What are the velocities of two objects after an elastic head-on collision?

[ace214]

A 10.0 g object moving to the right at 22.0 cm/s makes an elastic head-on collision with a 15.0 g object moving in the opposite direction at 32.0 cm/s. Find the velocity of each object after the collision.

First, I converted the masses to kg and the velocities to m/s.
I used m1v1i + m2v2i = m1v1f + m2v2f and solved for v1f in terms of v2f and plugged this into
v1i - v2i = v2f - v1f to attempt to get v2f first (yes, I am converting back to cm/s) and I'm not getting the right answer... I don't understand what I'm doing wrong...

Here's my numbers:
.01(.22) + .015(.32) = .01v1f + (.015)v2f
.007 = .01v1f + .015v2f
v1f = (.007 - .015v2f)/.01 = .7 - 1.5v2f

(.22) - (-.32) = v2f - v1f
.54 = v2f - (.7 - 1.5v2f)
.54 = v2f -.7 + 1.5v2f
.61 = 2.5v2f
v2f = .244 m/s = 24.4 cm/s

This is due at 8:30 AM. EST tomorrow.

 

[ace214]

I have also tried v2i - v1i = v1f - v2f with positive numbers and v1i +v2i = v1f + v2f with properly-signed numbers... Someone please help...

 

[Dick]

So far, you have one equation in two unknowns, v1f and v2f. There is no way you can solve that for either one. You need another equation. In an elastic collision the kinetic energy is conserved as well as the momentum. Use that to get another equation.

 

[ace214]

So far, you have one equation in two unknowns, v1f and v2f. There is no way you can solve that for either one. You need another equation. In an elastic collision the kinetic energy is conserved as well as the momentum. Use that to get another equation.

No there's two equations... I already said that.

I used m1v1i + m2v2i = m1v1f + m2v2f and solved for v1f in terms of v2f and plugged this into v1i - v2i = v2f - v1f

 

[TMM]

I solved it by writing an equation for the total momentum and for the total kinetic energy. From there you can just substitute.

 

[ace214]

I solved it by writing an equation for the total momentum and for the total kinetic energy. From there you can just substitute.

When I did that, I got imaginary numbers from the quadratic.

 

[Dick]

No there's two equations... I already said that.

I used m1v1i + m2v2i = m1v1f + m2v2f and solved for v1f in terms of v2f and plugged this into v1i - v2i = v2f - v1f

Ok, then the problem is that the second equation isn't true. Use KE.

 

[TMM]

I didn't.

I wrote:

mv(2) + mv(1) = -260

.5mv(2)^2 + .5mv(1)^2 = 11000

Then I just substituted momentum one into the energy one and solved .

I got 12.3 cm/s to the right for the larger particle and 44.4 cm/s to the left for the smaller one, which is correct.

If you're getting lost in the conversion, just leave it in cgs.

 

[ace214]

Ok, then the problem is that the second equation isn't true. Use KE.

When I did that, I got imaginary numbers from the quadratic.

Also, the book says that that equation is supposed to work.

 

[Dick]

Something like v1i + v2i = v2f + v1f will only work if the two masses are equal.

 

[ace214]

Ok, I redid the KE equation and got 32 and 24 for the larger mass. I've tried 24 already as I got it from the equation with just velocities above and it wasn't right... Gaaaaaah...

 

[ace214]

Ok, I was an idiot and didn't use a negative velocity in the original momentum equation... Wish somebody had caught it but oh well. Also the v1 - v2 = v2 - v1 does work for objects with different masses.