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Particle Collisions (Momentum)

  1. Nov 27, 2006 #1
    You are at the controls of a particle accelerator, sending beam of 2.70*10^7 m/s protons (mass m) at a gas target of an unknown element. Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. All such protons rebound with a speed of 2.40*10^7 m/s. Assume that the initial speed of the target nucleus is negligible and that the collision is elastic.

    (a) Find the mass of one of the nuclei of the unknown element. Express you answer in terms of the proton mass m.

    (b) What is the speed of the unknown nucleus immediately after such a collision?

    I really do not know how to start this problem. I know you somehow have to use momentum but i am not sure how to start it without the mass given?
  2. jcsd
  3. Nov 27, 2006 #2
    Hint: conservation of linear momentum.
  4. Nov 27, 2006 #3
    The mass is given if you have notice. It is m and you are suppose to leave it in such a way. Apply conservation of momentum and write out your equation for it then.
  5. Oct 24, 2007 #4
    i dont get it... so the mass m is 2.10×10^7...????
  6. Oct 24, 2007 #5
    i meant 2.70*10^7
  7. Oct 25, 2007 #6
    "Express your answer in terms of the proton mass m."
    Therefore, mass of the proton = m.
    Assume, mass of the unknown nucleus = k.m, where 'k' is some constant. We need to find 'k'.
    Assume, direction of proton before collision be +X.
    Initial velocity of the proton, u = +2.70*10^7 m/s.
    "Assume that the initial speed of the target nucleus is negligible."
    Thus, initial velocity of the nucleus = 0 m/s.
    Final velocity of the proton, u' = -2.40*10^7 m/s. [Note that it rebounds with this speed.]
    Let, final velocity of the proton be +v.
    Collision is elastic => |velocity of separation along the line of collision| = |velocity of approach along the line of collision|
    => |u' - v| = |u| => v + 2.40*10^7 m/s = 2.70*10^7 m/s.
    => v = +0.30*10^7 m/s.

    Further, apply linear momentum balance along X-direction:
    m.(u) + k.m.(0) = m.(u') + k.m.(v)
    Solve it to get the value of 'k' (= 17).
    Thus, mass of the unknown nucleus is 17m.
  8. Oct 25, 2007 #7
    17m is not the right answer.......
  9. Oct 26, 2007 #8
    could you, plz, tell the correct answer?
  10. Nov 6, 2009 #9
    Not to beat a dead thread, but, perhaps this thread would be more useful if it contained a complete explanation for people who stumble on it in the future. Here is another way of looking at the problem:

    It's an elastic collision where the target body is at rest.

    Let the fired particle be A and the target nucleus be B.

    The initial x-velocity of A is:

    [tex]} v_{\texttt{A1x}} = 2.70\times10^7[/tex]

    The final x-velocity of A is:

    [tex]v_{\texttt{A2x}} = -2.40\times10^7[/tex]

    Since B is initially at rest, it's x-velocity is:

    [tex]v_{\texttt{B1x}} = 0[/tex]

    When the target body is at rest, the momentum conservation equation is:

    [tex]m_{\texttt{A}}v_{\texttt{A1x}} = m_{\texttt{A}} v_{\texttt{A2x}} + m_{\texttt{B}} v_{\texttt{B2x}}[/tex]

    From this, we derive:

    Eq. 1: [tex]m_{\texttt{B}} v_{\texttt{B2x}} = m_{\texttt{A}}(v_{\texttt{A1x}} - v_{\texttt{A2x}})[/tex]

    When the target body is at rest, the kinetic energy equation is:

    Eq. 2: [tex]m_{\texttt{B}} v_{\texttt{B2x}}^2 = m_{\texttt{A}}(v_{\texttt{A1x}}^2 - v_{\texttt{A2x}}^2) = m_{\texttt{A}}(v_{\texttt{A1x}} - v_{\texttt{A2x}})(v_{\texttt{A1x}} + v_{\texttt{A2x}})[/tex]

    Dividing equation 2 by 1, we are left with:

    [tex]v_{\texttt{B2x}} = v_{\texttt{A1x}} + v_{\texttt{A2x}}[/tex]

    Note that we were given the initial x-velocity and final x-velocity of A, so we can use these velocities to calculate the final x-velocity of B. This answers part (a) of the question.

    Then, substitute your result for the final x-velocity of B into equation 1 and solve for the mass of B. This answers part (b) of the question.
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