# Funky Elastic Collisions

1. Oct 24, 2005

### ms. confused

Hey hey!
A 10g object moving to the right with a speed of 20cm/s makes an elastic head-on collision with a 15g object moving in the opposite direction with a speed of 30cm/s. Find the velocity of each object after the collision.
So I have to come up with two equations and solve for unknown's simultaneously with plugged in values. That's one way of doing it. My solutions manual sucks at explaining the steps it took to get certain values, so I'm kind of lost on that.
So you know I'm not just trying to get out of doing homework myself, this is what I tried:
m1v1i + m2v2i = m1v1f + m2v2f
(10g)(20m/s) + (-15g)(30m/s) = (10g)v1f + (15g)v2f
-250 = 10v1f + 15v2f eq'n#(1)
(10)(20) + (15)(30) = 10v1f + 15v2f
650 = 10v1f + 15v2f eq'n #(2)
I think one of these isn't quite right because for 10g, I get -45cm/s when it's actually -40cm/s and for 15g, I get 30cm/s when it's only supposed to be 10cm/s. Could someone please point me in the right direction?

2. Oct 25, 2005

### ehild

It is the eq. for conservation of momentum, and it is correct. But the KE is also conserved in an elastic collision, so the second equation should express conservation of the KE which is 1/2 * mv^2. So you should square the velocities in the second equation.
(10)(20)^2 + (15)(30)^2 = 10v1f^2 + 15v2f^2
ehild

3. Oct 25, 2005

### ms. confused

What happens to the 1/2 in the KE equation? Is that included too? And once I get this equation, do I simultaneously solve for the unknown with the other equation? There's a problem if I do that. I won't have like terms...unless I factor...because the variables are squared. I'm still not any closer to the actual answers, though. Are you sure this is what should be done?

4. Oct 25, 2005

### Diane_

The one-half is in all of the terms, so he simply cancelled it out. Think of it as multiplying both sides by 2, if you will.

You've got the right idea. You will end up with two equations - one linear in both v1 and v2, one quadratic in both. This can be solved. Probably the easiest way would be to solve the linear version for one of the speeds and plug the result into the quadratic. You'll end up with an equation quadratic in one of the speeds - multiply it out, collect similar terms, and solve it. The quadratic formula will come in handy.

Interestingly, you'll end up with two solutions. Can you interpret both of them physically?