Force and Motion Electron Problem

[hansel13]

Homework Statement

An electron with a speed of 1.2 x 10⁷ moves horizontally into a region where a constant vertical force of 4.5 x 10^-16N acts on it. The mass of the electron is 9.11 x 10^-31kg. Determine the vertical distance the electron is deflected during the time it has moved 30mm horizontally.

Homework Equations

F=ma

The Attempt at a Solution

4.5 x 10^-16N
^
|
|
| /|height=?
|/ |
|--|------------------> 1.2 x 10⁷
30mm

Honestly, I'm not exactly sure where to start. This seems like a rather simple problem but I don't know how to approach it...

 

[LowlyPion]

Homework Statement

An electron with a speed of 1.2 x 10⁷ moves horizontally into a region where a constant vertical force of 4.5 x 10^-16N acts on it. The mass of the electron is 9.11 x 10^-31kg. Determine the vertical distance the electron is deflected during the time it has moved 30mm horizontally.

Homework Equations

F=ma

The Attempt at a Solution

4.5 x 10^-16N
^
|
|
| /|height=?
|/ |
|--|------------------> 1.2 x 10⁷
30mm

Honestly, I'm not exactly sure where to start. This seems like a rather simple problem but I don't know how to approach it...

Welcome to PF.

First figure what the acceleration on the particle would be.

Then it's just like a rock thrown horizontally off a building and affected by gravity - only using the numbers of the problem - 30mm, a, v.

 

[Doc Al]

The force only acts vertically, so the horizontal motion is unaffected. Treat horizontal and vertical motion separately.

Edit: Oops... LowlyPion beat me to it!

 

[hansel13]

Alright so...

a=F/M = (4.5 x 10^-16N) / (9.11 x 10^-31kg) = 4.9x10¹⁴

t = d/v = 30mm/(1.2 x 10⁷) = 2.5 x 10^-9

Now that we have a and t, we can solve for vertical distance:

D = vt-1/2at²

= (1.2 x 10⁷)*(2.5 x 10^-9)-(1/2)*(4.9x10¹⁴)*(2.5 x 10^-9)²
= -1513.63 m
= -1.5mm

Is that right? Because that's very nearly the books answer.
One more question, why is the answer negative? The books answer is positive 1.5mm.

 

[LowlyPion]

Alright so...

a=F/M = (4.5 x 10^-16N) / (9.11 x 10^-31kg) = 4.9x10¹⁴

t = d/v = 30mm/(1.2 x 10⁷) = 2.5 x 10^-9

Now that we have a and t, we can solve for vertical distance:

D = vt-1/2at²

= (1.2 x 10⁷)*(2.5 x 10^-9)-(1/2)*(4.9x10¹⁴)*(2.5 x 10^-9)²
= -1513.63 m
= -1.5mm

Is that right? Because that's very nearly the books answer.
One more question, why is the answer negative? The books answer is positive 1.5mm.

Not quite. Because your initial velocity is normal to the deflection (the v in that direction is 0).

This is your D

(1/2)*(4.9x10¹⁴)*(2.5 x 10^-9)²

 

[hansel13]

Oh I see, so it should be like this:

D = v₀t+1/2at2

= 0*(2.5 x 10^-9)+(1/2)*(4.9x10¹⁴)*(2.5 x 10^-9)²
=.001543m
=1.5mm

Got it. Thanks a lot.