How Does Falling Water Affect Scale Readings?

[asura]

Homework Statement

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?

Homework Equations

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

The Attempt at a Solution

So I assumed that the water was already beginning to fill the bucket at t=0, since it can't reach the bucket in 2s.

First I found the velocity of the water right before it fills the bucket...
v_f²=v_i²+2ax
v_f²= 2(9.81 m/s²)(60m)
v_f= 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

 

[tiny-tim]

Hi asura!

First I found the velocity of the water right before it fills the bucket...
v_f²=v_i²+2ax
v_f²= 2(9.81 m/s²)(60m)
v_f= 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

Yes, your v is correct.

But your method after that is completely wrong.

The impulse momentum theorem is the correct principle, but you should use it to find the instantaneous force (because the scale only measures instantaneous force, not total force).

Then add the weight of the water already in the bucket

(and don't forget to convert from N back into g )