anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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)
anonymous
  • anonymous
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.