × 10⁴ Pa − (1000)(10)(3.0) Pa − (1/2)(1000)(4.0² − 2.0²) Pa = 5.0 × 10⁴ − 3.0 × 10⁴ − 6.0 × 10³ = 1.4 × 10⁴ Pa. (Gauge pressures can be used because atmospheric pressure adds the same amount to both sides.)
Water flows steadily through a completely filled horizontal pipe. In a wide section, where the pressure is P₁, the cross-sectional area is A₁ and the water's speed is v. The pipe then narrows to a section of area A₁/2, where the pressure is P₂. Treat the water as an ideal fluid of density ρ. Which expression gives P₁ − P₂?
Answer and reasoning
A+(1/2)ρv² A student who squares the change in speed instead of taking the change in the squares gets (1/2)ρ(2v − v)² = (1/2)ρv². The kinetic energy terms must be subtracted after squaring: (1/2)ρ(4v² − v²).
B−(3/8)ρv² A student who thinks the water slows down in the narrow section, halving its speed there because the area is halved, gets P₁ − P₂ = (1/2)ρ(v²/4 − v²) = −(3/8)ρv². Continuity, A₁v = (A₁/2)v₂, makes the water speed up to 2v in the narrow section, so P₁ is the larger pressure.
C−(3/2)ρv² A student who thinks faster water has a greater pressure puts P₂ above P₁ and gets −(3/2)ρv². The water speeds up entering the narrow section, which requires a net force in the direction of flow, so P₁ is the larger pressure.
D+(3/2)ρv²Correct Continuity gives the speed in the narrow section: A₁v = (A₁/2)v₂, so v₂ = 2v. Bernoulli's equation at one height then gives P₁ − P₂ = (1/2)ρ(v₂² − v²) = (1/2)ρ(4v² − v²) = +(3/2)ρv².
Working Continuity: A₁v = (A₁/2)v₂, so v₂ = 2v. Bernoulli with y₁ = y₂: P₁ + (1/2)ρv² = P₂ + (1/2)ρ(2v)², so P₁ − P₂ = (1/2)ρ(4v² − v²) = +(3/2)ρv².
The diagram shows water squirting from three small holes, A, B and C, in the side of a tall open bottle standing on the floor. The bottle is kept topped up so that the water level stays constant. Treat the water as an ideal fluid. From which hole does the water leave with the greatest speed?
Answer and reasoning
AHole C, since it is the farthest below the water surfaceCorrect By Torricelli's theorem, v = √(2gΔy), where Δy is the height of the water surface above the hole. Hole C is the lowest, so Δy is greatest and its water leaves fastest. Its jet does not land farthest because it starts close to the floor and has little time to fall.
BHole A, since its water leaves highest above the floor A student who thinks water leaves faster from a higher opening, because it has more potential energy there, picks this. The exit speed depends on how far the water has fallen from the surface to the hole, which is smallest for A, so A's water leaves most slowly.
CHole B, since its jet lands farthest from the bottle A student who judges speed by how far a jet lands picks this. Where a jet lands depends on its speed and on how long it falls: water from B leaves more slowly than water from C but falls from a greater height, so it spends longer in the air.
DAll three alike, since the pressure is the same throughout A student who thinks the pressure is the same everywhere in the water picks this. The pressure increases with depth, so the water at C is pushed out hardest; equivalently, Δy in v = √(2gΔy) is different for each hole.
Working Torricelli's theorem, v = √(2gΔy), with Δy the height of the water surface above each hole. Hole C is lowest, so its Δy is greatest and its exit speed is greatest. (With the surface at height H, holes at 0.8H, 0.5H and 0.2H above the floor give speeds in the ratio √0.2 : √0.5 : √0.8; the jet from B lands farthest because range = 2√(y(H − y)) is greatest for the middle hole.)
Water flows steadily through a completely filled garden hose of inner radius r and leaves through a nozzle whose circular opening has radius r/2. The water from the nozzle fills a bucket of volume V in a time t. Treat the water as an ideal fluid. Which expression gives the speed of the water as it leaves the nozzle?
Answer and reasoning
A2.00V/(πr²t) A student who thinks halving the radius halves the area uses A = πr²/2 and gets 2V/(πr²t). Area depends on the square of the radius, so the nozzle's area is π(r/2)² = πr²/4, one-quarter of the hose's.
B1.00V/(πr²t) A student who thinks the water moves at the same speed everywhere in the hose and nozzle divides V/t by the hose's area, πr². The same volume passes each second through the narrow opening as through the hose, so the water must move faster there: v = (V/t)/(πr²/4).
C4.00V/(πr²t)Correct Water leaves the nozzle at the rate V/t, and V/t = Av with the nozzle's area A = π(r/2)² = πr²/4. So v = (V/t)/(πr²/4) = 4V/(πr²t).
D0.25V/(πr²t) A student who thinks the narrow nozzle holds the water back scales the speed in the hose, V/(πr²t), down by the area ratio and gets V/(4πr²t). The same volume must pass each second through the smaller opening, so the speed there is four times the speed in the hose, not one-quarter of it.
Working Volume flow rate out of the nozzle: V/t. Nozzle area: A = π(r/2)² = πr²/4. From V/t = Av: v = (V/t)/A = (V/t)/(πr²/4) = 4V/(πr²t) = 4.00V/(πr²t).
Compiled from the AP Physics 1 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
