Solutions 060519

Question 1

The two fragments can be treated as point charges for purposes of calculating their potential energy. Using energy conservation, the potential energy is all converted to kinetic energy as the two fragments separate to a large distance.

$E_{\textrm {initial}}=E_{\textrm {final}}\rightarrow U_{\textrm {initial}}=K_{\textrm {final}}={\frac {1}{4\pi \epsilon _{0}}}{\frac {q_{1}q_{2}}{r}}V$

$=8.99\times 10^{9}{\textrm {N}}{\textrm {m}}^{2}/{\textrm {C}}^{2}{\frac {(38)(54)(1.60\times 10^{-19}{\textrm {C}})^{2}}{(5.5\times 10^{-15}{\textrm {m}})+(6.2\times 10^{-15}{\textrm {m}})}}{\frac {1{\textrm {eV}}}{1.60\times 10^{-19}{\textrm {J}}}}=250\times 10^{6}{\textrm {eV}}=250{\textrm {MeV}}$

This is about %25 greater than the observed kinetic energy of 200 MeV