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Elastic Collision of a gold nucleus

  1. Jun 19, 2008 #1
    1. The problem statement, all variables and given/known data
    A positively charged alpha particle is fired, in a direct line, at a positively charged gold nucleus:

    o--> O

    o: alpha, O: golden nucleus.

    Because of their mutual repulsion, the alpha particle does not actually hit the nuleus: it comes to rest, for a moment, some distance from the nucleus and then recedes. The alpha particle may be considered to have undergone an elastic "collision" with the golden nucleus.The golden nucleus has a mass of about 50 times the alpha particle. Compared to the gold nucleus, after the collision, the alpha particle has

    A. more momentum but less kinetic energy
    B. more momentum and more kinetic energy.
    C. less momentum and less kinetic energy
    D. less momentum but more kinetic energy
    E. the same momentum but more kinetic energy.

    2. Relevant equations

    Elastic collision: object 2 initially at rest. vi is the initial velocity of the alpha particle. The post-collision velocities are:
    v1= (m1 - m2)vi / (m1+ m2)
    v2= (2*m2)vi / (m1+ m2)

    3. The attempt at a solution

    This problem can be solved in two ways: either by using logic and doing no calculations or doing the calculations based on the elastic collision formulas. Logically, I know that the initial momentum is to the right, therefore post collision the overall momentum will be to the right. So the gold nucleus will definitely have more momentum. The question now remains about the kinetic energy. Logically, one would think that even though the net momentum is to the right, the velocity of the gold nucleus < velocity of alpha, due to the huge mass difference, therefore the KE of the alpha particle would be greater. However, to back this up, I wanted to prove it mathematically but it doesn't work out that way.

    Let the initial velocity of the alpha particle be vi.

    Velocity of gold nucleus post-elastic-collision:
    v2= (2*m2)vi / (m1+ m2)
    v2= (2*50m)vi / (m+ 50m)
    v2= (100m) vi/ (51m) = 100v/51

    Velocity of alpha nucleus post-elastic-collision:
    v1= (m1 - m2)vi / (m1+ m2)
    v1= (m - 50m)vi / (m+ 50m)
    v1= (-49m)vi / (51m)= -49/51 v

    From this I can see that the velocity of the alpha particle is much less than the velocity of the gold nucleus, which having the greater mass as well, will have the greater momentum. How is it that the math doesn't work out but logically answer D makes sense? Thanks in advance!
  2. jcsd
  3. Jun 19, 2008 #2


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    D doesn't make sense. Total kinetic energy is conserved since the collision is elastic. The alpha comes in with all of the kinetic energy and the gold nucleus with none. The gold nucleus leaves with some kinetic energy so that of the alpha MUST decrease. Momentum is a little trickier since it has a sign. But still, can you rethink this?
    Last edited: Jun 19, 2008
  4. Jun 19, 2008 #3
    This question is from a past midterm and the answer given in the answer key is D. I've thought of it in terms of momentum and since velocity is squared in the KE expression, larger velocity of the alpha particle gives it a greater KE. Besides this, I don't know what to conclude from this problem. Logic makes sense to me, but the math isn't working out, when technically it should be. :S I'm confused!
  5. Jun 20, 2008 #4


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    I think you're misreading the question.

    D says that momentum(alpha) < momentum(gold) and KE(alpha) > KE(gold). :wink:
  6. Jun 20, 2008 #5
    hehe. i do get that. the only discrepency is that the formulas give a greater velocity for the gold nucleus. Therefore, making it have a higher KE. Or should I not use the formula over here and it's not a REAL elastic collision.
  7. Jun 20, 2008 #6


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    Hi habibclan,

    The equation you have for v2 in your relevant equations section is not correct. Do you see what it needs to be?
  8. Jun 20, 2008 #7


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    I was definitely reading it wrong. And that's with absolute values on the momentum. habibclan, check your math. I'm getting v alpha=(-49/51)vi (just as you did) but for v gold I get (2/51)vi. Not (100/51)vi.
  9. Jun 20, 2008 #8
    My formula sheet has a weird way of writing the formula and they had m2 on there, thats why instead of having 2 in my numerator, i have 100. You guys saved me from using that formula on the exam! Thanks a lot!

    [Now my logical answer matches the math too! Yay!]
  10. Jun 20, 2008 #9
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