Finding velocity components of a charge leaving spectrometer

Click For Summary

Homework Help Overview

The discussion revolves around finding the velocity components of a charged particle as it exits a spectrometer. The original poster (OP) is attempting to determine the tangent of the angle related to the particle's trajectory, with initial conditions and assumptions about the motion being discussed.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the x and y components of motion, questioning how to express time in relation to the length of the tube. There is discussion about the independence of x and y motions and the implications of the particle's trajectory.

Discussion Status

Several participants have provided guidance on expressing the time variable and have pointed out potential misconceptions regarding the horizontal displacement. The OP has indicated some confusion about the relationship between the variables involved and has received feedback on the appropriateness of certain assumptions.

Contextual Notes

There is mention of constraints regarding the particle's trajectory, specifically that it does not impact the wall, which affects the assumptions about its horizontal displacement. The OP has expressed that the problem is misleading due to these assumptions.

trevorsternberg
Messages
3
Reaction score
0

Homework Statement


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
wttvly.png


Homework Equations


I thought vfx^2 =vix^2 +2a(xf-xi)
f=ma , f=eq

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:
Physics news on Phys.org
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.
 
Cutter Ketch said:
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.

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.
 
trevorsternberg said:
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.
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).
 
Your final equation looks right to me, just use E rather than e.
What are you suppose to work out?
Theta?
Should your answer contain L?
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 :smile: or alternatively you can just rotate your axes 90o anticlockwise :wink:
 
Last edited:
andrevdh said:
Your final equation looks right to me, just use E rather than e.
I don't believe that the answer should contain h. See my post (#4) above.
 
I get the impression that the op has the answer, but his/her solution do not match :smile:
 
andrevdh said:
I get the impression that the op has the answer, but his/her solution do not match :smile:
Well let's wait and see what the OP has to say about what he's getting versus what he expects to get.
 
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 into 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.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
15
Views
3K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K