## College Physics (4th Edition)

(a) The velocity of the two cars after the collision is $11.3~m/s$ (b) $208,600~J$ of kinetic energy was converted to another form during the collision.
(a) By conservation of momentum, the final momentum of the system is equal to the initial momentum. We can find the east component $p_x$ of the momentum: $p_x = (1500~kg)(17~m/s) = 25,500~kg~m/s$ We can find the south component $p_y$ of the momentum: $p_y = (1800~kg)(15~m/s) = 27,000~kg~m/s$ We can find the magnitude of the momentum: $p = \sqrt{p_x^2+p_y^2} = \sqrt{(25,500~kg~m/s)^2+(27,000~kg~m/s)^2} = 37,138~kg~m/s$ We can find the velocity of the two cars after the collision: $mv = p$ $v = \frac{p}{m}$ $v = \frac{37,138~kg~m/s}{1500~kg+1800~kg}$ $v = 11.3~m/s$ The velocity of the two cars after the collision is $11.3~m/s$. (b) We can find the initial kinetic energy: $KE_0 = \frac{1}{2}(1500~kg)(17~m/s)^2 + \frac{1}{2}(1800~kg)(15~m/s)^2$ $KE_0 = 419,250~J$ We can find the final kinetic energy: $KE_f = \frac{1}{2}(1500~kg+1800~kg)(11.3~m/s)^2$ $KE_f = 210,688.5~J$ We can find the amount of kinetic energy that was converted to another form during the collision: $KE_0-KE_f = (419,250~J)-(210,688.5~J) = 208,600~J$ $208,600~J$ of kinetic energy was converted to another form during the collision.