# Conservation of momentum in 2 dimensions, & with 2 unknowns

1. Nov 8, 2006

### xrotaryguy

Yikes, the algebra on this one is kinda scary!

The correct answer for V2 is 45.0m/s

Any idea why I'm getting this one wrong?

2. Nov 8, 2006

### OlderDan

If there is an angle θ for the y component of the initial velocity then it also applies to the x component. Unless you know that angle, you cannot find those two final velocities.

3. Nov 8, 2006

### xrotaryguy

Well, the angle for the y componant of the initial velocity is 90 degrees, and sin90=+1, so it doesn't really need to show up in the math does it?

The left side of the equal sign is before the seperation. The right side is after.

4. Nov 8, 2006

### OlderDan

If it is 90 degrees and you already know that, why are you carrying it around in your y equations? sinθ is an awfully obscure way of writing 1, especially when you never define θ. Your other equations are no longer vector equations. The v in those equations are the magnitudes you are looking for. You have 2 equations for the two unkowns. All the trig ratios you are carrying around are constants, and that is fine. Your approach is generally correct, except that in the x equation you are probably getting in trouble mixing vectors with scalars.

0 = v1_x + v2_x

means that one of either v1_x or v2_x is positive and one is negative. When you write

0 = m1v1sinθ1 + m2v2sinθ2 and
MV = m1v1cosθ1 + m2v2cosθ2

you are treating v1 and v2 as both positive, which makes your first equation inconsistent with your second equation. Don't forget the relationship between M, m1 and m2. Maybe you used it correctly, but I don't see it in your calculation.

5. Nov 8, 2006

### xrotaryguy

Ok, I guess I could just nix the sin theta stuff from the beginning. I meant to define theta as being 90 degrees in the drawing. I gess I accidently left that out. Thanks for the suggestions. I'll see what I can do.