A1445-kg car, c, moving east at 21.2m/s, collides with a 2625-kg car, d,moving south at 17.5 m/s, and the cars stick together. in what direction and with what speed do they move after the collision

After the collision, they move with a speed of 13.57 m/s at 56.30° south of east

Explanation:

Conservation Of Momentum

The total momentum of a system of particles with masses m1,m2,...mn that interact without the action of external forces, is conserved. It means that, being and the initial  and final momentums respectively:

Or equivalently, for a two-mass system

All the velocities are vectors. Let's express the velocities in magnitude v and direction as . It's rectangular components will be

The first car is moving east at 21.2 m/s. Its velocity is

Recall that East is the zero-degree reference for angles

Expressing in rectangular form:

The second car is moving south at 17.5 m/s. Its velocity is

The total initial momentum is

They collide and stick together in a common mass and velocity \vec v', thus

It must be equal to the initial momentum, thus

Solving for

The magnitude of \vec v' is

The direction angle is

After the collision, they move with a speed of 13.57 m/s at 56.30° south of east