Problem Involving Elastic Collision .

[bengaltiger14]

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?

 

[radou]

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]

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?

 

[bengaltiger14]

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

 

[bengaltiger14]

So..Conservation of Momentum yields me:

m1v1 + m2v2 = m2vx + m2vy

Where vx and vy are my unknowns

 

[bengaltiger14]

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]

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?

 

[bengaltiger14]

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]

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.)

 

[bengaltiger14]

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]

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.

 

[radou]

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'.

 

[bengaltiger14]

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.

 

[Doc Al]

That's right--they swap speeds.

 

[bengaltiger14]

Cool...Thanks for all your help everyone

 

[bengaltiger14]

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

 

[radou]

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" .

 

[bengaltiger14]

The link did not work...

 

[geoffjb]

The link did not work...

It should work for you.

 

[Doc Al]

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.