How Do You Calculate the Final Speed of Two Colliding Cars?

[shanshan]

Homework Statement

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 x south of east (as indicated in the figure). After the collision, the two-car system travels at speed vf at an angle east of north. Find the final speed of the two cars after the collision

Homework Equations

The Attempt at a Solution

We assume momentum is conserved, so that
p(i) = p(f)
p(f) = (2m)vf
p(east) = cosx*mv
vf(easte) = (vcosx)/2
p(north) = 2mv - mvsinx
vf(north) = (2v-vsinx)/2

So now that I've figured all these values out, I know that I can use pythagoras' theorum to solve for vf:

vf = [(vcosx/2)^2+((2v-vsinx)/2)^2)]^1/2
= [(v^2cos^2(x)+v^2sin^2(x)+4v^2+2vsinx)/2]^1/2
= [(v^2(1)+4v^2-2vsinx)/2]^1/2 ---- using trig identity sin^2x+cos^x = 1
= [(5v^2-2vsinx)/2]^1/2

But this is not the answer - is anyone able to find any problems in my work, or can I simplify it further?
I am pretty sure the mistake must be at some point after I use pythagoras' theorum, but I can't seem to find it.

 

[gneill]

Pull out the v/2 term first to make things cleaner. Take the square root at the end.

vf² = (v/2)²*(cos(x)² + (2 - sin(x))²)

= (v/2)²*( (1 - sin(x)²) + (2 - sin(x))² )

proceed.

 

[ashishsinghal]

vf = [(vcosx/2)^2+((2v-vsinx)/2)^2)]^1/2
= [(v^2cos^2(x)+v^2sin^2(x)+4v^2+2vsinx)/2]^1/2
= [(v^2(1)+4v^2-2vsinx)/2]^1/2 ---- using trig identity sin^2x+cos^x = 1
= [(5v^2-2vsinx)/2]^1/2

[(v^2cos^2(x)+v^2sin^2(x)+4v^2+2vsinx)/2]^1/2

here this should have been
[(v^2cos^2(x)/4+v^2sin^2(x)+4v^2-4vsinx)/2]^1/2

 

[ashishsinghal]

Now, check the answer.