A liquid of density $$\rho$$ is coming out of a hose pipe of radius $$a$$ with horizontal speed $$v$$ and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% loses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be:

We have a horizontal jet emerging from a hose of radius $$a$$. The cross-sectional area of the jet is

$$A=\pi a^{2}\;.$$

Density of the liquid is $$\rho$$ and its speed is $$v$$, so the total mass that leaves the pipe per second (the mass flow rate) is, by definition,

$$\dot m=\rho A v=\rho\pi a^{2}v\;.$$

The mesh divides this stream into three parts:

• 50 % of the mass passes straight through without any change in velocity.
• 25 % of the mass is completely stopped, that is, its final velocity becomes $$0$$.
• 25 % of the mass reverses direction and returns with the same speed $$v$$ (so its final velocity is $$-v$$, opposite to the incident direction).

The force exerted on the mesh equals the rate of change of linear momentum of the fluid. In symbols,

$$F=\sum \dot m_i\,(v_{f,i}-v_{i,i})$$

where the summation is taken over the three portions of the jet. We now evaluate each contribution separately.

1. 50 % passes through unchanged

The mass flow in this part is $$\dot m_1=0.50\,\dot m$$, the initial velocity is $$v$$, and the final velocity is also $$v$$. Therefore

$$v_{f,1}-v_{i,1}=v-v=0\quad\Longrightarrow\quad \dot m_1\,(v_{f,1}-v_{i,1})=0\;.$$

This part gives no force.

2. 25 % is stopped

Here the mass flow rate is $$\dot m_2=0.25\,\dot m$$. Initial velocity is $$v$$ and final velocity is $$0$$, hence

$$v_{f,2}-v_{i,2}=0-v=-v\;.$$

The momentum change per second for this portion is

$$\dot m_2\,(v_{f,2}-v_{i,2})=0.25\,\dot m\,(-v)=-0.25\,\rho A v^{2}\;.$$

The negative sign shows that the fluid loses forward momentum; the mesh receives an equal and opposite impulse, so the force magnitude from this part is

$$F_2=0.25\,\rho A v^{2}\;.$$

3. 25 % rebounds with speed $$v$$

This portion has $$\dot m_3=0.25\,\dot m$$, initial velocity $$v$$, and final velocity $$-v$$. Hence

$$v_{f,3}-v_{i,3}=-v-v=-2v\;,$$

and the momentum change per second is

$$\dot m_3\,(v_{f,3}-v_{i,3})=0.25\,\dot m\,(-2v)=-0.50\,\rho A v^{2}\;.$$

Again the negative sign merely signals direction; the force on the mesh has magnitude

$$F_3=0.50\,\rho A v^{2}\;.$$

Total force on the mesh

Adding the magnitudes (the contributions act in the same direction against the mesh), we get

$$F=F_2+F_3=\left(0.25+0.50\right)\rho A v^{2}=0.75\,\rho A v^{2}=\frac{3}{4}\,\rho A v^{2}\;.$$

Pressure on the mesh

Pressure is force divided by the area over which the force acts. The jet area is $$A=\pi a^{2}$$, so

$$P=\frac{F}{A}=\frac{0.75\,\rho A v^{2}}{A}=0.75\,\rho v^{2} =\frac{3}{4}\,\rho v^{2}\;.$$

Hence, the correct answer is Option B.