## College Physics (4th Edition)

(a) The speed of the VW after the collision is $43~m/s$ (b) The forces that the cars exert on each other are equal in magnitude.
(a) Let $v_{b0}$ be the initial velocity of the BMW. Let $v_{bf}$ be the final velocity of the BMW. Let $v_{v0}$ be the initial velocity of the Volkswagen. Let $v_{vf}$ be the final velocity of the Volkswagen. By conservation of momentum, the final momentum of the two cars is equal to the initial momentum of the two cars. We can find the final speed $v_{vf}$ of the Volkswagen after the collision: $m_b~v_{bf}+m_v~v_{vf} = m_b~v_{b0}+m_v~v_{v0}$ $m_v~v_{vf} = m_b~(v_{b0}-v_{bf})+m_v~v_{v0}$ $v_{vf} = \frac{m_b~(v_{b0}-v_{bf})+m_v~v_{v0}}{m_v}$ $v_{vf} = \frac{(2.0\times 10^3~kg)~(42~m/s-33~m/s)+(1.0\times 10^3~kg)(25~m/s)}{1.0\times 10^3~kg}$ $v_{vf} = 43~m/s$ The speed of the VW after the collision is $43~m/s$ (b) By Newton's third law, if the BMW exerts a force on the VW, then the VW exerts an equal and opposite force on the BMW. Therefore, the forces that the cars exert on each other are equal in magnitude.