Question

A jet of water hits a flat stationary plate perpendicular to its motion. The jet ejects 500 g of water per second with a speed of $1m{s}^{-1}$. Assuming that after striking, the water flows parallel to the plate, then the force exerted on the plate is:

• A. 500 N
• B. 1.0 N
• C. 0.5 N
• D. 1000 N

Answer 1

Answer: (C)

0.5 N

$v_i=$ initial velocity;
$v_f=$ final velocity$=0$, since water flows parallel to the plate after striking the plate.
$v_i=1$m/s
$v_f=0$
$m=500gm=0.5$kg

$\left|\overrightarrow{F}\right|=\left|\frac{\Delta p}{\Delta t}\right|=\frac{m(\overrightarrow{v_f}-\overrightarrow{v_i})}{\Delta t}=0.5$ kg m/s$=0.5$ N