Solution to 1992 Problem 75

In order for momentum to be conserved, the thorium nucleus and the helium nucleus must have ``equal and opposite" momenta. The relationship between momentum p and total energy E is E^2 = m^2 c^4 + p^2 c^2. The kinetic energy T is defined as T = E - m c^2. Therefore, the relationship between momentum and kinetic energy is (T + m c^2)^2 = m^2 c^4 + p^2 c^2 or
\begin{align*}T = \sqrt{m^2 c^4 + p^2 c^2} - m c^2\end{align*}
The derivative of T with respect to m is
\begin{align*}\frac{\partial T}{ \partial m} = \frac{ m c^4}{\sqrt{m^2 c^4 + p^2 c^2}}- c^2\end{align*}
which is always less than 0. Therefore, the helium nucleus will have more kinetic energy because it has smaller mass. Therefore, answer (E) is correct.

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