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Q5.28:
A stream of water flowing horizontally with a speed of $15 m s^{-1}$ gushes out of a tube of cross-sectional area $10^{-2} m^2$, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound ?

Solution :

Step 1: Given Variables & Theoretical Foundation

To determine the force exerted by a stream of water on a wall, we apply Newton’s Second Law of Motion in terms of momentum [Per Newton's Second Law of Motion, $F = \frac{dp}{dt}$].

  • Velocity of water ($v$): $15 \, ms^{-1}$
  • Cross-sectional area ($A$): $10^{-2} \, m^2$
  • Density of water ($\rho$): $10^3 \, kg \, m^{-3}$ [Standard physical constant]
  • Final velocity of water after impact ($v_f$): $0 \, ms^{-1}$ [Since the water does not rebound]

Step 2: Calculating the Mass Flow Rate

The mass ($m$) of the water hitting the wall per unit time ($t$) is determined by the volume of the water stream passing through the cross-section per second, multiplied by the density.

Volume of water per second ($V/t$) = $A \times v$

Mass flow rate ($\frac{dm}{dt}$) = $\rho \times A \times v$

Substituting the given values:

$\frac{dm}{dt} = (10^3 \, kg \, m^{-3}) \times (10^{-2} \, m^2) \times (15 \, ms^{-1})$

$\frac{dm}{dt} = 10 \times 15 = 150 \, kg \, s^{-1}$

Step 3: Determining the Force Exerted

According to the impulse-momentum theorem, the force ($F$) exerted on the wall is equal to the rate of change of momentum of the water stream.

$F = \frac{dp}{dt} = \frac{d(mv)}{dt} = v \frac{dm}{dt}$

Since the water stops upon hitting the wall, the change in velocity is $(\Delta v) = v - 0 = v$.

$F = \left(\frac{dm}{dt}\right) \times v$

$F = 150 \, kg \, s^{-1} \times 15 \, ms^{-1}$

$F = 2250 \, N$

Step 4: Visual Representation of the Impact

v = 15 m/s Wall

Step 5: Analytical Summary

Parameter Value
Density ($\rho$) $10^3 \, kg/m^3$
Mass Flow Rate ($dm/dt$) $150 \, kg/s$
Force ($F$) $2250 \, N$

Final Solution: The force exerted on the wall by the impact of water is 2250 N.


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