Answer
Since $K_i = K_f$, the collision is elastic.
Work Step by Step
We can find the total kinetic energy before the collision:
$K_i = \frac{1}{2}(1.6~kg)(5.5~m/s)^2+\frac{1}{2}(2.4~kg)(2.5~m/s)^2$
$K_i = 24.2~J+7.5~J$
$K_i = 31.7~J$
In part (a), we found that $v_{Af} = 1.9~m/s$
We can find the total kinetic energy of the system after the collision:
$K_f = \frac{1}{2}(1.6~kg)(1.9~m/s)^2+\frac{1}{2}(2.4~kg)(4.9~m/s)^2$
$K_i = 2.888~J+28.812~J$
$K_f = 31.7~J$
Since $K_i = K_f$, the collision is elastic.
