Solving Collisions Problem: Velocity of Third Fragment

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The discussion revolves around solving a physics problem involving momentum conservation after an explosion of a 600 kg cannon into three fragments. Two fragments have known velocities, and the task is to determine the velocity of the third fragment. The user attempts to apply vector addition and the Pythagorean theorem to find the missing velocity, arriving at a result of 100 m/s at an angle of 37 degrees North of West. However, there is confusion regarding the consistency of the diagram and equations used in the calculations. The conversation emphasizes the importance of correctly applying momentum conservation principles and ensuring accurate vector representation.
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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\theta = 6000/8000
\theta = 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\theta = 6000/8000
\theta = 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.
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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