Question

A stream of water flowing horizontally with the speed of \[{\mathbf{15}}{\text{ }}{\mathbf{m}}{{\mathbf{s}}^{ - {\mathbf{1}}}}\] gushes out of a tube of cross-sectional area\[{\mathbf{1}}{{\mathbf{0}}^{ - {\mathbf{2}}}}{\text{ }}{{\mathbf{m}}^{ - {\mathbf{2}}}}\], and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water? Assuming it does not rebound.
A. \[2250\;{\text{N}}\]
B. \[2000\;{\text{N}}\]
C. \[1500\;{\text{N}}\]
D. \[1000\;{\text{N}}\]

Answer 1

Hint: The formula for volume per second is obtained by,
\[V = Av\]. Mass of water flowing out through the pipe per second is obtained using the formula,
Mass = Density \[ \times \] Volume. Force is the rate of change of momentum.
\[F = \dfrac{P}{t}\]. Momentum p is defined as
\[p = mv\].

Complete step by step solution:
Given,
The horizontal speed of the stream is,
\[v = 15\;{\text{m/s}}\]
Cross-section area of the tube,
\[A = {10^{ - 2}}\;{{\text{m}}^{\text{2}}}\]
We know that, the density of water,
\[\rho = 1{0^3}\;{\text{kg/}}{{\text{m}}^{\text{3}}}\]

Now, the volume of water coming out from the pipe per second is obtained using the formula,
\[V = Av\] …… (i)
Now substitute the values of \[A\] and \[v\] in equation (i).
Therefore,
\[
  V = Av \\
   = {10^{ - 2}}\;{{\text{m}}^{\text{2}}} \times 15\;{\text{m/s}} \\
  {\text{ = 15}} \times {\text{1}}{{\text{0}}^{ - 2}}\;{{\text{m}}^{\text{3}}}{\text{/s}} \\
\]

Again the mass of water flowing out through the pipe per second is obtained using the formula,
Mass = Density \[ \times \] Volume …… (ii)
Now substitute the values of density, \[\rho = 1{0^3}\;{\text{kg/}}{{\text{m}}^{\text{3}}}\] and volume, \[V{\text{ = 15}} \times {\text{1}}{{\text{0}}^{ - 2}}\;{{\text{m}}^{\text{3}}}{\text{/s}}\] in equation (ii)

\[
  {\text{Mass}} = 1{0^3}\;{\text{kg/}}{{\text{m}}^{\text{3}}} \times {\text{15}} \times {\text{1}}{{\text{0}}^{ - 2}}\;{{\text{m}}^{\text{3}}}{\text{/s}} \\
  {\text{ = 150}}\;{\text{kg/s}} \\
\]
Because the water does not detach from the wall, the force exerted on the wall by the water is given by:
\[F = \] Rate of change of momentum, or,
\[F = \dfrac{P}{t}\] …… (iii)
Now, Linear momentum (brevity momentum) is calculated as the measure of the mass of a system, multiplied by its velocity. In signs linear momentum p is defined as \[p = mv\], where \[m\] is the system mass and \[v\] is its velocity.

Now, rearrange the equation (iii) as,
\[F = \dfrac{{mv}}{t}\] …… (iv)
Now
Let us consider the time as \[t = 1\;{\text{s}}\]
Place the values of \[m\], \[v\], and \[t\] in equation (iv)

Therefore,
\[
  F = \dfrac{{mv}}{t} \\
   = \dfrac{{150\;{\text{kg/s}}\; \times {\text{15}}\;{\text{m/s}}}}{{1\;{\text{s}}}} \\
   = 2250\;{\text{N}} \\
 \]

Hence, the required exerted force is\[2250\;{\text{N}}\].
Hence, option A is correct.

Note: In this question we are asked to determine the force. For this we use the formula, \[F = \dfrac{{mv}}{t}\]. Linear momentum is calculated as the measure of the mass of a system, multiplied by its velocity. In signs linear momentum p is defined as \[p = mv\]. The volumetric flow rate in physics and engineering, in particular fluid mechanics, is the volume of fluid that moves per unit time; expressed by \[Q\]. SI unit is given by cubic metres by second.