Question

A horizontal jet of water of cross sectional area 10 cm^2 hits a vertical wall with a velocity of 10m/sec. After having hit the wall, water moves parallel to the wall . The force exerted by the wall is

Answer 1

Answer:

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.