## University Physics with Modern Physics (14th Edition)

(a) The center of mass is 24.0 meters in front of the 1200-kg car. (b) $p = 50400~kg~m/s$ (c) $v_{cm} = 16.8~m/s$ (d) $p = 50400~kg~m/s$
(a) Let the 1200-kg car be at the origin. $x_{cm} = \frac{m_1x_1+m_2x_2}{m_1+m_2}$ $x_{cm} = \frac{0+(1800~kg)(40.0~m)}{1200~kg+1800~kg}$ $x_{cm} = 24.0~m$ The center of mass is 24.0 meters in front of the 1200-kg car. (b) $p = m_1v_1+m_2v_2$ $p = (1200~kg)(12.0~m/s)+(1800~kg)(20.0~m/s)$ $p = 50400~kg~m/s$ (c) $v_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}$ $v_{cm} = \frac{(1200~kg)(12.0~m/s)+(1800~kg)(20.0~m/s)}{1200~kg+1800~kg}$ $v_{cm} = 16.8~m/s$ (d) $p = (m_1+m_2)~v_{cm}$ $p = (1200~kg+1800~kg)(16.8~m/s)$ $p = 50400~kg~m/s$ As expected, this result is the same as part (b).