Question 40

[mvenus9292913]

I don't get how the answer was arrived at.

Initially, the velocity of the electron has only a non-zero x-component that is unaffected by the electric force in the y-direction. The final velocity has a y-component given by: vy = vx x tan(q) = 3 m/s.

I understand that the x component stays the same, but where do you get that equation for the y-component, and how do you arrive at the final velocity?

[mcat_premed3832]

OK, so you are comfortable that Vx is 3 m/s and remains that way.

Consider a 45-45-90 triangle (you must memorize that one and a 30-60-90 triangle for the MCAT; you can find both here: http://www.goldstandard-mcat.com/physics-equation-lists/); tan 45 is 1. Even if you did not know that, the ratio of the x to the y sides of the triangle is 1:1. Either way, if Vx is 3 then Vy is also 3.

Now that you have Vx and Vy, then you have a right angle triangle. So to get the hypotenuse (the final velocity), just use the Pythagorean theorem.

[Guest]

Okay so I think I understand why bc it is tan 45= 1 each side is equal to each other..correct? What is had been say a 30 degree angle,which brings us to the 30-60-90 triangle. This is where I get confused. I am aware of the triangle and it's sides but I am not sure what to do if the problem presented itself with the same numbers but a 30 or 60 degree angle instead of 45. Thank you

[Guest]

Okay so I think I may have it could you tell me if my reasoning is correct.
If it had been a 30 degree angle you would have set it up as:

tan 30= y(opp)/ x (adj)
xtan30= y
3tan30=y

Is that right? Also is the reason that the Vx is 3 m/s because it is traveling horizontally through the plates and is not affected by the plate charges or the field?
thank you

[admin]

[balak502987]

Shouldn't the initial horizontal velocity be equal to 3cos(45) since the electron is projected at an angle of 45degrees.

[priya.bork7031]

I don't think so. The electron only deflects after its passed a space where it has not been deflected (theta=0), it only deflects after entering the space between the capacitors

[nedaa.asba6809]

Before entering the capacitors the electron’s velocity in the x-direction was 3 m/s, but after entering the space between the capacitors it’s deflected and as a result the electron will have both a velocity in the x-direction and in the y-direction. Keeping in mind that the velocity in the x- direction stays constant, the velocity in the y- direction is (3 m/s tan 45).

[GraemeMatt8033]

hmm i understand the reasoning behind everything, I just don't understand why you can't use the cos of 45 and then solve for the hypotenuse??? it gives you a different answer

[jellywing_2058]

Actually it does work. Since
cos(angle)= adjacent/hypotenuse, you get
cos45= Vx/Vtotal
cos45= 3/Vtotal
Vtotal=3/cos45
Vtotal=4.24 m/s = 3x(square root of 2) m/s