Is my logic right for the answer to this inelastic car collision question?

[Lori]

Homework Statement

A 1000 kg car approaches an intersection traveling north at 20.0 m/s . A 1200 kg approaches this same intersection traveling east at 22.0 m/s. The two car collides and lock together. Ignoring any external forces , what is the velocity of the car immediately after the collision.

a)29.7 m/s, 47.7 degrees east of north
b)21.1, 47.7 west of south
c)15.1 m/s, 52.8 east of north
d) 21.1 m/s , 52.8 east of north
e) 21.1 m/s 47.7 east of north

Homework Equations

P = mv (momentum)
Mass 1 car = 1000
mass 2 car = 1200

The Attempt at a Solution

This is a inelastic collision so the velocity final would be the same for the two objects with mass 1 + mass 2
First thing i would do would be to find Py (momentum in y direction) and Px (in x direction). If i find the magnitude by squaring them and tying the sqrt, I would be magnitude of momentum total. Thus, if i solve for v in P = mv, where m is the total mass, i can get v. The Py and Px will also give me the direction right?

My work:
Py = 1000kg*20 = 20000
Px = 1200kg*22 m/s = 26400
P = 33120.37 = mv
33129.37 = (1000+1200kg)v
v = 15.1 m/s and the direction is east of north (north west)

 

[TSny]

Your work looks good to me.

v = 15.1 m/s and the direction is east of north (north west)

Were you able to get the specific angle in degrees?

I don't understand why you would say that "east of north" is the same as "north west".

 

[Lori]

oops, i meant north east!, yes, i took the arctan and got 52.8 degrees

 

[TSny]

oops, i meant north east!, yes, i took the arctan and got 52.8 degrees

Sounds good!