What force is exerted on a stationary flat plate held perpendicular to a jet of water as shown in Fig. 8-5? The horizontal speed of the water is $80 \mathrm{~cm} / \mathrm{s}$, and $30 \mathrm{~mL}$ of the water hit the plate each second. Assume the water moves parallel to the plate after striking it. One milliliter (mL) of water has a mass of $1.00 \mathrm{~g}$.
This question deals with speed, mass, time, and force, and that suggests impulse-momentum and Newton's Second Law. The plate exerts an impulse on the water and changes its horizontal momentum. The water exerts a counterforce on the plate. Taking the direction to the right as positive,
$(\text { Impulse })_{x}=$ Change in $x$ -directed momentum
$$
F_{x} \Delta t=\left(m v_{x}\right)_{\text {final }}-\left(m v_{x}\right)_{\text {initial }}
$$
Let $t$ be $1.00 \mathrm{~s}$ so that $m$ will be the mass that strikes in $1.00 \mathrm{~s}$, namely $30 \mathrm{~g}$. Then the above equation becomes
$$
F_{x}(1.00 \mathrm{~s})=(0.030 \mathrm{~kg})(0 \mathrm{~m} / \mathrm{s})-(0.030 \mathrm{~kg})(0.80 \mathrm{~m} / \mathrm{s})
$$
from which $F_{x}=-0.024 \mathrm{~N}$. This is the force exerted by the plate on the water. The law of action and reaction tells us that the jet exerts an equal but opposite force on the plate.