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.

Solution to 1992 Problem 75