1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Jan 24, 2014 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations

    ΔK = -qV
    Δk = mv^2/2

    3. 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.
     
  2. jcsd
  3. Jan 24, 2014 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  4. Jan 24, 2014 #3
    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.
     
  5. Jan 24, 2014 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

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

    Right you are. My bad. But again, the reduction is by 3/4 of the initial speed, not 1/4.
     
  6. Jan 24, 2014 #5
    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

    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
     
  7. Jan 24, 2014 #6

    rude man

    User Avatar
    Homework Helper
    Gold Member

    The general formula is 1/2 mvf2 = 1/2 mvi2 - qV. q and V are considered positive.

    For part b, vf = vi/4
    For part c, 1/2 mvf2 = (1/2 mvi2 )/4.

    The rest is just algebra.
     
    Last edited: Jan 24, 2014
  8. Jan 24, 2014 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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).
     
  9. Jan 24, 2014 #8
    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.
     
  10. Jan 24, 2014 #9

    rude man

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  11. Jan 28, 2014 #10
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: What potential difference is required to bring the proton to rest?
Loading...