Solving Collisions Problem: Velocity of Third Fragment

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Homework Statement



a 600 kg canon, initially at rest, exploded upon firing and broke into three fragments. One fragment of mass of 200 kg traveled east at 40m/s and a second fragment of mass 300 kg traveled due south at 20 m/s. What was the velocity of the third fragment?

Homework Equations



sum of P = sum of P'
0= P1 + P2 + P3

The Attempt at a Solution



I'm not sure whether I'm supposed to use trig or something to find the missing velocity.
Any help would be GREATLY appreciated.
 
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bondgirl007 said:

Homework Statement



a 600 kg canon, initially at rest, exploded upon firing and broke into three fragments. One fragment of mass of 200 kg traveled east at 40m/s and a second fragment of mass 300 kg traveled due south at 20 m/s. What was the velocity of the third fragment?

Homework Equations



sum of P = sum of P'
0= P1 + P2 + P3

The Attempt at a Solution



I'm not sure whether I'm supposed to use trig or something to find the missing velocity.
Any help would be GREATLY appreciated.
Momentum is a vector quantity, and as such, the sum of the momenta must obey the laws of vector additions. Are you familiar with resultants and vector sums?
 
This is what I have so far. Am I on the right track?
Will the resultant be the hypotenuse?

http://img408.imageshack.us/my.php?image=physicsay5.jpg
 

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I did Pythragoras to find the missing vector.

(8000)^2 + (6000)^2 = c^2
c=100 m/s

tan[tex]\theta[/tex] = 6000/8000
[tex]\theta[/tex] = 37

I got 100 m/s for my velocity and 37 degrees N of W. Is this right?
 
You know anythin about momentum conservation?
 
bondgirl007 said:
I did Pythragoras to find the missing vector.

(8000)^2 + (6000)^2 = c^2
c=100 m/s

tan[tex]\theta[/tex] = 6000/8000
[tex]\theta[/tex] = 37

I got 100 m/s for my velocity and 37 degrees N of W. Is this right?
Yes, that is correct, but your diagram and equation are not consistent with this answer. Looks like you combined a step when solving for 'c' , and the direction of the diagonal is wrong.