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Ackbar
 4 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?
Ackbar
 4 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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LeoMessi
 4 years ago
Best ResponseYou've already chosen the best response.0v_gf = p_gf/m_g = (p_netp_pf)/m_g = (m_p⋅v_pim_p⋅v_pf)/m_g = = (m_p⋅v_pim_p⋅v_pi⋅r_vp)/m_g = m_p⋅v_pi⋅(1r_vp)/m_g v_gf is the postdeflection velocity of the gold atom p_gf is the postdeflection 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 postdeflection momentum of the proton m_p is the mass of the proton, 1 amu v_pi is the predeflection velocity of the proton, 5.0×10⁶ m/s v_pf is the postdeflection velocity of the proton r_vp is the ratio of pre and postdeflection velocities of the proton, 0.90 (The predeflection velocity and momentum of the gold atom is omitted, because it contributes zero to the net momentum of the system)

LeoMessi
 4 years ago
Best ResponseYou've already chosen the best response.0v_gf = m_p⋅v_pi⋅(1r_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_gim_p⋅v_pi⋅r_vp)/m_g
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