# Homework Help: Head on Collision

1. Mar 13, 2005

### Jayhawk1

Can anyone help me with this problem?

Two billiard balls of equal mass undergo a perfectly elastic head-on collision. If one ball (ball A) is initially moving at 1.9 m/s in the +x direction, and the other (ball B) is initially moving at 3.9 m/s in the -x direction, what will be the velocities of (a) ball A and (b) ball B after the collision? (Indicate the direction by the sign of your answer, taking a positive velocity as being in the +x direction.)

I've tried using the conservation of momentum and energy... doesn't seem to work.

2. Mar 13, 2005

### Jameson

Show me what you did.

3. Mar 13, 2005

### Jayhawk1

First I tried mvA+MvB=MvA'+MvB' but that didn't work... then I got some long formulas from class like V(B)'=2M(A)(VA)/(MA+MB)+(M(B)-(MA)(VB)/(MA+MB)

4. Mar 13, 2005

### Tony Zalles

Hi Jaykawk1,

Ok this kind of problem requires more algebra than physics...

Ok you have to use two principles two solve this

1) (1/2)M(a)(V^2) + (1/2)M(b)(V^2) = (1/2)M(a)(V'(a)^2) + (1/2)M(b)(V'(b)^2)
[ Conservation of total kinetic energy for elastic collisions ]

2) M(a)V(a) + M(b)V(b) = M(a)V'(a) + M(b)V'(b)
[ Conservation of Momentum ]

Now what you need to see is that here you have two equations.

"what will be the velocities of (a) ball A and (b) ball B after the collision?"

So there are TWO velocities that need to be found thus two unkowns.

Therefore we have two equations;

1) (1/2)M(a)(V^2) + (1/2)M(b)(V^2) = (1/2)M(a)(V'(a)^2) + (1/2)M(b)(V'(b)^2)

2) M(a)V(a) + M(b)V(b) = M(a)V'(a) + M(b)V'(b)

And also two unkowns, V'(a) and V'(b), Now from here its just algebra.

Solve for one of the unkowns, V'(a) or V'(b), using equation (2) and substitute that in to equation (1), and then solve for the other unkown velocity (you may have to use the quadratic formula) and than once you get the velocity.

Then go back to what you came up with earlier when you were first trying to solve for one of the unkowns and substitute what you found from above in to that equation and solve for the final one.

Regards,

-Jose Antonio Zalles II

P.S. just reply if its still giving you issues.

Last edited: Mar 13, 2005
5. Mar 13, 2005

### Jayhawk1

Thanks so much!!

6. Mar 13, 2005

### Tony Zalles

No prob, best of luck with your studies.

-Jose Antonio Zalles II