Question

Sphere stuff

Answer 1

@Kainui

Answer 2

Last physics problem haha

Answer 3

I don't really know what it means to have an impact parameter of R, is there some kind of formula?

Answer 4

The problem is, I think this professor makes up his own problems, and he NEVER has any examples in class so we have to do self research a lot of the times.

Answer 5

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/impar.html I just googled impact parameter lol

Answer 6

Because I can't find such problems/ questions really related in the textbook.

Answer 7

Well it is 4:37am here and I am quite hungry and tired. I might come back in an hour and check this out but it reminds me of the derivation for the beer-lambert law.

Answer 8

It's about the same here, I think I'll call it a day as well, still got a whole day for this problem, will you be on later tomorrow?

Answer 9

Hi @phi could you help me out with this problem?

Answer 10

I'm not sure what kind of scattering they want, Compton scattering...? But that wouldn't make much sense right.

Answer 11

a thickness of 10 cm. This thickness exactly matches the diameter of the spheres.
this means the spheres form only one layer. They have a radius of 5 cm = 0.05 m

The "surface area" of a sphere is the area it presents looking straight-on, i.e. the area of a circle with radius 0.05 m = pi*.0025 m^2

They say " hard sphere scattering...greater than R, the radius of the sphere, then no scattering takes place"
I think that means there is no scattering unless a particle "hits" a sphere (gets with 5 cm of its center)

Answer 12

50% scattering means the sum of the areas of the spheres is 50% of the total area
(i.e. 1/2 of 1 m^2)

we have
N* pi * 0.05^2 = 0.5 m^2
solve for N

Answer 13

Mhm alright, I think I understand it somewhat,

\[N = \frac{ 0.5m^2 }{ 0.05^2*\pi }\]

Answer 14

Hey thanks @phi but there's another part haha, sorry.

The question asks, what fraction of the bullets striking the box are scattered at an angle greater than 90 degrees.

So would I just multiply that by sin90?

Answer 15

Oh wait, it's asking for greater than 90 degrees

Answer 16

The 0.05^2 in the formula N*pi*0.05^2 = 0.5m^2 that's the area of the circle right?

Answer 17

They don't say how big the bullets are. I assume they are "small" and we can treat them as point particles.
I am making this up.... but I am thinking it would reflect like this: