What potential difference is required to bring the proton to rest?

[Jbreezy]

Homework Statement

A proton has an initial speed of 4.9 105 m/s.
(a) What potential difference is required to bring the proton to rest?

(b) What potential difference is required to reduce the initial speed of the proton by a factor of 4?

(c) What potential difference is required to reduce the initial kinetic energy of the proton by a factor of 4?

Homework Equations

ΔK = -qV
Δk = mv^2/2

The Attempt at a Solution

For part a.)

Δk = (kf - Ki). Kf is 0 because your stopping it

[m(v_i)^2 ]/2q = V
I just pop in my values. Look OK?


For part b I'm unsure.

I said V_i = V/4 So I get

[m((v_i/4))^2 ]/2q = V I'm unsure.
The last one I would just take the answer from a and divide by 4.

Thanks,
I;m worried about b.

 

[rude man]

in part b you need to reduce the k.e. by 3/4, not just 1/4. Solve for the new v.
part c is solved by part b.

 

[Jbreezy]

in part b you need to reduce the k.e. by 3/4, not just 1/4. Solve for the new v.
part c is solved by part b.

Thanks for the reply. I'm sure I understand you though. Why am I reducing it by (3/4)?
And why would I solve for the new V? I want the potential difference.

 

[rude man]

Thanks for the reply. I'm sure I understand you though. Why am I reducing it by (3/4)?

Because that's what the problem asked. "To reduce by a factor of four" means to go from 1 to 1/4.

And why would I solve for the new V? I want the potential difference.

[/quote]

Right you are. My bad. But again, the reduction is by 3/4 of the initial speed, not 1/4.

 

[Jbreezy]

Because that's what the problem asked. "To reduce by a factor of four" means to go from 1 to 1/4.

Right you are. My bad. But again, the reduction is by 3/4 of the initial speed, not 1/4.[/QUOTE]

Its Ok. Keeps me thinking. Can you please explain to me why you reduce by 3/4? I feel confused because first you say

"To reduce by a factor of four" means to go from 1 to 1/4.

Which is what I thought so I reduced the velocity in part b by 1/4. Vi = v/4. But your saying 3/4? I'm not following you.

Wait are you saying this;
You want the change in velocity to be (1/4) of what it was. So


Can you say (1/2)m(Δv)^2 = qV
so, (1/2)m(Vf - (1/4)Vi)^2 = qV which is (1/2)m((3/4)V)^2 = qV ?If that is the case I don't get it . Might not be though.
Thanks

 

[rude man]

The general formula is 1/2 mv_f² = 1/2 mv_i² - qV. q and V are considered positive.

For part b, v_f = v_i/4
For part c, 1/2 mv_f² = (1/2 mv_i² )/4.

The rest is just algebra.

 

[haruspex]

Right you are. My bad. But again, the reduction is by 3/4 of the initial speed, not 1/4.

That's what rude man meant - reduce by 3/4 of the initial speed.
Reducing x by a factor of n means the new value is x/n. The change is (x-x/n) = x(1-1/n).

 

[Jbreezy]

So I'm right. It means divide but

That's what rude man meant - reduce by 3/4 of the initial speed.
Reducing x by a factor of n means the new value is x/n. The change is (x-x/n) = x(1-1/n).

The change is what part is confusing me. Why do you want the change and not simply x/n ?
Thank you.

 

[rude man]

So I'm right. It means divide but



The change is what part is confusing me. Why do you want the change and not simply x/n ?
Thank you.

Because V is what's needed to reduce (change) the speed (part b) or k.e. (part c). The reduction is 3/4 of the original speed or k.e.

 

[Jbreezy]

Actually it ended up being V/4 not (2/4)V like you guys had suggested. I think that this was a poorly written question. I asked my tutor and he agreed with you guys but we tried the answer and it was wrong. I'm going to say this question is poor. Thanks for the help