#### Answer

(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.

#### Work Step by Step

(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.