Determining Initial Speed of Electron in Head-On Collision

In summary, when two electrons are fired head-on at each other, the closest approach is 4e-14 m. The electron with twice the speed starts with that speed.
  • #1
Matt1234
142
0

Homework Statement



two electrons separated by a large distance are fired directly at each other. The closest approach in this head on collision is 4e-14 m. One electron starts with twice the speed of the other. Assuming there is no deflection from the original path, determine the initial speed of the electron.


Homework Equations



Ee = (k q1 q2) / r

ek = 0.5 m v^2





The Attempt at a Solution




see here:

http://img32.imageshack.us/i/attempt9.jpg/

Answer from book is:
5.3e7 m/s
 
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  • #2
bump please, I am sorry i have a test tom. I am taking in a lot of info today.
 
  • #3
I see a problem with your solution:

When the electrons are at their closest approach, their speeds (and hence KE) are not zero. However, you can use conservation of momentum to get their speed at that point.
 
  • #4
ok in that case what is a better way to go about it?

i don't think he wanted us to use momentum.

the only other euation i got with velocity is:
q/m = v / (B *r)

q= charge (got that)
m = mass (got that)

v = velocity, what I am looking for

b = (dont have that and can't determine without a current)

and r i do have

Ill ask him in the morn before the test, and will post back his soln.
 
Last edited:
  • #5
Redbelly98 said:
I see a problem with your solution:

When the electrons are at their closest approach, their speeds (and hence KE) are not zero. However, you can use conservation of momentum to get their speed at that point.

Oh wow, I didn't see that RB. Thanks. If their initial velocities were equal, that would be different. But the unequal initial velocities...
 
  • #6
Matt1234 said:
ok in that case what is a better way to go about it?

You can go about it the same way you did, using conservation of energy.

The difference is, you need to account for the kinetic energy when they are at their closest approach ... you seemed to assume it was zero, but it isn't.

An alternative approach is to transform to a frame of reference in which the initial speeds are equal, and solve the problem in that reference frame -- still using conservation of energy. Finally, transform back to the lab frame to find the velocities.
 

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