This is problem 5 A. 4.
We have a fire hydrant with the same cross section
everywhere.
Let's call the velocity of the water here v 2, and the
velocity of the water here v 1.
The magnitudes are the same, and they're
25 meters per second.
I have L kilograms of water per second flowing through
there, and the velocity of v 2 and v 1 is 25 meters per
second, and this number is given.
The velocity is changing from this to this--
the speed is not changing, but the velocity is changing.
This is v 1, and this is v 2.
The change in velocity--
this one is v 2 minus v 1, and that's call that for
simplicity delta v. If there is a change in velocity, there
must be a force on the system.
Remember the [UNINTELLIGIBLE], or let's write
down delta for now--
delta p delta t, which is a change of
momentum per unit time--
that was an external force.
What is delta p?
In the extreme case, of course, when the delta goes to
0, it becomes dp.
Delta p equals m delta v--
this is delta v--
and this delta t.
What is m divided delta t?
That's this L--
that's the number of kilograms per second.
What is the magnitude of delta v?
The magnitude of delta v equals 25 times the
square root of 2.
What does it mean?
It means that the wall must push on the water in this
direction, in the same direction as delta p, and you
can calculate the force.
Since action equals minus reaction, the water will push
back on the wall in exactly the opposite direction but
with the same magnitude.
I think you have enough ammunition
now to do this problem.