# Simple question on Conservation of Momentum in 2 directions

1. Apr 22, 2015

### mcapuchin

1. The problem statement, all variables and given/known data
A firework at rest with a mass of 0.90kg explodes into 3 pieces. One piece has a mass of 0.25kg and moves horizontally at 36m/s [N]. Another has a mas of 0.35kg and moves horizontally at 32m/s [S50°W]
Find the velocity of the third peice

2. Relevant equations
SOH CAH TOA
Conservation of Momentum
Pythagorean Theorem

3. The attempt at a solution
Let N and E be positive

I first split up both the x and y components
v1y=36m/s
v1x=0m/s
v2y-32cos50=-20.57m/s
v2x=-32sin50=-24.51m/s

Now I subtracted the 2 masses from .90kg which gave me a mass of .30kg.
Then I used the conservation of momentum for both the x and y components, which must equal zero as the object was at rest initially
PT=0
0=m1v1fx-m2v2fx+m3v3fx
0=(0.25kg)(0m/s)-(0.35kg)(24.51m/s)+(0.30kg)v3fx
v3fx=28.59m/s

0=m1v1fy-m2v2fy+m3v3fy
0=(0.25)(36m/s)-(0.35)(20.57m/s)+(0.30kg)v3fy
v3fy=6.0m/s

To find the final velocity with the x and y components
a2+b2=c2
62+28.592=820
√820=28.63
v3f=28.63m/s

Now we find the angle of the magnitude
tan-1=(6/28.59)
=11.85°

However, this is where I'm stuck. How do I know if it is north or south or west or east? I looked at the book and my calculations are correct, but they say it is S11.9W. How do I determine this?

2. Apr 22, 2015

### Staff: Mentor

Try drawing a diagram of the event and trajectories and see if things make sense from it.

3. Apr 22, 2015

### mcapuchin

Thanks for the suggestion, I think there's a typo in the text as on the diagram I drew it makes sense that it will be E11.85S. Will point it out to my teacher.