# Homework Help: Perfectly elastic collision, both velocities unknown

1. Jul 24, 2014

### chris787

1. The problem statement, all variables and given/known data

Ball 1, with a mass of 110g and traveling at 15m/s , collides head on with ball 2, which has a mass of 350g and is initially at rest.

What are the final velocities of each ball if the collision is perfectly elastic?

2. Relevant equations

Conservation of momentum:

m1u1 + m2u2 = m1v1 +m2v2

and energy conservation for kinetic energy:

1/2m1u1^2 + 1/2m2u2^2 = 1/2m1v1^2 + 1/2m2v2^2

3. The attempt at a solution

Its really the algebra that I am stuck at, at this point. Using the momentum equation, I got
v1 = (1.65kg m/s - 0.35kgv2) / 0.11kg

and I know that I have to plug that into this equation:
1/2 (0.11kg) (15m/s)^2 = 1/2 (0.11kg) v1^2 + (0.175kg)v2^2

I really get lost at this point, I keep end up adding or subtracting numbers to 0.

Any help would be greatly appreciated. Thank you!

Last edited: Jul 24, 2014
2. Jul 24, 2014

### Nathanael

You said it's a 0.12kg ball moving 12 m/s but your math says it's a 0.11kg ball moving at 15 m/s, which is it?

Anyway, the physics is virtually solved, you just need to plug the first equation in to the second (and then solve for v2).

Sorry, it's a little hard to help, because, as you said, your problem is an algebraic one. But you haven't shown your attempt (at the algebra) so it's difficult to pinpoint your mistake.

3. Jul 24, 2014

### chris787

sorry about that! i had two versions of the question and I copied the wrong one. It is .11kg moving at 15 m/s, I will change the main post to reflect that.

So far my attempt at plugging in the first equation into the second has been this: (I was told by my professor to drop the units at this point, so I have done that)

1/2 (0.11) (15)^2 = 1/2 (0.11)v1^2 + 1/2(0.35)v2^2

12.375 = 0.055 ((1.65 - 3.18v2)/0.11) ^2 + 0.175v2^2

the fraction with the addition and to the power of 2 is the part I am struggling with, this was my most recent attempt:

12.375 = 0.055 (15 - 3.18V2)^2 + 0.175V2^2

I guess where I am stuck at this point, is do I divide 12.375 by 0.055, and then factor out the middle section?
If so, I got

225 = 225 - 95.4v2 + 10.11V2^2 + 0.175V2^2

I have tried to proceed with this and use the quadratic equation but the number came out to be 0 / the denominator.

4. Jul 24, 2014

### Nathanael

Is the following true?

$(\frac{1.65-3.18V_2}{0.11})^2=(15-3.18V_2)^2$

Your math says it's true, but does it look correct to you?

Beyond that though, you should be able to rearrange it to the form $ax^2+bx+c=0$ where ($x=V_2$) and then you can use the quadratic equation to solve it.

5. Jul 24, 2014

### tms

First, you should always solve these problem symbolically, and only plug in number at the very last step. Every time you plug in numbers you lose information that could be helpful in solving the problem. Also, when you have a symbolic answer, you can easily examine it to see if it makes sense, and to see the physics (such as what would happen if one mass were much larger than the other, for instance).

For this problem, take the energy and momentum equations, and in each collect the terms in $m_1$ and $m_2$ on opposite sides for each. Then divide one equation by another. From there the path to the answer should be clear.