Pressure Problem: Solving for Average Pressure

[grandpa2390]

Homework Statement

During a hailstorm, hailstones with average mass of 2 grams and speed of 15## \frac {m}{s} ## strike a window pane at a 45 degree angle. The area of the window is .5##m^2## and the hailstone hit at a rate of 30 per second. What is the average Pressure do they exert on the window?

Homework Equations

##PV = Nmv^2_x##
N = number of particles

P = ## - \frac{\frac{Δv_x}{Δt}}{A}##

The Attempt at a Solution

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I solved it using the first formula. I solved that formula for P, and then plugged in everything because it seemed like I was given all of the info to use that formula.
##N = \frac{30}{s}##
m = 2g
v_x = 15*cos(45)
V = 1m * .5m^2 = .5m^3

force in the numerator = 636 g * m/s^2
area (or Volume?) in denominator = .5m^3so that
##P = \frac{\frac{30}{s}*2g*10.6066\frac{m}{s}}{.5m^3}##
which is:
##P = \frac{30*2g*10.6066\frac{m}{s^2}}{.5m^3}##

or 1272

chegg is giving a completely different answer and method that doesn't make much sense to me and could be wrong.

am I on the right track? Can I get a nudge into the right direction? can you help me with my units?

edit: I skipped down to the end and in the end it looks like we got the same answer. The only difference is that his/her velocity is 21 m/s and mine is 10.606.
for some reason, they multiplied the velocity I got by 2.

edit: In their method, F = change in velocity which uses a 2*v in calculating the acceleration since the change in acceleration is final - initial (positive - negative)

but in the method that I used, the velocity square is used. I get all of the way to the end and get the same formula, but my velocity is half of what it should be...

I lied. My velocity is also squared. so the issue is now more complicated. I am missing a factor of 2 and my velocity is squared

 

[gneill]

The hailstones are striking the window and bouncing off of it. In doing so they are changing their direction of travel. You've deduced that it's the component of their velocity that is normal to the window that is involved. What's the magnitude of that velocity change? It might help to draw the vectors.

 

[grandpa2390]

The hailstones are striking the window and bouncing off of it. In doing so they are changing their direction of travel. You've deduced that it's the component of their velocity that is normal to the window that is involved. What's the magnitude of that velocity change? It might help to draw the vectors.

I understand what you are saying I think. Change in velocity is twice the velocity. Because the final minus negative initial is 2 times. But the formula I tried using didn't ask for the change in the velocity. Am I misunderstanding the formula?

 

[grandpa2390]

The hailstones are striking the window and bouncing off of it. In doing so they are changing their direction of travel. You've deduced that it's the component of their velocity that is normal to the window that is involved. What's the magnitude of that velocity change? It might help to draw the vectors.

Btw, thnx for your help :) . And your patience.

 

[gneill]

The formula calls for the velocity squared. Not the change in velocity.

Okay, I'm not understanding the applicability of your formula to the problem at hand. The window represents a flat surface and not a volume, yet you've introduced a dimension of 1 m from somewhere:

V = 1m * .5m^2 = .5m^3

Where does that 1 m come from?

The "traditional" way to approach this type of problem is to look to the change in momentum that occurs when the particles are reflected from the surface and identify the force with the change in momentum of the particles with respect to time, F = Δp/Δt.

 

[grandpa2390]

Okay, I'm not understanding the applicability of your formula to the problem at hand. The window represents a flat surface and not a volume, yet you've introduced a dimension of 1 m from somewhere:
Where does that 1 m come from?

The "traditional" way to approach this type of problem is to look to the change in momentum that occurs when the particles are reflected from the surface and identify the force with the change in momentum of the particles with respect to time, F = Δp/Δt.

Because the formula I listed above, is PV=...
Solving for P you get (...) / V

And the final formula for both methods looks the same except the velocity.

I have accepted the method used on Chegg, I am just wondering why this formula doesn't work here.

I can't throw it away because it is the formula that is reached by the other method. It is just the velocity that is different. I really want to understand why it doesn't work

 

[gneill]

I have no knowledge of what occurred on Chegg or what problem they were addressing. But I don't recognize any applicable volume in the present problem. As far as I can tell, a formula like PV = (something)(velocity)² looks suspiciously like it would apply the the energy contained in a volume of gas.

 

[grandpa2390]

I have no knowledge of what occurred on Chegg or what problem they were addressing. But I don't recognize any applicable volume in the present problem. As far as I can tell, a formula like PV = (something)(velocity)² looks suspiciously like it would apply the the energy contained in a volume of gas.

Essentially the formula comes from the book where the problem is gas molecules hitting a piston. It is derived from that situation. Maybe it doesn't seem like it but being inside a volume makes a difference

 

[gneill]

Essentially the formula comes from the book where the problem is gas molecules hitting a piston. It is derived from that situation. Maybe it doesn't seem like it but being inside a volume makes a difference

Okay, but I can't see the justification for introducing an arbitrary volume here. It doesn't relate to anything physical in the problem.

If I might suggest, go with the change in momentum approach.

