A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 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?
anonymous
 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?

This Question is Closed

anonymous
 5 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)

anonymous
 5 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.