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

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.