# Force and Motion Problem

1. Oct 23, 2008

### hansel13

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
F=ma

3. 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...

2. Oct 23, 2008

### LowlyPion

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.

3. Oct 23, 2008

### Staff: Mentor

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

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

4. Oct 23, 2008

### 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.

5. Oct 23, 2008

### LowlyPion

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

6. Oct 23, 2008

### 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.

Last edited: Oct 23, 2008