Question

little doubt

Answer 1

a parallel beam of nitrogen molecules moving with velocity v=400m/s impinges on a wall at an angle \(\theta\)=30 degrees to its normal. The concentration of molecules in the beam is n=\(9 \times 10^{18} cm^{-3}\). Find the pressure exerted by the beam on the wall, assuming that collisions are perfectly elastic.

Answer 2

i know how to do this type of problems but i have a little doubt
so i did it like this-

heres the diagram- http://prntscr.com/bipvkp

lets say the area of cross section of that cylinder is \(A\)
so volume of the cylinder becomes \(Av \color{red}{cos\theta }\)

number of molecules striking \(A\) every second= \(Avcos\theta \times n\)

change in momentum per molecule= \(2mvcos \theta\)

pressure=force/area
pressure=\(\large \frac{(2mvcos\theta)(acos\theta \times n)}{Acos^2 \theta}\)


solving this we get our answer

Answer 3

so my doubt is-
->why did we multiply that "\(cos\theta\)" in the volume??
->why did we multiply the area in the final equation with "\(cos^2\theta\)" ??

Answer 4

@Michele_Laino @agent0smith

Answer 5

Cosine gives the horiz. comp. of momentum. The vert. comp. applies no pressure to the wall.

Answer 6

yes but the number of molecules that will pass through area A is 1 sec should be=Avn right? :/ why is there a need to take a component
andd why is there a "\(cos^{\color{red}2}\theta\)" in the area thing in the pressure equation

Answer 7

wait a sec
while calculating pressure we have to consider the force which is perpendicular to the surface! soo just a min lemme think

Answer 8

The circle for area would be oriented parallel to the wall. Which means the relevant speed is v*cos theta.

Answer 9

Think of it like an oblique cylinder

Answer 10

hey i made a typo here - http://prntscr.com/birmno

it must be this-

pressure=force/area =\(\large \frac{(2mvcos\theta)(\color{red}{Av}cos\theta \times n)}{Acos^2\theta}\)

okay i get the numerator part of this^ expression

but i don't know why they multiplied the denominator with that extra "\(cos\theta\)"
well \(Acos\theta\) is the area where the particles will collide with the wall so why did they write it as \(Acos^2\theta\)

Answer 11

Yeah I can't figure that part out. Because that will cancel off all cosines, making it seem like pressure does NOT depend on the angle, when it clearly does! Intuitively it has to :/

Answer 12

true!!!

Answer 13

Yeah that makes no sense. If cosines cancel, it implies that pointing the beam directly at the wall gives the same pressure as pointing it like 45 degrees to the wall. You know that's not true, imagine spraying a hose at a wall.

Answer 14

well another thing popping in my mind is-
we multiplied the change in momentum with the number of particles "hitting the wall"

we know that the change in momentum is \(2mvcos\theta\)

now i'm thinking that number of particles hitting the wall must not be \(Avcos\theta \times n\) .. my reasoning-

we know that the concentration of particles in beam is n and velocity with which they are travelling is v
now we multiply nv with area upon which the beam is falling
that area must look like this- http://prntscr.com/birvpe
so the actual area is more than \(Acos\theta\) therefore \(Avcos\theta \times n\) is wrong?

Answer 15

Yeah like I said earlier, you should think of this as an oblique cylinder.

Answer 16

umm something like this- http://prntscr.com/bis0m7 ?

Answer 17

Yep

Answer 18

but if we are considering it to be like that then why do we even take "\(cos\theta\) " in this expression- \(Avcos\theta \times n\)

here A is the area and the area vector and velocity vectors are already in the same direction so we don't need to take components so therefore it must be \(Avn\)

Answer 19

vcos theta gives the "height" of the cylinder. Vol of on oblique cylinder is area of base times the vert. height.

Answer 20

okay! i get it now :))

Answer 21

thanks so much!! :D

Answer 22

:D but what about the denominator?