 

[grandpa2390]

Okay, but I can't see the justification for introducing an arbitrary volume here. It doesn't relate to anything physical in the problem.

If I might suggest, go with the change in momentum approach.

Well the way they did it is starting with pressure and just break down every unit and derive the formula for the problem .

So pressure equals force over area. We know the area. Force equals mass times acceleration. We know The mass. Acceleration is the change in velocity over the change in time. And so the change in velocity is Final minus initial which gives you two times the velocity.

Then multiply that by 30 for the 30 pieces of hail and that gives you the answer.What is the change in momentum approach

I am in intermediate level physics. As my professor in mechanics said, it is time to get out of the plug and chug mentality . This served as a reminder.
Soln must be derived

 

[gneill]

Well the way they did it is starting with pressure and just break down every unit and derive the formula for the problem .

While the form of the required relationship can be obtained that way (unit analysis), it doesn't account for scaling factors (numerical constants) that might be involved. For example, kinetic energy is given by KE = (1/2)mv². Unit analysis will reveal the form mv², but not the (1/2) constant.

So pressure equals force over area. We know the area. Force equals mass times acceleration. We know The mass. Acceleration is the change in velocity over the change in time. And so the change in velocity is Final minus initial which gives you two times the velocity.

Then multiply that by 30 for the 30 pieces of hail and that gives you the answer.What is the change in momentum approach

It's almost what you've written above, only its not the number of pieces of hail that's important, it's the number of pieces of hail per second: the rate of delivery. The change in momentum approach looks at the change in momentum for a single particle in the stream of particles impacting the surface, then multiplies by the rate at which they are arriving to determine the overall rate of change of momentum per second. As you know, Δp/Δt yields the overall force (in fact it's the definition of force!).

I am in intermediate level physics. As my professor in mechanics said, it is time to get out of the plug and chug mentality . This served as a reminder.
Soln must be derived

It's a good idea to get away from plug and chug. Understanding the applicability and limitations of "canned" formulas is an important step to progressing.

 

[grandpa2390]

While the form of the required relationship can be obtained that way (unit analysis), it doesn't account for scaling factors (numerical constants) that might be involved. For example, kinetic energy is given by KE = (1/2)mv². Unit analysis will reveal the form mv², but not the (1/2) constant.

The book discussed this. in the derivation for the solution of one particle hitting a piston, the solution came to be
##PV = Nmv^2_x##
since the ideal gas law say PV = NkT we can substitute NkT for PV. after cancelling the Ns we get
##kT = mv^2_x##
so
##\frac{1}{2}kT = \frac{1}{2}mv^2_x##
so we have the average translational kinetic energy.

for the average pressure on the piston, the formula derived was PV = Mv^2, and then for every molecule you just add up the Mv^2 for each one.

is that what you mean by scaling. Otherwise, we haven't gotten to scaling factors. I am sure that will come at a more advanced level (maybe farther in the book)

It's almost what you've written above, only its not the number of pieces of hail that's important, it's the number of pieces of hail per second: the rate of delivery. The change in momentum approach looks at the change in momentum for a single particle in the stream of particles impacting the surface, then multiplies by the rate at which they are arriving to determine the overall rate of change of momentum per second. As you know, Δp/Δt yields the overall force (in fact it's the definition of force!).

It's a good idea to get away from plug and chug. Understanding the applicability and limitations of "canned" formulas is an important step to progressing.

That's what it was. acceleration is Δv / Δt
Δt is 1/30 s :)

 

[grandpa2390]

@gneill

I don't know much much about derivations, but I think it is probably like programming. You start out small writing a completely unsophisticated program that does one lame thing. And as your education proceeds, you can write more and more sophisticated programs.

Deriving formulas is probably like that. As you learn more and more, you can probably write more and more sophisticated formulas. ?

 

[gneill]

@gneill

I don't know much much about derivations, but I think it is probably like programming. You start out small writing a completely unsophisticated program that does one lame thing. And as your education proceeds, you can write more and more sophisticated programs.

Deriving formulas is probably like that. As you learn more and more, you can probably write more and more sophisticated formulas. ?

You definitely get better with practice and adding more mathematical tools to your toolkit let's you accomplish more complex manipulations and "describe" more things mathematically. Just like programming, you start with simple, familiar things like velocity and distance calculations and then layer on concepts like vectors.

When the basic math becomes second nature you start to get a feeling for what the math represents in the physics and vice versa. Rather than going immediately to numbers and plugging them into canned formulas you'll start to think and work in terms of symbols (variables). Often a lot of calculation work can be bypassed with a bit of algebra. What is most important though is getting a feeling for the fundamental concepts of physics, being able to recognize which of them are in play in a given scenario and being comfortable shaping the math to describe it.

 

[grandpa2390]

Thanks!
I hope to get better. One of these problems is asking me to do something Mathematically that is going over my head. the sad part is that the author believes I should be able to do it by now... Maybe one day in the future I will be able to come back and laugh about how it is so simple. I would ask for help, but I am too ignorant at this point to get help. I would just be asking for the soln. ;)

Thanks! I'll be back again... and again... and again ;)