Answer
(a) Before the collision, Sam's speed was 9.97 m/s and Abigail's speed was 2.26 m/s.
(b) The total kinetic energy decreased by 639 J during the collision.
Work Step by Step
(a) Let $S$ be Sam and let $A$ be Abigail.
$p_x = m_S~v_{S1}$
$m_S~v_{S1} = m_S~v_{S2x}+m_A~v_{A2x}$
$v_{S1} = \frac{m_S~v_{S2x}+m_A~v_{A2x}}{m_S}$
$v_{S1} = \frac{(80.0~kg)(6.00~m/s)~cos(37.0^{\circ})+(50.0~kg)(9.00~m/s)~cos(23.0^{\circ})}{80.0~kg}$
$v_{S1} = 9.97~m/s$
$p_y = m_A~v_{A1}$
$m_A~v_{A1} = m_S~v_{S2y}-m_A~v_{A2y}$
$v_{A1} = \frac{m_S~v_{S2y}-m_A~v_{A2y}}{m_A}$
$v_{A1} = \frac{(80.0~kg)(6.00~m/s)~sin(37.0^{\circ})-(50.0~kg)(9.00~m/s)~sin(23.0^{\circ})}{50.0~kg}$
$v_{S1} = 2.26~m/s$
Before the collision, Sam's speed was 9.97 m/s and Abigail's speed was 2.26 m/s.
(b) $K_1 = \frac{1}{2}m_Sv_{S1}^2+\frac{1}{2}m_Av_{A1}^2$
$K_1 = \frac{1}{2}(80.0~kg)(9.97~m/s)^2+\frac{1}{2}(50.0~kg)(2.26~m/s)^2$
$K_1 = 4104~J$
$K_2 = \frac{1}{2}m_Sv_{S2}^2+\frac{1}{2}m_Av_{A2}^2$
$K_2 = \frac{1}{2}(80.0~kg)(6.00~m/s)^2+\frac{1}{2}(50.0~kg)(9.00~m/s)^2$
$K_2 = 3465~J$
$\Delta K = K_2-K_1$
$\Delta K = 3465~J - 4104~J$
$\Delta K = -639$
The total kinetic energy decreased by 639 J during the collision.
