Answer
The total kinetic energy of the system decreases by $~~1.25~J$
Work Step by Step
We can find the total kinetic energy of the system before the collision:
$K_i = \frac{1}{2}(5.0~kg)(3.0~m/s)^2+\frac{1}{2}(10.0~kg)(2.0~m/s)^2$
$K_i = 22.5~J+20.0~J$
$K_i = 42.5~J$
In part (a), we found that the velocity of the 5.0-kg block is $2.0~m/s$ after the collision.
We can find the total kinetic energy of the system after the collision:
$K_f = \frac{1}{2}(5.0~kg)(2.0~m/s)^2+\frac{1}{2}(10.0~kg)(2.5~m/s)^2$
$K_f = 10.0~J+31.25~J$
$K_f = 41.25~J$
We can find the change in the kinetic energy of the system:
$\Delta K = K_f - K_i = 41.25~J - 42.5~J = -1.25~J$
The total kinetic energy of the system decreases by $~~1.25~J$
