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Homework Help: Neutron collides elastically with a helium nucleus

  1. Mar 7, 2010 #1
    "neutron collides elastically with a helium nucleus"

    1. The problem statement, all variables and given/known data

    A neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus is observed to rebound at an angle θ'2 = 40° from the neutron's initial direction. The neutron's initial speed is 5.0 e5 m/s. Determine the angle at which the neutron rebounds, θ'1, measured from its initial direction.

    2. Relevant equations

    (1/2)mu1^2 = (1/2)(4m)(v2^2) + (1/2)(m)(v1)^2 where u1 = 5.0e5m/s
    (1/2)u1^2 = (1/2)(4)(v2^2) + (1/2)(v1)^2
    u1^2 = (4)(v2^2) + (v1)^2

    m(u1)(cos40) = (4)(m)(v2) + (m)(v1x)
    (u1)(cos40) = (4)(v2) + (v1x)
    (u1)(.7660) = (4)(v2) + (v1x)
    (5.0e5)(.7660)
    383022.2216 = (4)(v2) + (v1x)
    v1x = 383022.2216 - (4)(v2)

    m(u1)(sin40) = (m)(v1y)
    (u1)(sin40) = (v1y)
    (5.0e5)(.6428) = (v1y)
    v1y = 321393.8048

    (v1^2) = (v1x)^2 + (v1y)^2
    (v1^2) = { (383022.2216 - (4)(v2) }^2 + ( 321393.8048)^2
    (v1^2) = (1.4671e11 - 16v2^2 + 1.0329e11)
    (v1^2) = 2.50e11 - 16v2^2
    (v1) = 5.0e5 - 4v2)

    u1^2 = (4)(v2^2) + (v1)^2
    (2.50e11) = (4)(v2^2) + (2.5e11+16v2^2)
    5.0e5 = 4v2 + 5.0e5+4v2
    0 = 8v2 NOOOO!!!



    Thanks, any help is appreciated!!!
     
  2. jcsd
  3. Mar 7, 2010 #2

    ehild

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    Re: "neutron collides elastically with a helium nucleus"

    Show a drawing, please. I do not see which is the 40° angle, and what is u and what is v.

    ehild.
     
  4. Mar 7, 2010 #3

    lightgrav

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    Re: "neutron collides elastically with a helium nucleus"

    you seem to be using u for initial velocities, v for final velocities.
    interesting to have used alpha's final direction as "x" .
    The last step in set 4 is bogus ...
     
  5. Mar 7, 2010 #4
    Re: "neutron collides elastically with a helium nucleus"

    Should it be v1 = 5.05e5 + 16v2?

    (v1^2) = (v1x)^2 + (v1y)^2
    (v1^2) = { (383022.2216 - (4)(v2) }^2 + ( 321393.8048)^2
    (v1^2) = (1.4671e11 - 16v2^2 + 1.0329e11)
    (v1^2) = 2.50e11 - 16v2^2
    (v1) = 5.0e5 - 4v2)
     
  6. Mar 7, 2010 #5

    lightgrav

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    Re: "neutron collides elastically with a helium nucleus"

    well, lets see ... I remember that 3^2 = 5^2 - 4^4 ...
    so you think that implies 3 = 5 - 4 ?
    you do the same thing again in set 5 ...

    If it were me, I would only analyze collisions in the center-of-momentum frame .
     
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