#### Answer

(a) After the collision, the final speed of the two cars is $0.20~m/s$
(b) Before the collision, the initial speed of the loaded car was $0.25~m/s$. Note that the loaded car was moving in the opposite direction from the empty car.

#### Work Step by Step

(a) By conservation of momentum, the final momentum of the two cars is equal to the initial momentum. We can find the final speed $v_f$ of the two cars:
$m_f~v_f = m~v_1+(4.0~m)~v_2$
$v_f = \frac{m~v_1+(4.0~m)~v_2}{m_f}$
$v_f = \frac{(m)(1.0~m/s)+(4.0~m)(0)}{5m}$
$v_f = 0.20~m/s$
After the collision, the final speed of the two cars is $0.20~m/s$
(b) By conservation of momentum, the final momentum of the two cars is equal to the initial momentum. We can find the initial speed $v_2$ of the loaded car:
$m_f~v_f = m~v_1+(4.0~m)~v_2$
$(5.0~m)(0) = (m)~(1.0~m/s)+(4.0~m)~v_2$
$(4.0~m)~v_2 = -(m)~(1.0~m/s)$
$v_2 = \frac{-(m)~(1.0~m/s)}{4.0~m}$
$v_2 = -0.25~m/s$
Before the collision, the initial speed of the loaded car was $0.25~m/s$. Note that the loaded car was moving in the opposite direction from the empty car.