# Water Fall Momentum Question

1. Nov 4, 2008

### asura

1. The problem statement, all variables and given/known data

Water falls at the rate of 250 g/s from a height of 60 m into a 780 g bucket on a scale (without splashing). If the bucket is originally empty, what does the scale read after 2 s?

2. Relevant equations

p=mv
F$$\Delta$$t=$$\Delta$$p

3. The attempt at a solution

So I assumed that the water was already beginning to fill the bucket at t=0, since it cant reach the bucket in 2s.

First I found the velocity of the water right before it fills the bucket...
vf2=vi2+2ax
vf2= 2(9.81 m/s2)(60m)
vf= 34.3 m/s

Then I used the impulse momentum theorem...
F$$\Delta$$t=$$\Delta$$p
F( 2 s) = .500 kg( 0 - 34.3 m/s)
F = 8.58 N

Weight of the bucket is mg, which is 7.65 N...
so 8.58 + 7.65 is 16.2 N

im not sure about this though... can someone double check my work, I only have one try left

Last edited: Nov 4, 2008
2. Nov 5, 2008

Hi asura!