1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Elastic head-on collision homework

  1. Dec 3, 2006 #1
    Here is a problem that I'm not sure of how to start. Please give me advice on the correct procedures I have to use. Thanks!

    Problem: A neutron in a reactor makes an elastic head-on collision with a carbon atom that is initially at rest. (The mass of the carbon nucleus is about 12 times that of the neutron.) a) What fraction of the neutron's kinetic energy is transferred to the carbon nucleus? b) If the neutron's initial kinetic energy is 1.6 x 10^-13 J, finds its final kinetic energy and the kinetic engery of the carbon nucleus after the collision.

    I'm confused on how to start this problem. They only give you 2 masses but you don't know any of the velocities.

    Well I used the KE(initial)=KE(final)
    and I used x as the mass for the neutron and 12x as the mass for the carbon.

    So then I have: (1/2)x(v1)^2=(1/2)x(v1')^2 + (1/2)12x(v2')^2

    Am I even going in the right direction?? Well I just divided both sides by (x), multiplied both sides by 2, and square rooted both sides and i got:

    v1=v1'+sqrt.(12)v2
     
  2. jcsd
  3. Dec 3, 2006 #2

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    You did not square root both sides. Your last step is invalid. What besides energy is conserved in this collision?
     
  4. Dec 3, 2006 #3
    Momentum is also conserved right? What do you mean my last step is invalid??

    After you divide the x and multiply by 2 you should get:

    v1^2=v1'^2+12(v2')^2 right?

    If you sqrt. both sides you get:

    v1=v1' + (sqrt.(12))(v2) correct?

    The answer is .28 or 28% for a) by the way.

    How would you achieve that answer? I've tried 1/(sqrt.12) and I got .288675 but I don't think the book would round down. Am I missing something here for part a)?
     
    Last edited: Dec 3, 2006
  5. Dec 4, 2006 #4

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    Think about the Pythagoren theorem

    c² = a² + b² is not the same thing as c = a + b

    If c = a + b then c² = a² +2ab + b² ≠ a² + b²

    You cannot take square roots term by term. You must add the squares before you take a square root.

    c = sqrt(a² + b²)

    Think of v1 as an unspecified but known quantity, and use your conservation of energy equation (your first equation; it is correct) AND conservation of momentum to solve for v1' and v2' in terms of v1
     
  6. Dec 4, 2006 #5
    Ah I see, well now I have the equations:

    v1^2=v1'^2+(12)v2'^2

    Now what equation should I use to find the fraction of the neutron's kinetic energy transfered to the carbon nucleus? Are you saying that I should use the conservation of momentum equation: m1v1=m1v1'+m2v2' ?? Do I solve for v1 and then find the finals?

    Can I use the equation: v1-v2=-(v1'-v2') ??

    I got v1'^2=(-5.5)v2'^2 .

    To find the fraction of the neturons KE transfered to the carbon KE I use the equation: KE carbon/KE neutron right?? Therefore would it be:

    ((1/2)m2v2'^2)/((1/2)m1v1'^2) ??

    Thanks a lot!! :smile:
     
  7. Dec 4, 2006 #6
    help someone??
     
  8. Dec 4, 2006 #7

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    You can use the diference of velocities equation for elastic collisions.

    You don't mean this

    I got v1'^2=(-5.5)v2'^2

    You mean v1' = (-5.5)v2'

    Squares are always positive, you cannot have one square being a negative multiple of another.

    ((1/2)m2v2'^2)/((1/2)m1v1'^2)

    Is not quite right. You want the fraction of the energy the neutron had before the collison. You still need to express the final velocities in terms of the intial velocity.
     
  9. Dec 4, 2006 #8
    I don't really get how to solve for the fraction of KE transfered. I know you get it from KE(initial)=KE(final)

    This right?: (1/2)x(v1)^2=(1/2)x(v1')^2 + (1/2)12x(v2')^2

    Then what are you suppose to do with the v1' that is solved for as:
    v1' = (-5.5)v2'

    I don't think I have v1 initial or v2'.. Do you even need them?

    I don't know what I'm suppose to do here =/ Please help. Thanks. :smile:
     
  10. Dec 5, 2006 #9

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    The fraction of the initial energy transferred to the carbon is
    ((1/2)m2v2'^2)/((1/2)m1v1^2) There is no prime in the denominator. It is the total energy since the carbon was initially at rest.

    You can either find v2' in terms of v1 using the the result relating v1' and v2' and a conservation or velocitiies difference equation, or since the denominator in the ratio is the total energy, express it as the sum of the two final klintic energies. Then use the v1' to v2' relationship to eliminate v1'. The v2' will then divide out.
     
  11. Dec 5, 2006 #10
    To solve for v2' can I use: v1' = (-5.5)v2'

    and just divide (-5.5) to get: v2'=v1'/(-5.5) ??

    If so, I just substitute it into the equation:
    ((1/2)m2v2'^2)/((1/2)m1v1^2)

    right?? but then that would leave me with v1'/v1..

    Seriously, I'm so confused on how to solve for differen variables and I don't even know if I'm doing it right. I can just keep solving and solving for v1,v1' and v2' that I don't know where to substitute them and solve for what I'm looking for.

    Am I suppose to solve for v2' and get an answer equal to v1 so that it can me divided out of the eqatuion? ((1/2)m2v2'^2)/((1/2)m1v1^2)
     
  12. Dec 5, 2006 #11

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    Did you ever write the conservation of momentum equation?

    m1v1 + m2v2 = m1v1' + m2v2' (with v2 = 0)

    You also used the velocity difference equation, which is valid for elastic collisions, or maybe you used this instead of momentum

    v1 - v2 = -(v1' - v2') (again v2 = 0)

    Now that you know how to express v1' in terms of v2' or vice versa, you could substitute one for the other in either of these equations and solve for v2' in terms of v1 (and also solve for v1' in terms of v1).
     
    Last edited: Dec 5, 2006
  13. Dec 5, 2006 #12
    So if I substitued v1'=(-5.5v2') into v1=-(v1'-v2') I would get:

    v1=6.5v2 right? am I going in the right direction here?

    Do I then plug it into: (1/2)x(v1)^2=(1/2)x(v1')^2 + (1/2)12x(v2')^2
    and cancel out the v2's??
     
  14. Dec 5, 2006 #13
    Wow, I got the correct answer! Thanks to you of course. =] Thanks a lot!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Elastic head-on collision homework
Loading...