Calculate Pressure of Sand on Paper: D=19.6 Pa

[Bgerst103]

Homework Statement

Calculate the pressure of sand grains falling on a 10cm X 10cm sheet of paper. Assume the grains are each 1 mg in mass, the ate pf the sand as it falls is 100 grains per second. The sane is falling from a height of 1m under gravity, and the sheet it falls to is sticky, so that the grains stick after they fall. Calculate the PRESSURE the sand exerts on the sheet of paper each second.

A) 5.0 Pa
B) -5.0 Pa
C) -19.6 Pa
D) 19.6 Pa

Homework Equations

P=F/A

The Attempt at a Solution

I tried finding F by doing mg of 100 grains and dividing that by .01 cm^2 but didn't come up with any of these answers.

 

[Chestermiller]

What is the velocity of each grain of sand when it hits the sheet of paper? What is the change in momentum of each grain of sand when it hits the sheet?

Chet

 

[Bgerst103]

What is the velocity of each grain of sand when it hits the sheet of paper? What is the change in momentum of each grain of sand when it hits the sheet?

Chet

Using Vf^2=Vi^2+2a(DeltaX) i got that the velocity of each grain is 19.6 m/s when it hits. How do I solve for change in momentum and how are these implemented into the solution?

 

[haruspex]

Using Vf^2=Vi^2+2a(DeltaX) i got that the velocity of each grain is 19.6 m/s when it hits. How do I solve for change in momentum and how are these implemented into the solution?

If a 1mg mass changes speed from 19.6m/s to 0, what is the magnitude of its change in momentum?

 

[Chestermiller]

Using Vf^2=Vi^2+2a(DeltaX) i got that the velocity of each grain is 19.6 m/s when it hits. How do I solve for change in momentum and how are these implemented into the solution?

The velocity drops to zero when the grain hits and sticks to the paper. So, what is its change in momentum? (Incidentally, I get a different velocity after it falls 1 meter. I get ~ 4.5 m/s. Did you remember to take the square root?) You are going to be trying to find the rate of change of momentum of the sand grains that hit the paper.

Chet

 

[Bgerst103]

The velocity drops to zero when the grain hits and sticks to the paper. So, what is its change in momentum? (Incidentally, I get a different velocity after it falls 1 meter. I get ~ 4.5 m/s. Did you remember to take the square root?) You are going to be trying to find the rate of change of momentum of the sand grains that hit the paper.

Chet

Oops, you're right. I forgot to take the square root. My professor gave us a bank of equations such as p=mv, pf=pi, deltaP=F(deltaT), where p is momentum. So, from the looks of it, I need to find the time using vf=vi + at to find the change in momentum? If that's right I got ~.46 s. Then to calculate F I did mg and plugged that back into deltaP=F(deltaT) which ends up looking like .000001 (9.8) x .46 = .000005. However, I'm rather confused why calculating mg for each grain of sand and dividing by the area doesn't work. How is momentum incorporated into the equation P= F/A.

 

[haruspex]

Then to calculate F I did mg

Then don't. As pointed out already, you need to calculate the rate of change in momentum of the sand as it strikes the sheet. If a 1mg of sand moving at 4.5 m/s comes to rest, what is its change in momentum? How often does that happen each second? What therefore is the rate of change of momentum of the stream of sand due to its impact with the sheet?

 

[Bgerst103]

The velocity drops to zero when the grain hits and sticks to the paper. So, what is its change in momentum? (Incidentally, I get a different velocity after it falls 1 meter. I get ~ 4.5 m/s. Did you remember to take the square root?) You are going to be trying to find the rate of change of momentum of the sand grains that hit the paper.

Chet

So I calculated the change in momentum to be 0.0443 which I then plugged into P=F/A to get 4.43 Pa. I'm hoping the right answer is 5.0 Pa and that they just rounded up.

 

[Chestermiller]

I get a change in momentum per grain of 0.000001 kg x 4.43 m/s = 0.00000443 kg m /s. If there are 100 grains/second hitting the paper, the rate of change of momentum of the grains is 100/s x 0.00000443 kg m/s = 0.000443 kg m/s²=0.000443 Newtons. This force is spread over an area of 0.01 m², so the pressure from the impact is 0.0443 N/m²= 0.0443 Pa. This doesn't match any of the answers.

In addition to the change in momentum of the grains, their weight is also building up on the paper. For 100 grains/s, the rate of mass buildup is 0.0001 kg/sec. The weight of this buildup is 0.00098 N/sec. The rate of increase of pressure resulting from this weight buildup is 0.00098/0.01 = 0.098 Pa/sec. So, the total pressure after 1 second is 0.143 Pa, and after 2 seconds, it is 0.24 Pa.

These results also do not match the answers provided.

Haruspex, what am I doing wrong?

Chet

 

[haruspex]

I get a change in momentum per grain of 0.000001 kg x 4.43 m/s = 0.00000443 kg m /s. If there are 100 grains/second hitting the paper, the rate of change of momentum of the grains is 100/s x 0.00000443 kg m/s = 0.000443 kg m/s²=0.000443 Newtons. This force is spread over an area of 0.01 m², so the pressure from the impact is 0.0443 N/m²= 0.0443 Pa. This doesn't match any of the answers.

In addition to the change in momentum of the grains, their weight is also building up on the paper. For 100 grains/s, the rate of mass buildup is 0.0001 kg/sec. The weight of this buildup is 0.00098 N/sec. The rate of increase of pressure resulting from this weight buildup is 0.00098/0.01 = 0.098 Pa/sec. So, the total pressure after 1 second is 0.143 Pa, and after 2 seconds, it is 0.24 Pa.

These results also do not match the answers provided.

Haruspex, what am I doing wrong?

Chet

No idea, Chet. Everything in your post looks right to me.

 

[Bgerst103]

I get a change in momentum per grain of 0.000001 kg x 4.43 m/s = 0.00000443 kg m /s. If there are 100 grains/second hitting the paper, the rate of change of momentum of the grains is 100/s x 0.00000443 kg m/s = 0.000443 kg m/s²=0.000443 Newtons. This force is spread over an area of 0.01 m², so the pressure from the impact is 0.0443 N/m²= 0.0443 Pa. This doesn't match any of the answers.

In addition to the change in momentum of the grains, their weight is also building up on the paper. For 100 grains/s, the rate of mass buildup is 0.0001 kg/sec. The weight of this buildup is 0.00098 N/sec. The rate of increase of pressure resulting from this weight buildup is 0.00098/0.01 = 0.098 Pa/sec. So, the total pressure after 1 second is 0.143 Pa, and after 2 seconds, it is 0.24 Pa.

These results also do not match the answers provided.

Haruspex, what am I doing wrong?

Chet

You did everything right. I emailed the professor who said they used 100m rather than 1m for the answers so everything is off by a factor of 100.

 

[Chestermiller]

You did everything right. I emailed the professor who said they used 100m rather than 1m for the answers so everything is off by a factor of 100.

It seems to me this would only account for a factor of 10.

Chet