Science and Technology Quiz 3

uppi · Apr 7, 2009

dutchfire said:
80 GeV?

It'd be 200 GeV of energy if they hit head on with the same speed. ymc^2=200 GeV -> y = 200 -> v=0.9999875 c in the original reference system. Then we Lorentztransform that with another 0.9999875 c. That gives 1 - 7.81261722 × 10^-11 c. That gives a y of 1 / sqrt(1 - (((0.9999875 + 0.9999875) / (1 + (0.9999875^2)))^2)) = 79 999.3998 = 80000, so the energy is 80000 GeV=80 TeV

80 TeV is right.

I calculated it with the four-momentum and the conservation of its inner product (note that c=1):

In the center of mass system the momentum is zero, and the energy is just the mass, so that the product is (400GeV)^2. This has to be conserved.

in the lab system the first particle has (E,p) and the second has (m=1GeV,0). Adding them together and taking the product results in E^2 + 2E*m + m^2 - p^2. E^2 - p^2 is m^2, resulting in 2m^2 which can be ignored, because they're small.

So because of the conservation:

(400GeV)^2 = 2E*m -> E = (400GeV)^2 / (2*1GeV) = 80 TeV

The last formula is for all highly relativistic collisions where the target is stationary. Because it is quadratic, the required energy rises fast compared to the case where two particles are colliding with the same speed, where it rises linearly with the energy.

Your turn.

dutchfire · Apr 7, 2009

Assume we have a sphere with a uniformly distributed charge Q>0. We enter an electron somewhere in the sphere. Assume the electron can move inside the sphere without any friction or other forces except the electromagnetic force working on it. Describe it's motion.

(Yeah, the assumptions are ridiculous)

ParadigmShifter · Apr 7, 2009

I'd say the motion would be most accurately described as bendy.

dutchfire · Apr 7, 2009

ParadigmShifter said:
I'd say the motion would be most accurately described as bendy.

You can do better than that!

Silly mathematicians :rolleyes:

uppi · Apr 7, 2009

dutchfire said:
Assume we have a sphere with a uniformly distributed charge Q>0. We enter an electron somewhere in the sphere. Assume the electron can move inside the sphere without any friction or other forces except the electromagnetic force working on it. Describe it's motion.

(Yeah, the assumptions are ridiculous)

is the charge uniformly distributed on the surface of the sphere or in the volume of the sphere? And what material is inside the sphere?

thetrooper · Apr 7, 2009

My understanding is that the charge is uniformly distributed on the surface of the sphere...
And perfect vacuum inside the sphere.

dutchfire · Apr 7, 2009

uppi said:
is the charge uniformly distributed on the surface of the sphere or in the volume of the sphere? And what material is inside the sphere?

In the volume of the sphere (so the sphere is made of an isolator)

peter grimes · Apr 7, 2009

I'd say the electron would come to orbit the center of the sphere, where the sum of the charge is equal all around it... but it would be pretty bendy getting there, sort of like dropping a marble into a curved funnel

Mise · Apr 7, 2009

I don't think it will do anything.

dutchfire · Apr 7, 2009

Hint: According to Gauss, to electromagnetic force (and thus the acceleration) goes with the charge inside a Gaussian surface, divided by the surface area of said Gaussian surface. How do these two quantities depend on the distance r from the center of the sphere?

aaglo · Apr 8, 2009

It leaves the sphere and destroys the universe?

Mise · Apr 8, 2009

Well if I'm wrong and it does do something, then it's motion will be simple harmonic, oscillating between +/-x where x is distance between initial position and the centre of the sphere. That's my other guess, I suppose.

dutchfire · Apr 8, 2009

Mise said:
Well if I'm wrong and it does do something, then it's motion will be simple harmonic, oscillating between +/-x where x is distance between initial position and the centre of the sphere. That's my other guess, I suppose.

d^2 r / dt^2 = a = F / m = E*q_electron/m= (Gauss) 4 pi k_e q_in*q_electron/ (A*m)= 4 pi k_eQ*r^3*q_electron/(R^3 * 4pi r^2*m)=k_eQq_electron*r/(R^3).

The second derivative of the distance r to the center of the sphere is equal to a constant times r, so the particle will oscillate harmonically.

Mise · Apr 9, 2009

Oh! Umm... Balls. Easy one then... If a stock price rises by 10% then falls by 10% then falls by 10% then rises by 10% how much money will I have gained?

First prize goes to the person who works it out without any electronic device.

ParadigmShifter · Apr 9, 2009

1.1*0.9*0.9*1.1 = 0.9801

So you will have gained nothing but lost 1.99% of your initial outlay.

Mise · Apr 9, 2009

I see you fell into my cunning trap!

...Oh no wait you got it right. Your go!

ParadigmShifter · Apr 9, 2009

Another maths one then.

Give the necessary and sufficient condition(s) such that if n is an integer > 0, n is a sum of 2 square integers

(i.e. n = a*a + b*b for integer a, b).

EDIT: Changed condition to conditions. There are 2 cases (or more if you phrase them differently).

Mise · Apr 9, 2009

"n = a*a + b*b for integer a, b" <-- surely that's the necessary and sufficient condition

ParadigmShifter · Apr 9, 2009

Come on you can do better than that

1, 2, 4, 5, 8, 9, 10, 13, ...

dutchfire · Apr 9, 2009

If you can divide it by 3, it is also divisible by 9.

Science and Technology Quiz 3

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Furioso!

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