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 * t`). Therefore, we need to determine the value of that the water is in air by using the second equation of motion,

`s=ut+(1)/(2) gt^2`

Here, u is initial velocity which is zero.

`s= (1)/(2) gt^2 \\\\ t = \sqrt{(2s)/(g) }`.

Thus, impulse

`= F * \sqrt{(2s)/(g) }`

From Newton`s second law,

`F =mg`

Therefore, impulse

`= mg * \sqrt{(2s)/(g) } = m √(2gs)`

Given,
`m = 7.3 kg` and
`s = 2.0 m`

Substituting these values, we get

Change in momentum = impulse

`= 7.3 \ kg √(2 * 9.8 \ m/s^2 * 2.0 \ m ) = 45 .7 \ Ns`.