Question

What is the magnitude of the momentum change of two gallons of water (inertia about 7.3 kg ) as it comes to a stop in a bathtub into which it is poured from a height of 2.0 m ?

Answer 1

We know that the change in momentum is equals to the product of force and time that is impulse (  $F \times t$). Therefore, we need to determine the value of that the water is in air by using the second equation of motion,

$s=ut+\frac{1}{2} gt^2$

Here, u is initial velocity which is zero.

$s= \frac{1}{2} gt^2 \\\\ t = \sqrt{\frac{2s}{g} }$.

Thus, impulse

$= F \times \sqrt{\frac{2s}{g} }$

From Newton`s second law,

$F =mg$

Therefore, impulse

$= mg \times \sqrt{\frac{2s}{g} } = m \sqrt{2gs}$

Given,  $m = 7.3 kg$ and $s = 2.0 m$

Substituting these values, we get

Change in momentum = impulse  

$= 7.3 \ kg \sqrt{2 \times 9.8 \ m/s^2 \times 2.0 \ m } = 45 .7 \ Ns$.