Answer
The change in the total kinetic energy is $~~-500~J$
Work Step by Step
In part (a), we found that the final velocity of B is $~~v = (10\hat{i} +15\hat{j})~m/s$
We can find the initial kinetic energy:
$K_0 = \frac{1}{2}(2.0~kg)(\sqrt{(15~m/s)^2+(30~m/s)^2})^2+\frac{1}{2}(2.0~kg)(\sqrt{(-10~m/s)^2+(5.0~m/s)^2})^2$
$K_0 = 1250~J$
We can find the final kinetic energy:
$K_f = \frac{1}{2}(2.0~kg)(\sqrt{(-5.0~m/s)^2+(20~m/s)^2})^2+\frac{1}{2}(2.0~kg)(\sqrt{(10~m/s)^2+(15~m/s)^2})^2$
$K_f = 750~J$
We can find the change in the total kinetic energy:
$\Delta K = K_f-K_0$
$\Delta K = 750~J-1250~J$
$\Delta K = -500~J$
The change in the total kinetic energy is $~~-500~J$
