Separating ions. Acceleration problem

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The discussion revolves around calculating the acceleration needed to separate fast and slow ions, which travel at different speeds of 1000 m/s and 100 m/s, respectively. The ions are subjected to an electric field between two charged plates, with a separation distance of 5 cm and a gap of 3 cm. The time each ion spends in the plates is critical, with the fast ion taking 0.00005 seconds and the slow ion taking 0.0005 seconds to traverse the 5 cm distance. The goal is to achieve a 2 cm separation between the two ions at a distance of 50 cm after exiting the plates, leading to the derived acceleration of approximately 7696 m/s². The calculations involve kinematic equations to determine the deflection of each ion based on their respective times and accelerations.
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Homework Statement


You are trying to separate 2 types of ions (fast and slow). The fast ones travel at 1000m/s and the slow ones travel at 100m/s. At the beginning you have 2 plates. One has + and the other one has negative charge. The size of the plates is 5cm and they are 3 cm apart. What will the acceleration have to be between the 2 plates that will put the fast and slow ions 2 cm apart at 50cm once out of the plates.

Remember once the ions are out of the plates, they will not accelerate any more.

The Attempt at a Solution



I drew it out.
The reds are the ions. The blue box at the bottom is the emitter. The 2 gray ones are the charged plates that will accelerate the ions.

Ions.gif


From what i know there will be 2 different times since the ions travel at different speeds
For the fast one i calculated
.5m / 1000m/s = .0005s
for the slow one i got
.5m / 100m/s = .005s

From here on out, I'm stuck and have no idea what to do next.
Thanks in advance.
 
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The key is how long the fast or the slow will be traveling through the plates..

By establishing a charge between the plates, they will be traveling across the 5 cm distance and will be accelerated (deflected) according to normal kinematics. The slower ones being between the plates will be subjected to an accelerating force for a longer interval and hence will be deflected more.

Start with how long each will be traveling across the plates and see what occurs to you remembering that as usual F = m*a.
 
Thank you for the feedback sir. So i got that. 5cm = .05 m

Since their initial velocity is in the Y direction during the acceleration towards X will not change the Y, I do this to find their times.
Fast = .05m / 1000m/s = .00005s
Slow = .05m / 100m/s = .0005s

Then i don't know where to plug it in or how. Where does the 2cm difference end up going?
I assume I'm going to have 2 equations. One for the acceleration here and another once its out of the plates but i have no idea what formulas to use.

Since both are going to curve, i don't know what the angle in between will be which is where I'm confused.

I've never done a problem like this before so I'm confused what to do next.
 
The 2 cm distance is at 50 cm.

What component of velocity is required to get the slow ion to move farther to the right by 2 cm by the time it reaches 50 cm than the fast one?

You know V = a*t where t is the time the particle is subjected to acceleration between the plates.

That velocity times the time T to get to 50 cm at the original speed then would be the deflection I'd think for each particle. The additional constraint will be that 1 is 2 cm further along when it gets there. One distance to the right will be 2 cm more than the other.
 
alright so here is what i got.

position after the .05m (between plates) is
Xf = 0ms + 0t + .5At^2

slow beam..
Xfs = .5a(.0005s^2)
fast beam
Xff = .5a(.00005s^2)

Next velocity at the end
Slow beam
Vfs = a*.0005s
Fast beam
Vff = a*.00005sNext, X2-X1 = .02m
so
.02m = (.5a(.0005s^2) + (a*.0005s)*.005) - (.5a(.00005s^2) + (a*.00005s)*.0005)
A = 7696.00 m/s^2
 
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