(N/A) The relationship between kinetic energy $(K.E.)$ and momentum $(p)$ is given by $K.E. = \frac{p^2}{2m}$.
Given that both objects have the same momentum $(p_1 = p_2 = p)$,
Let $m_1$ be the mass of the light object and $m_2$ be the mass of the heavy object,where $m_1 < m_2$.
The kinetic energy of the light object is $(K.E.)_1 = \frac{p^2}{2m_1}$.
The kinetic energy of the heavy object is $(K.E.)_2 = \frac{p^2}{2m_2}$.
The ratio of their kinetic energies is $\frac{(K.E.)_1}{(K.E.)_2} = \frac{\frac{p^2}{2m_1}}{\frac{p^2}{2m_2}} = \frac{m_2}{m_1}$.
Since $m_1 < m_2$,it follows that $\frac{m_2}{m_1} > 1$,which implies $(K.E.)_1 > (K.E.)_2$.
Therefore,the light object has a larger kinetic energy.