Answer
(a) The enmeshed cars were moving with a speed of 7.09 m/s after the collision.
(b) Car A was moving with a speed of 5.24 m/s just before the collision.
Work Step by Step
(a) Let south be the positive direction.
$p_y = (1500~kg)(15~m/s)$
$p_y = 22500~kg~m/s$
After the collision:
$m~v~sin(65^{\circ})= p_y$
$v=\frac{p_y}{m~sin(65^{\circ})}$
$v=\frac{22500~kg~m/s}{(3500~kg)~sin(65^{\circ})}$
$v = 7.09~m/s$
The enmeshed cars were moving with a speed of 7.09 m/s after the collision.
(b) $m_A~v_{A1} = p_x$
$v_{A1} = \frac{p_x}{m_A}$
$v_{A1} = \frac{m~v~cos(65^{\circ})}{m_A}$
$v_{A1} = \frac{(3500~kg)(7.09~m/s)~cos(65^{\circ})}{2000~kg}$
$v_{A1} = 5.24~m/s$
Car A was moving with a speed of 5.24 m/s just before the collision.
