Proton in a uniform electric field

[Feodalherren]

Homework Statement

A proton is projected in the positive x direction into a region of uniform electric field E = (-5.80 ✕ 10^5) N/C at t = 0. The proton travels 6.50 cm as it comes to rest.

Find the initial velocity and the time it takes for the proton to stop.

Homework Equations

The Attempt at a Solution

The charge of the proton multiplied by the electric field must equal the mass of the proton times the acceleration

qE=ma

hence

a= (-5.556x10^13)

Constant accelecration, integrate with respect to time to get velocity

V= Vi - 5.556x10^13 t

X= 0 + Vi(t) - 2.2778 t^2

Since initial position = 0.

Using the position function to find the time it takes to stop:

.065m = (5.556x10^13)t^2 - (2.2778x10^13)t^2

solving for t, t= 4.8x10^-8 s

Plugging this back to the velocity equation where at t= 4.8x10^-8 s the final velocity = 0 therefore:

Vi=(5.556x10^13 (m/s^2))(4.8x10^-8 s)

Clearly not right, it's faster than the SOL.
Where am I going wrong?

ps. I don't like to memorize stuff so please don't refer me to the kinematic equations. I prefer to derive them by myself.

 

[STEMucator]

You need to solve $v dv = a dx$ assuming that $v = v_i$ at $x = x_0 = 0$.

Or just use $v_f^2 = v_i^2 + 2a \Delta x$ to save some time.

Also that acceleration should be positive I think.

 

[Feodalherren]

That's exactly what I did? It doesn't seem to be working. I missed something along the way.

Acceleration can't be positive since it's slowing down and moving in the positive x-direction.

 

[STEMucator]

That's exactly what I did? It doesn't seem to be working. I missed something along the way.

Acceleration can't be positive since it's slowing down and moving in the positive x-direction.

Ahh so the electric field is in the $(- \hat i)$ direction.

You seem to be solving $a = \frac{dv}{dt}$.

 

[Feodalherren]

Well yes, but only as an initial step. Once I have that I integrate once more to solve for the position function. I can't see anything wrong with it :/.

 

[STEMucator]

You can find a relationship between velocity and position because you have the constant acceleration.$$v dv = a dx$$
$$\int_{v_i}^{0} v dv = \int_{0}^{0.065} a dx$$

This let's you solve for the initial velocity. The time taken can now be found from $a = \frac{dv}{dt}$, or simply use $v_f = v_i + at$.

 

[Feodalherren]

But that's exactly what I did...

 

[STEMucator]

But that's exactly what I did...

Look at the integral limits carefully and ask yourself where you went wrong.

 

[Feodalherren]

I honestly have no clue. This is how I did all of my mechanics in Physics I. I never bothered to remember the kinematic equations as just integrating / taking derivatives worked for everything. I'm missing something fundamental.

Why can't I just integrate from acceleration to velocity to position?

 

[STEMucator]

Everything looks correct to me too, I'm not sure then.

 

[ehild]

Vi=(5.556x10^13 (m/s^2))(4.8x10^-8 s)

Clearly not right, it's faster than the SOL.
Where am I going wrong?

What numerical value do you get for Vi? And what is the speed of light?

ehild

Spoiler

Vi≈3*10⁶

 

[Feodalherren]

Vi=(5.556x10^13 (m/s^2))(4.8x10^-8 s)= 2,666,880 m/s

... my bad I thought the SOL was about 3 x 10^6 m/s but that's km/s.

Thanks ehild.

 

[ehild]

Vi=(5.556x10^13 (m/s^2))(4.8x10^-8 s)= 2,666,880 m/s

... my bad I thought the SOL was about 3 x 10^6 m/s but that's km/s.

Thanks ehild.

No. So what is the SOL? You need to know the magnitude...

Spoiler

3x10^5 km/s = 3x10^8 m/s

ehild

 

[Feodalherren]

Don't worry about it, I solved it. Thanks again :).