- #1

grandpa2390

- 474

- 14

## 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

[/B]

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

Last edited: