A hole is punched in a full milk carton, 10cm below the top. What's the initial velocity of outflow?
P2 = P1 + ρgh
Conservation of energy of fluids: P2 + 1/2ρv22 + ρgy2 = P1 + 1/2ρv12 + ρgy1
The Attempt at a Solution
Set the top of the carton as P1, and the point 10cm below the surface as P2.
Assume that milk has the same density as water: 1000kg/m3
P1 = atmospheric pressure = 101325Pa
P2 = 101325 + 1000(9.8)(0.1) = 102305Pa
P2 + 1/2ρv22 + ρgh2 = P1 + 1/2ρv12 + ρgy1
The milk doesn't have kinetic energy at the top of the carton, so K=1/2ρv22 =0
At 10cm below the surface, y=0 (although this is still elevated above the ground, we're trying to find the change in energy between the top of the carton and hole punched in it); therefore U = ρgy1 = 0
P2 + ρgh2 = P1 + 1/2ρv12
102305 + 1000(9.8)(0.1) = 101325 + 1/2(1000)v12
103285 - 101325 = 500v12
V1 = 1.98m/s
Is my method correct? The actual answer is V=1.4m/s, but perhaps it's because I assumed that the density of milk is the same as water....