Question

(3 points) A proton makes a head-on elastic collision with a nucleus at rest and rebounds with a speed which is nine-tenths of its initial speed. The mass of the proton s m = 1.673 × 10-27 kg and its charge is e = 1.6 × 10-19 C. (a) What is the mass of the nucleus? (b) If the nucleus has a charge +9e and the proton has an initial energy of 1 MeV, what is the distance of closest approach between the proton and nucleus?
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Answer 1

By momentum conservation Total Initial Momentum = Total Final momentum mV = mv_1 + mV_2 because v_1 = -9/10 u therefore mv_2 = 19/10 mv By Energy conservation Initial Kinetic Energy = Final Kinetic Energy (mv)^2/2m = (mv_1)^2/2m + (mv_2)^2/2 m put mv_1 = -9/10 mv & mv_2 = 19/10 mv we get 1/2m = 1/2m [-9/10]^2 + [19/10]^2 1/2m doublerightarrow M = 19 m = 19 times 1.673 times 10^-27 M = 3.1787 times 10^-26 kg (c) when proton and nucleus are closest, All of kinetic Energy that has been lost is transformed into Electrostatic potential Energy By momentum conservation mv = mv_c + MV_c doublerightarrow V_c = m/M + m v KE. Lost = 1/2 (mv)^2/m - ((M + m)v_c)^2/2(M + m) = 1/2 (mv)^2 [1/m - 1/M + m] = 1/2 (mv)^2/m M/M + m P.E gained = K q_1 q_2/d^2 = k middot g_e middot C/d^2 because Initial Energy \r\n