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.
