Help with collisions?

Question

I'm having a hard time with this problem (and I want to finally go to sleep plz)

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 (as indicated in the figure). After the collision, the two-car system travels at speed v_final at an angle theta east of north.

Please see the graph http://session.masteringphysics.com/problemAsset/1007938/22/6318.jpg

Thank you

I forgot about the question:
Find the final velocity

and express the answer in terms of v and the angle phi

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-north, 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)]
----------------
Had phi = 90 (going south), then V_ = v/2 and thita = 0 (System going up north due to 2v magnitude).

Answer 2

This is conservation of momentum in two dimensions.

Mass times velocity before = mass times velocity after.

They give you initial masses and velocities. You just need to use trigonometry to break the one velocity into components. x component = magnitude cos theta. y component = minus magnitude sine theta (minus because the angle is given S of E instead of N like usual)

The masses don't change and they stick together, so just divide the total momentum by the total mass.

At the end, you have to use trigonometry again to turn the component velocities and momenta into magnitude and angle.

Magnitude: use pythagorus
Angle = arctangent of y component over x (you'll have to take the complement of the ange (90 -) to get from the normal way of measuring) N of E to the funny way the ask you to do it (E of N).

Good luck!

Answer 3

I don't understand why we have 2m {[V_ sin (thita)] i + [V_ sin (thita)] j} in calculation
m { (2v) j + [v cos (phi)] i – [v sin (phi)] j} = 2m {[V_ sin (thita)] i + [V_ sin (thita)] j}
I think it must be:
m { (2v) j + [v cos (phi)] i – [v sin (phi)] j} = 2m {[V_ cos (thita)] i + [V_ sin (thita)] j}