Answer
(a) v = 5.00 m/s
(b) v = 5.71 m/s
(c) v = 3.78 m/s
Work Step by Step
We can use conservation of momentum to solve these questions. Let $m_c$ be the final mass of the car and let $m_m$ be the mass of the object.
(a) Since the object is thrown sideways, it still has a velocity to the east of 5.00 m/s.
$m_cv_c + m_mv_m = m_0v_0$
$m_cv_c = m_0v_0 - m_mv_m$
$v_c = \frac{m_0v_0 - m_mv_m}{m_c}$
$v_c = \frac{(200~kg)(5.00~m/s) - (25.0~kg)(5.00~m/s)}{175~kg}$
$v_c = 5.00~m/s$
(b) Since the object is thrown backwards with a velocity of 5.00 m/s relative to the initial velocity of the car, the object will have a velocity of 0 relative to the ground.
$m_cv_c + m_mv_m = m_0v_0$
$m_cv_c = m_0v_0 - m_mv_m$
$v_c = \frac{m_0v_0 - m_mv_m}{m_c}$
$v_c = \frac{(200~kg)(5.00~m/s) - (25.0~kg)(0)}{175~kg}$
$v_c = 5.71~m/s$
(c) $m_cv_c = m_0v_0 + m_mv_m $
$v_c = \frac{m_0v_0 + m_mv_m}{m_c}$
$v_c = \frac{(200~kg)(5.00~m/s) + (25.0~kg)(-6.00~m/s)}{225~kg}$
$v_c = 3.78~m/s$
