# Problem Involving Elastic Collision .

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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Homework Helper
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.

Doc Al
Mentor
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?

Ok.....Kinetic Energy is conserved and so is momentum......

So..Conservation of Momentum yields me:

m1v1 + m2v2 = m2vx + m2vy

Where vx and vy are my unknowns

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?

Doc Al
Mentor
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?

yes... Than would be easier. I just named them different because they were different equations. But, I am going on the right track now?

Doc Al
Mentor
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.)

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.

Doc Al
Mentor
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. Homework Helper
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'.

So, is my answer correct. After doing calculations, I determined that ball 1 transfered 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.

Doc Al
Mentor
That's right--they swap speeds.

Cool...Thanks for all your help everyone

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

Homework Helper
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" [Broken].

Last edited by a moderator:

bengaltiger14 said:
It should work for you.

Doc Al
Mentor
bengaltiger14 said:
Just for general knowledge. If the collison were inelastic and the formula would be:
Vfinal = (m1v1 +m2v2)/(m1 + m2)