A uniform rod of mass m and length l is struck at one end by a...

The linear impulse applied to the rod is $J = F t$.

This impulse provides a translational velocity to the center of mass, given by $v = \dfrac{J}{m} = \dfrac{F t}{m}$.

The angular impulse about the center of mass is the product of the linear impulse and the perpendicular distance from the center of mass, which is $L = J \times \dfrac{l}{2} = \dfrac{F t l}{2}$.

The moment of inertia of the uniform rod about its center of mass is $I = \dfrac{m l^2}{12}$.

The translational kinetic energy is $K_{trans} = \dfrac{p^2}{2m} = \dfrac{(F t)^2}{2m} = \dfrac{F^2 t^2}{2m}$.

The rotational kinetic energy is $K_{rot} = \dfrac{L^2}{2I} = \dfrac{\left(\dfrac{F t l}{2}\right)^2}{2 \left(\dfrac{m l^2}{12}\right)} = \dfrac{\dfrac{F^2 t^2 l^2}{4}}{\dfrac{m l^2}{6}} = \dfrac{3 F^2 t^2}{2m}$.

The total kinetic energy of the rod after the force has stopped acting is the sum of the translational and rotational kinetic energies:

$K = K_{trans} + K_{rot} = \dfrac{F^2 t^2}{2m} + \dfrac{3 F^2 t^2}{2m} = \dfrac{4 F^2 t^2}{2m} = \dfrac{2 F^2 t^2}{m}$.