Analyzing Inelastic Collisions in 2-D: Solving for Final Velocity and Angle

[Hurricane3]

Homework Statement

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.

Homework Equations

p = mv
\sum p before = \sum after

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...

 

[azatkgz]

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

Just put \sin\theta and \cos\theta

And unknown left is v(f)

 

[nrqed]

Homework Statement

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.

Homework Equations

p = mv
\sum p before = \sum after

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...

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.

 

[Hurricane3]

ahh okay thanks.

i never did it this way before :shy: