# Homework Help: Momentum 2

1. Oct 20, 2007

### aligass2004

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

Similar to the other one...
On billiard ball is shot east at 2.6m/s. A second, identical billiard ball is shot west at .8m/s. The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90 degrees and sending it north at 1.58m/s. What are the speed and direction of the first ball after the collision?

2. Relevant equations

3. The attempt at a solution
I tried using (P1x)f + (P2x)f = (P1x)i + (P2x)i ----> (m1xv1x)f + (m2xv2x)f = (m1xv1x)i + (m2xv2x)i. I tried solving for (v1x)f. I assumed since the billiard balls are identical that the mass cancels out.

2. Oct 20, 2007

### learningphysics

yes, the masses cancel. Looks like you have the right idea. can you show your calculations?

3. Oct 20, 2007

### aligass2004

V1f = V1i + V2i - V2f
= 2.6 + .8 - 1.58 = 1.82

4. Oct 20, 2007

### learningphysics

but 1.58 is not in the x-direction... what is the Vxf of the second ball?

5. Oct 20, 2007

### aligass2004

I didn't realize that. The Vxf of the 2nd ball is zero since it isn't moving in the x direction anymore.

6. Oct 20, 2007

### learningphysics

exactly.

7. Oct 20, 2007

### aligass2004

Ok, so I tried V1f = 2.6 + .8 = 3.4, but that wasn't right.

8. Oct 20, 2007

### learningphysics

Oh, sorry I didn't notice... you should have -0.8, not 0.8.

use east positive. west negative. north positive. south negative.

9. Oct 20, 2007

### aligass2004

Alright, this time I tried V1f = 2.6 - .8 = 1.8, but it again wasn't right.

10. Oct 20, 2007

### learningphysics

that's only in the east west direction... you also need to consider the north south direction.

I thought the question was asking for the components separately. I just noticed now it only asks for speed. I'm sorry... wish I realized before you submitted again...

get the north south component, same way you got the east/west component... then get the speed using pythagorean theorem.

11. Oct 20, 2007

### aligass2004

Blah....freaking components. Ok, for the north/south component, V1f = V1i + V2i - V2f -----> V1f = 0 + 0 -1.58. There's only one problem with using the pythagorean theroem. With the last problem, I figured out the angle really easily since both components were the same. I don't know the angle with this one.

12. Oct 20, 2007

### learningphysics

you can get the angle using arctan(vy/vx). be careful before you submit though... how do they ask for the angle? Do they want the angle from the east axis etc..?

13. Oct 20, 2007

### aligass2004

They ask for degrees in the south east direction.

14. Oct 20, 2007

### aligass2004

So the angle would be -41.276 degrees, but it would be positive since they want it in the SE direction, right?

15. Oct 20, 2007

### learningphysics

yeah, but it is East 41.276 degrees South...

Do they just say SE... because it could be east of south or south of east... that makes a difference...

16. Oct 20, 2007

### aligass2004

It says south of east

17. Oct 20, 2007

### learningphysics

Cool. Then 41.276 is right.

18. Oct 20, 2007

### aligass2004

Ok so the angle was right, but I don't know the velocity....and I'm out of tries. I tried the -1.58 = vcos(theta).

19. Oct 20, 2007

### learningphysics

The 1.58 is the sin. so vsin(41.276) = 1.58. v = 2.395m/s.

But it's best to just use the pythagorean theorem when you need the speed... you don't need to worry about the angles. sqrt(1.58^2 + (1.8)^2) = 2.395m/s

20. Oct 20, 2007

Alright.