Inelastic collision at an angle with two cars

[huh]

almost there-inelastic collision at an angle

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.

I want to find v(final) in terms of v and phi.

Only momentum is conserved, so would I just use m1(2v1)+m2v2cos(phi)=(m1+v1)v(final)sin(theta)

or could it be (m1+m2)2vcos(phi) divided by (m1+m2)

I'm not sure how to set up the equation...I've been confusing myself...

 

[huh]

Please help me understand.

 

[radou]

You mixed the sines and cosines up - it should be: m(2v)+m(v)sin(phi) =(2m)v(final)cos(theta)

 

[huh]

Thanks. okay, so...

m2v+ mv sin(phi) divided by 2m cos(theta) is v(final)?

how can I get v(final) with only v and Phi? I can't have m or theta in the equation.

 

[radou]

Well, to get rid of m, you can obviously divide the whole equation by m.

 

[huh]

Wouldn't there be an extra m on the bottom, and how could I get rid of theta and be left with phi, or does theta count for much (is it negligible)?

 

[radou]

Wouldn't there be an extra m on the bottom, and how could I get rid of theta and be left with phi, or does theta count for much (is it negligible)?

Since momentum is a vector quantity, you can write the equation of conservation of momentum for the other direction, too. (The x-direction.) That should allow you to elliminate the angle phi.

 

[soupastupid]

what does v final look like?

what do u mean write the equation of momentum for the other direction?