Question

A stream of water flowing horizontally with a speed of 15 m s¯¹ gushes out of a tube of cross-sectional area 10¯² m², and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound ?

Answer 1

Given, the initial velocity of the water stream is 15 m/s and the cross sectional area of the wall that the water hits is 10 −2   m 2 .

In one second, the distance travelled by the water stream is equal to the velocity of the water.

The equation for the volume of water hitting the wall per second is,

V=av

Here, a is the cross sectional area of the tube and v is the speed of the water coming out of the tube.

Substitute the value in the above equation.

V=( 10 −2   m 2 )( 15 m/s ) =15× 10 −2   m 3 /s

The equation for the mass of the water hitting the wall per second is,

M=ρV

Here, ρ is the density of the water.

Substitute the value in the above equation.

M=( 10 3 )( 15× 10 −2 ) =150  kg/s

The initial momentum of the water hitting the wall per second is,

P 1 =MV

Substitute the value in the above equation.

P 1 =( 150  kg/s )( 15 m/s ) =2250  kg⋅m/ s 2

The equation for the final momentum of the water stream is,

P 2 =M V 0

The final velocity is zero so the corresponding value of the momentum is also zero.

The equation to find the value of force exerted is,

F= P 2 − P 1

Substitute the value in the above equation.

F=0−2250 =−2250 N =−2.25× 10 3 N

Thus, the magnitude of force exerted on the wall is 2.25× 10 3 N.