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Proton Bombardment Work and Energy

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data

    A proton with mass 1.67 x 10^-27 kg is propelled at an initial speed of 3.00x10^5 m/s directly toward a uranium nucleus 5.00 away. The proton is repelled by the uranium nucleus with a force of magnitude F=α/x^2, where x is the separation between the two objects and α = 2.12 x 10^-26 N*m^2. Assume that the uranium nucleus remains at rest.

    A)What is the speed of the proton when it is from the uranium nucleus?

    B)As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, after which the proton moves away from the uranium nucleus. How close to the uranium nucleus does the proton get?

    2. Relevant equations

    v22=vv[2/SUB]+2ad
    F=α/x^2 (given)
    Fd=KE2-KE1
    KE = .5mv2

    3. The attempt at a solution
    For a), I got that the distance equaled 5-8*10^-10 = 5 m, and I tried using that distance times the force (found with the given equation), to find the work and I set that equal to the the change in Kinetic energy, and tried to find the second velocity. But this did not get me the right answer. For b) I tried the sae approach, but plugged in the known velocities, canceling out KE2 because v = 0 there, and still didn't get the right answer. What am I doing wrong? Thanks.
     
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  3. Oct 18, 2009 #2

    rl.bhat

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    In A) at what distance from the nucleus the speed is needed?
     
  4. Oct 18, 2009 #3
    In a), the proton is 8.00 x 10-10m from the uranium nucleus
     
  5. Oct 18, 2009 #4

    rl.bhat

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    F = m*a = α/x^2
    So a = 1/m( α/x^2)
    or dv/dx = 1/m( α/x^2)
    dv = 1/m( α/x^2)*dx.
    Find the integration. To find the constant of integration apply the initial condition. i.e. when x = 0, vo is the initial velocity of the proton.
     
  6. Oct 18, 2009 #5
    before I took the derivative of the equation, I plugged in the mass, since it's a constant, as well as what "α" is, and I got that a=(1.27*10^-19)(x^-2). I took the derivative of it and got that velocity equals -1.27*10^-19(x^-1) on the integral of 5 to 8.00*10^-10 m. I got the answer to be -1.59*10^-10, which is the wrong answer. Can you see what I did wrong?
     
  7. Oct 18, 2009 #6

    rl.bhat

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    Here you are not taking derivative anywhere. Don't put the values initially.
    Find the integration of dv to find v. Now put the values.
     
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