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sharra.26428
Joined: 12 Feb 2008
Posts: 8
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 Posted: Wed Apr 02, 2008 5:02 pm Post subject: Question 39 |
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could you please explain why the horizontal velocity is constant. Is it because the force due to the electric field is pointed in the Y direction?
thanks |
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admin
Site Admin
Joined: 08 Dec 2003
Posts: 2176
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| Posted: Thu Apr 17, 2008 2:05 pm Post subject: |
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| To begin with, the last sentence in the passage specifically states that the horizontal velocity is constant. Additionally, you are correct to state that the only force acting on the electron is in the Y direction. And, according to Newton's 2nd Law, if there is no net force (ie in the X direction) then there is no acceleration. Keep in mind the similarity of this setup and that of projectile motion where, neglecting wind resistance, velocity is always constant in the horizontal direction. |
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maxsati7049
Joined: 02 May 2009
Posts: 10
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| Posted: Tue Jun 16, 2009 10:25 pm Post subject: |
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| Could we have treated this problem as if it was a projectile with an initial Vx velocity and Zero initial Vy velocity?...if so, then would the equation Vyf^2=Vyi^2+2aL work?? that's one of the Big Five constant acceleration equations...because when solved for the final vertical velocity of the electron, the equation becomes Vyf= sq(2aL)...does that work? |
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enachbe18198
Joined: 24 Feb 2010
Posts: 9
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| Posted: Sun Apr 04, 2010 12:51 pm Post subject: |
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| I don't understand the vertical component here... why is it just aL/v? isn't final velocity: Vf = Vi + at... here it seems as though Vf just equals at..... which is just the change in velocity |
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dnpgr16513
Joined: 14 Jun 2010
Posts: 75
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| Posted: Sun Jul 11, 2010 11:02 am Post subject: |
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I agree with maxsati7049. Can we use one of those equations to solve the problem?
thanks. |
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jellywing_2058
Joined: 04 May 2009
Posts: 179
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| Posted: Tue Jul 13, 2010 9:06 am Post subject: |
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| maxsati7049 wrote: | | Could we have treated this problem as if it was a projectile with an initial Vx velocity and Zero initial Vy velocity?...if so, then would the equation Vyf^2=Vyi^2+2aL work?? that's one of the Big Five constant acceleration equations...because when solved for the final vertical velocity of the electron, the equation becomes Vyf= sq(2aL)...does that work? |
Yes, we can treat it as a projectile problem, with an initial horizontal velocity Vx, and initial vertical velocity Vy = 0.
The Equation Vyf^2=Vyi^2+2aL would work; however with one change.
The distance L included here is the horizontal distance traveled, whereas to find the final vertical velocity we have to include the vertical distance traveled, which is surely not L.
Let us find the vertical distance traveled (h) by the time (t) the electron reaches the far right.
t = L/v
Where L is the Horizontal distance covered with the constant horizontal velocity v.
With an initial vertical velocity (Vyi) of 0 m/s, and a constant vertical acceleration a, the vertical distance covered in time t = L/v would be,
Using
S = vyi t + (1/2) a t^2
h = 0 x t + (1/2) a (L/v)^2
h = (1/2) x a x L^2/v^2
Where v is the constant horizontal velocity
Now, substituting in your suggested Equation
Vyf^2=Vyi^2+2aL,
h for L, with h = (1/2) x a x L^2/v^2
Vyf^2 = 0 + 2 x a x h
Vyf^2 = 2 x a x (1/2) x a x L^2/v^2
Vyf^2 = a^2 x L^2/v^2
OR,
Vyf = a L/v
Thus, the Correct Answer is D:
vx = v and vy = aL/v |
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