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Show particles are scattered at right angles in the CM frame

  1. Jan 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Suppose that in a fixed-target proton-proton collision, the two protons are scattered with equal energies at equal angles θ/2 to the beam. Show that in the CM frame, this implies they are scattered at right angles to the direction of the beam. Calculate the energy and momentum of the scattered protons in the CM frame.


    2. Relevant equations
    E(cm)= √2mc^2(E1+mc^2) (E(cm) is the energy in the centre of momentum frame, derived using energy and momentum conservation, E1 is the energy of the initial proton beam)
    E'(cm)=1/2 E(cm) (as they have equal energy after it must be half the initial centre of momentum frame energy)


    3. The attempt at a solution
    I can't figure out how I'm going to yield an angle from this because in theory they will be coming off at equal angles and the modulus of the 2 protons momentum after will be equal (as they are equal and opposite) so I've tried to solve using conservation of momentum perpendicular to the beam energy which would give:

    |P3|sin(θ3)=|P4|sin(θ4)
    where P3 and θ3 represent the scattering angle from the beam energy for one of the protons and P4, θ4 the other proton. But these would just cancel due to them being equal and opposite so the angles are equal?

    Thanks in advance, I've been tearing my hair out over this for a while!
     
  2. jcsd
  3. Jan 10, 2016 #2

    mfb

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    2016 Award

    Staff: Mentor

    If they get scattered at the same angle relative to the beam, what can you say about their relative energies? How does the motion look like in the center of mass system?

    You don't need any formulas here, symmetry arguments are sufficient.
     
  4. Jan 10, 2016 #3
    Thank you, I knew I could justify it using the information in a wordy explanation but I assumed the 'Show that' nature of the question meant I'd need to prove it mathematically using the angles in the CM, but it does make more sense to just explain.
    Thanks again!
     
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