A proton and an \"positron\" (identical to an electron, except positively charge

Question

A proton and an "positron" (identical to an electron, except positively charged) are brought 6 pm apart and released from rest. What is the initial potential energy stored by this system? In all of the previous problems on this homework, the system's collective potential energy has been assigned to a single charge which is moveable. In this problem, we'll let BOTH charges move. The collective kinetic energy of the charges (K_sys) will be drawn from U_sys. However, since both charges are moving, they will need to "share" this energy. Once in motion, what percentage of K_sys does each charge have at any given moment? (In this scenario, how valid is our usual approximation to assign the entire U_sys to one particle? Which particle should it be assigned to?) What is the speed of each charge after they have repelled a long distance apart?

Explanation / Answer

a)

Initial PE = k*q*q/r = k*q^2/r

= 9*10^9*(1.6*10^-19)^2/(6*10^-6)

= 3.84*10^-23 J

b)

Momentum needs to be conserved:

So, m1*v1 + m2*v2 = 0

So, m1*v1 = -m2*v2 = p

So, KE of the proton, KE1 = 0.5*(m1)*v1^2 = 0.5*(m1*v1)^2/m1 = 0.5(p)^2/m1

Similarly, KE of the positron, KE2 =0.5*(m2)*v2^2 = 0.5*(m2*v2)^2/m2 = 0.5(p)^2/m2

So, total KE of the system , KE,sys = KE1 + KE2 = 0.5*(p)^2*(1/m1 + 1/m2)

So, percentage of K,sys proton has = ( 0.5*p^2/m1 /(0.5*p^2 *(1/m1 + 1/m2)) ) *100

= ( m2 / (m1+m2) )*100 = (9.1*10^-31)/(9.1*10^-31 + 1.67*10^-27) *100 = 0.05 percent <----answer

Percentage of K,sys positron has =  (1.67*10^-27)/(9.1*10^-31 + 1.67*10^-27) *100 = 99.95 percent <-----answer

All the initital PE is converted to KE

So, KE,proton = 0.05*(3.84*10^-23)/100 = 0.5*(1.67*10^-27)*vf^2

So, vf = 4.8 m/s <----answer

Now, for positron :

99.95*(3.84*10^-23)/100 = 0.5*(9.1*10^-31)*vf^2

So, vf = 9184 m/s <-----answer

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