#### Answer

(a) The other glider is moving at a velocity of $0.30~m/s$ to the left.
(b) The total kinetic energy of the gliders is greater after the collision because the elastic potential energy stored in the spring was converted into kinetic energy.

#### Work Step by Step

(a) Let $M$ be the mass of each glider. By conservation of momentum, the final momentum is equal to the initial momentum. We can find the velocity $v_f$ of the other glider after the spring is uncompressed:
$p_f = p_0$
$Mv_f+M~(1.30~m/s) = (2M)(0.50~m/s)$
$v_f = (2)(0.50~m/s) - (1.30~m/s)$
$v_f = -0.30~m/s$
The other glider is moving at a velocity of $0.30~m/s$ to the left.
(b) We can find the initial kinetic energy of the gliders:
$KE_0 = \frac{1}{2}(2M)v_0^2$
$KE_0 = \frac{1}{2}(2M)(0.50~m/s)^2$
$KE_0 = (0.25~M)~J$
We can find the final kinetic energy of the gliders:
$KE_f = \frac{1}{2}M(1.30~m/s)^2+\frac{1}{2}M(-0.30~m/s)^2$
$KE_f = (0.89~M)~J$
The total kinetic energy of the gliders is greater after the collision because the elastic potential energy stored in the spring was converted into kinetic energy.