Answer
(a) The magnitude of the net momentum is 22.5 kg m/s and the direction is to the left.
(b) K = 838 J
Work Step by Step
(a) $p = m_1v_1+m_2v_2$
$p = (110~kg)(2.75~m/s)+(125~kg)(-2.60~m/s)$
$p = -22.5~kg~m/s$
The magnitude of the net momentum is 22.5 kg m/s and the direction is to the left.
(b) $K = \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$
$K = \frac{1}{2}(110~kg)(2.75~m/s)^2+\frac{1}{2}(125~kg)(2.60~m/s)^2$
$K = 838~J$
