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Homework Help: Colliding Particles using energy

  1. Jun 19, 2010 #1
    this question is from my previous assignment which was due 2 weeks back. just revising for mid-term not sure how to solve it. if someone explains it thoroughly it would be grate

    1. The problem statement, all variables and given/known data
    An alpha particle is a nucleus of Helium. It has twice the charge and four times the mass of the proton.

    A proton and an alpha particle headed directly toward each other, had each initial speed of 3.9×10−3 c when they were far away.

    Here, as is customary when describing processes involving nuclear targets, the speed is expressed as a fraction of c, the speed of light.

    What is the distance of closest approach between the proton and the alpha particle?

    Hint: There are two conserved quantities. Make use of both.



    2. Relevant equations
    U=Eqd or KqQ/r
    K=1/2*(mv^2)

    3. The attempt at a solution

    .5m(v_i)^2 + Eq(d_i)= .5m(v_f)^2 + Eq(d_f)

    where (V_f)=0 then i solved for (d_f)
    and got the 1.87*10^-27 but the answer should be d_f=1.3*10^-13

    could some 1 explain this to me and the reasoning!! please
     
    Last edited: Jun 19, 2010
  2. jcsd
  3. Jun 19, 2010 #2

    Doc Al

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    Staff: Mentor

    You are assuming that V_f = 0. Make use of the hint that you were given.
     
  4. Jun 19, 2010 #3
    this question is from my previous assignment which was due 2 weeks back. just revising for mid-term not sure how to solve it. if someone explains it thoroughly it would be grate
     
  5. Jun 19, 2010 #4

    Doc Al

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    Staff: Mentor

    Why don't you make an attempt using the hints I just gave?
     
  6. Jun 19, 2010 #5
    v_f is the final velocity , at minimum distance it can reach it will be zero.no?
     
  7. Jun 19, 2010 #6

    Doc Al

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    Staff: Mentor

    No it won't. Use the hint that the problem gave you! What else is conserved besides energy?
     
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