Physics homework · AP Physics C
12 physics problems, and the numbers are real.
This page is for high-school physics students and their teachers: a side room, with nothing on it a resident needs. Every quantity in these problems is one our flood model actually uses, and every answer is computed from the model rather than typed in, so if somebody changes the opening or the size of the basin, these answers change with it. Nothing here is a textbook fiction with plausible units.
The level is AP Physics C, and everything you need is on that syllabus: conservation of mass, the work–energy theorem, one separable differential equation, superposition and beats, and the exponential approach you already know from an RC circuit. The physics section of Sources & limits shows where each lives in the model; read it first if a problem does not make sense.
Nothing answered yet. 12 problems.
01
Work–energy theorem
The water behind the sluice gate stands 1.20 m above the bay. Treating the culvert as an orifice, how fast does water leave through it?
- Head, H
- 1.20 m
- g
- 9.81 m/s²
Marked correct within 2%, so rounding along the way is fine.
02
Torricelli · unit conversion
With 2.00 m of head, what discharge does the model send through the culvert? Give your answer in cubic feet per second, the unit the agencies publish.
- Culvert opening, A_c
- 1.765 m²
- Discharge coefficient, C_d
- 0.6
- Head, H
- 2.00 m
- Conversion
- 1 m³/s = 35.3147 cfs
Marked correct within 2%, so rounding along the way is fine.
03
Conservation of mass
The gate is shut, so nothing leaves. Runoff arrives at 6.00 m³/s. How fast does the water level in the basin rise, in meters per hour?
- Basin area, A
- 12,950 m² (3.2 acres)
- Inflow, Q_in
- 6.00 m³/s
- Outflow
- 0 — the gate is shut
Marked correct within 2%, so rounding along the way is fine.
04
Separable differential equation
Hold the bay at a fixed level and let the basin drain from 1.20 m of head down to 0.30 m. How long does it take, in hours?
- Basin area, A
- 12,950 m²
- C_d · A_c
- 1.059 m²
- From / to
- h₀ = 1.20 m, h₁ = 0.30 m
Marked correct within 3%, so rounding along the way is fine.
05
Superposition and beats
Two of the model's tidal constituents are the lunisolar diurnal K₁ at 23.9345 h and the solar diurnal P₁ at 24.0659 h. What is the beat period of the two, in days?
- T₁
- 23.9345 h
- T₂
- 24.0659 h
Marked correct within 2%, so rounding along the way is fine.
06
Exponential approach · RC analogy
Rain begins at a steady rate on a catchment modeled as a linear reservoir. How long until it is delivering 90% of the steady-state runoff, in hours?
- Time constant, τ
- 1.4 h
- Response
- Q(t) = Q∞ (1 − e^(−t/τ))
Marked correct within 2%, so rounding along the way is fine.
07
Volume · unit conversion
How much water does the basin hold between its silted invert and the level at which it spills over its bank? Give your answer in acre-feet.
- Basin area
- 12,950 m², taken as constant
- Invert
- 0.91 m NAVD88
- Spill
- 2.286 m NAVD88
- Conversion
- 1 acre-foot = 1,233.48 m³
Marked correct within 3%, so rounding along the way is fine.
08
Continuity · multi-step
A 25 mm/h storm falls on the whole watershed with the gate shut. Ignoring the lag, how long until the basin fills from its invert to the spill level, in hours?
- Watershed area
- 1,392,120 m² (344 acres)
- Runoff coefficient, C
- 0.6
- Intensity, i
- 25 mm/h
- Storage to fill
- 17,819 m³
Marked correct within 3%, so rounding along the way is fine.
09
Continuity · rates
The proposed pump station is rated at 50 cfs. Working alone against a shut gate, how long would it take to draw the basin down by 0.30 m, in hours?
- Pump
- 50 cfs = 1.416 m³/s
- Basin area
- 12,950 m²
- Drawdown
- 0.30 m
Marked correct within 3%, so rounding along the way is fine.
10
Rates · reverse flow
The bay stands 0.50 m above the basin and the gate is shut. The gate leaks. At what rate does the bay push water into the basin, in m³/s?
- Culvert opening, A_c
- 1.765 m²
- Leak fraction
- 0.18 of the open area
- Reverse head
- 0.50 m
Marked correct within 3%, so rounding along the way is fine.
11
Concept
Emptying the basin from 2.0 m of head takes about twice as long as emptying it from 0.5 m — not four times. Why?
12
Concept · limits
The model solves conservation of mass and uses energy to get a flow rate through the culvert. Which conservation law does it never solve — and what does that cost?
Two of these are worth arguing with
Where the physics stops being clean
Problem 2 gives a discharge, and you should check it against somebody else. The answer comes out between the 10-year and 100-year flows for this culvert published by BKF, the engineering firm whose hydraulic model appears in Caltrans' environmental document, and that agreement is exactly why our model uses this opening. An earlier version used an opening about a third of the size, which no possible tide could have made pass those flows. A problem whose answer agrees with an engineering firm's hydraulic model is testing more than arithmetic.
Problem 4 asks for a drain time the model refuses to use. The closed form is correct for a fixed , but the tide moves the bay while the basin drains, so the real problem has no clean solution and has to be integrated step by step. That gap between the exam problem and the real one is not a flaw in either; it is the reason numerical methods exist.
If you find an error, that is worth more to us than a correct answer. The working is printed for every problem so you can check it line by line, and the constants all live in one file in our code, which is not public yet (see What you can do). An arithmetic slip here is a slip in the model too: they are the same numbers. Sources & limits keeps the full accounting of what is measured, what is fitted, and what is still a guess.