Question

a horizontal jet of water coming out of a pipe of area of cross - section 20cm^2 hits a vertical wall with a velocity of 10m/s and rebounds with the same speed. The force exerted by water on the wall is

Answer 1

volume of water coming out of the pipe = A * v
     = 20 cm² * 10 m/sec
   = 0.02  m³/sec

mass of water hitting the wall  = m = 0.02 m³/sec * 1000 kg/m³
         = 20 kg/sec

As the water hits the wall normal to the wall and gets rebounded at the same speed, the change of momentum of water hitting the wall is:
            = m (v  - (-v))  =  2 m v
           = 2 * 20 kg /sec *  10 m/sec
           =  400 kg-m/sec²

Since, the change of momentum obtained is per unit time,  it is same as the force.  Thus the force acting on the wall by water = force experienced by water jet.

         = 400 Newtons.

Answer 2

F = P A V*2
= 1(2(10*-4)) (10)*2
= 400N