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

In summary, after the collision, the car has a final speed of 16 m/s and the minivan has a final speed of 21 m/s.
  • #1
Hurricane3
16
0

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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  • #2
[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)
 
  • #3
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.
 
  • #4
ahh okay thanks.

i never did it this way before :shy:
 

What is an inelastic collision in 2-D?

An inelastic collision in 2-D is a type of collision in which two objects collide and stick together after the collision, losing some of their initial kinetic energy.

How is momentum conserved in an inelastic collision in 2-D?

In an inelastic collision in 2-D, momentum is conserved in both the x and y directions. This means that the total momentum in the x direction before the collision is equal to the total momentum in the x direction after the collision, and the same applies for the y direction.

What is the equation for calculating the final velocities of objects in an inelastic collision in 2-D?

The equation is m1v1i + m2v2i = (m1 + m2)vfinal, where m1 and m2 are the masses of the two objects, v1i and v2i are their initial velocities, and vfinal is their final velocity after the collision.

Can there be a perfectly inelastic collision in 2-D?

Yes, in a perfectly inelastic collision in 2-D, the two objects stick together and move with the same final velocity. This means that all of the initial kinetic energy is lost and there is no rebound or separation between the objects.

What is the difference between an inelastic collision and an elastic collision in 2-D?

In an elastic collision in 2-D, both momentum and kinetic energy are conserved. This means that the objects may bounce off each other after the collision and have different final velocities than their initial velocities. In an inelastic collision, only momentum is conserved and some kinetic energy is lost.

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