Answer
(a) The magnitude of Daniel's velocity is 7.20 m/s at an angle of $38.0^{\circ}$ from Rebecca's initial direction.
(b) The change in kinetic energy is -678 J.
Work Step by Step
(a) Let's assume that Rebecca's initial direction was in the +x direction.
$p_x = m_R~v_{R1}$
$m_R~v_{Rx}+ m_D~v_{Dx} = m_R~v_{R1}$
$m_D~v_{Dx} = m_R~v_{R1}-m_R~v_{Rx}$
$v_{Dx} = \frac{m_R~v_{R1}-m_R~v_{Rx}}{m_D}$
$v_{Dx} = \frac{(45.0~kg)(13.0~m/s)-(45.0~kg)(8.00~m/s)~cos(53.1^{\circ})}{65.0~kg}$
$v_{Dx} = 5.67~m/s$
$p_y = 0$
$m_D~v_{Dy} = m_R~v_{Ry}$
$v_{Dy} = \frac{m_R~v_{Ry}}{m_D}$
$v_{Dy} = \frac{(45.0~kg)(8.00~m/s)~sin(53.1^{\circ})}{65.0~kg}$
$v_{Dy} = 4.43~m/s$
We can find the magnitude of Daniel's velocity.
$v_D = \sqrt{(5.67~m/s)^2+(4.43~m/s)^2}$
$v_D = 7.20~m/s$
We can find the angle $\theta$ from Rebecca's initial direction.
$tan(\theta) = \frac{4.43~m/s}{5.67~m/s}$
$\theta = arctan(\frac{4.43~m/s}{5.67~m/s})$
$\theta = 38.0^{\circ}$
The magnitude of Daniel's velocity is 7.20 m/s at an angle of $38.0^{\circ}$ from Rebecca's initial direction. (Note that Daniel's angle is in the opposite direction from Rebecca's angle of $53.1^{\circ}$.)
(b) $K_1 = \frac{1}{2}m_R~v_{R1}^2$
$K_1 = \frac{1}{2}(45.0~kg)(13.0~m/s)^2$
$K_1 = 3802.5~J$
$K_2 = \frac{1}{2}m_R~v_{R2}^2+\frac{1}{2}m_D~v_{D}^2$
$K_2 = \frac{1}{2}(45.0~kg)(8.00~m/s)^2+\frac{1}{2}(65.0~kg)(7.20~m/s)^2$
$K_2 = 3124.8~J$
We can find the change in kinetic energy.
$\Delta K = K_2-K_1$
$\Delta K = 3124.8~J-3802.5~J$
$\Delta K = -678~J$
The change in kinetic energy is -678 J.
