Problem Involving Elastic Collision .

In summary: Absolutely! If v1 is positive, v2 must be negative. Momentum is a vector--direction matters. :)In summary, the two balls collide and each loses half of its original velocity.
  • #1
bengaltiger14
138
0
Problem Involving Elastic Collision...

There are two ball that are both 10kg. The velocity of ball 1 is 10 m/s and the velocity of ball 2 is 40 m/s in the opposite direction of ball 1. They collide to form an elastic collision. Determine speed of both balls after collision?

Am I starting off on the right foot?

1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v3^2 + 1/2m2v4^2

=1/2(10kg)(10m/s)^2 + !/2 (10kg)(40m/s)^2 = 1/2(10)V3^2 + 1/2(10)V4^2

Solve for V3 and V4 and that should be my answers correct?
 
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  • #2
You can't solve for v3 and v4 without using the fact that momentum is conserved. Then you have two equations with two unknowns, which is solvable.
 
  • #3
So far, so good. But how can you solve for two variables with only one equation? What else is conserved that will give you a second equation?
 
  • #4
Ok...Kinetic Energy is conserved and so is momentum...
 
  • #5
So..Conservation of Momentum yields me:

m1v1 + m2v2 = m2vx + m2vy

Where vx and vy are my unknowns
 
  • #6
Do I solve for vx in the momentum equation and then plug that anwer in for my V3 on my original or something like that?
 
  • #7
bengaltiger14 said:
So..Conservation of Momentum yields me:

m1v1 + m2v2 = m2vx + m2vy

Where vx and vy are my unknowns
Why not use the same symbols that you used in your conservation of energy equation?
 
  • #8
yes... Than would be easier. I just named them different because they were different equations. But, I am going on the right track now?
 
  • #9
By using conservation of energy and conservation of momentum, you are on the right track. (Your notation of Vx and Vy bothers me a bit--I hope you are not thinking in terms of x and y components.)
 
  • #10
Doc Al said:
By using conservation of energy and conservation of momentum, you are on the right track. (Your notation of Vx and Vy bothers me a bit--I hope you are not thinking in terms of x and y components.)


No. I am not thinking in terms of x and y components. I just simply changed the notations because I was looking at two different formulas.
 
  • #11
Cool. But the same quantities should have the same notation, even if they appear in different equations. That's how you'll end up with two equations and two unknowns--not four unknowns. :wink:
 
  • #12
bengaltiger14 said:
No. I am not thinking in terms of x and y components. I just simply changed the notations because I was looking at two different formulas.

No problemo. But it's a bit more practical to name the velocities after the collision as v1' and v2'.
 
  • #13
So, is my answer correct. After doing calculations, I determined that ball 1 transferred it velocity to ball 2 and vise versa. Ball 2 is now at 10 m/s and ball 1 is now at 40 m/s.
 
  • #14
That's right--they swap speeds.
 
  • #15
Cool...Thanks for all your help everyone
 
  • #16
Just for general knowledge. If the collison were inelastic and the formula would be:
Vfinal = (m1v1 +m2v2)/m1 + m2

Would I need to set v1 or v2 to negative since they are going in the opposite direction?? When I set v2 to negative, the final V= -15m/s.
When I leave it positive, Vfinal = 25 m/s
 
  • #17
bengaltiger14 said:
Just for general knowledge. If the collison were inelastic and the formula would be:
Vfinal = (m1v1 +m2v2)/m1 + m2

This should help: http://hyperphysics.phy-astr.gsu.edu/hbase/inecol.html" .
 
Last edited by a moderator:
  • #18
The link did not work...
 
  • #19
bengaltiger14 said:
The link did not work...
It should work for you.
 
  • #20
bengaltiger14 said:
Just for general knowledge. If the collison were inelastic and the formula would be:
Vfinal = (m1v1 +m2v2)/(m1 + m2)
Yes. (See my added brackets.)

Would I need to set v1 or v2 to negative since they are going in the opposite direction??
Absolutely! If v1 is positive, v2 must be negative. Momentum is a vector--direction matters.
 

Related to Problem Involving Elastic Collision .

1. What is an elastic collision?

An elastic collision is a type of collision between two objects in which there is no loss of kinetic energy. This means that the total kinetic energy of the objects before and after the collision remains the same.

2. What is the difference between an elastic and inelastic collision?

In an elastic collision, there is no loss of kinetic energy and the objects bounce off each other. In an inelastic collision, some of the kinetic energy is lost and the objects stick together after the collision.

3. How is momentum conserved in an elastic collision?

In an elastic collision, momentum is conserved because the total momentum of the objects before the collision is equal to the total momentum after the collision. This means that the mass and velocity of the objects must be taken into account.

4. What is the formula for calculating the velocities of two objects after an elastic collision?

The formula for calculating the velocities of two objects after an elastic collision is:
v1f = (m1-m2)v1i + 2m2v2i / (m1+m2)
v2f = 2m1v1i + (m2-m1)v2i / (m1+m2)
where m1 and m2 are the masses of the objects, v1i and v2i are the initial velocities, and v1f and v2f are the final velocities.

5. How is an elastic collision different from a perfectly elastic collision?

An elastic collision is a general term that describes any collision in which there is no loss of kinetic energy. A perfectly elastic collision is a type of elastic collision in which the objects have the same mass and the same velocity before and after the collision. This means that the objects bounce off each other with the same speed and direction after the collision.

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