Question.
A stream of water flowing horizontally with a speed of $15 \mathrm{~m} \mathrm{~s}^{-1}$ gushes out of a tube of cross-sectional area $10^{-2} \mathrm{~m}^{2}$, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?
A stream of water flowing horizontally with a speed of $15 \mathrm{~m} \mathrm{~s}^{-1}$ gushes out of a tube of cross-sectional area $10^{-2} \mathrm{~m}^{2}$, 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:
Speed of the water stream, $v=15 \mathrm{~m} / \mathrm{s}$
Cross-sectional area of the tube, $A=10^{-2} \mathrm{~m}^{2}$
Volume of water coming out from the pipe per second,
$V=A v=15 \times 10^{-2} \mathrm{~m}^{3} / \mathrm{s}$
Density of water, $\rho=10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
Mass of water flowing out through the pipe per second $=\rho \times V=150 \mathrm{~kg} / \mathrm{s}$
The water strikes the wall and does not rebound. Therefore, the force exerted by the water on the wall is given by Newton’s second law of motion as:
$F=$ Rate of change of momentum $=\frac{\Delta P}{\Delta t}$
$=\frac{m v}{t}$
$=150 \times 15=2250 \mathrm{~N}$
Speed of the water stream, $v=15 \mathrm{~m} / \mathrm{s}$
Cross-sectional area of the tube, $A=10^{-2} \mathrm{~m}^{2}$
Volume of water coming out from the pipe per second,
$V=A v=15 \times 10^{-2} \mathrm{~m}^{3} / \mathrm{s}$
Density of water, $\rho=10^{3} \mathrm{~kg} / \mathrm{m}^{3}$
Mass of water flowing out through the pipe per second $=\rho \times V=150 \mathrm{~kg} / \mathrm{s}$
The water strikes the wall and does not rebound. Therefore, the force exerted by the water on the wall is given by Newton’s second law of motion as:
$F=$ Rate of change of momentum $=\frac{\Delta P}{\Delta t}$
$=\frac{m v}{t}$
$=150 \times 15=2250 \mathrm{~N}$
