Answer
(a) The lighter skater will travel at a speed of 1.74 m/s in the opposite direction.
(b) The amount of kinetic energy created is 180 J. This energy comes from the work done by the skaters during the push.
Work Step by Step
$p_1+p_2 = 0$
$p_2 = -p_1$
$m_2v_2 = -m_1v_1$
$v_2 = \frac{-m_1v_1}{m_2}$
$v_2 = \frac{-(725~N/9.80~m/s^2)(1.50~m/s)}{625~N/9.80~m/s^2}$
$v_2 = -1.74~m/s$
The lighter skater will travel at a speed of 1.74 m/s in the opposite direction.
(b) $K = \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$
$K = \frac{1}{2}(725~N/9.80~m/s^2)(1.50~m/s)^2+\frac{1}{2}(625~N/9.80~m/s^2)(1.74~m/s)^2$
$K = 180~J$
The amount of kinetic energy created is 180 J. This energy comes from the work done by the skaters during the push.
