# Finding velocity components of a charge leaving spectrometer

1. Feb 2, 2017

### trevorsternberg

1. The problem statement, all variables and given/known data
I need to find the tangent of the angle, but I don't think that is the part I'm messing up... Assume the particle doesn't hit the wall. Initial velocity is v. It seems that Vix = 0m/s and ay=0 m/s

2. Relevant equations
I thought vfx^2 =vix^2 +2a(xf-xi)
f=ma , f=eq
3. The attempt at a solution
ax=a=f/m=eq/m

vfx2=vix2+2a(xf-xi)
vfx=(vix2 +2a(xf-xi))1/2
vfx=(2a*h/2)1/2
vfx=(eqh/m)1/2
Since vix=0, v=vfy
tanθ=(eqh/m)1/2/v

I've spent way too long on this.

Last edited: Feb 2, 2017
2. Feb 2, 2017

### Cutter Ketch

Tangent is a ratio. What is it a ratio of in this problem? Also remember the x and y motions are independent. Start by writing an expression for the time to travel the length of the tube in termas of v and L.

3. Feb 2, 2017

### trevorsternberg

tanθ = Vfx/Vfy is what I showed above.
Time is not a variable that we can answer with, are you saying that I should use a kinematic equation involving time and then substitute another equation in for it?
The answer is in terms of e, q, h, m, v, L.

4. Feb 2, 2017

### Staff: Mentor

Yes, find an expression for the time to traverse the device in terms of the given values.

You were only told that the particle does not impact the plate ("wall"), so you don't know how close it comes to the plate before it exits (so you cannot assume that the particle moves distance h/2 in the x-direction).

5. Feb 2, 2017

### andrevdh

Your final equation looks right to me, just use E rather than e.
What are you suppose to work out?
Theta?
The problem is similar to a horizontally launched projectile, so you can use all of its formulas, where your x is y in the projectile formulas or alternatively you can just rotate your axes 90o anticlockwise

Last edited: Feb 2, 2017
6. Feb 2, 2017

### Staff: Mentor

I don't believe that the answer should contain h. See my post (#4) above.

7. Feb 2, 2017

### andrevdh

I get the impression that the op has the answer, but his/her solution do not match

8. Feb 2, 2017

### Staff: Mentor

Well let's wait and see what the OP has to say about what he's getting versus what he expects to get.

9. Feb 2, 2017

### trevorsternberg

I figured it out, thanks for the help, guys. In case anyone is wondering:
You must solve for t using a kinematic equation in the y direction, and then plug it in to a horizontal kinematic equation to find Vfx.
It is worth noting that the horizontal displacement is NOT h/2, which is what was tricking me.
h is never needed.
I found this problem to be very misleading for this reason and due to it appearing that the horizontal displacement is h/2.
Take care.