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

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Homework Help Overview

The problem involves analyzing an inelastic collision between a car and a minivan at an intersection, where both vehicles stick together after the collision. The objective is to determine the final speed and direction of the combined wreckage, using principles of momentum conservation.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster describes the steps taken to set up momentum equations for both the x and y components. They express confusion about the method of dividing the y momentum equation by the x momentum equation to eliminate the final velocity and solve for the angle.

Discussion Status

Participants are engaged in clarifying the reasoning behind the algebraic manipulation used to eliminate variables. Some guidance has been provided regarding the validity of dividing equations under certain conditions, but no consensus on the approach has been reached.

Contextual Notes

The discussion includes a note about the original poster's notation confusion regarding subscripts and superscripts in their equations.

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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