AP Physics 1 · Free-response skill

Predict in words, then check with numbers

The move. Make and justify a claim about a fluid’s speed and pressure in words, derive the matching expression, calculate values, and show that the words and the expression agree.

How to do it

  1. When a part asks for an answer in words only, name the physical principle that settles it and connect that principle to a specific feature of the setup, such as a change in size, height or speed.
  2. To derive, start from the general equations, cancel what is the same on both sides, and solve for the requested quantity using only the symbols the question gives you.
  3. Then put in numbers with units, keeping the powers of ten straight, and give the result to sensible precision.
  4. Last, check the expression against your words: does each factor make the result larger or smaller in the way you claimed, and does its sign match?

Practice prompt Try it

A gardener holds a large plastic syringe level and uses it to squirt water onto seedlings. The barrel has a cross-sectional area A1=4.8×10−4A_1 = 4.8 \times 10^{-4} m², and the short nozzle at its tip has an opening of area A2=1.6×10−5A_2 = 1.6 \times 10^{-5} m². The gardener pushes the plunger at a steady speed v1=0.060v_1 = 0.060 m/s, and the water leaves the nozzle as a jet into the open air. Treat the water as an ideal fluid with density ρ = 1000 kg/m³, and treat the barrel and the nozzle as being at the same height.

The questions below are about the steady flow of water through the syringe.

  1. In words only, with no equations: is the water faster inside the barrel or as it leaves the nozzle? Give the reason.
  2. Again in words only: compare the water pressure inside the barrel with the pressure of the air at the nozzle opening (greater, less or equal), and give the reason.
  3. Combine the two flow equations to get a formula for the gauge pressure of the water in the barrel that uses only ρ, v1v_1, A1A_1 and A2A_2.
  4. Calculate the speed of the water as it leaves the nozzle.
  5. Calculate the gauge pressure of the water in the barrel.
  6. Show that the sign of your part (c) formula fits what you said in part (b).

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 8 points

Model answer, annotated

(a) The water moves faster as it leaves the nozzle. Water does not compress, so each second the same volume that the plunger pushes along the wide barrel must get out through the much smaller nozzle opening, and it can do that only by moving faster.

  • Point 1 States which place has the greater speed.
  • Point 2 Uses the same volume per second through a smaller opening as the reason.

(b) The pressure in the barrel is greater than the air pressure at the nozzle. The water speeds up as it moves from the barrel into the nozzle, so something must push it forward harder from behind than from in front: the pressure behind it, in the barrel, must be higher than the pressure where it leaves.

  • Point 3 States the comparison plainly.
  • Point 4 Ties the higher pressure to the force needed to speed the water up.

(c) Continuity: A1v1=A2v2A_1v_1 = A_2v_2, so v2=v1A1/A2v_2 = v_1A_1/A_2. Bernoulli at one height: P1+12ρv12=Patm+12ρv22P_1 + \frac{1}{2}\rho v_1^2 = P_{atm} + \frac{1}{2}\rho v_2^2. So the gauge pressure is P1−Patm=12ρ(v22−v12)=12ρv12[(A1/A2)2−1]P_1 - P_{atm} = \frac{1}{2}\rho(v_2^2 - v_1^2) = \frac{1}{2}\rho v_1^2\left[(A_1/A_2)^2 - 1\right].

  • Point 5 Combines the two equations and leaves only ρ, v₁, A₁ and A₂.

(d) v2=v1A1/A2=(0.060)(4.8×10−4)/(1.6×10−5)=(0.060)(30)=1.8v_2 = v_1A_1/A_2 = (0.060)(4.8 \times 10^{-4})/(1.6 \times 10^{-5}) = (0.060)(30) = 1.8 m/s.

  • Point 6 Uses the area ratio of 30 from continuity.

(e) P1−Patm=12(1000)[(1.8)2−(0.060)2]=500(3.24−0.0036)=1.6×103P_1 - P_{atm} = \frac{1}{2}(1000)\left[(1.8)^2 - (0.060)^2\right] = 500(3.24 - 0.0036) = 1.6 \times 10^3 Pa, about 1620 Pa.

  • Point 7 Puts the numbers into the part (c) result with the right units.

(f) Because A1A_1 is larger than A2A_2, the ratio A1/A2A_1/A_2 is greater than 1, so (A1/A2)2−1(A_1/A_2)^2 - 1 is positive. Every other factor is positive too, so the gauge pressure is positive: the barrel pressure is above atmospheric, just as part (b) claimed.

  • Point 8 Links the sign of the expression to the claim in part (b).