Inelastic Collisions in 2-D

1. Nov 20, 2007

Hurricane3

1. The problem statement, all variables and given/known data
A car with mass of 950 kg and speed of 16 m/s approches an intersection in the positive x direction. A 1300-kg minivan traveling at 21m/s is heading towards the same intersection in the positive y direction. The car and minivan collide and stick together. Fine the speed and direction of the wrecked vehicles just after the collision, assuming external forces can be ignored.

2. Relevant equations
p = mv
$$\sum$$ p before = $$\sum$$ after

3. The attempt at a solution

Well this question was grabbed from my physics textbook as an example question.

Here are the steps they preformed:

1. Set the initial x component of momentum equal to the final x component of momentum
m$$_{1}$$v$$_{1}$$ = (m$$_{1}$$ + m$$_{2}$$)v$$_{f}$$cos$$\vartheta$$

2. Do the same of the y component of momentum.
m$$_{2}$$v$$_{2}$$ = (m$$_{2}$$ + m$$_{2}$$)v$$_{f}$$sin$$\vartheta$$

I understand everything up to this part, but I got confused when they divide the y momentum equation by the x momentum equation. They did it to eliminate the final velocity, giving an equation solving for $$\vartheta$$ alone.

My question is: Why is it that you can divide to eliminate variables??

Oh, and those are suppose to be subscripts, not superscripts.....

Last edited: Nov 20, 2007
2. Nov 20, 2007

azatkgz

$$\sin^2\theta+\cos^2\theta=1$$

Just put $$\sin\theta$$ and $$\cos\theta$$

And unknown left is v(f)

3. Nov 20, 2007

nrqed

You can always divide two equations. If A = B and C=D, then the ratio A/C must equal B/D as long as C and D are not zero. This is a common trick.

4. Nov 22, 2007

Hurricane3

ahh okay thanks.

i never did it this way before :shy: