Physics: Car Collision at an Angle, Trying to figure out, Please help!?

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

What is the angle theta with respect to north made by the velocity vector of the two cars after the collision?

Express your answer in terms of phi. Your answer should contain an inverse trigonometric function.

I was trying to solve it by breaking it down the components in x & y direction.

so i did in x-direction: (2m)v_i cos (0) - (2m) 2v_i sin (theta) = (2m)v_i tan (theta)

Now i used the formula m1v1+m2v2= (m1+m2)v3

Since this is inelastic i thought the 2nd direction would be subtracted since energy isn't conserved.

Please help i tried to do the algebra and im confused. Am i even supposed to be using this equation?

http://session.masteringphysics.com/problemAsset/1007938/22/6318.jpg

Answer 1

You have the right idea, but you're using wrong values.

I would also recommend drawing a diagram, if you haven't already done so.

The equation in the x direction (assuming north points in the +x direction) should be

m(2v) + m(-v sin(phi)) = (2m) v_final cos(theta)

The equation in the y-direction is

-mv cos(phi) = -2m v_final sin(theta)

(I write both with a minus sign because both components point in the negative y direction.)

m divides out everywhere, so we're left with

2v - v sin(phi) = 2 v_final cos(theta)
v cos(phi) = 2 v_final sin(theta)

So

sin(theta) = v cos(phi)/(2 v_final)
cos(theta) = v(2 - sin(phi))/(2 v_final)

Divide the first of these equations by the second equation to get an expression for tan(theta). Note that v and v_final drop out. You're almost done.

Answer 2

sqrt[(vcosϕ2)^2+(v−vsinϕ2)^2]

Answer 3

Here's a representation of what Ron W above said.

Answer 4

I want to ask the same question as the person above.