## College Physics (4th Edition)

(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.
(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.