Answer
The amount of kinetic energy acquired by the smaller block is 3300 J
The amount of kinetic energy acquired by the larger block is 2200 J
Work Step by Step
Let the masses of the pieces be $m$ and $1.5m$
We can use conservation of momentum to set up an equation.
$m~v_A = 1.5~m~v_B$
$v_A = 1.5~v_B$
We can find the kinetic energy $KE_1$ of the smaller block.
$KE_1 = \frac{1}{2}mv_A^2 = \frac{1}{2}m(1.5v_B)^2$
$KE_1 = \frac{2.25}{2}m~v_B^2$
We can find the kinetic energy $KE_2$ of the larger block.
$KE_2 = \frac{1}{2}(1.5~m)v_B^2 = \frac{1.5}{2}m~v_B^2$
We can find the total kinetic energy $KE$ of the two blocks.
$KE = KE_1 + KE_2$
$KE = \frac{2.25}{2}m~v_B^2 + \frac{1.5}{2}m~v_B^2$
$KE = \frac{3.75}{2}m~v_B^2$
We can find the fraction of the total energy acquired by the smaller block.
$\frac{KE_1}{KE} = \frac{ \frac{2.25}{2}mv_B^2}{\frac{3.75}{2}mv_B^2} = \frac{2.25}{3.75} = 0.6$
The amount of kinetic energy acquired by the smaller block is $(0.6)(5500~J) = 3300~J$
The amount of kinetic energy acquired by the larger block is $5500~J - 3300~J = 2200~J$