Answer
The collision is elastic.
Work Step by Step
Let $M$ be the mass of each ball.
We can find an expression for the kinetic energy before the collision:
$K_i = \frac{1}{2}M(2.2)^2$
$K_i = (2.4~M)~J$
In part (b), we found that the horizontal component of the velocity of ball 1 is $1.65~m/s$ after the collision.
In part (b), we found that the vertical component of the velocity of ball 1 is $-0.95~m/s$ after the collision.
We can find the speed of ball 1 after the collision:
$v = \sqrt{(1.65~m/s)^2+(-0.95~m/s)^2}$
$v = 1.9~m/s$
We can find an expression for the kinetic energy after the collision:
$K_f = \frac{1}{2}M(1.1)^2+\frac{1}{2}M(1.9)^2$
$K_f = (0.605~M)~J+(1.805~M)~J$
$K_f = (2.4~M)~J$
Since the kinetic energy after the collision is the same as the kinetic energy before the collision, the collision is elastic.
