Question

Water flowing through a pipe of area of cross-section \( 2 \times 10^{-3} \, \text{m}^2 \) hits a vertical wall horizontally with a velocity of \( 12 \, \text{m/s} \). If the water does not rebound after hitting the wall, then the force acting on the wall due to water is

• A. \( 24 \, \text{N} \)
• B. \( 144 \, \text{N} \)
• C. \( 288 \, \text{N} \)
• D. \( 72 \, \text{N} \)

Hint:

When a fluid jet impacts a surface, the force exerted by the fluid on the surface can be calculated using the principle of impulse-momentum. If the fluid does not rebound, the change in momentum is simply the initial momentum of the fluid that impacts the surface in a given time interval. The force is then the mass flow rate (\( \dot{m} = \rho A v \)) multiplied by the initial velocity (\( v \)), i.e., \( F = \rho A v^2 \). If the fluid rebounds with some velocity, the change in momentum will be different. Always consider the direction of the velocity vectors for change in momentum calculations.

Answer 1

The Correct Option is C

Step 1: Understand the principle.
The force acting on the wall is due to the change in momentum of the water. According to Newton's second law, force is equal to the rate of change of momentum.
\( F = \frac{\Delta p}{\Delta t} \)
Here, \( \Delta p \) is the change in momentum and \( \Delta t \) is the time interval.
Step 2: Determine the mass of water hitting the wall per unit time.
Let \( \rho \) be the density of water. \( \rho = 1000 \, \text{kg/m}^3 \).
The volume of water flowing per second through the pipe is given by:
Volume per second = Area of cross-section \( \times \) Velocity
\( V_{\text{rate}} = A \cdot v \)
Given: Area \( A = 2 \times 10^{-3} \, \text{m}^2 \)
Given: Velocity \( v = 12 \, \text{m/s} \)
\( V_{\text{rate}} = (2 \times 10^{-3} \, \text{m}^2) \times (12 \, \text{m/s}) = 24 \times 10^{-3} \, \text{m}^3 \, \text{s}^{-1} \).
The mass of water hitting the wall per second (mass flow rate) is:
Mass per second, \( \dot{m} = \rho \cdot V_{\text{rate}} \)
\( \dot{m} = (1000 \, \text{kg/m}^3) \times (24 \times 10^{-3} \, \text{m}^3 \, \text{s}^{-1}) = 24 \, \text{kg/s} \).
Step 3: Calculate the change in momentum per unit time.
Initial momentum of the water hitting the wall per unit mass = \( m \cdot v \).
Since the water does not rebound, its final velocity after hitting the wall is \( 0 \, \text{m/s} \).
Therefore, the final momentum is \( 0 \).
The change in momentum for a mass \( \Delta m \) is \( \Delta p = m_{\text{water}} \cdot v_{\text{initial}} - m_{\text{water}} \cdot v_{\text{final}} = \Delta m \cdot v - \Delta m \cdot 0 = \Delta m \cdot v \).
The rate of change of momentum is \( \frac{\Delta p}{\Delta t} = \frac{\Delta m}{\Delta t} \cdot v = \dot{m} \cdot v \).
Step 4: Calculate the force.
The force \( F \) acting on the wall is equal to the rate of change of momentum of the water:
\( F = \dot{m} \cdot v \)
\( F = (24 \, \text{kg/s}) \times (12 \, \text{m/s}) \)
\( F = 288 \, \text{N} \).