(b) What mass of water hits the surface in a time t? (c) Given the geometry show
ID: 1885039 • Letter: #
Question
(b) What mass of water hits the surface in a time t?
(c) Given the geometry shown in the figure, determine the water jet’s velocity vector, w after the jet has bounced off the wall.
(d) What is the change in the velocity vector of the water jet as a result of hitting the surface.
(e) What is the change in momentum of the water jet (call it p) in a time t?
(f) What is the force (FS) on the surface? Sketch this force on the figure.
(g) What is the magnitude of this force? Comment on the dependence of the force.
A water jet of cross-sectional area A, in which the water's velocity is initially vi, hits a stationary, flat surface that is inclined at an angle 0 to the jet. The water bounces off as shown in the figure. Its velocity after bouncing off the surface has the same magnitude as before, but its direction is changed, as shown. The (mass) density of water is (Greek rho). (Ignore gravity in this problem.) Cvoss -sechom Sutace velocat 1sExplanation / Answer
given water jet of cross section area A
initial velocity = vi
angle of incline with horizonta = theta
final velocity magnitude is the same
mass density = rho
hence
a. in time dT
volume of water hitting = A*vi*dT
b. mass of water in dT = rho*A*vi*dT
c. after the jet bounced off, reflected jects velocity vector is given by
vf = vi(cos(2*theta)i + sin(2*theta)j)
d. dv = vf - vi = vi((cos(2*theta) - 1)i + sin(2*theta)j)
e. change in momentum dp = dm*dv = rho*A*dT*vi^2((cos(2*theta) - 1)i + sin(2*theta)j)
f. force on surface = dp/dt = rho*A*vi^2((cos(2*theta) - 1)i + sin(2*theta)j )
g. |F| = rho*A*vi^2*sqrt(2 - 2*cos(2*theta))
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