- #1

mcapuchin

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## Homework Statement

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[/B]

## Homework Equations

SOH CAH TOA

Conservation of Momentum

Pythagorean Theorem

## The Attempt at a Solution

Let N and E be positive[/B]

I first split up both the x and y components

v

_{1y}=36m/s

v

_{1x}=0m/s

v

_{2y}-32cos50=-20.57m/s

v

_{2x}=-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=m

_{1}v

_{1fx}-m

_{2}v

_{2fx}+m

_{3}v

_{3fx}

0=(0.25kg)(0m/s)-(0.35kg)(24.51m/s)+(0.30kg)v

_{3fx}

v

_{3fx}=28.59m/s

0=m

_{1}v

_{1fy}-m

_{2}v

_{2fy}+m

_{3}v

_{3fy}

0=(0.25)(36m/s)-(0.35)(20.57m/s)+(0.30kg)v

_{3fy}

v

_{3fy}=6.0m/s

To find the final velocity with the x and y components

a

^{2}+b

^{2}=c

^{2}

6

^{2}+28.59

^{2}=820

√820=28.63

v

_{3f}=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?**