A liquid of density rho is coming out of a hose pipe of radius a with horizontal speed v and hits a

Q: Why is the mass flow rate $\rho A v$?

Imagine a cylinder of fluid moving at velocity $v$. In one second, a cylinder of length $v$ passes through area $A$. The volume is $A \cdot v$, and multiplying by density $\rho$ gives the mass per second.

Q: Why doesn't the 50% that passes through exert any force?

Force is caused by a *change* in momentum. Since this fluid passes through unaffected, its velocity doesn't change, meaning it transfers no momentum to the mesh.

Q: Why is the change in velocity $2v$ for the bouncing fluid?

Velocity is a vector (it has direction). If the fluid hits the mesh at $+v$ and bounces back at $-v$, the change is $v_{final} - v_{initial} = -v - (+v) = -2v$. The magnitude of this change is $2v$.

Q: Why do we divide the mass flow rate by 4 for the stopping and bouncing parts?

The problem states that 25% of the fluid stops and 25% bounces. 25% is exactly one-fourth ($\frac{1}{4}$) of the total mass flow rate.

Q: What happened to the area $A$ in the final answer?

We calculated the total force, which had an $A$ in the numerator. But the question asked for pressure, which is Force divided by Area. When we divided by $A$, it canceled out!

Question

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\% losses all of its momentum and, 25\% comes back with the same speed. The resultant pressure on the mesh will be

Options:

(A) $\rho v^2$

(B) $\frac{1}{2}\rho v^2$

(C) $\frac{1}{4}\rho v^2$

(D) $\frac{3}{4}\rho v^2$

Sigma Percentile · Accepted Answer

Have you ever held a hose pipe against a wall and felt the pushback? That push is the physical manifestation of momentum transfer. In this fascinating problem, we are looking at a stream of liquid hitting a wire mesh, but with a twist: the liquid splits into three distinct behaviors upon impact. Let's break down the physics step-by-step.

### The Master Equation: Mass Flow Rate

Before we can calculate any forces, we need to know how much liquid is actually hitting the mesh every second. Imagine a cylindrical chunk of water of length $v$ (the distance it travels in one second) and cross-sectional area $A$. The volume of this chunk is $A \cdot v$.

To find the mass, we multiply this volume by the density of the liquid, $\rho$. This gives us the **mass flow rate**, which is the mass of liquid passing a point per unit time:
$$ \frac{dm}{dt} = \rho A v $$
This equation is the beating heart of fluid dynamics problems involving momentum.

### The Ghost Fluid: 50% Passes Through

The problem states that half of the liquid (50%) passes through the mesh completely unaffected.

Think about what this means according to Newton's Second Law. Force is the rate of change of momentum. If the liquid passes through unaffected, its velocity remains $v$. The change in velocity is zero, which means the change in momentum is zero.
$$ \Delta p_{pass} = 0 $$
Therefore, this 50% of the liquid acts like a ghost—it exerts absolutely **zero force** on the mesh.

### The Splatting Fluid: 25% Loses Momentum

Now, let's look at the next 25% of the liquid. The problem tells us it "loses all of its momentum." Physically, this means the liquid hits the mesh and comes to a dead stop.

Its initial velocity was $v$, and its final velocity is $0$. The change in velocity magnitude is simply $v$. To find the force exerted by this fraction, we take 25% (or $\frac{1}{4}$) of our mass flow rate and multiply it by this change in velocity:
$$ F_1 = \frac{1}{4} \left( \frac{dm}{dt} \right) \cdot v $$
Substituting our master equation, we get:
$$ F_1 = \frac{1}{4} (\rho A v) \cdot v = \frac{1}{4} \rho A v^2 $$

### The Bouncing Fluid: 25% Reflects Back

Here is where many students make a critical mistake! The final 25% of the liquid bounces back with the *same speed*.

Velocity is a vector. If the initial velocity is $v$ (to the right), the final velocity after bouncing is $-v$ (to the left). The total change in velocity is:
$$ \Delta v = v - (-v) = 2v $$
Because it bounces, it transfers **twice as much momentum** as the liquid that merely stopped. Let's calculate the force for this bouncing fraction:
$$ F_2 = \frac{1}{4} \left( \frac{dm}{dt} \right) \cdot (2v) $$
$$ F_2 = \frac{1}{4} (\rho A v) \cdot (2v) = \frac{1}{2} \rho A v^2 $$

### The Grand Finale: Total Pressure

We are now ready to find the total pressure on the mesh. First, we find the net force by adding the forces from the splatting and bouncing fractions:
$$ F_{net} = F_1 + F_2 $$
$$ F_{net} = \frac{1}{4} \rho A v^2 + \frac{1}{2} \rho A v^2 = \frac{3}{4} \rho A v^2 $$

But the question asks for **pressure**, not force. Pressure is defined as force per unit area:
$$ p = \frac{F_{net}}{A} $$
When we divide our net force by the area $A$, the area terms beautifully cancel out, leaving us with a pure, elegant expression:
$$ p = \frac{3}{4} \rho v^2 $$

And there we have it! By carefully tracking the momentum of each fraction of the fluid, we've arrived at the correct answer. Always remember: when things bounce, they hit twice as hard!

Select Answer:

Visualized Solution

$M a ss F l o w R a t e$

$U na f f ec t e d L i q u i d$

$S t o pp e d L i q u i d$

$B o u n c in g L i q u i d$

$T o t a l P r ess u r e$

The Sigma Insight: Flow of Fluid

The Master Equation

The Ghost Fluid: 50% Passes Through

The Splatting Fluid: 25% Loses Momentum

The Bouncing Fluid: 25% Reflects Back

The Grand Finale

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