Answer
(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) $p = 50400~kg~m/s$
Work Step by Step
(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).
