Finding the Angle of Velocity After a Collision at an Angle

[gods_laughing]

Two cars, both of mass , collide and stick together. Prior to the collision, one car had been traveling north at speed , while the second was traveling at speed at an angle south of east (as indicated in the figure). After the collision, the two-car system travels at speed at an angle east of north.

What is the angle with respect to north made by the velocity vector of the two cars after the collision?
Express your answer in terms of . Your answer should contain an inverse trigonometric function.

I have already tried tan(theta)=cos(phi)/(2-sin(phi)) and arctan(cos(phi)/(2-sin(phi)) and tan(cos(phi)/(2-sin(phi)) but none seem to be the right answer

someone help please!

 

[Delphi51]

Make two headings for "north/south" and "east/west".
Under each write "momentum before = momentum after".
Use only the component of velocity in the direction specified by the heading in each mv calculation. You should be able to calculate the two components of velocity for the attached vehicles. Then you can use trigonometry to find their combined velocity, magnitude and direction.

 

[gods_laughing]

soooo for x I have v_finalx = (2m)v_final * cos (phi) and for y i have v_finaly = (2m)v_final * sin (phi). This would mean that tan (theta) = cos (phi)/sin(phi). I've already tried atan (cos(phi)/sin(phi)...right?

 

[Delphi51]

Sorry, I am not quite able to see your final answers. Certainly the m should have canceled out somewhere along the line so the answers will have the units of velocity rather than momentum. Also the initial velocities for the two cars should appear in the expressions for the final velocities. Maybe show your work so I can follow . . .