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Ackbar

  • 5 years ago

A proton (mass 1 amu) is shot at a speed of 5.0 x 10^6 m/s toward a gold target. The nucleus of a gold atom (mass 197 amu) repels the proton and deflects it straight back toward the source with 90% of its initial speed. What is the recoil speed of the gold nucleus?

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  1. LeoMessi
    • 5 years ago
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    v_gf = p_gf/m_g = (p_net-p_pf)/m_g = (m_p⋅v_pi-m_p⋅v_pf)/m_g = = (m_p⋅v_pi-m_p⋅v_pi⋅r_vp)/m_g = m_p⋅v_pi⋅(1-r_vp)/m_g v_gf is the post-deflection velocity of the gold atom p_gf is the post-deflection momentum of the gold atom m_g is the mass of the gold atom, 197 amu p_net is the conserved net momentum of the system p_pf is the post-deflection momentum of the proton m_p is the mass of the proton, 1 amu v_pi is the pre-deflection velocity of the proton, 5.0×10⁶ m/s v_pf is the post-deflection velocity of the proton r_vp is the ratio of pre- and post-deflection velocities of the proton, -0.90 (The pre-deflection velocity and momentum of the gold atom is omitted, because it contributes zero to the net momentum of the system)

  2. LeoMessi
    • 5 years ago
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    v_gf = m_p⋅v_pi⋅(1-r_vp)/m_g = (1 amu)⋅(5.0×10⁶ m/s)⋅(1-(-0.90))/(197 amu) = 4.8223×10⁴ m/s v_gf = (1 amu)⋅v_pi⋅(1-(-0.90))/(197 amu) = (1.9/197)⋅v_pi = 0.009645⋅v_pi ... including the initial state of the gold atom: v_gf = v_gi+(m_p⋅v_pi+m_g⋅v_gi-m_p⋅v_pi⋅r_vp)/m_g

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