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

Hurricane3
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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
[tex]\sum[/tex] p before = [tex]\sum[/tex] 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[tex]_{1}[/tex]v[tex]_{1}[/tex] = (m[tex]_{1}[/tex] + m[tex]_{2}[/tex])v[tex]_{f}[/tex]cos[tex]\vartheta[/tex]

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

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 [tex]\vartheta[/tex] 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:
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[tex]\sin^2\theta+\cos^2\theta=1[/tex]

Just put [tex]\sin\theta[/tex] and [tex]\cos\theta[/tex]

And unknown left is v(f)
 
Hurricane3 said:

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
[tex]\sum[/tex] p before = [tex]\sum[/tex] 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[tex]_{1}[/tex]v[tex]_{1}[/tex] = (m[tex]_{1}[/tex] + m[tex]_{2}[/tex])v[tex]_{f}[/tex]cos[tex]\vartheta[/tex]

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

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 [tex]\vartheta[/tex] 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.
 
ahh okay thanks.

i never did it this way before :shy:
 

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