1. Oct 18, 2007

### MrTaylor

Ive been working this problem for the past hours, and i keep getting the wrong answer... can someone check to see what I am doing wrong, and please correct me..

Question:
An 18g object moving to the right at 26cm/s overtakes and collides elastically with a 33g object moving in the same direction at 16cm/s.

I need to find the velocity of the slower object and then the faster object.

My attempt at this question:
m1v1 + m2v2 = m1v1f + m2v2f
(.018)(26) + (.033)(16) = (.018)v1f + (.033)v2f
(.468) + (.528) = (.018)v1f + (.033)v2f
(.996)= (.018)v1f + (.033)v2f

then KE cons.
26-16 = -(v1f-v2f)
10=v1f + v2f
v1f = -10 + v2f

and subsituted (v1f = -10 + v2f) in v1f in the first eq.

(.996)= (.018)(-10 + v2f) + (.033)v2f
(.996)= (-.180) + (.018)v2f + (.033)v2f
(.996)= (-.180) + (.051)v2f
(1.176)= (.051)v2f
(1.176/.051) = v2f
v2f= 23.05882 (((<<< this is for the slower object)))

For the faster object I work the problem the same way, only using the substitution for v2f this time.

2. Oct 18, 2007

### Rahmuss

You're trying to take a shortcut for the K.E.

Use : $$K.E. = \frac{1}{2}mv^{2}$$

Then get $$v_{1f}$$ and $$v_{2f}$$ using that and you've already got it for the first equation, so plug one of the found velocities into the other equation and it should work out.

*gulps* I sure hope I'm not messing this up. I should really check before posting; but I'm sure someone will correct me if I'm wrong.