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Alpha collision with unknown nucleus

  1. Dec 7, 2014 #1
    1. The problem statement, all variables and given/known data
    calculate the mass of an unknown nucleus of mass [itex] M_X[/itex] (initially at rest) if it it is hit by an alpha particle of mass [itex]m_x[/itex]and is deflected by 40 degrees, and the alpha particle is deflected by an angle of 70 degrees.
    Assume elastic collisions




    2. Relevant equations

    [itex]p_x+p_X=p_y+p_Y[/itex]

    [itex]Q=K_y+K_Y-K_x+0[/itex]



    3. The attempt at a solution




    From the conservation of momentum

    [itex]p_Ycos\theta_Y=p_x-p_y cos \theta_y[/itex]
    and
    [itex]p_Ysin\theta_Y=p_ysin\theta_y[/itex]

    squaring and adding these gives

    [itex](p_Ycos\theta_Y)^2+(p_Ysin\theta_Y)^2= (p_x-p_y cos \theta_y)^2+ (p_ysin\theta_y)^2[/itex]

    using some trig identities gives
    [itex]p_Y^2=p_x^2+p_y^2-2p_xp_ycos\theta_y[/itex]

    Then i'm not sure where to go next, velocities or energies aren't given so I can't use this equation directly and any other equation I can get from this involve kinetic energies or q value, so can't be used
    (I got

    [itex]Q=K_y+(\frac{m_x}{M_Y}K_x+\frac{m_y}{M_Y}K_y-\frac{2}{M_Y}\sqrt{m_xm_yK_xK_ycos\theta_y}) -K_x[/itex])

    A pointer in the right direction would be appreciated,

    Thanks.
     
    Last edited: Dec 7, 2014
  2. jcsd
  3. Dec 7, 2014 #2

    mfb

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    Okay, I guess capital letters are the nucleus and small letters are the alpha particle, with x before the collision and y afterwards, and you choose x as direction of the incoming alpha particle (please write those things down, that makes it easier for others).

    The conservation of momentum gives two equations - in addition to the one you got, you can use one of the original equations because it is still useful. Conservation of energy gives the third equation, and the absolute momenta cancel out (you can work with ratios).
     
  4. Dec 7, 2014 #3
    Oops, forgot to add that part in.

    So conservation of energy gives [itex]\frac{p_x^2}{2m_x}=\frac{p_y^2}{2m_y}+\frac{p_Y^2}{2M_Y}[/itex] , i'm not quite sure I follow what you mean about the ratios.

    Thanks.
     
  5. Dec 7, 2014 #4

    mfb

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    You do not have to, and you cannot, solve for the individual momenta.
    You can divide all equations by ##p_x##or ##p_x^2##, then you are left with ratios of momenta only. Give those ratios new names, and you have one variable less to care about.
     
  6. Dec 7, 2014 #5
    Ok, let me see if I've got this straight, if one defines the ratio [itex]\frac{p_Y}{p_x}=A[/itex] and [itex]\frac{p_y}{p_x}=B[/itex] ,and divides the momentum equations by [itex]p_x[/itex] and the energy conservation equation by [itex]p_x^2[/itex] as you said, then one gets [itex]Acos\theta_Y=1-Bcos\theta_y[/itex] , [itex]Asin\theta_Y=Bsin\theta_y[/itex] and [itex]\frac{1}{2m_x}=\frac{A^2}{2M_Y}+\frac{B^2}{2m_y}[/itex] am I on the right line?

    Thank you.
     
  7. Dec 7, 2014 #6

    mfb

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    Looks good so far. You can do the same for the masses and get rid of another parameter (also, the alpha particle does not change its mass).

    Then you have three variables and three equations to find them.
     
  8. Dec 8, 2014 #7
    Ok, so that means [itex]m_x=m_y[/itex] and dividing the energy equation by
    [itex]m_x[/itex] gives [itex]\frac{1}{2}=\frac{m_xA^2}{2M_y}+\frac{B^2}{2}[/itex] and if one defines [itex]\frac{m_x}{M_Y}=C[/itex] then is it simply a case of solving the 3 equations for C and subbing in this ratio back in to solve for [itex]M_y[/itex]? Thanks
     
  9. Dec 8, 2014 #8

    mfb

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    Right.
     
  10. Dec 4, 2016 #9
    Why have you used My for the last post? Is this the mass of the nucleus or alpha particle? I'm confused
     
  11. Dec 4, 2016 #10

    mfb

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    It is the mass of the unknown nucleus.
    This thread is from 2014.
     
  12. Dec 4, 2016 #11
    right ok, what would the final equation be for the unknown mass of the nucleus? would you use kg or u when putting in the mass of the helium nucleus?
     
  13. Dec 4, 2016 #12

    mfb

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    That is part of the homework problem, and we don't post full solutions here.
    It does not matter as long as you keep the units consistent.

    Please start a new thread if you have a similar homework.
     
  14. Dec 5, 2016 #13
    where does the conservation of energy equation come from in Purple Baron's second post?
     
  15. Dec 5, 2016 #14

    mfb

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    Every fraction is the energy of one particle. p=mv, therefore ##\frac{p^2}{2m} = \frac 1 2 mv^2##.
     
  16. Mar 21, 2017 #15
    I'm still lost as to where the conservation of energy equation comes from in Purple Baron's second post?
     
  17. Mar 21, 2017 #16

    mfb

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    ##\frac{p^2}{2m}## is the nonrelativistic kinetic energy of a particle, the left side is the energy of the incoming particle (the other particle doesn't move), the right side is the kinetic energy of the two outgoing particles.
     
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