A flat plate moves normally with a speed v_(1) towards a horizontal jet of water of uniform area of cross section. The jet discharges water at the rate of volume V per second at a speed of v_(2). The density of water is rho. Assume that water splashes along the surface of the plate at right angles to the original motion. The magnitude of the force acting on the plate due to the jet is

A flat plate moves normally with a speed v_(1) towards a horizontal je

1.	A flat plate moves normally with a speed v_(1) towards a horizontal jet of water of uniform area of cross section. The jet discharges water at the rate of volume V per second at a speed of v_(2). The density of water is rho. Assume that water splashes along the surface of the plate at right angles to the original motion. The magnitude of the force acting on the plate due to the jet is
Answer» SOLUTION :`"Force acting on the PLATE F"=(DP)/(dt)=u_(r)=(dm)/(dt)` `"Since, "Av_(2)=V rArr (dm)/(dt)=A(v_(1)+v_(2))rho=(V)/(v_(2))(v_(1)+v_(2))rho` `(u_(r)=v_(1)+v_(2)="VELOCITY of water COMING out of jet w.r.t plate")` `F=(v_(1)+v_(2)).(V)/(v_(2))(v_(1)+v_(2))rho=(V)/(v_(2))rho=(V)/(v_(2))(v_(1)+v_(2))^(2)rhoN`