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2. Originally Posted by the_sensai
hey folks,
im really stuck on this question and its been bugging me all week and its due in on monday. So any help on this would be great. Okay so heres the problem(dont worry its not physics its just a story question):

A sub- atomic particle is travelling in a straight line through a tubular cloud-chamber that is 28cm long. The particle is subjected to an electromagnetic field that reverses the direction of the particle so that it disappears from the cloud chamber. Almost instantaneously with the disappearence of the first particle, a second particle enters the chamber from the other direction. This particle is subjected to the same electromagnetic field so that its direction is also reversed and it disappears from the cloud chamber after a period of time along the same trajectoy as the first particle would have done if the electromagnetic field had not been applied. It is possible to model approximately these events with the function:

s(t) = 4t + (2/t-3) + (2/3)

where s represents the posistion of either particle in the tubular cloud chamber measured in centimetres, while t represents the time in nano-seconds.

right finally for the questions,

a) when is the time of transistion between the disappearence of the first particle and the appearence of the second one?

b) when does the first particle actually leave the cloud chamber according to the model
[HINT: consider s(t) = 0]

c)Both particles change directions. When do they do this?
[HINT : use differential calculus]

d) what, eventually, is the function that describes the initial trajectory of the first particle which is taken up by the second one?

and thats it. I know it seems an essay but its not i had another 9 questions just like this one. I just could not get my head around this. Its really annoyed me in a way and any help on this would be great, and a load of my mind.

thankyou guys hope you can help.
s(t) = 4t + 2/(t-3) + (2/3)

(a) The first particle shoots off to -infinity ar t=3 (since this is the singular
point in the equation for its position). This is the time required for this part.

Note with this equation for position at t=3 the second particle flies in from
+ infinity, and this does not make much sense to me.

(b) The first particle leaves the chamber when s(t) = 0 for the second time,
so we need to solve:

4t + 2/(t-3) + (2/3) = 0

multiply through by (t-3) to get:

4t^2 -12t +2 + 2/3 t -2 = 0

or:

4 t^2 -34/3 t =0

which has roots t=0, and t=17/6 ns.

So it leave the chamber at 16/6 ns.

(c) The change directions when ds/dt=0.

ds/dt = 4 + 2 (-1)/(t-3)^2

so they change direction at the roots of:

4 - 2/(t-3)^2 = 0.

Multiply through by (t-3)^2 (we can do this as we know the roots we seek
do not occur at t=3) to get:

4(t-3)^2 - 2 = 0

or:

2 t^2 - 12 t + 17 =0

which from the quadratic formula has roots 3-1/sqrt(2) and 3+1/sqrt(2).

Hence the particles change directions at 3-1/sqrt(2) and 3+1/sqrt(2) ns.

(d) The composite curve is a hyperbola.

RonL