Physics- car collision at an angle?

Question

Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at speed 2v, while the second was traveling at speed v at an angle phi south of east. After the collision, the two-car system travels at speed v_final at an angle theta east of north.

What is the speed v_final of the joined cars after the collision

Answer 1

Let i be unit vector pointing to east, and j that to north. (+x,+y) sense
Before collision:
Car with [m, 2v], heading north, with speed 2v has velocity vector as
v1 = (2v) j

Car with [m, v], traveling with speed v at angle “phi” south-of-east, has velocity vector as
v2 = [v cos (phi)] i – [v sin (phi)] j

After collision:
Cars stick together [2m, V_]
and travel with speed V_ at angle “thita” east-of-east, has velocity vector as
Vf = [V_ sin (thita)] i + [V_ sin (thita)] j

Momentum conservation
m(v1) + m(v2) = 2m Vf

m { (2v) j + [v cos (phi)] i – [v sin (phi)] j} = 2m {[V_ sin (thita)] i + [V_ sin (thita)] j}

{ (2v) j + [v cos (phi)] i – [v sin (phi)] j}=2 {[V_ sin (thita)] i + [V_ sin (thita)] j} comparing coefficients of I and j unit vectors:

(2v) – [v sin (phi)] = 2 [V_ sin (thita)] (1)
[v cos (phi)] = 2 [V_ sin (thita)] (2)

these give tan (thita) = cos (phi) / [2 – sin (phi)]
sin (thita) = cos (phi) / sqrt [5 – 4 sin (phi)]
(2) gives

V_ = (v/2) sqrt [5 – 4 sin (phi)]
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Had phi = 90 (going south), then V_ = v/2 and thita = 0 (System going up north due to 2v magnitude).

Answer 2

m1u1+m2u2 = (m1+m2)V
m1=m2=m
v* unit vector north= j
v* unit vector east = i

m(2)j + m(sin phi (-j) + (cos phi (i)) = 2m (V)
m[j(2-sin phi) + cos phi i) = 2m V
V = j (2-sin phi)/2 + i (1/2 cos phi) (remember to multiply by the initial velocity though))

I may have mixed up the sin and cos, it depends on how you draw phi.